Topological Data Analysis

Problem: Compute the Betti numbers for (a) a circle, (b) a sphere, (c) a torus, (d) a figure-eight.

Solution:

(a) Circle $S^1$:

• $\beta_0 = 1$ (one connected component)
• $\beta_1 = 1$ (one loop)
• $\beta_k = 0$ for $k \geq 2$

Euler characteristic: $\chi = 1 - 1 = 0$

(b) Sphere $S^2$:

• $\beta_0 = 1$ (connected)
• $\beta_1 = 0$ (no loops—any loop can be contracted to a point)
• $\beta_2 = 1$ (one void/enclosed cavity)
• $\beta_k = 0$ for $k \geq 3$

Euler characteristic: $\chi = 1 - 0 + 1 = 2$

(c) Torus $T^2 = S^1 \times S^1$:

• $\beta_0 = 1$ (connected)
• $\beta_1 = 2$ (two independent loops: meridian and longitude)
• $\beta_2 = 1$ (one void)
• $\beta_k = 0$ for $k \geq 3$

Euler characteristic: $\chi = 1 - 2 + 1 = 0$

(d) Figure-eight (wedge of two circles):

• $\beta_0 = 1$ (connected)
• $\beta_1 = 2$ (two loops)
• $\beta_k = 0$ for $k \geq 2$

Euler characteristic: $\chi = 1 - 2 = -1$

Result: The Betti numbers distinguish these shapes: the torus has $\beta_1 = 2$ while the sphere has $\beta_1 = 0$. The figure-eight and torus share $\beta_1 = 2$ but differ in $\beta_2$.

Problem: Given points $A = (0,0)$, $B = (1,0)$, $C = (0.5, 0.866)$ (vertices of an equilateral triangle with side length 1), construct the Vietoris-Rips complex for $\epsilon = 0.5$, $\epsilon = 1.0$, and $\epsilon = 1.5$.

Solution:

Step 1: Compute pairwise distances. $d(A,B) = 1$, $d(A,C) = 1$, $d(B,C) = 1$ (equilateral triangle).

Step 2: $\epsilon = 0.5$: No pair of points has distance $\leq 0.5$.

• 0-simplices: $\{A\}, \{B\}, \{C\}$ (3 vertices)
• 1-simplices: none
• $\beta_0 = 3$ (three components), $\beta_1 = 0$

Step 3: $\epsilon = 1.0$: All pairwise distances $\leq 1$.

• 0-simplices: $\{A\}, \{B\}, \{C\}$
• 1-simplices: $\{A,B\}, \{A,C\}, \{B,C\}$ (all edges)
• 2-simplices: $\{A,B,C\}$ (since all edges present, triangle filled)
• $\beta_0 = 1$ (connected), $\beta_1 = 0$ (loop filled by triangle)

Step 4: $\epsilon = 1.5$: Same as $\epsilon = 1.0$ (no new features).

• Same complex as $\epsilon = 1.0$

Result: The persistence diagram has one connected component born at $\epsilon = 0$ (never dies), and no other features. The triangle forms at $\epsilon = 1.0$.

Problem: 20 points sampled from a unit circle with small Gaussian noise. Describe the expected persistence diagram.

Solution:

Step 1: As $\epsilon$ increases from 0:

• At $\epsilon \approx 0$: Each point is a separate component ($\beta_0 = 20$).
• As $\epsilon$ grows: Nearby points connect, reducing $\beta_0$.

Step 2: At $\epsilon \approx 0.3$ (half the typical inter-point spacing):

• Most points connect into a single ring.
• $\beta_0 = 1$ (one component).
• $\beta_1 = 1$ (one loop appears—the circle).

Step 3: The loop persists until $\epsilon \approx 1.0$ (diameter of circle), when the interior fills in and the loop dies.

Step 4: Expected persistence diagram:

• $\beta_0$: 19 "noise" components born at $\epsilon = 0$, dying at various small $\epsilon$ values (short bars). 1 "signal" component born at $\epsilon = 0$, never dying (infinite bar).
• $\beta_1$: 1 "signal" loop born at $\epsilon \approx 0.3$, dying at $\epsilon \approx 1.0$ (long bar). Possible small "noise" loops born and dying at similar scales (short bars).

Result: The persistent homology clearly identifies the circle: one long-lived $\beta_1$ feature with persistence $\approx 0.7$, while noise features have short persistence. The bottleneck distance to the exact circle's diagram is small.

Problem: Compute the Euler characteristic of a complete graph $K_5$ (5 vertices, all edges) and a tree with 5 vertices.

Solution:

Complete graph $K_5$:

• $\alpha_0 = 5$ (vertices), $\alpha_1 = \binom{5}{2} = 10$ (edges), $\alpha_k = 0$ for $k \geq 2$
• $\chi = \alpha_0 - \alpha_1 = 5 - 10 = -5$

Alternatively: $\beta_0 = 1$, $\beta_1 = \alpha_1 - \alpha_0 + 1 = 10 - 5 + 1 = 6$ (independent cycles). $\chi = \beta_0 - \beta_1 = 1 - 6 = -5$ ✓

Tree with 5 vertices:

• $\alpha_0 = 5$ (vertices), $\alpha_1 = 4$ (edges, since a tree on $n$ vertices has $n-1$ edges)
• $\chi = 5 - 4 = 1$

Alternatively: $\beta_0 = 1$ (connected), $\beta_1 = 0$ (no cycles in a tree). $\chi = 1 - 0 = 1$ ✓

Result: $K_5$ has $\chi = -5$ (many cycles), while the tree has $\chi = 1$ (no cycles). The Euler characteristic distinguishes these graphs: trees always have $\chi = 1$, while graphs with cycles have $\chi \leq 0$.

Problem: Determine the connected components and compactness of $A = \{(x,y) : x^2 + y^2 \leq 1\} \cup \{(x,y) : (x-3)^2 + y^2 \leq 1\}$.

Solution:

Step 1: $A$ is the union of two closed disks: $D_1$ centered at $(0,0)$ with radius 1, and $D_2$ centered at $(3,0)$ with radius 1.

Step 2: The distance between centers is 3, and the sum of radii is 2. Since $3 > 2$, the disks do not overlap (their intersection is empty).

Step 3: Connected components: Each disk is connected (convex sets are connected), and since they are disjoint, $A$ has exactly 2 connected components.

Step 4: Compactness: Each disk is closed and bounded in $\mathbb{R}^2$, hence compact by Heine-Borel. The finite union of compact sets is compact, so $A$ is compact.

Result: $\beta_0(A) = 2$ (two connected components), $A$ is compact.

Problem: Show that a circle $S^1$ is homotopy equivalent to $\mathbb{R}^2 \setminus \{(0,0)\}$ (the punctured plane).

Solution:

Step 1: Define $f: S^1 \to \mathbb{R}^2 \setminus \{(0,0)\}$ as the inclusion $f(z) = z$ (view $S^1 \subset \mathbb{R}^2$).

Step 2: Define $g: \mathbb{R}^2 \setminus \{(0,0)\} \to S^1$ as the radial projection $g(x) = x/\|x\|$.

Step 3: Check $g \circ f = \text{id}_{S^1}$: For $z \in S^1$, $g(f(z)) = g(z) = z/\|z\| = z$ (since $\|z\| = 1$). ✓

Step 4: Check $f \circ g \simeq \text{id}$: Define $H(x, t) = (1-t) \cdot \frac{x}{\|x\|} + t \cdot x$.

For $t \in [0,1]$, $H(x, t) = (1-t)\frac{x}{\|x\|} + tx$. This is continuous and never zero (since $(1-t)/\|x\| + t > 0$ for $x \neq 0$).

$H(x, 0) = x/\|x\| = g(x)$, $H(x, 1) = x$.

Result: $f$ and $g$ are homotopy inverses, so $S^1 \simeq \mathbb{R}^2 \setminus \{(0,0)\}$. Both have $\beta_0 = 1$, $\beta_1 = 1$.

Problem: Is the closed unit ball $B = \{f \in C[0,1] : \|f\|_\infty \leq 1\}$ compact in the supremum norm?

Solution:

Step 1: Consider the sequence $f_n(x) = x^n$ in $B$. Each $f_n$ is continuous with $\|f_n\|_\infty = 1$.

Step 2: Pointwise limit: $f(x) = \lim_{n \to \infty} x^n = \begin{cases} 0 & x \in [0,1) \\ 1 & x = 1 \end{cases}$

This limit is NOT continuous, so $f \notin C[0,1]$.

Step 3: Any subsequence of $\{f_n\}$ converges pointwise to the same discontinuous function, so no subsequence converges in the supremum norm.

Step 4: Therefore $\{f_n\}$ has no convergent subsequence in $B$.

Result: The closed unit ball in $C[0,1]$ is NOT compact. This is a consequence of the Riesz theorem: in an infinite-dimensional Banach space, the closed unit ball is never compact. Contrast with finite-dimensional $\mathbb{R}^n$ where the closed unit ball IS compact (Heine-Borel).

Problem: Describe the persistence diagram for 30 points sampled from two concentric circles of radii 1 and 2.

Solution:

Step 1: At small $\epsilon$: 30 connected components ($\beta_0 = 30$).

Step 2: As $\epsilon$ increases, points on each circle connect. The inner circle (radius 1) forms a loop first (at $\epsilon \approx 0.2$, the typical spacing). The outer circle (radius 2) forms a loop at a slightly larger $\epsilon$.

Step 3: At $\epsilon \approx 0.5$: Both circles are fully connected.

• $\beta_0 = 2$ (two connected components: inner and outer rings)
• $\beta_1 = 2$ (two loops)

Step 4: As $\epsilon$ grows toward 1.0: The inner circle's loop fills in (dies). The outer circle's loop persists longer.

Step 5: At $\epsilon \approx 2.0$: The outer loop fills in.

Persistence diagram:

• $\beta_0$: 1 infinite bar (everything connected eventually), 28 short bars (noise components)
• $\beta_1$: 2 long bars. Inner loop: born $\approx 0.2$, dies $\approx 1.0$ (persistence $\approx 0.8$). Outer loop: born $\approx 0.4$, dies $\approx 2.0$ (persistence $\approx 1.6$).

Result: The two loops are clearly distinguished by their persistence. The outer loop has higher persistence because it is larger. TDA correctly identifies the "shape" of two concentric circles.

Problem: For a set of points forming a regular hexagon, compare the Čech and Vietoris-Rips complexes at different scales.

Solution:

Step 1: Consider a regular hexagon with side length 1. Vertices: $v_0, \ldots, v_5$.

Step 2: For $\epsilon < 1/2$: No edges in either complex.

Step 3: For $\epsilon = 1/2$:

• Čech complex: Two adjacent balls intersect iff the distance between centers $\leq 1$ (diameter of intersection region). Adjacent vertices are distance 1 apart, so edges appear.
• Vietoris-Rips: Same edges (distance 1 $\leq 2\epsilon = 1$).
• Both give the cycle graph $C_6$.

Step 4: For $\epsilon = 1/\sqrt{3} \approx 0.577$:

• Čech complex: Three consecutive vertices $v_0, v_1, v_2$ have balls intersecting at the center of the triangle they form. A 2-simplex $\{v_0, v_1, v_2\}$ appears.
• Vietoris-Rips: All pairwise distances $\leq 2\epsilon \approx 1.15$. Since the hexagon's diameter is 2, only adjacent and next-adjacent pairs (distance $\leq \sqrt{3} \approx 1.73$) are connected.

Step 5: Key difference: The Čech complex requires a common intersection point for all balls in a simplex, while Vietoris-Rips only requires pairwise distances. The Čech complex is "smaller" (fewer simplices) but captures the true topology by the Nerve Theorem.

Result: For the same data, the Čech complex is a subcomplex of the Vietoris-Rips complex. Čech is more topologically accurate but computationally harder (requires checking ball intersections). Vietoris-Rips depends only on pairwise distances, making it computationally efficient.

Problem: Describe the expected persistence diagram for points sampled from a torus embedded in $\mathbb{R}^3$.

Solution:

Step 1: A torus has $\beta_0 = 1$, $\beta_1 = 2$, $\beta_2 = 1$.

Step 2: At small scales, the point cloud looks like scattered points ($\beta_0 = n$).

Step 3: As $\epsilon$ increases, points connect into a "skeleton" of the torus:

• $\beta_0$ drops to 1 (one connected component)
• Two loops appear ($\beta_1 = 2$): the meridian and longitude circles

Step 4: The two loops persist over a wide range of $\epsilon$ values. Eventually:

• The meridian loop fills in (dies first, at $\epsilon \approx r$ where $r$ is the tube radius)
• The longitude loop fills in later (at $\epsilon \approx R$ where $R$ is the major radius)
• The void ($\beta_2 = 1$) appears when the surface is fully reconstructed and dies when the interior fills

Step 5: Expected persistence diagram:

• $\beta_0$: Many short bars (noise), one infinite bar
• $\beta_1$: Two long bars (the two independent loops of the torus)
• $\beta_2$: One bar (the enclosed void)

Result: Persistent homology correctly identifies the torus topology: $\beta_0 = 1$, $\beta_1 = 2$, $\beta_2 = 1$. The two $\beta_1$ features have different persistence values reflecting the different radii of the torus.

Problem: How does persistent homology help determine the number of clusters in a point cloud?

Solution:

Step 1: Consider 3 clusters of points, each cluster forming a tight ball, with large gaps between clusters.

Step 2: At small $\epsilon$: Each point is separate ($\beta_0 = n$).

Step 3: As $\epsilon$ increases: Points within each cluster connect first. At $\epsilon$ equal to the intra-cluster spacing, $\beta_0 = 3$ (three components).

Step 4: The three components persist until $\epsilon$ reaches the inter-cluster distance, at which point they merge into one component ($\beta_0 = 1$).

Step 5: The persistence diagram for $\beta_0$ shows:

• Many short bars (noise within clusters)
• Three medium bars (the three clusters), dying at the inter-cluster distance
• One infinite bar (everything eventually connected)

Result: The number of long-lived $\beta_0$ features (after filtering short noise bars) indicates the number of clusters. This is the basis of topological clustering methods that use persistent homology to determine cluster count automatically.

Problem: For 4 points forming a square with side length 1, determine at what value of $\epsilon$ the Čech complex first contains a 2-simplex (triangle).

Solution:

Step 1: Points: $A = (0,0)$, $B = (1,0)$, $C = (1,1)$, $D = (0,1)$.

Step 2: For a 2-simplex $\{A, B, C\}$, we need balls of radius $\epsilon$ around $A$, $B$, $C$ to have a common intersection point.

Step 3: The circumcenter of triangle $ABC$ is at $(0.5, 0.5)$ with circumradius $r = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.5} = 1/\sqrt{2} \approx 0.707$.

Step 4: The three balls intersect when $\epsilon \geq r = 1/\sqrt{2}$.

Step 5: At $\epsilon = 1/\sqrt{2}$, the point $(0.5, 0.5)$ is equidistant from $A$, $B$, $C$ at distance $1/\sqrt{2}$, so it lies in all three balls.

Result: The Čech complex first contains a 2-simplex at $\epsilon = 1/\sqrt{2} \approx 0.707$. This is less than the Vietoris-Rips threshold ($\epsilon = 1$ for the same simplex), illustrating that Čech is "tighter" than Rips.

Problem: Given a persistence diagram with features $(0.2, 0.8)$ and $(0.3, 0.6)$ for $\beta_1$, compute the first persistence landscape.

Solution:

Step 1: For each feature $(b_i, d_i)$, define the tent function:

Step 2: For feature 1: $(0.2, 0.8)$, midpoint $= 0.5$:

Peak value: $0.3$ at $t = 0.5$.

Step 3: For feature 2: $(0.3, 0.6)$, midpoint $= 0.45$:

Peak value: $0.15$ at $t = 0.45$.

Step 4: The first persistence landscape is $\lambda_1(t) = \max_i f_i(t)$:

• For $t \in [0.2, 0.3]$: $\lambda_1(t) = t - 0.2$ (only feature 1 active)
• For $t \in [0.3, 0.45]$: $\lambda_1(t) = \max(t-0.2, t-0.3) = t-0.2$ (feature 1 dominates)
• For $t \in [0.45, 0.5]$: $\lambda_1(t) = \max(t-0.2, 0.6-t) = t-0.2$ (feature 1 still dominates)
• For $t \in [0.5, 0.6]$: $\lambda_1(t) = \max(0.8-t, 0.6-t) = 0.8-t$ (feature 1 dominates)
• For $t \in [0.6, 0.8]$: $\lambda_1(t) = 0.8-t$ (only feature 1 active)

Result: The first persistence landscape $\lambda_1(t)$ is the upper envelope of the tent functions. It can be used as a feature vector for statistical analysis (averaging landscapes across samples, computing confidence bands, etc.).

Problem: How would you use persistent homology to classify periodic vs. non-periodic time series?

Solution:

Step 1: Given a time series $x(t)$, create a delay embedding in $\mathbb{R}^d$:

where $\tau$ is the delay and $d$ is the embedding dimension.

Step 2: For a periodic signal, the delay embedding traces a closed curve (a loop in $\mathbb{R}^d$).

Step 3: Compute persistent homology of the point cloud $\{\mathbf{y}(t)\}$.

Step 4: Classification criteria:

• Periodic signal: $\beta_1 = 1$ with high persistence (long-lived loop)
• Non-periodic signal: $\beta_1 = 0$ or $\beta_1 = 1$ with low persistence (short-lived feature)

Step 5: Use the persistence of the dominant $\beta_1$ feature as a classification feature: threshold at some value to separate periodic from non-periodic.

Result: TDA provides a principled, noise-robust method for detecting periodicity. The delay embedding + persistent homology approach works even for noisy, irregularly sampled time series, making it valuable for applications in astronomy (variable star classification), medicine (heartbeat analysis), and finance (cycle detection).