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.
De Rham cohomology of $U(2)$
de-rham-cohomologyhomology-cohomology
Related Solutions
We cut $T^{3}$ into two parts, each part is homotopic to a torus. One visualize this by considering $T^{3}=T^{2}\times \mathbb{S}^{1}$, and the two parts are $\mathbb{T}^{2}\times I$ respectively, with the $I$ coming out of considering $\mathbb{S}^{1}$ as gluing two intervals together. The intersection of the two parts is again homotopic to the torus. Nowe we have:
$$\rightarrow H^{2}(X)\rightarrow H^{2}(\mathbb{T}^{2}\times I)\oplus H^{2}(\mathbb{T}^{2}\times I)\rightarrow H^{2}(\mathbb{T}^{2}\times I)\rightarrow H^{3}(X)\rightarrow0$$
We know $H^{2}(\mathbb{T}^{2})=\mathbb{R}^{1}$ via induction. So we have
$$H^{0}(X)\rightarrow \mathbb{R}^{2}\rightarrow \mathbb{R}\rightarrow H^{1}(X)\rightarrow \mathbb{R}^{4}\rightarrow \mathbb{R}^{2}\rightarrow H^{2}(X)\rightarrow \mathbb{R}^{2}\rightarrow \mathbb{R}\rightarrow \mathbb{R}\rightarrow0$$
This gives $H^{2}(X)=\mathbb{R}^{3}$ because last map is an isomorphism and the map $H^{1}(\mathbb{T}^{2}\times I)\rightarrow H^{2}(X)$ has a one dimensional image. Consider a closed one-form $w$ on $\mathbb{T}^{2}$, if we use partition of unity to split it into two parts, then no matter which choice we use we would end up with the same class in $H^{2}(X)$ if one thinks geometrically. This together with the first part gives us $H^{1}(X)=\mathbb{R}^{3}$, $H^{2}(X)=\mathbb{R}^{3}$.
Its easy to visualize when looking at the fundamental domain of the torus:
$U$ is a neighbourhood of $x$, $V$ is a neighbourhood of $y$. Removing $x$ and $y$ from the torus is homotopy equivalent to removing the whole neighbourhoods $U$ and $V$. Further we can choose $U,V$ so large, that they fill the entire triangle they lie in.
So the fundamental domain is homotopy equivalent to the union of the boundary of the square with the diagonal. But some parts of the boundary are identified with each other. Doing the identifications, we obtain the wedge sum of three circles (their common point is $A$). One circle corresponds to the left side = right side of the bundary, one circle corresponds to the top side = bottom side and one circle corresponds to the diagonal.
More precisely, first identifying top with bottom, we get two distinct edges connecting the bottom/top left vertex with the bottom/top right vertex (corresponding to the bottom/top edge and the diagonal) and two circles (corresponding to the left and right edges).
Then we identify the left and right edges. The two circles we already had get identified, and as the bottom/top left edge gets identified with the bottom/top right edge, the edges which previously connected these two vetices become circles.
So in the end we get three circles which are connected in their common point $A$, the bottom/top now left/right vertex.
I hope the detailled description did not make things more confusing. I probably should've also drawn this.
Best Answer
Here are two facts that you can use to compute the cohomology of $U(2)$:
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 $\deg\beta = 1$.