Why the rank of a locally free sheaves is well defined? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:27:16Z http://mathoverflow.net/feeds/question/5897 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5897/why-the-rank-of-a-locally-free-sheaves-is-well-defined Why the rank of a locally free sheaves is well defined? Ying Zhang 2009-11-18T02:01:43Z 2010-01-31T01:26:49Z <p>In Hartshorne P109 he defines a sheaf $\mathcal{F}$ of $O_X$-modules to be locally free if there is an open cover of X, s.t. on each U, $\mathcal{F}|U$ is a free $O_X|U$ module of rank $I$. Then if X is connected, rank $I$ is globally well-defined. Here $(X,O_X)$ is any ringed topological space(e.g. not necessarily the structure sheaf of a ring). A similar definition is here <a href="http://eom.springer.de/L/l060450.htm" rel="nofollow">link text</a></p> <p>However, it didn't seem obvious to me that if $V$ is a smaller open set included in $U$(say $U$ connected), then the number of copies $J$ of $(O_X|V)^J=\mathcal{F}|V$ would remain the same, because in general the restriction map of the sheaf $O_X$ or $\mathcal{F}$ from $U$ to $V$ need be neither subjective nor injective, why would the index $J$ stay the same as $I$?</p> http://mathoverflow.net/questions/5897/why-the-rank-of-a-locally-free-sheaves-is-well-defined/5903#5903 Answer by Eric Wofsey for Why the rank of a locally free sheaves is well defined? Eric Wofsey 2009-11-18T02:22:59Z 2009-11-18T02:22:59Z <p>Locally free means not just that $F(U)$ is a free $O_X(U)$-module for an open cover of $U$'s, but that <em>as a sheaf</em> the restriction $F|_U$ is isomorphic to a direct sum of copies of $O_X|_U$. In particular, this gives an isomorphism of $F(V)$ with a corresponding sum of copies of $O_X(V)$ for <em>every</em> open subset $V\subseteq U$.</p> http://mathoverflow.net/questions/5897/why-the-rank-of-a-locally-free-sheaves-is-well-defined/5905#5905 Answer by Alberto García-Raboso for Why the rank of a locally free sheaves is well defined? Alberto García-Raboso 2009-11-18T02:34:16Z 2009-11-28T15:22:08Z <p>Actually, there are two different restriction maps:</p> <ol> <li>The first one (the one you correctly say is neither surjective nor injective in general) is that on <b>sections</b>: for $\mathcal{F}$ a sheaf on a scheme $X$ and two open subsets $V \subseteq U$, there is a map $\mathcal{F}(U) \to \mathcal{F}(V)$. </li> <li>On the other hand, the inclusion map $i_{VU}: V \to U$ induces a functor $i_{VU}^{-1}$ from the category of sheaves on $U$ to the category of sheaves on $V$. This is called restriction of <b>sheaves</b>. By functoriality, $W \subseteq V \subseteq U$ yields $i_{WV}^{-1}\circ i_{VU}^{-1} = i_{WU}^{-1}$, so the notation is usually shortened to just $-|_{W}$. </li> </ol> <p>The second one is the one that you want to look at: the statement is then that $\mathcal{F}|_{U} \cong \mathcal{O}_X^{\oplus I}|_{U}$ implies $\mathcal{F}|_{V} \cong \mathcal{O}_X^{\oplus I}|_{V}$ for $V \subseteq U$.</p> http://mathoverflow.net/questions/5897/why-the-rank-of-a-locally-free-sheaves-is-well-defined/5927#5927 Answer by Adam Topaz for Why the rank of a locally free sheaves is well defined? Adam Topaz 2009-11-18T05:07:38Z 2009-11-18T07:31:27Z <p>Let $\mathcal{F}$ be a locally free sheaf on $X$. For any $x$ in $X$ there exists $x \in U \subset_{open} X$ such that </p> <p> $\mathcal{F}|_U \cong \mathcal{O}_X|_U^{(I)}$ $\ \ \ \ (\star)$. </p> <p> In particular, for each $y$ in this particular $U$, one has $\mathcal{F}_y \cong \mathcal{O}_{X,y}^{(I)}$ (which is given by the isomorphism above!!!). </p> <p>Suppose now $X$ is connected and $\mathcal{F}$ is locally free (we need this). Fix an indexing set $I$ (and I think I need to take this $I$ to be one of the indexing sets from $(\star)$ above). The properties of $\mathcal{F}$ show that the set </p> <p> $S_I = \left(x \in X : \mathcal{F}_x \cong \mathcal{O}_{X,x}^{(I)}\right)$ </p> <p>is both closed and open in $X$. We know that there exists </p> <p> $x$ in $X$ with $\mathcal{F}_x \cong \mathcal{O}_{X,x}^{(I)}$, </p> <p>we have $S_I = X$.</p> <p> In particular, $\text{rank}_{\mathcal{O}_{X,x}}(\mathcal{F}_x)$ is constant as $x$ varies in $X$. </p> http://mathoverflow.net/questions/5897/why-the-rank-of-a-locally-free-sheaves-is-well-defined/13524#13524 Answer by Martin Brandenburg for Why the rank of a locally free sheaves is well defined? Martin Brandenburg 2010-01-31T01:26:49Z 2010-01-31T01:26:49Z <p>The question is answered, but I think a comment has to be made:</p> <p>When $(X,\mathcal{O}_X)$ is an arbitrary ringed space, the sheaf $\mathcal{O}_X$ could be supported on some subset of $X$. Outside this support, the stalks are zero and don't have the invariant dimension property. The rank is only defined (as a locally constant function on $X$) when the support of $O_X$ is the whole $X$, which is the case when $(X,\mathcal{O}_X)$ is a locally ringed space such as a scheme.</p>