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Let $d \ge 2$ be an integer and let $X=\mathcal S_{d-1}$ the unit-sphere in $\mathbb R^d$. Let $\tau_d$ be the uniform distribution on $X$. Define a function $K:X \times X \to \mathbb R$ by $K(x,y) := e^{-c\|x-y\|^\beta}$, for some scale parameter $c>0$ and $\beta \ge 1$. I'm particularly interested in the case $\beta \in \{1,2\}$. The case $\beta = 1$ corresponds to the Laplace kernel while the case $\beta=2$ corresponds to the Gaussian kernel.

Now, consider the kernel integral operator $T_K:L^2(X,\tau_d) \to L^2(X,\tau_d)$ defined by

$$ T_Kf:X \to \mathbb R,\,T_K f(x) = \int_X K(x,x')f(x')d\tau_d(x'). $$

This is a compact positive operator has a countable spectrum / eigenvalues $\lambda_1 \ge \lambda_2 \ge \ldots $ with $\lambda_n \ge 0$ for all $n$.

Question. Is there an exact analytic formula for the $\lambda_n$'s (as a function of $c$, $\beta$, $d$, and $n$) ? Same question for the trace of $T_K$, namely the sum $\sum_{n=1}^\infty \lambda_n$ (as a function of $c$, $\beta$, and $d$).

Edit: The trace equals $1$

It turns out that $\mbox{Tr}(T_K) = \mathbb E_{x \sim \tau_d} [K(x,x)] = 1$ since $K(x,x) = e^{-c\|x-x\|^\beta} = e^0 = 1$ for all $\beta \ge 1$.

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  • $\begingroup$ Why is $T_K$ positive? By the way, the trace is 1, since $K(x,x)=1$. $\endgroup$
    – Giorgio Metafune
    Commented Apr 26, 2021 at 19:12
  • $\begingroup$ Indeed see that was a typo, see corrected computation of trace under the question. $T_K$ is positive in the sense that $\sum_{i=1}^n\sum_{j=1}^n a_i a_j K(x_i,x_j) \ge 0 $ for all $a_1,\ldots,a_n \in \mathbb R$ and all $x_1,\ldots,x_n \in X$. $\endgroup$
    – dohmatob
    Commented Apr 26, 2021 at 19:36
  • $\begingroup$ For $d=2$, $\beta=2$ one can write $e^{-c|x-y|^2}=e^{-2c(1-\cos (\theta-\phi))}$, $x=e^{i\theta}, y=e^{i\phi}$ and expand the kernel in Fourier series using $e^{-2c(1-\cos \theta)}=e^{-2c}|e^{-2c e^{i\theta}}|$. One gets the eigenvalues in the form (modulo some mistake that can occur) $$\lambda_k=(2c)^k\sum_{m \geq 0, m+k \geq 0} \frac{(2c)^{2m}}{m! (m+k)!}.$$ $\endgroup$
    – Giorgio Metafune
    Commented Apr 27, 2021 at 13:49

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