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My problem is perhaps a general lack of understanding but it occurred in a special case of a theorem in Eisenbud's and Harris' "The geometry of schemes" (Theorem VI-29). Let $K$ be a field and $n\in\mathbb{N}$.

Given a closed subscheme $X$ of $\mathbb{P}^n$ with Hilbert Polynomial $P$. The tangent space $T_{[X]}$ to the Hilbert scheme $\mathcal{H}_{P}$ at the point corresponding to $X$ is the space of global sections of the normal sheaf $\mathcal{N}_{\mathbb{P}^n/X}$.

I don't know what "is" precisely means here. $T_{[X]}$ has a natural $K([X])$-vector space structure and I would like to have an isomorphism of $K([X])$-vector spaces but my problem begins much earlier in the definition of $\mathcal{H}om$-sheaves:

Let $J$ be the ideal sheaf of $X$. The presheaf $$ U \mapsto \mathcal{H}om_{\mathcal{O}_X|_U}(J/J^2|_U,\mathcal{O}_X|_U) $$ is already a sheaf of $\mathcal{O}_X$-modules (this does not depend on the special arguments here). This sheaf $\mathcal{N}_{\mathbb{P}^n/X}$ is called the normal sheaf of $X$ in $\mathbb{P}^n$. The global sections $$ \mathcal{N}_{\mathbb{P}^n/X}(X)=\mathcal{H}om_{\mathcal{O}_X}(J/J^2|_X,\mathcal{O}_X) $$ of this normal sheaf is an $\mathcal{O}_X(X)$-module. Is there an isomorphism of $\mathcal{N}_{\mathbb{P}^n/X}(X)$ to some $hom$ of rings or modules where I can work with? What is a possible $K([X])$-vector space structure on it?

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Since the crux of your question seems to be the action of K([X]) on the tangent bundle in the Hilbert scheme, perhaps you could define K([X])? Because I don't understand the question at all. T_[X] and the sections of the normal bundle are finite-dimensional K vector spaces, and canonically isomorphic as such. –  Ben Webster Mar 22 '10 at 16:24
    
Ben, maybe it is trivial but I am still struggling with the basic definitions in algebraic geometry. So I hope you are patient with my ignorance. Let $X$ be a scheme, $x$ a point of $X$ and $m_x$ the maximal ideal of the residue field $K(x)$ at $x$. $T_x$ is the dual of the $K(x)$-vector space $m_x/m_x^2=m_x\otimes_{\mathcal{O}_{X,x}}K(x)$. –  roger123 Mar 22 '10 at 17:28
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2 Answers

up vote 4 down vote accepted

Dear roger123,

This is largely a response to your question aksed as a comment below Charles Siegel's answer, but it won't fit in the comment box. Since $\mathcal N_{\mathbb P^n/X}$ is a sheaf of modules over the sheaf of rings $\mathcal O_{\mathbb P^n}$ (the structure sheaf of projective space), its global sections $\mathcal N_{\mathbb P^n/X}(\mathbb P^n)$ are a module over the ring $\mathcal O_{\mathbb P^n}(\mathbb P^n)$, which in turn are just $k$ (the ground field).
In short, the global sections of $\mathcal N_{\mathbb P^n/X}$ form a $k$-vector space.

Maybe you are being confused by the fact that $\mathcal N_{\mathbb P^n/X}$ is a sheaf on $\mathbb P^n$ that is supported on $X$, so that people often simultaneously regard it as a sheaf on either $X$ or $\mathbb P^n$. This is okay, because if $U$ is any open in $\mathbb P^n$ and $\mathcal F$ is a sheaf supported on $X$, then the sections over an open subset $U$ of $\mathbb P^n$ (when it is regarded as a sheaf on $\mathbb P^n$) will coincide with the sections over $U\cap X$ (when it is regarded as a sheaf on $X$).

In particular, one has the equation $\mathcal N_{\mathbb P^n/X}(X) = \mathcal N_{\mathbb P^n/X}(\mathbb P^n)$ (an abuse of notation if taken literally; however one is supposed to regard $\mathcal N_{\mathbb P^n/X}$ as a sheaf on $X$ on the left-hand side, and as a sheaf on $\mathbb P^n$ on the right-hand side).

One more thing: If the residue field of the Hilbert scheme at the point $P$ is $k(P)$, then $P$ is a $k(P)$-valued point of the Hilbert scheme, and so the corresponding closed subscheme $X$ lies in $\mathbb P^n_{k(P)}$. Thus, in the above discussion, $k$ can (and should) be taken to be $k(P)$. Thus the above discussion explains why $\mathcal N_{\mathbb P^n/X}(X)$ is a $k(P)$-vector space, as it should be.

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Thank you, Emerton. The domain of the sheaf was the source of my confusion. –  roger123 Mar 23 '10 at 11:53
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As for what "is" means here, it means "is naturally isomorphic to" that is, you should be able to find an isomorphism that doesn't depend on any choices you make (that is, it'll be functorial). As for if there's an isomorphism of sheaf hom to some hom of rings, keep in mind that $\mathcal{H}om_{\mathcal{O}_X}(J/J^2|_X,\mathcal{O}_X)$ is a sheaf, whereas between rings you get a set and between modules a group (in fact, module) of homomorphisms.

For understanding this theorem, you don't really want to see the normal bundle as something else. Really, you want to think about what is a tangent vector at a point of the Hilbert scheme? It's a morphism from $k[\epsilon]/\epsilon^2$ to it, which means that if you take the pullback along it, you'll be looking at infinitesimal embedded deformations. Think about why these are locally local sections of the normal bundle, and then glue them together to see that a global infinitesimal embedded deformation will be a global section quite naturally, without needing to see the global sections as something else at all.

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What do you mean functorial in this context? Anyway, I think the point of the question was to realize what is the structure of vector space on the global sections of the normal bundle. –  Andrea Ferretti Mar 22 '10 at 15:01
    
Thank you, Charles. But $\mathcal{N}_{\mathbb{P}^n/X}(X)$ have to be a ring somehow. The sheaf $\mathcal{N}_{\mathbb{P}^n/X}$ is a contravariant functor from a certain category to rings, isn't it? Andrea, you are right :). –  roger123 Mar 22 '10 at 15:05
    
No, $\mathcal{N}_{\mathbb{P}^n/X}$ takes values in modules, not in rings. I guess what's confusing you is the fact that $J/J^2$ is a sheaf of non unital rings, afetr all (as it is $J$). But, in (usual) algebraic geometry, rings are always assumed to be unital. –  Qfwfq Mar 22 '10 at 19:20
    
Yes, I am sorry, modules! But this does not change the question substantially. How does $\mathcal{N}_{\mathbb{P}^n/X}(X)$ admit algebraic structure? Is it isomorph to some $\hom$ of modules, abelian groups, etc? –  roger123 Mar 22 '10 at 19:36
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