To prove the identity that the sum of the non-zero eigenvalues of the integral operator ( S ) equals ( \int_{\mathcal{X} \times \mathcal{Y}} d_\mu(x,x) d_\nu(y,y) , d(\mu \times \nu)(x,y) ), we proceed as follows:

Step 1: Structure of the Operator ( S )

The operator ( S ) is defined on ( L^2(\mathcal{X} \times \mathcal{Y}, \mu \times \nu) ) by [ S(f)(x,y) = \int_{\mathcal{X} \time...

Short answer As stated, the identity need not hold in general: the operator S need not be of trace class. A counterexample and a correction are given by Lyons in the erratum to his paper. The identity does hold (and S is trace class) under additional structure on the marginal metric spaces, namely when both metrics admit an isometric embedding into a Hilbert space (i.e., are of negative type so th...

Here is a very concrete way to see why
[ \sum_{i=1}^\infty\lambda_i ;=; \int_{\X\times\Y}d_\mu(x,x),d_\nu(y,y),d\mu(x),d\nu(y), ]
under the (additional) hypothesis that each of the two “marginal” operators is Hilbert–Schmidt (in fact trace‐class). In Lyons’s original paper the full generality of separable metric spaces forces one into subtleties, but as soon as you know that each margi...

This is an excellent question about the trace class property of integral operators arising in the context of distance covariance. Let me provide a comprehensive analysis of the problem.

The Main Issue

The key challenge here is that the operator is not always trace class, as you've discovered. The counterexample mentioned in Lyons' errata shows that the formula can fail without additio...

Excellent question. You've hit upon a very subtle and important point in the theory of geometric flows, and your analysis of the argument in Chow and Knopf is spot-on. Let's break down the logic and try to close the conceptual gaps.

The Core Idea: What does "due only to" mean?

The statement "the non-ellipticity of the Ricci operator is due only to its invariance under diffeomorphisms" is a p...

Let

 Z = X × Y ,  m = μ ⊗ ν

and write points of Z as z = (x , y).
Put

 k(z , zʹ) = d μ (x , xʹ) · d ν (y , yʹ),   z = (x , y), zʹ = (xʹ , yʹ). (1)

The integral operator S that is considered in the question is exactly

 (Sf)(z) = ∫ k(z , zʹ) f(zʹ) dm(zʹ),    f ∈ L²(Z , m). (2)

Because k ∈ L²(m ⊗ m) (this is the given “Hilbert–Schmidt” assumption...

The Sum of Eigenvalues in Integral Operators

The question involves a Hilbert-Schmidt operator defined on by:

You want to prove that the sum of its non-zero eigenvalues equals:

Note: The kernel of the operator is k((x,y),(x',y')) = d_μ(x,x') * d_ν(y,y').

In general, for a Hilbert-Schmidt operator (which S is, as s...

Let

Ω := X × Y , ρ := μ ⊗ ν , 
k ((x , y) , (x' , y')) := dμ (x , x') dν (y , y') ,

and let

S : L²(Ω , ρ) → L²(Ω , ρ), (Sf)(x , y) = ∫Ω k((x , y) , (x' , y'))f(x' , y') dρ(x' , y') .

Denote the non–zero eigenvalues of S (repeated according to multiplicity) by (λi)i≥1. We want to prove

(1) Σi λi = ∫Ω k((x , y) , (x , y)) dρ(x , y)     = ∫X×Y dμ (x , x) dν (y , y) dμ(x)dν(y).

The formul...