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Consider a Riemannian manifold $(M,g_0)$ which is the interior of a compact manifold $(\overline{M}, \overline{g})$.

I'm interested in a kind of conformal variation of the background metric $g_0$. So for $\tau \in [0,T] \subset \mathbb{R}$ let $\omega_\tau \in C^{\infty}(\overline{M})$ and $g_{\tau} := e^{\omega(\tau)} g_0$

In addition there is a family of operators $\{H_\tau \}$ where $H_\tau: L^2(M, g_\tau) \rightarrow L^2(M, g_\tau)$. My goal is to talk about the $(L²-)$ trace of these operators with respect to the background $L^2$-space $L^2(M,g_0)$.

Is this possible? Are there suitable conditions for $\omega$ to get a statement like \begin{equation} L^2(M,g_0)= L^2(M, g_\tau) \ \forall \tau \in [0,T] \end{equation} so that the operators can be viewed as acting on the same space?

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up vote 1 down vote accepted

If you take $\omega_\tau \in C^{\infty}(\bar{M};\mathbb{R})$, then by compactness we have that there exist a function $\Omega:[0,T]\to\mathbb{R}_+$ such that $-\Omega(\tau) < \omega(\tau,x) < \Omega(\tau)$ for all $(\tau,x)\in [0,T]\times \bar{M}$.

This means that there exists a constant $k$ depending only on the dimension of the manifold $M$ such that in a fixed coordinate charge you have $$ \sqrt{|g_0|}e^{-k\Omega(\tau)} \leq \sqrt{|g_\tau|} \leq \sqrt{|g_0|}e^{k\Omega(\tau)} $$ or, in other words, the volume forms of the two Riemannian metrics are compatible. This immediately implies that the two $L^2$ spaces are compatible.

I suspect, however, you may also be interested in the case where $\omega_\tau\in C^\infty(M,\mathbb{R})$, which would be the case if you were to consider $M$ as an asymptotically hyperbolic manifold with conformal boundary $\partial\bar{M}$. In this case $\omega_\tau$ is allowed (in fact encouraged) to blow-up at the boundary. On any open subset with compact closure $\bar{G}\Subset M$ the above argument still works: $L^2(G,g_0) = L^2(G,g_\tau)$. However degeneracies at the boundary can make big differences. If $g_0$ is induced from $\bar{g}$, then it has finite total volume; so all the constant functions are in $L^2(M,g_0)$. But if $g_\tau$ has infinite volume (as in the asymptotically hyperbolic cases usually considered), then clearly the constant functions cannot be in $L^2(M,g_\tau)$.

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Hello Willie Wong, you're absolutely right, I'm definitely interested in the asymptotic hyperbolic case. In fact it's always $g_0 = \frac{dx^2+d\theta^2}{x^2}$ for $d\theta^2$ the round metric on $S^1$ and $x$ a boundary defining function. In the paper where the notation comes from there is $\omega_\tau \in C^{\infty}(\overline{M})$ so that one can follow your first argumentation, right? I'm a bit confused because you handle the asymptotic hyperbolic case explicitly different? – Robin Neumann May 24 '12 at 12:40
@Robin: right, if $\omega_\tau$ is bounded above and below, then you are fine. I mentioned the AH case separately because I wasn't sure if $g_0$ is a separate metric or if it is the induced metric from $\bar{g}$. (In the latter case $L^2(M,g_0) = L^2(\bar{M},\bar{g})$ by definition.) As you indicated your $g_0$ already degenerates near $x = 0$, and hence is not the induced metric; you cannot compare $L^2(M,g_\tau)$ with $L^2(\bar{M},\bar{g})$, but I suspect that is not a problem for you. – Willie Wong May 24 '12 at 13:51

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