Consider a closed Riemannian manifold $(M,g)$ of dimension n and let $K(t,x,y)$ be its heat kernel. Then it is known that the heat kernel has an asymptotic expansion as $t\downarrow 0$:
$$K(t,x,x)\sim (4\pi t)^{-n/2}\sum\limits_{k=0}^{\infty}u_k(x,x)t^k.$$
Often it is said that the functions $u_k(x.x)$ can be written as $O(n)-$invariant polynomials in the components of $R_x, (\nabla R)_x, (\nabla^2 R)_x,..., (\nabla^{2k-4} R)_x$ where $R$ denotes the Riemannian curvature tensor. The components of these polynomials do only depend on the dimension of $M$.
I would really like to understand the above description for the functions $u_k(x,x)$ and need some help with it.
My first naive interpretation would be the following: In order to evaluate $u_k(x,x)$, first choose an orthonormal basis $e_1,...,e_n$ of $T_xM$ and consider the 'components' of all the tensors above with respect to this orthonormal basis. The components would be:
\begin{align} R_x &: \lbrace g(R(e_i,e_j)e_k, e_m) \vert i,j,k,m=1,...,n\rbrace \\ \nabla R_x&: \lbrace g(\nabla R(e_i,e_j,e_k,e_p), e_m)\vert i,j,k,m,p=1,...,n \rbrace ...\text{etc. up to $(\nabla^{4k-4}R)_x$} \end{align}
There exist a polynomial $P_k(x)$ (only depending on $k$), $x\in\mathbb{R}^N$ with $N=N(k)$ sufficiently large and
$$p(x)=\sum_{\alpha\in \mathbb{N_0}^N} a_{\alpha}x^{\alpha} $$
such that:
1) the coefficients $a_{\alpha}\in\mathbb{R}$ do only depend on the dimension $n$ (but not the particular manifold $M$)
2) If we replace formally the variables $x_1,...,x_N$ by all of the components above, we get the value $u_k(x,x)$
3) The value of the polynomial does not depend on the choice of orthonormal basis in $T_xM$. A different choice of orthonormal basis would obviously give rise to different components, but the values of the polynomial will not be effected by such a change of basis. (This would correspond to the $O(n)$-invariance of the polynomial)
I am wondering if my interpretation is correct?
Even if my suggestions are right, I am sure my interpretation is not the only possible or best one. So I would be glad if you could share your knowledge and can tell me different or better ways to handle this situation.
Best wishes
