Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

First question: For a semisimple invertible $n \times n$ matrix with entries over a field K, its characteristic polynomial completely describes the similarity class of the matrix. For non-semisimple elements, the characteristic polynomial is no longer a complete description of the similarity class, but there exists the rational canonical form, which is a complete invariant.

I'm interested in whether we can find invariants that help determine the similarity class of elements of $GL(n,\mathbb{Z})$ (where $\mathbb{Z}$ is the ring of integers) under the $GL(n,\mathbb{Z})$-conjugation action. Clearly, the invariants for similarity class over $\mathbb{Q}$, but there are semisimple matrices in the same similarity class over $\mathbb{Q}$ but not conjugate over $\mathbb{Z}$:

1 0

0 -1

and

0 1

1 0

These are clearly conjugate in $GL(2,\mathbb{Q})$ but not in $GL(2,\mathbb{Z})$. To see why they aren't conjugate in the latter, note that on reducing mod 2, the first matrix becomes the identity matrix and the second matrix becomes a non-identity matrix, so they cannot be conjugate mod 2, and hence cannot be conjugate in $GL(2,\mathbb{Z})$.

Second question: For a finite group G, call representations $\alpha, \beta: G \to GL(n,R)$ "locally conjugate" if $\alpha(g)$ is conjugate to $\beta(g)$ for every $g \in G$ via some element of $GL(n,R)$ depending on g.

We say that $\alpha$, $\beta$ are equivalent as representations if we can choose a single element of $GL(n,R)$ that conjugates $\alpha(g)$ to $\beta(g)$ for every $g \in G$.

My question is: does locally conjugate imply equivalent when $R = \mathbb{Z}$?

NOTE 1: When R is a field of characteristic not dividing the order of G, then locally conjugate implies equivalent, and we can prove this by noting that $\alpha$, $\beta$ have the same character (SORRY, the "same character" is enough to complete the proof only in characteristic zero -- in prime characteristics, even those that don't divide the order of the group, having the same character isn't good enough to conclude the representations are equivalent. However, the "locally equivalent implies equivalent" seems to still hold when the characteristic does not divide the order of G using more indirect arguments). However, for $\mathbb{Z}$, the character no longer determines the representation, so the proof used for fields (taking the character) does not work for $\mathbb{Z}$.

NOTE 2: When the characteristic of R divides the order of G, there exist examples of locally conjugate representations that are not equivalent -- in fact, we can construct such examples for the Klein four-group with the field of four elements. I haven't had success with using these to generate counter-examples over $\mathbb{Z}$, though there may be a way.

share|improve this question
2  
I think you are awfully close with the Klein 4 example. The idea, as you probably know already, would be to construct a representation which is free for every cyclic subgroup, but not free for the Klein 4-group itself. Then the representation would locally conjugate, bout not equivalent, to a free representation. –  Geoff Robinson Jul 5 '11 at 23:04
    
There are examples of outer automorphisms $\varphi : G \to G$ such that for all $g \in G$, $\varphi(g)$ is conjugate to $g$. Maybe composing a faithful representation with such an automorphism could give a counterexample to Question 2 ? –  François Brunault Jul 6 '11 at 0:09
add comment

2 Answers

up vote 9 down vote accepted

Here is the requested example of two representations of the Klein 4 group over $\mathbb{Z}$, locally conjugate but not conjugate.

Let $K:= \mathbb{Z}/2 \times \mathbb{Z}/2$ act on $\mathbb{Z}^4$ by permuting the coordinates. Inside $\mathbb{Z}^4$, consider the following two lattices:

$$L_1 := \{ (a,b,c,d) \in \mathbb{Z}^4 : a \equiv b \equiv c \equiv d \mod 2 \}$$

$$L_2 := \{ (a,b,c,d) \in \mathbb{Z}^4 : a + b + c + d \equiv 0 \mod 2 \}$$

Verification that $L_1$ and $L_2$ are locally isomorphic:

Let $\sigma$ be the element of $K$ which switches the first two and the last two coordinates. Consider the following bases for $L_1$ and $L_2$: $$(1,1,1,1),\ (1,-1,1,-1),\ (2,0,0,0), (0,2,0,0)$$ and $$(1,1,0,0),\ (1,-1,0,0),\ (1,0,1,0),\ (0,1,0,1)$$ In both cases, $\sigma$ fixes the first basis element, negates the second and switches the last two. So $L_1$ and $L_2$ are isomorphic modules for $\mathbb{Z}[\sigma]/\langle \sigma^2-1 \rangle$.

Verification that $L_1$ and $L_2$ are not isomorphic:

For each character $\chi$ of $K$, let $L_i^{\chi}$ be the sublattice of $L_i$ on which $K$ acts by $\chi$. Let $M_i = \bigoplus_{\chi} L_i^{\chi}$. Then $L_1/M_1$ has order $2$, and $L_2/M_2$ has order $8$.

share|improve this answer
add comment

In each conjugacy class you can always find a representative which is block upper triangular, and the diagonal blocks have irreducible characteristic polynomials. This gives a partial reduction to the case when the characteristic polynomial is irreducible.

If you fix an irreducible monic polynomial $p$ with integer coefficients, then there is a one-to-one correspondence between conjugacy classes of integer matrices with characteristic polynomial $p$ and ideal classes of the ring $\mathbf{Z}(\theta)$ where $\theta$ is a root of $p$.

The proofs are given e.g. in the book "Integral Matrices" by Morris Newman, Chapter III, sections 14-16.

share|improve this answer
    
Is there a way to do better than "partial reduction" to the case where the characteristic polynomial is irreducible? In the example outlined in the original question, the characteristic polynomial splits completely over Z, so the information on irreducible constituents doesn't seem to explain the splitting of the conjugacy class over Z. –  Vipul Naik Jul 10 '11 at 15:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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