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I'm trying to calculate the de Rham cohomology of $U(2)$, but I don't know how to do this. I'd like to avoid Mayer-Vietoris if possible. I'm doing this in preparation for an exam in my topology course next week, so any help would be appreciated.

Michael Albanese
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Sorey
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1 Answers1

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Here are two facts that you can use to compute the cohomology of $U(2)$:

  • $U(n)$ is diffeomorphic to $SU(n)\times S^1$; see this answer for example.
  • $SU(2)$ is diffeomorphic to $S^3$, see here for example.

Therefore $U(2)$ is diffeomorphic to $S^3\times S^1$. It follows from the Künneth Theorem that $H_{\text{dR}}^*(U(2)) \cong \mathbb{R}[\alpha, \beta]/(\alpha^2, \beta^2)$ where $\deg\alpha = 3$ and $\deg\beta = 1$.

Michael Albanese
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  • My knowledge when it comes to applying the Kunneth formula is mostly concerned with using the Poincare polynomials. So, does this mean that $P_{U(2)}(t) = P_{S^1}(t) \cdot P_{S^3}(t)$? So we'd have $H^k_{dR}(U(2)) = \mathbb{R}$ for $k\neq 2$ and $H^2_{dR}(U(2)) = 0$? – Sorey Apr 10 '19 at 22:03
  • Yes, $H^k_{\text{dR}}(U(2)) \cong \mathbb{R}$ for $k = 0, 1, 3, 4$ and zero otherwise. – Michael Albanese Apr 10 '19 at 22:23