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?

  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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

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 $V/K$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.