Let $G$ be a Lie group.
I am wondering if there is a way to say that the map $(g, h)\mapsto dL_g|_h$ defined on $G\times G$ is a smooth map (Here $L_g$ is the left translation map from $G$ to $G$ and by $dL_g|_h$ I mean the differential of $L_g$ at $h$).
The challenge here is to make the set $\bigsqcup_{g, h\in G}\mathcal L(T_hG, T_{gh}G)$ into a smooth manifold in a fruitful way since $dL_g|_h\in \mathcal L(T_gG, T_{gh}G)$.
I was reading the proof of the fact that the Lie algebra of a Lie group is finite dimensional from where the above question is motivated. In the proof there was this fact used that the map $g\mapsto dL_g|_ev:G\to TG$ is a smooth map, where $v$ is a fixed vector in $T_eG$.
Thank you.
I think the right way is using pullback bundles.
You need to know in advance that:
$(1)$ pullback bundle is an embedded submanifold of the product of its factors. (Hence it always carries a natural structure of a smooth manifold "inherited from the factors"). For details you can look at this answer.
$(2)$ Given two vector bundles over the same base manifold, you can create their Hom-bundle. The point is that the Hom-bundle can also be given a natural structure of a smooth manifold.
(I do not know an immediate way of showing this, but perhaps it's easier to show this for the tensor product of bundles and use the identification $E^* \otimes F \cong \text{Hom}(E,F)$). Again, as far as I know the literature in this domain is far from optimal).
Assuming for the moment, these two "basic" (and supposedly well-know) facts, we can see the following:
(See here for details). Consider the following two maps $G \times G \to G$:
$m(g,h)=gh \, \,\text{(multiplication)} \, \, , \, \, \pi(g,h)=h$, and their respective pullback bundles: $m^*(TG),\pi^*(TG)$. These are vector bundles over the base space $G \times G$.
Note that fiberwise:
$\big(m^*(TG)\big)_{(g,h)}=T_{gh}G,\big(\pi^*(TG)\big)_{(g,h)}=T_{h}G,$
So now if you take their Hom-bundle: $E=\text{Hom}(\pi^*(TG),m^*(TG))$
, then $E$ is a vector bundles over $G \times G$, with the required fiber:
$E_{(g,h)}=\text{Hom}(T_{h}G,T_{gh}G)$.
Now your $dL_g|_h$ is just a section of $E$. $E$ is a smooth manifold, so the smoothness of a map $G \times G \to E$ is well defined. You probably want to prove $dL$ is a smooth section of $E$.
$(1)$ You might see a discussion similar in spirit here.
$(2)$ You might also want to consult Lee's book "Introduction to smooth manifolds", page 192 (chapter 8 theorem 8.37), where he proves that every left translation of a vector from $T_eG$ is a smooth vector field.
(He does this without using the language of vector bundles, but it is a very powerful language which is worth knowing anyway...)