Timeline for What is known about the algebraic variety defined by the group determinant?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 16, 2016 at 16:48 | vote | accept | CommunityBot | ||
| Jan 16, 2016 at 15:29 | answer | added | Jason Starr | timeline score: 4 | |
| Jan 16, 2016 at 14:48 | comment | added | Neil Strickland | @JasonStarr: you should promote your comments to an answer. | |
| Jan 16, 2016 at 13:35 | history | edited | user6671 | CC BY-SA 3.0 |
added 81 characters in body
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| Jan 14, 2016 at 21:41 | comment | added | Jason Starr | So the correct statement is that your variety $X_G$ is isomorphic to the variety of $r$-tuples $(a_1,\dots,a_r)\in A_1\times \dots \times A_r$ such that $\text{det}_{A_1}(a_1)^{n_1}\cdot \dots \cdot \text{det}_{A_r}(a_r)^{n_r}$ equals $1$. There is a morphism $X_G \to \mathbb{G}_m^r$ sending $(a_1,\dots,a_r)$ to $(\text{det}_{A_1}(a_1),\dots,\text{det}_{A_r}(a_r))$. The image is the set of $(t_1,\dots,t_r)$ such that $t_1^{n_1}\cdot \dots \cdot t_r^{n_r} = 1$. The fibers are each isomorphic to $\textbf{SL}(A_1)\times \dots \times \textbf{SL}(A_r)$. | |
| Jan 14, 2016 at 20:28 | comment | added | Jason Starr | There is a mistake in what I wrote. If $A_i$ is isomorphic to $\text{Mat}_{n_i\times n_i}(k)$, then the determinant on $A_i$ is not the usual determinant $\text{det}_{A_i}$, it is $\text{det}_{A_i}^{n_i}$. So probably there should be some product of copies of groups of unity, $\mu_{n_1}\times \dots \mu_{n_r}$ in there somewhere . . . | |
| Jan 14, 2016 at 20:24 | comment | added | Jason Starr | There is a "regular representation" morphism of associative, unital $k$-algebras, $\rho:k[G] \to \text{Hom}_k(k[G],k[G])$. Your determinant is the composition of $\rho$ with the determinant on $\text{Hom}_k(k[G],k[G]) \cong \text{Mat}_{n\times n}(k)$, $ n = \#G$. If memory serves, so long as $k$ is algebraically closed of characteristic prime to $n$, then $k[G]$ is isomorphic to a product of matrix algebras, $A_1\times \dots \times A_r$. So your variety fibers over $\mathbb{G}_m^{r-1}$ with fibers isomorphic to $\textbf{SL}(A_1)\times \dots \times \textbf{SL}(A_r)$. | |
| Jan 14, 2016 at 18:59 | history | asked | user6671 | CC BY-SA 3.0 |