How does a differential form looks on a matrix manifold?

Question

I want to know how does a differential form looks in a matrix manifold. For example, given that the special linear group $$SL(n,R)$$ of all matrices with determinant 1 is a manifold, how looks a 1-form on it? All closed forms are exact?How can I prove it?

Answer 1

There are many possible ways to compute $H^1_{\text{dR}}(\text{SL}(n, \mathbf{R}))$ but arguably one that uses the least machinery goes as follows:

There is a diffeomorphism $\text{SL}(n, \mathbf{R}) \to \mathbf{R}^{(n+2)(n-1)/2} \times \text{SO}(n)$ given by the $QR$-decomposition, factoring a matrix in the domain into an upper triangular matrix with positive eigenvalues and determinant $+1$, and an orthonormal matrix with determinant $+1$. Note that this is effectively performing a Gram-Schmidt process on the columns of $\text{SL}(n, \mathbf{R})$. $\text{SO}(n)$ is therefore called the "maximal compact subgroup" of $\text{SL}(n, \mathbf{R})$. As this in particular implies $\text{SL}(n, \mathbf{R})$ deformation retracts to $\text{SO}(n)$, $H^1(\text{SL}(n, \mathbf{R})) \cong H^1(\text{SO}(n))$.

If $G$ is a compact connected Lie group, then for any $k$-form $\omega$ on $G$ one can perform an averaging operation to get a left-invariant $k$-form as follows: Let $L_g$ denote the diffeomorphism $G \to G$, $h \mapsto gh$ given by left multiplication by $g$. Then define $$\eta(X_1, \cdots, X_k) = \displaystyle \int_G L_g^*\omega(X_1, \cdots, X_k) \; d\mu(g)$$ where $\mu$ is the Haar measure on $G$, which is a unique bi-invariant measure. By construction $\eta$ satisfies $L_g^*\eta = \eta$, i.e., it is left-invariant. We shall show that $\eta$ is cohomologous to $\omega$. First of all, for any $g \in G$, let $\{g_t\}_{t \in I}$ be a path from the identity of the group to $G$. Then we have a homotopy $L_{g_t} : G \times I \to G$ between the identity and $L_g$. Since pullback of a form by homotopic maps are cohomologous, $L_g^*\omega$ is cohomologous to $\omega$. As all the translates of $\omega$ are cohomologous to $\omega$, the integral over them must be cohomologous to $\omega$ as well.

Therefore any $1$-form $\omega$ on $\text{SO}(n)$ is cohomologous to a left-invariant $1$-form $\eta$ on $\text{SO}(n)$. If $\omega$ is a closed form, then so is $\eta$, but then $0 = d\eta(X, Y) = X\eta(Y) - Y\eta(X) - \eta([X, Y])$. If $X$ and $Y$ are both left-invariant, then $\eta(X)$ and $\eta(Y)$ are both constant, hence has zero directional derivative. Hence $\eta([X, Y]) = 0$ for all pairs of left-invariant vector fields $X, Y$. Let $\mathfrak{f} : \mathfrak{so}(n) \to \mathbf{R}$ be the linear functional corresponding to $\eta$ at the identity; the previous condition implies $\mathfrak{f}([v, w]) = 0$ for all $v, w\in \mathfrak{so}(n)$ - i.e., $\mathfrak{f}$ vanishes on the commutator ideal $[\mathfrak{so}(n),\mathfrak{so}(n)]$, giving rise to a functional on $\mathfrak{so}(n)/[\mathfrak{so}(n), \mathfrak{so}(n)]$. Conversely every such functional can be extended to a left-invariant $1$-form on $\text{SO}(n)$ by translating. Note as well that if $\omega$ is exact, so is $\eta$, in which case it's the zero form.

So (no pun intended) $H^1(\text{SO}(n))$ is isomorphic to the space of closed left-invariant $1$-forms on $\text{SO}(n)$ which is in turn $(\mathfrak{so}(n)^{ab})^*$. $\mathfrak{so}(n)$ is a perfect Lie algebra for all $n \geq 3$, therefore $H^1(\text{SO}(n)) = Z^1(\text{SO}(n)) = 0$ for all $n \geq 3$. For $n = 2$, we know by hand that $H^1(\text{SO}(2)) = \mathbf{R}$ as $\text{SO}(2) \cong S^1$.