# Hodge numbers of 2n-dimensional torus

1. Apr 21, 2004

### Paul Cook

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!

2. Apr 21, 2004

### lethe

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

$$b^{p,q}=\binom{n}{p}\binom{n}{q}$$

Last edited: Apr 23, 2004
3. Aug 10, 2004

### mathwonk

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.