## Non-conjugate words with the same trace

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Let n>=2, p a large prime, G = SL_n(Z/pZ).

If n=2, there are words that, while not conjugate in the free group, do have identical trace in G. For example, tr(g h^2 g^2 h)= tr(g^2 h^2 g h) for all g, h in SL_2(Z/pZ).

```Question: does the same happen for n>=3?
Could it even be possible that there are words w_1, w_2 that are not conjugate even in G=SL_n(Z/pZ), yet always have the same trace: 
tr(w_1(g_1,g_2,..,g_k)) = tr(w_2(g_1,g_2,...,g_k)) for all g_1,...,g_k in SL_n(Z/pZ)?```

This would be extremely helpful.

Actually, I just need a weaker statement:

Wild guess.- Let g, h be elements of SL_n(K), n>2. There is a constant k (which may depend on n but not on K) such that there are two elements a, b of the ball ({g,h,g^{-1},h^{-1},e})^k for which (1) tr(a)=tr(b) and (2) a is not conjugate to b (for g and h generic).

Does anybody have a clue as to whether this is or isn't true?

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It would be easy to check this in Magma for small values of n and p. – David Zureick-Brown♦ Oct 15 2009 at 17:08
You mean for words up to a given length? – H A Helfgott Oct 16 2009 at 10:38

## 5 Answers

I don't think you asked the question you intended. Even though this question is old, I'll say a bit:

For any fixed $n$ and $p$, there are a finite number of homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$.

Hence, there are only a finite number of possibilities for the trace of a group element as a function of the representation. Therefore, there are many duplicates.

I think a better question is to ask whether there are pairs $(g_1, g_2)$ of non-conjugate elements of $F_2$ such that for all primes $p$ and all homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$ the traces are equal. In this form, you can pass to a limit $p \rightarrow \infty$ and conclude if nonconjugate elements can be distinguished by traces for an infinite sequence of $SL_n(Z/pZ)$, they could be distinguished by a homomorphisms to $SL_n(\mathbb C)$. Conversely, if there are no trace identities in $SL_n(\mathbb C)$, then there are no trace identities true in all $SL_n(Z/pZ)$, by finding $Z/pZ$ quotients of the ring of coefficients.

Martin Kassabov has been interested in the latter question, and in discussions he and I have had, we've both come to the opinion that there are probably no trace identitites for $SL_n(C)$ when $n > 2$, but it's not easy to find a proof. One possible strategy is to first characterize all trace identities in $SL_2$, and then construct representations in $SL_3$ where they break down. This is interesting to me in any case because of its meaning in 2 and 3-manifold topology -- the trace identities give collections of distinct elements in $\pi_1$ that are forced to have the equal length in any hyperbolic structure.

A weaker question is whether the characteristic polynomials for representations in $SL_n$ can distinguish conjugacy classes in $F_2$.

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It seems like this paper might have interesting things to say about trace identities in SL_n(C) arxiv.org/abs/math.RT/9806016 – Peter Samuelson Sep 12 2010 at 16:55
Yes, I did mean n was fixed and p goes to infinity. – H A Helfgott Sep 14 2010 at 1:49

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Since James seems to have found the group theory lemma we need, I'm going to rewrite this post to be clearer about what it proves.

Theorem: If g and h are two non-conjugate elements of the free group Fk, then there is some n, some p, and a map r:Fk --> GLn(Z/p) such that Tr r(g) \neq Tr r(h).

Proof: By James's post, there is a finite group G and a map a: Fk --> G so that a(g) and a(h) are not conjugate. As characters distinguish conjugacy classes, there is a cyclotomic field K=Q(e^{2 pi i/q}) and a representation b: G --> GLn(K) such that Tr b(a(g)) \neq Tr b(a(h)). Let p be a prime which is (1) large enough not to appear in the denominators of any of the matrices occurring in b(G) (2) congruent to 1 modulo q and (3) large enough not to divide Tr b(a(g)) - Tr b(a(h)). By condition (2), there is a prime P of K, living over p, such that O(K)/P = Z/p. By condition (1), we can reduce the entries of the matrices of b(G) modulo p. This gives us a map c : G --> GLn(Z/p) which is morally the composite of G --> GLn(K) --> GLn(O(K)/P). (The second map of this sequence does not actually exist.)

Composing a and c gives the map r. Condition (3) insures that Tr r(g) \neq Tr r(h). QED

Note that I have not solved the following more difficult question: given n, do there exist any pair of nonconjugate elements g and h in Fk so that Tr r(g) = Tr r(h) for all maps r: Fk --> GLn(Z/p).

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Restatement: Does the canonical map from F_k to its profinite completion merge any conjugacy classes? – S. Carnahan♦ Oct 16 2009 at 2:42

I'm not sure I follow the reduction to the assertion about non-conjugate elements in free groups. However, it is true that, if two members of a free group are not conjugate, then there is a finite quotient in which their images are not conjugate. This property is known as conjugacy separability, and it is well known that free groups are conjugacy separable. There are a number of ways to prove this. One can, for example, reduce to finitely generated nilpotent groups via the lower central series in the free group (which is residually a torsion free nilpotent groups) and then invoke the result of Formanek/Remeslennikov that f.g. torsion-free nilpotent groups are conjugacy separable. Alternatively, free products of conjugacy separable groups are conjugacy separable (Remeslennikov, 1971), and infinite cyclic groups obviously are.

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I don't think Critch's reply above answers Harald's question; it seems to presume that the map F_2 -> SL_n(Z/pZ) factors through a chosen inclusion of SL_2(Z/pZ), while Harald wants pairs of elements in F_2 (or, more generally, F_k) which have the same trace after applying ANY homomorphism F_k > SL_n(Z/pZ).

EDITED to comment on David's answer below:

"Also, if there is any finite quotient G of F_k, in which the two elements stay nonconjugate, then there is some representation of G in which they have different traces."

Yeah, but not an n-dimensional representation.

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Right. What I meant to say was that the question for all SL_n(Z/p) is equivalent to the question for all finite groups. – David Speyer Oct 16 2009 at 5:32
Exactly. I want two words w(g,h), w'(g,h) (say) such that tr(w(g,h))=tr(w'(g,h)) for all g, h in SL_n(Z/pZ), not just for some g,h in SL_n(Z/pZ) (such as, say, g and h in a copy of SL_2(Z/pZ) contained in SL_n(Z/pZ)). – H A Helfgott Oct 16 2009 at 10:38

Yes, if I understand your question correctly. Consider `SL_2(Z/pZ)` included in `SL_n(Z/pZ)` as those elements with a 2x2 block in the top left, 1's on the the remaining diagonal, and 0's elsewhere. Also consider The free group `F_2` on `g_1, g_2` included into the free group `F_n `on `g_1, g_2, ..., g_n`.

Two elements of `F_2` are conjugate in `F_n` iff they are conjugate in `F_2` (since if you conjugate by a word involving some `g_i, i>2`, then you land outside `F_2`). So whatever pairs of elements work in your `n=2` example also work for n arbitrary: they will have the same trace in `SL_n(Z/pZ)`, namely [their trace in `SL_2(Z/pZ)] + [n-2]`, the extra diagonal elements we added, but the words expressing them won't be conjugate in `F_n` because they weren't in `F_2`.

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I thought he intended to require that the two traces be equal for all maps from F_g to SL_n(Z/p). That matches his example that Tr(g*h*g^2*h^2) = Tr(h^2*g^2*h*g) in SL_2. – David Speyer Oct 16 2009 at 2:05
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