I am currently trying to read the T. Kotake's paper "An Analytical Proof of the Classical Riemann Roch Theorem", in which he defines a differential operator which acts on smooth sections of a tensored vector bundle. I have trouble understanding the notation, because I am not very familiar with tensor notation. I'll try to describe my problem and post questions as I go along:

We have a compact Riemann surface $\Sigma$ with metric locally given by $$h(z,\overline{z}) \,dz\otimes d\overline{z}.$$ Question 1: This metric confuses me a little - should'nt we have a metric depending on one complex variable only, if we are working with Riemann surfaces ?

Furthermore we are given a holomorphic vector bundle $\mathcal{E} \to \Sigma$ of rank $N$. It carries a hermitian metric which is determined by a system of locally defined functions $A \colon V \to \text{GL}(N,\mathbb{C})$ whose value at each point is a hermitian matrix (here $V$ denotes a coordinate chart for $\Sigma$.) These are then patched together via a partition of unity.

If we let $K = \Lambda^{1,0}\,T^*\Sigma$ denote the canonical bundle, and $\mathcal{E}^*$ the dual bundle of $\mathcal{E}$, we also have the vector bundle $K \otimes \mathcal{E}^*$. It has an induced metric, namely $$ h^{-1} \otimes(A^T)^{-1} \colon V \to \mathbb{C}^* \otimes \text{GL}(N,\mathbb{C}). $$

Question 2: I am very new to tensor bundles - is this the correct way of writing the metric structure?

Now the differential operator is defined on smooth sections of the bundle $K \otimes \mathcal{E}^*$, as it stands in the paper it acts locally as $$ \triangle = - A \frac{\partial}{\partial z} (hA)^{-1} \frac{\partial}{\partial \overline{z}} $$

Question 3: How do I underrstand the action of $\triangle$ on a section $\omega \otimes s$ ? The problem I have is I don't know what is supposed to act on $\omega$, and what acts on $s$. If I expand the expression, above, I get $$ \triangle = h^{-1}\frac{\partial^2}{\partial z\partial\overline{z}} - h^{-1}A\frac{\partial A^{-1}}{\partial z} \frac{\partial}{\partial \overline{z}} - \frac{\partial h^{-1}}{\partial z}\frac{\partial}{\partial \overline{z}} $$ and so my guess is in tensor notation I would write this as $$ \triangle = h^{-1} \otimes \frac{\partial^2}{\partial z\partial\overline{z}} - h^{-1} \otimes A\frac{\partial A^{-1}}{\partial z} \frac{\partial}{\partial \overline{z}} - \frac{\partial h^{-1}} {\partial z} \otimes \frac{\partial}{\partial \overline{z}} $$

I am really not sure here, somehow it looks wrong. Any comments on my attempts would be hugely helpful!

I hope these questions make sense, I shall try my best to clarify if not!

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    $\begingroup$ A Riemann surface may have complex dimension $1$, but it still has real dimension $2$ – there's no way a metric can be a differential $1$-form! $\endgroup$ – Zhen Lin Jul 16 '12 at 13:11
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    $\begingroup$ Question 1 is clear once you expand $z = x+iy$ and $\bar z = x - iy$. Then, the metric you wrote down becomes $h ( dx^2 + dy^2 - i dx \wedge dy)$, which is precisely $h$ times the standard Hermitian metric. The real part of this is a Riemannian metric. I think that's where your confusion is from. $\endgroup$ – Sam Lisi Jul 17 '12 at 19:28

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