Timeline for Grassmannians of planes isotropic with respect to general tensors
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 26, 2018 at 12:41 | comment | added | Robert Bryant | @TobiasShin: Sorry, your correction didn't show up in my feed until after I posted my reply to your question. | |
| Nov 26, 2018 at 1:49 | comment | added | Tobias Shin | I explained it above, there was a typo. My mistake. | |
| Nov 26, 2018 at 0:47 | comment | added | Robert Bryant | @TobiasShin: I'm not familiar with this result, but you must be misunderstanding it. If what you claim were true, then an elliptic curve could not appear embedded in a $6$-dimensional projective space (which is clearly simply connected, compact, and irreducible), and that is obviously false, since an elliptic curve can appear embedded in a projective space of any dimension at least 2. Until you explain why this result doesn't apply in this simpler and obvious case of a non-simply connected curve in a simply connected compact variety of high dimension, I don't really have anything to do. | |
| Nov 25, 2018 at 23:47 | comment | added | Tobias Shin | Ack sorry I think there might be a typo... they might mean surjective onto $\pi_1(X)$ | |
| Nov 25, 2018 at 21:34 | comment | added | Tobias Shin | how do you reconcile the above with the following theorem (in Gunning and Rossi's Analytic Functions of Several Complex Variables): if $X$ is an irreducible complex variety whose universal cover $\tilde{X}$ is an irreducible analytic space, then for any closed analytic subspace $Z$ inside $X$, we have $\pi_1(X-Z)$ surjects onto $\pi_1(Z)$. It seems like in the above situation, your $X$ is the Grassmannian of 2-planes in 5-space, and your $Z$ is of codimension 5. But removing a submanifold of real codimension at least 3 should not change the fundamental group. | |
| Nov 21, 2018 at 21:28 | comment | added | Robert Bryant | @TobiasShin: I believe that the answer to the question over $\mathbb{C}$ is 'no'. The first example I can think of is the generic case when $M$ has dimension $10$ and $D$ has dimension $5$. In that case, generically, there will be no $3$-dimensional integral elements, and, at each point, the space of $2$-dimensional integral elements will have dimension $1$. In fact, it will be a smooth curve of degree $5$ lying in a projective space of dimension $4$ and it will have genus $1$, so it will be an elliptic curve, which is topologically a (real) $2$-torus. Hence, it will not be simply connected. | |
| Nov 21, 2018 at 19:33 | comment | added | Tobias Shin | Can one say anything about the fundamental group in the above case? eg, if we are considering complex planes, is it simply connected? I would imagine so since any reasonable Schubert cell decomposition should only involve even dimensional cells ... | |
| Nov 20, 2018 at 17:11 | vote | accept | Tobias Shin | ||
| Nov 20, 2018 at 9:52 | history | answered | Robert Bryant | CC BY-SA 4.0 |