Timeline for Connectedness of the linear algebraic group SO_n
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2017 at 22:39 | answer | added | fherzig | timeline score: 5 | |
| Jun 8, 2012 at 15:03 | comment | added | Robert Bryant | @Tom: Thanks for the explanation. I understand your notation now. | |
| Jun 7, 2012 at 8:21 | history | edited | Tom De Medts | CC BY-SA 3.0 |
added a few sentences about connectedness
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| Jun 7, 2012 at 8:10 | comment | added | Tom De Medts | @Robert: Note that I'm referring to $\mathsf{SO}_2$, and not to its group of $k$-rational points $\mathsf{SO}_2(k)$, i.e. I'm considering the linear algebraic groups as group functors. I'll add a few sentences in the question to make this clear. | |
| Jun 6, 2012 at 18:53 | comment | added | Robert Bryant | @Tom: I have to confess that I'm still puzzled. It appears that you are only avoiding characteristic $2$, so that you would appear to be claiming that $\mathsf{SO}_2(k)$ is connected in the Zariski topology even when $k$ is finite, and I don't understand this. After all, when $k=\mathbb{Z}_3$, doesn't ${\mathsf{SO}}_2(k)$ have $4$ elements and isn't the complement of any element open? Why then is this group not disconnected? Looking at the answers below, it seems that you at least need $k$ to be infinite for those kinds of arguments to work. | |
| Jun 6, 2012 at 8:16 | vote | accept | Tom De Medts | ||
| Jun 6, 2012 at 8:10 | history | edited | Tom De Medts | CC BY-SA 3.0 |
added 208 characters in body
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| Jun 6, 2012 at 8:04 | comment | added | Tom De Medts | @Robert: As Igor points out, I meant indeed connected in the Zariski topology. @George: You are right, that argument is only valid in characteristic 0. I will slightly edit the question accordingly. | |
| Jun 6, 2012 at 0:28 | comment | added | Jim Humphreys | @Robert, George: I think Tom was starting out in characteristic 0, which is one reason for my attempted clarifications below. | |
| Jun 5, 2012 at 23:39 | comment | added | George McNinch | I'm slightly confused by your second paragraph: the finite group $\mathbf{Z}/p\mathbf{Z}$ may be viewed as a linear algebraic group. Over a field of char. p, it is generated by unipotent elements. But it is not connected. | |
| Jun 5, 2012 at 23:00 | answer | added | Jim Humphreys | timeline score: 13 | |
| Jun 5, 2012 at 20:10 | comment | added | Igor Rivin | Presumably the OP means connected as a scheme. | |
| Jun 5, 2012 at 18:48 | comment | added | Robert Bryant | What do you mean by 'connected' when $k$ is not $\mathbb{R}$ or $\mathbb{C}$? (I'm not doubting that you have a definition; you wouldn't be asking the question if you didn't. I just don't know exactly what you are trying to prove, since I'm having trouble figuring out what you'd mean by 'connected' in the case where $k$ is a finite field.) | |
| Jun 5, 2012 at 18:45 | answer | added | Will Sawin | timeline score: 19 | |
| Jun 5, 2012 at 17:54 | comment | added | Igor Rivin | I am a little confused by the question: Skew matrices form an affine space, and your set of nonsingular matrices is an affine hypersurface (which is presumably easy to show is irreducible). $O_n$ is a much more complicated variety, so this Cayley transform trick seems to have simplified your life. | |
| Jun 5, 2012 at 15:44 | history | edited | Tom De Medts | CC BY-SA 3.0 |
added characteristic assumption
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| Jun 5, 2012 at 15:43 | comment | added | Tom De Medts | @Jim: Thanks for your commments! The version of the Cayley transform that I've used is an involutory formula, so it doesn't really matter whether it is from the skew-symmetric matrices to the special orthogonal matrices or conversely, does it? I will edit my question to make clear I'm assuming $\operatorname{char}(k) \neq 2$ here. | |
| Jun 5, 2012 at 15:29 | comment | added | Jim Humphreys | Your formulation of the Cayley transform seems to be backward, since usually A is an arbitrary skew-symmetric matrix; adding the identity matrix automatically yields a special orthogonal matrix. It's also important to specify which algebraically closed field you work over, since the connectedness theorem for algebraic groups is true even in characteristic 2. | |
| Jun 5, 2012 at 15:07 | history | asked | Tom De Medts | CC BY-SA 3.0 |