On the cohomology ring and upper characteristic rank of Grassmannian of oriented 3-planes

Abstract

In this paper we study the mod 2 cohomology ring of the Grasmannian $\widetilde{G}_{n,3}$ of oriented 3-planes in ${\mathbb {R}}^n$. We determine the degrees of the indecomposable elements in the cohomology ring. We also obtain an almost complete description of the cohomology ring. This description allows us to provide lower and upper bounds on the cup length of $\widetilde{G}_{n,3}$. As another application, we show that the upper characteristic rank of $\widetilde{G}_{n,3}$ equals the characteristic rank of $\widetilde{\gamma }_{n,3}$, the oriented tautological bundle over $\widetilde{G}_{n,3}$ if n is at least 8.

1 Introduction

Grassmann manifolds have been a central object in topology and geometry, especially in the study of classifying spaces and vector bundles. Understanding the cohomology ring, therefore, is of importance. If we consider the Grassmann manifold of k-planes in real n-dimensional Euclidean space, denoted by $G_{n,k}$, then its cohomology ring with mod 2 coefficients is well-known by classical results of Borel [1]. These manifolds come naturally equipped with a rank k bundle $\gamma _{n,k}\rightarrow G_{n,k}$, called the tautological bundle. The associated characteristic classes $w_i$ are the universal Stiefel–Whitney classes. The manifold $G_{n,k}$ has a double cover $\widetilde{G}_{n,k}$, the Grassmannian of oriented k-planes in ${\mathbb {R}}^n$. This cover is also the universal cover when $(n,k)\ne (2,1)$. The direct limit of $\widetilde{G}_{n,k}$ over n is the classifying space for oriented real vector bundles of rank k. However, very little is known about the cohomology ring (with mod 2 coefficients) of $\widetilde{G}_{n,k}$ for general values of n and k.

When we work with $G_{n,k}$ and its universal cover, we will assume that $n\ge 2k$ as there is a diffeomorphism between $G_{n,k}$ and $G_{n,n-k}$. We have a complete description of $H^*(\widetilde{G}_{n,2};{\mathbb {Z}}_2)$ due to [6]. In fact, we observe that an alternate (and geometric) proof of the description of $H^*(\widetilde{G}_{n,2};{\mathbb {Z}}_2)$ for n even is possible using a stronger result of Lai [7] on even-dimensional complex quadrics. We use algebraic topological methods to analyze $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$; this is different from the more algebraic methods used in the literature. We introduce the space

$$\begin{aligned} W_{2,1}^n:=\left\{ (\widetilde{P},v)\in \widetilde{G}_{n,2}\times S^{n-1}\,|\,\widetilde{P}\hbox { and }v\hbox { are orthogonal}\right\} \end{aligned}$$

and consider the $S^{n-3}$-bundle given by $W_{2,1}^n \rightarrow \widetilde{G}_{n,2}$. This helps us in characterizing the mod 2 cohomology ring of $W_{2,1}^n$ (cf. Theorem 2.9). We also analyze the map

$$\begin{aligned} \iota ^*:H^*\left( W^{n+1}_{2,1};{\mathbb {Z}}_2\right) \rightarrow H^*\left( W^{n}_{2,1};{\mathbb {Z}}_2\right) \end{aligned}$$

(cf. Theorem 2.10). Exploiting the existence of the 2-sphere bundle , given by the span of the two plane and the vector, we are able to analyze the mod 2 cohomology ring of $\widetilde{G}_{n,3}$.

Given a graded cohomology ring $H^*=H^*(X;{\mathbb {Z}}_2)$, let $H^+$ denote the elements of $H^*$ in positive degrees. The elements of $H^+/(H^+ \cdot H^+)$ will be called indecomposables. Consider the covering map $\pi :\widetilde{G}_{n,3}\rightarrow G_{n,3}$. The pullback bundle $\widetilde{\gamma }_{n,3}$ of the tautological bundle $\gamma _{n,3}$ via this covering map is called the oriented tautological bundle. Let $\widetilde{w}_2:=\pi ^*(w_2),\widetilde{w}_3:=\pi ^*(w_3)$ denote the 2nd and 3rd Stiefel–Whitney classes of $\widetilde{\gamma }_{n,3}$. These two classes are the indecomposables in degree 2 and 3 respectively. The degree of the first indecomposable element (after degree 3) of $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ is known due to [4, 11]. In this article we determine the complete set of degrees of indecomposables in $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ (cf. Theorem 3.6).

Theorem A

• (a) There is one indecomposable in each of the degrees $2,~3,~ 2^t-1$ in $H^*(\widetilde{G}_{2^t,3};{\mathbb {Z}}_2)$.

• (b) There is one indecomposable in each of the degrees $2,~3,~ 2^t-4$ in $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ for $n=2^t-1,~2^t-2,~2^t-3$.

• (c) Let $2^{t-1}<n\le 2^t-4$. Then the only indecomposables in $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ are in degrees $2,~3,~3n-2^t-1,~2^t-4$ with one indecomposable in each degree.

We note that Theorem A (cf. Theorem 3.6 in this article) gives a different proof of Theorem 1.1 of [11].

We do not have any specific choice of an indecomposable in degree i when $i>3$. Let us denote an indecomposable of degree $i>3$ by $w_i\in H^i(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$.

Remark 1.1

These elements $w_i$ depend on n but for notational convenience, and kee** in mind the tedious algebraic computations that require the use of these elements and its powers, we shall stick to this notation. Usually, the positive integer n will be clear in the context.

Such a choice of an indecomposable element in $H^i(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ will satisfy some relations stated in the following result (Theorem 3.9 and part of Theorem 3.10).

Theorem B

(1) Let $n=2^t$, then

$$\begin{aligned} H^*\left( \widetilde{G}_{2^t,3};{\mathbb {Z}}_2\right) \cong \frac{\frac{{\mathbb {Z}}_2[\widetilde{w}_2,\widetilde{w}_3]}{\langle g_{2^t-2}, g_{2^t-1}\rangle }\otimes {\mathbb {Z}}_2[w_{2^t-1}]}{\left\langle w_{2^t-1}^2-P w_{2^t-1}\right\rangle } \end{aligned}$$

for some $P(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^*(G_{2^t,3})).$

(2) Let $n=2^t-1,~2^t-2,~2^t-3.$ Then

$$\begin{aligned} H^*\left( \widetilde{G}_{n,3};{\mathbb {Z}}_2\right) \cong \frac{\frac{{\mathbb {Z}}_2[\widetilde{w}_2,\widetilde{w}_3]}{\langle g_{n-2},g_{n-1},g_n\rangle }\otimes {\mathbb {Z}}_2[w_{2^t-4}]}{\left\langle w_{2^t-4}^2-P_1 w_{2^t-4}-P_2\right\rangle } \end{aligned}$$

for some $P_1(\widetilde{w}_2,\widetilde{w}_3),P_2(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^*(G_{n,3}))$ with $P_2=0$ for $n=2^t-2,~2^t-3.$

(3) Let $2^{t-1}<n\le 2^t-4$. Then we have the following relations in $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$:

(a) If $3n-2^t-1<2^t-4$ then

$$\begin{aligned} w_{3n-2^t-1}^2 = R_1+w_{2^t-4}R_2+w_{3n-2^t-1}R_3,\,\,w_{2^t-4}^2=0 \end{aligned}$$

while if $3n-2^t-1>2^t-4$ then

$$\begin{aligned} w_{2^t-4}^2 = Q_1+w_{2^t-4}Q_2+w_{3n-2^t-1}Q_3,\,\,w_{3n-2^t-1}^2=0 \end{aligned}$$

where $R_i,Q_i$’s are polynomials in $\widetilde{w}_2$ and $\widetilde{w}_3$.

(b) There exists $v_{2^t-8}\in \pi ^*(H^{2^t-8}(G_{n,3};{\mathbb {Z}}_2))$ and $v_{3n-2^t-5}\in \pi ^*(H^{3n-2^t-5}(G_{n,3};{\mathbb {Z}}_2))$ which when paired with $w_{3n-2^t-1}$ and $w_{2^t-4}$ respectively give the top cohomology class of $\widetilde{G}_{n,3}$. Moreover, these satisfy

$$\begin{aligned} v_{2^t-8} w_{2^t-4}=0,\,\,\,\,v_{3n-2^t-5} w_{3n-2^t-1}=0. \end{aligned}$$

We shall prove an extended result (cf. Theorem 3.10) where $v_{2^t-8},v_{3n-2^t-5}$ will be studied extensively.

Remark 1.2

Poincaré duality implies that the cup product on $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ defines a non-degenerate pairing. Thus, given a non-zero element $\alpha \in H^i(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$, there exists $\beta \in H^{3n-9-i}(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ such that $\alpha \beta$ is the generator of the top cohomology group $H^{3n-9}(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$. We call $\beta$dual to $\alpha$ with respect to the cup product pairing.

As illustrated by Theorem B 3(b), there are natural choices for dual(s) $v_j$ of the indecomposable(s), which themselves are not canonically obtainable. It is expected that these dual classes enjoy certain stability properties.

As a corollary (see Corollary 3.12 of Theorem 3.10) of Theorem B, we obtain a lower bound and an upper bound for the cup-length. Recall that ${\mathbb {Z}}_2$-cup-length of a path-connected space X, denoted by $\texttt {cup}(X)$, is defined as the maximum r such that there exist classes $x_1,x_2,\ldots ,x_r\in H^{+}(X;{\mathbb {Z}}_2)$ with non-trivial cup product, i.e., $x_1\cup x_2\cup \cdots \cup x_r\ne 0$. Fukaya in [2, Conjecture 1.2] conjectured about the ${\mathbb {Z}}_2$-cup-length of oriented Grassmann manifolds $\widetilde{G}_{n,3}.$ In [2, 4, 6, 12] the authors have proved the conjecture for $n=2^t-1,~2^t,~2^t+1,~2^t+2,~2^t+2^{t-1}+1,~2^t+2^{t-1}+2$ ($t\ge 3$). In [11] the authors have given the exact values of $\texttt {cup}(\widetilde{G}_{n,3})$ for $2^{t-1}<n\le 2^{t-1}+\frac{2^{t-1}}{3}-1$ and $t\ge 3$. We note that for [2, 11] one has to make a slight adjustment; in these two papers $\widetilde{G}_{n,3}$ consists of the oriented 3-dimensional subspaces of ${\mathbb {R}}^{n+3}$. The exact value of $\texttt {cup}(\widetilde{G}_{n,3})$ is not known for $n\in [2^{t-1}+\frac{2^{t-1}}{3},2^t-2]$ ($t\ge 3$). When $t\ge 4$ an upper bound for $\texttt {cup}(\widetilde{G}_{n,3})$ is given in [12, Corollary 3.2] for the sub-interval $[2^{t-1}+2^{t-2},2^t-2]$ of $[2^{t-1}+\frac{2^{t-1}}{3},2^{t}-2]$. In the following result (cf. Corollary 3.12) we show that the same upper bound will also work for the whole interval $[2^{t-1}+\frac{2^{t-1}}{3},2^t-2]$. We also provide a lower bound for $\texttt {cup}(\widetilde{G}_{n,3})$. These bounds are consistent with Fukaya’s conjecture.

Observation b

Let $t\ge 3$ and $n\in [2^{t-1}+\frac{2^{t-1}}{3},2^t-2]$. Then the ${\mathbb {Z}}_2$-cup-length $\texttt {cup}(\widetilde{G}_{n,3})$ is bounded above by $\frac{3(n-3)}{2}-2^{t-1}+3$ and is bounded below by the maximum of $2^{t-1}-3$ and $\frac{4(n-3)}{3}-2^{t-1}+3.$

In another direction we study the characteristic rank of $\widetilde{G}_{n,k}$ ($k\ge 5$) and the upper characteristic rank of $\widetilde{G}_{n,3}.$ If X is a connected finite CW complex and $\xi$ is a real (finite rank) vector bundle over X, recall from [10] that the characteristic rank of $\xi$ over X, denoted by $\texttt {charrank}_X(\xi )$, is by definition the largest integer $k \le \texttt {dim}(X)$ such that every cohomology class $x\in H^j (X;{\mathbb {Z}}_2)$, $0\le j\le k,$ is a polynomial in the Stiefel–Whitney classes $w_i(\xi )$. The upper characteristic rank of X, denoted by $\texttt {ucharrank}(X)$, is the maximum of $\texttt {charrank}_X (\xi )$ as $\xi$ varies over vector bundles over X. In [4] it was shown that

$$\begin{aligned} \texttt {charrank}_{\,\widetilde{G}_{n,k}}(\widetilde{\gamma }_{n,k})\ge n-k+1\,\,\text {for }k\ge 5, \end{aligned}$$

where $\widetilde{\gamma }_{n,k}$ is the oriented tautological bundle over $\widetilde{G}_{n,k}$. Our result (cf. Theorem 3.13, Corollary 3.17) is the following.

Theorem C

If $k\ge 5$ and $n\ge 2k$, then characteristic rank of $\widetilde{\gamma }_{n,k}$ is at least $n-k+2$.

The upper characteristic rank $\texttt {ucharrank}(\widetilde{G}_{n,3})=\texttt {charrank}(\widetilde{\gamma }_{n,3})$ for $n\ge 8$.

Unless otherwise mentioned, all cohomology is taken with ${\mathbb {Z}}_2$-coefficients.

2 Cohomology rings of relevant spaces

2.1 Preliminaries

We start by recalling some known results. The ${\mathbb {Z}}_2$-cohomology ring of $G_{n,k}$ is classically well-known due to Borel [1]. We assume that $n\ge 2k$ as there is a diffeomorphism between $G_{n,k}$ and $G_{n,n-k}$. There is an isomorphism of rings

$$\begin{aligned} H^*(G_{n,k};{\mathbb {Z}}_2)\cong \frac{{\mathbb {Z}}_2[w_1,\ldots ,w_k]}{\langle {\bar{w}}_{n-k+1},\ldots , {\bar{w}}_{n}\rangle } \end{aligned}$$

where $\langle {\bar{w}}_{n-k+1},\ldots , {\bar{w}}_{n}\rangle$ is the ideal generated by ${\bar{w}}_{n-k+1},\ldots ,\!{\bar{w}}_n.$ Here $w_i$ is the $i{\text {th}}$ Stiefel–Whitney class of the tautological k-plane bundle $\gamma$ over $G_{n,k}$. Moreover, ${\bar{w}}_i$ is the homogeneous component of $(1+w_1+\cdots + w_k)^{-1}$ in degree i. Therefore, for $i\in \{n-k+1,\ldots ,n\}$, we have

$$\begin{aligned} {\bar{w}}_i=w_1 {\bar{w}}_{i-1}+w_2 {\bar{w}}_{i-2}+\cdots +w_k {\bar{w}}_{i-k}. \end{aligned}$$

(2.1)

Let $g_i(w_2,\ldots ,w_k)$ (abbreviated to $g_i$ here after) be the reduction of ${\bar{w}}_i$ modulo $w_1$. We recall the following result from [4], which will be used in the proofs of our results.

Lemma 2.1

(i) If $k=3$ then $g_i(w_2,w_3)=0$ if and only if $i=2^r-3$ for some $r\ge 3.$

(ii) If $k\ge 5$, then for $i\ge 2$, the expression $g_i(w_2,\ldots ,w_k)$ is never zero.

Recall the following well-known result. Consider the Gysin sequence for the sphere bundle $S^{n-1}\hookrightarrow Y\xrightarrow {p}X$ associated to a real vector bundle ${\mathbb {R}}^n\hookrightarrow \xi \rightarrow X$ (cf. [9, p. 144]).

$$\begin{aligned} \cdots \rightarrow H^i(X)\xrightarrow {\cup \,w_n}H^{i+n}(X)\xrightarrow {p^*}H^{i+n}(Y)\rightarrow H^{i+1}(X)\xrightarrow {\cup \,w_n}H^{i+n+1}(X)\rightarrow \cdots \end{aligned}$$

where $w_n\in H^n(X;{\mathbb {Z}}_2)$ is the $n\text {th}$ Stiefel–Whitney class of $\xi$. It is well known that the above Gysin sequence is the degeneration of the Serre spectral sequence associated to the sphere bundle. The only possible non-zero differential in the spectral sequence is $d^n:E_n^{p,n-1}\rightarrow E_n^{p+n,0}$. Hence, we need only compute $d^n:E_n^{0,n-1}\rightarrow E_n^{n,0}$. Let $X^l$ denote the l-skeleton of X. There is a filtration $0\subset F_l^l\subset F_{l-1}^l\subset \cdots \subset F_0^l=H^l(Y)$, where

$$\begin{aligned} F^l_{j}:=\text {ker}\left( H^l(Y){\mathop {\longrightarrow }\limits ^{i^*}} H^l(p^{-1}(X^{j-1}))\right) . \end{aligned}$$

We know from [1] that

$$\begin{aligned} H^n(X)\xrightarrow {i_*} H^n(X;H^0(S^{n-1}))=E_2^{n,0}\rightarrow E_{n+1}^{n,0}=E_\infty ^{n,0}=F^n_n \hookrightarrow H^n(Y) \end{aligned}$$

is the map $p^*.$ From the Gysin sequence, we know that the kernel of the map is ${\mathbb {Z}}_2 w_n$. Therefore, $d^n(1)=w_n$ where $1\in H^0(X;H^{n-1}(S^{n-1}))$ is the generator.

We shall introduce three fibre bundles which will be important in what follows.

(i) (double cover) The universal covering map is a principal ${\mathbb {Z}}_2$-bundle of the form

$$\begin{aligned} {\mathbb {Z}}_2\hookrightarrow \widetilde{G}_{n,3}\xrightarrow {\pi } G_{n,3}. \end{aligned}$$

The associated Gysin sequence (with ${\mathbb {Z}}_2$-coefficients) is written below:

$$\begin{aligned}&\cdots \rightarrow H^{j-1}(G_{n,3})\xrightarrow {\cup w_1}H^j(G_{n,3})\xrightarrow {\pi ^*}H^j(\widetilde{G}_{n,3})\rightarrow H^j(G_{n,3})\nonumber \\&\quad \xrightarrow {\cup w_1}H^{j+1}(G_{n,3})\rightarrow \cdots \end{aligned}$$

(2.2)

We have that

$$\begin{aligned} \pi ^*(w_1)=0,\,\,\widetilde{w}_2:=\pi ^*(w_2)\ne 0,\,\,\widetilde{w}_3:=\pi ^*(w_3)\ne 0. \end{aligned}$$

For $j+1<n-2$, $H^j(G_{n,3})\xrightarrow {\cup \,w_1} H^{j+1}(G_{n,3})$ is injective, so $H^j(\widetilde{G}_{n,3})$ consists of all the homogeneous polynomials of degree j built out of $\widetilde{w}_2$ and $\widetilde{w}_3$. As $n\ge 6$, $j=2$, we conclude that $j+1$ is always less than $n-2$. So $H^2(\widetilde{G}_{n,3})=\widetilde{w}_2 {\mathbb {Z}}_2$. The above Gysin sequence implies that for $n\ge 7$, $H^3(\widetilde{G}_{n,3})=\widetilde{w}_3{\mathbb {Z}}_2$. When $n=6$, we still can check that $H^3(\widetilde{G}_{6,3})=\widetilde{w}_3{\mathbb {Z}}_2$. Thus, $H^3(\widetilde{G}_{n,3})=\widetilde{w}_3 {\mathbb {Z}}_2$ for any $n\ge 6$.

In [4, Theorem 2.1(1)] it has been shown that if $n\ne 2^t,~2^t-1,~2^t-2,~ 2^t-3$ then the map $H^j(G_{n,3})\xrightarrow {\pi ^*}H^j(\widetilde{G}_{n,3})$ is surjective for $j\le n-2$.

Proposition 2.2

Let n be an integer not equal to $2^t,~2^t-1,~2^t-2$ or $2^t-3$ for any t. Then the cohomology map $H^j(G_{n,3};{\mathbb {Z}}_2)\xrightarrow {\pi ^*}H^j(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ is surjective for $j\le n-1$.

Proof

Consider the Gysin sequence (2.2). It follows that $H^j(G_{n,3})\xrightarrow {\pi ^*}H^j(\widetilde{G}_{n,3})$ is surjective if and only if the subgroup

$$\begin{aligned} \text {ker}\left( H^j(G_{n,3})\xrightarrow {\cup w_1}H^{j+1}(G_{n,3})\right) \end{aligned}$$

(2.3)

vanishes. So, we only need to show that when $n\ne 2^t,~2^t-1,~2^t-2,~2^t-3$, the subgroup defined in (2.3) vanishes up to degree $n-1$. Let $x\ne 0\in H^{n-1}(G_{n,3})$ be such that $w_1\cup x$ vanishes in $H^n(G_{n,3}).$ Then

$$\begin{aligned} w_1\cup x=a_1w_1^2{\bar{w}}_{n-2}+a_2w_2{\bar{w}}_{n-2}+a_3w_1{\bar{w}}_{n-1}+a_4{\bar{w}}_n. \end{aligned}$$

Reducing the above equation modulo $w_1$, we get $0=a_2w_2g_{n-2}+a_4g_n.$ Therefore, if we set $a_2=0$ then we get $a_4g_n=0$. As $n\ne 2^t-3$ implies $g_n\ne 0$ (cf. Lemma 2.1), we conclude that $a_4=0$. Similarly, $a_4=0$ implies $a_2=0.$ Also, $a_2=a_4=1\in {\mathbb {Z}}_2$ gives $g_n+w_2g_{n-2}=0$. Now (2.1) with $i=n$ and $k=3$ gives us

$$\begin{aligned} {\bar{w}}_n=w_1 {\bar{w}}_{n-1}+w_2 {\bar{w}}_{n-2}+w_3 {\bar{w}}_{n-3}. \end{aligned}$$

The above equation when read modulo $w_1$ along with $g_n+w_2g_{n-2}=0$ implies $w_3g_{n-3}=0$. This is a contradiction as $n\ne 2^t.$ Therefore, $a_2=a_4=0$ and

$$\begin{aligned} x=a_1w_1{\bar{w}}_{n-2}+a_3{\bar{w}}_{n-1}. \end{aligned}$$

Thus, $x=0\in H^{n-1}(G_{n,3}),$ which proves that $\text {ker}(H^{n-1}(G_{n,3})\xrightarrow {\cup w_1}H^n(G_{n,3}))$ vanishes. $\square$

We will also require the following related result.

Lemma 2.3

The map

$$\begin{aligned} \cup \,\widetilde{w}_3:H^j(\widetilde{G}_{n,3};{\mathbb {Z}}_2)\longrightarrow H^{j+3}(\widetilde{G}_{n,3};{\mathbb {Z}}_2) \end{aligned}$$

is injective if $j\le n-6$. If n is even then the above map is injective also for $j\le n-5$.

Proof

It follows from the commutative diagram

and (2.2) that we have an induced commutative diagram

(2.4)

where the vertical arrows are injective. For $j\le n-4$ we know that $\pi ^*:H^j(G_{n,3})\rightarrow H^{j}(\widetilde{G}_{n,3})$ is surjective. Therefore, the map $\bar{\pi }^*$ is an isomorphism. As the cohomology ring of $G_{n,3}$, up to degree $n-3$, is freely generated by monomials in $w_2$ and $w_3$, the map

$$\begin{aligned} \cup \,w_3:H^j(G_{n,3})/(w_1)\longrightarrow H^{j+3}(G_{n,3})/(w_1) \end{aligned}$$

is injective whenever $j+3\le n-3$, or, equivalently $j\le n-6$. We can now conclude that the top horizontal map in (2.4) is injective for $j\le n-6$. Let n be even and $j=n-5$. If $x\in H^{j}(G_{n,3})$ such that $x+ (w_1)$ belongs to the kernel of the map above then $x w_3=w_1 \alpha + {\bar{w}}_{n-2}$ for some $\alpha \in H^{n-3}(G_{n,3})$. As n is even, the coefficient of $w_2^{n/2-1}$ in ${\bar{w}}_{n-2}$ is one. Thus, the lower horizontal arrow in (2.4) is injective, whence the top horizontal arrow is also injective. $\square$

(ii) (2-sphere bundle) We have introduced an intermediate space for our purposes of studying $\widetilde{G}_{n,3}$. Towards that we consider the following space

$$\begin{aligned} W_{2,1}^n:=\left\{ (\widetilde{P},v)\in \widetilde{G}_{n,2}\times S^{n-1}\,|\,\widetilde{P}\hbox { and }v\hbox { are orthogonal}\right\} . \end{aligned}$$

The space above is a flag manifold of the type (2, 3). Consider the bundle

(2.5)

where $\langle \widetilde{P},v\rangle$ is the oriented 3-plane generated by $\widetilde{P}$ and v, in order. This is the sphere bundle of the oriented tautological bundle $\widetilde{\gamma }_{n,3}\rightarrow \widetilde{G}_{n,3}$. The fiber over an oriented 3-plane is given by the unit vectors in it, i.e., it is $S^2$. The space $W_{2,1}^n$ is simply connected. Consider the associated Gysin sequence:

where $\widetilde{w}_3\in H^3(\widetilde{G}_{n,3})$ is the (mod 2 reduction of the) Euler class of the bundle. As $W_{2,1}^n$ is simply connected, $H^1(W_{2,1}^n)=0$. We also conclude that where . We shall denote by $\text {w}_2$.

(iii) (principalSO(3)-bundle) To introduce the following bundle, let us recall Stiefel manifolds. The real Stiefel manifold $V_k({\mathbb {R}}^n)$ is the space of all orthonormal k-tuples in ${\mathbb {R}}^n$. It is a classical result [1], due to Borel, that the cohomology ring of $V_k({\mathbb {R}}^n)$ with ${\mathbb {Z}}_2$-coefficients can be described explicitly:

$$\begin{aligned} H^*(V_k({\mathbb {R}}^n);{\mathbb {Z}}_2)\cong {\mathbb {Z}}_2[a_{n-k},a_{n-k+1},\ldots ,a_{n-1}]/I \end{aligned}$$

(2.6)

with $a_i\in H^i(V_k({\mathbb {R}}^n);{\mathbb {Z}}_2)$ and I being the ideal generated by $a_i^2=a_{2i}$ when $2i\le n-1$ and $a_i^2=0$ otherwise. Let us consider

$$\begin{aligned} SO(3)\hookrightarrow V_3({\mathbb {R}}^n)\xrightarrow {p_1} \widetilde{G}_{n,3},\,\,\,\,p_1(v_1,v_2,v_3):= (\langle v_1,v_2 ,v_3\rangle , v_1\wedge v_2\wedge v_3) \end{aligned}$$

(2.7)

where $\langle v_1,v_2,v_3\rangle$ is the 3-plane generated by the oriented bases $\{v_1,v_2,v_3\}$, i.e., the plane has the orientation given by $v_1\wedge v_2\wedge v_3$.

Definition 2.4

(Decomposability)  We say that the cohomology ring of a space at a particular degree r is decomposable with respect to a given filtration F if

$$\begin{aligned} \bigoplus _{i+j=r, p+q=s,i,j>0}F_p^i\otimes F_q^j\rightarrow F_s^r \end{aligned}$$

is surjective for all s. Moreover, we say that the $E_\infty$-page at degree r is decomposable if

$$\begin{aligned} \bigoplus _{i+j=r^\prime ,p+q=s,(i,p)\ne (0,0),(j,q)\ne (0,0))}E_\infty ^{i,p}\otimes E_\infty ^{j,q}\rightarrow E_\infty ^{r^\prime ,s} \end{aligned}$$

is surjective for all $r^\prime +s=r$.

Proposition 2.5

The Serre spectral sequence for cohomology of a fiber bundle comes with a filtration F. The cohomology ring of the total space is decomposable at degree r with respect to the filtration F if and only if $E_\infty$-page is decomposable at degree r.

Proof

Suppose the cohomology ring is decomposable at degree r with respect to the filtration F. Then we have

$$\begin{aligned} \bigoplus _{i+j=r, p+q=s,i,j>0}F_p^i\otimes F_q^j\rightarrow F_s^r \text{ is } \text{ surjective } \text{ for } \text{ all } \text{ s. } \end{aligned}$$

This implies that

$$\begin{aligned} \bigoplus _{i+j=r, p+q=s,i,j>0}F_p^i/F_{p+1}^i\otimes F_q^j/F_{q+1}^j\rightarrow F_s^r/F_{s+1}^r\hbox { is surjective for all }s.\\ \qquad \Rightarrow \bigoplus _{i+j=r, p+q=s,i,j>0}E_\infty ^{p,i-p}\otimes E_\infty ^{q,j-q}\rightarrow E_\infty ^{s,r-s}\hbox { is surjective for all } s. \end{aligned}$$

Thus the $E_\infty$-page at degree r is decomposable.

To prove the converse, suppose that the $E_\infty$-page at degree r is decomposable. Choose $x\in F_s^r-F_{s+1}^r$. Then, $x\ne 0\in E_\infty ^{s,r-s}.$ Then ${\bar{x}}=\sum _k{\bar{x}}_k\cup {\bar{y}}_k$ where $x_k\in F_p^i$, $y_k\in F_q^j$ and $i,j>0$ with $p+q=s$, $i+j=r$. So, $x-\sum _k x_k\cup y_k\in F_{s+1}^r$. Now, by induction $F_{s+1}^r$ is decomposable. So, $x\in F_s^r$ is also decomposable. Therefore, the cohomology ring of the total space is decomposable at degree r with respect to the filtration F. $\square$

It follows from the definition that if the cohomology ring is decomposable at degree r with respect to the filtration F, then the cohomology ring is decomposable at degree r.

2.2 The space $\widetilde{G}_{n,2}$ and its cohomology ring

The cohomology ring of $\widetilde{G}_{n,2}$ with ${\mathbb {Z}}_2$-coefficients has been characterized completely in [5].

Theorem 2.6

(Korbaš-Rusin) Let $n\ge 4$ and consider the Grassmannian $\widetilde{G}_{n,2}$ of oriented 2-planes in ${\mathbb {R}}^n$. Let $\widetilde{w}_2$ denote $\pi ^*w_2$, where $\pi :\widetilde{G}_{n,2}\rightarrow G_{n,2}$ is the covering map.

• (i) If n is odd then there is an isomorphism of rings

  $$\begin{aligned} H^*(\widetilde{G}_{n,2};{\mathbb {Z}}_2)\cong {\mathbb {Z}}_2[\widetilde{w}_2]/\left( \widetilde{w}_2^{\frac{n-1}{2}}\right) \otimes _{{\mathbb {Z}}_2} \Lambda _{{\mathbb {Z}}_2}(a_{n-1}) \end{aligned}$$

  where $a_{n-1}\in H^{n-1}(\widetilde{G}_{n,2};{\mathbb {Z}}_2)$.

• (ii) If $n\equiv 0\,(\text {mod}\,4)$ then

  $$\begin{aligned} H^*(\widetilde{G}_{n,2};{\mathbb {Z}}_2)\cong {\mathbb {Z}}_2[\widetilde{w}_2]/\left( \widetilde{w}_2^{\frac{n}{2}}\right) \otimes _{{\mathbb {Z}}_2} \Lambda _{{\mathbb {Z}}_2}(b_{n-2}) \end{aligned}$$

  where $b_{n-2}\in H^{n-2}(\widetilde{G}_{n,2};{\mathbb {Z}}_2)$.

• (iii) If $n\equiv 2\,(\text {mod}\,4)$ then

  $$\begin{aligned} H^*(\widetilde{G}_{n,2};{\mathbb {Z}}_2)\cong \frac{{\mathbb {Z}}_2[\widetilde{w}_2]/\left( \widetilde{w}_2^{\frac{n}{2}}\right) \otimes _{{\mathbb {Z}}_2} {{\mathbb {Z}}_2}[b_{n-2}]}{\left( b_{n-2}^2-\widetilde{w}_2^{\frac{n-2}{2}}b_{n-2}\right) }, \end{aligned}$$

  where $b_{n-2}\in H^{n-2}(\widetilde{G}_{n,2};{\mathbb {Z}}_2)$.

We note that (ii) and (iii) of the above theorem can be deduced from a much stronger result of Lai [7]. Let $\widetilde{\gamma }\rightarrow \widetilde{G}_{2n+2,2}$ be the oriented tautological bundle.

Theorem 2.7

(Lai) The integral cohomology groups of $\widetilde{G}_{2n+2,2}$ are isomorphic to those of $\mathbb {CP}^n\times S^{2n}$. Moreover, the cohomology ring is generated by $\widetilde{\Omega }:=e(\widetilde{\gamma })\in H^2(\widetilde{G}_{2n+2,2};{\mathbb {Z}})$ and $\Omega :=e(\widetilde{\gamma }^\perp ), \kappa \in H^{2n}(\widetilde{G}_{2n+2,2};{\mathbb {Z}})$ with the relations

$$\begin{aligned} \widetilde{\Omega }^{n+1}= & {} 2\kappa \cup \widetilde{\Omega },\,\,\kappa \cup \widetilde{\Omega }^n=(-1)^n,\,\,\widetilde{\Omega }^{2n}=2(-1)^n,\,\,\kappa \cup \kappa =(1+(-1)^n)/2\\ \Omega \cup \Omega= & {} 2,\,\,\kappa \cup \Omega =1,\,\,\Omega +\widetilde{\Omega }^n=2\kappa . \end{aligned}$$

Proof of Theorem 2.6(ii),(iii) using Theorem 2.7. It follows from universal coefficient theorem in cohomology and homology that there is an isomorphism of rings

$$\begin{aligned} H^*(\widetilde{G}_{2n+2,2};{\mathbb {Z}})\otimes {\mathbb {Z}}_2\xrightarrow {\cong } H^*(\widetilde{G}_{2n+2,2};{\mathbb {Z}}_2). \end{aligned}$$

Note that the isomorphism above identifies $\widetilde{\Omega }\otimes 1$ with $\widetilde{w}_2$, the second Stiefel–Whitney class of $\widetilde{\gamma }$. Let b denote $\kappa \otimes 1\in H^{2n}(\widetilde{G}_{2n+2,2};{\mathbb {Z}}_2)$. Therefore, the relations among generators in $H^*(\widetilde{G}_{2n+2,2};{\mathbb {Z}}_2)$ become

$$\begin{aligned} \widetilde{w}_2^{n+1}=0,\,\,b\cup \widetilde{w}_2^n=1,\,\,\widetilde{w}_2^{2n}=0,\,\,b\cup b=\left\{ \begin{array}{ll} 0 &{}\quad \text {if }\,2n+2\, \equiv \, 0\text { mod }4\\ 1 &{}\quad \text {if }\,2n+2\, \equiv \, 2\text { mod }4 \end{array}\right. \end{aligned}$$

The third relation is redundant as it follows from the first while the second and fourth combine to give the relation $b^2=0$ or $b^2+b\widetilde{w}_2^n=0$ depending on whether $2n+2$ is 0 or 2 module 4. $\square$

It follows that when n is odd, $H^*(\widetilde{G}_{n,2})$ has only 2 indecomposables $\widetilde{w}_2,a_{n-1}$ in degrees 2 and $n-1$ respectively while if n is even, $H^*(\widetilde{G}_{n,2})$ has only 2 indecomposables $\widetilde{w}_2,b_{n-2}$ in degree 2 and $n-2$ respectively. We wish to analyze the map induced, in cohomology with ${\mathbb {Z}}_2$-coefficients, by the natural inclusion $\widetilde{i}:\widetilde{G}_{n,2}\hookrightarrow \widetilde{G}_{n+1,2}$.

Proposition 2.8

The induced map $\widetilde{i}^*:H^*(\widetilde{G}_{n+1,2};{\mathbb {Z}}_2)\rightarrow H^*(\widetilde{G}_{n,2};{\mathbb {Z}}_2)$ on the cohomology rings is described by the following action on the generators:

$$\begin{aligned}&\widetilde{w}_2\mapsto \widetilde{w}_2,\,b_{n-1}\mapsto a_{n-1} \quad \mathrm{if}\;n\;\mathrm{is\,odd}\\&\widetilde{w}_2\mapsto \widetilde{w}_2,\,a_{n}\mapsto \widetilde{w}_2 b_{n-2} \quad \mathrm{if}\;n\;\mathrm{is\, even}. \end{aligned}$$

Proof

There is map of covering spaces

The tautological bundle over $G_{n+1,2}$ pulls back under i to the tautological bundle over $G_{n,2}$. Similar considerations hold for all the other three maps in the diagram above. This implies that $\widetilde{i}^*(\widetilde{w}_2)=\widetilde{w}_2$. The above diagram also induces a map between the Gysin sequences associated to the double covers:

We now analyze the two cases separately, depending on the parity of n.

Case 1  Let n be odd, $j=n$ and note that

$$\begin{aligned} \pi ^*(H^{n-1}(G_{n+1,2}))={\mathbb {Z}}_2 \widetilde{w}_2^{\frac{n-1}{2}},\,\,\pi ^*(H^{n-1}(G_{n,2}))=0 \end{aligned}$$

The diagram of long exact sequences now looks like

The non-zero element $\varphi _{n+1}(b_{n-1})$ is mapped to zero in $H^n(G_{n+1,2})$ when multiplied by $w_1$. We know that $H^j(G_{n+1,2})$ is freely generated by monomials in $w_1,w_2$ if $j\le n-1$. In $H^n(G_{n+1,2})$ we have ${\bar{w}}_n=0$. As n is odd, let $p_{n-1}\in H^{n-1}(G_{n+1,2})$ such that ${\bar{w}}_n=w_1\cup p_{n-1}$. It follows that $\varphi _{n+1}(b_{n-1})=p_{n-1}$. To prove that $\widetilde{i}^*(b_{n-1})=a_{n-1}$, it suffices to show that $i^*(p_{n-1})\ne 0$ in $H^{n-1}(G_{n,2})$. Note that

$$\begin{aligned} {\bar{w}}_n=w_1{\bar{w}}_{n-1}+w_2{\bar{w}}_{n-2}=w_1({\bar{w}}_{n-1}+w_2q_{n-3}) \end{aligned}$$

as ${\bar{w}}_{n-2}$ is non-zero and can be written as $w_1 q_{n-3}$. Thus, $p_{n-1}={\bar{w}}_{n-1}+w_2q_{n-3}$ and

$$\begin{aligned} i^*(p_{n-1})=i^*({\bar{w}}_{n-1})+i^*(w_2 q_{n-3})=w_2 q_{n-3} \end{aligned}$$

where the last equality follows from the fact that ${\bar{w}}_{n-1}=0\in H^{n-1}(G_{n,2})$ and $i^*({\bar{w}}_{n-2})={\bar{w}}_{n-2}$. Since $w_2q_{n-3}$ has no term of the form $w_1^{n-1}$, it cannot equal ${\bar{w}}_{n-1}$, whence $w_2 q_{n-3}\ne 0$ in $H^{n-1}(G_{n,2})$. This implies that $i^*(p_{n-1})\ne 0$.

Case 2  Let n be even and we shall show that $\widetilde{i}^*(a_n)=\widetilde{w}_2 b_{n-2}$. Since $H^n(\widetilde{G}_{n+1,2})={\mathbb {Z}}_2 a_n$ and $H^n(\widetilde{G}_{n,2})={\mathbb {Z}}_2 \widetilde{w}_2b_{n-2}$, it suffices to show that $\widetilde{i}^*$ is injective in the following diagram

As both the maps $\pi ^*$ are zero, the map $\varphi _n$ and $\varphi _{n+1}$ are injective. Let $\varphi _{n+1}(a_n)=p_n$ where $w_1 p_n={\bar{w}}_{n+1}$. Now,

$$\begin{aligned} {\bar{w}}_{n+1}=w_1 {\bar{w}}_n+w_2 {\bar{w}}_{n-1}=w_1 {\bar{w}}_n + w_2 w_1 q_{n-2}=w_1 (w_2 q_{n-2})\in H^{n+1}(G_{n+1,2}) \end{aligned}$$

for some non-zero $q_{n-2}\in H^{n-2}(G_{n,2})$. Therefore, $p_n=w_2 q_{n-2}$ and $i^*(p_n)=w_2 q_{n-2}$. It follows from the definition of ${\bar{w}}_{n-1}$ that

$$\begin{aligned} q_{n-2}= & {} \sum _{r=0}^{\frac{n-2}{2}} {n-1-r \atopwithdelims ()r} w_2^r w_1^{n-2-2r}\nonumber \\= & {} \sum _{r=0}^{\frac{n-2}{2}} {n-2-r \atopwithdelims ()r} w_2^r w_1^{n-2-2r}+\sum _{r=0}^{\frac{n-2}{2}} {n-2-r \atopwithdelims ()r-1} w_2^r w_1^{n-2-2r}\nonumber \\= & {} {\bar{w}}_{n-2} + \left( w_2 w_1^{n-4}+\cdots + \frac{n-2}{2} w_2^{\frac{n-2}{2}}\right) . \end{aligned}$$

(2.8)

It can be verified that if $w_2q_{n-2}=0\in H^n(G_{n,2})$ then $w_2q_{n-2}=\lambda w_1 {\bar{w}}_{n-1}+\mu {\bar{w}}_n$, where $\lambda ,\mu \in {\mathbb {Z}}_2$. Note that $w_1^n$ occurs as a term in ${\bar{w}}_n$ as well as $w_1{\bar{w}}_{n-1}$ while no such term occurs in $w_2 q_{n-2}$. Therefore, $\lambda =\mu =1$ which implies $q_{n-2}={\bar{w}}_{n-2}$. This contradicts the equality (2.8). Thus, $i^*(p_n)\ne 0$ and $\widetilde{i}^*(a_n)=\widetilde{w}_2 b_{n-2}$. $\square$

2.3 The space $W_{2,1}^n$ and its cohomology ring

Let us consider the sphere bundle

$$\begin{aligned} S^{n-3}\hookrightarrow W_{2,1}^n\xrightarrow {p}\widetilde{G}_{n,2},\,\,(\widetilde{P},v)\mapsto \widetilde{P} \end{aligned}$$

Note that if $\widetilde{\gamma }\rightarrow \widetilde{G}_{n,2}$ denotes the oriented tautological bundle then $W^n_{2,1}$ is the sphere bundle of $\widetilde{\gamma }^\perp$. Thus, the Euler class of $W_{2,1}^n$ is the Euler class $\Omega :=e(\widetilde{\gamma }^\perp )\in H^{n-2}(\widetilde{G}_{n,2};{\mathbb {Z}})$. Using results proved in Sect. 2.2 we can completely describe the cohomology ring of $W^n_{2,1}$ with ${\mathbb {Z}}_2$-coefficients.

Theorem 2.9

(a) The cohomology ring of $W^{2n+2}_{2,1}$ with ${\mathbb {Z}}_2$-coefficients is isomorphic to that of $\mathbb {CP}^{n-1}\times S^{2n}\times S^{2n+1}$. There are generators $\text {w}_2:=p^*\widetilde{w}_2, c_{2n}:=p^*b_{2n}$ and $c_{2n+1}$ in degrees 2n, 2 and $2n+1$ respectively, such that

$$\begin{aligned} H^*\left( W^{2n+2}_{2,1};{\mathbb {Z}}_2\right) \cong {\mathbb {Z}}_2[\text {w}_2]/(\text {w}_2^n)\otimes \Lambda _{{\mathbb {Z}}_2}(c_{2n})\otimes \Lambda _{{\mathbb {Z}}_2}(c_{2n+1}). \end{aligned}$$

(b) The cohomology ring of $W^{2n+1}_{2,1}$ with ${\mathbb {Z}}_2$-coefficients is isomorphic to that of $\mathbb {CP}^{n-1}\times S^{2n-2}\times S^{2n}$. There are generators $\text {w}_2:=p^*\widetilde{w}_2, d_{2n}:=p^*a_{2n}$ and $d_{2n-2}$ in degrees 2n, 2 and $2n-2$ respectively, such that

$$\begin{aligned} H^*\left( W^{2n+1}_{2,1};{\mathbb {Z}}_2\right) \cong \frac{\big ({\mathbb {Z}}_2[\text {w}_2]/(\text {w}_2^n)\otimes \Lambda _{{\mathbb {Z}}_2}(d_{2n})\big )[d_{2n-2}]}{\left\langle d_{2n-2}^2-\lambda \text {w}_2^{n-2}d_{2n}-\mu \text {w}_2^{n-1}d_{2n-2}\right\rangle } \end{aligned}$$

for some $\lambda ,\mu \in {\mathbb {Z}}_2$.

Proof

We prove (a) first. It follows from the Serre spectral sequence in (integral cohomology) for $S^{2n-1}\hookrightarrow W_{2,1}^{2n+2}\xrightarrow {p}\widetilde{G}_{2n+2,2},\,\,(\widetilde{P},v)\mapsto \widetilde{P}$ that the only non-zero differential is $d^{2n}$ for the $E_{2n}$-page. Moreover, $d^{2n}$ is multiplication by $\Omega$. Using $\Omega =2\kappa -\widetilde{\Omega }^n$, it can be verified that the integral cohomology groups of $W^{2n+2}_{2,1}$ are the same as that of $\mathbb {CP}^{n-1}\times S^{2n}\times S^{2n+1}$. In $E_{2n+1}$-page we see that $E_{2n+1}^{0,2n-1}=0$ and $E_{2n+1}^{2,2n-1}={\mathbb {Z}}$. Let this integer be generated by an element $c\in H^{2n+1}(W^{2n+2}_{2,1};{\mathbb {Z}})$. Similarly, $E_{2n+1}^{4n,0}=0$, i.e., the top cohomology class of $\widetilde{G}_{2n+2,2}$ is killed by $d^{2n}(\kappa \otimes 1)=\kappa \cup \Omega$.

A similar consideration with ${\mathbb {Z}}_2$-coefficients implies that $H^*(W^{2n+2}_{2,1};{\mathbb {Z}}_2)$ is isomorphic as groups to that of $H^*(\mathbb {CP}^{n-1}\times S^{2n}\times S^{2n+1};{\mathbb {Z}}_2)$. Here the relation $\Omega =2\kappa -\widetilde{\Omega }^n$ becomes $\Omega =\widetilde{\Omega }^n$ as we are working with ${\mathbb {Z}}_2$-coefficients. If $1\otimes 1\in E^{0,2n-1}_{2n}$ then

$$\begin{aligned} d^{2n}(1\otimes 1)=\Omega =\widetilde{\Omega }^n=\widetilde{w}_2^n \end{aligned}$$

and we conclude that $E_\infty ^{2n,0}=E_{2n+1}^{2n,0}={\mathbb {Z}}_2 p^*\kappa$. Since $E_{2n+1}^{4n,0}=E_\infty ^{4n,0}=0$ we conclude that $p^*\kappa \cup p^*\kappa =0$. Let $c_{2n+1}$ denote the generator of $E_\infty ^{2,2n-1}=E_{2n+1}^{2,2n-1}={\mathbb {Z}}_2$. It can be shown using Proposition 2.5 that the only indecomposables occur in degrees $2, 2n, 2n+1$. Since $H^{4n+2}(W^{2n+2}_{2,1};{\mathbb {Z}}_2)=0$ it follows that $c_{2n+1}^2=0$. These relations along with non-degeneracy of cup product for $W^{2n+2}_{2,1}$ imply that the cohomology ring is generated by $\text {w}_2,c_{2n},c_{2n+1}$ with the relations

$$\begin{aligned} \text {w}_2^{n}=0,\quad c_{2n}^2=0,\quad c_{2n+1}^2=0. \end{aligned}$$

This completes the proof of (a).

Consider the Serre spectral sequence for the sphere bundle $S^{2n-2}\hookrightarrow W_{2,1}^{2n+1}\xrightarrow {p}\widetilde{G}_{2n+1,2}$. The only possible non-zero differential is $d^{2n-1}$ which is given by multiplying by (the mod 2 reduction of) $e(\widetilde{\gamma }^\perp )\in H^{2n-1}(\widetilde{G}_{2n+1,2};{\mathbb {Z}}_2)=0$. Therefore,

$$\begin{aligned} E_\infty ^{p,q}=H^p(\widetilde{G}_{2n+1,2})\otimes H^{q}(S^{2n-2}) \end{aligned}$$

and $W^{2n+1}_{2,1}$ has the same cohomology groups as that of $\mathbb {CP}^{n-1}\times S^{2n-2}\times S^{2n}$. According to Proposition 2.5, $H^*(W_{2,1}^{2n+1})$ is decomposable in degrees other than $2,2n-2, 2n$. From the ring structure of $E_\infty$ we observe that there are indecomposables only in degree $2, 2n-2, 2n$, denoted by $\text {w}_2,d_{2n-2}$ and $d_{2n}$ where $\text {w}_2=p^*\widetilde{w}_2$ and $d_{2n}=p^*(a_{2n})$ are in $p^*(H^*(\widetilde{G}_{2n+1,2}))$ and are the only indecomposables in these degrees. However, $d_{2n-2}$ is not uniquely determined as an indecomposable. In $H^*(\widetilde{G}_{2n+1,2})$, $a_{2n}^2=0$ and this implies that $d_{2n}^2=0$. Similarly, it follows from Theorem 2.6 (i) that $\text {w}_2^{n-1}\ne 0$ but $\text {w}_2^n=0$. It follows from non-degeneracy of cup product that $\text {w}_2^{n-1} d_{2n} d_{2n-2}$ is the top cohomology class. Note that $\text {w}_2^j d_{2n}\ne \text {w}_2^{j+1} d_{2n-2}$ for $j\le n-2$; if these were equal then

$$\begin{aligned} 0\ne \text {w}_2^{n-1} d_{2n-2} d_{2n}=\text {w}_2^{n-2-j}\left( \text {w}_2^{j+1} d_{2n-2}\right) d_{2n}=\text {w}_2^{n-2-j}\text {w}_2^{j} d_{2n}^2=0 \end{aligned}$$

which is a contradiction. We note that

$$\begin{aligned} d_{2n-2}^2=\lambda \text {w}_2^{n-2}d_{2n}+\mu \text {w}_2^{n-1}d_{2n-2} \end{aligned}$$

where $\lambda ,\mu \in {\mathbb {Z}}_2$. $\square$

Let $\iota :W_{2,1}^n\hookrightarrow W_{2,1}^{n+1}$ be the canonical inclusion. We shall compute $\iota ^*:H^*(W_{2,1}^{n+1})\rightarrow H^*(W_{2,1}^n)$, by evaluating $\iota ^*$ on the indecomposables, in the following result.

Theorem 2.10

The map $\iota ^*:H^*(W_{2,1}^{n+1};{\mathbb {Z}}_2)\rightarrow H^*(W_{2,1}^n;{\mathbb {Z}}_2)$ is the following:

$$\begin{aligned}&\iota ^*:H^*\left( W_{2,1}^{2k+2}\right) \rightarrow H^*\left( W_{2,1}^{2k+1}\right) , \quad \text {w}_2\mapsto \text {w}_2,\, c_{2k}\mapsto d_{2k},\,c_{2k+1}\mapsto 0\\&\iota ^*:H^*\left( W_{2,1}^{2k+1}\right) \rightarrow H^*\left( W_{2,1}^{2k}\right) , \quad \text {w}_2\mapsto \text {w}_2,\, d_{2k-2}\mapsto c_{2k-2},\,d_{2k}\mapsto \text {w}_2 c_{2k-2}. \end{aligned}$$

Proof

We consider the following map of fibre bundles:

It is clear that $\iota ^*(\text {w}_2)=\text {w}_2$ because $\text {w}_2$ in $H^*(W_{2,1}^n)$ and $H^*(W_{2,1}^{n+1})$ are the images of the same $\widetilde{w}_2$ from $H^*(\widetilde{G}_{n,2})$ and $H^*(\widetilde{G}_{n+1,2})$ respectively. Now, we consider the even and odd cases separately.

Case 1  Let $n=2k+1$ be odd. We know that $H^*(W_{2,1}^{2k+2})$ is generated by $\text {w}_2, c_{2k}=p^*(b_{2k}), c_{2k+1}$ while $H^*(W_{2,1}^{2k+1})$ is generated by $\text {w}_2, d_{2k}=p^*(a_{2k}), d_{2k-2}$. It follows from Proposition 2.8 that $\widetilde{i}^*(b_{2k})=a_{2k}$. This implies that $\iota ^*(c_{2k})=d_{2k}$. As $H^*(W_{2,1}^{2k+1})$ is non-zero in even degrees only, we have $\iota ^*(c_{2k+1})=0$.

Case 2  Let $n=2k$ be even. As $H^*(W_{2,1}^{2k+1})$ is generated by $\text {w}_2, d_{2k-2}, d_{2k}=p^*(a_{2k})$ and $H^*(W_{2,1}^{2k})$ is generated by $\text {w}_2, c_{2k-2}=p^*(b_{2k-2}), c_{2k-1}$ and $\widetilde{i}^*(a_{2k})=\widetilde{w}_2 b_{2k-2}$ (cf. Proposition 2.8), it follows that $\iota ^*(d_{2k})=\text {w}_2 c_{2k-2}$. It remains to compute $\iota ^*(d_{2k-2})$. To prove this, we need to consider the map between Serre spectral sequences (in ${\mathbb {Z}}_2$-cohomology) related to the following map between fibre bundles (cf. (2.5)).

(2.9)

Note that either of the fibre bundles are unit sphere bundles of the oriented tautological 3-plane bundle $\widetilde{\gamma }_{m,3}$ over the base $\widetilde{G}_{m,3}$. Therefore, in both the spectral sequences the only non-zero differential is $d^3$, which is given by multiplication by cup product with $\widetilde{w}_3\in H^3(\widetilde{G}_{n,3})$. This class is actually the mod 2 reduction of the Euler class of $\widetilde{\gamma }$.

We first assume that $n=2k\ne 2^t-2$ for any t. Let us consider the Serre spectral sequence for the fibre bundle . Note that $H^{2k-2}(W_{2,1}^{2k})\cong {\mathbb {Z}}_2$ generated by $c_{2k-2}$. The only non-zero differential is in the $E_3$-page, given by cup product with $\widetilde{w}_3$. Thus, the $E_2$-page equals the $E_3$-page and $E_3^{j,2}=H^j(\widetilde{G}_{2k,3})\otimes H^2(S^2)$. Let denote the generator of $H^2(S^2)\cong {\mathbb {Z}}_2$. Thus, in the $E_3$-page

is injective whenever $j\le 2k-5$ due to Lemma 2.3. Now consider ${\bar{w}}_{2k-1}$; it is a polynomial of degree $2k-1$ in $w_1, w_2$ and $w_3$. As $g_{2k-1}$ is the reduction of ${\bar{w}}_{2k-1}$ modulo $w_1$, $g_{2k-1}=\pi ^*({\bar{w}}_{2k-1})=0\in H^{2k-1}(\widetilde{G}_{2k,3})$. Observe that $g_{2k-1}$, as a polynomial in $\widetilde{w}_2,\widetilde{w}_3$ is non-zero (cf. Lemma 2.1 (i)) because otherwise $2k-1=2^t-3$, which is impossible by our assumption on n. Moreover, as $2k-1$ is odd, $g_{2k-1}$ cannot have $\widetilde{w}_2^k$ as a term. We may write $g_{2k-1}=\widetilde{w}_3 \widetilde{P}_{2k-4}$, where $\widetilde{P}_{2k-4}=f(\widetilde{w}_2,\widetilde{w}_3)$ is non-zero polynomial in $\widetilde{w}_2,\widetilde{w}_3$ of degree¹$2k-4$. As $H^{2k-4}(G_{2k,3})$ is generated freely by polynomials in $w_1,w_2,w_3$ of degree $2k-4$, we conclude that $P_{2k-4}:=f(w_2,w_3)\ne 0$ in $H^{2k-4}(G_{2k,3})$. Now $\bar{\pi }^*:H^{j}(G_{2k,3})/(w_1)\rightarrow H^j(\widetilde{G}_{2k,3})$ is injective. Therefore, $\widetilde{P}_{2k-4}=\bar{\pi }^*(P_{2k-4})\ne 0$ and

$$\begin{aligned} E_\infty ^{2k-4,2}=E_4^{2k-4,2}=\text {ker}(E_3^{2k-4,2}\xrightarrow {\cup \, \widetilde{w}_3}E_3^{2k-1,0})={\mathbb {Z}}_2, \end{aligned}$$

generated by . As

$$\begin{aligned} 1=\dim _{{\mathbb {Z}}_2}H^{2k-2}(W^{2k}_{2,1})= \dim _{{\mathbb {Z}}_2}\big ( E_\infty ^{2k-2,0}\oplus E_\infty ^{2k-4,2}), \end{aligned}$$

this implies that $E_\infty ^{2k-2,0}=0$. Thus, the element in $E_\infty ^{2k-4,2}$ is identified with $c_{2k-2}\in H^{2k-2}(W^{2k}_{2,1})$.

We now consider the Serre spectral sequence related to the sphere bundle Like the last analysis, we conclude that $E_\infty ^{j,2}=0$ for $j\le (2k+1)-6=2k-5$. Thus, the first non-zero $E_\infty ^{j,2}$ occurs when $j=2k-4$. Even here, and this element is identified with either $d_{2k-2}$ or $d_{2k-2}+\text {w}_2^{k-1}$. The map between spectral sequences implies that $E_\infty ^{2k-4,2}\rightarrow E_\infty ^{2k-4,2}$ is non-zero. It follows that

$$\begin{aligned} \iota ^*:H^{2k-2}\left( W_{2,1}^{2k+1}\right) \rightarrow H^{2k-2}\left( W_{2,1}^{2k}\right) ,\,\,d_{2k-2}\rightarrow c_{2k-2},\,\text {w}_2^{k-1}\rightarrow 0. \end{aligned}$$

We now assume that $n=2k=2^t-2$ for some $t\ge 3$. Then $H^{2k-2}(\widetilde{G}_{2k,3})$ has the first indecomposable element after degree 3 due to [4]. The image of this element under will be $c_{2k-2}+\mu \text {w}_2^{k-1}$ for some $\mu \in {\mathbb {Z}}_2$. Similarly, $H^{2k-2}(\widetilde{G}_{2k+1,3})$ has got the first indecomposable element after degree 3. The image of this element will be $d_{2k-2}+\lambda \text {w}_2^{k-1}$ for some $\lambda \in {\mathbb {Z}}_2.$ From the naturality of the Gysin sequence, we find that $\iota ^*(d_{2k-2})+ \lambda \text {w}_2^{k-1}=c_{2k-2}+\kappa \text {w}_2^{k-1}$ for some $\kappa \in {\mathbb {Z}}_2$. So, $\iota ^*(d_{2k-2})=c_{2k-2}+(\kappa +\lambda ) \text {w}_2^{k-1}=c_{2k-2}$. $\square$

3 Main results and applications

3.1 Main theorems

We shall start by discussing some relations in the cohomology ring of $G_{n,3}$.

Lemma 3.1

For $2^{t-1}< n\le 2^t$, we have $w_2^{n-2^{t-1}}g_{n-3}=g_{3n-2^t-3}+P,$ where P is a polynomial in $w_2,~g_i$’s with $i\ge n-1$ and each monomial contains exactly one $g_i$.

Proof

We begin by noting that the result holds, by a direct check, when $n=6,7,8$ (for $t=3$). We shall now prove the result by induction. We assume that $t\ge 4$. When $n=2^{t-1}+2$, we have

$$\begin{aligned} w_2^2g_{2^{t-1}-1}= & {} w_2(g_{2^{t-1}+1}+w_3g_{2^{t-1}-2})\\= & {} g_{2^{t-1}+3}+w_3g_{2^{t-1}}+w_2w_3g_{2^{t-1}-2}\\= & {} g_{2^{t-1}+3}+w_3^2g_{2^{t-1}-3}+2w_2w_3g_{2^{t-1}-2}\\= & {} g_{2^{t-1}+3} \end{aligned}$$

as $g_{2^{t-1}-3}=0$. When $n=2^{t-1}+1$ we obtain $w_2 g_{2^{t-1}-2}=g_{2^{t-1}}.$ Let us assume that the statement is true for n. We shall prove it for $n+2$. Observe that

$$\begin{aligned} w_3^2g_{3n-2^t-3} = w_3^2\left( w_2^{n-2^{t-1}}g_{n-3}+P\right) = w_2^{n-2^{t-1}}w_3\left( g_n+w_2g_{n-2}\right) +w_3P_1, \end{aligned}$$

where $P_1$ is a polynomial in $w_2, g_i$ with $i\ge n$ and each monomial contains exactly one $g_i$. Hence,

$$\begin{aligned} w_3^2g_{3n-2^t-3}= & {} w_2^{n-2^{t-1}}\left( g_{n+3}+w_2g_{n+1}+w_2g_{n+1}+w_2^2g_{n-1}\right) +P_2\\= & {} w_2^{n-2^{t-1}}\left( g_{n+3}+w_2^2g_{n-1}\right) +P_2 \end{aligned}$$

where $P_2$ is a polynomial in $w_2, g_i$ with $i\ge n+1$ and each monomial contains exactly one $g_i$. Therefore,

$$\begin{aligned} w_2^{n-2^{t-1}+2}g_{n-1}= & {} w_3^2g_{3n-2^t-3}+w_2^{n-2^{t-1}}g_{n+3}+P_2\\= & {} g_{3n-2^t+3}+w_2^2g_{3n-2^t-1}+w_2^{n-2^{t-1}}g_{n+3}+P_2\\= & {} g_{3n-2^t+3}+P_3 \end{aligned}$$

where $P_3$ is a polynomial in $w_2, g_i$ with $i\ge n+1$ and each monomial contains exactly one $g_i$ (as $n\ge 2^{t-1}+1$ implies $3n-2^t-1\ge n+1$). $\square$

Consider the Serre spectral sequence related to the sphere bundle . Here $E_2^{*,j}=0$ if $j\ne 0,2$, and the differentials $d^i=0$ if $i\ne 3$ and $d^3:E_3^{0,2}\rightarrow E_3^{3,0}$ is given by cup product with $\widetilde{w}_3$. Note that $E_2^{i,j}\ne 0$ implies $j=0,2$.

Let n be even and $n\ne 2^t,~2^t-2$. Since $n-1$ is odd, if $i\le n-3$ then $\text {ker}(H^i(\widetilde{G}_{n,3})\xrightarrow {\cup \widetilde{w}_3}H^{i+3}(\widetilde{G}_{n,3}))$ is non-zero only when $i=n-4$ and $i=n-3$. In each case the dimension of the kernel is 1; call these elements $p_{n-4}$ and $p_{n-3}$ respectively, where

$$\begin{aligned} \widetilde{w}_3 p_{n-4} = g_{n-1},\,\,\,\widetilde{w}_3 p_{n-3} = g_n +\widetilde{w}_2 g_{n-2}. \end{aligned}$$

(3.1)

So, $E_\infty ^{n-4,2}\ne 0$ and $E_\infty ^{n-3,2}\ne 0$ and as up to degree $n-1$, $H^*(G_{n,3})\xrightarrow {\pi ^*}H^*(\widetilde{G}_{n,3})$ is surjective, we have $c_{n-2}\not \in F_{n-2}^{n-2}$ and $c_{n-1}\not \in F_{n-1}^{n-1}.$ So, and .

Similarly, when n is odd and $n\ne 2^t-1,~2^t-3$ we have that $d_{n-3}\not \in F_{n-3}^{n-3}$ and $d_{n-1}\not \in F_{n-1}^{n-1}$. We also have

where the following hold

$$\begin{aligned} \widetilde{w}_3 p_{n-5} = g_{n-2},\,\,\,\widetilde{w}_3 p_{n-3} = g_n. \end{aligned}$$

(3.2)

Recall (2.9) in the proof of Theorem 2.10; we get maps $E_\infty ^{*,*}(W_{2,1}^{n+1})\xrightarrow {\iota ^*}E_\infty ^{*,*}(W_{2,1}^n).$ In fact, as is implied by Theorem 2.10, when n is odd

$$\begin{aligned} \bar{\text {w}}_2\rightarrow \bar{\text {w}}_2, ~{\bar{c}}_{n-1}\rightarrow {\bar{d}}_{n-1}, ~{\bar{c}}_n\rightarrow 0 \end{aligned}$$

(3.3)

while when n is even

$$\begin{aligned} \bar{\text {w}}_2\rightarrow \bar{\text {w}}_2,~{\bar{d}}_{n-2}\rightarrow {\bar{c}}_{n-2},~{\bar{d}}_n\rightarrow \bar{\text {w}}_2 {\bar{c}}_{n-2}. \end{aligned}$$

(3.4)

Let n be odd such that $n\ne 2^t-1,~2^t-3$. As $E_2^{n-3,2}(W_{2,1}^{n+1})\xrightarrow {\iota ^*}E_2^{n-3,2}(W_{2,1}^n)$, we observe that is ${\bar{d}}_{n-1}$. Again we know that is $\bar{\text {w}}_2 {\bar{d}}_{n-3}$. Therefore,

(3.5)

where

$$\begin{aligned} q_{n-3}=p_{n-3}+\widetilde{w}_2 p_{n-5}. \end{aligned}$$

(3.6)

Proposition 3.2

Let $2^{t-1}<n\le 2^t-4$ and consider the Serre spectral sequence regarding the sphere bundle .

• (i) If n is even then the least integer i such that $\text {w}_2^i c_{n-2}\in F_{2i+n-2}^{2i+n-2}$ is $2^{t-1}-\frac{n}{2}-1$ and the least integer i such that $\text {w}_2^i c_{n-1}\in F_{2i+n-1}^{2i+n-1}$ is $n-2^{t-1}.$

• (ii) If n is odd then

  • (a) the least integer i such that $\text {w}_2^i d_{n-1}\in F_{2i+n-1}^{2i+n-1}$ is $2^{t-1}-\frac{n+3}{2}$,

  • (b) the least integer i such that $\text {w}_2^i d_{n-3}\in F_{2i+n-3}^{2i+n-3}$ is $\max \{2^{t-1}-\frac{n+1}{2},n-2^{t-1}+1\}$,

  • (c) the least integer i such that $\text {w}_2^{i+1}d_{n-3}+\text {w}_2^i d_{n-1}\in F_{2i+n-1}^{2i+n-1}$ is $n-2^{t-1}.$

Proof

Consider the ring $H^*(W_{2,1}^{2^t-4})$ generated by $\text {w}_2, ~c_{2^t-6},~c_{2^t-5}$. Let $p_{2^t-8}\in H^{2^t-8}(\widetilde{G}_{2^t-4,3})$ be such that $\widetilde{w}_3 p_{2^t-8}=g_{2^t-5}.$ Then

As $g_{2^t-3}=0$ (cf. Lemma 2.1) we have

$$\begin{aligned} 0&=g_{2^t -3}=\widetilde{w}_2 g_{2^t-5}+\widetilde{w}_3 g_{2^t-6} =\widetilde{w}_2 \widetilde{w}_3 p_{2^t-8}\\&\quad +\,\widetilde{w}_3 g_{2^t-6} \Longrightarrow \widetilde{w}_2 p_{2^t-8}+g_{2^t-6} = 0. \end{aligned}$$

Therefore, $\widetilde{w}_2 p_{2^t-8} = 0\in H^*(\widetilde{G}_{2^t-4,3})$ and . This implies that

$$\begin{aligned} \bar{\text {w}}_2 {\bar{c}}_{2^t-6}=0 \end{aligned}$$

(3.7)

in $E_4^{2^t-6,2}$. Now we make use of the following observations which are proved later. These proofs make use of (3.7).

Lemma 3.3

If n is even and $2^{t-1}<n\le 2^t-4$ then we have the following:

$$\begin{aligned}&\text {w}_2^{2^{t-1}-\frac{n}{2}-1}c_{n-2} \in F_{2^t-4}^{2^t-4}\end{aligned}$$

(3.8)

$$\begin{aligned}&\text {w}_2^{n-2^{t-1}}c_{n-1} \in F_{3n-2^t-1}^{3n-2^t-1}\end{aligned}$$

(3.9)

$$\begin{aligned}&\text {w}_2^i c_{n-2} \not \in F_{2i+n-2}^{2i+n-2}\quad \text {for all}\,\,i\le 2^{t-1}-\frac{n}{2}-2\end{aligned}$$

(3.10)

$$\begin{aligned}&\text {w}_2^i c_{n-1} \not \in F_{2i+n-1}^{2i+n-1}\quad \text {for all}\,\,i\le n-2^{t-1}-1. \end{aligned}$$

(3.11)

Lemma 3.4

If n is odd and $2^{t-1}<n\le 2^t-4$ then we have the following:

$$\begin{aligned}&\text {w}_2^{2^{t-1}-1-\frac{n+1}{2}}d_{n-1}\in F_{2^t-4}^{2^t-4}\end{aligned}$$

(3.12)

$$\begin{aligned}&\text {w}_2^{n-2^{t-1}}(d_{n-1}+\text {w}_2 d_{n-3})\in F_{3n-2^t-1}^{3n-2^t-1}\end{aligned}$$

(3.13)

$$\begin{aligned}&\text {w}_2^i d_{n-1}\not \in F_{2i+n-1}^{2i+n-1}\quad \text {for all}\,\,i\le 2^{t-1}-\frac{n+3}{2}-1\end{aligned}$$

(3.14)

$$\begin{aligned}&\text {w}_2^i d_{n-3}\not \in F_{2i+n-3}^{2i+n-3}\quad \text {for all}\,\,i\le 2^{t-1}-\frac{n+3}{2}\end{aligned}$$

(3.15)

$$\begin{aligned}&\text {w}_2^i d_{n-3}\not \in F_{2i+n-3}^{2i+n-3}\quad \text {for all}\,\,i\le n-2^{t-1}\end{aligned}$$

(3.16)

$$\begin{aligned}&\text {w}_2^i d_{n-3}+\text {w}_2^{i-1}d_{n-1}\not \in F_{2i+n-3}^{2i+n-3}\quad \text {for all}\,\,i\le n-2^{t-1}. \end{aligned}$$

(3.17)

We note that (3.8), (3.10) prove the first part of (i) while (3.9), (3.11) prove the second part of (i). Similarly, (3.12),(3.14), (3.15) prove (ii)(a) while (3.13) and (3.16)-(3.17) imply (ii)(b) and (ii)(c) respectively. $\square$

Proof of Lemma 3.3

Proof of (3.8): It follows by an iterated application of (3.4) and (3.3) to the term in (3.7) successively that $\bar{\text {w}}_2^{1+s}{\bar{c}}_{n-2}=0\in E_{\infty }^{2^t-6,2}$ where $2s=2^t-4-n$. Therefore,

$$\begin{aligned} \bar{\text {w}}_2^{2^{t-1}-\frac{n}{2}-1}{\bar{c}}_{n-2}=0\in E_\infty ^{2^t-6,2} \end{aligned}$$

and the claim follows.

Proof of (3.9): As $E_\infty ^{n-3,2}$ is generated by , where $\widetilde{w}_3 p_{n-3}=\widetilde{w}_2g_{n-2}+g_n$, we conclude that $p_{n-3}=g_{n-3}$. So, we observe from Lemma 3.1 that $\widetilde{w}_2^{n-2^{t-1}}g_{n-3}=0$ in $H^*(\widetilde{G}_{n,3})$, whence

Therefore, $\bar{\text {w}}_2^{n-2^{t-1}}{\bar{c}}_{n-1}=0\in E_{\infty }^{3n-2^t-3,2}$ and this proves the claim.

Proof of (3.10): We have seen in (3.9) that $\text {w}_2^{n-2^{t-1}}c_{n-1}\in F_{3n-2^t-1}^{3n-2^t-1}$. We also know that $\text {w}_2^{\frac{n}{2}-2}c_{n-2}c_{n-1}\not \in F_{3n-7}^{3n-7}$ as $H^{3n-7}(\widetilde{G}_{n,3})=0$. Thus,

$$\begin{aligned} \text {w}_2^{\frac{n}{2}-2}c_{n-2}c_{n-1}=\left( \text {w}_2^{n-2^{t-1}}c_{n-1}\right) \left( \text {w}_2^{2^{t-1}-\frac{n}{2}-2}c_{n-2}\right) \end{aligned}$$

implies $\text {w}_2^{2^{t-1}-\frac{n}{2}-2}c_{n-2}\not \in F_{2^t-6}^{2^t-6}$ and the claim follows.

Proof of (3.11): We have shown in (3.8) that $\bar{\text {w}}_2^{1+s}{\bar{c}}_{n-2}=0\in E_{\infty }^{2^t-6,2}$, where $2s=2^t-4-n$. Therefore, $\text {w}_2^{1+s}c_{n-2}\in F_{2^t-4}^{2^t-4}.$ We see that $\text {w}_2^{\frac{n}{2}-2}c_{n-2}c_{n-1}\not \in F_{3n-7}^{3n-7}$ because $H^{3n-7}(\widetilde{G}_{n,3})=0.$ Therefore,

$$\begin{aligned} \text {w}_2^{\frac{n}{2}-2}c_{n-2}c_{n-1}=\left( \text {w}_2^{1+s}c_{n-2}\right) \left( \text {w}_2^{\frac{n}{2}-3-s}c_{n-1}\right) \end{aligned}$$

implies our claim. $\square$

Proof of Lemma 3.4

Proof of (3.12): An iterated application of (3.4) and (3.3) to the term in (3.7) implies that $\bar{\text {w}}_2^{1+s}{\bar{d}}_{n-1}=0\in E_{\infty }^{2^t-6,2}$ where $2s=2^t-5-n$. Therefore,

$$\begin{aligned} \bar{\text {w}}_2^{2^{t-1}-1-\frac{n+1}{2}}{\bar{d}}_{n-1}=0\in E_\infty ^{2^t-6,2} \end{aligned}$$

and the claim follows.

Proof of (3.13): As , where $\widetilde{w}_3 q_{n-3}=\widetilde{w}_2 g_{n-2}+g_n$, it follows that $q_{n-3}=g_{n-3}$. As in (3.9) previously, we see that $\widetilde{w}_2^{n-2^{t-1}}g_{n-3}=0$ in $H^*(\widetilde{G}_{n,3})$, whence

As we did for (3.9), we can show (using (3.5)) that the claim holds.

Proof of (3.14)–(3.15): It follows from the cohomology ring, as described in Theorem 2.9, that

$$\begin{aligned} \left( \text {w}_2^{\frac{n-3}{2} - n +2^{t-1}} d_{n-3}\right) \left( \text {w}_2^{n-2^{t-1}}d_{n-1}\right)= & {} \left( \text {w}_2^{\frac{n-3}{2} - n +2^{t-1}} d_{n-3}\right) \\&\times \left( \text {w}_2^{n-2^{t-1}}d_{n-1}+\text {w}_2^{n-2^{t-1}+1}d_{n-3}\right) \\= & {} \text {w}_2^{\frac{n-3}{2}}d_{n-1}d_{n-3}. \end{aligned}$$

The Eq. (3.15) now follows from (3.13) and $\text {w}_2^{\frac{n-3}{2}}d_{n-1}d_{n-3}\not \in F_{3n-7}^{3n-7}$ (as $H^{3n-7}(\widetilde{G}_{n,3})=0$). Theorem 2.9 implies that

$\left( \text {w}_2^{\frac{n-3}{2} - n +2^{t-1}-1} d_{n-1}\right) \left( \text {w}_2^{n-2^{t-1}+1}d_{n-3}+\text {w}_2^{n-2^{t-1}}d_{n-1}\right) =\text {w}_2^{\frac{n-3}{2}}d_{n-1}d_{n-3}$.

Combining this with (3.13) we conclude Eq. (3.14).

Proof of (3.16)–(3.17): We have seen in (3.12) that $\text {w}_2^{2^{t-1}-1-\frac{n+1}{2}}d_{n-1}\in F_{2^t-4}^{2^t-4}$. Therefore,

$$\begin{aligned} \text {w}_2^{\frac{n-3}{2}}d_{n-3}d_{n-1}= & {} \left( \text {w}_2^{2^{t-1}-1-\frac{n+1}{2}}d_{n-1}\right) \left( \text {w}_2^{n-2^{t-1}}d_{n-3}\right) \\= & {} \left( \text {w}_2^{2^{t-1}-1-\frac{n+1}{2}}d_{n-1}\right) \left( \text {w}_2^{n-2^{t-1}}d_{n-3}+\text {w}_2^{n-2^{t-1}-1}d_{n-1}\right) \end{aligned}$$

implies that the following hold

$$\begin{aligned} \text {w}_2^{n-2^{t-1}}d_{n-3}\not \in F_{3n-2^t-3}^{3n-2^t-3},\,\,\text {w}_2^{n-2^{t-1}}d_{n-3}+\text {w}_2^{n-2^{t-1}-1}d_{n-1}\not \in F_{3n-2^t-3}^{3n-2^t-3}. \end{aligned}$$

Both claims now follow. $\square$

Let $F_{s}^{r}$ be the filtration of $H^r(W_{2,1}^n)$ obtained with the Serre spectral sequence. It follows that $F_{0}^{r}=F_{r-2}^r$. We conclude that $F_{i-2}^i\otimes F_{j-2}^j\rightarrow F_{i+j-2}^{i+j}.$ We denote by ${\mathcal {P}}_r$ the following

$$\begin{aligned} {\mathcal {P}}_r:=\frac{\text {Image of }{\oplus }_{q+j\le r, p\ne 0, i\ne 0, p+i=r}F_q^p\otimes F_j^i \text { in }F_r^r}{\text {Image of }{\oplus }_{p+i=r,p\ne 0, i\ne 0}F_p^p\otimes F_i^i \text { in }F_r^r}. \end{aligned}$$

(3.18)

It is seen that ${\mathcal {P}}_r$ is a quotient of a subspace of $F_r^r$. There is a natural map

(3.19)

Moreover, when $r\ge n$, it follows from Theorem 2.9 that

$$\begin{aligned} F_r^r = \left( \text {Image of }\underset{q+j\le r, p\ne 0, i\ne 0}{\oplus }F_q^p\otimes F_j^i \text { in }F_r^r\right) . \end{aligned}$$

Lemma 3.5

The map as defined in (3.19) is a vector space isomorphism for $r>3$.

Proof

It is clear that the map is a linear morphism. The fact that is onto follows from . Let $x\in H^*(\widetilde{G}_{n,3})$ be such that . It follows that

Note that and are surjective maps. Therefore, there exists $P\in H^+(\widetilde{G}_{n,3})\cdot H^+(\widetilde{G}_{n,3})$ such that . Hence, and this implies that $x-P=w_3 Q$ for some $Q\in H^*(\widetilde{G}_{n,3})$. As x has degree $r>3$, the degree of Q is at least 1. As a consequence, $x=P+w_3 Q \in H^+(\widetilde{G}_{n,3})\cdot H^+(\widetilde{G}_{n,3}).$$\square$

In the following theorem we shall compute all the indecomposables in $H^*(\widetilde{G}_{n,3})$.

Theorem 3.6

• (a) The only indecomposables in $H^*(\widetilde{G}_{2^t,3};{\mathbb {Z}}_2)$ are in degrees $2,~3,~2^t-1$ with one indecomposable in each degree.

• (b) The only indecomposables in $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ for $n=2^t-1,~2^t-2,~2^t-3$ are in degrees $2,~3,~2^t-4$ with one indecomposable in each degree.

• (c) Let $2^{t-1}<n\le 2^t-4$. Then the only indecomposables in $H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ are in degrees $2,~3,~3n-2^t-1,~2^t-4$ with one indecomposable in each degree.

Proof

(a) By [4] we know that apart from $\widetilde{w}_2,\widetilde{w}_3$ the first indecomposable in $H^*(\widetilde{G}_{2^t,3})$ is in degree $2^t-1$. Therefore, from Lemma 3.5 we will get one non-zero element in

$$\begin{aligned} \frac{F_{2^t-1}^{2^t-1}}{\text {Image of }{\oplus }_{p+i=2^t-1,p\ne 0,i\ne 0}F_p^p\otimes F_i^i\text { in }F_{2^t-1}^{2^t-1}}. \end{aligned}$$

As $H^{2^t-1}(W_{2,1}^{2^t})=\langle c_{2^t-1}\rangle$, it follows that $c_{2^t-1}\in F_{2^t-1}^{2^t-1}$. Therefore, if for some $i\ge 0$ we have $\text {w}_2^i c_{2^t-2}\in F_{2i+2^t-2}^{2i+2^t-2}$ then $\text {w}_2^{2^{t-1}-2}c_{2^t-2}c_{2^t-1}\in F_{3 \cdot 2^t-7}^{3 \cdot 2^t-7}=0$, which is a contradiction. Therefore, for any $0\le i \le 2^{t-1}-2$,

$$\begin{aligned} \text {w}_2^i c_{2^t-2}\not \in F_{2i+2^t-2}^{2i+2^t-2}. \end{aligned}$$

Similar arguments imply that $\text {w}_2^i c_{2^t-2}c_{2^t-1}\not \in F_{2i+2^{t+1}-3}^{2i+2^{t+1}-3}$. Therefore, it is clear that

$$\begin{aligned} \frac{F_{r}^{r}}{\text {Image of }{\oplus }_{p+i=r,p\ne 0,i\ne 0}F_p^p\otimes F_i^i\text { in }F_{r}^{r}}\ne 0 \end{aligned}$$

in degrees $r=2,3$ and $2^t-1$ only. Therefore, the indecomposables in $H^*(\widetilde{G}_{2^t,3})$ are $\widetilde{w}_2,~\widetilde{w}_3,~w_{2^t-1}.$

For (b) the proofs for the three values of n will be similar. We will prove it for $n=2^t-3$. According to [4], the degree of the next indecomposable after $\widetilde{w}_2,~\widetilde{w}_3$ in $H^*(\widetilde{G}_{2^t-3,3})$ is $2^t-4.$ Observe that

where $\widetilde{w}_3 p_{2^t-8}=g_{2^t-5}$ and $p_{2^t-8}\ne 0$. We also observe $\widetilde{w}_2 p_{2^t-8}\ne 0\in H^*(\widetilde{G}_{2^t-3,3})$. The ring structure obtained from the spectral sequence implies that $\text {w}_2 d_{2^t-6}\not \in F_{2^t-4}^{2^t-4}.$ Therefore, exactly one of $d_{2^t-4}$ or $d_{2^t-4}+\text {w}_2 d_{2^t-6}\in F_{2^t-4}^{2^t-4}$.

If $d_{2^t-4}\in F_{2^t-4}^{2^t-4}$ then $\text {w}_2^i d_{2^t-6}\not \in F_{2i+2^t-6}^{2i+2^t-6}$ for $i\le \frac{2^t-6}{2}$ and $\text {w}_2^i (d_{2^t-4}+\text {w}_2 d_{2^t-6})\not \in F_{2i+2^t-4}^{2i+2^t-4}$ for $i<\frac{2^t-6}{2}$. We also have

$$\begin{aligned} \text {w}_2^{\frac{2^t-6}{2}}(d_{2^t-4}+\text {w}_2 d_{2^t-6})=\text {w}_2^{\frac{2^t-6}{2}}d_{2^t-4}. \end{aligned}$$

If $d_{2^t-4}+\text {w}_2 d_{2^t-6}\in F_{2^t-4}^{2^t-4}$ then $\text {w}_2^i d_{2^t-6}\not \in F_{2i+2^t-6}^{2i+2^t-6}$ for $i\le \frac{2^t-6}{2}$ and $\text {w}_2^i d_{2^t-4}\not \in F_{2i+2^t-4}^{2i+2^t-4}$ for $i<\frac{2^t-6}{2}$. Again we have

$$\begin{aligned} \text {w}_2^{\frac{2^t-6}{2}}(d_{2^t-4}+\text {w}_2 d_{2^t-6})=\text {w}_2^{\frac{2^t-6}{2}}d_{2^t-4}. \end{aligned}$$

Therefore, in either cases, we observe that

$$\begin{aligned} \frac{F_{r}^{r}}{\text {Image of }{\oplus }_{p+i=r,p\ne 0,i\ne 0}F_p^p\otimes F_i^i\text { in }F_{r}^{r}}=\left\{ \begin{array}{ll} 0 &{}\quad \text {if }r>2^t-4\\ {\mathbb {Z}}_2 &{}\quad \text {if }r=2^t-4. \end{array}\right. \end{aligned}$$

Thus, by Lemma 3.5, $H^*(\widetilde{G}_{2^t-3,3})$ has only one indecomposable in degree $2^t-4$ and no indecomposable in higher degrees.

(c)  In Proposition 2.2, we have seen that the degree of the first indecomposable after $\widetilde{w}_2,\widetilde{w}_3$ is at least n. To obtain the indecomposables of degree at least n, we shall use Lemma 3.5. We shall first do this for an even integer n. The case when n is odd will follow similarly.

We see from the proof of Proposition 3.2, $\text {w}_2^i c_{n-2}\not \in F_{2i+n-2}^{2i+n-2}$ for $i\le 2^{t-1}-\frac{n}{2}-2$ and $\text {w}_2^i c_{n-1}\not \in F_{2i+n-1}^{2i+n-1}$ for $i\le n-2^{t-1}-1$. We observe that $\text {w}_2^i c_{n-2}c_{n-1}\not \in F_{2i+2n-3}^{2i+2n-3}$ as if $\text {w}_2^i c_{n-2}c_{n-1}\in F_{2i+2n-3}^{2i+2n-3}$ then $\text {w}_2^{\frac{n}{2}-2}c_{n-2}c_{n-1}\in F_{3n-7}^{3n-7}$, which is a contradiction. Therefore, via Proposition 3.2, the only possible indecomposables will correspond to

$$\begin{aligned} \text {w}_2^i c_{n-2}\,\,\,\text {if}\,\,i\ge 2^{t-1}-\textstyle {\frac{n}{2}}-1\quad \text {and}\quad \text {w}_2^i c_{n-1}\,\,\,\text {if}\,\,i\ge n-2^{t-1}. \end{aligned}$$

If $i>2^{t-1}-\frac{n}{2}-1$ then

$$\begin{aligned} \text {w}_2^i c_{n-2}=\text {w}_2^{i-(2^{t-1}-\frac{n}{2}-1)}\text {w}_2^{2^{t-1}-\frac{n}{2}-1}c_{n-2}\in \underset{p+i=2i+n-2,p\ne 0,i\ne 0}{\oplus }F_p^p\otimes F_i^i. \end{aligned}$$

Similarly, for $i>n-2^{t-1}$ we have

$$\begin{aligned} \text {w}_2^i c_{n-1}\in \underset{p+i=2i+n-2,p\ne 0,i\ne 0}{\oplus }F_p^p\otimes F_i^i. \end{aligned}$$

Thus, the only two possible indecomposables correspond to $\text {w}_2^{2^{t-1}-\frac{n}{2}-1}c_{n-2}$ and $\text {w}_2^{n-2^{t-1}}c_{n-1}.$

We shall now prove that these two elements are non-zero in ${\mathcal {P}}_{2^t-4}$ and ${\mathcal {P}}_{3n-2^t-1}$ (cf. (3.18)). Let the degree of these two elements $\alpha _{i_1},\alpha _{i_2}$ be $i_1,i_2$ where $i_1<i_2.$ If $\alpha _{i_1}$ is in the image of ${\oplus }_{{p+i=i_1,p\ne 0,i\ne 0}}F_p^p\otimes F_i^i$ in $F_{i_1}^{i_1}$ then it is a polynomial in $\text {w}_2$ in $H^*(W_{2,1}^n)$. As any monomial of degree at least n in $\text {w}_2$ is zero, the element $\alpha _{i_1}$ is indecomposable. Now, if $\alpha _{i_2}$ is in the image of ${\oplus }_{p+i=i_2,p\ne 0,i\ne 0}F_p^p\otimes F_i^i$, then we shall get relations of the form

$$\begin{aligned} \alpha _{i_2}= & {} \text {w}_2^{2^{t-1}-\frac{n}{2}-1}c_{n-2} = \text {w}_2^{i_2-i_1} \text {w}_2^{n-2^{t-1}}c_{n-1}\,\,\text {if } i_2=2^t-4>3n-2^t-1=i_1 \\ \alpha _{i_2}= & {} \text {w}_2^{n-2^{t-1}}c_{n-1} = \text {w}_2^{i_2-i_1} \text {w}_2^{2^{t-1}-\frac{n}{2}-1}c_{n-2} \,\,\text {if } i_1=2^t-4<3n-2^t-1=i_2. \end{aligned}$$

In both cases, it leads to a contradiction as one side is odd while the other is even in degree.

The case when $2^{t-1}<n\le 2^t-4$ and n is odd, can be proved similarly with necessary changes. $\square$

We do not have any natural choice for the indecomposables in degrees higher than 3. Thus, if there is an indecomposable in degree i, we choose an arbitary indecomposable element in this degree and denote it by $w_{i}$. We refer the reader to Remark 1.1 for the implicit dependence of $w_i$ on n. Note that $w_{2^t-4}w_{3n-2^t-1}=0$ as this element is in degree $3n-5$ which is beyond the dimension of $\widetilde{G}_{n,3}$.

Definition 3.7

Let $t\ge 4$ and let $n\in [2^{t-1},2^t-1]$ be an integer.

If n is even then we define

$$\begin{aligned} v_{2^t-8}:=\widetilde{w}_2^{2^{t-1}-2-\frac{n}{2}}p_{n-4},\,\,v_{3n-2^t-5}:=\widetilde{w}_2^{n-2^{t-1}-1}p_{n-3} \end{aligned}$$

where $\widetilde{w}_3 p_{n-4} = g_{n-1}$ and $\widetilde{w}_3 p_{n-3} = g_n +\widetilde{w}_2 g_{n-2}$ (cf. (3.1)). It is implicit that $v_{2^t-8}$ or $v_{3n-2^t-5}$ is defined to be zero whenever the exponent of $\widetilde{w}_2$ in the expression defining it is negative. In particular, when $n=2^t-2$ we have $v_{2^t-8}=0$ and when $n=2^{t-1}$ we have $v_{3n-2^t-5}=0$.

If n is odd then we define

$$\begin{aligned} v_{2^t-8}:=\widetilde{w}_2^{2^{t-1}-2-\frac{n+1}{2}}p_{n-3},\,\,v_{3n-2^t-5}:=\widetilde{w}_2^{n-2^{t-1}-1}q_{n-3} \end{aligned}$$

where $\widetilde{w}_3 p_{n-3}=g_n$ and $\widetilde{w}_3 q_{n-3}=g_n+\widetilde{w}_2g_{n-2}$ (cf. (3.2) and (3.6)). We note that when $n=2^t-1,2^t-3$ we have $v_{2^t-8}=0$.

It is clear that $v_{2^t-8}$ and $v_{3n-2^t-5}$ depend on n although the notation may suggest otherwise. From Proposition 3.2 and the discussion preceeding it, we conclude that

$$\begin{aligned} \widetilde{w}_2 v_{2^t-8}=0,\quad \widetilde{w}_3 v_{2^t-8}=0,\quad \widetilde{w}_2 v_{3n-2^t-5}=0,\quad \widetilde{w}_3 v_{3n-2^t-5}=0. \end{aligned}$$

(3.20)

It is crucial that $v_{2^t-8}$ and $v_{3n-2^t-5}$ are both non-zero elements in the cohomology ring.

Remark 3.8

It is expected that the cohomology classes $v_{2^t-8}$ are semi-stable in the sense that if ${\tilde{i}}:\widetilde{G}_{n,3}\hookrightarrow \widetilde{G}_{n+1,3}$ then ${\tilde{i}}^*v_{2^t-8}=v_{2^t-8}$ whenever $2^{t-1}<n\le 2^t-5$. A similar property is expected for $w_{2^t-4}$.

We are ready to state the two main results concerning the cohomology ring of $\widetilde{G}_{n,3}$. The first is as follows.

Theorem 3.9

(a) If $n=2^t$ then

$$\begin{aligned} H^*(\widetilde{G}_{2^t,3};{\mathbb {Z}}_2)=\frac{\frac{{\mathbb {Z}}_2[\widetilde{w}_2,\widetilde{w}_3]}{\langle g_{2^t-2}, g_{2^t-1}\rangle }\otimes {\mathbb {Z}}_2[w_{2^t-1}]}{\left\langle w_{2^t-1}^2-P w_{2^t-1}\right\rangle } \end{aligned}$$

for some $P(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^{2^t-1}(G_{2^t,3};{\mathbb {Z}}_2))$.

(b) If $n=2^t-1,~2^t-2,~2^t-3$ then

$$\begin{aligned} H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)=\frac{\frac{{\mathbb {Z}}_2[\widetilde{w}_2,\widetilde{w}_3]}{\langle g_{n-2},g_{n-1},g_n\rangle }\otimes {\mathbb {Z}}_2[w_{2^t-4}]}{\left\langle w_{2^t-4}^2-P_1 w_{2^t-4}-P_2 \right\rangle } \end{aligned}$$

for some $P_1(\widetilde{w}_2,\widetilde{w}_3),P_2(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^*(G_{n,3};{\mathbb {Z}}_2))$ with $P_2=0$ for $n=2^t-2,~2^t-3$.

Proof

(a) From the proof of Theorem 3.6, it follows that maps $w_{2^t-1}$ to $c_{2^t-1}.$ As $c_{2^t-1}^2=0$, it follows that $w_{2^t-1}^2=\widetilde{w}_3 P(\widetilde{w}_2,\widetilde{w}_3,w_{2^t-1})$ for some polynomial P. Observe that

where $p_{2^t-4}\ne 0\in H^{2^t-4}(\widetilde{G}_{2^t,3})$ satisfies $\widetilde{w}_3 p_{2^t-4}=g_{2^t-1}.$ So, from the proof of Theorem 3.6 (a), we note that $\widetilde{w}_2^{2^{t-1}-2}p_{2^t-4}\ne 0$ and

$$\begin{aligned} \widetilde{w}_2 \widetilde{w}_2^{2^{t-1}-2}p_{2^t-4}=\widetilde{w}_3 \widetilde{w}_2^{2^{t-1}-2}p_{2^t-4}=0\in H^*(\widetilde{G}_{2^t,3}). \end{aligned}$$

Denote $\widetilde{w}_2^{2^{t-1}-2}p_{2^t-4}$ as $v_{2^{t+1}-8}.$ Also note that $v_{2^{t+1}-8}\in \pi ^*(H^*(G_{2^t,3})).$

Now, consider the spectral sequence corresponding to the fibre bundle $SO(3)\hookrightarrow V_3({\mathbb {R}}^{2^t})\xrightarrow {p_1}\widetilde{G}_{2^t,3}.$ As observed earlier, $w_{2^t-1}^2=\widetilde{w}_3 P(\widetilde{w}_2,\widetilde{w}_3,w_{2^t-1})$ for some polynomial P. This implies that ${\bar{w}}_{2^t-1}^i=0$ in $E_4^{*,0}$ when $i\ge 2$. Thus, $E_4^{j,0}$ is nonzero only when $j=2^t-1$ and $E_4^{2^t-1,0}=\langle {\bar{w}}_{2^t-1}\rangle .$ So, $E_4^{i,3}$ is zero when $i\ne 2^{t+1}-8$ and $E_4^{2^{t+1}-8}$ has dimension at most 1. We have already noticed that $v_{2^{t+1}-8}\otimes a^3\in E_4^{2^{t+1}-8,3}.$ Thus, $v_{2^{t+1}-8}w_{2^t-1}=1\in H^{3(2^t-3)}(\widetilde{G}_{2^t,3})$ and $v_{2^{t+1}-8}$ is the element of highest degree in $\pi ^*(H^*(G_{2^t,3})).$ As we did for (a), we again get if $Q(\widetilde{w}_2,\widetilde{w}_3)\ne 0$ and $Q(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^*(G_{2^t,3}))$, then there is a monomial $Q_1(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^*(G_{2^t,3}))$ such that $Q(\widetilde{w}_2,\widetilde{w}_3)Q_1(\widetilde{w}_2,\widetilde{w}_3)=v_{2^{t+1}-8}$. This will again imply that $Q(\widetilde{w}_2,\widetilde{w}_3) w_{2^t-1}\not \in \pi ^*(H^*(G_{2^t,3})).$ Therefore, we get the expected result.

(b) The proof is similar to the proof of (a). $\square$

The second result, for the values of n not covered in Theorem 3.9, is given below.

Theorem 3.10

Let $2^{t-1}<n\le 2^t-4$. Then we have the following relations:

• (i) $v_{2^t-8}^2=0,\,v_{3n-2^t-5}^2=0,\,v_{2^t-8} v_{3n-2^t-5}=0$;

• (ii) $v_{2^t-8} w_{2^t-4}=0,\,v_{3n-2^t-5} w_{3n-2^t-1}=0,\,w_{2^t-4}w_{3n-2^t-1}=0$;

• (iii) $H^{3n-9}(\widetilde{G}_{n,3};{\mathbb {Z}}_2)=\langle v_{2^t-8} w_{3n-2^t-1}\rangle =\langle v_{3n-2^t-5} w_{2^t-4}\rangle$;

• (iv) the following hold:

  $$\begin{aligned}&w_{2^t-4}^2=0,\,\,w_{3n-2^t-1}^2 = R_1+w_{2^t-4}R_2+w_{3n-2^t-1}R_3\,\,\,\,\text {if } 3n-2^t-1<2^t-4\\&w_{3n-2^t-1}^2=0,\,\,w_{2^t-4}^2 = Q_1+w_{2^t-4}Q_2+w_{3n-2^t-1}Q_3\,\,\,\,\text {if } 3n-2^t-1>2^t-4 \end{aligned}$$

  where $R_i,Q_i$’s are polynomials in $\widetilde{w}_2,\widetilde{w}_3$;

• (v) For any $P(\widetilde{w}_2,\widetilde{w}_3)\ne 0 \in H^*(\widetilde{G}_{n,3};{\mathbb {Z}}_2)$ we have either $P|v_{3n-2^t-5}$ or $P|v_{2^t-8}$, i.e., there exists a monomial $Q(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^*(G_{n,3};{\mathbb {Z}}_2))$ such that $P\cdot Q=v_{3n-2^t-5}$ or $P\cdot Q=v_{2^t-8}$ respectively;

• (vi) Let $P=P(\widetilde{w}_2,\widetilde{w}_3)$. If $P|v_{3n-2^t-5}$, then $P w_{2^t-4}\not \in \pi ^*(H^*(G_{n,3};{\mathbb {Z}}_2))$. Similarly, if $P|v_{2^t-8}$, then $P w_{3n-2^t-1}\not \in \pi ^*(H^*(G_{n,3};{\mathbb {Z}}_2))$.

Proof

(a) (i): It follows from the definition of $v_{2^t-8}$ and the fact $\widetilde{w}_2 v_{2^t-8}=0$ that $v_{2^t-8}^2=0$ if $v_{2^t-8}$ has $\widetilde{w}_2$ as a factor, i.e., if $2^t-5>n$. When $n=2^t-4,2^t-5$ we get that $v_{2^t-8}$ is a polynomial (of degree $2^t -8$) in $\widetilde{w}_2$ and $\widetilde{w}_3$. Due to (3.20), it follows that $v_{2^t-8}^2=0$. Identical reasons imply that $v_{3n-2^t-5}^2=0$ as well as $v_{2^t-8} v_{3n-2^t-5}=0$.

(ii)–(iii): Case I  $3n-2^t-1<2^t-4$

Due to degree reasons (the degree of the classes being higher than the dimension $3n-9$ of $\widetilde{G}_{n,3}$) it follows that

$$\begin{aligned} w_{2^t-4} w_{3n-2^t-1}=0,\,\,w_{2^t-4}^2=0,\,\,v_{2^t-8} w_{2^t-4}=0. \end{aligned}$$

(3.21)

By non-degeneracy of cup product, there exists $P+\lambda w_{3n-2^t-1}\in H^{3n-2^t-1}(\widetilde{G}_{n,3})$, where P is a polynomial in $\widetilde{w}_2,\widetilde{w}_3$ and $\lambda \in {\mathbb {Z}}_2$, such that $v_{2^t-8}(P+\lambda w_{3n-2^t-1})=1\in H^{3n-9}(\widetilde{G}_{n,3}).$ The fact that any polynomial in $\widetilde{w}_2,\widetilde{w}_3$ of dimension $3n-9$ in $H^*(\widetilde{G}_{n,3})$ is 0 implies that $\lambda =1$. Therefore, $v_{2^t-8}w_{3n-2^t-1}=1\in H^{3n-9}(\widetilde{G}_{n,3}).$ By non-degeneracy of cup product, there will be an element

$$\begin{aligned} {\check{v}}:=S_1(\widetilde{w}_2,\widetilde{w}_3)+S_2(\widetilde{w}_2,\widetilde{w}_3)w_{3n-2^t-1}+\lambda w_{2^t-4}\in H^{2^t-4}(\widetilde{G}_{n,3}) \end{aligned}$$

such that ${\check{v}}v_{3n-2^t-5}=1\in H^{3n-9}(\widetilde{G}_{n,3})$. Here $S_1, S_2$ are polynomials in $\widetilde{w}_2,\widetilde{w}_3$ and $\lambda \in {\mathbb {Z}}_2.$ Observe that there is no term $S_3(\widetilde{w}_2,\widetilde{w}_3)w_{3n-2^t-1}^2$ as $2(3n-2^t-1)>2^t-4$. Due to (3.20), $S_1,S_2$ annihilates $v_{3n-2^t-5}$, whence $\lambda =1$ and $w_{2^t-4}v_{3n-2^t-5}=1$.

It remains to prove that $v_{3n-2^t-5} w_{3n-2^t-1}=0$. If not then there will be an element $u\in H^*(\widetilde{G}_{n,3})$ such that $v_{3n-2^t-5} w_{3n-2^t-1} u=1\in H^{3n-9}(\widetilde{G}_{n,3})$. From the previous paragraph, we conclude that $w_{3n-2^t-1} u=S_1+S_2 w_{3n-2^t-1}+w_{2^t-4}$. This is a contradiction as $w_{2^t-4}$ is an indecomposable in $H^*(\widetilde{G}_{n,3})$. Therefore, $v_{3n-2^t-5} w_{3n-2^t-1}=0.$

(ii)–(iii): Case II  $3n-2^t-1>2^t-4$

Due to degree reasons (the degree of the classes being higher than the dimension $3n-9$ of $\widetilde{G}_{n,3}$) it follows that

$$\begin{aligned} w_{2^t-4} w_{3n-2^t-1}=0,\,\,w_{3n-2^t-1}^2=0,\,\,v_{3n-2^t-5} w_{3n-2^t-1}=0. \end{aligned}$$

(3.22)

By non-degeneracy of cup product, there exists $P+\lambda w_{2^t-4}\in H^{2^t-4}(\widetilde{G}_{n,3})$, where P is a polynomial in $\widetilde{w}_2,\widetilde{w}_3$ and $\lambda \in {\mathbb {Z}}_2$, such that $v_{3n-2^t-5}(P+\lambda w_{2^t-4})=1\in H^{3n-9}(\widetilde{G}_{n,3}).$ The fact that any polynomial in $\widetilde{w}_2,\widetilde{w}_3$ of dimension $3n-9$ in $H^*(\widetilde{G}_{n,3})$ is 0 implies that $\lambda =1$. Therefore, $v_{3n-2^t-5}w_{2^t-4}=1\in H^{3n-9}(\widetilde{G}_{n,3}).$ By non-degeneracy of cup product, there will be an element

$$\begin{aligned} {\check{v}}:=S_1(\widetilde{w}_2,\widetilde{w}_3)+S_2(\widetilde{w}_2,\widetilde{w}_3)w_{2^t-4}+\lambda w_{3n-2^t-1}\in H^{3n-2^t-1}(\widetilde{G}_{n,3}) \end{aligned}$$

such that ${\check{v}}v_{2^t-8}=1\in H^{3n-9}(\widetilde{G}_{n,3})$. Here $\lambda \in {\mathbb {Z}}_2$, $S_1, S_2$ are polynomials in $\widetilde{w}_2,\widetilde{w}_3$ and there is no term of the form $S_3(\widetilde{w}_2,\widetilde{w}_3)w_{2^t-4}^2$ as $2(2^t-4)>3n-2^t-1$. Due to (3.20), $S_1,S_2$ annihilates $v_{2^t-8}$, whence $\lambda =1$ and $w_{3n-2^t-1}v_{2^t-8}=1$.

It remains to prove that $v_{2^t-8} w_{2^t-4}=0$. Suppose that $v_{2^t-8}w_{2^t-4}\ne 0$. Then there will be an element $u\in H^*(\widetilde{G}_{n,3})$ such that $v_{2^t-8} w_{2^t-4} u=1\in H^{3n-9}(\widetilde{G}_{n,3})$. From the previous paragraph, we conclude that $w_{2^t-4} u=S_1+S_2 w_{2^t-4}+w_{3n-2^t-1}$. This is a contradiction as $w_{3n-2^t-1}$ is an indecomposable in $H^*(\widetilde{G}_{n,3})$. Therefore, $v_{2^t-8} w_{2^t-4}=0.$

(iv)–(v)  Consider the Serre spectral sequence corresponding to the fibre bundle $SO(3)\hookrightarrow V_3({\mathbb {R}}^n)\xrightarrow {p_1} \widetilde{G}_{n,3}$, as introduced in (2.7). Let the cohomology of $SO(3)\cong \mathbb {RP}^3$ be ${\mathbb {Z}}_2[a]/\langle a^4\rangle$. As $n\ge 9$ we note that (cf. (2.6))

$$\begin{aligned} H^*(V_3({\mathbb {R}}^n))\cong {\mathbb {Z}}_2 [a_{n-3},a_{n-2},a_{n-1}]/\left\langle a_{n-3}^2, a_{n-2}^2, a_{n-1}^2\right\rangle \end{aligned}$$

contains no non-zero elements in cohomology of degree 2 or 3. Therefore, in the $E_2$-page, $d^2(1\otimes a)=\widetilde{w}_2$ and in the $E_3$-page $d^3(1\otimes a^2)=\widetilde{w}_3$. So, $v_{3n-2^t-5}\otimes a^3$ and $v_{2^t-8}\otimes a^3$ will survive to the $E_4$-page.

Case I  $3n-2^t-1<2^t-4$

There is no non-zero cohomology in degree $3n-2^t-1$ for $V_3({\mathbb {R}}^n)$. Due to (3.21), the only non-zero elements in $E_4^{*,0}$ are ${\bar{w}}_{3n-2^t-1}^i~(i\ge 0)$ and ${\bar{w}}_{2^t-4}$. Note that here ${\bar{w}}_{2^t-4}$ denotes its image in the $E_4$-page and is not to be confused with the notation used at the outset. Therefore, we conclude that $d^4(v_{3n-2^t-5}\otimes a^3)={\bar{w}}_{3n-2^t-1}$ and

$$\begin{aligned} {\bar{w}}_{3n-2^t-1}^2= & {} d^4((v_{3n-2^t-5}\otimes a^3)(w_{3n-2^t-1}\otimes 1))\\= & {} d^4(v_{3n-2^t-5} w_{3n-2^t-1}\otimes a^3)=d^4(0)=0. \end{aligned}$$

In particular, it follows that

$$\begin{aligned} w_{3n-2^t-1}^2=R_1+w_{2^t-4}R_2+w_{3n-2^t-1}R_3 \end{aligned}$$

(3.23)

where $R_i, i=1,2,3$ are polynomials in $\widetilde{w}_2,\widetilde{w}_3$. It now follows that $E_4^{j,0}$ is non-zero only when $j=3n-2^t-1,~2^t-4$ and $E_4^{3n-2^t-1,0}=\langle {\bar{w}}_{3n-2^t-1}\rangle$, $E_4^{2^t-4,0}=\langle {\bar{w}}_{2^t-4}\rangle .$ We claim that $E_4^{j,3}$ is nonzero only for $j=3n-2^t-5, 2^t-8$ and in each case the dimension is 1. If possible, let $c_i\otimes a^3\in E_4^{i,3}$. There exists $c_j\in H^j(\widetilde{G}_{n,3})$ such that $i+j=3n-9$ and $c_i c_j=1.$ So, in the $E_4$- page $\overline{c_i\otimes a^3}\cdot \bar{c_j}\ne 0.$ Therefore, for $\bar{c_j}$ to be non-zero we need $j=3n-2^t-1,2^t-4.$ So, $i=3n-2^t-5, 2^t-8.$ If $E_4^{3n-2^t-5,3}$ has rank at least 2 then let $v_{3n-2^t-5},z$ be two linearly independent elements in it. As $z\in H^{3n-2^t-5}(\widetilde{G}_{n,3})$ is a non-zero element, by non-degeneracy of cup product and the fact that $2(3n-2^t-1)>2^t-4$ (as $n>2^{t-1}$) there exists polynomials $T_1,T_2$ (in $\widetilde{w}_2$ and $\widetilde{w}_3$) and $\lambda \in {\mathbb {Z}}_2$ such that

$$\begin{aligned} z (T_1+T_2 w_{3n-2^t-1}+\lambda w_{2^t-4})=1\in H^{3n-9}(\widetilde{G}_{n,3}). \end{aligned}$$

(3.24)

Due to degree reasons, all the polynomials $T_i$’s, if they are non-zero, are of positive degree. If $z w_{2^t-4}=0$ then we may set $\lambda =0$. Otherwise, $z w_{2^t-4}=1$. Now $\lambda =0$ gives a contradiction when we consider (3.24) in $E_4$; the right hand side is non-zero while the co-factor of z in the left hand side is zero. If $\lambda =1$ (and $z w_{2^t-4}=1$) then consider the Poincaré dual of $v_{3n-2^t-5}+z$ as in (3.24). As $(v_{3n-2^t-5}+z)w_{2^t-4}=0$, the dual can be chosen such that $\lambda =0$ and we again arrive at a contradiction. Thus, the rank of $E^{3n-2^t-5,3}_4$ is one. Similar arguments apply to $E_4^{2^t-8,3}$, i.e., the only non-zero elements in $E_4^{*,3}$ are $v_{3n-2^t-5}\otimes a^3$ and $v_{2^t-8}\otimes a^3$.

Case II  $3n-2^t-1>2^t-4$

There is no non-zero cohomology in degree $2^t-4$ for $V_3({\mathbb {R}}^n)$. Due to (3.22), the only non-zero elements in $E_4^{*,0}$ are ${\bar{w}}_{2^t-4}^i~(i\ge 0)$ and ${\bar{w}}_{3n-2^t-1}$. Therefore, we conclude that $d^4(v_{2^t-8}\otimes a^3)=w_{2^t-4}$ and

$$\begin{aligned} {\bar{w}}_{2^t-4}^2=d^4\left( \left( v_{2^t-8}\otimes a^3\right) \left( w_{2^t-4}\otimes 1\right) \right) =d^4\left( v_{2^t-8} w_{2^t-4}\otimes a^3\right) =d^4(0)=0. \end{aligned}$$

In particular, it follows that

$$\begin{aligned} w_{2^t-4}^2=Q_1+w_{2^t-4}Q_2+w_{3n-2^t-1}Q_3 \end{aligned}$$

(3.25)

where $Q_i, i=1,2,3$ are polynomials in $\widetilde{w}_2,\widetilde{w}_3$. Note that (3.23) and (3.25) imply (iv). It now follows that $E_4^{j,0}$ is non-zero only when $j=3n-2^t-1,~2^t-4$ and $E_4^{3n-2^t-1,0}=\langle {\bar{w}}_{3n-2^t-1}\rangle$, $E_4^{2^t-4,0}=\langle {\bar{w}}_{2^t-4}\rangle .$ We claim that $E_4^{j,3}$ is nonzero only for $j=3n-2^t-5, 2^t-8$ and in each case the dimension is 1. If possible, let $c_i\otimes a^3\in E_4^{i,3}$. There exists $c_j\in H^j(\widetilde{G}_{n,3})$ such that $i+j=3n-9$ and $c_i c_j=1.$ So, in the $E_4$- page $\overline{c_i\otimes a^3}\cdot \bar{c_j}\ne 0.$ Therefore, for $\bar{c_j}$ to be non-zero we need $j=3n-2^t-1,2^t-4.$ So, $i=3n-2^t-5, 2^t-8.$ If $E_4^{3n-2^t-5,3}$ has rank at least 2 then let $v_{3n-2^t-5},z$ be two linearly independent elements in it. As $z\in H^{3n-2^t-5}(\widetilde{G}_{n,3})$ is a non-zero element, by non-degeneracy of cup product and the fact that $2(2^t-4)>3n-2^t-1$ (as $n\le 2^{t}-4$) there exists a polynomial $T_1$ (in $\widetilde{w}_2$ and $\widetilde{w}_3$) and $\lambda \in {\mathbb {Z}}_2$ such that

$$\begin{aligned} z (T_1+\lambda w_{2^t-4})=1\in H^{3n-9}(\widetilde{G}_{n,3}). \end{aligned}$$

(3.26)

Due to degree reasons, the polynomial $T_1$, if non-zero, are of positive degree. If $z w_{2^t-4}=0$ then we may set $\lambda =0$. Otherwise, $z w_{2^t-4}=1$. Now $\lambda =0$ gives a contradiction when we consider (3.26) in $E_4$; the right hand side is non-zero while the co-factor of z in the left hand side is zero. If $\lambda =1$ (and $z w_{2^t-4}=1$) then consider the Poincaré dual of $v_{3n-2^t-5}+z$ as in (3.26). As $(v_{3n-2^t-5}+z)w_{2^t-4}=0$, the dual can be chosen such that $\lambda =0$ and we again arrive at a contradiction. Thus, the rank of $E^{3n-2^t-5,3}_4$ is one. Similar arguments apply to $E_4^{2^t-8,3}$, i.e., the only non-zero elements in $E_4^{*,3}$ are $v_{3n-2^t-5}\otimes a^3$ and $v_{2^t-8}\otimes a^3$.

Now let us consider both cases (either $3n-2^t-1<2^t-4$ or $3n-2^t-1>2^t-4$) together. Let $P(\widetilde{w}_2,\widetilde{w}_3)\ne 0\in H^*(\widetilde{G}_{n,3})$. If P is either $v_{3n-2^t-5}$ or $v_{2^t-8}$, then we are done. Otherwise $P\otimes a^3\not \in E_4^{*,3}$ implies either $\widetilde{w}_2 P\ne 0$ or $\widetilde{w}_3 P\ne 0.$ If $\widetilde{w}_2 P\ne 0$, then $\widetilde{w}_2 P$ is either $v_{3n-2^t-5}$ or $v_{2^t-8}$ or $(\widetilde{w}_2 P)\otimes a^3 \not \in E_4^{*,3}$. Thus, either $\widetilde{w}_2 \widetilde{w}_2 P\ne 0$ or $\widetilde{w}_3 \widetilde{w}_2 P\ne 0$. After repeating this process finitely many times (as $H^*(\widetilde{G}_{n,3})$ is finite dimensional) we shall get $\widetilde{w}_2^r \widetilde{w}_3^s P$ is either $v_{3n-2^t-5}$ or $v_{2^t-8}$ for some r, s. Therefore, either $P|v_{3n-2^t-5}$ or $P|v_{2^t-8}.$

(vi)  Let $P|v_{3n-2^t-5}$ and suppose if possible that $P w_{2^t-4}\in \pi ^*(H^*(G_{n,3}))$. Let Q be a monomial in $\widetilde{w}_2,\widetilde{w}_3$ such that $PQ=v_{3n-2^t-5}$. As $Q\in \pi ^*(H^*(G_{n,3}))$, it follows that

$$\begin{aligned} PQw_{2^t-4}=v_{3n-2^t-5} w_{2^t-4}=1\in \pi ^*(H^{3n-9}(G_{n,3})), \end{aligned}$$

which is a contradiction as $\pi ^*(H^{3n-9}(G_{n,3}))=0$. $\square$

Remark 3.11

Results similar to Theorem 3.10 (a) (vi) hold in the following cases:

• (a) When $n=2^t$ there exists $v_{2^{t+1}-8}\ne 0\in \pi ^*(H^*(G_{n,3}))$ such that for any $P(\widetilde{w}_2,\widetilde{w}_3)\ne 0\in \pi ^*(H^*(G_{n,3}))$, we have a monomial $Q(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^*(G_{n,3}))$ with the property $P Q=v_{2^{t+1}-8}$ and $v_{2^{t+1}-8}w_{2^t-1}=1$.

• (b) When $n=2^t-1,~2^t-2,~2^t-3$ there exists $v_{3n-2^t-5}\ne 0\in \pi ^*(H^*(G_{n,3}))$ such that for any $P(\widetilde{w}_2,\widetilde{w}_3)\ne 0\in \pi ^*(H^*(G_{n,3}))$, we have a monomial $Q(\widetilde{w}_2,\widetilde{w}_3)\in \pi ^*(H^*(G_{n,3}))$ with the property $PQ=v_{3n-2^t-5}$ and $v_{3n-2^{t}-5}w_{2^t-4}=1$

Corollary 3.12

For any integer $n\in [2^{t-1}+\frac{2^{t-1}}{3},2^t-2]$, the ${\mathbb {Z}}_2$-cup-length $\texttt {cup}(\widetilde{G}_{n,3})$ is bounded as follows:

$$\begin{aligned} \max \left\{ 2^{t-1}-3,\textstyle {\frac{4(n-3)}{3}}-2^{t-1}+3\right\} \le \texttt {cup}(\widetilde{G}_{n,3})\le \frac{3(n-3)}{2}-2^{t-1}+3. \end{aligned}$$

Proof

The lower bound on n implies that $3n-2^t-5> 2^t-8$. This further implies that

$$\begin{aligned} n-2^{t-1}-1+{\textstyle \frac{n-3}{3}}> 2^{t-1}-2-{\textstyle \frac{n}{2}+\frac{n-4}{3}}. \end{aligned}$$

(3.27)

From Theorem 3.10, (3.25) and Remark 3.11, we observe that the monomial attaining $\texttt {cup}(\widetilde{G}_{n,3})$ will be of the form:

• 1. $\widetilde{w}_2^{a_1}\widetilde{w}_3^{b_1}w_{2^t-4}$ or $\widetilde{w}_2^{a_2}\widetilde{w}_3^{b_2}w_{3n-2^t-1}$ if $n\in [2^{t-1}+\frac{2^{t-1}}{3},2^t-4]$;

• 2. $\widetilde{w}_2^{a_1}\widetilde{w}_3^{b_1}w_{2^t-4}$ if $n=2^t-3,~2^t-2$.

It follows from [3] that $\widetilde{w}_2^{2^{t-1}-4}\ne 0$ in $H^{2^{t}-8}(\widetilde{G}_{2^{t-1},3})$. Note that $\widetilde{i}:\widetilde{G}_{n,3}\rightarrow \widetilde{G}_{n+1,3}$ induces an isomorphism $\widetilde{i}^*:H^2(\widetilde{G}_{n+1,3})\xrightarrow {\cong }H^2(\widetilde{G}_{n,3})$ that maps $\widetilde{w}_2$ to $\widetilde{w}_2$. Thus,

$$\begin{aligned} \widetilde{w}_2^{2^{t-1}-4}\ne 0\,\,\text {in }H^{2^t-8}(\widetilde{G}_{n,3}) \hbox { for any integer }n\in [2^{t-1},2^t-1]. \end{aligned}$$

Thus, due to Theorem 3.10 (v), we have

$$\begin{aligned} \widetilde{w}_2^{2^{t-1}-4}=v_{2^t -8}\,\,\,\text {or}\,\,\,\widetilde{w}_2^{2^{t-1}-4}\widetilde{w}_2^{a_1-2^{t-1}+4}\widetilde{w}_3^{b_1}=v_{3n-2^t-5}. \end{aligned}$$

Thus, the cup-length is at least $2^{t-1}-3$. The description of $v_{3n-2^t-5}$ and $v_{2^t-8}$ (cf. Definition 3.7) implies that

$$\begin{aligned} n-2^{t-1}-1+\frac{n-3}{3}\le & {} a_1+b_1\le \frac{3n-2^t-5}{2}\quad \text {for any }n\ne 2^t-3\\ n-2^{t-1}-1+\frac{n-2}{3}\le & {} a_1+b_1\le \frac{3n-2^t-5}{2}\quad \text {for }n=2^t-3\\ 2^{t-1}-2-\frac{n}{2}+\frac{n-4}{3}\le & {} a_2+b_2\le 2^{t-1}-4\quad \text {for }n\text { even}\\ 2^{t-1}-2-\frac{n+1}{2}+\frac{n-3}{3}\le & {} a_2+b_2\le 2^{t-1}-4\quad \text {for }n\text { odd}. \end{aligned}$$

Therefore, combining the above with (3.27), we obtain

$$\begin{aligned} \max \left\{ 2^{t-1}-3,n-2^{t-1}-1+\frac{n-3}{3}+1\right\} \le \texttt {cup}(\widetilde{G}_{n,3})\le \frac{3n-2^t-5}{2}+1 \end{aligned}$$

which simplifies to the bounds as claimed. $\square$

3.2 Characteristic rank and upper characteristic rank

In [4] it has been proved that $\texttt {charrank}(\widetilde{\gamma }_{n,k})\ge n-k+1$ for $k\ge 5$ and $n\ge 2k.$ Here we improve the previously known result.

Theorem 3.13

If $k\ge 5$ and $n\ge 2k$, then characteristic rank of $\widetilde{\gamma }_{n,k}$ is at least $n-k+2$.

To prove this theorem we first need to prove some basic results. Let q be a non-negative integer and q can be uniquely written as a finite sum $\sum _i a_i 2^i$ in the binary expansion, where $a_i=0$ or 1. Let S(q) be the set of non-negative integers i for which $a_i=1$, written in decreasing order. For example, let us choose 23. We can write 23 as $23=2^4+2^2+2^1+2^0$. Then $S(23)=\{4,2,1,0\}$. So, $S(q)=\varnothing$ if and only if $q=0.$ If $\{p_1,p_2,p_3\}$ be a set of three non-negative integers with the property that $S(p_1)\cap S(p_2)=S(p_2)\cap S(p_3)=S(p_1)\cap S(p_3)=\varnothing$, then we say that $\{p_1,p_2,p_3\}$satisfies property P. Now, we shall state the required fact (see [8]).

Fact 3.14

If $p_1,p_2$ are two positive integers such $S(p_1)\cap S(p_2)=\varnothing$, then $p_1+p_2\atopwithdelims ()p_1$ is odd.

The above statement and the following proposition will be used to prove Theorem 3.13.

Proposition 3.15

If $i\ne 1,2,7$, then i can be written as $i=3p_1+4p_2+5p_3$ such that $\{p_1,p_2,p_3\}$ satisfies property P.

Proof

First we prove the statement for $i=5.2^s+r$ for $r=1,2,7$ and for some non-negative integer s. We assume that $i=5.2^s+1$. Now we observe that $5.2^{s}+1-(5.2^{s-2}+1)=5.2^{s-2}.3$. Then putting $s=0$, we get $5.2^0+1=6$, which is divisible by 3. So, for an even integer s, $5.2^s+1=3.\frac{5.2^s+1}{3}$. Here $\{p_1,p_2,p_3\}=\{\frac{5.2^s+1}{3},0,0\}$ which satisfies property P. When $s=1,$$i=11=3.2+5.1$ where $\{2,0,1\}$ has property P. So, for any odd integer s, $5.2^s+1=3.(2+5.(2^{s-2}+2^{s-4}+\cdots +2))+5$ where $\{2+5.(2^{s-2}+2^{s-4}+\cdots +2),0,1\}$ has property P as $2+5.(2^{s-2}+2^{s-4}+\cdots +2)$ is even.

The other cases can be proved similarly using the facts that $5.2+2=3.4$, $5.2^2+2=3.6+4.1$, $5.2+7=3.4+5.1$, $5.2^0+7=3.4$. For $i=3,4,5,6,8,9,10$, this statement is obvious. Now, we apply induction to prove the statement for any i with $i>10.$ Fix an integer i and assume that j can be written as $3p_1+4p_2+5p_3$ where $\{p_1,p_2,p_3\}$ has the property P for $j<i$ and $j\ne 1,2,7$. Let $s\ge 1$ be an integer such that $5.2^s\le i<5.2^{s+1}$. If $i-5.2^s=1,2,7$, then this case is already done. Otherwise, by induction $i-5.2^s$ can be written as $3p_1+4p_2+5p_3$ where $\{p_1,p_2,p_3\}$ has property P and $p_3<2^s$.

Case 1  If both $p_1,p_2<2^s$, then i can be written as $3p_1+4p_2+5(p_3+2^s)$. Then $S(p_3+2^s)=S(p_3)\cup \{s\}$. So, $S(p_1)\cap S(p_2)=S(p_2)\cap S(p_3+2^s)=S(p_1)\cap S(p_3+2^s)=\varnothing$. Hence, $\{p_1,p_2,p_3+2^s\}$ has the property P.

Case 2  Assume $p_1\ge 2^s.$ Then $p_1<2^{s+1}$ as if $p_1\ge 2^{s+1}$, then $i-5.2^s-3.2^{s+1}\ge 0,$ which gives a contradiction to $i<5.2^{s+1}.$ Similarly we can show that $p_2<2^s$, $p_3<2^s.$ Let $S(p_1)=\{s,s_1,\ldots \}$, $S(p_2)=\{t_1,\ldots \}$ and $S(p_3)=\{q_1,\ldots \}$. As $s_1,t_1,q_1<s$ we have

$$\begin{aligned} i-5.2^s= & {} 3(2^s+2^{s_1}+\cdots )+4(2^{t_1}+\cdots )+5(2^{q_1}+\cdots )\\ i= & {} 3(2^{s_1}+\cdots )+4(2^{s+1}+2^{t_1}+\cdots )+5(2^{q_1}+\cdots )\\ i= & {} 3(p_1-2^s)+4(p_2+2^{s+1})+5 p_3. \end{aligned}$$

We have $S(p_1-2^s)=S(p_1)-\{s\}$ and $S(p_2+2^{s+1})=S(p_2)\cup \{s+1\}$. Therefore, the statement is true for i.

Case 3  Assume $p_2\ge 2^s.$ Then, as we did in Case 2, we can show here too that $p_2<2^{s+1}$, $p_1<2^s$ and $p_3<2^s.$ Let $S(p_1)=\{s_1,\ldots \}$, $S(p_2)=\{s,t_1,\ldots \}$ and $S(p_3)=\{q_1,\ldots \}$. As $s_1,t_1,q_1<s$ we have

$$\begin{aligned} i-5.2^s= & {} 3(2^{s_1}+\cdots )+4(2^s+2^{t_1}+\cdots )+5(2^{q_1}+\cdots )\\ i= & {} 3(2^{s+1}+2^s+2^{s_1}+\cdots )+4(2^{t_1}+\cdots )+5(2^{q_1}+\cdots )\\ i= & {} 3(p_1+2^s+2^{s+1})+4(p_2-2^s)+5 p_3. \end{aligned}$$

We have $S(p_1+2^s+2^{s+1})=S(p_1)\cup \{s,s+1\}$ and $S(p_2-2^s)=S(p_2)-\{s\}.$ Therefore, the statement is true for i. $\square$

Now we shall give a proof of Theorem 3.13.

Proof

As ${\bar{w}}_i$ is the $i\text {th}$ degree term of the formal inverse of $(1+w_1+\cdots +w_k)$, we obtain from Proposition 3.15, ${\bar{w}}_i$ contains a monomial $w_3^{p_1}w_4^{p_2}w_5^{p_3}$ with coefficient ${p_1+p_2+p_3\atopwithdelims ()p_1+p_2}\cdot {p_1+p_2\atopwithdelims ()p_1}$ where $\{p_1,p_2,p_3\}$ satisfies property P. Using Fact 3.14 we conclude that the coefficient ${p_1+p_2+p_3\atopwithdelims ()p_1+p_2}\cdot {p_1+p_2\atopwithdelims ()p_1}$ of $w_3^{p_1}w_4^{p_2}w_5^{p_3}$ is $1\in {\mathbb {Z}}_2$. The Gysin sequence of the covering map ${\mathbb {Z}}_2\hookrightarrow \widetilde{G}_{n,k}\xrightarrow {\pi } G_{n,k}$ is the following:

$$\begin{aligned}&\cdots \rightarrow H^{j-1}(G_{n,k})\xrightarrow {\cup w_1}H^j(G_{n,k})\xrightarrow {\pi ^*}H^j(\widetilde{G}_{n,k})\rightarrow H^j(G_{n,k}) \xrightarrow {\cup w_1}H^{j+1}(G_{n,k})\rightarrow \cdots . \end{aligned}$$

By Theorem 2.1 of [4] and the discussion following its statement, it suffices to show that

$$\begin{aligned} \text {ker}\left( H^{n-k+2}(G_{n,k})\xrightarrow {\cup w_1}H^{n-k+3}(G_{n,k})\right) =0. \end{aligned}$$

Let $\alpha _{n-k+2}\in \text {ker}(H^{n-k+2}(G_{n,k})\xrightarrow {\cup w_1}H^{n-k+3}(G_{n,k}))$, i.e.,

$$\begin{aligned} w_1 \alpha _{n-k+2}=a_1 w_1^2{\bar{w}}_{n-k+1}+a_2w_2{\bar{w}}_{n-k+1}+a_3w_1{\bar{w}}_{n-k+2}+a_4{\bar{w}}_{n-k+3}. \end{aligned}$$

(3.28)

As $n-k+3>7$, ${\bar{w}}_{n-k+3}$ contains a monomial $w_3^{p_1}w_4^{p_2}w_5^{p_3}$ with non-zero coefficient. So, comparing the coefficients of $w_3^{p_1}w_4^{p_2}w_5^{p_3}$ in (3.28), we get $a_4=0$. Now we get the modified equation

$$\begin{aligned} w_1 \alpha _{n-k+2}=a_1w_1^2{\bar{w}}_{n-k+1}+a_2w_2{\bar{w}}_{n-k+1}+a_3w_1{\bar{w}}_{n-k+2}. \end{aligned}$$

Reducing modulo $w_1$ and using Lemma 2.1 (ii) (which says that $g_{n-k+1}\ne 0$), we deduce that $a_2=0$. Therefore, $w_1 \alpha _{n-k+2}=a_1w_1^2{\bar{w}}_{n-k+1}+a_3w_1{\bar{w}}_{n-k+2}$ which implies that $\alpha _{n-k+2}=0\in H^{n-k+2}(G_{n,k}).$$\square$

Next we prove a result which gives a correspondence between the characteristic rank of the oriented tautological k-plane bundle $\widetilde{\gamma }_{n,k}$ over $\widetilde{G}_{n,k}$ and the upper characteristic rank of $\widetilde{G}_{n,k}$.

Theorem 3.16

If $\texttt {charrank}(\widetilde{\gamma }_{n,k})=r-1$, where $r\ne 2^t$ for some t, then $\texttt {ucharrank}(\widetilde{G}_{n,k})= r-1.$

Proof

As $r-1$ is the characteristic rank of $\widetilde{\gamma }_{n,k}$, it follows that $H^i(G_{n,k})\xrightarrow {\pi ^*} H^i(\widetilde{G}_{n,k})$ is surjective for $i< r$ and not surjective for $i=r$. We also know that $\texttt {ucharrank}(\widetilde{G}_{n,k})$ is at least $r-1$. Let $\xi$ be any vector bundle over $\widetilde{G}_{n,k}$. If we can show that $w_r(\xi )\in \pi ^*(H^r(G_{n,k}))$, then we can conclude that $\texttt {ucharrank}(\widetilde{G}_{n,k})$ is exactly $r-1.$

Let $r=2^ts$ where s is an odd number with $s>1.$ Then by Wu’s formula we have

$$\begin{aligned} Sq^{2^t}(w_{2^ts-2^t}(\xi ))= & {} \sum _{l=0}^{2^t}{{2^ts-2^t+l-2^t-1}\atopwithdelims (){l}}w_{2^t-l}(\xi ) w_{2^t s - 2^t+l}(\xi )\\= & {} A+{{2^ts-2^t-1}\atopwithdelims (){2^t}}w_{2^ts}(\xi ), \end{aligned}$$

where $A\in \pi ^*(H^*(G_{n,k}))$. Let $s=2l+1$ for some $l\ge 1$. Note that

$$\begin{aligned} 2^ts-2^t-1=2^{t+1}(l-1)+2^t+2^{t-1}+\cdots +1. \end{aligned}$$

Now we set $p_1=2^t$ with $S(p_1)=\{t\}$ and

$$\begin{aligned} p_2=(2^ts-2^t-1)-2^t=2^{t+1}(l-1)+2^{t-1}+2^{t-2}+\cdots +1 \end{aligned}$$

with $S(p_2)\cap S(p_1)=\varnothing$. Using Fact 3.14 we conclude that ${p_1+p_2 \atopwithdelims ()p_1}$ is odd, whence

$$\begin{aligned} Sq^{2^t}(w_{2^ts-2^t}(\xi ))=A+w_{2^ts}(\xi ). \end{aligned}$$

Due to the naturality of $Sq^i,$ we have $Sq^{2^t}(w_{2^ts-2^t}(\xi ))\in \pi ^*(H^r(G_{n,k}))$. Therefore, $w_{2^ts}(\xi )\in \pi ^*(H^r(G_{n,k}))$. As this is true for any vector bundle $\xi$, we infer that $\texttt {ucharrank}(\widetilde{G}_{n,k})=r-1.$$\square$

We shall give one immediate implication of Theorem 3.16.

Corollary 3.17

The upper characteristic rank $\texttt {ucharrank}(\widetilde{G}_{n,3})=\texttt {charrank}(\widetilde{\gamma }_{n,3})$ for $n\ge 8.$

Proof

By [4, 11] we have the full description of the characteristic rank of $\widetilde{G}_{n,3}.$ Consider the following observations:

• (i) $2^r-4$ is not of the form $2^s$ when $r>3$, and

• (ii) when $n>2^{t-1}$ and $3n-2^t-1<2^t-4$, then $2^{t-1}<3n-2^t-1<2^t-4,$ so $3n-2^t-1$ is also not of the form $2^s$.

The above two statements when combined together implies the result. $\square$