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.
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$\begingroup$ 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. $\endgroup$– ThiKuCommented Mar 9, 2019 at 11:22
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$\begingroup$ Yes， the harmonic forms should be $\Gamma$-invariant. $\endgroup$– user94640Commented Mar 9, 2019 at 11:37
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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.
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$\begingroup$ 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 $\tilde{X}$ ? $\endgroup$ Commented Mar 9, 2019 at 17:59
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$\begingroup$ A metric on X is the same as an invariant metric on the covering. $\endgroup$– ThiKuCommented Mar 9, 2019 at 21:09