Timeline for How should we define $\mathrm{PSL}_2$ of a Clifford group?
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17 events
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| Feb 13, 2017 at 23:09 | vote | accept | j0equ1nn | ||
| Feb 13, 2017 at 23:09 | answer | added | j0equ1nn | timeline score: 0 | |
| Feb 9, 2017 at 20:30 | history | edited | j0equ1nn | CC BY-SA 3.0 |
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| Feb 9, 2017 at 8:45 | history | edited | j0equ1nn | CC BY-SA 3.0 |
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| Feb 9, 2017 at 8:05 | history | edited | j0equ1nn | CC BY-SA 3.0 |
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| Feb 9, 2017 at 7:35 | history | edited | j0equ1nn | CC BY-SA 3.0 |
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| Feb 9, 2017 at 7:19 | answer | added | j0equ1nn | timeline score: 0 | |
| Feb 7, 2017 at 11:30 | comment | added | Vít Tuček | @j0equ1nn Yeah, on a second look it seems that it is me who's off by one. Sorry. | |
| Feb 6, 2017 at 21:04 | comment | added | j0equ1nn | @VítTuček I'm looking at "Möbius transformations and Clifford numbers" by Ahlfors (1984) and for him the $n$ in $\mathscr{C}_n$ starts at $1$ rather than $0$, as you suggested. On Wikipedia though Ahlfors' $\mathscr{C}_1$ is their $\mathscr{C}_{0,0}(\mathbb{R})$. I think there are a lot of disagreeing conventions going on here in general, and that might be the whole cause of the confusion. | |
| Feb 5, 2017 at 23:21 | comment | added | j0equ1nn | @VítTuček No, the theorem is that $\mathrm{PSL}_2(C_n)$ is the group of Möbius transformations of $\mathbb{R}^{n+1}$, which is also $\mathrm{Iso}^+(\mathfrak{H}^{n+2})$. For example, $\mathrm{PSL}_2(\mathbb{R})$ is the group of Möbius transformations of $\mathbb{R}$, or equivalently, is isomorphic to $\mathrm{Iso}^+(\mathfrak{H}^{2})$. | |
| Feb 5, 2017 at 22:48 | comment | added | Vít Tuček | @j0equ1nn If $C_0 = \mathbb{R}$ then it shouldn't act on elements of $\mathbb{R}^0$, should it? You are right except the Clifford group does not contain zero. But this minute detail shouldn't matter in these issues. | |
| Feb 4, 2017 at 23:05 | comment | added | j0equ1nn | @VítTuček I think the dimension indices are right, remember that $C_0=\mathbb{R}$ and $C_1=\mathbb{C}$. Also, the Clifford groups is the same as the Clifford algebra in the case of $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$ (but no other time). Vahlen matrices do seem to be what I'm looking for though, I will look into that, thanks. | |
| Feb 4, 2017 at 21:20 | comment | added | Vít Tuček | @Venkataramana I'm no expert but the literature suggest that Vahlen matrices do correspond to Möbius transformations. But there is some extra condition on the entries. | |
| Feb 4, 2017 at 21:18 | comment | added | Vít Tuček | It seems to me that your dimension indices are wrong. According to Wikipedia, the Clifford group for $(\mathbb{R}^n, -\|x\|^2)$ gives the Möbius transformations over $\mathbb{R}^n$. In your definition of Clifford group I think you have off by one error. Also, Clifford group is certainly not the same object as the Clifford algebra, so writing $PSL_2(C_2) \simeq PSL_2(\mathbb{H})$ seems suspicious. | |
| Feb 4, 2017 at 12:13 | comment | added | Venkataramana | I have not thought everything through, but it seems that this holds only in low dimensions. If you tensor with $\mathbb{C}$, then the Clifford algebra $C^*$ is just the matrix algebra (or two copies of it). Hence the units form a group of type $A$. Same for $SL_2(C^*)$, which is a group of type $A$. This it cannot be of type $B$ or $D$ (except in low dimensions) | |
| Feb 3, 2017 at 23:10 | history | edited | j0equ1nn | CC BY-SA 3.0 |
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| Feb 3, 2017 at 21:56 | history | asked | j0equ1nn | CC BY-SA 3.0 |