Timeline for Topology of the projective symplectic group
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2021 at 9:07 | vote | accept | Puzzled | ||
| Mar 8, 2020 at 18:41 | comment | added | Puzzled | I just checked. All the matrices in the component of degree 20 have zero determinant. Thanks for the hint. However, I still do not understand why this component shows up in the computation. It seems to me that those five equations select exactly symplectic matrices modulo scalar. | |
| Mar 8, 2020 at 18:31 | comment | added | Puzzled | The two components are both 10-dimensional. One is of degree 12 and the other of degree 20. | |
| Mar 8, 2020 at 17:41 | comment | added | YCor | About the edit: the 10-dimensional component is the right one. The computation might detect another component consisting of matrices of determinant zero. | |
| Mar 8, 2020 at 10:20 | history | edited | Puzzled | CC BY-SA 4.0 |
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| Mar 7, 2020 at 8:54 | answer | added | YCor | timeline score: 3 | |
| Mar 7, 2020 at 8:40 | comment | added | YCor | The symplectic group is connected. Otherwise there would be a special name for its unit component... | |
| Mar 7, 2020 at 8:35 | comment | added | YCor | OP: I guess that $n$ is an odd number (so $n+1$ is even)? | |
| Mar 7, 2020 at 8:33 | history | edited | YCor | CC BY-SA 4.0 |
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| Mar 6, 2020 at 20:08 | comment | added | LSpice | @MikhailBorovoi, yes, good point. I tend to think of the algebraic group $\operatorname{PGSp}_{2n}$ as "the (implicitly connected) adjoint group with root datum $\mathsf C_n$", so I glossed over that important issue. | |
| Mar 6, 2020 at 18:08 | comment | added | Mikhail Borovoi | @LSpice: Yes, any connected semisimple group is generated by unipotent subgroups. Maybe we should understand the question as follows: Is the group ${\rm PSp}(2n,{\Bbb C})$ connected? | |
| Mar 6, 2020 at 18:00 | comment | added | Mikhail Borovoi | Yes, the projective symplectic group $G={\rm PSp}(2n,{\Bbb C})$ is an irreducible algebraic variety. Indeed, it is simple as an abstract group; see e.g. E. Artin, Geometric Algebra, 1957, Theorem 5.1. On the other hand, it is easy to see that the irreducible component of the identity in $G$ is a normal subgroup of $G$, and hence, coincides with $G$. | |
| Mar 6, 2020 at 17:56 | comment | added | LSpice | Like any semisimple group over $\mathbb C$, your group (which I, concerned with rationality issues, would prefer to call PGSp) is generated by irreducible unipotent subgroups, hence is irreducible. | |
| Mar 6, 2020 at 17:37 | comment | added | Sasha | What is the point of using $n+1$ instead of $n$ (or $2n$, if you want to emphasize parity). | |
| Mar 6, 2020 at 17:30 | comment | added | Puzzled | Exactly. That's it. | |
| Mar 6, 2020 at 17:29 | comment | added | LSpice | Irreducible in the Zariski topology, I guess? | |
| Mar 6, 2020 at 17:22 | history | asked | Puzzled | CC BY-SA 4.0 |