I'm trying to show that $so(3) = \{A\in M(3; \mathbb{R}): A = -A^T \} $ is the lie algebra of $SO(3)$. For this i am using the following fact:"The tangent space at the identity to a Lie subgroup of $GL(n,\mathbb{R})$, endowed with the matrix commutator, is isomorphic to its Lie algebra". I defined $\phi:GL(3, \mathbb{R}) \to Sym$ (Sym is the subset of symmetric matrices) $\phi(A)=AA^T$. I want to check that the kernel of $d\phi(I): gl(3,R) → Sym $ is the subspace of skew-symmetric matrices in $Gl(3,R)$. I'm having trouble finding $d \phi (I)$, how do I proceed?
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1You can either use the chain rule, or the fact that for differentiable $\phi$, we have for all $X$, $d\phi_I[X] = \dfrac{d}{dt}\bigg|_{t=0}\phi(I+tX)$ (this is just the familiar relationship between total derivatives and directional derivatives). In this case, the RHS should be pretty easy to calculate. Another way to calculate $d\phi_I$, is to recognze that $\phi$ is the composition of two maps, namely the product map $(A,B)\mapsto A\cdot B$, and the map $A \mapsto (A, A^T)$. The first is a bilinear map, so it's derivative is easy to calculate, and the second is a linear map. – peek-a-boo Jun 09 '20 at 23:48
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The derivative at a matrix $A$ applied to a matrix $H$ is, by definition $$ d\phi_A(H)=\lim_{t\rightarrow 0}\frac{(A+tH)(A+tH)^T-AA^T}{t} = HA^T+AH^T $$ In particular, putting $A=I$ we get $$ \ker d\phi_I=\{H:H+H^T=0\} $$
OnlyDay
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2Note that while this computation will work whenever the function in question is differentiable, this formulation is not equivalent to the usual definition of the derivative. – Ben Grossmann Jun 09 '20 at 23:55
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$\phi(A_0+tA)=(A_0+tA)(A_0+tA)^T=A_0A_0^T+t(A_0A^T+AA^T_0)+t^2AA^T)$ implies that $d\phi(A_0)(A)=A_0A^T+AA_0^T$, we deduce that $\phi(I)(A)=A+A^T$ and $d\phi(I)$ is the set of antisymmetric matrices.
Tsemo Aristide
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