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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!

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up vote 2 down vote accepted

The nullspace of the Laplacian on a compact manifold is just the space of constant functions, which can be identified with $\mathbb R$. A function is orthogonal to the constants if and only if it has average value zero. Thus the orthogonal projection onto the nullspace is the map that takes each function to its average value, which is exactly what the map $\frac{1}{A}$ does.

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