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Jul 30, 2020 at 10:15 comment added მამუკა ჯიბლაძე Under the identification $T_I\operatorname{Aut}(V)\cong\operatorname{Hom}(V,V)$ the subspace $T_I\operatorname{Aut}(V,W)\subset T_I\operatorname{Aut}(V)$ becomes identified with those $\alpha:V\to V$ which restrict to $\alpha|_W:W\to W$ (hence induce a map $V/W\to V/W$). And an $\alpha$ is such if and only if $\pi(\alpha)$ is zero.
Jul 30, 2020 at 10:13 comment added მამუკა ჯიბლაძე @rmdmc89 Maybe the following is easier (but essentially the same): fix $W$ and consider $\varphi:\operatorname{Aut}(V)\to G_k(V)$ defined by $\varphi(A)=AW$. Differentiating $\varphi$ gives precisely the map $\mu_*$. The same $\varphi$ identifies $G_k(V)$ with the homogeneous space $\operatorname{Aut}(V)/\operatorname{Aut}(V,W)$, and (something omitted) gives an isomorphism$$T_WG_k(V)\cong T_I\operatorname{Aut}(V)/T_I\operatorname{Aut}(V,W).$$
Jan 19, 2019 at 13:45 comment added rmdmc89 How can we identify $T_{(I,W)}(\text{Aut}(V)\times G_k(V))$ with $\text{Hom}(V,V)$?
Jul 6, 2013 at 8:24 history answered Neil Strickland CC BY-SA 3.0