Computing persistent Stiefel–Whitney classes of line bundles

Abstract

We propose a definition of persistent Stiefel–Whitney classes of vector bundle filtrations. It relies on seeing vector bundles as subsets of some Euclidean spaces. The usual Čech filtration of such a subset can be endowed with a vector bundle structure, that we call a Čech bundle filtration. We show that this construction is stable and consistent. When the dataset is a finite sample of a line bundle, we implement an effective algorithm to compute its first persistent Stiefel–Whitney class. In order to use simplicial approximation techniques in practice, we develop a notion of weak simplicial approximation. As a theoretical example, we give an in-depth study of the normal bundle of the circle, which reduces to understanding the persistent cohomology of the torus knot (1,2). We illustrate our method on several datasets inspired by image analysis.

A Supplementary material for Sect. 5

1.1 A.1 Study of Example 4

We consider the set

$$\begin{aligned}&X = \bigg \{ \bigg ( \begin{pmatrix} \cos (\theta ) \\ \sin (\theta ) \end{pmatrix} , \begin{pmatrix} \cos (\frac{\theta }{2})^2 &{} \cos (\frac{\theta }{2}) \sin (\frac{\theta }{2}) \\ \cos (\frac{\theta }{2}) \sin (\frac{\theta }{2}) &{} \sin (\frac{\theta }{2})^2 \end{pmatrix} \bigg ), \theta \in [0, 2\pi ) \bigg \}. \end{aligned}$$

In order to study the Čech filtration of X, we shall apply the following affine transformation: let Y be the subset of ${\mathbb {R}}^2 \times M({\mathbb {R}}^2)$ defined as

$$\begin{aligned}&Y = \bigg \{ \bigg ( \begin{pmatrix} \cos (\theta ) \\ \sin (\theta ) \end{pmatrix} , \gamma \begin{pmatrix} \cos (\frac{\theta }{2})^2 &{} \cos (\frac{\theta }{2}) \sin (\frac{\theta }{2}) \\ \cos (\frac{\theta }{2}) \sin (\frac{\theta }{2}) &{} \sin (\frac{\theta }{2})^2 \end{pmatrix} \bigg ), \theta \in [0, 2\pi ) \bigg \}. \end{aligned}$$

and let $\mathbb {Y} = (Y^t)_{t\ge 0}$ be the Čech filtration of Y in ${\mathbb {R}}^2 \times M({\mathbb {R}}^2)$ endowed with the norm $\left\| (x,A)\right\| _1 = \sqrt{ \left\| x\right\| ^2 + \left\| A\right\| _\mathrm {F} ^2}$. We recall that the Čech filtration of X, denoted $\mathbb {X} = (X^t)_{t\ge 0}$, is defined with respect to the norm $\left\| \cdot \right\| _\gamma$. It is clear that, for every $t \ge 0$, the thickenings $X^t$ and $Y^t$ are homeomorphic via the application

$$\begin{aligned} h :{\mathbb {R}}^2 \times M({\mathbb {R}}^2)&\longrightarrow {\mathbb {R}}^2 \times M({\mathbb {R}}^2) \\ (x, A)&\longmapsto (x, \gamma A). \end{aligned}$$

As a consequence, the persistence cohomology modules associated to $\mathbb {X}$ and $\mathbb {Y}$ are isomorphic.

Next, notice that Y is a subset of the affine subspace of dimension 2 of ${\mathbb {R}}^2 \times M({\mathbb {R}}^2)$ with origin O and spanned by the vectors $e_1$ and $e_2$, where

$$\begin{aligned} O&= \left( \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \frac{\gamma }{2}\begin{pmatrix} 1 &{} 0 \\ 0 &{} 1 \end{pmatrix} \right) , ~~~~~ e_1 = \left( \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \frac{\gamma }{2} \begin{pmatrix} 1 &{} 0 \\ 0 &{} -1 \end{pmatrix} \right) , ~~~~~ e_2 = \left( \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \frac{\gamma }{2} \begin{pmatrix} 0 &{} 1 \\ 1 &{} 0 \end{pmatrix} \right) . \end{aligned}$$

Indeed, using the equality

$$\begin{aligned} \begin{pmatrix} \cos (\frac{\theta }{2})^2 &{} \cos (\frac{\theta }{2}) \sin (\frac{\theta }{2}) \\ \cos (\frac{\theta }{2}) \sin (\frac{\theta }{2}) &{} \sin (\frac{\theta }{2})^2 \end{pmatrix} \bigg ) = \frac{1}{2}\begin{pmatrix} 1 &{} 0 \\ 0 &{} 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} \cos (\theta ) &{} \sin (\theta ) \\ \sin (\theta ) &{} -\cos (\theta ) \end{pmatrix}, \end{aligned}$$

we obtain

$$\begin{aligned} Y = O + \left\{ \cos (\theta ) e_1 + \sin (\theta ) e_2, ~~\theta \in [0, 2\pi ) \right\} . \end{aligned}$$

We see that Y is a circle, of radius $\left\| e_1\right\| = \left\| e_2\right\| = \sqrt{1 + \frac{\gamma ^2}{2}}$.

Let E denotes the affine space with origin O and spanned by the vectors $e_1$ and $e_2$. Lemma 10, stated below, shows that the persistent cohomology of Y, seen in the ambient space ${\mathbb {R}}^2 \times M({\mathbb {R}}^2)$, is the same as the persistent cohomology of Y restricted to the subspace E. As a consequence, Y has the same persistence as a circle of radius $\sqrt{1 + \frac{\gamma ^2}{2}}$ in the plane. Hence its barcode can be described as follows:

• one $H^0$-feature: the bar $[0, +\infty )$,

• one $H^1$-feature: the bar $\left[ 0, \sqrt{1 + \frac{\gamma ^2}{2}}\right)$.

Lemma 10

Let $Y \subset {\mathbb {R}}^n$ be any subset, and define $\check{Y} = Y \times \{(0, \ldots , 0)\} \subset {\mathbb {R}}^n \times {\mathbb {R}}^m$. Let these spaces be endowed with the usual Euclidean norms. Then the Čech filtrations of Y and $\check{Y}$ yields isomorphic persistence modules.

Proof

Let $\mathrm {proj}_n:{\mathbb {R}}^n \times {\mathbb {R}}^m \rightarrow {\mathbb {R}}^n$ be the projection on the first n coordinates. One verifies that, for every $t \ge 0$, the map $\mathrm {proj}_n : \check{Y} ^t \rightarrow Y^t$ is a homotopy equivalence. At cohomology level, these maps induce an isomorphism of persistence modules. $\square$

Let us now study the Čech bundle filtration of Y, denoted $( \mathbb {Y} , \mathbb {p} )$. According to Eq. (11), its filtration maximal value is $t^{\mathrm {max}}\left( Y \right) = t_{\gamma }^{\mathrm {max}}\left( X \right) = \frac{\gamma }{\sqrt{2}}$. Note that $\frac{\gamma }{\sqrt{2}}$ is lower than $\sqrt{1+ \frac{\gamma ^2}{2}}$, which is the radius of the circle Y. Hence, for $t < t^{\mathrm {max}}\left( Y \right)$, the inclusion $Y \hookrightarrow Y^t$ is a homotopy equivalence. Consider the following commutative diagram:

It induces the following diagram in cohomology:

The horizontal arrow is an isomorphism. Hence the map $(p^t)^*:H^*(Y^t) \leftarrow H^*( {\mathscr {G}}_{1}({\mathbb {R}}^2) )$ is equal to $(p^0)^*$. We only have to understand $(p^0)^*$.

Remark that the map $p^0:Y \rightarrow {\mathscr {G}}_{1}({\mathbb {R}}^2)$ can be seen as the tautological bundle of the circle. It is then a standard result that $(p^0)^*:H^*(Y) \leftarrow H^*( {\mathscr {G}}_{1}({\mathbb {R}}^2) )$ is nontrivial. Alternatively, $p^0$ can be seen as a map between two circles. It is injective, hence its degree (modulo 2) is 1. We still deduce that $(p^0)^*$ is nontrivial. As a consequence, the persistent Stiefel–Whitney class $w_1^t(X)$ is nonzero for every $t< t^{\mathrm {max}}\left( Y \right)$.

1.2 A.2 Study of Example 5

We consider the set

$$\begin{aligned}&X = \bigg \{ \bigg ( \begin{pmatrix} \cos (\theta ) \\ \sin (\theta ) \end{pmatrix} , \begin{pmatrix} \cos (\theta )^2 &{} \cos (\theta ) \sin (\theta ) \\ \cos (\theta ) \sin (\theta ) &{} \sin (\theta )^2 \end{pmatrix} \bigg ), \theta \in [0, 2\pi ) \bigg \}. \end{aligned}$$

As we explained in the previous subsection, the Čech filtration of X with respect to the norm $\left\| \cdot \right\| _\gamma$ yields the same persistence as the Čech filtration of Y with respect to the norm $\left\| \cdot \right\| _1$, where

$$\begin{aligned}&Y = \bigg \{ \bigg ( \begin{pmatrix} \cos (\theta ) \\ \sin (\theta ) \end{pmatrix} , \gamma \begin{pmatrix} \cos (\theta )^2 &{} \cos (\theta ) \sin (\theta ) \\ \cos (\theta ) \sin (\theta ) &{} \sin (\theta )^2 \end{pmatrix} \bigg ), \theta \in [0, 2\pi ) \bigg \}. \end{aligned}$$

Notice that Y is a subset of the affine subspace of dimension 4 of ${\mathbb {R}}^2 \times M({\mathbb {R}}^2)$ with origin $O = \left( \begin{matrix} 0 \\ 0 \end{matrix}, \frac{1}{2}\begin{matrix} 1 &{} 0 \\ 0 &{} 1 \end{matrix} \right)$ and spanned by the vectors $e_1, e_2, e_3$ and $e_4$, where

$$\begin{aligned} e_1 =&\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 &{} 0 \\ 0 &{} 0 \end{pmatrix} \right) , ~~~~~~~~~~~~~~~~~e_2 = \left( \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 &{} 0 \\ 0 &{} 0 \end{pmatrix} \right) , \\ e_3 = \frac{1}{\sqrt{2}}&\left( \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 &{} 0 \\ 0 &{} -1 \end{pmatrix} \right) , ~~~~~~~e_4 = \frac{1}{\sqrt{2}} \left( \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 &{} 1 \\ 1 &{} 0 \end{pmatrix} \right) . \end{aligned}$$

Indeed, Y can be written as

$$\begin{aligned} Y = O + \left\{ \cos (\theta ) e_1 + \sin (\theta ) e_2 + \frac{\gamma }{\sqrt{2}}\cos (2\theta ) e_3 + \frac{\gamma }{\sqrt{2}}\sin (2\theta ) e_4, ~~\theta \in [0, 2\pi ) \right\} . \end{aligned}$$

This comes from the equality

$$\begin{aligned} \begin{pmatrix} \cos (\theta )^2 &{} \cos (\theta ) \sin (\theta ) \\ \cos (\theta ) \sin (\theta ) &{} \sin (\theta )^2 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 &{} 0 \\ 0 &{} 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} \cos (2\theta ) &{} \sin (2\theta ) \\ \sin (2\theta ) &{} -\cos (2\theta ) \end{pmatrix}. \end{aligned}$$

Observe that Y is a torus knot, i.e. a simple closed curve included in the torus ${\mathbb {T}}$, defined as

$$\begin{aligned} {\mathbb {T}}= O + \left\{ \cos (\theta ) e_1 + \sin (\theta ) e_2 + \frac{\gamma }{\sqrt{2}}\cos (\nu ) e_3 + \frac{\gamma }{\sqrt{2}}\sin (\nu ) e_4, ~~\theta , \nu \in [0, 2\pi ) \right\} . \end{aligned}$$

The curve Y winds one time around the first circle of the torus, and two times around the second one, as represented in Fig.  34. It is known as the torus knot (1, 2).

Let E denotes the affine subspace with origin O and spanned by $e_1, e_2, e_3, e_4$. Since Y is a subset of E, it is equivalent to study the Čech filtration of Y restricted to E (as in Lemma 10). We shall denote the coordinates of points $x \in E$ with respect to the orthonormal basis $(e_1, e_2, e_3, e_4)$. That is, a tuple $(x_1, x_2, x_3, x_4)$ shall refer to the point $O+x_1 e_1 + x_2e_2 + x_3e_3 + x_4 e_4$ of E. Seen in E, the set Y can be written as

$$\begin{aligned} Y = \left\{ \left( \cos (\theta ), \sin (\theta ), \frac{\gamma }{\sqrt{2}}\cos (2\theta ), \frac{\gamma }{\sqrt{2}}\sin (2\theta )\right) , \theta \in [0, 2\pi )\right\} . \end{aligned}$$

Moreover, for every $\theta \in [0, 2\pi )$, we shall denote $y_\theta = \left( \cos (\theta ), \sin (\theta ), \frac{\gamma }{\sqrt{2}} \cos (2\theta ), \frac{\gamma }{\sqrt{2}} \sin (2\theta )\right)$.

Fig. 34

Representations of the set Y, lying on a torus, for a small value of $\gamma$ (left) and a large value of $\gamma$ (right)

We now state two lemmas that will be useful in what follows.

Lemma 11

For every $\theta \in [0, 2\pi )$, the map $\theta ' \mapsto \left\| y_{\theta } - y_{\theta '}\right\|$ admits the following critical points:

• $\theta '-\theta = 0$ and $\theta '-\theta = \pi$ if $\gamma \le \frac{1}{\sqrt{2}}$,

• $\theta '-\theta = 0$, $\pi$, $\arccos (-\frac{1}{2\gamma ^2})$ and $-\arccos (-\frac{1}{2\gamma ^2})$ if $\gamma \ge \frac{1}{\sqrt{2}}$.

They correspond to the values

• $\left\| y_{\theta } - y_{\theta '}\right\| = 0$  if  $\theta '-\theta = 0$,

• $\left\| y_{\theta } - y_{\theta '}\right\| = 2$  if  $\theta '-\theta = \pi$,

• $\left\| y_{\theta } - y_{\theta '}\right\| = \sqrt{2}\sqrt{1 + \gamma ^2 + \frac{1}{4 \gamma ^2}}$  if  $\theta '-\theta = \pm \arccos (-\frac{1}{2\gamma ^2})$.

Moreover, we have $\sqrt{2}\sqrt{1 + \gamma ^2 + \frac{1}{4 \gamma ^2}} \ge 2$ when $\gamma \ge \frac{1}{\sqrt{2}}$.

Proof

Let $\theta , \theta ' \in [0, 2\pi )$. One computes that

$$\begin{aligned} \left\| y_{\theta } - y_{\theta '}\right\| ^2 = 4 \sin ^2\left( \frac{\theta -\theta '}{2}\right) + 2 \gamma ^2 \sin ^2(\theta -\theta '). \end{aligned}$$

Consider the map $f:x \in [0, 2\pi ) \mapsto 4 \sin ^2\left( \frac{x}{2}\right) + 2 \gamma ^2 \sin ^2(x)$. Its derivative is

$$\begin{aligned} f'(x)&= 4 \cos \left( \frac{x}{2}\right) \sin \left( \frac{x}{2}\right) + 4 \gamma ^2 \cos (x) \sin (x) \\&= 2 \sin (x)\left( 1+2\gamma ^2 \cos (x)\right) . \end{aligned}$$

It vanishes when $x = 0$, $x = \pi$, or $x = \pm \arccos (-\frac{1}{2\gamma ^2})$ if $\gamma \ge \frac{1}{\sqrt{2}}$. To conclude, a computation shows that $f(0) = 0$, $f(\pi ) = 4$ and $f\left( \pm \arccos \left( -\frac{1}{2\gamma ^2}\right) \right) = 2\left( 1 + \gamma ^2 + \frac{1}{4\gamma ^2}\right)$. $\square$

Lemma 12

For every $x \in E$ such that $x \ne 0$, the map $\theta \mapsto \left\| x-y_\theta \right\|$ admits at most two local maxima and two local minima.

Proof

Consider the map $g:\theta \in [0, 2\pi ) \mapsto \left\| x-y_\theta \right\| ^2$. It can be written as

$$\begin{aligned} g(\theta )&= \left\| x\right\| ^2 + \left\| y_\theta \right\| ^2 - 2 \left\langle x, y_\theta \right\rangle \\&= \left\| x\right\| ^2 + 1 + \frac{\gamma ^2}{2} - 2 \left\langle x, y_\theta \right\rangle . \end{aligned}$$

Let us show that its derivative $g'$ vanishes at most four times on $[0,2\pi )$, which will prove the result. Using the expression of $y_\theta$, we see that $g'$ can be written as

$$\begin{aligned} g'(\theta ) = a \cos (\theta ) + b \sin (\theta )+ c \cos (2\theta )+ d \sin (2\theta ), \end{aligned}$$

where $a,b,c,d \in {\mathbb {R}}$ are not all zero. Denoting $\omega = \cos (\theta )$ and $\xi = \sin (\theta )$, we have $\xi ^2 = 1-\omega ^2$, $\cos (2 \theta ) = \cos ^2(\theta ) - \sin ^2(\theta ) = 2\omega ^2 -1$ and $\sin (2\theta ) = 2 \cos (\theta ) \sin (\theta ) = 2\omega \xi$. Hence

$$\begin{aligned} g'(\theta ) = a \omega + b \xi + 2 c \omega ^2 + 2d \omega \xi . \end{aligned}$$

Now, if $g'(\theta ) = 0$, we get

$$\begin{aligned} a \omega + 2 c \omega ^2 = -( b + 2d \omega ) \xi \end{aligned}$$

(17)

Squaring this equality yields $\left( a \omega + 2 c \omega ^2\right) ^2 = \left( b + 2d \omega \right) ^2 (1-\omega ^2)$. This degree four equation, with variable $\omega$, admits at most four roots. To each of these w, there exists a unique $\xi = \pm \sqrt{1-w^2}$ that satisfies Eq. (17). In other words, the corresponding $\theta \in [0,2\pi )$ such that $\omega = \cos (\theta )$ is unique. We deduce that $g'$ vanishes at most four times on $[0,2\pi )$. $\square$

Before studying the Čech filtration of Y, let us describe some geometric quantities associated to it. Using a symbolic computation software, we see that the curvature of Y is constant and equal to

$$\begin{aligned} \rho = \frac{\sqrt{1+8\gamma ^2}}{1+2\gamma ^2}. \end{aligned}$$

In particular, we have $\rho \ge 1$ if $\gamma \le 1$, and $\rho < 1$ if $\gamma > 1$. We also have an expression for the diameter of Y:

$$\begin{aligned} \frac{1}{2} \mathrm {diam}\left( Y \right) = \left\{ \begin{array}{ll} 1 &{} \text{ if } \gamma \le \frac{1}{\sqrt{2}}, \\ \frac{1}{\sqrt{2}} \sqrt{1 + \gamma ^2 + \frac{1}{4 \gamma ^2}} &{} \text{ if } \gamma \ge \frac{1}{\sqrt{2}}. \end{array} \right. \end{aligned}$$

It is a consequence of Lemma 11. We now describe the reach of Y:

$$\begin{aligned} \mathrm {reach}(Y) = \left\{ \begin{array}{ll} \frac{1+2\gamma ^2}{\sqrt{1+8\gamma ^2}} &{} \text{ if } \gamma \le 1, \\ 1 &{} \text{ if } \gamma \ge 1. \end{array} \right. \end{aligned}$$

(18)

To prove this, we first define a bottleneck of Y as pair of distinct points $(y,y') \in Y^2$ such that the open ball ${\mathscr {B}} \left( \frac{1}{2}(y+y'),\frac{1}{2} \left\| y-y'\right\| \right)$ does not intersect Y. Its length is defined as $\frac{1}{2} \left\| y-y'\right\|$. According to the results of (Aamari et al. 2019, Theorem 3.4), the reach of Y is equal to

$$\begin{aligned} \mathrm {reach}(Y) = \min \left\{ \frac{1}{\rho }, \delta \right\} , \end{aligned}$$

where $\frac{1}{\rho }$ is the inverse curvature of Y, and $\delta$ is the minimal length of bottlenecks of Y. As we computed, $\frac{1}{\rho }$ is equal to $\frac{1+2\gamma ^2}{\sqrt{1+8\gamma ^2}}$. Besides, according to Lemma 11, a bottleneck $(y_\theta , y_{\theta '})$ has to satisfy $\theta '-\theta = \pi$ or $\pm \arccos (-\frac{1}{2\gamma ^2})$. The smallest length is attained when $\theta '-\theta = \pi$, for which $\frac{1}{2} \left\| y_\theta - y_{\theta '}\right\| =1$. It is straightforward to verify that the pair $(y_\theta , y_{\theta '})$ is indeed a bottleneck. Therefore we have $\delta =1$, and we deduce the expression of $\mathrm {reach}(Y)$.

Last, the weak feature size of Y does not depend on $\gamma$ and is equal to 1:

$$\begin{aligned} \mathrm {wfs}\left( Y \right) = 1. \end{aligned}$$

(19)

We shall prove it by using the characterization of Boissonnat et al. (2018): $\mathrm {wfs}\left( Y \right)$ is the infimum of distances $\mathrm {dist}\left( x, Y\right)$, where $x \in E$ is a critical point of the distance function $d_Y$. In this context, x is a critical point if it lies in the convex hull of its projections on Y. Remark that, if $x \ne 0$, then x admits at most two projections on Y. This follows from Lemma 12. As a consequence, if x is a critical point, then there exists $y, y' \in Y$ such that x lies in the middle of the segment $[y,y']$, and the open ball ${\mathscr {B}} \left( x, \mathrm {dist}\left( x, Y\right) \right)$ does not intersect Y. Therefore $y'$ is a critical point of $y' \mapsto \left\| y-y'\right\|$, hence Lemma 11 gives that $\left\| y-y'\right\| \ge 2$. We deduce the result.

We now describe the thickenings $Y^t$. They present four different behaviours:

• $0\le t<1$: $Y^t$ is homotopy equivalent to a circle,

• $1 \le t < \frac{1}{2} \mathrm {diam}\left( Y \right)$: $Y^t$ is homotopy equivalent to a circle,

• $\frac{1}{2} \mathrm {diam}\left( Y \right) \le t < \sqrt{1+\frac{\gamma ^2}{2}}$: $Y^t$ is homotopy equivalent to a 3-sphere,

• $t \ge \sqrt{1+\frac{\gamma ^2}{2}}$: $Y^t$ is homotopy equivalent to a point.

Recall that, in the case where $\gamma \le \frac{1}{\sqrt{2}}$, we have $\frac{1}{2} \mathrm {diam}\left( Y \right) = 1$. Consequently, the interval $\left[ 1,\frac{1}{2} \mathrm {diam}\left( Y \right) \right)$ is empty, and the second point does not appear in this case.

Study of the case $0\le t<1$. For $t \in [0,1)$, let us show that $Y^t$ deform retracts on Y. According to Eq. (19), we have $\mathrm {wfs}\left( Y \right) = 1$. Moreover, Eq. (18) gives that $\mathrm {reach}(Y) > 0$. Using the results of Boissonnat et al. (2018), we deduce that $Y^t$ is isotopic to Y.

Study of the case $1 \le t < \frac{1}{2} \mathrm {diam}\left( Y \right)$. Denote $z_\theta = \left( 0,0,\frac{\gamma }{\sqrt{2}}\cos (2\theta ), \frac{\gamma }{\sqrt{2}}\sin (2\theta )\right)$, and define the circle $Z = \left\{ z_\theta , \theta \in [0,\pi ) \right\}$. It is repredented in Fig. 35.

Fig. 35

Representation of the set Y (black) and the circle Z (red) (colour figure online)

We claim that $Y^t$ deform retracts on Z. To prove so, we shall define a continuous application $f:Y^t \rightarrow Z$ such that, for every $y\in Y^t$, the segment [y, f(y)] is included in $Y^t$. This would lead to a deformation retraction of $Y^t$ onto Z, via

$$\begin{aligned} (s,y) \in [0,1]\times Y^t \mapsto (1-s)y + s f(y). \end{aligned}$$

Equivalently, we shall define an application $\varTheta :Y^t \rightarrow [0,\pi )$ such that the segment $[y, z_{\varTheta (y)}]$ is included in $Y^t$.

Let $y \in Y^t$. According to Lemma 12, y admits at most two projection on Y. We start with the case where y admits only one projection, namely $y_{\theta }$ with $\theta \in [0, 2\pi )$. Let ${\overline{\theta }} \in [0, \pi )$ be the reduction of $\theta$ modulo $\pi$, and consider the point $z_{{\overline{\theta }}}$ of Z. A computation shows that the distance $\left\| y_\theta - z_{{\overline{\theta }}}\right\|$ is equal to 1. Besides, since $y \in Y^t$, the distance $\left\| y_\theta - y\right\|$ is at most t. By convexity, the segment $\left[ y, z_{{\overline{\theta }}}\right]$ is included in the ball ${\overline{ {\mathscr {B}} }}\left( y_\theta ,t\right)$, which is a subset of $Y^t$. We then define $\varTheta (y)= {\overline{\theta }}$.

Now suppose that y admits exactly two projection $y_\theta$ and $y_{\theta '}$. According to Lemma 11, these angles must satisfy $\theta '-\theta = \pi$. Indeed, the case $\left\| y_{\theta } - y_{\theta '}\right\| = \sqrt{2}\sqrt{1 + \gamma ^2 + \frac{1}{4 \gamma ^2}}$ does not occur since we chose $t < \frac{1}{2} \mathrm {diam}\left( Y \right) = \frac{\sqrt{2}}{2}\sqrt{1 + \gamma ^2 + \frac{1}{4 \gamma ^2}}$. The angles $\theta$ and $\theta '$ correspond to the same reduction modulo $\pi$, denoted ${\overline{\theta }}$, and we also define $\varTheta (y)= {\overline{\theta }}$.

Study of the case $t\in \left[ \frac{1}{2} \mathrm {diam}\left( X \right) , \sqrt{1+\frac{\gamma ^2}{2}}\right)$. Let ${\mathbb {S}}_3$ denotes the unit sphere of E. For every $v = (v_1,v_2,v_3,v_4) \in {\mathbb {S}}_3$, we shall denote by $\langle v \rangle$ the linear subspace spanned by v, and by $\langle v \rangle _+$ the cone $\{\lambda v, \lambda \ge 0\}$. Moreover, we define the quantity

$$\begin{aligned} \delta (v) = \min _{y \in Y} \mathrm {dist}\left( y, \langle v \rangle _+\right) . \end{aligned}$$

and the set

$$\begin{aligned} S = \left\{ \delta (v) v, v \in {\mathbb {S}}_3 \right\} . \end{aligned}$$

The situation is depicted in Fig. 36. We claim that S is a subset of $Y^t$, and that $Y^t$ deform retracts on it. This follows from the two following facts: for every $v \in {\mathbb {S}}_3$,

• 1. $\delta (v)$ is not greater than $\frac{1}{2} \mathrm {diam}\left( Y \right)$,

• 2. $\langle v \rangle _+ \cap Y^t$ consists of one connected component: an interval centered on $\delta (v)v$, that does not contain the point 0.

Suppose that these assertions are true. Then one defines a deformation retraction of $Y^t$ on S by retracting each fiber $\langle v \rangle _+ \cap Y^t$ linearly on the singleton $\{ \delta (v)v \}$. We shall now prove the two items.

Fig. 36

Representation of the set Y (dashed), lying on a 3-sphere of radius $\sqrt{1+\frac{\gamma ^2}{2}}$

Item 1. Note that Item 1 can be reformulated as follows:

$$\begin{aligned} \max _{v \in {\mathbb {S}}_3} \min _{y \in Y} \mathrm {dist}\left( y, \langle v \rangle _+\right) \le \frac{1}{2} \mathrm {diam}\left( Y \right) . \end{aligned}$$

(20)

Let us justify that the pairs (v, y) that attain this maximum-minimum are the same as in

$$\begin{aligned} \max _{v \in {\mathbb {S}}_3} \min _{y \in Y} \left\| y-v\right\| . \end{aligned}$$

(21)

From the definition of $Y = \{y_\theta , \theta \in [0,2\pi )\}$, we see that $\min _{y \in Y} \mathrm {dist}\left( y, \langle v \rangle _+\right) = \min _{y \in Y} \mathrm {dist}\left( y, \langle v \rangle \right)$. A vector $v \in {\mathbb {S}}_3$ being fixed, let us show that $y\mapsto \mathrm {dist}\left( y, \langle v \rangle \right)$ is minimized when $y \mapsto \left\| v-y\right\|$ is. Let $y\in Y$. Since v is a unit vector, the projection of y on $\langle v \rangle$ can be written as $\left\langle y, v\right\rangle v$. Hence $\mathrm {dist}\left( y, \langle v \rangle \right) ^2 = \left\| \left\langle y, v\right\rangle v - y\right\| ^2$, and expanding this norm yields

$$\begin{aligned} \mathrm {dist}\left( y, \langle v \rangle \right) ^2 = \left\| y\right\| ^2- \left\langle y, v\right\rangle ^2. \end{aligned}$$

Expanding the norm $\left\| y-v\right\| ^2$ and using that $\left\| y\right\| ^2 = 1 + \frac{\gamma ^2}{2}$, we get $\left\langle y, v\right\rangle = \frac{1}{2}\left( 2 + \frac{\gamma ^2}{2} - \left\| y-v\right\| ^2\right)$. We inject this relation in the preceding equation to obtain

$$\begin{aligned} \mathrm {dist}\left( y, \langle v \rangle \right) ^2 = -\left( \frac{\gamma }{2}\right) ^4+\gamma ^2 + \frac{1}{4} \left\| y-v\right\| ^2\left( 4+\gamma ^2 - \left\| y-v\right\| ^2\right) . \end{aligned}$$

Now we can deduce that $y \mapsto \mathrm {dist}\left( y, \langle v \rangle \right) ^2$ is minimized when $y \mapsto \left\| y-v\right\|$ is minimized. Indeed, the map $\left\| y-v\right\| \mapsto \frac{1}{4} \left\| y-v\right\| ^2\left( 4+\gamma ^2 - \left\| y-v\right\| ^2\right)$ is increasing on $\left[ 0, \frac{1}{2}(4+\gamma ^2)\right]$. But $\left\| y-v\right\| \le \left\| y\right\| + \left\| v\right\| = \frac{1}{2}(4+\gamma ^2)$.

We deduce that studying the left hand term of Eq. (20) is equivalent to studying Eq.  (21). We shall denote by $g:{\mathbb {S}}_3 \rightarrow {\mathbb {R}}$ the map

$$\begin{aligned} g(v) = \min _{y \in Y} \left\| y-v\right\| . \end{aligned}$$

(22)

Let $v \in {\mathbb {S}}_3$ that attains the maximum of g, and let y be a corresponding point that attains the minimum of $\left\| y-v\right\|$. The points v and y attains the quantity in Eq. (20). In order to prove that $\mathrm {dist}\left( y, \langle v \rangle \right) \le \frac{1}{2} \mathrm {diam}\left( Y \right)$, let p(y) denotes the projection of y on $\langle v \rangle$. We shall show that there exists another point $y' \in Y$ such that p(y) is equal to $\frac{1}{2}(y+y')$ Consequently, we would have $\left\| y - p(y)\right\| = \frac{1}{2} \left\| y' - y\right\| \le \frac{1}{2} \mathrm {diam}\left( Y \right)$, i.e.

$$\begin{aligned} \mathrm {dist}\left( y, \langle v \rangle \right) \le \frac{1}{2} \mathrm {diam}\left( Y \right) . \end{aligned}$$

Remark the following fact: if $w \in {\mathbb {S}}_3$ is a unit vector such that $\left\langle p(y)-y, w\right\rangle >0$, then for $\varepsilon >0$ small enough, we have

$$\begin{aligned} \mathrm {dist}\left( y, \langle v+\varepsilon w \rangle \right) > \mathrm {dist}\left( y, \langle v \rangle \right) . \end{aligned}$$

Equivalently, this statement reformulates as $0 \le \left\langle y, \frac{1}{ \left\| v+\varepsilon w\right\| }(v+\varepsilon w)\right\rangle < \left\langle y, v\right\rangle$. Let us show that

$$\begin{aligned} \left\langle y, \frac{1}{ \left\| v+\varepsilon w\right\| }(v+\varepsilon w)\right\rangle = \left\langle y, v\right\rangle - \varepsilon \kappa + o(\varepsilon ) , \end{aligned}$$

(23)

where $\kappa = \left\langle p(y)-y, w\right\rangle > 0$, and where $o(\varepsilon )$ is the little-o notation. Note that $\frac{1}{ \left\| v+\varepsilon w\right\| } = 1-\varepsilon \left\langle v, w\right\rangle + o(\varepsilon )$. We also have

$$\begin{aligned} \frac{1}{ \left\| v+\varepsilon w\right\| }(v+\varepsilon w)&= v + \varepsilon \left( w- \left\langle v, w\right\rangle v \right) + o(\varepsilon ) . \end{aligned}$$

Expanding the inner product in Eq. (23) gives

$$\begin{aligned} \left\langle y, \frac{1}{ \left\| v+\varepsilon w\right\| }(v+\varepsilon w)\right\rangle&= \left\langle y, v\right\rangle + \varepsilon \bigg ( \left\langle y, w\right\rangle - \left\langle v, w\right\rangle \left\langle y, v\right\rangle \bigg ) + o(\varepsilon ) \\&= \left\langle y, v\right\rangle + \varepsilon \bigg \langle y- \left\langle y, v\right\rangle v , w \bigg \rangle + o(\varepsilon ) \\&= \left\langle y, v\right\rangle + \varepsilon \left\langle y-p(y) , w\right\rangle + o(\varepsilon ) , \end{aligned}$$

and we obtain the result.

Next, let us prove that y is not the only point of Y that attains the minimum in Eq. (22). Suppose that it is the case by contradiction. Let $w \in {\mathbb {S}}_3$ be a unit vector such that $\left\langle p(y)-y, w\right\rangle >0$. For $\varepsilon$ small enough, let us prove that the vector $v' = \frac{1}{ \left\| v+\varepsilon w\right\| }(v+\varepsilon w)$ of ${\mathbb {S}}_3$ contradicts the maximality of v. That is, let us prove that $g(v') > g(v)$. Let $y' \in Y$ be a minimizer $\left\| y'-v'\right\|$. We have to show that $\left\| y'-v'\right\| > \left\| y-v\right\|$. This would lead to $g(v') > g(v)$, hence the contradiction.

Expanding the norm yields

$$\begin{aligned} \left\| v'-y'\right\| ^2 = \left\| v'-v+v-y'\right\| ^2&\ge \left\| v'-v\right\| ^2 + \left\| v-y'\right\| ^2 -2 \left\langle v'-v, v-y'\right\rangle . \end{aligned}$$

Using $\left\| v'-v\right\| ^2 \ge 0$ and $\left\| v-y'\right\| ^2 \ge \left\| v-y\right\| ^2$ by definition of y, we obtain

$$\begin{aligned} \left\| v'-y'\right\| ^2 \ge \left\| v-y\right\| ^2 -2 \left\langle v'-v, v-y'\right\rangle . \end{aligned}$$

We have to show that $\left\langle v'-v, y-y'\right\rangle$ is positive for $\varepsilon$ small enough. By writing $v-y' = v-y+(y-y')$ we get

$$\begin{aligned} \left\langle v'-v, v-y'\right\rangle&= \left\langle v'-v, v\right\rangle - \left\langle v'-v, y\right\rangle + \left\langle v'-v, y-y'\right\rangle \end{aligned}$$

According to Eq. (23), $- \left\langle v'-v, y\right\rangle = \varepsilon \kappa + o(\varepsilon )$. Besides, using $v' - v = \varepsilon (w- \left\langle v, w\right\rangle v) + o(\varepsilon )$, we get $\left\langle v'-v, v\right\rangle = o(\varepsilon )$. Last, Cauchy-Schwarz inequality gives $| \left\langle v'-v, y-y'\right\rangle | \le \left\| v'-v\right\| \left\| y-y'\right\|$. Therefore, $\left\langle v'-v, y-y'\right\rangle = O(\varepsilon ) \left\| y-y'\right\|$, where $O(\varepsilon )$ is the big-o notation. Gathering these three equalities, we obtain

$$\begin{aligned} \left\langle v'-v, v-y'\right\rangle&= o(\varepsilon ) + \varepsilon \kappa + O(\varepsilon ) \left\| y-y'\right\| . \end{aligned}$$

As we can read from this equation, if $\left\| y-y'\right\|$ goes to zero as $\varepsilon$ does, then $\left\langle v'-v, v-y'\right\rangle$ is positive for $\varepsilon$ small enough. Observe that $v'$ goes to v when $\varepsilon$ goes to 0. By assumption y is the only minimizer in Eq.  (22). By continuity of g, we deduce that $y'$ goes to y.

By contradiction, we deduce that there exists another point $y'$ which attains the minimum in g(v). Note that it is the only other one, according to Lemma 12. Let us show that p(y) lies in the middle of the segment $[y,y']$. Suppose that it is not the case. Then $p(y)-y$ is not equal to $-(p(y')-y')$, where $p(y')$ denotes the projection of $y'$ on $\langle v \rangle$. Consequently, the half-spaces $\{w \in E, \left\langle p(y)-y, w\right\rangle > 0\}$ and $\{w \in E, \left\langle p(y')-y', w\right\rangle > 0\}$ intersects. Let w be any vector in the intersection. For $\varepsilon >0$, denote $v' = \frac{1}{ \left\| v+\varepsilon w\right\| }(1+\varepsilon w)$. If $\varepsilon$ is small enough, the same reasoning as before shows that $v'$ contradicts the maximality of v. The situation is represented in Fig.  37.

Fig. 37

Left: Representation of the situation where y and $y'$ are minimizers of Eq. (22). Right: Representation in the plane passing through the points y, $y'$ and p(y). The dashed area corresponds to the intersection of the half-spaces $\{w \in E, \left\langle p(y)-y, w\right\rangle > 0\}$ and $\{w \in E, \left\langle p(y')-y', w\right\rangle > 0\}$

Item 2. Let $v \in {\mathbb {S}}_3$. The set $\langle v \rangle _+ \cap Y^t$ can be described as

$$\begin{aligned} \langle v \rangle _+ \cap \bigcup _{y \in Y} {\overline{ {\mathscr {B}} }}\left( y,t\right) . \end{aligned}$$

Let $y \in Y$ such that $\langle v \rangle _+ \cap {\overline{ {\mathscr {B}} }}\left( y,t\right) \ne \emptyset$. Denote by p(y) the projection of y on $\langle v \rangle _+$. It is equal to $\left\langle y, v\right\rangle v$. Using Pythagoras’ theorem, we obtain that the set $\langle v \rangle _+ \cap {\overline{ {\mathscr {B}} }}\left( y,t\right)$ is equal to the interval

$$\begin{aligned} \left[ p(y) \pm \sqrt{t^2 - \mathrm {dist}\left( y, \langle v \rangle \right) ^2} v \right] . \end{aligned}$$

Using the identity $\mathrm {dist}\left( y, \langle v \rangle \right) ^2 = \left\| y\right\| - \left\langle y, v\right\rangle ^2 = 1 + \frac{\gamma ^2}{2} - \left\langle y, v\right\rangle ^2$, we can write this interval as

$$\begin{aligned} \big [I_1(y) \cdot v, ~ I_2(y) \cdot v\big ], \end{aligned}$$

where $I_1(y) = \left\langle y, v\right\rangle - \sqrt{ \left\langle y, v\right\rangle ^2 - (1 + \frac{\gamma ^2}{2} -t^2)}$ and $I_2(y) = \left\langle y, v\right\rangle + \sqrt{ \left\langle y, v\right\rangle ^2 - (1 + \frac{\gamma ^2}{2} -t^2)}$. Seen as functions of $\left\langle y, v\right\rangle$, the map $I_1$ is decreasing, and the map $I_2$ is increasing (see Fig. 38). Let $y^* \in Y$ that minimizes $\mathrm {dist}\left( y, \langle v \rangle \right)$. Equivalently, $y^*$ maximizes $\langle y,v\rangle$. It follows that the corresponding interval $\big [I_1(y^*) \cdot v, ~ I_2(y^*) \cdot v\big ]$ contains all the others. We deduce that the set $\langle v \rangle _+ \cap Y^t$ is equal to this interval.

Fig. 38

Left: Representation of two intervals $\langle v \rangle _+ \cup {\overline{ {\mathscr {B}} }}\left( y,t\right)$ and $\langle v \rangle _+ \cup {\overline{ {\mathscr {B}} }}\left( y',t\right)$. Right: Representation of the maps $x\mapsto x \pm \sqrt{x^2-1}$

Study of the case $t \ge \sqrt{1+\frac{1}{2}\gamma ^2}$. For every $y \in Y$, we have $\left\| y\right\| = \sqrt{1+\frac{1}{2}\gamma ^2}$. Therefore, if $t \ge \sqrt{1+\frac{1}{2}\gamma ^2}$, then $Y^t$ is star shaped around the point 0, hence it deform retracts on it.

To close this subsection, let us study the Čech bundle filtration $( \mathbb {Y} , \mathbb {p} )$ of Y. According to Eq. (11), its filtration maximal value is $t^{\mathrm {max}}\left( Y \right) = t_{\gamma }^{\mathrm {max}}\left( X \right) = \frac{\gamma }{\sqrt{2}}$. Note that $\frac{\gamma }{\sqrt{2}}$ is lower than $\frac{1}{2} \mathrm {diam}\left( Y \right)$. Consequently, only two cases are to be studied: $t \in [0,1)$, and $t \in \left[ 1, \frac{1}{2} \mathrm {diam}\left( Y \right) \right)$.

The same argument as in Subsect. A.2 yields that for every $t \in [0,1)$, the persistent Stiefel–Whitney class $w_1^t(Y)$ is equal to $w_1^0(Y)$. Accordingly, for every $t \in \left[ 1, \frac{1}{2} \mathrm {diam}\left( Y \right) \right)$, the class $w_1^t(Y)$ is equal to $w_1^1(Y)$. Let us show that $w_1^0(Y)$ is zero, and that $w_1^1(Y)$ is not.

First, remark that the map $p^0:Y \rightarrow {\mathscr {G}}_{1}({\mathbb {R}}^2)$ can be seen as the normal bundle of the circle. Hence $(p^0)^*:H^*(Y) \leftarrow H^*( {\mathscr {G}}_{1}({\mathbb {R}}^2) )$ is nontrivial, and we deduce that $w_1^0(Y)=0$. As a consequence, the persistent Stiefel–Whitney class $w_1^t(X)$ is nonzero for every $t < 1$.

Next, consider $p^1:Y^1 \rightarrow {\mathscr {G}}_{1}({\mathbb {R}}^2)$. Recall that $Y^1$ deform retracts on the circle

$$\begin{aligned} Z = \left\{ \left( 0,0,\frac{\gamma }{\sqrt{2}}\cos (2\theta ), \frac{\gamma }{\sqrt{2}}\sin (2\theta )\right) , \theta \in [0,\pi ) \right\} . \end{aligned}$$

Seen in ${\mathbb {R}}^2 \times M({\mathbb {R}}^2)$, we have

$$\begin{aligned} Z = \bigg \{ \bigg ( \begin{pmatrix} 0 \\ 0 \end{pmatrix} , \gamma \begin{pmatrix} \cos (\theta )^2 &{} \cos (\theta ) \sin (\theta ) \\ \cos (\theta ) \sin (\theta ) &{} \sin (\theta )^2 \end{pmatrix} \bigg ), \theta \in [0, \pi ) \bigg \}. \end{aligned}$$

Notice that the map $q:Z \rightarrow {\mathscr {G}}_{1}({\mathbb {R}}^2)$, the projection on ${\mathscr {G}}_{1}({\mathbb {R}}^2)$, is injective. Seen as a map between two circles, it has degree (modulo 2) equal to 1. We deduce that $q^*:H^*(Z) \leftarrow H^*( {\mathscr {G}}_{1}({\mathbb {R}}^2) )$ is nontrivial. Now, remark that the map q factorizes through $p^1$:

It induces the following diagram in cohomology:

Since $q^*$ is nontrivial, this commutative diagram yields that the persistent Stiefel–Whitney class $w_1^1(Y)$ is nonzero. As a consequence, the persistent Stiefel–Whitney class $w_1^t(Y)$ is nonzero for every $t \ge 1$.