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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 \begin{equation}\omega_g=(L_{g^{-1}}^*\omega)_g.\end{equation} 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.

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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.

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  • $\begingroup$ What does the notation $T_g((L_{g^{-1}})(\xi)$ means. I have never seen this type of notationd. $\endgroup$
    – Uncool
    Commented Jul 25, 2023 at 7:09
  • $\begingroup$ @Uncool $T_gL_{g^{-1}}$ is the differential of $L_{g^{-1}}$ at $g$. Using your notation, $T_gL_{g^{-1}}(\xi)$ would be $dL_{g^{-1}}(\xi)$. $\endgroup$
    – Frank
    Commented Jul 25, 2023 at 7:13
  • $\begingroup$ Oh.. yes this silly me, I forgot to use the left-invariant frame. The use of a left-invariant frame is indeed much easier. $\endgroup$
    – Uncool
    Commented Jul 25, 2023 at 7:41

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