Recovering the Homology of Immersed Manifolds

Abstract

Given a sample of an abstract manifold immersed in some Euclidean space, we describe a way to recover the singular homology of the original manifold. It consists in estimating its tangent bundle—seen as subset of another Euclidean space—from a measure theoretical point of view, and in applying measure-based filtrations for persistent homology. The construction we propose is consistent and stable, and does not require the dimension of the manifold.

Appendices

A Notations

We adopt the following notations:

• $n, d > 0$ are integers.

• If $x, y \in {\mathbb {R}}$, $\min {(x,y)}$ is the minimum of x and y.

• I is the interval $[0, +\infty )$ or [0, T] for $T \ge 0$.

• $E = {\mathbb {R}}^n$ is the Euclidean space, $\textrm{M}(E)$ the vector space of $n \times n$ matrices, $\mathscr {G}_{d}(E)$ the Grassmannian of d-planes in E.

• A is a subset of E, $\textrm{med}\left( A \right)$ denotes its medial axis, ${\text {reach}}(A)$ its reach. For every $x\in E$, ${\text {dist}}(x, A)$ is the distance from x to A.

• For $x,y \in E$, $x\,\bot \,y$ denotes the orthogonality of x and y.

• If $x,y\in E$, $x \otimes y = x {}^ty \in \textrm{M}(E)$ is the outer product, and ${x}^{\otimes 2} = x \otimes x$.

• $\Vert \,{\cdot }\,\Vert$ is the Euclidean norm on E and $\langle \,{\cdot }\,, \,{\cdot }\,\rangle$ the corresponding inner product, $\left\| \,{\cdot }\,\right\| _\textrm{F}$ the Frobenius norm on $\textrm{M}(E)$, $\Vert \,{\cdot }\,\Vert _\gamma$ the $\gamma$-norm on $E \times \textrm{M}(E)$ (defined in Sect. 4.1).

• $\textrm{W}_{p}$ is the p-Wasserstein distance between measures on E, $\textrm{W}_{p,\gamma }$ is the $(p,\gamma )$-Wasserstein distance between measures on $E \times \textrm{M}(E)$ (defined in Sect. 4.1).

• $\mathscr {H} ^d$ is the d-dimensional Hausdorff measure on E or on a subspace $T \subset E$ (not renormalized).

• If $\mu$ is a measure of positive finite mass, $|\mu |$ denotes its mass, $\overline{\mu }={\mu }/{|\mu |}$ is the associated probability measure, denotes the associated lifted measure (introduced in Sect. 4.1).

• $1_A$ is the indicator function of a measurable set A.

• If T is a subspace of E, $p_T$ denotes the orthogonal projection matrix on T.

• $\mathscr {B} (x,r)$ and $\overline{ \mathscr {B} }\left( x,r\right)$ denote the open and closed balls of E, $\partial \mathscr {B}\left( x,r\right)$ the sphere. $V_{d}$ and $S_{d-1}$ denote $\mathscr {H} ^d( \mathscr {B} (0,1) )$ and $\mathscr {H} ^{d-1}( \partial \mathscr {B}\left( 0,1\right) )$ (note that $S_{d-1} = d V_{d}$).

• $\mathscr {M}_0$ is a Riemannian manifold, and $\mathscr {B} _{ \mathscr {M}_0 }(x,r)$ and $\overline{ \mathscr {B} }_{ \mathscr {M}_0 }(x,r)$ denote the open and closed geodesics balls. For $x_0, y_0 \in \mathscr {M}_0$, $d_{ \mathscr {M}_0 }(x_0, y_0)$ denotes the geodesic distance.

• If T is a subspace of E, $\mathscr {B} _{T}(x,r)$ and $\overline{ \mathscr {B} }_{T}(x,r)$ denote the open and closed balls of T for the Euclidean distance.

• If f is a map with values in ${\mathbb {R}}$ and $t \in {\mathbb {R}}$, $f^t$ denotes the sublevel set $f^t = f^{-1}((-\infty , t])$.

B Table of Constants

In the following table, each constant is preceded by the result where it appeared first. If a constant is defined from the others, it is indicated here. The indices are arbitrary and only reflect the order of apparition of each result.

Index	Result	Constant
1	Corollary 2.9	a,   $c_{1} = 8{\text {diam}}{({\text {supp}}(\mu ))} + 5$,   $c_{1}' = 2a^{-{1}/{d}}$
2	Hypothesis 2	$\rho$
3	Hypothesis 3	$L_0$,   $f_{\min }$,   $f_{\max }$
4	Hypothesis 4	$c_{4}$,   $r_{4}$
5	Lemma 3.5	$c_{5}:t \mapsto \bigl ({1 - \sqrt{1 - 2t}}\bigr )/ t$
6	Lemma 3.10	$J_{\min }=({23}/{24})^d$,   $J_{\max }= ({5}/{4})^d$
7	Lemma 3.12	$c_{7} = 4 L_0 J_{\max }+d\rho f_{\max } /2$
8	Proposition 3.14	$c_{8} = c_{7} + f_{\max } J_{\max }d 2^{d} \rho$
9	Proposition 3.14, Hypothesis 5	$c_{9} = f_{\min } J_{\min } V_{d}$
10	Proposition 3.14, Hypothesis 6	$c_{10} = d 2^{d} f_{\max } J_{\max } V_{d}$
11	Proposition 3.15, Hypothesis 7	$c_{11}= ({ f_{\max } J_{\max }}/({ f_{\min } J_{\min }}))\bigl ({\rho }/{\sqrt{4 - \sqrt{13}}}\bigr )^d d 2^{2d} \sqrt{3}$
12	Section 3.3	$\varDelta$,   $\varDelta _0$,   $\varTheta$
13	Proposition 3.20	$r_{13} = \min {\bigl ({\sin \varTheta }/({8\rho }),{(\sin \varTheta )^2}/{4}, {\varDelta _0\sin \varTheta }/{4},\varDelta \bigr )}$
		$c_{13} =({2}/{\sin \theta })^\alpha V_{\alpha } f_{\max } \mathscr {H} ^{d'}_{ \mathscr {M}_0 }( \mathscr {N}_0 )$
14	Proposition 4.3	$c_{14} = 6 \rho + ( 4c_{7} + f_{\max } 2^d d \rho + c_{8})/( f_{\min } J_{\min })$,
15	Lemma 4.7	$c_{\textrm{15}} = 2(1+4\cdot 5^{d-1}/{3^d})({c_{10}}/{c_{9}})$
16	Lemma 4.8	$c_{\textrm{16}} = (2 + {2^{5/2}5^{d-{1/2}}}/{3^d})({c_{11}}/{c_{9}})$
17	Lemma 4.9	$c_{17} ={2^{d-1}}/c_{9}+ 2({12 \cdot 5^{d-1}c_{10}+1})/({3^d c_{9}})$
		$+\,2^{d+3}({(3/{2})^{d-1}c_{10}+1})/{c_{9}}$
18	Lemma 4.10	$c_{18} ={2^{d-2}}/{c_{9}} + ({4 \cdot 3^{1/2}5^{d-{1/2}} c_{11}+4^{d-{1/2}}})/({3^d c_{9}})$
		$+\,2\cdot 4^d({2 c_{11} (3/2)^{d-1/2}+1})/({3^d c_{9}})$
19	Proposition 4.13	$c_{19} = 4(1 + c_{20})$,   $c_{19}'=4c_{18}$
20	Lemma 4.15	$c_{20} = 3 +c_{15} +c_{16} + c_{17}$
21	Theorem 4.16	$c_{21} = 2 +c_{19}'/2= 2(1 + c_{18})$
22	Corollary 4.19	$c_{22} = c_{21}(c_{4})^{1/p} + c_{19} + c_{14}$
23	Section 5.2	,   ,
24	Corollary 5.5	,

C Supplementary Material for Sect. 2

Proof of Lemma 2.4

The proof is based on the following observations. We can use the triangle inequality, then the Pythagorean Theorem with $\langle v, y-x\rangle = 0$ and Lemma 2.3 (i) to get

$$\begin{aligned}{} & {} \Vert \gamma (t) - x\Vert \le \Vert (y+tv) - x\Vert + \Vert \gamma (t) - (y+tv)\Vert \\{} & {} \qquad \qquad \qquad \le \sqrt{ \Vert tv\Vert ^2+\Vert y-x\Vert ^2} + \frac{\rho }{2}t^2= \sqrt{t^2+l^2} + \frac{\rho }{2}t^2. \end{aligned}$$

For any $r \le {1}/{\rho }$, consider the equation

$$\begin{aligned} \sqrt{t^2+l^2} + \frac{\rho }{2}t^2 = r. \end{aligned}$$

(49)

By squaring this equality, we get $({\rho }/{2})^2t^4 - (1+\rho r)t^2+(r^2-t^2)=0$. By considering the polynomial $T\mapsto ({\rho }/{2})^2 T^2 - (1+\rho r)T +r^2-t^2$, whose discriminant is $1 + 2\rho r + (\rho t)^2>0$, we see that the solutions of (49) are

$$\begin{aligned} T_1 = \frac{\sqrt{2}}{\rho }\sqrt{1+\rho r - \sqrt{ 1 + 2\rho r + \rho ^2 l^2}}\,,\quad T_1' = \frac{\sqrt{2}}{\rho }\sqrt{1+\rho r + \sqrt{ 1 + 2\rho r + \rho ^2 l^2}}\,. \end{aligned}$$

Following the same ideas, one obtains

$$\begin{aligned}\Vert \gamma (t) - x\Vert \ge \sqrt{t^2+l^2} - \frac{\rho }{2}t^2.\end{aligned}$$

Moreover, the equation

$$\begin{aligned} \sqrt{t^2+l^2} - \frac{\rho }{2}t^2 = r \end{aligned}$$

(50)

admits the following roots:

$$\begin{aligned} T_2 = \frac{\sqrt{2}}{\rho }\sqrt{1-\rho r - \sqrt{ 1 - 2\rho r + \rho ^2 l^2}}\,,\quad T_2' = \frac{\sqrt{2}}{\rho }\sqrt{1-\rho r + \sqrt{ 1 - 2\rho r + \rho ^2 l^2}}\,. \end{aligned}$$

We now prove the five points successively.

(i) Observe that $\dot{\phi }(t) = 2\langle \dot{\gamma }(t), \gamma (t) - x\rangle$, and that

$$\begin{aligned}\ddot{\phi }(t) = 2\langle \dot{\gamma }(t), \dot{\gamma }(t)\rangle + 2\langle \ddot{\gamma }(t), \gamma (t) - x\rangle .\end{aligned}$$

By Cauchy–Schwarz inequality, $\langle \ddot{\gamma }(t), \gamma (t) - x\rangle \ge -\Vert \ddot{\gamma }(t)\Vert \cdot \Vert \gamma (t) - x\Vert$. Note that $\langle \dot{\gamma }(t), \dot{\gamma }(t)\rangle = 1$ since $\gamma$ is parametrized by arc-length, and that $\Vert \ddot{\gamma }(t)\Vert \le \rho$ by (1). Hence we get

$$\begin{aligned} \ddot{\phi }(t) \ge 2(1-\rho \Vert \gamma (t) - x\Vert ). \end{aligned}$$

(51)

Consider (49) with $r ={1}/{\rho }$. We see that $\Vert \gamma (t) - x\Vert \le {1}/{\rho }$ when t is lower than

$$\begin{aligned} T_1 = \frac{\sqrt{2}}{\rho } \sqrt{2-\sqrt{3+\rho ^2 l^2}}\,. \end{aligned}$$

In this case, $\ddot{\phi }(t) \ge 0$ according to (51). Since $\dot{\phi }(0) = 0$, we deduce that $\phi$ is increasing on $[0,T_1]$.

(ii) As we have seen with (50), we have $\Vert \gamma (t)-x\Vert > r$ when $t \in (T_2, T_2')$. In order to give an upper bound on $T_2$, we use the inequality $\sqrt{b} - \sqrt{a} = (b-a)/\bigl ({\sqrt{a}+\sqrt{b}}\bigr ) \le (b-a)/{\sqrt{b}}$, where $a<b$, to get

$$\begin{aligned}1-\rho r - \sqrt{ 1 - 2\rho r + \rho ^2 l^2 }\le \frac{\rho ^2(r^2 - l^2)}{1-\rho r}\end{aligned}$$

and we conclude that $T_2 \le {\sqrt{2}}\sqrt{r^2 - l^2}/{\sqrt{1-\rho r}}$. Since $r \le {1}/({2\rho })$, we obtain $T_2 \le 2 \sqrt{r^2 - l^2}$.

(iii) When $l = 0$, algebraic manipulations show that $T_2 =(1-\sqrt{1-2\rho r})/\rho$ and $T_2' =(1+\sqrt{1-2\rho r})/\rho$.

(iv) Consider the map $\phi :t \mapsto \Vert \gamma (t) - x\Vert ^2$. By definition of b, for $t \in (0,b)$, we have $\Vert \gamma (t) - x\Vert \le r$. Hence (51) gives $\ddot{\phi }(t) \ge 2(1-\rho r)$. It follows that $\dot{\phi }(t) \ge 2(1-\rho r)t$, and that

$$\begin{aligned}\phi (b)-\phi (a)= \int _{a}^{b} \dot{\phi }(t) \, \textrm{d} t\ge \int _{a}^{b} 2(1-\rho r)t\, \textrm{d} t= (1-\rho r)(b^2-a^2).\end{aligned}$$

Note that $r^2 = \phi (b)$. Besides, $s^2 = \phi (a)$ or $s^2 < \phi (a)$, depending on whether $s \ge l$ or $s < l$. In both cases, we have $r^2 - s^2 \ge \phi (b)-\phi (a)$, and we deduce that

$$\begin{aligned}r^2 - s^2 \ge (1-\rho r)(b^2-a^2).\end{aligned}$$

Writing $r^2-s^2 = (r+s)(r-s)$ and $b^2-a^2 =(b+a)(b-a)$ leads to

$$\begin{aligned} b-a \le \frac{r+s}{b+a}\cdot \frac{r-s}{1-\rho r}. \end{aligned}$$

(52)

Now, let us give a lower bound on b. According to (49), b is lower bounded by

$$\begin{aligned} T_1 = \frac{\sqrt{2}}{\rho }\sqrt{1+\rho r - \sqrt{ 1 + 2\rho r + \rho ^2 l^2}}. \end{aligned}$$

Using the inequality $\sqrt{b} - \sqrt{a} =(b-a)/\bigl ({\sqrt{b}+\sqrt{a}}\bigr ) \ge (b-a)/{2\sqrt{b}}$, where $a<b$, we get

$$\begin{aligned}1+\rho r - \sqrt{ 1 + 2\rho r + \rho ^2 l^2 }\ge \frac{\rho ^2(r^2 - l^2)}{2(1+\rho r)},\end{aligned}$$

and $b \ge (1+\rho r)^{-{1}/{2}}\sqrt{r^2 - s^2}$. Inserting $b+a \ge b \ge (1+\rho r)^{-{1}/{2}}\sqrt{r^2 - s^2}$ in (52) yields

$$\begin{aligned}b(v) - a(v) \le \frac{(1+\rho r)^{{1}/{2}}}{1-\rho r} \sqrt{r^2-s^2}.\end{aligned}$$

Under the hypothesis $r \le {1}/({2\rho })$, we get $b-a \le \sqrt{6}\sqrt{r^2-s^2}$.

(v) When $l=0$, we have $b(v)+a(v)\ge r+s$. Hence (52) yields $b(v)-a(v)\le (r-s)/({1-\rho r})$. Using $r \le 1/(2\rho )$, we obtain $b(v)-a(v)\le 2(r-s)$. $\square$

Proof of Corollary 2.9

We shall first study an intermediate quantity. Let $\mu$ be a probability measure on $E={\mathbb {R}}^n$, $m\in (0,1)$, and ${\text {d}}_{\mu ,m}$ the corresponding DTM. Consider the quantity $c(\mu , m)$ is defined as

$$\begin{aligned}c(\mu , m)\, =\! \sup _{x\in {\text {supp}}(\mu )} {\text {d}}_{\mu , m} (x).\end{aligned}$$

Suppose that $\mu$ satisfies the following for $r <({m}/{a})^{1/d}$: for all $x \in {\text {supp}}(\mu )$, $\mu ( \mathscr {B} (x,r) ) \ge a r^d$. Let us show that $c(\mu , m) \le C m^{1/d}$ with $C = a^{-{1/d}}$. By definition,

$$\begin{aligned}\delta _{\mu , t}(x) = \inf {\{r\ge 0 \mid \mu ( \overline{ \mathscr {B} }\left( x,r\right) )>t\}}\quad \text { and }\quad {\text {d}}_{\mu ,m} ^2(x) = \frac{1}{m}\int _0^{m} \delta _{\mu ,t}^2(x)\, \textrm{d} t.\end{aligned}$$

Using the assumption $\mu ( \mathscr {B} (x,r) ) \ge a r^d$ for all $x \in {\text {supp}}(\mu )$, we get $\delta _{\mu , t}(x) \le ({t}/{a})^{1/d}$, and a simple computation yields

$$\begin{aligned} {\text {d}}_{\mu ,m} ^2(x)\le \frac{d}{d+2} \biggl (\frac{t}{a}\biggr )^{\!2/d}\!\le \,\biggl (\frac{t}{a}\biggr )^{\!{2/d}},\end{aligned}$$

which yields the result.

We can now prove the corollary. Let $\pi$ be an optimal transport plan for $w=W _2(\mu ,\nu )$. Denote $\alpha = w^{1/2}$ and $D = {\text {diam}}{({\text {supp}}(\mu ))}$. Define $\pi '$ to be $\pi$ restricted to the set $\{x,y \in E \mid \Vert x-y\Vert <\alpha \}$. We denote its marginals $\mu '$ and $\nu '$. By the Markov inequality, $1-|\pi '| \le {w^2}/{\alpha ^2} = w$, where we recall that $|\pi '|$ denotes the total mass of $\pi '$. Consider the probability measures $\overline{\mu '}=\mu '/{|\mu '|}$ and $\overline{\nu '}=\nu '/{|\nu '|}$. Let us show that we have

$$\begin{aligned}{} & {} W _2(\mu ,\overline{\mu '})= 2D \alpha ,\quad W _2({\overline{\mu '}},{\overline{\nu '}} )\le \alpha ,\quad \text {and}\nonumber \\{} & {} W _2(\nu ,\overline{\nu '})\le 2(1+D)\alpha . \end{aligned}$$

(53)

The first inequality is an application of Lemma 4.6:

$$\begin{aligned}W _2(\mu ,\overline{\mu '})\le 2(1-|\mu '|)^{1/2} D = 2(1-|\pi '|)^{1/2} D \le 2 w^{1/2} D.\end{aligned}$$

To obtain the second inequality, we write

$$\begin{aligned} \textrm{W}_{2} ^2(\overline{\mu '}, \overline{\nu '})= & {} \int \Vert x-y\Vert ^2 \textrm{d} \overline{ \pi '}(x,y)\\= & {} \int \Vert x-y\Vert \,\frac{ \textrm{d} \pi '(x,y)}{|\pi '|}\le \frac{1}{|\pi '|}\int \Vert x-y\Vert \, \textrm{d} \pi (x,y).\end{aligned}$$

Hence the Jensen inequality leads to $W _2(\overline{\mu '},\overline{\nu '})\le {w}/{|\pi '|^{1/2}}$. Since $1-|\pi '| \le w$, we have ${w}/{|\pi '|^{1/2}} \le {w}/({1-w})$, and the assumption $w \le {1}/{4}$ yields ${w}/({1-w}) \le \alpha$. This proves the second point. Finally, we obtain the third inequality by applying the triangle inequality:

$$\begin{aligned}W _2(\nu ,\overline{\nu '}) \le W _2(\nu ,\mu )+ W _2(\mu ,\overline{\mu '})+ W _2(\overline{\mu '},\overline{\nu '}).\end{aligned}$$

Next, let us deduce that

$$\begin{aligned} \begin{aligned}{} & {} c(\overline{\mu '},m) \le c(\mu ) + m^{-{1}/{2}} 2D \alpha \quad \text {and} \\{} & {} c(\overline{\nu '},m) \le c(\mu ,m) + \bigl (m^{-{1}/{2}}+m^{-{1}/{2}}2D+1\bigr )\alpha . \end{aligned} \end{aligned}$$

(54)

The first inequality follows from the stability of the DTM (see (3)):

$$\begin{aligned}c(\overline{\mu '},m)\,=\! \sup _{x \in {\text {supp}}(\overline{\mu '})} \! {\text {d}}_{\overline{\mu '}} (x)\,\le \! \sup _{x \in {\text {supp}}(\overline{\mu '})}\! {\text {d}}_{\mu } (x) + m^{-{1}/{2}}W _2(\overline{\mu '},\mu ),\end{aligned}$$

and we conclude with $W _2(\mu ,\overline{\mu '})= 2D \alpha$. In order to prove the second inequality, we also use (3):

$$\begin{aligned}c(\overline{\nu '},m)\,= \!\sup _{x \in {\text {supp}}(\overline{\nu '})}\! {\text {d}}_{\overline{\nu '}} (x)\,\le \!\sup _{x \in {\text {supp}}({\overline{\nu '}})} \! {\text {d}}_{\overline{\mu '}} (x) + m^{-{1}/{2}}W _2(\overline{\mu '},\overline{\nu '}).\end{aligned}$$

Since $\pi '$ has support included in $\{x,y \in E \mid \Vert x-y\Vert <\alpha \}$, we use the fact that the DTM is 1-Lipschitz to obtain

$$\begin{aligned}\sup _{x \in {\text {supp}}({\overline{\nu '}})}\! {\text {d}}_{\overline{\mu '}} (x)\,\le \! \sup _{x \in {\text {supp}}({\overline{\mu '}})}\! {\text {d}}_{\overline{\mu '}} (x) + \alpha = c(\mu ',m) + \alpha \end{aligned}$$

and we deduce

$$\begin{aligned}c(\overline{\nu '},m)\le c(\mu ',m) + \alpha + m^{-{1}/{2}}W _2(\overline{\mu '},\overline{\nu '})\le c(\mu ,m) + (m^{-{1/2}}+m^{-{1/2}}2D+1)\alpha .\end{aligned}$$

We can now conclude with Theorem 2.8. In our context, it reads

$$\begin{aligned} d_i( W[\mu ] , W[\nu ] )&\le m^{-1/2}W _2(\mu ,\overline{\mu '}) + m^{-{1/2}} W _2(\overline{\mu '},\overline{\nu '}) + m^{-1} W _2(\nu ,\overline{\nu '})\\&\quad + c(\overline{\mu '},m) + c(\overline{\nu '},m) \\&\le (m^{-{1}/{2}}(4D + 1)+4(D+1))\alpha + 2 c(\mu ,m), \end{aligned}$$

where we used (53) and (54) in the last line. Since $m \le 1$, we can simplify this expression into

$$\begin{aligned}\textrm{d}_\textrm{i}( W[\mu ] , W[\nu ] )\le m^{-1/2}(8D+5)\alpha + 2 c(\mu ,m).\end{aligned}$$

We conclude the proof by using the inequality $c(\mu ,m) \le a^{-{1/d}} m^{1/d}$ shown at the beginning of the proof. $\square$