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
1
$\begingroup$
abstract-algebra
differential-geometry
-
0What is the $d*$ map? – 2012-10-11
-
0It is a determinant map. The determinant of $T_{I_n}$, $n \times n$ identity matrix goes to $T_1$ – 2012-10-11
-
0Sorry. I don't understand what the symbol $T_{I_n}GL(n,R)$ means. I suppose that $R$ is a ring with unity. But what is $T_1R$? Sorry. – 2012-10-11
-
1I think he means $T_1 \mathbb{R}$, but that's just $\mathbb{R}$. – 2012-10-11
-
0$d*$ here is supposed to be the differential $d_{*} = Td$ of the determinant map $d$. The proof is an application of the chain rule – 2012-10-12
-
0@Yuri: Yes, I think you're right. $d*$ is the push forward of d. Could you give me more details about how to use chain rule here? thanks. – 2012-10-12
-
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