The metric completion of the space of vector-valued one-forms

Abstract

The space of full-ranked one-forms on a smooth, orientable, compact manifold (possibly with boundary) is metrically incomplete with respect to the induced geodesic distance of the generalized Ebin metric. We show a distance equality between the induced geodesic distances of the generalized Ebin metric on the space of full-ranked one-forms and the corresponding Riemannian metric defined on each fiber. Using this result, we immediately have a concrete description of the metric completion of the space of full-ranked one-forms. Additionally, we study the relationship between the space of full-ranked one-forms and the space of all Riemannian metrics, leading to quotient structures for the space of Riemannian metrics and its completion.

Appendices

Appendix A Geodesics in the space of full-ranked one-forms

We present an explicit formula for geodesics in the space of full-ranked one-forms $\Omega ^1_+(M, \mathbb {R}^{n})$, which refines the one given in [1], offering detailed expressions for the variables within the formula.

Proposition A.1

Let $\alpha \in \Omega ^1_+(M,\mathbb {R}^{n})$ and $\zeta \in T_{\alpha }\Omega ^1_+(M,\mathbb {R}^{n})$. Then the geodesic in $\Omega ^1_+(M,\mathbb {R}^{n})$ starting at $\alpha$ in the direction of $\zeta$ is the curve

$$\begin{aligned} \alpha (t) = f(t)^{\frac{1}{m}}e^{s(t)(Z_0 - Z_0^\textrm{T})}e^{s(t)Z_0^\textrm{T}\alpha \alpha ^+}\alpha , \end{aligned}$$

(A1)

where $Z_0 = Z - \frac{{{\,\textrm{tr}\,}}(Z)}{m}\alpha \alpha ^+$ is the traceless part of $Z = \zeta \alpha ^+$, and where

$$\begin{aligned} f(t)&= \frac{m}{4}{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})t^2 + \left( 1+\frac{{{\,\textrm{tr}\,}}(Z)}{2} t\right) ^2\\ s(t)&={\left\{ \begin{array}{ll} \frac{2}{\sqrt{m{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})}}\arctan \left( \frac{\sqrt{m{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})}t}{2+{{\,\textrm{tr}\,}}(Z)t}\right) &{} Z_0 \ne 0\\ \frac{t}{1 + \frac{t}{2}{{\,\textrm{tr}\,}}(Z)} &{} Z_0 = 0. \end{array}\right. } \end{aligned}$$

In particular, the change in the induced volume element is given by

$$\begin{aligned} \sqrt{\det (\alpha (t)^\textrm{T}\alpha (t))} = f(t)\sqrt{\det (\alpha ^\textrm{T}\alpha )}. \end{aligned}$$

Proof

This theorem is basically a reformulation of [1, Theorem 3.6] of the geodesic equation on the space of full rank matrices $M_+(n,m)$. The pointwise nature of the metric (3) allows one to translate the result directly to the space of full-ranked one-forms $\Omega ^1_+(M,\mathbb {R}^{n})$. Using $\alpha ^+\alpha = I_{m\times m}$ and $Z\alpha \alpha ^+ = Z$, with the same $f(t), s(t), \omega _0$ and $P_0$ as in [1, Theorem 3.6] we compute

$$\begin{aligned} \delta&= {{\,\textrm{tr}\,}}(Z^\textrm{T}Z) = {{\,\textrm{tr}\,}}\left( \left( Z_0 + \frac{tr(Z)}{m}\alpha \alpha ^+\right) \left( Z_0^\textrm{T} + \frac{tr(Z)}{m}\alpha \alpha ^+\right) \right) \\&= {{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T}) + \frac{2{{\,\textrm{tr}\,}}(Z)}{m}{{\,\textrm{tr}\,}}(Z_0\alpha \alpha ^+) + \frac{({{\,\textrm{tr}\,}}(Z))^2}{m}\\&= {{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T}) + \frac{({{\,\textrm{tr}\,}}(Z))^2}{m}. \end{aligned}$$

Then

$$\begin{aligned} f(t)&= \frac{m\delta }{4}t^2+\tau t+1\nonumber \\&= \frac{m}{4}\left( {{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T}) + \frac{({{\,\textrm{tr}\,}}(Z))^2}{m}\right) t^2 + {{\,\textrm{tr}\,}}(Z) t + 1\nonumber \\&= \frac{m}{4}{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})t^2 + \frac{({{\,\textrm{tr}\,}}(Z))^2}{4} t^2 + {{\,\textrm{tr}\,}}(Z) t + 1\nonumber \\&= \frac{m}{4}{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})t^2 + \left( 1+\frac{{{\,\textrm{tr}\,}}(Z)}{2} t\right) ^2, \end{aligned}$$

(A2)

and thus

$$\begin{aligned} s(t)&= \int \limits _0^t\frac{d\sigma }{f(\sigma )} = \int \limits _0^t\frac{1}{\frac{m}{4}{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})\sigma ^2 + \left( 1+\frac{{{\,\textrm{tr}\,}}(Z)}{2} \sigma \right) ^2}d\sigma \nonumber \\&= 4\int \limits _0^t \frac{d\sigma }{m{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})\sigma ^2 + (2+{{\,\textrm{tr}\,}}(Z)\sigma )^2}\nonumber \\&={\left\{ \begin{array}{ll} \frac{2}{\sqrt{m{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})}}\arctan \left( \frac{\sqrt{m{{\,\textrm{tr}\,}}(Z_0Z_0^\textrm{T})}t}{2+{{\,\textrm{tr}\,}}(Z)t}\right) &{} Z_0 \ne 0\\ \frac{t}{1 + \frac{t}{2}{{\,\textrm{tr}\,}}(Z)} &{} Z_0 = 0. \end{array}\right. } \end{aligned}$$

(A3)

Note that for any arbitrary $m\times m$ matrix A and full rank $n\times m$ matrix a,

$$\begin{aligned} e^{aAa^+} = I_{n\times n} - \alpha \alpha ^+ + ae^Aa^+. \end{aligned}$$

It follows that $\alpha e^{s(t)P} = e^{s(t)\alpha P\alpha ^+}\alpha$. By computation

$$\begin{aligned} \alpha P\alpha ^+&= \alpha \left( (\alpha ^\textrm{T}\alpha )^{-1}(\zeta ^\textrm{T}\alpha ) - \frac{\tau }{m}I_{m\times m}\right) \alpha ^+\\&= Z^\textrm{T}\alpha \alpha ^+ - \frac{{{\,\textrm{tr}\,}}(Z)}{m}\alpha \alpha ^+\\&= Z_0^\textrm{T}\alpha \alpha ^+. \end{aligned}$$

It is easy to see that $\omega = Z - Z^\textrm{T} = Z_0 - Z_0^\textrm{T}$. We then have a reformulation of the geodesic formula

$$\begin{aligned} \alpha (t)&= f(t)^{1/m}e^{-s(t)\omega }\alpha e^{s(t)P}\\&= f(t)^{1/m}e^{s(t)(Z_0 - Z_0^\textrm{T})}e^{s(t)Z_0^\textrm{T}\alpha \alpha ^+}\alpha , \end{aligned}$$

where f(t) and s(t) are as in (A2) and (A3). In addition, let $\zeta _0 = \zeta - \frac{{{\,\textrm{tr}\,}}(Z)}{m}\alpha$ and we compute

$$\begin{aligned} \alpha (t)^\textrm{T}\alpha (t)&= f^{2/m}\alpha ^\textrm{T}e^{s(t)\alpha \alpha ^\textrm{T}Z_0}e^{s(t)Z_0^\textrm{T}\alpha \alpha ^+}\alpha \\&= f^{2/m}e^{s(t)\alpha ^\textrm{T}\zeta _0(\alpha ^\textrm{T}\alpha )^{-1}}(\alpha ^\textrm{T}\alpha )e^{s(t)(\alpha ^\textrm{T}\alpha )^{-1}\zeta _0^\textrm{T}\alpha \alpha }. \end{aligned}$$

Using the property of the matrix exponential that $\det (e^{A}) = e^{{{\,\textrm{tr}\,}}(A)}$, we obtain

$$\begin{aligned} \det (\alpha (t)^\textrm{T}\alpha (t))&= f(t)^2e^{{{\,\textrm{tr}\,}}(s(t)\alpha ^\textrm{T}\zeta _0(\alpha ^\textrm{T}\alpha )^{-1})}\det (\alpha ^\textrm{T}\alpha )e^{{{\,\textrm{tr}\,}}(s(t)(\alpha ^\textrm{T}\alpha )^{-1}\zeta _0^\textrm{T}\alpha )}\\&=f(t)^2e^{s(t){{\,\textrm{tr}\,}}(Z_0)}\det (\alpha ^\textrm{T}\alpha )e^{s(t){{\,\textrm{tr}\,}}(Z_0)}\\&=f(t)^2\det (\alpha ^\textrm{T}\alpha ). \end{aligned}$$

The second statement follows immediately. $\square$

It is easy to see from the geodesic formula (A1) that the geodesic is only defined for $t\in [0, t_0]$ with $t_0 = -\frac{2}{-{{\,\textrm{tr}\,}}(Z)}$ for the points of M where $Z_0 = 0$ and ${{\,\textrm{tr}\,}}(Z)<0$. As a direct consequence, we have geodesically incompleteness and the metric incompleteness of the space of full-ranked one-forms $\Omega ^1_+(M,\mathbb {R}^{n})$ and the space of full rank matrices $M_+(n,m)$.

Appendix B Theorem 1.1 for the Ebin metric

The aim of this appendix is to show an alternative proof of the equivalent of Theorem 1.1 for the Ebin metric on the space of Riemannian metrics of a compact, connected, orientable manifold without boundary, which simplifies the argument of [13, Theorem 3.8]. In the sequel, M is a compact, connected, orientable, m-dimensional manifold without boundary. The space Met(M) is by definition the set of smooth sections of the fiber bundle $E=S^2_+T^*M$ of positive definite symmetric (0, 2)-tensors. It is equipped with the Riemannian structure $(\cdot ,\cdot )_{L^2}$ defined by (12). It can be expressed equivalently with the help of a fixed Riemannian metric $g_0$ of volume 1 of M, as observed for instance in [13]. Indeed to each fiber $E_x = S^2_+T_x^*M$ of the fiber bundle E is associated the Riemannian metric

$$\begin{aligned} \langle a,b \rangle _{h,x} = \frac{1}{4}{{\,\textrm{tr}\,}}(h^{-1}ah^{-1}b)\sqrt{\det (g_0(x)^{-1}h)}, \end{aligned}$$

where $h \in S^2_+T_x^*M$ and $a,b \in T_h S^2_+T_x^*M \cong S^2T_x^*M$. For this appendix, we denote this metric by $g_x$ and its induced distance by $d_x$. The metric (12) can be equivalently defined as

$$\begin{aligned} (h,k)_g = \frac{1}{4}\int \limits _M \langle h(x),k(x)\rangle _{g(x),x} {\text {d}}\mu _{g_0}(x). \end{aligned}$$

Recall that the induced distance on $\textrm{Met}(M)$ is denoted by ${\text {dist}}_{\textrm{Met}}$.

Theorem B.1

In the situation above, it holds

$$\begin{aligned} {\text {dist}}_{\textrm{Met}}(g,g')^2 = \int \limits _M d_x(g(x),g'(x))^2 {\text {d}}\mu _{g_0}(x) \end{aligned}$$

for all $g,g'\in \textrm{Met}(M)$.

We need a bit of preparation, in the same spirit of Sect. 2.

A local trivialization $\varphi$ of the bundle E induced by a local chart on an open set U sending the volume form $\mu _{g_0}$ to the Euclidean one is called standard. Let $\varphi \vert _x :E_x \rightarrow {\text {Sym}}_+(m)$ be the restriction of $\varphi$ to the fiber at $x\in U$. Then the pushforward of the metric on $E_x$, namely $(\varphi \vert _x)_*(\langle \cdot ,\cdot \rangle _{\cdot ,x})$, defines the Riemannian structure (13) on ${\text {Sym}}_+(m)$. We denote it by $g_{m,+}$. Recall that its induced distance is denoted by ${\text {dist}}_{m\times m}$. Therefore, if $\varphi$ is a standard local trivialization around $x\in M$ then $\varphi \vert _x$ is an isometry between $d_x$ and ${\text {dist}}_{m\times m}$.

A curve $c:[0,1] \rightarrow {\text {Sym}}_+(m)$ is said to be piecewise linear if it is the concatenation of k linear segments ${\text {lin}}(A_0, A_1)$, ${\text {lin}}(A_1, A_2), \ldots$, ${\text {lin}}(A_{k-1}, A_k)$ and c is parametrized proportionally to arc-length with respect to $g_{m,+}$ in such a way that $\left\| \frac{{\text {d}}}{{\text {d}}t}c(t)\right\| _{g_{m,+}} = L_{g_{m,+}}(c)$. Here the right-hand side is the length of the curve with respect to the Riemannian metric $g_{m,+}$. Because of this parametrization, we always have:

$$\begin{aligned} \int \limits _0^1 \left\| \frac{{\text {d}}}{{\text {d}}t}c(t)\right\| _{g_{m,+}}^2 {\text {d}}t = L_{g_{m,+}}(c)^2. \end{aligned}$$

(B4)

The distance ${\text {dist}}_{m\times m}$ can be computed using piecewise linear curves.

Lemma B.2

Let $c:[0,1]\rightarrow {\text {Sym}}_+(m)$ be a piecewise $C^1$-curve and $\varepsilon > 0$. Then there exists a piecewise linear curve $c_\varepsilon$ with same endpoints of c and such that $\vert L_{g_{m,+}}(c_\varepsilon ) - L_{g_{m,+}}(c) \vert < \varepsilon$.

The proof is the same of Lemma 2.4.

Proof of Theorem B.1

As observed in [13, Theorem 2.1], the left-hand side is always bigger than or equal to the right-hand side. So what we need to prove is

$$\begin{aligned} {\text {dist}}_{\textrm{Met}}(g, g')^2 \le \int \limits _M d_x(g(x),g'(x))^2{\text {d}}\mu _{g_0}(x). \end{aligned}$$

For all $x\in M$ and all $\varepsilon > 0$, we can find a neighborhood $U_x^\varepsilon$ of x supporting a standard trivialization $\varphi _x^\varepsilon$ such that

• (i) the segments ${\text {lin}}(\varphi _x^\varepsilon \vert _x(g(x)), \varphi _x^\varepsilon \vert _y(g(y)))$ and ${\text {lin}}(\varphi _x^\varepsilon \vert _x(g'(x)), \varphi _x^\varepsilon \vert _y(g'(y)))$ are contained in ${\text {Sym}}_+(m)$ for all $y\in U_x^\varepsilon$, by convexity of ${\text {Sym}}_+(m)$;

• (ii) the lengths of the segments above is smaller than $\varepsilon$ for all $y\in U_x^\varepsilon$, namely

  $$\begin{aligned} L_{g_{m,+}}({\text {lin}}(\varphi _x^\varepsilon \vert _x(g(x)), \varphi _x^\varepsilon \vert _y(g(y)))) < \varepsilon \end{aligned}$$

  and the same for $g'$.

For every $x\in M$, we can apply Lemma B.2 to find a piecewise linear curve $c_x^\varepsilon \subseteq {\text {Sym}}_+(m)$ with endpoints $\varphi _x^\varepsilon \vert _x(g(x))$ and $\varphi _x^\varepsilon \vert _x(g'(x))$ and such that

$$\begin{aligned} L_{g_{m,+}}(c_x^\varepsilon )< {\text {dist}}_{m\times m}(\varphi _x^\varepsilon \vert _x(g(x)),\varphi _x^\varepsilon \vert _x(g'(x))) + \varepsilon . \end{aligned}$$

By compactness, we extract a finite covering $\lbrace U_i^\varepsilon \rbrace$ from the covering $\lbrace U_x^\varepsilon \rbrace$. Let us call $\varphi _i^\varepsilon$ the trivializing chart for $U_i^\varepsilon$ and $c_i^\varepsilon = {\text {lin}}(A_{0,i},\ldots ,A_{k(i),i})$ the piecewise linear curve associated to this neighborhood. By finiteness, we can suppose without loss of generality that $k(i)=k$ for each i, maybe adding some constant subpath. We define $\Gamma _i^\varepsilon :[0,1] \rightarrow Met (U_i^\varepsilon )$ by

$$\begin{aligned} \begin{aligned} \Gamma _i^\varepsilon (\cdot )(y)&= {\text {lin}}(g(y), \varphi _i^{\varepsilon }\vert _y^{-1}(A_{0,i}),\ldots ,\varphi _i^{\varepsilon }\vert _y^{-1}(A_{k,i}),g'(y))\\&= \varphi _i^{\varepsilon }\vert _y^{-1}({\text {lin}}(\varphi _i^{\varepsilon }\vert _y(g(y)), A_{0,i}, \ldots , A_{k,i}, \varphi _i^{\varepsilon }\vert _y(g'(y)))). \end{aligned} \end{aligned}$$

Observe that for each fixed $y\in U_i^\varepsilon$ this is a piecewise linear curve living on $S^2T_y^*M$.

Condition (i) guarantees that this curve is positive-definite. Moreover, $\Gamma _i^\varepsilon$ is piecewise $C^1$ and satisfies $\Gamma _i^\varepsilon (0) = g\vert _{U_i^\varepsilon }$, $\Gamma _i^\varepsilon (1) = g'\vert _{U_i^\varepsilon }$ and

$$\begin{aligned} L_{g_y}(\Gamma _i^\varepsilon (\cdot )(y)) < d_y(g(y),g'(y)) + 3\varepsilon \end{aligned}$$

(B5)

for all $y \in U_i^\varepsilon$ by (ii).

Let $\lbrace \rho _i^\varepsilon \rbrace$ be a partition of unity associated to the covering $\lbrace U_i^\varepsilon \rbrace$. For $y\in M$, we define $I_y$ to be the set of indices i such that $\rho _i^\varepsilon (y)>0$, in particular $y\in U_i^\varepsilon$.

We define the map $\Gamma ^\varepsilon :[0,1] \rightarrow Met (M)$ by $t\mapsto (y\mapsto \Sigma _i\rho _i^\varepsilon (y)\Gamma _i^\varepsilon (t)(y))$. It is piecewise $C^1$ and $\Gamma ^\varepsilon (t)(y)$ is positive definite for all $(t,y)\in [0,1]\times M$ because ${\text {Sym}}_+(m)$ is convex. Moreover $\Gamma ^\varepsilon (0) = g$ and $\Gamma ^\varepsilon (1) = g'$. We can now estimate the distance between g and $g'$ using this curve:

$$\begin{aligned} d_{L^2}(g, g')^2&\le \left( \int \limits _0^{1} \left\| \frac{{\text {d}}}{{\text {d}}t}\Gamma ^\varepsilon (t)(\cdot )\right\| {\text {d}}t\right) ^2 \\&\le \int \limits _0^{1}\left\| \frac{{\text {d}}}{{\text {d}}t}\Gamma ^\varepsilon (t)(\cdot )\right\| ^2 {\text {d}}t\\&= \int \limits _0^{1} \left\| \Sigma _{i}\rho _i^\varepsilon (\cdot )\frac{{\text {d}}}{{\text {d}}t}\Gamma _i^\varepsilon (t)(\cdot )\right\| ^2 {\text {d}}t \\&=\int \limits _0^{1}\int \limits _M \left\| \Sigma _{i}\rho _i^\varepsilon (y)\frac{{\text {d}}}{{\text {d}}t}\Gamma _i^\varepsilon (t)(y)\right\| _{g_y}^2{\text {d}}\mu _{g_0}(y)\\&\le \int \limits _0^{1} \int \limits _M (\Sigma _{i\in I_y}\rho _i^\varepsilon (y))^2 \cdot \max _{i\in I_y}\left\| \frac{{\text {d}}}{{\text {d}}t}\Gamma _i^\varepsilon (t)(y)\right\| _{g_y}^2 {\text {d}}\mu _{g_0}(y) {\text {d}}t \\&= \int \limits _M \int \limits _0^{1} \max _{i\in I_y}\left\| \frac{{\text {d}}}{{\text {d}}t}\Gamma _i^\varepsilon (t)(y)\right\| _{g_y}^2 {\text {d}}t {\text {d}}\mu _{g_0}(y). \end{aligned}$$

For all $i\in I_y$, we have

$$\begin{aligned} \int \limits _0^{1}\left\| \frac{{\text {d}}}{{\text {d}}t}\Gamma _i^\varepsilon (t)(y)\right\| _{g_y}^2 \le L_{g_y}(\Gamma _i^\varepsilon (t)(y))^2 \le (d_y(g(y),g'(y))+ 3\varepsilon )^2 \end{aligned}$$

because of (B4) and (B5). Therefore,

$$\begin{aligned} d_{L^2}(g,g') \le \int \limits _M (d_y(g(y),g'(y))+ 3\varepsilon )^2 {\text {d}}\mu _{g_0}(y). \end{aligned}$$

The thesis follows by taking $\varepsilon$ going to 0. $\square$