Which sections of $T^*M\odot T^*M$ have reproducing kernel "primitives"?

Question

Given a smooth reproducing kernel $\kappa:M\times M\rightarrow \mathbb{R}$ on a manifold $M$, we can construct a section, $\alpha_{\kappa}$, of the symmetric tensor product $T^*M\odot T^*M$ by taking the exterior derivative of $\kappa$ in each of its arguments and then restricting to the diagonal. More precisely, we first construct a section, $\alpha_{M\times M}$, of $T^*M\boxtimes T^*M$ using the formula \begin{align} \alpha_{M\times M}(v_1,v_2)=\frac{d}{d\epsilon_1}\bigg|_0\frac{d}{d\epsilon_2}\bigg|_0\kappa(c_1(\epsilon_1),c_2(\epsilon_2)), \end{align} where $v_j\in T_{m_j}M$ and $c_j$ is a smooth curve that passes through $m_j$ at $\epsilon_j=0$ with velocity $v_j$. $\alpha_{\kappa}$ is then given by \begin{align} \alpha_{\kappa}(v_m,w_m)=\alpha_{M\times M}(v_m,w_m), \end{align} where $v_m,w_m\in T_mM$. When a section of $T^*M\odot T^*M$ arises in this manner, I'd like to say that the section has a reproducing kernel primitive.

How can we chracterize the space of sections of $T^*M\odot T^*M$ with reproducing kernel primitives? That is, given an $\alpha\in \Gamma(T^*M\odot T^*M)$, how can we determine if there is some kernel $\kappa$ such that $\alpha=\alpha_{\kappa}$?

One straightforward consequence of the positive-semidefinite property of $\kappa$ is that $\alpha_{\kappa}|_m\in T_m^*M\odot T_m^*M$ must be a positive semi-definite bilinear form for each $m\in M$. Are there any other special properties $\alpha_{\kappa}$ must have? I want to somehow use the fact that $\mathbf{d}^2=0$, but I'm having trouble seeing where I can usefully apply it.

update 4/28: I think that when $M=\mathbb{R}$ every positive semi-definite $\alpha$ has a reproducing kernel primitive. The argument is based on the fact that each $\alpha$ must be of the form $\alpha=a\, dx\,dx$ for some non-negative smooth scalar function $a:M\rightarrow\mathbb{R}$. We can write $\sqrt{a}\,dx=d\phi$, where $\phi(x)=\int_0^x\sqrt{a}(s)\,ds$. A reproducing kernel primitive for $\alpha$ is therefore $\kappa(x,y)=\phi(x)\phi(y)$. Things are easy on $\mathbb{R}$ because, there, all $1$-forms are exact.

Answer 1

This is not a complete answer, but maybe it will be helpful.

When $H^1_{dR}(M)=0$, I think I can determine when an $\alpha$ with rank $1$ has a reproducing kernel primitive. By ``rank $1$" I mean the following. For each $x\in M$, let $T_{ox}M\subset T_xM$ be the linear subspace consisting of those $v_x\in T_xM$ that annihilate $\alpha_x$, i.e. $\forall w_x\in T_xM,~\alpha_x(v_x,w_x)=0$. $\alpha$ has rank $1$ when $\dim(T_xM)-\dim(T_{ox}M)=1$ for each $x\in M$.

When $\alpha$ has rank $1$, it must be of the form $\alpha=\omega^2(=\omega\odot \omega)$ for some $1$-form $\omega$ with the property $\omega_x\in (T_{ox}M)^{\perp}\subset T^*_{x}M$ for each $x\in M$. Moreover this $\omega$ is determined uniquely up to multiplication by $-1$.

In order to determine if $\alpha$ has a reproducing kernel primitive, first I'll use the fact that $\alpha_\kappa$ must have rank $1$ to constrain $\kappa$. If $\{\phi_j\}$ is an o.n. basis for the RKHS associated with $\kappa$, $\kappa$ can be written $\kappa(x,y)=\sum_j\phi_j(x)\phi_j(y)$. In order for $\alpha_\kappa$ to be rank $1$, each of the differentials $d\phi_j$ must be colinear.

Fix an $x\in M$ and suppose $\alpha=\alpha_\kappa$. For some $j_o$, we must have $d\phi_{j_o}(x)\neq 0$ because $\alpha$ is rank $1$. Therefore there is an open neighborhood of $x$ with $d\phi_{j_o}\neq 0$. Restrict attention to this neighborhood. By the collinearity of the $d\phi_j$ and the non-vanishing of the differential $d\phi_{j_o}$, each $\phi_j=a_j(\phi_{j_o})$ for some function $a_j:\mathbb{R}\rightarrow\mathbb{R}$. Thus, $$\alpha_{\kappa}=f(\phi_{j_o})\,d\phi_{j_o}^2$$ where $f=(\sum_j(\frac{da_j}{d\phi_{j_o}})^2)$ is a non-negative smooth function of $\phi_{j_o}$. By the argument I gave earlier when $M=\mathbb{R}$, it must therefore be the case that $$\alpha_\kappa=dg^2=\omega^2,$$ for some smooth function $g:M\rightarrow\mathbb{R}$. In other words, $\omega$ must be exact in order for a rank $1$ $\alpha$ to have a reproducing kernel primitive. Conversely, if a rank $1$ $\alpha$ has $\omega=dg$, then a reproducing kernel primitive is given by $\kappa(x,y)=g(x)g(y)$.