Proof of a Theorem in the paper "Construction of bundles on P^n" by Horrocks - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T06:40:47Z http://mathoverflow.net/feeds/question/20593 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20593/proof-of-a-theorem-in-the-paper-construction-of-bundles-on-pn-by-horrocks Proof of a Theorem in the paper "Construction of bundles on P^n" by Horrocks Fei YE 2010-04-07T04:42:23Z 2010-04-15T06:54:28Z <p>I am trying to understand Horrocks's <a href="http://dl.dropbox.com/u/3849644/Construction%20of%20Bundles.pdf" rel="nofollow"> construction of vector bundles</a>. However I have been stuck on the proof the first theorem in the paper.</p> <p>In the paper, a trivial bundle is a direct sum of Hopf bundles $\mathcal{O}(p)$. </p> <p>Theorem: Let $E$ be a vector bundle without a trivial direct summand. Then there exist a trivial bundle $T$ such that $E\oplus T$ has a filtration $$E\oplus T=F^0 \supseteq F^1\supseteq F^2\supseteq\cdots\supseteq F^N=0$$ with $F^i/F^{i+1}$ a twisted exterior power of the tangent bundle $T_{\mathbb{P}^n}$.</p> <p>Here is his proof:</p> <p>Take a resolution $L$ of the dual of $E$ by trivial sheaves which is exact as a resolution of graded modules. The dual $L^*$ can be dismantled into Koszul complexes. </p> <p>Here are my questions. How to break up $L^*$ into Koszul complexes? What are the Koszul complexes? Where does the $T$ come from?</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/20593/proof-of-a-theorem-in-the-paper-construction-of-bundles-on-pn-by-horrocks/21420#21420 Answer by Hailong Dao for Proof of a Theorem in the paper "Construction of bundles on P^n" by Horrocks Hailong Dao 2010-04-15T06:01:04Z 2010-04-15T06:54:28Z <p>That is a pretty terse proof! Let me give an outline of a proof that I know. First, one could deduce the statement from a more general:</p> <p><strong>Theorem 1</strong>: Let $R$ be a regular local ring, $E$ be a reflexive $R$-module locally free on $U_R$, the punctured spectrum such that $E$ has no free direct summand. Then one can find a free module $T$ and a filtration:</p> <p>$$E\oplus T = F_0 \supseteq F_1 \supseteq \cdots F_N =0$$</p> <p>with $F_i/F_{i+1}$ a syzygy of $k=R/m$. </p> <p>Why is this local statement implies what you want? </p> <p>Let $A=k[x_0,\cdots, x_n],m=(x_0,\cdots,x_n), X=Proj(A)=\mathbb P^n, R=A_m$. There is natural functor from the category of vector bundles on $X$ to that of vector bundles on $U_R$, which is the same as the category of reflexive $R$-modules which are locally free on $U_R$. This is used by Horrocks all the time and is explained in Section 9 of his <a href="http://plms.oxfordjournals.org/cgi/pdf_extract/s3-14/4/689" rel="nofollow">paper</a>: "Vector bundles on punctured spectrum of a regular local ring". </p> <p>A proof of Theorem 1 can be found in Chapter 5 (theorem 5.2) of the book "Syzygy" by Evans-Griffith. A brief outline in case you can't find the book:</p> <p>As suggested in the paper you quoted, one starts with a minimal resolution of <code>$E^*$</code>. Then dualizing gives a complex (remember that $E^{**} \cong E$ as $E$ is reflexive):</p> <p>$0 \to E \to L_0 \to L_1 \cdots$ </p> <p>whose cohomologies are $Ext^i(E^*,R)$. Let $i>0$ be the smallest number such that <code>$X=Ext^i(E^*,R) \neq 0$</code> Break the l.e.s in to the exact sequences:</p> <p>$0\to E \to L_0 \to L_1 \cdots \to L_i \to N \to 0 (*)$ </p> <p>and $0 \to X \to N \to N/X \to 0$. Now build free resolutions for $X$ and $N/X$ and map them onto $(*)$ as in <a href="http://en.wikipedia.org/wiki/Horseshoe_lemma" rel="nofollow">Horseshoe Lemma</a>, stopping at the spot $E$, one gets a s.e.s: </p> <p>$0 \to B \to E\oplus T \to C \to 0$ here $T$ is free and $C$ is a syzygy of <code>$X = Ext^i(E^*,R)$</code>. Repeat if necessary and you have a filtration whose quotient are syzygies of various <code>$Ext^i(E^*,R)$</code>. But each of this $Ext$ modules has finite length (as $E$ is locally free on $U_R$), so they can be filtered by copies of $k$. Now use the same trick to build a finer filtration whose quotients are syzygies of $k$. Since $R$ is regular, the resolution of $k$ is the Koszul complex, answering your second question. </p>