Identity regarding the differential of a smooth action of a Lie group

Question

Hey I am currently following a course on Lie groups and I have followed a course on smooth manifolds. The question I have is the following,

Let $G$ be a Lie group acting smoothly on a manifold $M$ by the action $A:G\times M\to M$. Since $A$ is an action the following equality holds; $$A(g,A(h,x)) = A(gh, x).$$ Now we are asked to take the differential on both sides. However I seem to struggle with this rather easy sounding task. What I have done till now;

Write $$\tilde A: G\times G\times M\to M,\quad (g,h,x)\mapsto A(g,A(h,x)) = A(gh,x).$$ Now I want to take the differential, so I should end up with a map $$d\tilde A:TG\times TG\times TM \to TM,\quad (v,w,u)\mapsto d\tilde A(v,w,u).$$ I think this should be the differential when evaluated in $(g,h,x)$ $$d(A(g,A(h,x))) = (dA)_{(g,A(h,x))}\circ (dA)_{(h,x)}.$$

Answer 1

It seems to me that your solution has the wrong domain: You correctly identify that your tangent map $D\tilde{A}$ has tangent vectors of the smooth product manifold $G\times G \times M$ as inputs, the dimension of these tangent spaces is $\dim G +\dim G+\dim M$. $DA$ itself is a map on the $\dim G + \dim M$-dimensional tangent space of $G\times M$, so your map cannot be correct.

Maybe this helps:

The right hand side can be written as $$\tilde{A} = A\circ (m_G\times \operatorname{id}_M)\colon, (G\times G)\times M \to M,$$ where $m_G\colon G\times G\to G$ is the multiplication map, i.e. $gh = m_G(g,h)$. The left hand side can be written as $$\tilde{A} = A\circ (id_G \times A) \colon G\times G\times M,$$ where the inner map $A$ acts on the second factor $G$ of the product of manifolds. The product of maps just tells us on what factors of the manifold the factors of the maps act. Notice that both domains are diffeomorphic, $$ G\times G\times M \simeq (G\times G)\times M.$$ We can canonically identify the tangent bundles in a similar fashion $$T(G\times G\times M) \simeq T(G\times G)\times TM \simeq TG\times TG \times TM.$$ Now applying chain rule to both sides yields $$D\tilde{A} = DA\circ (Dm_G \times\operatorname{id}_{TM})\quad\text{(RHS)}$$ $$D\tilde{A} = DA\circ (\operatorname{id}_{TG}\times DA)\quad\text{(LHS)}.$$

Now we can move to evaluating at specific tangent vectors: Let $g,h \in G$ and $x\in M$, aswell as $u \in T_gG$, $v\in T_hG$ and $w \in T_xM$. Then $(u,v) \in T_{(g,h)}(G\times G)$. We calculate: $$D\tilde{A}_{g,h,x}(u,v,w) = DA_{(m_g(g,h),x)} \left((Dm_{G})_{(g,h)}(u,v)\times \operatorname{id}_{TM}(w)\right) =DA_{(m_g(g,h),x)} \left((Dm_{G})_{(g,h)}(u,v)\times w \right) \quad\text{(RHS)}$$ $$D\tilde{A}_{g,h,x}(u,v,w) = DA_{(g,A(h,x))} \circ (id_{TG}(u) \times DA_{(h,x)}(v,w)) =DA_{(g,A(h,x))} (u\times DA_{(h,x)}(v,w)) \quad \text{(LHS)}$$

By definition of a smooth action, the map $A$ has the same image for both base points of the outer differential $DA$, so both expressions for the differential of $D\tilde{A}$ land in the same tangent space. But in general the base point $(gh,x)$ and $(g,A(h,x))$ are not the same, so both maps can not be compared directly without further information. Both maps have the right domain and codomain. To me, this completes the calculation.

For some further properties of the tangent maps of maps between Lie-groups and ideas for proofs I recommend Hilgert-Neeb, Ch. 9.1. For example, there is a rather useful identity for the differential of the multiplication map involving the sum of the differential of left and right multiplication.