Posted by: Steven Sam | August 17, 2009

## Tangent bundle of the Grassmannian

This post will be an exercise in algebraic (differential) geometry: I’ll calculate the tangent bundle of the Grassmannian $X = \mathbf{Gr}(r,E)$ of r-dimensional subspaces of a fixed vector space E defined over a field k. We’ll deduce from this that Grassmannians are Fano varieties.

The two methods will be similar, but working in the differential category requires less machinery, so we’ll assume k is the real numbers first. In this case, we’ll use the definition of tangent spaces at a point x as the tangent vectors of curves starting at x. We’ll need the tautological bundle R on X, which is the subbundle of E x X given by $\{(x,W) \mid x \in W\}$. The quotient bundle $(E \times X) / R$ will be denoted Q. Recall that the Plücker embedding is a map $X \to \mathbf{P}(\bigwedge^r E) = P$ sends a subspace W of E to its rth exterior power $\bigwedge^r W$. Under this embedding, the line bundle $\mathcal{O}(1)$ on P restricts to $\bigwedge^r R^*$ since $\mathcal{O}(-1)$ is the tautological subbundle of $\bigwedge^r E \times P$.

The image of X under the Plücker embedding are precisely all lines in P which have a generator of the form $e_1 \wedge \cdots \wedge e_r$ for $e_i \in E$. So pick a point $W = \langle e_1 \wedge \cdots \wedge e_r \rangle$ of X. Complete $\{e_1, \dots, e_r \}$ to a basis $\{e_1, \dots, e_n\}$ of E. Given a map $\phi \colon W \to E$, we can define a curve $\phi_c \colon [-1,1] \to X$ by $\phi_c(t) = \langle (e_1 + t\phi(e_1)) \wedge \cdots \wedge (e_r + t\phi(e_r)) \rangle$. Since $\phi_c(0) = W$, this determines a tangent vector $\phi'_c(0) = \langle \sum_{i=1}^r (e_1 \wedge \cdots \wedge \phi(e_i) \wedge \cdots \wedge e_r) \rangle$. Two curves given by $\phi_c, \psi_c$ determine the same tangent vector if and only if the image of their difference $\phi - \psi$ lies in $W' = \langle e_1, \dots, e_r \rangle$, so the tangent space naturally contains Hom(W’, E/W’) as a subspace. But this subspace has dimension r(n-r), which is the dimension of X, so in fact they are equal. Hence we conclude that the tangent bundle of X is Hom(R, Q).

How do we turn the above analytic argument into an algebraic one for an arbitrary field? Recall that the Zariski tangent space of a scheme X at a point p is defined to be $\mathrm{Hom}_{\kappa(p)}(\mathfrak{m}_p / \mathfrak{m}_p^2, \kappa(p))$ where $\mathfrak{m}_p$ is the maximal ideal of the local ring $\mathcal{O}_{X,p}$, and $\kappa(p)$ is the residue field $\mathcal{O}_{X,p} / \mathfrak{m}_p$. In the case that X is a scheme over a field k, and p is a k-rational point (i.e., its residue field is k), tangent vectors at p are equivalent to maps $\text{Spec }(k[\varepsilon] / (\varepsilon)^2) \to X$ whose image is p. For simplicity of notation, set $S = k[\varepsilon] / (\varepsilon)^2$. This definition seems harder to understand, but we’ll see that it’s the natural one to use in the case of the Grassmannian.

This is true because the functor of points Hom(–, X) of the Grassmannian has a nice description. To be specific, let F be the functor which sends a k-scheme Y to the set of all rank r locally free subsheaves $L \subseteq \mathcal{O}_Y \otimes E$ whose quotient is also locally free. Then F is isomorphic to Hom(–, X). (Given a map $Y \to X$, we get such a subsheaf L by pulling back the tautological sequence $0 \to R \to E \times X \to Q \to 0$ from X, and conversely, every choice of L induces a map from Y to X). All closed points of the Grassmannian are k-rational. Using this language, X = F(Spec k), and for a closed point p of X, the tangent space is the fiber of p under the map $T \colon F(\text{Spec } S) \to F(\text{Spec } k)$, which is obtained by applying $F \circ \text{Spec}$ to the quotient map $S \to k$.

A closed point of F(Spec k) is just a dimension r subspace of W of E whose quotient is free (this condition is automatic), and so the fiber under T is just all rank r locally free subsheaves of $\mathcal{O}_{\text{Spec } S}^n$ whose quotient is locally free, and whose pullback under the map $\text{Spec } k \to \text{Spec } S$ is W. Locally free sheaves on Spec S are equivalent to finitely generated projective S-modules, and since S is local, all such modules are free. Hence we have reduced to the following situation: find all free S-submodules M of $S^n$ whose quotient is free, and such that upon specializing $\varepsilon = 0$, M becomes W.

This is very similar to writing down tangent vectors for curves: let $\phi \colon W \to E$ be any linear map. Identifying $S^n$ and $E \oplus \varepsilon E$ as vector spaces, we can define $M_\phi = \{ x + \varepsilon \phi(x), \varepsilon x \mid x \in W \} \subset S^n$. This is a free S-submodule of $S^n$ with a basis given by any basis of W, and $M_\phi = M_{\psi}$ if and only if the image of the difference $\phi - \psi$ lies in W. So again we’ve concluded that Hom(W, E/W) is a subspace of the Zariski tangent space of W, which again must be whole thing by dimension counting since the Grassmannian is nonsingular.

Conclusion: the tangent bundle $T_X$ of the Grassmannian is $\text{Hom}(R, Q) = R^* \otimes Q$, and hence the cotangent bundle is $\Omega^1_X = R \otimes Q^*$. Taking the highest exterior power (determinant), the canonical bundle is $\omega_X = (\det R)^{\otimes (n-r)} \otimes (\det Q^*)^{\otimes r}$. But using the short exact sequence $0 \to R \to E \times X \to Q \to 0$, we see that det R and det Q are dual bundles. Recall from above also that det R is O(-1) under the Plücker embedding, so the canonical bundle simplifies to $\omega_X = \mathcal{O}(-n)$. In particular, the anticanonical bundle is $\omega^*_X = \mathcal{O}(n)$, which is very ample, and hence Grassmannians are Fano varieties (by definition, this means that the anticanonical bundle is ample).

-Steven

## Responses

1. Respected Sir

Thanks for the post.
May I ask, what is E in ExX given by
{(x,W)|x belogs to W}

2. E is the fixed vector space that we started with, so $E \times X$ just means the trivial product bundle over X with fiber E.