Similar to a previous question here, I wonder if cyclic permutations are the only relations amongst traces of (non-commutative) monomials. Since the evaluations $\operatorname{tr}:k\langle x,y,\dots \rangle \to k$ take an infinite dimensional vector space to a one-dimensional vector space there must be quite a few relations, but I wonder if any of them are on binomials other than the cyclic permutations.

At any rate, for small dimensions, we probably get some extra relations.

It appears that $\operatorname{tr}(AABABB−AABBAB) = 0$ for all $2×2$ matrices. Is this true? How does one prove it?

  • $\begingroup$ This is the only extra relation on 2×2 matrices (over sufficiently large rings) of word length up to 6, so the extra relations are a little rare at least. $\endgroup$ – Jack Schmidt Mar 25 '11 at 21:44
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    $\begingroup$ Either there's a typo or I'm missing something. You've got four $A$s and two $B$s in one product and three each in the other, so if you multiply $A$ by some factor, the traces will get multiplied by different factors, so they could only be the same if they're both zero, which they are clearly not? $\endgroup$ – joriki Mar 25 '11 at 21:55
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    $\begingroup$ Or when $A$ is invertible just set $B = A^{-1}$.. Tr$(A^2) = $ Tr$(I) = 2$ which obviously isn't going to be true for all $A$. $\endgroup$ – Zarrax Mar 25 '11 at 22:09
  • $\begingroup$ Typo fixed, sorry. $\endgroup$ – Jack Schmidt Mar 25 '11 at 22:15
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    $\begingroup$ Direct computation? $\endgroup$ – Alexei Averchenko Mar 26 '11 at 9:03

First if $N$ is $2\times2$ matrix with $\operatorname{tr}(N)=0$, then $N^2$ is scalar (either $N$ is nilpotent or its eigenvalues are opposite and have same square). This implies that $\operatorname{tr}(N^3)=0$.

And $$\begin{eqnarray} (AB-BA)^3 &=& (ABABAB - BABABA) + (ABBABA + BABAAB + BAABBA) \\ && - (ABABBA + ABBAAB + BAABAB) \end{eqnarray} $$

taking the trace this gives

$$0 = 0 + 3\operatorname{tr}(ABBABA) - 3 \operatorname{tr}(ABABBA) = 3 \operatorname{tr}(AABBAB - AABABB).$$

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    $\begingroup$ Computing the last trace explicitely (and thanking von Neumann in the process...) one can see that the trace is actually zero integrally, so this is true also in characteristic 3. $\endgroup$ – Mariano Suárez-Álvarez Mar 25 '11 at 23:45
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    $\begingroup$ Mariano, I don't understand your comment, but what about the following argument: The above proof works also with matrices over integral domains (consider the field of fractions), in particular polynomial rings over $\mathbb{Z}$, where we can cancell $3$. So if we plug in generic matrices $A,B$ and evaluate them in any ring, we get our identity. $\endgroup$ – Martin Brandenburg Mar 26 '11 at 9:30
  • $\begingroup$ @Martin Brandenburg: I had the same question as yourself (although my doubts were greater, I think) and asked a question about it that was answered affirmative by a few people. $\endgroup$ – Myself Mar 26 '11 at 11:29

Also relevant is the Amitsur-Levitki theorem.


I recall reading long ago that such trace identities generally arise from those associated with the Cayley-Hamilton theorem (by multilinearizing the characteristic polynomial[2]). A quick web search on related keywords turned up the following paper[1]. In the introduction it is stated that "we prove that all trace identies of the full matrix algebra of order n over a field of characteristic zero are consequences of one corresponding to the Hamilton-Cayley theorem". There is also independent seminal work of Procesi, who obtains the trace identities via multilinear invariants of tensor products of vector spaces. No doubt much work has been done the in three decades since this seminal work appeared. A search on Razmyslov / Procesi and "trace identities" reveals much.

[1] Razmyslov. Trace identities of full matrix algebras over a field of characteristic zero.
Math Ussr Izv, 1974, 8 (4), 727-760.

[2] Formanek. Polynomial identities and the Cayley-Hamilton Theorem.
The Mathematical Intelligencer. Vol. 11, 1, 1989, 37-39.

  • $\begingroup$ Is the equation in this question actually a trace identity? It does not seem to fit the definition. $\endgroup$ – Jack Schmidt Mar 26 '11 at 16:00
  • $\begingroup$ @Jack: Which definition do you refer to? $\endgroup$ – Bill Dubuque Mar 26 '11 at 16:34
  • $\begingroup$ The trace identities in Formanek and several other papers required the variables to be distinct. Apparently this is because all trace identities are polynomials in such restricted identities. I believe user7406's solution expresses my identity in terms of restricted identities. $\endgroup$ – Jack Schmidt Mar 26 '11 at 16:48
  • $\begingroup$ @Jack: I'm fairly sure that your identites are covered by the standard theory. But, alas, I don't have the time to try to jog my decades-old memory right now. It might help to check more recent papers. $\endgroup$ – Bill Dubuque Mar 26 '11 at 16:58

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