Hodge numbers of 2n-dimensional torus

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The Hodge numbers for a 2n-dimensional torus can be determined using the Hodge-de Rham cohomology. Specifically, the Hodge numbers are given by the formula bp,q = C(n, p) * C(n, q), where C(n, k) denotes the binomial coefficient. The triviality of the tangent bundle of a complex torus simplifies the computation, allowing for the analysis of the tangent space at the origin. This leads to the conclusion that the dimension of the Hodge (p,q) bundle is directly related to the wedge products of the vector space Cn.

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Paul Cook
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Hi,

A small but exceptionally annoying algebraic topology question:

I'm trying to find the Hodge numbers (from the Hodge-de Rham cohomology) for a 2n-dimensional torus (that is, n complex dimensions).

Anyone have any ideas? It's a rather technical question, but I don't really want to deduce it from first principles.

Thanks!
 
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well, off the top of my head, i would say, isn't H^{p,q} spanned by dz_{i_1}\wedge\dotsb\wedge dz_{i_p}\wedge d\overline{z}_{j_1}\wedge\dotsb\wedge d\overline{z}_{j_q} on the torus?

then we would have

b^{p,q}=\binom{n}{p}\binom{n}{q}
 
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I believe lethe is right. the thing that makes this case so easy is that the tangent bundle of a complex torus is trivial, so you just need to look only at the tangent space at the origin.

I.e. the wedge product of the trivial bundle is the trivial bundle whose fibres are the wedge product of the given vector space. Thus lethe has just computed the pth wedge of C^n times the qth wedge of C^n, (or their duals).

The dimension of this last space is the number of sections of the (dual) hodge (p,q) bundle, because that bundle is trivial, and that is the definition of the hodge number h^(p,q).

So this is actually computed from first principles.
 

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