Proposition 50

Let G be a Lie Group (assumed to be finite-dimensional if K is of characteristic > 0). Let E be the vector space of continuous alternating multilinear forms of degree k on $\mathrm{T}_e(G)$. For all $u \in E$, let $\omega^u$ be the differential form of degree k on G such that $(\omega^u)_g$ is the multilinear form on $\mathrm{T}_g(G)$ derived from u by the translation $h \mapsto gh$ (resp. $h \mapsto hg$). Then $\omega^u$ is analytic and left (resp. right) invariant on G. The mapping $u \mapsto \omega^u$ is an isomorphism of E onto the vector space of left (resp. right) invariant differential forms of degree k on G.

This is a special case of what we have said above.

Let F be a complete normable space. Proposition 50 remains true if differential forms on G with values in K are replaced by differential forms on G with values in F. For every continuous linear mapping $u$ of $\mathrm{T}_e(G)$ into F, there exists a differential form $\omega^u$ of degree 1 on G, with values in F, such that $(\omega^u)_g = u \circ \mathrm{T}_g(\gamma(g)^{-1})$. In particular, take $F = \mathrm{T}_e(G)$ and $u = \mathrm{Id}_{\mathrm{T}_e(G)}$. We then obtain the differential form $\omega$ on $G$ such that $\omega_g = T_g(\gamma(g^{-1}))$; this differential form is left invariant and analytic; it is called the left canonical differential form of $G$. $\omega_g(t) = g^{-1}t$ for all $t \in T_g(G)$.

If $F$ is again an arbitrary complete normable space and $u \in \mathscr{L}(T_e(G), F)$, then $\omega^u = u \circ \omega$. In particular (taking $F = K$), the mapping $v \mapsto v \circ \omega$ is a linear bijection of the dual of $T_e(G)$ onto the vector space of differential forms of degree 1 with values in $K$ which are left invariant under $G$.

Similarly, the differential form $\omega'$ on $G$ such that $\omega'_g = T_g(\delta(g))$ is called the right canonical differential form of $G$. There are analogous properties to those of $\omega$, which we leave to the reader to state. The mapping $g \mapsto g^{-1}$ of $G$ onto $G$ transforms $\omega$ into $\omega'$.