Quiz: Lecture 16 — Riemannian Manifolds: Fréchet Means – Lectures on Non-Euclidean Data Analysis

Q1. The first-order condition characterizing a Fréchet mean \(\mu\) on a Riemannian manifold (away from the cut locus) is:

A. \(\mathbb{E}[X] = \mu\) B. \(\mathbb{E}[d(X, \mu)] = 0\) C. \(\mathbb{E}\{\log_\mu(X)\} = 0\) — the expected tangent vector from \(\mu\) to \(X\) is zero D. \(\mathbb{E}[\exp_\mu(X)] = 0\)

Q2. Why is the cut locus a key technical obstacle for Fréchet means on manifolds?

A. Because the manifold ends at the cut locus B. Because the squared-distance function \(d^2(x, \cdot)\) is not differentiable at the cut locus of \(x\) — the gradient formula \(\nabla d^2(x, y) = -2\log_y(x)\) only holds when \(x\) is not in the cut locus of \(y\) C. The cut locus has no effect on Fréchet means D. Because distances cannot be measured beyond the cut locus

Q3. The Bhattacharya–Patrangenaru (2003) existence theorem states that a Fréchet mean exists:

A. Only on Euclidean spaces B. Only on positively curved manifolds C. On any complete connected Riemannian manifold — existence is guaranteed under very mild conditions D. Only when the data is Gaussian

Q4. Uniqueness of Fréchet means on Hadamard manifolds (nonpositive curvature) is:

A. Global — the Fréchet function is strictly convex on Hadamard manifolds, so the minimizer is unique whenever it exists B. Only local — there can be multiple minima C. Never holds — Fréchet means are always non-unique D. Not guaranteed — uniqueness is the same as on positively curved manifolds

Q5. On positively curved manifolds (e.g., \(\mathbb{S}^2\)), Fréchet means are:

A. Never unique B. Globally unique regardless of data spread C. Unique only when the sample size is large D. Unique only locally — Afsari’s theorem guarantees uniqueness when the data is concentrated in a sufficiently small geodesic ball (the “small-ball” condition)

Q6. On Hadamard manifolds, the convergence rate of the sample Fréchet mean is:

A. The rate is unknown B. \(O_P(n^{-1/2})\) — the parametric rate, matching the Euclidean central limit theorem rate C. \(O_P(n^{-1/4})\) — slower than parametric D. \(O_P(n^{-1})\) — faster than parametric

Q7. Riemannian gradient descent (RGD) for computing Fréchet means differs from Euclidean gradient descent by:

A. Using only the distance function, not the gradient B. Using a larger step size C. RGD is identical to Euclidean gradient descent D. Using the exponential map to update: \(\mu_{k+1} = \exp_{\mu_k}(-\alpha_k \nabla F(\mu_k))\), where the gradient is a tangent vector and the update moves along the manifold via \(\exp\)

Q8. The Fréchet function on a Riemannian manifold is \(F(x) = \mathbb{E}[d^2(x, X)]\). Its gradient (away from the cut locus) is:

A. \(\nabla F(x) = 0\) by definition of Fréchet mean B. \(\nabla F(x) = -2\,\mathbb{E}\{\log_x(X)\}\) — derived from the gradient of the squared-distance function C. \(\nabla F(x) = \mathbb{E}[d(x, X)]\) D. \(\nabla F(x) = \mathbb{E}[X] - x\)

Q9. How does the manifold approach to Fréchet means differ from the metric-space approach (Lectures 4–5)?

A. The manifold approach exploits the smooth structure (tangent spaces, exponential/logarithmic maps, gradients) for explicit first-order conditions, gradient descent, and CLTs — the metric-space approach only uses distances and works in any metric space B. They are identical in all respects C. The manifold approach only works for Euclidean data D. The metric-space approach always gives better results

Q10. On the SPD manifold \(\mathcal{S}_{++}^m\) with the affine-invariant metric, Riemannian gradient descent computes Fréchet means using:

A. Standard Euclidean averaging of the matrices B. Principal component analysis of the matrices C. Iterative updates using the matrix exponential and logarithm: \(\mu_{k+1} = \mu_k^{1/2} \exp(-\alpha_k \sum_i \log(\mu_k^{-1/2} X_i \mu_k^{-1/2})) \mu_k^{1/2}\) D. Element-wise median of the matrix entries