Indexing the line bundles over a Grassmannian. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:28:24Z http://mathoverflow.net/feeds/question/7604 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7604/indexing-the-line-bundles-over-a-grassmannian Indexing the line bundles over a Grassmannian. Jean Delinez 2009-12-02T19:52:12Z 2009-12-03T17:55:10Z <p>As is well known, the line bundles over <strong>CP</strong>$^1$ are indexed by the integers. My question is how are the line bundles over <strong>CP</strong>$^n$, $n > 1$, and <strong>Gr</strong>$(n,k)$ indexed? Moreover, do there exist any other interesting classifications of line bundles over spaces (I remember something about Atiyah and elliptic curves)?</p> http://mathoverflow.net/questions/7604/indexing-the-line-bundles-over-a-grassmannian/7607#7607 Answer by Spinorbundle for Indexing the line bundles over a Grassmannian. Spinorbundle 2009-12-02T20:01:08Z 2009-12-03T09:44:12Z <p>First of all: this is from the "differential-geometric point of view"</p> <p>If you want to "classify" vector bundles chern classes are a very helpful tool. Well, it can happen that two bundles are "different" but the chern classes coincide, but if the Chern classes of a pair of vector bundles do not agree, then the vector bundles are different.</p> <p>In the case of line bundles, the only non-trivial chern class is the first chern class and it turns out that the first chern class is a complete invariant which classifies complex line bundles. (In other words: there is a bijection between the isomorphism classes of line bundles over a topological space $X$ and the elements of $H^2(X;\mathbb{Z})$)<br> But note: line bundles can have equivalent first Chern class, but different holomorphic structures. In this case, see the posts of Evgeny,Ilya...</p> <p>Smooth real line bundles are classified (in the same way) by the first Stiefel-Whitney class. As in the complex case, the only non-vanishing Stiefel-Whitney class is $w_1$. </p> <p>For references see: <a href="http://en.wikipedia.org/wiki/Line_bundle" rel="nofollow">Wikipedia: Line bundle</a><br> <a href="http://en.wikipedia.org/wiki/Chern_class#Chern_classes_of_line_bundles" rel="nofollow">Wikipedia: Chern class</a><br> <a href="http://en.wikipedia.org/wiki/Stiefel-Whitney_class" rel="nofollow">Wikipedia: Stiefel-Whitney class</a><br> <a href="http://books.google.de/books?id=5zQ9AFk1i4EC&amp;printsec=frontcover#v=onepage&amp;q=&amp;f=false" rel="nofollow">Milnor, Stasheff, Characteristic classes</a> (google books link)<br> </p> http://mathoverflow.net/questions/7604/indexing-the-line-bundles-over-a-grassmannian/7608#7608 Answer by Ilya Nikokoshev for Indexing the line bundles over a Grassmannian. Ilya Nikokoshev 2009-12-02T20:08:22Z 2009-12-02T22:09:45Z <p>If your question is about <strong>complex line bundles up to an algebraic isomorphism</strong>, such bundles are classified by <a href="http://en.wikipedia.org/wiki/Divisor%5F%28algebraic%5Fgeometry" rel="nofollow">the group of divisors</a> for any algebraic variety. There is also only one definition of the group of divisors for smooth varieties, as in your case.</p> <p>It's not hard to see that group of divisors of $\mathbb P^n$ is generated by a <strong>hyperplane</strong> $H$. The key observation is that a variety defined by a homogenous polynomial of degree $n$ is rationally equivalent to $n$ copies of $H$ (because you can deform the coefficients of polynomial so that it becomes just $x_1^n = 0$).</p> <p>Therefore the complex line bundles on any $\mathbb P^n$ are classified by integer numbers: $\mathcal O,\ \mathcal O(1),\ \mathcal O(2)$ etc.</p> <p><hr /></p> <p><strong>Update:</strong> see also the answer by Evgeny who explains the same things as me, but better.</p> <p><hr /></p> <p>Grassmannians have a cell decomposition into <strong><a href="http://en.wikipedia.org/wiki/Schubert%5Fvariety" rel="nofollow">Schubert_cells</a></strong> where each cell is just an affine space $\mathbb A^l$. This particularly simple structure means that divisors up to rational equivalence will be the same as algebraic cycles up to rational equivalence and the same as topological $H^2$. Moreover, it will be generated by cells in complex codimension one.</p> <p>Now it's an exercise to find out from the definition of Schubert cells that there's <strong>only 1 cell</strong> with this property. To see it, fix a flag $(F_i), F_i\in V$. Then the cell is given by the conditions $$\text{dim}\, V\cap F_{n-k-1+i} = i \quad \text{for}\ 0\le i\le k$$ for subspaces $V$ of dimension $k$.</p> <p>Therefore the bundles on $Gr(n, k)$ are <strong>also numbered by integers</strong>.</p> http://mathoverflow.net/questions/7604/indexing-the-line-bundles-over-a-grassmannian/7612#7612 Answer by Leonid Positselski for Indexing the line bundles over a Grassmannian. Leonid Positselski 2009-12-02T20:59:01Z 2009-12-03T09:50:36Z <p>Algebraic line bundles over generalized flag varieties of semisimple groups should be classifiable by characters of the Cartan subgroup of the group modulo the Cartan subgroup of the semisimple part of the parabolic subgroup. This is because all line bundles over a generalized flag variety are equivariant with respect to the semisimple group action, so they correspond to the characters of the parabolic subgroup (see David's answer). For a Grassmannian Gr(n,k), my guess is that line bundles are indexed by the integers, the generator line bundle being the k-th exterior power of the tautological k-dimensional bundle. I hope better educated people would come up with references and/or corrections to the above.</p> http://mathoverflow.net/questions/7604/indexing-the-line-bundles-over-a-grassmannian/7614#7614 Answer by Evgeny Shinder for Indexing the line bundles over a Grassmannian. Evgeny Shinder 2009-12-02T21:01:41Z 2009-12-03T17:55:10Z <p>Algebraic line bundles on a smooth variety $X$ are classified by the Picard group $Pic(X) = H^1(X, \mathbf O_X^*)$. This is an exercise in Hartshorne's book, basically every line bundle is mapped to it's gluing cocycle. The group $Pic(X)$ is also equal to the group $CH^1(X)$ of divisors modulo rational equivalence. The map sends a line bundle to it's first Chern class.</p> <p>Now for a projective space Picard group is $\mathbf Z$ generated by $\mathbf O(1)$. Picard group of a Grassmannian is also $\mathbf Z$. I believe that the generator is a pullback of $\mathbf O(1)$ for a Plucker embedding $Gr(n,k) \subset \mathbf P^N$.</p> <p>One can prove it as follows: both $\mathbf P^n$ and $Gr(n,k)$ are algebraically cellular, meaning that they consist of pieces isomorphic to affine spaces: for $\mathbf P^n$ this is obvious, and for $Gr(n,k)$ the cells are Schubert cells.</p> <p>For such varieties Chow groups coincide with cohomology groups and are generated by cells (it's like computing cohomology of CW complex without cells of odd dimension). Since there is only one cell of complex codimension 1 for projective spaces and Grassmannians, we get $Pic(X) = H^2(X, \mathbf Z) = \mathbf Z$.</p> <p>EDIT: </p> <ul> <li><p>Actually, a projective space IS a Grassmannian. :)</p></li> <li><p>As other people commented, the generator of the $Pic(Gr(n,k))$ is the n'th wedge power of the canonical vector bundle of rank n.</p></li> </ul> http://mathoverflow.net/questions/7604/indexing-the-line-bundles-over-a-grassmannian/7640#7640 Answer by David Bar Moshe for Indexing the line bundles over a Grassmannian. David Bar Moshe 2009-12-03T05:17:48Z 2009-12-03T05:17:48Z <p>As a compact homogeneous Kahler manifold the Grassmannian is a coset space of the general linear group by a maximal parabolic subgroup. Homogeneous line bundles over the Grassmannian are in a one to one correspondence with the character representations of the maximal parabolic, which are indexed by one integer. According to the Bott-Borel-Weil theorem, the space of holomorphic sections of the line bundle carries an irreducible representation of the special unitary group SU(n). In the case of the Grassmannian Gr(n,k), these representations correpond to Young tableaux with k rows. The integer characterizing the line bundle is just the number of columns in this tableaux. </p>