Consider a closed Riemannian manifold $(M,g)$ of dimension 2, so we're talking about a surface. To its Laplacian $\Delta_g$ one can consider the trace of the heat kernel given by \begin{equation} \operatorname{Tr}(e^{-t\Delta_g}) = \sum\limits_{k=0}^\infty e^{-t \lambda_k} \end{equation} where the $\lambda_k$ are the eigenvalues of the Laplacian.
In the famous paper of Osgood, Phillips and Sarnak [OPS87]
it is said, that (here $\lambda_0$ = $0$) \begin{equation} \operatorname{Tr}(e^{-t\Delta_g}) - 1 = \operatorname{Tr}(e^{-t\Delta_g}) - e^{-t\lambda_0} = \operatorname{Tr}\left(e^{-t\Delta_g}-\frac{1}{A}\right) \end{equation} where $A$ denotes the area of the surface. In a book I found that $\frac{1}{A}$ means the map $f \mapsto \frac{1}{A}\int_M f \mathrm{dvol_g}$. How can one show that this map is related to the eigenvalue $\lambda_0 = 0$?
Sometimes it is also stated as $\operatorname{Tr}(e^{-t\Delta_g}-P)$ where $P$ is the projection to the nullspace. But that's exactly what I don't understand: Why is $\frac{1}{A}$ the projection to the nullspace?
Thanks for your help!