# How to prove that every left-invariant differential form on a Lie group is smooth

Let $$G$$ be a Lie group. Recall that a differential $$k$$-form $$\omega \in \Omega^k(G)$$ is called left-invariant if its pullback by left multiplication maps is itself: $$L_g^*\omega=\omega$$ for all $$g\in G$$. Such forms are determined by its value at the identity $$e\in G$$, because for any $$g\in G$$, we have that $$$$\omega_g=(L_{g^{-1}}^*\omega)_g.$$$$ I believe that such differential forms are always smooth. I tried searching for this result from Google but could not get satisfactory answers. So I thought maybe the proof was too easy and tried proving it. My idea is to use take any smooth vector fields $$X_1,\ldots, X_k$$ and show that $$\omega(X_1,\ldots, X_k): G\to \mathbb{R}$$ is a smooth function. Now by definition, the function $$\omega(X_1,\ldots,X_k)$$ is defined by $$\omega(X_1,\ldots,X_k)(g)=\omega_g(X_1|_g,\ldots,X_k|_g),$$ using the above expression for $$\omega_g$$ we get that $$\omega(X_1,\ldots,X_k)(g)=\omega_e(dL_{g^{-1}}(X_1|_g), \ldots, dL_{g^{-1}}(X_k|_g)).$$ This shows that $$\omega(X_1,\ldots,X_k)$$ is equal to the composition: $$\require{AMScd} \begin{CD}G @>dL_{g^{-1}}(X_1|_g)\times \cdots \times dL_{g^{-1}}(X_1|_g)>> T_eG\times\cdots\times T_eG @>\omega_e>> \mathbb{R}. \end{CD}$$ Since $$\omega_e$$ is multilinear, so it is smooth. Therefore showing that $$\omega$$ is smooth boils down to showing that the map $$G\to T_eG: g\mapsto dL_{g^{-1}}(X_i|_g) (i=1,\ldots, k)$$ is smooth. This is where I get stuck. I don't have any idea how to show that this map is smooth.

Your line of argument is correct, although there are simpler ways to see this. To complete your argument, you can use the (standard) result that the so-called left trivialization, which sends a tangent vector $$\xi$$ at $$g\in G$$ to $$(g,T_gL_{g^{-1}}(\xi))$$ defines a diffeomorphism $$\Phi:TG\to G\times T_eG$$. A vector field then is a smooth map $$X_i:G\to TG$$, so $$\Phi\circ X_i$$ is smooth and its second component is the $$i$$th component of the map $$G\to T_eG\times\dots\times T_eG$$ that you consider.
The simpler way to get the result is to observe that smoothness of $$\omega$$ follows if for each $$g\in G$$ you find a local frame for $$TG$$ defined on an open neighborhood of $$g$$ such that inserting any frame elements into $$\omega$$ one gets a smooth function on the domain of definition of the frame. Now you can take a global frame formed by the left invariant vector fields corresponding to a basis of $$T_eG$$. Inserting frame elements into $$\omega$$ you always get functions that are constant and hence smooth.
• What does the notation $T_g((L_{g^{-1}})(\xi)$ means. I have never seen this type of notationd. Commented Jul 25, 2023 at 7:09
• @Uncool $T_gL_{g^{-1}}$ is the differential of $L_{g^{-1}}$ at $g$. Using your notation, $T_gL_{g^{-1}}(\xi)$ would be $dL_{g^{-1}}(\xi)$. Commented Jul 25, 2023 at 7:13