Let $d : GL(n,R) \rightarrow R$ be the determinant map. I don't know how to prove that if the map $d* : T_{I_n}GL(n,R) \rightarrow T_1R$ is surjective, then the map $d* : T_AGL(n,R) \rightarrow T_1R$ is also surjective for any matrix $A \in GL(n,R)$ with $det(A) = 1$. Please help me. Thank you.
Lie group GL(n,R) and the determinant map
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abstract-algebra
differential-geometry
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0Try to think of matrix $A$ as *acting* on elements of $GL(n,\mathbb{R})$ and notice that $d(A)=1$, this is why you arrive to $T_1\mathbb{R}$, but $T_1\mathbb{R} \cong \mathbb{R}$ as @Christopher pointed. – 2012-10-12