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Let $X$ be a smooth projective variety and let $D$ be an effective divisor on $X$. Is there a natural way to describe the tangent space to $|D|$ (or $|D|^\vee$, of course) at a divisor $D'$? Preferably as some sort of cohomology group, again ideally on $X$. I would prefer to avoid using the fact that the linear system is a projective space, and do things naturally.

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up vote 8 down vote accepted

The linear system $|D|$ is the projectivization of $H^0(X,\mathcal O(D)).$ The divisor $D'$ corresponds to a line $\ell_{D'} \subset H^0(X,\mathcal O(D)).$ (This line consists of all the sections whose zero loci are equal to $D'$.) The space $Hom(\ell_{D'},H^0(X,\mathcal O(D))/\ell_{D'})$ is a vector space of the same dimension as $|D|$, and is naturally isomorphic to the tangent space of $|D|$ at $D'$.

(Here is am using the general fact that if $V$ is an vector space, and $\ell \subset V$ a line through the origin, then the tangent space to the projectivization $\mathbb P(V)$ at the point corresponding to the line $\ell$ is identified with $Hom(\ell,V/\ell)$.)

[Thanks to Georges Elencwajg for correcting an earlier misstatement here.]

One can say a little more; before doing so, it's convenient to note that $D$ can be any divisor in the linear system, and so it is no loss of generality to set $D = D'$; this eases the notation somewhat. We also fix a section of $\mathcal O(D)$ cutting out $D$, i.e. a basis of $\ell_D$, which gives an identification $\ell_D = k = H^0(X,\mathcal O).$; this allows us to rewrite $Hom(\ell_D,H^0(X,\mathcal O(D))/\ell_D)$ simply as $H^0(X,\mathcal O(D))/\ell_D$.

Our choice of basis for $\ell_D$ gives a short exact sequence $$0 \to \mathcal O \to \mathcal O(D) \to \mathcal O(D)\_{| D} \to 0,$$ and taking global sections gives $$0 \to \ell_{D} \to H^0(X,\mathcal O(D)) \to H^0(D, \mathcal O(D)\_{| D}),$$ and hence an injection $$H^0(X,\mathcal O(D))/\ell_{D} \hookrightarrow H^0(D,\mathcal O(D)\_{| D}).$$ But this is not going to be an isomorphism in general, I guess.

Indeed, the cokernel embeds into $H^1(X,\mathcal O)$, which is the tangent space to Pic $X$, while $H^0(D,\mathcal O(D)\_{|D})$ is the tangent space to the Hilbert scheme of $X$ at $D$. [Note: To see this, observe that our choice of section cutting out $D$ corresponds to a choice of isomorphism $\mathcal O(D) \cong \mathcal I_D^{-1},$ and it is $(\mathcal I_{D}^{-1})\_{| D}$ that is canonically the normal bundle to $D$.] The map $H^0(D,\mathcal O(D)_{|D}) \to H^1(X,\mathcal O)$ then measures the extent to which the deformations of $D$ in $X$ fill up the component of the Picard scheme containing $D$. I imagine that if $D$ is sufficiently positive then this map is surjective; at least when $X$ is a surface, this is the main result of Mumford's "Lectures on curves on an algebraic surface" (if I am remembering correctly).

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Dear Matt, the tangent space to $\mathbb P(V)$ at the point corresponding to $\ell$ should rather be $\mathcal L(\ell,V/\ell)$, shouldn't it ? – Georges Elencwajg Oct 9 '10 at 7:20
Dear Georges, Yes, you're right. I'll correct the above when I have more time. – Emerton Oct 9 '10 at 13:17
Thanks! Now, of course, I'm feeling a bit stupid, because somehow it didn't occur to me that projective space is a Grassmannian, and I've seen this for Grassmannians...though seeing the rest of it out was also helpful, and I definitely need to check out that Mumford book. – Charles Siegel Oct 9 '10 at 22:01
Dear Matt, I am sure I am not the only one here to apppreciate this crystal-clear note: thank you. – Georges Elencwajg Oct 9 '10 at 22:42
Dear Georges, Thank you very much for your kind words. – Emerton Oct 10 '10 at 0:00

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