Left invariant Affine connections I just started to study Affine connections and I'm trying to prove the following;
Let $\nabla$ be an Affine connection on a Lie group $G$, and $X_1, X_2, \dots, X_n$ be a basis of the Lie algebra $\mathfrak{g}$ of $G$. (Note that $X_i$ are left-invariant vector fields on $G$ by definition.) If $\nabla_{X_i} X_j$ is a left-invariant vector field for any $i, j = 1, 2, \dots, n$, then $\nabla$ is left-invariant.
To show that $\nabla$ is left-invariant, I need to show that for every vector fields $Z$ and $Z'$, it satisfies that $\nabla_{dL_{\sigma}Z} (dL_{\sigma}Z') = dL_{\sigma} \nabla_Z Z'$ for each $\sigma \in G$. Since $X_1, X_2, \dots, X_n$ is a basis, we may write $Z = \sum_{i=1}^n f_i X_i$ and $Z' = \sum_{j=1}^n g_j X_j$ for some smooth functions $f_i, g_j \in C^{\infty}(G)$. And then by linearity I can write
$\nabla_{Z^{L_{\sigma}}}(Z')^{L_{\sigma}} = \nabla_{\sum_{i=1}^n f_i^{L_{\sigma}} X_i^{L_{\sigma}}} \sum_{j=1}^n g_j^{L_{\sigma}} X_j^{L_{\sigma}} = \sum_{i=1}^n f_i^{L_{\sigma}} \nabla_{X_i^{L_{\sigma}}} \sum_{j=1}^n g_j^{L_{\sigma}} X_j^{L_{\sigma}}$
where $Z^{L_{\sigma}}$ is the push-forward $dL_{\sigma}Z$, and same for $f^{L_{\sigma}}$.
But then I stuck here, since the axioms on Affine connections does not tell me that $\nabla$ is linear in the second variable. I want to apply the Leibniz rule on $g_j^{L_{\sigma}} X_j^{L_{\sigma}}$, but I should first separate those summands somehow.
How can I resolve this situation?
 A: Let $X_1,\dots, X_n$ be a basis of left invariant vector fields on $G$ and let $\nabla$ be a connection on $G$ such that
$$
\nabla_{X_i} X_j
$$
is left invariant.  We want to show that for all $g\in G$ and all vector fields $A$ and $B$ on $G$ we have
$$
\nabla_{dL_gA}(dL_gB)=dL_g \nabla_AB.
$$
Write (using Einstein summation notation)
$$
A=f^iX_i,~B=h^iX_i
$$
Let $k\in G$.  Then
$$
\begin{align}
(dL_gA)_{k}&=dL_g~A_{g^{-1}k}\\
&=f^i(g^{-1}k)dL_g(X_i)_{g^{-1}k}\\
&=f^i(g^{-1}k)(X_i)_k
\end{align}
$$
Hence,
$$
dL_g A= (f^i\circ L_{g^{-1}})X_i
$$
Likewise,
$$
dL_g B= (h^i\circ L_{g^{-1}})X_i
$$
Write $\tilde{f}^i:=f^i\circ L_{g^{-1}}$ and $\tilde{h}^i:=h^i\circ L_{g^{-1}}$.
Then
$$
\begin{align}
\nabla_{dL_gA}(dL_gB)&=\tilde{f}^i\nabla_{X_i}(\tilde{h}^jX_j)\\
&=\tilde{f}^i(X_i\tilde{h}^j) X_j+\tilde{f}^i\tilde{h}^j\nabla_{X_i} X_j
\end{align}
$$
On the other hand,
$$
\nabla_{A}B=f^i(X_ih^j) X_j+f^ih^j\nabla_{X_i} X_j
$$
Using the fact that $\nabla_{X_i} X_j$ is left invariant, we apply $dL_g$ to $\nabla_{A}B$
$$
\begin{align}
dL_g \nabla_{A}B&=(f^i\circ L_{g^{-1}})[(X_ih^j)\circ L_{g^{-1}}] X_j+(f^i\circ L_{g^{-1}})(h^j\circ L_{g^{-1}})\nabla_{X_i} X_j\\
&=\tilde{f}^i[(X_ih^j)\circ L_{g^{-1}}] X_j+\tilde{f}^i \tilde{h}^j\nabla_{X_i} X_j
\end{align}
$$
The left invariance of $\nabla$ is proved if we can show that the function
$$
(X_ih^j)\circ L_{g^{-1}}=X_i\tilde{h}^j
$$
Apply $k\in G$ to the left side and using the left invariance of $X_i$ gives:
$$
\begin{align}
(X_ih^j)\circ L_{g^{-1}}(k)&=(X_ih^j)(g^{-1}k)\\
&=(X_i)_{g^{-1}k}h^j\\
&=[dL_{g^{-1}}(X_i)_k]h^j\\
&=(X_i)_k (h^j\circ L_{g^{-1}})\\
&=(X_i)_k \tilde{h}^j\\
&=(X_i \tilde{h}^j)(k)
\end{align}
$$
Hence, $(X_ih^j)\circ L_{g^{-1}}= (X_i \tilde{h}^j)$ which proves the left invariance of $\nabla$.
