Timeline for tannakian description of vector bundles
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2017 at 6:23 | answer | added | Artur Jackson | timeline score: 3 | |
| Mar 27, 2017 at 21:45 | comment | added | user36254 | One nice thing you can do is use Borel-Weil-Bott to realize all the vector bundles coming from these Schur functors geometrically. Specifically, you can take the bundle of complete flag varieties associated to $E$. This will have a bunch of tautological line bundles on it. If you push forward these line bundles to $S$, you will get all the vector bundles coming from Schur functors applied to $E$. | |
| Mar 27, 2017 at 18:06 | comment | added | jojo | Ah ! In fact it is not hard to see that Remy's suggestion works. Just use the fact that every irreducible representation of GL_n can be obtained as the image by a schur functor of the standard representation. Now it is easy to see that a tensor exact functor has to commute with schur functors so you win : $\eta$ is indeed determined by it's image on the standard representation. | |
| Mar 25, 2017 at 23:02 | comment | added | jojo | Thanks guys for your comments. Remy I really like your suggestion. Does it really work ? I know that the category of representations of GL_n is generated by the standard representation. So I guess this means that if we now $\eta(V)$ we should now $\eta(W)$ for most $W$ but it's not clear that it works for all no ? I mean since $V$ is irreducible we don't have sub-representations which is useful but I'm not sure we really can get everything. To be (maybe) clearer can we really get the value of $\eta$ on representations inside the tensors ? | |
| Mar 25, 2017 at 17:03 | answer | added | Qiaochu Yuan | timeline score: 4 | |
| Mar 25, 2017 at 5:00 | comment | added | მამუკა ჯიბლაძე | @R.vanDobbendeBruyn Thanks, that was helpful! I now realized - in fact $P$ is (among other things) the "universal isomorphiser" of $G$ with $GL_n$. So, there you get $(\text{lift of $E$})\otimes_{(\text{lift of $G$})\cong GL_n}V$, and then you must push it down back over $S$ and that is your $\eta(V)$. It becomes even more burning how to get rid of $P$, after all it is a canonical thing - the object of isomorphisms, it should be eliminable... | |
| Mar 24, 2017 at 22:12 | comment | added | R. van Dobben de Bruyn | @მამუკაჯიბლაძე: It is not true that $G = GL_{n, S}$. For example, if $\mathcal E = \mathcal O(1) \oplus \mathcal O$ on $S = \mathbb P^1$, then $\mathscr End_S(\mathcal E) = \mathscr Hom(\mathcal O(1) \oplus \mathcal O, \mathcal O(1) \oplus \mathcal O) = \mathcal O \oplus \mathcal O(1) \oplus \mathcal O(-1) \oplus \mathcal O$, which is not the trivial bundle. Then $\mathscr Aut_S(\mathcal E)$ is some open subfunctor, and it's not hard to see that it cannot be globally isomorphic to $GL_2$ (e.g. consider global sections). | |
| Mar 24, 2017 at 9:19 | comment | added | მამუკა ჯიბლაძე | Unfortunately I lack knowledge, and could not find, the basic object related to this: the scheme $G:=\operatorname{Aut}_S(E)$ over $S$. It must be locally isomorphic to $GL_n$; note that $P$ is in fact a $G$-$GL_n$-bitorsor. Maybe $G$ is $GL_n$ over $S$? In that case, the answer should be $\eta(V)=E\otimes_GV$ | |
| Mar 24, 2017 at 5:53 | comment | added | R. van Dobben de Bruyn | A half answer is that any finite-dimensional representation of $GL_n$ can be constructed inside suitable tensor powers of the defining representation $V$. Clearly $V^{\otimes d}$ corresponds to the vector bundle $\mathcal E^{\otimes d}$, and it might be possible (?) to somehow deduce the general case from this. But this is a bit disappointing because it doesn't give a nice formula, but rather relies on explicit knowledge of the representation theory of $GL_n$. | |
| Mar 23, 2017 at 19:12 | review | First posts | |||
| Mar 23, 2017 at 19:23 | |||||
| Mar 23, 2017 at 19:09 | history | asked | jojo | CC BY-SA 3.0 |