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Let $A$ be a local ring with the unique maximal ideal $\mathfrak{m}$. The punctured spectrum of $A$ is the open subset $\text{Spec}(A)\setminus \{\mathfrak{m}\}$. I have seen many papers (for instance Horrocks' papers) studying vector bundles over algebraic varieties (in particular, projective spaces) by putting them over a punctured spectrum of a local ring.

However, I am wondering why punctured spectrum is better than varieties in this satiation. I feel that geometric pictures of varieties are clearer than that of a punctured spectrum. More essential, are the categories of coherent sheaves over a variety and its punctured spectrum equivalent? How much information about coherent sheave (in particular, vector bundles) can be recovered from punctured spectrums? For second question, I am thinking examples that reflect some relations between those two gadgets.

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One reason to study punctured spectra is to study projective varieties. If $S$ is homogenous with $\mathfrak n=S_+$ and $A=S_{\mathfrak n}$, $\mathfrak m=\mathfrak n_{\mathfrak n}\subset A$ then a vector bundle over $\mathrm{Proj} S$ pulls back to give one over $\mathrm{Spec} A\setminus\mathfrak m$. –  Sándor Kovács Mar 26 '12 at 14:40
    
Is there any relation (such as an isomorphism) between a projective variety and the completion of its punctured spectrum? –  Fei YE Mar 26 '12 at 15:37
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For projective varieties $X$ pulling back a vector bundle $E$ on $X$ to the punctured affine cone $C(X)$ gives you the equivalence between graded locally free sheaves, possibly with a twist, and vector bundles on $X$. By using local cohomology of $C(X)$ with supports at the vertex and an excision move, you can sometimes relate it directly to the cohomology of the punctured spectrum where you have tools such as depth available. –  Ray Hoobler Mar 26 '12 at 22:23
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Problems over the punctured spectrum of an arbitrary local ring are usually more general (therefore harder) than the same problem over the punctured spectrum of cone of a projective variety. If you can solve a problem (e.g., splitting of vector bundles) over the punctured spectrum of an arbitrary regular local ring, then you get it for free for the projective space, but the converse is not necessarily true. There are theorems about vector bundles over projective spaces that are not proved for vector bundles over the punctured spectrum of a regular local ring. –  Mahdi Majidi-Zolbanin Mar 27 '12 at 2:24
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Dear Fei: If I get a chance, I will, soon. –  Mahdi Majidi-Zolbanin Mar 28 '12 at 1:39

3 Answers 3

up vote 4 down vote accepted

Here is a general principle. If a problem is formulated over a general (say noetherian) base, by localization arguments one can often reduce to the case of a local base. Arguing by induction on the dimension of the base, we can exploit the fact that the punctured spectrum, while often not affine, is of smaller dimension! Then we can try to "extend" the solution over the punctured spectrum to a solution over the local spectrum (by taking into account local cohomology or other tools). I think this brilliant idea is due to Grothendieck (truly exploiting the concept of a scheme).

For example, observe that the punctured spectrum of a 2-dimensional regular local ring is a (typically non-affine!) Dedekind scheme, over which flatness is a hands-on concept. Much deeper, the proof of Zariski's Main Theorem in EGA IV$_3$ goes via dimension induction on the base using the above trick with a punctured spectrum having smaller dimension.

For the study of coherent sheaves, the categories on the spectrum and punctured spectrum are not at all equivalent, though the situation for vector bundles is more accessible; the problem is governed by local cohomology and depth considerations (so the situation is more tractable over a regular base). This is studied in depth (no pun intended) in the middle of SGA2.

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Thanks for your answer. I don't understand why the punctured spectrum is of smaller dimension. Can you explain it? Thanks! –  Fei YE Mar 29 '12 at 8:20
    
Dear F.Y.: This requires nothing beyond understanding basic scheme-theoretic definitions, so it is better that you spend more time thinking about it by yourself. I will just point out a key feature which makes this genuinely different from the variety situation: a local scheme has only one closed point (and beware that this does not mean the punctured scheme has no closed points of its own, just that they're not closed in the local scheme; think about local rings at rational points on irreducible curves and surfaces, drawing some pictures for yourself). Good luck. Regards, q-c. –  user22479 Mar 29 '12 at 12:39

This is not directly responsive to your question, but it does show that if you want to study vector bundles, you can't always replace the punctured spectrum with the full spectrum:

Let $A$ be a regular local ring of dimension 2, and $U$ its punctured spectrum. Then every vector bundle over $Spec(A)\times {\bf A}^1=Spec(A[X])$ is trivial, but there are certainly vector bundles over $U\times{\bf A}^1$ that are non-trivial.

To construct one, let the maximial ideal of $A$ be generated by elements $a,b$, let ${\cal O}$ be the structure sheaf on $U\times{\bf A}^1$ and let ${\cal F}$ be the kernel of the map ${\cal O}^3\rightarrow{\cal O}$ given by $(a,b,X)$.

It's not too hard to check that ${\cal F}$ can't be trivial. If it were, then (extending to the full spectrum and observing that global sections don't change) we'd have a sequence of $A[x]$ modules

$$0\rightarrow A[X]^2\rightarrow A[X]^3\rightarrow A[X] \rightarrow k \rightarrow 0$$

which gives an $A[X]$-projective resolution of $k$ of length 2, contradicting the fact that, over $A[X]$, $k$ has projective dimension 3.

I am almost sure I learned this from Chuck Weibel,and almost equally sure that the argument is due to Dick Swan.

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I think we can adapt the argument from my other answer to give an answer that is directly responsive to your question:

Let $A$ be the local ring of the origin in 3-space over your favorite field $k$. Let $X,Y,Z$ be the coordinates on 3-space. Map $A^3\rightarrow A$ by $(X,Y,Z)$. Let $M$ be the kernel. Restricting $M$ to the punctured spectrum, we get a vector bundle. But this vector bundle can't be free, or we'd have a projective resolution of $k$ by length 2 over $A$, whereas the projective dimension of $k$ is 3.

Therefore the punctured spectrum supports a non-trivial vector bundle, wherease the full spectrum doesn't.

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One needs a tiny bit more work here, as a non-free module can still give a trivial bundle. –  Hailong Dao Mar 26 '12 at 21:57

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