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Question I'm given a Laplacian $\Delta_n=-4y^2 \cdot \frac{\partial^2}{\partial\bar{z} \partial z} + 4 iny \cdot \frac{\partial}{\partial\bar{z}}$, and I want it to be the Laplace operator associated to a Cauchy-Riemann operator $\bar\partial:\Omega^{0,0}(E) \rightarrow \Omega^{0,1}(E)$, which is a differential operator of the form $\bar\partial=(\partial_{\bar{z}} + \alpha(z))d\bar{z}$. How to find a suitable Cauchy-Riemann operator?

Here $E$ is a smooth vector bundle over a compact Riemann surface $X$, and $\alpha(z)$ is a smooth function.

Since the problem is dependent on the metric we have on $X$, we have to specify it. Let $X$ be the quotient of the upper half plane $\mathbb{H}$ by a cocompact Fuchsian subgroup of the first kind, and the metric is the quotient of the hyperbolic metric on $\mathbb{H}$ by the action of the group.

How to associate a Generalized Laplacian to a Cauchy-Riemann operator

The first step is to define an $L^2$-scalar product in $\Omega^{0,0}(E)$ and $\Omega^{0,1}(E)$. The hyperbolic metric on $X$ corresponds to an Hermitian metric $h$ on the holomorphic tangent space of $X$ , i.e. the subbundle $T^{1,0}(X)$ of the complex tangent bundle of $X$. This metric induces a metric on the dual of $T^{1,0}(X)$ , i.e. differential forms of type $(1,0)$, and, by complex conjugation, on forms of type $(0,1)$. By taking the exterior powers of this metric, and by tensoring with the Hermitian metric on $E$ we get a pointwise scalar Hermitian product $(s(x),t(x))$ for two sections of $\Omega^{0,i}(E)=\Omega^{0,i}(X) \otimes_{C^\infty(X)} \Gamma(X,E)$. On the other hand, let $\omega_0$ be the normalized Kähler form:

$\omega_0=\frac{i}{2\pi}h(\frac{\partial}{\partial z}, \frac{\partial}{\partial z}) dz d\bar{z}$

Then the $L^2$ scalar product of two sections $s,t \in \Omega^{0,i}(E)$ is defined by: $(s,t)_{L^2}=\int_X(s(x),t(x))\omega_0$.

The second step consists in defining the operator $\bar{\partial}^*$ in such a way that it is the $L^2$-adjoint of $\bar{\partial}$. So $\bar\partial^*:\Omega^{0,1}(E) \rightarrow \Omega^{0,0}(E)$ and $(s,\bar\partial^* t)_{L^2}=(\bar\partial s, t)_{L^2}$. Then the Laplace operator is defined by: $\Delta_n=\bar\partial^*\bar\partial:\Omega^{0,0}(E) \rightarrow \Omega^{0,0}(E)$.

  • This blockquote is a, more or less faithfully, quote from "Lectures on Arakelov Geometry" of Soulé, Abramovich, Burnol, Kramer. Chapter V, §2.2.*

My Approach My idea to find the right Cauchy-Riemann operator is to compute in full generality it's $L^2$-adjoint and find a general expression for the induced Laplacian.

My Computations As Willie kindly pointed out I was working on the wrong definition of Laplacian, so all my previous computations were useless. I will post some progress here as soon as I will have them.

Thank you very much!

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You may have copied something down wrong. If $\bar\partial:\Omega^{0,0}\to\Omega^{0,1}$ and $\bar\partial*:\Omega^{0,1}\to\Omega^{0,0}$, the composition $\triangle$ should be $\Omega^{0,0}\to\Omega^{0,0}$. –  Willie Wong May 25 '12 at 7:59
    
Note that the $L^2$ adjoint of $\bar{\partial}$, since you are working with a Hermitian metric, would be roughly speaking "the complex conjugate of the Hodge dual". Pictorially if $\bar\partial:\Omega^{0,0}\to\Omega^{0,1}$, and we know that $*:\Omega^{0,1}\to\Omega^{0,1}$, what we want is roughly to hit it with $\partial:\Omega^{0,1}\to\Omega^{1,1}$ and take the Hodge dual to transform $\Omega^{1,1}\to\Omega^{0,0}$. There are some more details and computations involved, but this should be the very rough picture you keep in mind. –  Willie Wong May 25 '12 at 8:03
    
@Willie, Thank you again, I corrected the typo as in your first comment. At the moment I still don't see the picture you described, but I'm starting now to work on it! Do you have a reference where this relation between $L^2$ adjoint and Hodge dual is more carefully described? That would be very helpful! –  Giovanni De Gaetano May 25 '12 at 8:15
    
For real manifolds, it should be described in most textbooks that cover Hodge theory. Look at the definition for the codifferential on Wikipedia. The complex/Hermitian case is formally the same, except where "symmetric" is used you have to replace by "Hermitian" and hence an extra complex conjugate. The first place I would check would be Morita's book, but it is possibly not in there (my copy isn't handy at the moment). –  Willie Wong May 25 '12 at 8:20
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Dear Giovanni, we only in extreme circumstances delete questions which have answers (and even less when tese have been upvoted) and/or significant comments. –  Mariano Suárez-Alvarez May 29 '12 at 19:55

2 Answers 2

up vote 2 down vote accepted

Some obvious problems:

$*dy = - dx$ so $*(dz) = *(dx + i dy) = dy - i dx = -i dz$. This is actually what you expect, since $dz$ is null and is supposed to be either self-dual or anti-self-dual. (Also, on a general complex manifold one has the the Hodge star operator take $\Omega^{p,q}$ to $\Omega^{n-q,n-p}$ where $n$ is the complex dimension. You really should not have $dz$ transformed into $d\bar{z}$.

Secondly, when you apply $df$, you should get $\left( \partial_{\bar{z}}f(z,\bar{z}) + \alpha(z)f(z,\bar{z})\right) \mathrm{d}\bar{z}$ if I understand your notation right.

Thirdly, if you write $f = f(z)$, it is holomorphic, and thus $\partial_{\bar{z}} f = 0$. You should really be carrying around both $z$ and $\bar{z}$ coordinates.

Lastly, you should not be trying to compute the Hodge Laplacian, at least not from what you've started with. What you wrote as $d$ is probably more properly written as $\bar{\partial}^\alpha$ (or something like that). It is not the exterior differential, which is $d = \partial + \bar{\partial}$. By definition $*\bar{\partial}f = i \bar{\partial}f$, so $\bar{\partial}*\bar{\partial}f = 0$ using that $\bar{\partial}^2 = 0$.

I think you may want to double check your definitions to make sure you are asking the question you want to ask. Or perhaps you can ask a new question giving full details of your motivation (with preferably linked references) so we can address the above issues. In particular, it would be nice if you specified what it means to be "the Laplacian associated to a Cauchy-Riemann operator".

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thank you for the useful comments. I'm going to check my definitions and to correct the question consequently! In the meantime thank you for your time! –  Giovanni De Gaetano May 24 '12 at 16:25
    
I corrected the "minor details" you pointed out at the beginning. But then I discovered that, as you suggested, I was working with the wrong Laplacian!!! So I have to start again from the scratch! In any case I posted the construction of "Generalized Laplacian" I'm using and I changed the notation for the Cauchy-Riemann operator from $d$ to $\bar\partial$. Thank you very much again! –  Giovanni De Gaetano May 24 '12 at 17:20

A suitable Cauchy-Riemann operator is: $\bar \partial= 2 \partial_{\bar z}$. Indeed if the metric on $X$ is given by $h(s(z)dz, t(z)dz)= s(z) \overline{t(z)}\rho (z)$ the its $L^2$ dual is $\bar \partial ^*=-\rho(z)^{1-n} 2\partial_z \rho(z)^n$.

Using the hyperbolic metric, i.e. $\rho(z)= y^2$, and taking the product $\bar \partial ^*\cdot \bar \partial $ we have the Laplacian as in the question.

This answer has been posted in such a way to have an explicit formula associated to the question, because the accepted answer is very useful but not a complete answer.

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