# Simultaneous “Monomialization” of a set of operators.

We all know that a set of commuting diagonalizable matrices can be simultaneously put in diagonal form. My general question is:

Under what conditions can a set of (diagonalizable) matrices be simultaneously put in monomial form?

That is, when can find a basis with respect to which a collection of operators $\{A_i\}$ take the form $D_iP_i$ where $D_i$ is a diagonal matrix and $P_i$ is a permutation matrix? Clearly they must be diagonalizable, but I would like sufficient conditions.

For example:

Is it enough if (a) there exist $n_i$ so that $A_i^{n_i}$ pairwise commute and are diagonalizable? Or (b) if the set $\{A_i\}$ is closed under conjugation, i.e. $A_iA_jA_i^{-1}=A_{i(j)}$ (which obviously implies (a))?

Edit: As Will Sawin points out, this is not enough as stated, so let me add another condition to (b): suppose that $|\{A_i\}|\leq n$ where $A_i\in \mathbb{C}^{n\times n}$, so the number of generators is less than the dimension of the representation.

As you might guess this has something to do with representations of knot groups, but I am interesting in the general problem.

-

$(a)$ and $(b)$ are both true for finite group representations, but some of them, such as the $4$-dimensional representation of $A_5$, cannot be put in this form.

For $(a)$, the group generated by the matrices need not even be virtually abelian. Take $A_1 = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$ and $A_2 = \left(\begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array}\right)$. Take $n_1=4$ and $n_2=3$. Then $A_1^{n_1}=A_2^{n_2}=I$, but the group generated by $A_1$ and $A_2$ is $SL_2(\mathbb Z)$, not virtually abelian. Not sure about $b$.

I can't think of a better condition.

-
Thanks Will! Including (a) was an afterthought that should have been removed upon further reflection. – Eric Rowell Nov 13 '12 at 16:22
One potential problem with the new condition is that every finite group representation satisfies it when you add enough trivial summands. I can't immediately think of an example of a finite group representation that cannot be put in this form after adding enough trivial summands, but it seems likely that one exists. – Will Sawin Nov 13 '12 at 17:04
I think the automorphism group of a nice lattice, like the $E_8$ lattice or the Leech lattice, with the standard representation, should give a counterexample. – Will Sawin Nov 13 '12 at 17:42

As Will Sawin suggests, the answer to the last version of the question is negative. If we tak $G$ to be a non-Abelian finite simple group which is not a doubly transitive permutation group, and $\chi$ to be a non-trivial complex irreducible character of last degree o $G,$ then it is impossible to write $\chi$ + any multiple of the trivial character as a sum of characters each induced from linear characters of (not necessarily proper) subgroups. For if $\lambda$ is a non-trivial linear character of a proper subgroup $H$ of $G,$ then ${\rm Ind}_{H}^{G}(\lambda)$ does not contain the trivial character by Frobenius reciprocity. Neither can it be $\chi$, for otherwise ${\rm Ind}_{H}^{G}(1)$ would have a trivial constituent, and at least one non-trivial irreducible constituent $\mu$. But then $\mu(1) < \chi(1)$, contrary to the choice of $\chi.$ On the other hand,if $H$ is a proper subgroup of $G$, the permutation character ${\rm Ind}_{H}^{G}(1)$ only contains the trivial character once, but it can't be $1 + \chi,$ for otherwise $G$ would be a doubly transitive permutation character on the cosets of $H$ in $G,$ contrary to the choice of $G.$

-

There seems to be a stronger condition than your (b) (the original one, with a closed set), i.e.:

Two invertible diagonalizable matrices $X, Y$ are simultaneously monomializable (in your sense) iff $Y X Y^{-1}$ and $X$ commute (i.e. are diagonalizable in the same basis) and have the same spectrum (including degeneracies).

"$\Rightarrow$"

Let $X = C A C^{-1}$ and $Y=CPBC^{-1}$, where $A$ and $B$ are diagonal matrices and $P$ is a permutation matrix.

Then $$Y X Y^{-1} = C PBAB^{-1}P^{-1}C^{-1} = CPAP^{-1}C^{-1},$$ so both parts are diagonal in the basis $C$, with permuted diagonal elements.

"$\Leftarrow$"

Let $X=CAC^{-1}$ and $Y X Y^{-1} = C P A P^{-1} C^{-1}$, again, for $A$ diagonal and $P$ - a permutation matrix.

Then $$C^{-1}YCAC^{-1}Y^{-1}C=PAP^{-1}.$$ Now, let's define $B = P^{-1}C^{-1}YC$, which converts the above in the commutation condition $BAB^{-1}=A$. Hence*), $B$ is diagonal (as $A$ is diagonal).

So, we have $PB = C^{-1}YC$, what we wanted to show.

*) There is one fragile part of which I'm aware of: if some eigenvalues are degenerate, then there is freedom in choice of $C$. So then $B$ needs not to be diagonal, just there is a matrix $D$ (non diagonal), commuting with $A$, such that $D^{-1} BD$ is diagonal.

(I'm posting it anyway, as perhaps there is a simple fix of that.)

-
In "$\Rightarrow$" you assume there is a basis with respect to which $X$ is diagonal and $Y$ is a monomial matrix. This is stronger than "simultaneously monomializable." Indeed, take $X,Y$ to be the $3\times 3$ permutation matrices corresponding to $(1\/2)$ and $(1\/2\/3)$. Then they are monomial in the standard basis, but $YXY^-1$ corresponds to $(2\/3)$, which does not commute with $(1\/2)$. – Eric Rowell Nov 20 '12 at 17:28
@Eric Thanks. (I see, I had in mind a different, stronger condition.) – Piotr Migdal Nov 21 '12 at 16:05