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The Ray-Singer paper "R-torsion and the Laplacian on Riemannian manifolds" claimed that one may prove the metric invariance of analytical torsion by forming a homotopy between metric $\rho_{0},\rho_{1}$ using $\rho_{\mu}=\mu\rho_{0}+(1-\mu)\rho_{1}$. Then one differentiate $$ \frac{\partial }{\partial u}f(u,s)=\frac{1}{2}\sum^{N}_{q=0}(-1)^{q}q\int^{\infty}_{0}t^{s-1}Tr(e^{t\Delta_{q}(u)}\dot{\Delta_{q}})dt $$ here the $\frac{1}{\Gamma(s)}$ is omitted as it is entire, and if we know the differential of original function with $u$ is $0$, then with respect to $s$ should be $0$ as well. Now the differential of the fundamental solution of the initial value problem ("heat kernel") is given by $$ \frac{d}{d\mu}Tr e^{t\Delta_{q}(\mu)}=-tTr((\delta \alpha d-d\alpha\delta-\alpha \delta d+\alpha d\delta))e^{t\Delta_{q}(\mu)} $$ hence in the first equation we actually have $$ \dot{\Delta}_{q}=\alpha \delta d-\delta \alpha d+d\alpha \delta-d\delta\alpha,\alpha=*^{-1}\dot{*}=*^{-1}(\frac{\partial}{\partial \mu}*) $$ The proof then proceed by substituting the later formula into the above identity, using the fact that the trace of trace-class operator and bounded operator commutes to reduce the sum to a telescoping sum, which then reduces to the following: $$ \frac{1}{2}\sum^{N}_{q=0}(-1)^{q}\int^{\infty}_{0}t^{s}\frac{d}{dt}Tr(e^{t\Delta_{q}(\mu)}\alpha)dt $$ using integration by parts, we gain an additional $s$ on the front, and the desired result $\frac{\partial}{\partial \mu}f(\mu,s)=0$ now follows by regularity of the heat kernel for $\operatorname{Re}(s)$ large enough.

I noticed a seemingly discrepancy in the above proof,namely the setting of the theorem is when $W$ is a closed manifold, and the formula (Proposition 6.1 of the paper) is derived in the context of a manifold with two boundary components $M_1,M_2$, on which the Laplacian satisfies certain boundary conditions. In the case of the closed manifold, the formula does not easily reduce from the manifold with boundary case (see the remark in page 195), and the same strategy seems no longer works (possibly) due to the failure of using integration by parts in the second to last step. It seems the proof works mainly because in the detailed computation in page 196 we actually do not have any integral over the boundary components, hence the same strategy would work trivially for a closed manifold.

My question is: Is there a simpler proof for this in the case of a closed manifold? I am sure Ray and Singer must have tried and thought about it, but somehow still decided to use the strategy presented in paper to finish the proof. After all the proof presented in page 195-199 is really involved. So if there is a "clear cut" solution, it most likely would not be easy. I checked the literature but I am not sure if someone else has done any work related to it. It really isn't clear to me how to analyze the regularity of the differential of the Laplacian $\frac{\partial}{\partial \mu}\Delta_{q}(\mu)$ when $\operatorname{Re}{s}\rightarrow \infty$. I am not sure if the strategy in page 154 still works in this context.

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Ray and Singer state their independence result in Thm 2.1 for closed manifolds and give a full proof. This uses Proposition 6.1, which is formulated for manifolds with boundary, and the result slightly differs from the one used in chapter 2. However, the only difference is that $\alpha d\delta$ occurs instead of $d\delta\alpha$. Their remark says that one can rotate inside the trace to get back to the formula of chapter 2 only in the case that $W$ is closed.

There is a different formula for the analytic torsion in Bismut-Lott, Thm 3.29: $$-\int_0^\infty\left(\operatorname{tr}\left((1-2t\Delta)e^{-t\Delta}\right)-\frac{\chi'(M;F)}2-\left(\frac{\dim M\,\operatorname{rk} F\,\chi(M)}4-\frac{\chi'(M;F)}2\right)\right)\frac{dt}t\;.$$ In fact, this is the degree $0$ component of the analytic torsion form (Def 3.22). Here, $\Delta=(d+\delta)^2$, $\chi(M)$ is the Euler number, and $\chi'(M;F)=\sum (-1)^qq\,b_q(M;F)$ is a "secondary" Euler number for $M$ and the flat vector bundle $F$.

The main difference to Ray-Singer's construction is that the integral above converges without further regularisation. Moreover, the formula for the exterior differential of the torsion form implies a variational formula, see Thm 3.24. It implies that the Ray-Singer torsion is independent of the metric as long as $F$ is acyclic and $\dim M$ is odd.

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  • $\begingroup$ Hi! The proof of Theorem 2.1 hinges upon their proof of Proposition 6.1. See page 152 of the paper. I am not familiar with the Bismut-Lott paper, but $\dim M=2k+1$ is not needed as analytic torsion vanishes in all even dimensions. Thanks for the pointer nevertheless. $\endgroup$ Commented Feb 4, 2017 at 23:25
  • $\begingroup$ I see. I modified the answer accordingly. $\endgroup$ Commented Feb 5, 2017 at 13:50
  • $\begingroup$ Hi! I agree the $\alpha d\ delta$ and $d\delta \alpha$ difference does not really matter in the proof consider the trace. I need sometime to check the papers. One thing I am confused is the formula you have written seems to contain analytic torsion in its degree 0 component, not the other way around. Thanks a lot for the feedback! $\endgroup$ Commented Feb 6, 2017 at 4:34

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