Demystifying Latschev’s Theorem: Manifold Reconstruction from Noisy Data

Abstract

For a closed Riemannian manifold $\mathcal {M}$ and a metric space S with a small Gromov–Hausdorff distance to it, Latschev’s theorem guarantees the existence of a sufficiently small scale $\beta >0$ at which the Vietoris–Rips complex of S is homotopy equivalent to $\mathcal {M}$. Despite being regarded as a stepping stone to the topological reconstruction of Riemannian manifolds from noisy data, the result is only a qualitative guarantee. Until now, it had been elusive how to quantitatively choose such a proximity scale $\beta$ in order to provide sampling conditions for S to be homotopy equivalent to $\mathcal {M}$. In this paper, we prove a stronger and pragmatic version of Latschev’s theorem, facilitating a simple description of $\beta$ using the sectional curvatures and convexity radius of $\mathcal {M}$ as the sampling parameters. Our study also delves into the topological recovery of a closed Euclidean submanifold from the Vietoris–Rips complexes of a Hausdorff close Euclidean subset. As already known for Čech complexes, we show that Vietoris–Rips complexes also provide topologically faithful reconstruction guarantees for submanifolds.

Appendix A: Additional Proofs

Proof of Proposition 4.3

We divide the proof into the following two cases depending on the sign of $\kappa :=\kappa (\mathcal {M})$.

Case 1 ($\kappa \le 0$) From (2), we get

$$\begin{aligned} \textrm{diam}_{\mathcal {M}}\left( A\right) \ge 2\mathfrak {R}(A)\sqrt{\frac{n+1}{2n}}\ge 2\mathfrak {R}(A)\frac{1}{\sqrt{2}}\ge \frac{4}{3}\mathfrak {R}(A). \end{aligned}$$

Case 2 ($\kappa >0$) First note that the closed ball $\overline{{\mathbb B}_\mathcal {M}(p,\textrm{diam}_{\mathcal {M}}\left( A\right) )}$ for any $p\in A$ contains A entirely, implying that $\mathfrak {R}(A)\le \textrm{diam}_{\mathcal {M}}\left( A\right)$. Since $\textrm{diam}_{\mathcal {M}}\left( A\right) <\Delta (\mathcal {M})\le \frac{\pi }{4\sqrt{\kappa }}$, we therefore have $\mathfrak {R}(A)\in \left[ 0,\frac{\pi }{4\sqrt{\kappa }}\right]$.

Let us now define the function $J(r){\mathop {=}\limits ^{def}}\frac{2}{\sqrt{\kappa }}\sin ^{-1}\left( \sqrt{\frac{n+1}{2n}}\sin {\sqrt{\kappa }r}\right)$, and show that J(r)/r is a strictly decreasing function for $r\in \left( 0,\frac{\pi }{4\sqrt{\kappa }}\right)$. We take its derivative:

$$\begin{aligned} \frac{d}{dr}(J(r)/r)=\frac{rJ'(r)-J(r)}{r^2}=\frac{f(r)}{r^2}, \end{aligned}$$

(A1)

where $f(r){\mathop {=}\limits ^{def}}rJ'(r)-J(r)$. We observe that f(r) is strictly decreasing, since

$$\begin{aligned} f'(r)=rJ''(r)=r\frac{2\sqrt{\kappa }\sqrt{\frac{n+1}{2n}}\sin {\sqrt{\kappa }r}}{\left( 1-\frac{n+1}{2n}\sin ^2{\sqrt{\kappa }r}\right) ^{3/2}}\left( \frac{1-n}{2n}\right) <0 \end{aligned}$$

for $r\in \left( 0,\frac{\pi }{4\sqrt{\kappa }}\right)$ and $n\ge 2$. Then it follows from $f(0)=0$ that $f(r)<0$ on $r\in \left( 0,\frac{\pi }{4\sqrt{\kappa }}\right)$. When $n=1$, we remark that the result can be established using Case 1, since $\mathcal {M}$ must be flat.

So, (A1) implies that the function J(r)/r is a strictly decreasing function for $r\in \left( 0,\frac{\pi }{4\sqrt{\kappa }}\right)$. Its minimum, as a result, is attained at $r=\frac{\pi }{4\sqrt{\kappa }}$. So, for $r\in \left( 0,\frac{\pi }{4\sqrt{\kappa }}\right]$ we get

$$\begin{aligned} \frac{J(r)}{r}\ge \frac{2}{\sqrt{\kappa }}\sin ^{-1}\left( \sqrt{\frac{n+1}{2n}}\sin {\frac{\pi }{4}}\right) \frac{4\sqrt{\kappa }}{\pi } \ge \frac{8}{\pi }\sin ^{-1}\left( \sqrt{\frac{1}{2}}\sin {\frac{\pi }{4}}\right) =\frac{8}{\pi }\frac{\pi }{6}=\frac{4}{3}. \end{aligned}$$

On the other hand, (2) implies that $\textrm{diam}_{\mathcal {M}}\left( A\right) \ge J(\mathfrak {R}(A))$ for $\mathfrak {R}(A)\in \left[ 0,\frac{\pi }{4\sqrt{\kappa }}\right]$. Hence, $\textrm{diam}_{\mathcal {M}}\left( A\right) \ge \frac{4}{3}\mathfrak {R}(A)$ as desired.

$\square$

Proposition 4.1

(Path-connectedness) Let $(S,d_S)$ be a metric space and $\beta >0$ a number such that $d_{GH}(S,\mathcal {M})<\beta$, then for any positive $\alpha$, the geometric complex of $\mathcal {R}_{\alpha +2\beta }(S)$ is path-connected. The result also holds if $M\subset {\mathbb R}^d$ and $S\subset {\mathbb R}^d$ with $d_H(\mathcal {M},S)<\beta$.

Proof

Let $a,b\in S$, then there exist points $a',b'\in \mathcal {M}$ such that $(a',a),(b',b)\in \mathcal {C}$, where $\mathcal {C}\in \mathcal {C}(\mathcal {M},S)$ is a correspondence with $\textrm{dist}(\mathcal {C})<2\beta$. Since $\mathcal {M}$ is assumed to be path-connected, so is $\mathcal {R}_\alpha (\mathcal {M})$. As a result, there exists a sequence $\{p_i'\}_{i=0}^{m+1}\subset \mathcal {M}$ forming a path in $\mathcal {R}_\alpha (\mathcal {M})$ joining $a'$ and $b'$. In other words, $p_0'=a'$, $p_{m+1}'=b'$, and $d_\mathcal {M}(p_i',p_{i+1}')<\alpha$ for $0\le i\le m$. There is also a corresponding sequence $\{x_i\}_{i=0}^{m+1}\subset S$ such that $x_0=a$, $x_{m+1}=b$, and $(p_i',x_i)\in \mathcal {C}$ for all i. We note that

$$\begin{aligned} d_S(x_i,x_{i+1})\le d_\mathcal {M}(p_i',p_{i+1}')+2\beta <\alpha +2\beta . \end{aligned}$$

So, the sequence $\{x_i\}$ produces a path in $\mathcal {R}_{\alpha +2\beta }(S)$ joining a and b. We conclude that the geometric complex of $\mathcal {R}_{\alpha +2\beta }(S)$ is path-connected.

The Hausdorff case uses the exact same argument. $\square$

Proof of Proposition 4.2

In [22, Prop. 6.1], it has been shown that $\left\Vert L_\eta \right\Vert \le 1/\tau (\mathcal {M})$ for any unit norm $\eta \in T^\perp _p(\mathcal {M})$.

• (i) Take $\eta =B(u,v)/\left\Vert B(u,v) \right\Vert$, to get $\left\Vert B(u,v) \right\Vert =\big \langle \eta , B(u,v)\big \rangle =\big \langle u, L_\eta (v)\big \rangle$. Since u, v are unit norm, applying the Cauchy–Schwarz inequality we get

  $$\begin{aligned} \left\Vert B(u,v) \right\Vert \le \left\Vert u \right\Vert \left\Vert L_\eta \right\Vert \left\Vert v \right\Vert =\left\Vert L_\eta \right\Vert \le \frac{1}{\tau (\mathcal {M})}. \end{aligned}$$

• (ii) Use (6), then apply the Cauchy–Schwarz inequality:

  $$\begin{aligned} \kappa _p(u,v)\le \langle B(u,u),B(v,v)\rangle \le \left\Vert B(u,u) \right\Vert \cdot \left\Vert B(v,v) \right\Vert \le \frac{1}{\tau (\mathcal {M})^2}. \end{aligned}$$

  So, $\kappa (\mathcal {M})\le 1/\tau (\mathcal {M})^2$. Similarly from (6), we get

  $$\begin{aligned} -\kappa _p(u,v)\le \left\Vert B(u,v) \right\Vert ^2\le \frac{1}{\tau (\mathcal {M})^2}. \end{aligned}$$

  So, $\kappa (\mathcal {M})\ge -1/\tau (\mathcal {M})^2$.

• (iii) From [5, Prop. 95], we have

  $$\begin{aligned} \rho (\mathcal {M})\ge {\left\{ \begin{array}{ll} \min \left\{ \frac{\pi }{2\sqrt{\kappa (\mathcal {M})}},\frac{1}{2}\;\textrm{inj}(\mathcal {M})\right\} , &{}\text { for }\kappa (\mathcal {M})>0\\ \frac{1}{2}\;\textrm{inj}(\mathcal {M}),&{}\text { for }\kappa (\mathcal {M})\le 0. \end{array}\right. } \end{aligned}$$

  See [5, Defn. 23] for the definition of injectivity radius $\textrm{inj}(\mathcal {M})$. On the other hand, [3, Corr. 4] shows that $\textrm{inj}(\mathcal {M})\ge \pi \tau (\mathcal {M})$. This, combining with (ii), implies that $\rho (\mathcal {M})\ge \pi \tau (\mathcal {M})/2$.

• (iv) Follows directly from (1) due to (iii).

$\square$

Proof of Proposition 4.3

Let $\gamma :[0,L]\rightarrow \mathcal {M}$ be an arc-length parametrized, unit speed, length-minimizing geodesic joining p and q, where $L=d_\mathcal {M}(p,q)$. From the fundamental theorem of calculus, we have

$$\begin{aligned} \gamma (L)-\gamma (0)=\int _0^L \gamma '(t)dt, \end{aligned}$$

and

$$\begin{aligned} \gamma '(t)=\gamma '(0)+\int _0^t \gamma ''(s)ds. \end{aligned}$$

Combining, we can write

$$\begin{aligned} \gamma (L)-\gamma (0)=\int _0^L\left[ \gamma '(0)+\int _0^t \gamma ''(s)ds\right] dt =L\gamma '(0) + \int _0^L\left[ \int _0^t \gamma ''(s) ds\right] dt. \end{aligned}$$

From the triangle inequality and the fact that $\gamma$ is unit speed,

$$\begin{aligned} \left\Vert \gamma (L)-\gamma (0) \right\Vert \ge L\left\Vert \gamma '(0) \right\Vert - \left\Vert \int _0^L\left[ \int _0^t \gamma ''(s) ds\right] dt \right\Vert \ge L-\int _0^L\left[ \int _0^t\left\Vert \gamma ''(s) \right\Vert ds\right] dt. \end{aligned}$$

We note that $\left\Vert p-q \right\Vert =\left\Vert \gamma (L)-\gamma (0) \right\Vert$ and $L=d_\mathcal {M}(p,q)$. Since $\gamma$ is geodesic, $\gamma ''(s)=B\left( \gamma '(s),\gamma '(s)\right)$ for all s. So,

$$\begin{aligned} \left\Vert p-q \right\Vert \ge L-\int _0^L\left[ \int _0^t\left\Vert B(\gamma '(s),\gamma '(s)) \right\Vert ds\right] dt. \end{aligned}$$

On the other hand, $\left\Vert B(u,u) \right\Vert \le 1/\tau (\mathcal {M})$ from Proposition 4.2 implies that

$$\begin{aligned} \left\Vert p-q \right\Vert \ge L-\frac{1}{\tau (\mathcal {M})}\int _0^L\left[ \int _0^t ds\right] dt =L-\frac{1}{\tau (\mathcal {M})}\frac{L^2}{2} =d_\mathcal {M}(p,q)-\frac{1}{2\tau (\mathcal {M})}[d_\mathcal {M}(p,q)]^2. \end{aligned}$$

The inequality is satisfied only if $\left\Vert p-q \right\Vert \le \tau (\mathcal {M})/2$ and

$$\begin{aligned} d_\mathcal {M}(p,q)\le \tau (\mathcal {M})-\tau (\mathcal {M})\sqrt{1-\frac{2\left\Vert p-q \right\Vert }{\tau (\mathcal {M})}}. \end{aligned}$$

(A2)

Let $f(r){\mathop {=}\limits ^{def}}\tau (\mathcal {M})-\tau (\mathcal {M})\sqrt{1-\frac{2r}{\tau (\mathcal {M})}}$ for $0< r<2\left( \frac{\xi -1}{\xi ^2}\right) \tau (\mathcal {M})$. We note by taking derivative that $\frac{f(r)}{r}$ is an increasing function. The maximum is attained for $r=2\left( \frac{\xi -1}{\xi ^2}\right) \tau (\mathcal {M})$. Therefore,

$$\begin{aligned} \frac{f(r)}{r} \le \frac{1-\frac{\sqrt{(2-\xi )^2}}{\xi }}{\frac{2(\xi -1)}{\xi ^2}} =\xi \frac{\xi -\sqrt{(2-\xi )^2}}{2(\xi -1)} =\xi \frac{\xi -(2-\xi )}{2(\xi -1)} =\xi ,\text { since }1<\xi <2. \end{aligned}$$

Hence, (A2) implies

$$\begin{aligned} d_\mathcal {M}(p,q)\le f(\left\Vert p-q \right\Vert ) \le \xi \left\Vert p-q \right\Vert ,\text { since }\left\Vert p-q \right\Vert \le 2\left( \frac{\xi -1}{\xi ^2}\right) \tau (\mathcal {M})\le \tau (\mathcal {M})/2. \end{aligned}$$

$\square$

Proposition 4.2

(Commuting Diagram) Let $\mathcal {K}$ be a pure m-complex and $\mathcal {L}$ a flag complex. Let $f:~\mathcal {K}\xrightarrow {\quad \quad }\mathcal {L}$ and $g:\textrm{sd}^{}\left( \mathcal {K}\right) \xrightarrow {\quad \quad }\mathcal {L}$ be simplicial maps such that

• (a) $g(v)=f(v)$ for every vertex v of $\mathcal {K}$,

• (b) $f(\sigma )\cup g(\widehat{\sigma })$ is a simplex of $\mathcal {L}$ whenever $\sigma$ is a simplex of $\mathcal {K}$.

Then, the following diagram commutes up to homotopy:

where h is a linear homeomorphism sending each vertex of $\textrm{sd}^{}\left( \mathcal {K}\right)$ to the corresponding point of $\big |\mathcal {K} \big |$.

Proof

We show that the maps $\big |g \big |$ and $\left( \big |f \big |\circ h\right)$ are homotopic by constructing an explicit homotopy $H:~\big |\textrm{sd}^{}\left( \mathcal {K}\right) \big |\times [0,1]\xrightarrow {\quad \quad }\big |\mathcal {L} \big |$ with $H(\cdot ,0)=g(\cdot )$ and $H(\cdot ,1)=\left( \big |f \big |\circ h\right) (\cdot )$. The commutativity of the diagram then follows, since h is a homeomorphism.

The complex $\mathcal {K}$ is taken to be m-dimensional. Without any loss of generality, we can assume that every point of $\textrm{sd}^{}\left( \mathcal {K}\right)$ belongs to an m-simplex. Take an arbitrary $x\in \big |\textrm{sd}^{}\left( \mathcal {K}\right) \big |$. We can write x in its barycentric coordinates as $x=\sum _{i=0}^m\zeta _i\widehat{\sigma }_i$, where $\sigma _m=[a_0,a_1,\ldots ,a_m]$ is an m-simplex of $\mathcal {K}$ and $\sigma _i=[a_0,a_1,\ldots ,a_i]$ for $0\le i\le m$. Consider a partition

$$\begin{aligned}0=t_0<t_1<\cdots <t_{m}=1\end{aligned}$$

of [0, 1]. We first define the homotopy $H(\cdot ,t)$ at $t=t_i$ for $i=0,1,\ldots ,m$, and then show that we can interpolate H continuously (using the straight-line homotopy) for any $t\in [t_{i-1},t_i]$. For any $0\le i\le m$, define

$$\begin{aligned} H(x,t_i)=\sum _{j=0}^i\zeta _j\left( \big |f \big |\circ h\right) (\widehat{\sigma }_j)+\sum _{j=i+1}^m\zeta _j\big |g \big |(\widehat{\sigma }_j). \end{aligned}$$

(A3)

From the above definition, we note for $i=0$ that

$$\begin{aligned} H(x,0)=\zeta _0\left( \big |f \big |\circ h\right) (\widehat{\sigma }_0)+\sum _{j=1}^m\zeta _j\big |g \big |(\widehat{\sigma }_j)&=\zeta _0\big |g \big |(\widehat{\sigma }_0) +\sum _{j=1}^m\zeta _j\big |g \big |(\widehat{\sigma }_j) \\ {}&=\sum _{j=0}^m\zeta _j\big |g \big |(\widehat{\sigma }_j)=\big |g \big |(x). \end{aligned}$$

The second equality is due to condition (a) and fact that $\sigma _0$ is a vertex of $\mathcal {K}$. Also, for $i=m$ we note that

$$\begin{aligned} H(x,1)=\sum _{j=0}^m\zeta _j\left( \big |f \big |\circ h\right) (\widehat{\sigma }_j) =\left( \big |f \big |\circ h\right) \left( \sum _{j=0}^m\zeta _j\widehat{\sigma }_j\right) =\left( \big |f \big |\circ h\right) (x). \end{aligned}$$

The second equality is due to the fact that h is a linear homeomorphism.

We now fix $0\le i\le m$, and extend the definition of H(x, t) for any $t\in [t_{i-1},t_i]$ by using the straight-line joining $H(x,t_{i-1})$ and $H(x,t_{i})$. Such an extension is justified, we show now both $H(x,t_{i-1})$ and $H(x,t_{i})$ belong to $\big |\eta \big |$ for some simplex $\eta$ of $\mathcal {L}$.

Let us take $\eta$ to be the set $f(\sigma _i) \cup g([\widehat{\sigma }_i,\widehat{\sigma }_{i+1},\ldots ,\widehat{\sigma }_m])$. Since g is a simplicial map, $\eta$ can be written as $f(\sigma _i)\cup [g(\widehat{\sigma }_i),g(\widehat{\sigma }_{i+1}),\ldots ,g(\widehat{\sigma }_m)]$. Now, note from condition (b) that $f(\sigma _j)\cup g(\widehat{\sigma }_j)\in \mathcal {L}$ for any $i\le j\le m$. Since f is a simplicial map, $f(\sigma _i)\cup g(\widehat{\sigma }_j)$ is a face of $f(\sigma _j)\cup g(\widehat{\sigma }_j)$, hence a simplex of $\mathcal {L}$. Consequently, any pair of elements in the set $\eta =f(\sigma _i)\cup [g(\widehat{\sigma }_i),g(\widehat{\sigma }_{i+1}),\ldots ,g(\widehat{\sigma }_m)]$ is a simplex of $\mathcal {L}$. Since $\mathcal {L}$ is a flag complex, it implies that $\eta$ is a simplex of $\mathcal {L}$.

From the definition (A3) and the fact that $h(\widehat{\sigma }_{l})\in \big |\sigma _i \big |$ for all $0\le l\le i$, we conclude that both $H(x,t_{i-1})$ and $H(x,t_{i})$ belong to $\big |\eta \big |$. So, H(x, t) is well-defined and continuous for $t\in [t_{i-1},t_i]$. Moreover, the images agree at the endpoints as noted from (A3). Therefore, H defines the desired homotopy. $\square$