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Let $\mu$ be a (1,1)-form on a compact connected Riemann surface $X$ of genus $g$. (Assume $\mu$ to be a real positive smooth $(1,1)$-form if necessary.)

The space $H^0(X,\Omega^1)$ of holomorphic differential forms is a $g$-dimensional complex vector space.

Does $\mu$ induce a metric on $ H^0(X,\Omega^1)$?

Does $\mu$ induce a metric on $\det H^0(X,\Omega^1)$?

(I know that a metric on $H^0(X,\Omega^1)$ induces a metric on $\Lambda^g H^0(X,\Omega^1) = \det H^0(X,\Omega^1)$.)

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1 Answer 1

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You need $\mu$ to be real and positive definie (i.e. $\mu$ needs to be a metric) to get an inner product on the cohomology group $H^0(X,\Omega^1)$. There is a standard construction for doing this:

We use the Hodge isomorphism theorem and harmonic forms. The metric $\mu$ on $T_X$ induces a dual metric on the holomorphic vector bundle $\Omega^1 = T_X^*$. Given two classes $[u]$ and $[v]$ in $H^0(X,\Omega^1)$, pick their $\mu$-harmonic representants $u$ and $v$, and define

$$ \langle [u], [v] \rangle = \int_X \langle u,v \rangle_\mu \, dV_\mu $$

where the inner product of $u$ and $v$ is the dual metric, and $dV_\mu$ is the volume form defined by $\mu$. The details of this construction can be found in Jean-Pierre Demailly's book on complex differential geometry, but this really is just a matter of linear algebra. An alternative source are Andrei Moroianu's notes on Kähler geometry.

In this specific case one could also proceed as follows: given $[u]$ and $[v]$ one can get an inner product by setting

$$ \langle [u], [v] \rangle = \int_X [u \wedge \overline v]. $$

One needs to check that the wedge product is well defined, but once that is done everything works because $[u \wedge \overline v]$ is a class of top degree on $X$, so we can integrate it over the curve. However, this inner product doesn't depend on the metric $\mu$.

The difference between the two constructions is that the first works for any bundle you can make out of the tangent bundle and its dual, while the second only works on cohomology groups of degree $n$ on a manifold of dimension $n$.

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Nice. Just one question. We don't need $\mu$ to be smooth right? –  shaye Sep 26 '11 at 16:09
    
Hmm... I'm not sure. I think we need some regularity conditions on $\mu$ for the integral and the inner product on $T_X^*$ to make sense, but one might be able to get away with $\mu$ being smooth (or $C^1$ even) on a dense open set. Maybe less. What kind of $\mu$ do you have in mind? –  Gunnar Magnusson Sep 27 '11 at 5:31
    
My Riemann surfaces could arise as quotients of the complex upper-half plane by some subgroup of SL_2(Z). These might have elliptic and parabolic elements. The hyperbolic metric is a metric outside of the elliptic fixed points and singular at the cusps. Could I use this metric above? –  shaye Sep 28 '11 at 6:15
    
Any metric will do, but of course the more you know about your metric the more you'll be able to prove about the induced metric on the cohomology group. Theoretically using the hyperbolic metric would be a good idea, even if it has singularities at isolated points, but I'll refrain from making any promises without knowing more about your situation. Btw, are you working on moduli of elliptic curves? They arise as quotients of the half-plane by $SL_2(\mathbb Z)$, and we have the Weil-Petersson metric on that space. –  Gunnar Magnusson Sep 29 '11 at 19:17
    
Thnx alot for this answer! Just to answer your question, I work on subgroups of finite index in SL_2(Z). They aren't always congruence subgroups though. –  shaye Sep 30 '11 at 14:16

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