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let $Gr(k,V)$ be the grassmannian of k-dimensional subspaces of the complex vector space $V$ of dimension $n>k$.

I know that, given $K\in Gr(k,V)$, $T_{Gr(k,V),K}\simeq Hom(K,V/K)$, but i want to prove that this isomorphism is canonical.

I'm following Claire Voisin's book "Hodge theory and complex algebraic geometry I". She proceeds this way:

We choose a basis $\sigma_1,\cdots,\sigma_k$ of $K$ and let be $\widetilde{\sigma_1},\cdots,\widetilde{\sigma_k}$ sections of the canonical bundle such that $\widetilde{\sigma}_i(K)=\sigma_i$, $1\le i\le k$. To the tangent vector $u\in T_{Gr(k,V),K}$ we associate the linear map $h_u:K\rightarrow V/K$ defined as

$h_u(\sigma_i)=u(\widetilde{\sigma_i})$ $mod$ $K$

Where $u(\widetilde{\sigma_i})$ is the derivative with respect to $u$ of the section $\widetilde{\sigma_i}$ considered as a function on $G$ with values in $W$. Voisin writes that this identification is canonical because, if $\alpha$ is a section of the canonical bundle which vanishes on $K$, then locally we can write $\alpha=\sum_if_i\widetilde{\sigma_i}$ ($f_i$ holomorphic functions which vanish on $K$) and $u(\alpha)=\sum_i u(f_i)\widetilde{\sigma_i}(K)\in K$ and so $u(\alpha)=0$ in $V/K$.

Maybe I am missing something very basic, but how does this tell me that the association $u\mapsto h_u$ is indipendent to the choice of a base?

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    $\begingroup$ It seems to me that the easiest way to understand the Grassmannian as a manifold and its tangent bundle is by viewing it as a homogeneous space, i.e, the quotient of $GL(v)$ by the appropriate subgroup. And it's worth working everything out for projective space first. $\endgroup$ – Deane Yang Jul 6 '13 at 4:40
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    $\begingroup$ Given $K \in G(k,V)$ and a transversal subspace $L$, there is a natural map $\Phi_L: Hom(K,L) \rightarrow G(k,V)$ by $\Phi(\phi) = \{ k + \phi(k)\ :\ k \in K\}$. Check that 1) this is a diffeomorphism onto an open neighborhood of $K$ and 2) $\Phi_{L'} = \Phi_L$, if $[L] = [L'] \in V/K$. Therefore, $\Phi_L$ can be viewed as a map $\Phi: Hom(K,V/K) \rightarrow G(k,V)$. It follows from this that $T_KG(k,V) = Hom(K,V/K)$. $\endgroup$ – Deane Yang Nov 15 '17 at 22:09
  • $\begingroup$ @DeaneYang, when you talk about an "open neighbourhood of $K$", you're assuming a topology on $G(k,V)$. What topology is it? $\endgroup$ – rmdmc89 Jan 19 '19 at 13:34
  • $\begingroup$ It's the topology, where $\Phi_L$ is continuous (in fact, smooth) for every $K$ and $L$. $\endgroup$ – Deane Yang Jan 19 '19 at 20:20
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There is a natural map $\mu:\text{Aut}(V)\times G_k(V)\to G_k(V)$ sending $(A,W)$ to $AW$. Differentiating this at $(I,W)$ gives a natural map $\mu_*:\text{Hom}(V,V)\to T_WG_k(V)$. There is also a natural map $\pi:\text{Hom}(V,V)\to\text{Hom}(W,V/W)$, given by $$ \pi(\alpha) = (W \xrightarrow{\text{inc}} V \xrightarrow{\alpha} V \xrightarrow{\text{proj}} V/W). $$ I claim that there is a unique map $\nu:\text{Hom}(W,V/W)\to T_WG_k(V)$ with $\mu_*=\nu\circ\pi$, and that $\nu$ is an isomorphism. To prove this, we introduce the subgroup $\text{Aut}(V,W)=\{A\in \text{Aut}(V): AW=W\}$, and check that the tangent space to $\text{Aut}(V,W)$ at $I$ is the kernel of $\pi$; the rest follows easily from this.

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    $\begingroup$ How can we identify $T_{(I,W)}(\text{Aut}(V)\times G_k(V))$ with $\text{Hom}(V,V)$? $\endgroup$ – rmdmc89 Jan 19 '19 at 13:45
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If you choose another basis, $\alpha_1,\dots ,\alpha_k$, where each $\alpha_i$ can be represented by linear combaination of those $\sigma_i$. Suppose we have $\alpha_i=\sum_{i=1}^{k}a_{ij}\sigma_j$, then we can choose $\widetilde{\alpha_i}=\sum_{i=1}^{k}a_{ij}\widetilde{\sigma_j}$. Since the $u$ acts on the section linearly. We have $h_u(\alpha_i)=u(\widetilde{\alpha_i})$ give the same linear map from $V$ to $V/K$.

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