# $L^{2}$ Betti number

Let $$\tilde{X}$$ be a non-compact oriented, Riemannian manifold adimits a smooth metric $$\tilde{g}$$ on which a discrete group $$\Gamma$$ of orientation-preserving isometrics acts freely so that the quotient $$X=\tilde{X}/\Gamma$$ is compact. We denote $$b_{2}^{k}$$ by the dimension of $$L^{2}$$ harmonic $$k$$-form w.r.t. metric $$\tilde{g}$$. If $$f$$ is a smooth function on $$\tilde{X}$$, we consider the $$L^{2}$$ harmonic $$k$$-form on $$(\tilde{X},e^{f}\tilde{g})$$. we denote by $$b_{2,f}^{k}$$ the dimension of $$L^{2}$$ harmonic $$k$$-form w.r.t. metric $$e^{f}\tilde{g}$$. Is $$b_{2}^{k}=b_{2,f}^{k}$$ correct for all k? One can see that its correct when $$f$$ is bounded, since the metrics are quasi-isometric. Thanks very much.

• Do you want your harmonic forms to be $\Gamma$-invariant? Otherwise the dimension of $L^2$-harmonic forms has nothing to do with the $L^2$-Betti numbers. – ThiKu Mar 9 at 11:22
• Yes， the harmonic forms should be $\Gamma$-invariant. – user94640 Mar 9 at 11:37

Dodziuk: de Rham-Hodge theory for L2-cohomology of infinite coverings proves that the class of the representation of $$\Gamma$$ on the space of $$L^2$$-harmonic forms is a homotopy invariant of $$X$$. In particular the $$\Gamma$$-dimension (in the sense of von Neumann) of the space of $$L^2$$-harmonic forms does not depend on the chosen $$\Gamma$$-invariant metric.
• Thanks for your answer. At last, I has a puzzle about the "$\Gamma$-invariant metric". Is this metric on the base manifold $X$ or on the lifted manifold $\tilde{X}$ ? – user94640 Mar 9 at 17:59