I might have a new understanding to an old question of mine: Is the differential of left multiplication still left multiplication? I didn't seem to understand the question very well.
This is now how I understand the question: Let $G$ be the real n-dimensional general linear group. Let $A \in G$. Consider left multiplication $l: G \to G$, $l(B) = AB$ and its differential at identity $l_{\{*, I\}}: T_I(G) \to T_A(G)$. Because $G$ is open in $\mathbb R^{n \times n}$, the tangent spaces $T_I(G)$ and $T_A(G)$ are $\mathbb R$-vector spaces isomorphic to $\mathbb R^{n^2}$ and thus to each other and to $\mathbb R^{n \times n}$. Let $\alpha: \mathbb R^{n \times n} \to T_I(G)$ and $\beta: \mathbb R^{n \times n} \to T_I(G)$ be isomorphisms. Thus, saying $l_{\{*, I\}}: T_I(G) \to T_A(G)$ is left multiplication is equivalent to saying that if $X_I \in T_I(G)$ and if $Y_A \in T_A(G)$, with $l_{\{*, I\}}(X_I) = Y_A$, then
$$(l \circ \alpha^{-1}) (X_I) = \beta^{-1}(Y_A) \tag{1}$$
This is the same idea behind the question, In what sense is the differential of a linear map is itself?.
In my new approach to answering this question, I do not consider a smooth curve $s$. Are these statements correct?
Essentially: I consider left multiplication in $\mathbb R^{n \times n}$ and show its differential at $I$ or any matrix is itself, by the question, In what sense is the differential of a linear map is itself? (which I cannot apply directly to $l$ since $G$ is not an $\mathbb R$-vector space) and thus is left multiplication. Then, I say that the differential of left multiplication in $G$ is also left multiplication because the differential of left multiplication in $\mathbb R^{n \times n}$ is also left multiplication. (I think this is what Xipan Xiao is talking about in the old question Is the differential of left multiplication still left multiplication?)
Consider the maps $L: \mathbb R^{n \times n} \to \mathbb R^{n \times n}$, $L(B)=AB$, for the same $A \in G$ and the inclusion $i: G \to \mathbb R^{n \times n}$. Then $L|_G = L \circ i = i \circ l$.
Because $L$ is a smooth and $\mathbb R$-linear map under the identification of $\mathbb R^{n \times n}$ with $\mathbb R^{n^2}$ (Earlier, I created isomorphisms $\alpha$ and $\beta$. Now, I won't create another isomorphism for this.), for each $M \in \mathbb R^{n \times n}$, every $L_{\{*, M\}}: T_M(\mathbb R^{n \times n}) \to T_{AM}(\mathbb R^{n \times n})$ is the same as $L$ itself, by, again, the question In what sense is the differential of a linear map is itself?, again:
$$(L \circ \gamma_M^{-1})(Z_M) = \delta_{AM}^{-1}(W_{AM}) \tag{2}$$
for $L_{\{*, M\}}(Z_M)=W_{AM}$ and for isomorphisms $\gamma_M: \mathbb R^{n \times n} \to T_M(\mathbb R^{n \times n})$ and $\delta_{AM}: \mathbb R^{n \times n} \to T_{AM}(\mathbb R^{n \times n})$.
Next, $i$ is smooth, as I asked about here. Thus, its differential each point $g$ of $G$ is defined. We have $i_{\{*, g\}}: T_g(G) \to T_g(\mathbb R^{n \times n})$. $i$ is also an immersion, and so each $i_{\{*, g\}}: T_g(G) \to T_g(\mathbb R^{n \times n})$ is an injective map. By identifying each $T_g(G)$ with its image under $i_{\{*, g\}}$, we can consider $T_g(G)$ as a subset of $T_{g}(\mathbb R^{n \times n})$. Thus, for $g=M=I$, we have $\alpha^{-1} = \gamma_I^{-1}|_{T_I(G)}$ and $\beta^{-1} = \delta_I^{-1}|_{T_I(G)}$. Also, $L|_G = L \circ i = i \circ l$ tells us that for each $h \in G$, $l_{\{*, h\}} = (L \circ i)_{\{*, h\}}$ and thus the image of $T_h(G)$ under $L_{\{*, I\}}$ is a subset of $T_{Ah}(G)$. Again, let $h=I$.
The preceding statements enable us to choose $Z_M = X_I$ and $W_{AM} = Y_A$. Therefore, $(2)$ becomes
$$(L \circ \gamma_I^{-1})(Z_I) = \delta_{A}^{-1}(W_{A}) \tag{3}$$
- Finally, $(L \circ \gamma_I^{-1})(Z_I) = (L \circ \gamma_I^{-1})(X_I) = (L \circ \alpha^{-1})(X_I) = (l \circ \alpha^{-1})(X_I)$ and $\delta_{A}^{-1}(W_{A}) = \delta_{A}^{-1}(Y_A) = \beta^{-1}(Y_A)$, proving $(1)$.
Thanks in advance!
Yes. Left multiplication by a fixed matrix is the restriction to an open subset (i.e., $GL(n,\mathbb R)$) of a linear map on the vector space of all $n \times n$ real matrices.
