I'm currently working through an independent study on Lie groups, and I am a bit unsure on this problem:
Let $G$ be a Lie group and $X$ a vector field on $G$, viewed as a derivation of $C^\infty(G)$. Show that $X$ is left invariant if and only if $$(id\otimes X)\circ m_G^* = m_G^* \circ X,$$ where $m_G^* f(g,h) = f(gh)$ and $(id\otimes X)F(g,h) = (XF_g)(h)$ for $f\in C^\infty(G),\ G\in C^\infty(G\times G)$, and $F_g(h) = F(g,h)$.
I am mostly stuck on how to deal with the main equation. I figured that I would have these things act on the same object and see what needs to be true from there, but the notation is tripping me up a bit. I constructed the following diagram to gain some insight: $\require{AMScd}$ \begin{CD} \Omega^0(G) @>m_G^*>> \Omega^0(G\times G)\\ @V X V V @VV id\otimes X V\\ \Omega^0(G) @>>m_G^*> \Omega^0(G\times G). \end{CD}
I was able to work out that for any smooth function $f\in \Omega^0(G)$, $(m_G^*\circ X)(f)(g,h) = Xf(gh)$, but I'm struggling with how these operations work for $((id\otimes X)\circ m_G^*)(f)(g,h)$ and how I can use the fact that $X$ is left-invariant to arrive at the same formula. Any help is greatly appreciated. Thank you!
While it may not be the most insightful approach identities like these are often best manipulated by unraveling the definitions, inserting arbitrary arguments, and using parentheses liberally, or otherwise keeping careful track of which maps act in which order. We eventually want to reach the expression $$ X=dL_gX $$ or any equivalent insertion of dummy arguments. We can start with the expression $$ (id\otimes X)\circ m_G^*=m_G^*\circ X $$ Which holds iff the two derivations act equivalently on $C^\infty G$. $$ \iff \left[(id\otimes X)\circ m_G^*\right](f)=\left(m_G^*\circ X\right)(f)\ \ \ \ \ \forall f\in C^\infty M $$ $$ \iff (id\otimes X)(m_G^* f)=m_G^*(X f)\ \ \ \ \ \forall f\in C^\infty M $$ Likewise, these must be equal pointwise on $G\times G$. $$ \iff \left[(id\otimes X)(m_G^* f)\right](g,h)=\left[m_G^*(X f)\right](g,h)\ \ \ \ \ \forall f\in C^\infty M,\ g,h\in G $$ Applying the difinitions of $m_G^*$ and $id\otimes X$, $$ \iff \left[X(m_G^* f)_g\right](h)=(X f)(gh)\ \ \ \ \ \forall f\in C^\infty M,\ g,h\in G $$ Finally, we note the partial evaluation is just a left translation $(m_G^*f)_g(h)=f(gh)=(f\circ L_g)(h)$ $$ \iff \left[X(f\circ L_g)\right](h)=(X f)(gh)\ \ \ \ \ \forall f\in C^\infty M,\ g,h\in G $$ Using the definition of the differential of $L_g$, $$ \iff \left[(dL_gX)f\right](gh)=(X f)(gh)\ \ \ \ \ \forall f\in C^\infty M,\ g,h\in G $$ And we obtain something clearly equivalent to the standard statement of left invariance. This process also gives us a hint as to what's going on with the initial expression; in particular the pullback $m_G^*$ gives a function in $C^\infty$ "extra argument" which has the effect of left translating the input, so $m_G^*\circ X$ is effectively "act with $X$ first, then left translate the input of the resulting function", while $(id\otimes X)\circ m_G^*$ is the reverse.