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.



Tangent Space to a Linear System
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).


