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The de-Rham cohomology ring of U(n) is the exterior algebra generated by the odd-dimensional classes x_1, x_3, ..., x_(2n-1). Moreover, on a Lie group every cohomology class is represented by a unique invariant form (both left and right). I ask two questions: 1) if we represent U(n) with matrices U = [z_ij], what is an explicit expression of a generator as an invariant form of U(n), in terms of the differentials dz_ij, for each odd degree between 1 and 2n-1? 2) in the 1-dimensional space of the invariant forms of a fixed degree (multiples of x_i), which are the two (opposite to each other) which represent the real image of a generator of the integral cohomology?

Fabio
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    Your description of the cohomology is not quite right, the cohomology is the exterior algebra on one generator in each odd degree between $1$ and $2n-1$. – Torsten Ekedahl Nov 11 '10 at 15:43
  • I'm sorry, of course the cohomology is the exterior algebra, I was thinking to the generators and I got confused. – Fabio Nov 11 '10 at 15:58

1 Answers1

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Here's an answer to the first question.

Let $\theta = U^{-1} dU$ be the left-invariant Maurer-Cartan one-form: it is a matrix of one-forms. Then $$\omega_{2n+1} = \mathrm{Tr} \theta^{2n+1}$$ is the desired bi-invariant form. Here $\theta^{2n+1}$ stands both for the wedge and matrix product of $\theta$ with itself $2n+1$ times.

I'm not sure I understand the second question: what "two" representatives are you talking about? I think the canonical representative will be certain multiple of $\omega_{2n+1}$. I'll try to fish out a reference later on.

José Figueroa-O'Farrill
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  • Thanks a lot! For the second question, in fact I have not been very clear, the two forms I was thinking to are the canonical generator and its opposite, which correspond to the two generators of Z, 1 and -1. – Fabio Nov 11 '10 at 16:02
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    Of course there is a striking similarity to the formula for Chern classes in terms of curvature here, which is no coincidence, since the latter arise from the above classes via transgression along the universal bundle. – Tobias Hartnick Nov 12 '10 at 01:02
  • It's similar to Chern classes, in fact I suppose that the integral form is (2\pi i)^(-n) * \omega_(2n+1), but I cannot prove it. For n=1 it seems to work, since, if we embed U(1) in U(n) and integrate \omega_1, if my computation is not wrong the result is 2 \pi i. In the article of Borel "Sur la cohomologie de Espaces fibres...", p.145-146, it is shown how to find an integral representative for each degree, but not invariant. For example, for degree 2n-1, if we fix a point on S^(2n-1), inducing a map f: U(n) -> S^(2n-1), the pull-back of the volume form via f is integral, but not invariant. – Fabio Nov 12 '10 at 12:49