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Consider this problem on my assignment on manifolds:

Let A be a matrix on $GL(n,\mathbb{R})$. Consider the map $L_A: GL(n,\mathbb{R})\to GL(n, \mathbb{R})$ defined by $L_A(B) =AB$. After identifying $T_I GL(n,\mathbb{R})$ with $M(n,\mathbb{R})$ prove that $(dL_A)_I(X)=AX$.

Attempt: $T_I(GL(n,\mathbb{R})$ is the linear map from $\mathbb{C}^{\infty} (M) $ to $\mathbb{R}$ satisfying Liebnitz rule at I.The problem I am facing is that how should I relate it with $M(n,\mathbb{R})$ and I have no intuition on how it can be used to prove that $(dL_A)_I (X) =AX$.

I am sorry for not giving much in attempt because no other results came to my mind.

So, how should I proceed?

1 Answers1

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A possible attempt is to use the definition of differential of a map through its "derivative along curves". If $M$ and $N$ are two differentiable manifolds and $f:M\to N$ is a smooth map, then for $p\in M$ we can define $d_p f$ (the differential of $f$ at $p$) as follows: if $X_p \in T_p M$ and $\alpha:(-\epsilon, \epsilon)\to M$ is a $C^1$ curve s.t. $\alpha(0)=p$ and $\alpha'(0)=X_p$, then \begin{align*} d_p f X_p := \frac{d}{dt} \Big(f\circ \alpha\Big)\Bigg|_{t=0} \end{align*}

In your case $M=N=GL_n(\mathbb{R})$ and $L_A$ is a linear map defined on an open subset of a finite dimensional vector space (and thus continuos), so if $\alpha:(-\epsilon, \epsilon)\to GL_n(\mathbb{R})$ is a $C^1$ curve satisfying $\alpha(0)=I$ and $\alpha'(0)=X$, then \begin{align*} d_I L_A (X)&=\frac{d}{dt} \Big(L_A\circ \alpha \Big)\Bigg|_{t=0}=[N\ \rm{is\ an\ open\ subset\ of\ a\ vector\ space}]\\ &=\lim_{t\to 0} \frac{L_A(\alpha(t))-L_A(\alpha(0))}{t}=[M\ \rm{is\ an\ open\ subset\ of\ \ a\ vector\ space}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rm{and\ linearity\ of}\ L_A]\\ &=L_A \Bigg[\lim_{t\to 0} \frac{\alpha(t)-\alpha(0)}{t} \Bigg]=L_A[\alpha'(0)]=L_AX \end{align*}

Andrea Bisterzo
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