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What is known about the algebraic variety $V_G$ defined by $det(X_G) = 1$ where $X_G$ is the group matrix $(x_{g_ig_j^-1})$ of a finite group $G$? It is known that two finite groups having the same determinant are isomorphic.

Edit: Are there any results concernig the group structure on $V_G$?

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  • $\begingroup$ 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 $\textbf{SL}(A_1)\times \dots \times \textbf{SL}(A_r)$. $\endgroup$ Jan 14 '16 at 20:24
  • $\begingroup$ 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 . . . $\endgroup$ Jan 14 '16 at 20:28
  • $\begingroup$ 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 $\textbf{SL}(A_1)\times \dots \times \textbf{SL}(A_r)$. $\endgroup$ Jan 14 '16 at 21:41
  • $\begingroup$ @JasonStarr: you should promote your comments to an answer. $\endgroup$ Jan 16 '16 at 14:48
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Per the suggestion of Neil Strickland, here is an answer. There are some excellent notes of Keith Conrad about all of this.

If $k$ is an algebraically closed field of characteristic prime to $n$, the order of $G$, then classical results about representations of finite groups imply an isomorphism of associative, unital $k$-algebras, $$\phi:k[G]\to A_1\times \dots \times A_r,$$ between the group ring $k[G]$ is isomorphic and a product $A_1\times \dots \times A_r$ of matrix algebras $A_i \cong \text{Mat}_{n_i\times n_i}(k)$ (if $k$ is not algebraically closed, each $A_i$ may be a matrix algebra over a division ring). Here the integers $n_1\leq \dots \leq n_r$ are the ranks of the finitely many (non-isomorphic) irreducible representations of $G$ over $k$. In particular, because there is a trivial one-dimensional represenation, $n_1$ equals $1$.

For any associative, unital $k$-algebra $A$, multiplication on the right in $k[G]$ defines a "regular representation", i.e., a homomorphism of associative, unital $k$-algebras, $$ \rho_A: A \to \text{Hom}_k(A,A),\ a \mapsto (b\mapsto b\cdot a).$$ If $A$ is a finite dimensional $k$-vector space, then there is the standard determinant polynomial on $\text{Hom}_k(A,A)$. The composition with $\rho_A$ is the algebra norm, $\Theta_A$, which is some power of the reduced norm.

For $A=k[G]$, for the standard $k$-basis of $k[G]$, $(\mathbf{b}_\gamma)_{\gamma \in G}$, and for a general element $a = \sum_{\gamma} x_\gamma \mathbf{b}_\gamma$ of $k[G]$, the matrix entry of $\rho(a)$ with respect to $\mathbf{b}_\gamma$ (in the domain of $\rho(a)$) and $\mathbf{b}_\delta$ (in the target of $\rho(a)$) is $$ \rho(a)_{\delta,\gamma} = x_{\delta\cdot \gamma^{-1}}.$$ Thus the determinant $\text{det}(x_{\delta\cdot \gamma^{-1}})_{\gamma,\delta}$ equals $\Theta_G(a)$, i.e., composition of $\rho$ with the usual determinant on $\text{Hom}_k(A,A) \cong \text{Mat}_{n\times n}(k)$.

The algebra norm is compatible with isomorphism of associative, unital $k$-algebras. Thus, via $\phi$, $\Theta_G$ is the same as $\Theta_{A_1\times \dots \times A_r}$. Also the algebra norm diagonalizes, i.e., for $(a_1,\dots,a_r)\in A_1\times\dots \times A_r$, $$\Theta_{A_1\times \dots \times A_r}(a_1,\dots,a_r) = \Theta_{A_1}(a_1) \cdots \Theta_{A_r}(a_r).$$ Finally, $\Theta_{\text{Mat}_{n\times n}(k)}$ equals $\text{det}^n$, where $\text{det}$ is the usual determinant polynomial on $\text{Mat}_{n\times n}(k)$. Now consider the multiplicative morphism, $$ \text{det}:A_1\times \dots \times A_r \to k^r, \ (a_1,\dots,a_r) \mapsto (\text{det}(a_1),\dots,\text{det}(a_r)).$$ The algebra norm is the composite of this morphism with the morphism, $$k^r \mapsto k, \ (t_1,\dots,t_r) \mapsto t_1^{n_1}\cdots t_r^{n_r}.$$ Since $n_1$ equals $1$, for every $t\in \mathbb{G}_m$, the fiber of $\Theta_{k[G]}$ over $t$ is isomorphic to $\textbf{GL}(A_2)\times \dots \times \textbf{GL}(A_r)$ via the isomorphism, $$\textbf{GL}(A_2)\times \dots \times \textbf{GL}(A_r) \to \Theta_{k[G]}^{-1}(t), \ (a_2,\dots,a_r) \mapsto (t\cdot\text{det}(a_2)^{-n_2}\cdots \text{det}(a_r)^{-n_r}, a_2,\dots,a_r).$$ Thus, the fiber of $\Theta_{k[G]}$ over $1$ is isomorphic, as a $k$-group scheme, to the product $\textbf{GL}(A_2)\times \dots \times \text{GL}(A_r)$.

In particular, the structure of $\text{Ker}(\Theta_{k[G]})$ as a $k$-group scheme is equivalent to the information of the sequence of integers $(n_1=1,n_2,\dots,n_r)$ of ranks of the irreducible $k$-representations of $G$.

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