Jtdylktuy

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?

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 $$eta(X_1, cdots, X_k) = displaystyle int_G L_g^*omega(X_1, cdots, X_k) ; dmu(g)$$ 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 $0 = deta(X, Y) = Xeta(Y) - Yeta(X) - eta([X, Y])$. 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 $v, win mathfrak{so}(n)$ - 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 $text{SO}(2) cong S^1$.