Left invariance on m-forms

Question

Given a Lie Group $G$ of dimention $n$, how can we show that left invariant $m$-forms form a vector space of dimension $\binom{n}{m}$?

I know that, for a vector field of dimension $n$, these $m$-forms form a vector space of dimension $\binom{n}{m}$ by a simple counting argument, but what difference does this left invariance impose?

Hint: Show that a left-invariant tensor is determined by its value at the identity. — Kajelad, May 20 '21 at 17:20

score 1 · Accepted Answer · answered May 20 '21 at 17:21

Left-invariant forms are characterized by its value at the identity. So the space of left-invariant $m$-forms is naturally isomorphic to $(\mathfrak{g}^*)^{\wedge m}$, whose dimension is ${n \choose m}$. Namely, $\omega \mapsto \omega_e$ is an isomorphism.

Left invariance on m-forms

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