Asymptotic Expansion for heat operator $e^{-t\triangle}$ I'm afraid the question below might turn out to be very stupid - 
I just don't know how to make sense of two asymptotic expansions,
given the heat operator $e^{-t\triangle}$ with $\triangle$ a Laplace type operator on smooth functions over an n-dimensional manifold we have an asymptotic expansion of the kernel
\begin{equation}
\text{k}(e^{-t\triangle}) \backsim \sum_{j \geq 0} c_{\frac{-n - j}{2}} \,t^{\frac{-n - j}{2}}
\end{equation}
and we also have an asymptotic expansion for the trace,
\begin{equation}
\text{Tr}(e^{-t\triangle}) \backsim \sum_{j \geq 0} c_{\frac{-n - j}{2}} \,t^{\frac{-n - j}{2}}
\end{equation}
Now, where is the difference in these two expansions ? 
I realize this question suggests I don't know enough about both the kernel and the definition of the trace - which is absolutely true. The problem though is I don't know how to find an answer since all sources I have go on a large detour on how to construct the heat kernel, and on general definitions and properties of trace structures on operator algebras. I am totally aware I need to eventually understand these concepts, what I hope for with this question is to get a first idea in order to proceed. If that's not possible I'd also appreciate a reference to literature that starts at a low level so that I can get an overview.
Hope this makes sense, thanks a lot for your help!
 A: You just have to sort this a little bit. Let's assume we're talking about an closed Riemannian manifold $(M,g)$ with its Laplace-Beltrami-Operator $\Delta_g$. Then you have the heat kernel as fundamental solution of the heat equation:
$$ \mathcal{K} \in C^{\infty}(M \times M \times \mathbb{R}^+)$$
Note that the notation $e^{-t\Delta}$ sometimes refers to both,
the kernel itself and the heat operator, which is
defined as integral operator acting on functions $f \in C^{\infty}(M)$
via integration against the heat kernel:
$$e^{-t\Delta}: C^{\infty}(M) \rightarrow C^{\infty}(M)$$
$$e^{-t\Delta}(f)(x) := \int_M \mathcal{K}(t,x,y)f(y) \mathrm{dy}$$
The heat operator can be expanded to an operator on $L^2(M)$. This allows to consider the heat trace, defined as the $L^2-$trace of
the heat operator. For the heat trace, you have the identity:
$$\text{Tr}_{L^2}(e^{-t\Delta}) = \int_M \mathcal{K}(t,x,x) \mathrm{dx}$$ 
which relates your objects to each other. Now you're interested in asymptotic expansions for $t \downarrow 0$ for the heat kernel and for the heat trace. 
$$ \mathcal{K}(x,x,t) \sim \sum\limits_{k=0}^{\infty} u(x,x) t^k $$ 
for some (very interesting) smooth maps $u(x,y)$ and the latter gives you the expansion for the trace of the heat operator by the above identity:
$$ \text{Tr}_{L^2}(e^{-t\Delta}) = \int_M \mathcal{K}(t,x,x) \mathrm{dx} \sim \int_M \sum\limits_{k=0}^{\infty} u(x,x) t^k $$
I hope this is still useful for you. A great book including this topic is The Laplacian on a Riemannian manifold by S.Rosenberg.
A: if you know the Hamiltonian of the system then
$ Tr(exp(-t \Delta)\sim \int dq \int dp exp(-t\sum_{ab}p^{2}_{a}p^{2}_{b}-tV(q))$
the integeral is taken over ALL the p's and the q's momenta and position of the particle.
if the operator is a Laplacian operator then 
$ Tr(exp(-t \Delta)\sim \int dq \int dp exp(-t\sum_{ab}p^{2}_{a}p^{2}_{b}g^{ab}(q))$
here the metric $ ds^{2}= \sum_{ab}g_{ab}dx_{a}^{2}dx_{b}^{2} $
