Do Carmo Riemanniam Geometry Page 46 quesion 7

Question

In Do Carmo Riemannian Geometry, page 46 question 7, we show that if $G$ is a compact connected Lie group, it has a bi-invariant Riemannian metric. in the first part we are required to show that if $w$ is a left invariant differential n-form, it is also right invariant. I've shown that for any $a\in G$ we have $\textbf{R}^*_aw$ is left invariant. Do Carmo then says

$\cdots$ it follows that $\textbf{R}^*_aw=f(a)w$.

Is that an obvious relation which I'm missing, or is it something that requires proof?

Answer 1

There is, up to a constant factor, only one left-invariant $n$-form: If $\omega_1 (e) = c \omega_2 (e)$ for some $c$, where $e\in G$ is the identity, then for any $g\in G$,

$$ w_1(g) = (L_g^*)^{-1} \omega _1 (e) =c (L_g^*)^{-1} \omega _2 (e) = c\omega_2 (g).$$

Now $R_a^* \omega$ and $\omega$ are both left-invariant, then they are of by a constant $f(a)$.

Answer 2

John's answer is excellent, but it lacks one detail that may be non-trivial for anyone who is new to differential geometry:

Every $n$-form on an $n$-dimensional manifold $M$ is a section of a line bundle. This means that if $\omega$ and $\omega'$ are two non-vanishing $n$-forms on $M$, then they differ by a function; there exists $g:M\to\mathbb{R}$, such that $\omega'=g\omega$. (Note that this is true for any manifold, not necessarily a Lie group). Now, if $M$ is a Lie group and the differential forms in question are left-invariant, the function $g$ is in fact constant, as already explained by John.