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Let $(M,g)$ be an $n$-dimensional pseudo-Riemannian manifold and $\Box_g$ be the Laplace-Beltrami operator on $M$. Consider $z \in \mathbb{C}$ such that $\mathrm{ Im}(z)>0$, and we define $P_0 := \Box_g - z$. Now let us consider the product manifold $N := M \times \mathbb{R}$ equipped with the product metric $h := g + dy^2$ and its Laplace-Beltrami operator $\Box_h := \Box_g + \partial y^2$.

My goal is to find potential explicit relations between the kernel of the operator $P_0^{-1}$ and the kernel of the operator $(P_0 + \partial y^2)^{-1} = (\Box_h - z)^{-1} $ (specifically on the diagonal). To be clear, these are the (inverses of the) Laplace-Beltrami operators on $M$ and $M \times \mathbb{R}$, respectively.

I have tried two different methods:

First try : By using, amongst others, the relation between a kernel and its corresponding operator, i.e. $P_0^{-1}(x,x’) = (P_0^{-1}(\delta (\cdot - x’ ))(x)$ where $\delta$ is the Dirac delta function, I find that $$ P_0^{-1}(x,x’) = \int_{\mathbb{R}} (P_0+ \partial y^2)^{-1}(x,y,x’,y’) dy’, $$ with no dependance in $y$. I could detail the few lines of the proof but the problem is that the end result is not really satisfying, given that this integral seems a bit annoying.

Second try : We know that in dimension 1, for $\mathrm{ Im}(\lambda)>0$, the kernel of the operator $(-\lambda + \partial y^2)^{-1}$ is given by $$ (-\lambda + \partial y^2)^{-1}(y,y’) = \frac{1}{2(-\lambda)^{1/2}} e^{-(-\lambda)^{1/2}\vert y - y’ \vert}. $$

Now, by functional calculus (since $\mathrm{ Im}(z)>0$), the same formula hold if we replace $\lambda$ by $P_0 = \Box_g - z$ :

$$ (-P_0 + \partial y^2)^{-1}(y,y’) = \frac{1}{2(-P_0 )^{1/2}} e^{-(-P_0 )^{1/2}\vert y - y’ \vert}. $$ In particular, by taking $y = y’$, we get $$ (-P_0 + \partial y^2)^{-1}(y,y) = \frac{1}{2(-P_0 )^{1/2}} .$$

Thus $$ P_0^{-1}(x,x’) = -4(P_0 + \partial y^2)^{-2}(x,y,x’,y) .$$

As you can see it is not the right power on the right, we want a power $-1$. I have tried to express the right hand side with a contour integral of $(P_0 + \partial y^2 - \mu)^{-1}$ but it does not seem right.

Here’s my work. What do you think about these lines? Do you have any other ideas on how to link these two kernels? Any help would be nice.

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    $\begingroup$ The “first try” is perfectly fine. I fail to understand the transformations of the kernel in the “second try”: I think you can make it rigorous until the $(-P_0+\partial y^2)^{-1}(y,y)=\frac{1}{2(-P_0)^{1/2}}$ line, but the last step (“Thus…”) does not seem right. $\endgroup$ Commented Aug 27, 2024 at 9:26
  • $\begingroup$ @MateuszKwaśnicki Thank you very much. Yes I guess we cannot square the last line but I do not really see why. However, from there, would it exist a way to reach $P_0^{-1}(x,x’)$? $\endgroup$ Commented Aug 29, 2024 at 11:00
  • $\begingroup$ I do not think there is a simpler expression than the one that you get in the “first try”, but, of course, this is just intuition. $\endgroup$ Commented Aug 29, 2024 at 13:23
  • $\begingroup$ @MateuszKwaśnicki Alright I’ll try to work on that, thanks again :) $\endgroup$ Commented Sep 2, 2024 at 8:55

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