Let $\lambda_1, \ldots, \lambda_n$ be roots of unity in $\mathbb{C}$. Can we always find a finite group $G$, an element $g \in G$ and an irreducible representation $\rho: G \to \mathrm{GL}_n(\mathbb{C})$ such that $\rho(g)$ has eigenvalues $\lambda_1,\ldots,\lambda_n$?

What if $\mathbb{C}$ is replaced by another algebraically closed field? What if $\mathbb{C}$ is replaced by another field, not necessarily algebraically closed, and "irreducible" replaced by "indecomposable"?

Thanks in advance!

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    $\begingroup$ I have no idea why two people voted to close; this seems like a fine question to me. $\endgroup$ – Qiaochu Yuan Dec 31 '17 at 5:07
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    $\begingroup$ @Qiaochu I agree. It seems like a very natural question to me. And the answer does not seem clear off the top of my head. $\endgroup$ – Tobias Kildetoft Dec 31 '17 at 8:56
  • $\begingroup$ I don't understand the motivation to vote to close either. I did see the answer right away, because I have used John's trick earlier. Happy New Year to y'all! $\endgroup$ – Jyrki Lahtonen Jan 1 '18 at 0:30

Let $\lambda_{1},\ldots,\lambda_{n}$ be roots of unity in $\mathbb{C}$ and let $g$ be the diagonal matrix with $\lambda_{i}$ in the $(i,i)$ entry. Let $H$ be the subgroup of monomial matrices whose nonzero entries are always $\pm 1$ (so a matrix in $H$ has exactly one nonzero entry in each row and each column, and this entry is either $1$ or $-1$). Finally, let $G$ be the group generated by $g$ and $H$ and let $\rho:G\to GL_{n}(\mathbb{C})$ be the embedding representation.

First note that $G$ is finite. To see this, let $m$ denote the order of $g$ and let $X$ be the group of diagonal matrices with $m^{\text{th}}$ roots of unity along the diagonal. The group $H$ normalizes $X$ and $g\in X$, so $G\leq HX$. Since $H$ and $X$ are both finite, so is $HX$; consequently, $G$ is finite too.

The representation $\rho$ is irreducible because its restriction to $H$ is irreducible---see Jyrki Lahtonen's comment below. Here's another reason why the restriction of $\rho$ to $H$ is irreducible: a useful theorem of Burnside states that a representation $\varphi:G\to GL_{n}(\mathbb{C})$ of a finite group $G$ is irreducible if and only if the associated algebra homomorphism $\varphi:\mathbb{C}G\to M_{n}(\mathbb{C})$ from the group ring into the matrix algebra is surjective (the associated algebra homomorphism extends the original group representation, which explains the abuse of notation). Now, the restriction of $\rho$ to $H$ is just the embedding of $H$ in $GL_{n}(\mathbb{C})$. In order to apply the theorem of Burnside and deduce that this restriction is an irreducible representation, we need only show that the algebra generated by $H$ in $M_{n}(\mathbb{C})$ is the entire matrix algebra.

Claim: The $\mathbb{C}$-algebra generated by $H$ contains the set $S$ of matrices that have exactly one nonzero entry.

Since every matrix in $M_{n}(\mathbb{C})$ is a linear combination of matrices in $S$, if the claim is true then the algebra generated by $H$ is the whole matrix algebra. So then: let $A\in S$, and say that the unique nonzero entry of $A$ is $a$ and lives in the $(i,j)$ entry. Let $I^{(i,j)}$ be the identity matrix with columns $i$ and $j$ swapped. Note that $I^{(i,j)}\in H$ and has a $1$ in the $(i,j)$ entry. Let $J$ be the diagonal matrix with $-1$'s along the diagonal except in the $(i,i)$ entry, where we place a $1$. Let $J^{(i,j)}$ be the matrix $J$ with columns $i$ and $j$ swapped. Note that $J^{(i,j)}\in H$ and that this matrix looks like $I^{(i,j)}$ except that all entries besides the $(i,j)^{\text{th}}$ have been multiplied by $-1$. Finally, the claim follows because \begin{equation*} A=\dfrac{a}{2}\left(I^{(i,j)}+J^{(i,j)}\right) \end{equation*} and the term on the right is an element of the $\mathbb{C}$-algebra generated by $H$. (I prefer Jyrki Lahtonen's explanation, but sometimes it's nice to have more than one for reassurance.)

By construction, the eigenvalues of $\rho(g)$ are precisely $\lambda_{1},\ldots,\lambda_{n}$.

I believe the same construction or a similar one would work over any field, but I don't have time to check the details right now.

  • $\begingroup$ Thanks for the answer! Could you please explain or provide reference for why the restriction to $H$ is irreducible in general? $\endgroup$ – Cameron Zhao Dec 31 '17 at 22:24
  • $\begingroup$ @CameronZhao The group John described has as a subgroup the group of permutation matrices (=monomial matrices with non-zero entries all equal to $1$). That group is $\simeq S_n$, and this representation is the usual $n$-dimensional rep of $S_n$. The only submodules of that $S_n$-module are the zero sum subspace (of dimension $n-1$) and the 1-dimensional rep spanned by $(1,1,\ldots,1)$ (the orthogonal complement of the other). Neither of those is a subrep for $H$ though, so we can conclude. May be there is an even easier way, but that's how I would do this. +1, of course $\endgroup$ – Jyrki Lahtonen Jan 1 '18 at 0:26
  • $\begingroup$ Hi Cameron. I meant to include more details as to why that representation is irreducible but was pressed for time. Jyrki has provided a nice explanation. I'm going to add one more in an edit to my answer. $\endgroup$ – John McHugh Jan 1 '18 at 0:55
  • $\begingroup$ @JyrkiLahtonen thanks for the explanation! $\endgroup$ – Cameron Zhao Jan 1 '18 at 2:08
  • $\begingroup$ @JohnMcHugh Thanks! Does this theorem of Burnside work over positive characteristic? If it does, then this method works for all characteristic $p \neq 2$; and for characteristic $2$ I believe we can change $H$ to be the group consisting of monomial matrices with entries are $m$-th roots of unity $(m \neq 2)$. $\endgroup$ – Cameron Zhao Jan 1 '18 at 2:16

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