2
$\begingroup$

Let $G$ be a finite group, $F$ a field of characteristic $0$ or of relatively prime to $|G|$.

Let $\rho:G\rightarrow {\rm GL}_n(F)$ be an irreducible representation of $G$.

Extend $\rho$ to algebra homomorphism $\rho_1:F[G]\rightarrow M_n(F)$.

(*) If $F$ is algebraically closed then $\rho_1$ is surjective..

(**) If $F$ is not algebraically closed, then $\rho_1(F[G])$ is not necessarily $M_n(F)$ but it is certainly a subspace of $M_n(F)$.

Example: Let $\rho:\langle x:x^4\rangle\rightarrow {\rm GL}_2(\mathbb{Q})$ be given by $$\rho(x)=\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}.$$ We can see that $\rho(1)=I, \rho(x^2)=-I$ and $\rho(x^3)=-\rho(x)$. Thus, among the images of $\rho$,only $\rho(1)$ and $\rho(x)$ are independent in $M_2(\mathbb{Q})$, and so spans a subspace of dimension $2$ in $M_2(\mathbb{Q})$.

This forced me to think about: What this $2$ has to do with $G=\langle x:x^4=1\rangle$?

To state question:

If $\rho_1(F[G])$ is proper subspace of $M_n(F)$, how is its dimension is related to group $G$? Is this dimension of image of $F[G]$ well studied for irreducible representations?

$\endgroup$
1
$\begingroup$

In this case we have the dimension of the second $4$th cyclotomic polynomial $\Phi_4(X)=X^2+1$. For cyclic groups of order $n$, one can construct a representation with dimension is the degree of the $k$th cyclotomic polynomial for any $k|n$. This follows from factoring $$X^n-1=\prod_{k|n}\Phi_k(X).$$

In the case at hand, observe that $$\mathbb{Q}G\cong\mathbb{Q}[X]/(X^4-1)\cong\mathbb{Q}[X]/(X-1)\oplus\mathbb{Q}[X]/(X+1)\oplus\mathbb{Q}[X]/(X^2+1),$$ where the last isomorphism is the Chinese remainder theorem. The representation in question is obtained by letting $x\in G$ act on the last factor by multipication by $X$. It is two dimensional because $\deg(X^2+1)=2$. (The other two representations of this group are one dimensional with $x$ acting $\pm1$ and come from the other two factors.)

Observe that when $G=\langle x\mid x^3=1\rangle$, then $$\mathbb{Q}G\cong\mathbb{Q}[X]/(X^3-1)\cong\mathbb{Q}[X]/(X-1)\oplus\mathbb{Q}[X]/(X^2+X+1).$$ Let $G$ act on the last factor $V=\mathbb{Q}[X]/(X^2+X+1)$ so that the generator acts by left multiplication by $X$. A basis for $V$ is $\{1,X\}$ and $X^2=-X-1$, so the matrix for left multiplication by $X$ is $$\rho(x)= \begin{bmatrix}0&-1\\1&-1\end{bmatrix}.$$ Sure enough, $$\rho(x^2)=\begin{bmatrix}-1&1\\-1&0\end{bmatrix}=-\rho(x)-\rho(1),$$ and $\rho(x^3)=I$. Hence, we have obtained a two dimensional representation of $G$, the dimension corresponding to the degree of $\Phi_3(X)=X^2+X+1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.