# Are the trace relations among matrices generated by cyclic permutations?

Let $$X_1,\dots,X_n$$ be non commutative variables such that $$\operatorname{tr} f(X_1,\dots,X_n) = 0$$ whenever the $$X_i$$ are specialized to square matrices in $$M_r(k)$$ for any $$r \geq 1$$. Does this imply that $$f$$ is in the ideal generated by cyclic permutations: $$g_1\dots g_k - g_2\dots g_k g_1$$ for any polynomials $$g_i$$ in the $$X_i$$ and $$k \geq 2$$?

(And if I have missed any obvious relations, is the statement true up to adding in those relations to the ideal?)

• Is there a specific reason why you want to consider the ideal generated by these $f$ (I'm asking because the $f$ with $\textrm{tr }f=0$ don't seem to form an ideal). May 1 at 19:41
• I think you want just the linear span as @ChristianRemling suggests and then I think it should be generated by all differences of two words that differ by a cyclic permutation May 1 at 20:02
• $f$ is a non-commutative polynomial, right? I can reduce the problem (re-phrased as @BenjaminSteinberg and ‍ChristianRemling suggested, in terms of a linear span) to showing that, if $f = X_1 g$ for some other non-commutative polynomial $g$, then $f = 0$. But I can't seem to establish that yet. May 1 at 21:17
• I was able to show that if our field is the complex numbers, if $\text{Tr}(f(\phi(g_{1}),\dots,\phi(g_{n})))=0$ for each finite group G, irreducible representation $\phi:G\rightarrow U(r)$, and elements $g_{1},\dots,g_{n}\in G$, then $f$ is in the subspace spanned by the cyclic permutations. The proof uses the fact that the characters of irreducible representations form a basis for the class functions, so the proof is not to unexpected. May 5 at 14:37
• @JosephVanName That is more or less the context that inspired this question! I had an infinite sequence of products of matrices (over a polynomial ring) that seemed to converge $\ell$-adically and to prove this, I worked with every term in the expanded polynomial and showed that the coefficients converge. May 6 at 20:31

The reformulation suggested by Christian Remling and Benjamin Steinberg is true (at least over a field $$k$$ of characteristic zero):

If $$\operatorname{tr} f(X_1,\dots, X_n)=0$$ for all $$X_1,\dots, X_n$$ in $$M_r(k)$$ then $$f$$ is a linear combination of differences of cyclically permuted words.

An equivalent, linearly dual, statement:

For $$m$$ words $$w_1,\dots, w_m$$ in the variables $$X_1,\dots, X_n$$, if none is the cyclic permutation of the other, then for any $$a_1,\dots, a_n$$ in $$k$$, there exist $$r$$ and $$X_1,\dots X_n \in M_r(k)$$ such that $$\operatorname{tr}(w_i)=a_i$$ for all $$i$$.

This implies the previous statement because, for each word appearing in $$f$$, we can add up the coefficients by which all cyclic permutations of that word appear in $$f$$. If this sum is zero for all words then $$f$$ is a linear combination of differences of cyclic permutations, and if it is nonzero for some word we can take $$a_i=1$$ for that word and $$a_i=0$$ for all remaining words.

Proof: Without loss of generality, assume that the words $$w_1,\dots, w_n$$ are in nondecreasing order of length.

It suffices to find for each $$i$$ matrices $$Y_{1,i},\dots, Y_{n,i}$$ such that $$\operatorname{tr} w_j( Y_{1,i}, \dots, Y_{n,i})=0$$ for $$j and $$\neq 0$$ for $$j=i$$. Then we can construct $$X_i$$ as a direct sum of suitable scalar multiples of the $$Y_{j,i}$$.

Suppose the word $$w_i$$ is $$X_{j_1} X_{j_2} \dots X_{j_\ell}$$. Define $$Y_{j,i}$$ as the $$\ell \times \ell$$ matrix whose entry in the $$a$$th row and $$b$$th column is

$$(Y_{j,i})_{ab} =\begin{cases} 1 & b\equiv a+1 \textrm{ mod }\ell \textrm{ and } j_b=j \\ 0 & \textrm{otherwise} \end{cases}$$

Because the only nonzero enties in these matrices have $$b\equiv a+1\textrm{ mod }\ell$$, the only nonzero entries in the product of $$d$$ such matrices have $$b \equiv a+d\textrm{ mod }\ell$$, so the trace of such a product is nonzero only if $$d$$ is divisible by $$\ell$$. Thus the trace vanishes for all shorter words.

For a word of length exactly $$\ell$$, the only contribution to the trace of the product is from products of entries that follow the cyclic math around, i.e.

$$\operatorname{tr} Y_{s_1,i} Y_{s_2,i} \dots Y_{s_{\ell-1},i} Y_{s_\ell,i} = \sum_{c=1}^{\ell} (Y_{s_1,i} )_{c (c+1)} (Y_{s_2,i} )_{(c+1)(c_2)} \dots (Y_{j_{\ell-1},i} )_{(c-2)(c-1)} (Y_{s_\ell,i} )_{(c-1)c}$$ which is equal to the number of cyclic permutations of $$s_1,\dots s_\ell$$ that equals $$j_1,\dots, j_\ell$$ and thus vanishes unless $$s_1,\dots, s_\ell$$ is a cyclic permutation of $$j_1,\dots ,j_\ell$$ and is nonzero in case it is a cyclic permutation, as desired.

• The characteristic-$0$ restriction is essential. For example, over a finite field with $q$ elements, $\operatorname{tr}(X^q - X) = 0$ for every square matrix $X$. May 1 at 21:47
• @LSpice Good point! In general one has the semilinear relation $\operatorname{tr}(X^p) = \operatorname{Frob}_p( \operatorname{tr}(X))$. So the trace of any word that is a $p$-fold repetition of a shorter word is determined by the trace of that shorter word. I think my argument shows that any assignment of elements of $k$ to words that is cyclic permutation invariant and satisfies this additional rule arises from a trace, because the number of cyclic permutations preserving the word is nonzero mod $p$ if and only if the word is not a $p$-fold repetition. May 2 at 2:12
• Over an infinite field, the $\operatorname{Frob}_p^n( x)$ are linearly independent as functions of $x$, and over a finite field, the only linear relations are the obvious ones. From this, I believe one can check that the only linear relations satisfied are generated by the cyclic permutation relation and, over finite fields, the one you gave. May 2 at 2:15
• Thanks, this is great! May 2 at 3:54

For simplicity, let us assume that $$K=\mathbb{C}$$ (though, these results apply to any field of characteristic zero). We will use capital letters like $$X,Y,Z$$ to denote matrices while lower case letters like $$x,y,z$$ shall denote variables.

We will use the fact that the trace of matrices produces an inner product on $$M_{n}(\mathbb{C})$$ defined by $$\langle A,B\rangle=\operatorname{Tr}(AB^{*})$$.

If $$F$$ is a field, then let $$F\langle x_{1},\dots,x_{n}\rangle$$ denote the ring of non-commutative polynomials over $$F$$ in the variables $$x_{1},\dots,x_{n}$$.

We will now go over a few lemmas.

Lemma: Suppose that $$f,g\in F\langle x_{1},\dots,x_{n}\rangle$$. Then $$f=g$$ if and only if whenever $$r\geq 1$$ and $$X_{1},\dots,X_{n}\in M_{r}(F)$$, we have $$f(X_{1},\dots,X_{n})=g(X_{1},\dots,X_{n})$$.

Proof: Suppose that $$f\neq g$$, and $$u>\text{Deg}(f)+\text{Deg}(g)$$. Let $$V$$ be a finite dimensional vector space over $$F$$ with linearly independent set $$(e_{i_{1},\dots,i_{k}}|0\leq k\leq u,i_{1},\dots,i_{k}\in\{1,\dots,n\})$$ You can let $$X_{1},\dots,X_{n}:V\rightarrow V$$ be linear maps such that $$X_{i}(e_{i_{1},\dots,i_{k}})=e_{i,i_{1},\dots,i_{k}}$$ whenever $$k and $$X_{i}(e_{i_{1},\dots,i_{k}})=0$$ whenever $$k=u$$. Then $$f(X_{1},\dots,X_{n})\neq g(X_{1},\dots,X_{n})$$. Q.E.D.

Lemma: Suppose that $$f,g_{1},\dots,g_{n}\in\mathbb{C}\langle x_{1},\dots,x_{n},y_{1},\dots,y_{n}\rangle$$ and that $$f(x_{1},\dots,x_{n},y_{1},\dots,y_{n})=g_{1}(x_{1},\dots,x_{n})y_{1}+\dots +g_{n}(x_{1},\dots,x_{n})y_{n}$$. Then the following are equivalent.

1. $$f=0$$.

2. $$g_{1}=0,\dots,g_{n}=0$$.

3. $$\operatorname{Tr}(f(X_{1},\dots,X_{n},Y_{1},\dots,Y_{n}))$$ whenever $$r\geq 1$$ and $$X_{1},\dots,X_{n},Y_{1},\dots,Y_{n}\in M_{r}(\mathbb{C})$$.

Proof: The directions $$1\leftrightarrow 2,1\rightarrow 3$$ are clear.

$$3\rightarrow 2$$ Suppose that 3 holds. We have $$0=\operatorname{Tr}(f(X_{1},\dots,X_{n},Y_{1}^{*},\dots,Y_{n}^{*}))$$ $$=\operatorname{Tr}(g_{1}(X_{1},\dots,X_{n})Y_{1}^{*}+\dots +g_{n}(X_{1},\dots,X_{n})Y_{n}^{*})$$ $$=\langle g_{1}(X_{1},\dots,X_{n}),Y_{1}\rangle+\dots+\langle g_{n}(X_{1},\dots,X_{n}),Y_{n}\rangle$$ for each choice of $$X_{1},\dots,X_{n},Y_{1},\dots,Y_{n}$$. Using basic facts about inner product spaces, we can conclude that $$g_{1}(X_{1},\dots,X_{n})=\dots=g_{n}(X_{1},\dots,X_{n})=0$$ for each choice of matrices $$X_{1},\dots,X_{n}$$. Therefore, by using the above lemma, we can conclude that $$g_{1}=0,\dots,g_{n}=0$$. Q.E.D.

For all $$k$$, let $$\phi_{k}:\mathbb{C}\langle x_{1},\dots,x_{n}\rangle\rightarrow \mathbb{C}\langle x_{1},\dots,x_{n}\rangle$$ be the $$\mathbb{C}$$-linear mapping such that $$\phi_{k}(x_{a_{1}}\dots x_{a_{v}})=\sum(x_{a_{i+1}}\dots x_{a_{v}}x_{a_{1}}\dots x_{a_{i-1}}\mid 1\leq i\leq v,a_{i}=k)$$ and $$\phi_{k}(c)=0$$ for each constant term $$c$$. For example, $$\phi_{3}(x_{2}x_{3}x_{3})=x_{2}x_{3}+x_{3}x_{2}$$. One should intuitively think of the functions $$\phi_{k}$$ as a formal derivative of the trace operator.

Lemma: (product rule for matrix multiplication) Suppose that $$\{E,F\}\subseteq\{\mathbb{R},\mathbb{C}\}$$ and $$A_{1},\dots,A_{n}:E\rightarrow M_{n}(F)$$ are differentiable. Then $$\frac{d}{dt}(A_{1}(t)\dots A_{n}(t))=\sum_{k=1}^{n}A_{1}(t)\dots A_{k-1}(t)A_{k}'(t)A_{k+1}(t)\dots A_{n}(t)$$.

In particular, from the cyclicity of the trace, we have $$\operatorname{Tr}(\frac{d}{dt}(A_{1}(t)\dots A_{n}(t))) =\frac{d}{dt}\operatorname{Tr}(A_{1}(t)\dots A_{n}(t))$$ $$=\sum_{k=1}^{n}\operatorname{Tr}[A_{k}'(t)A_{k+1}(t)\dots A_{n}(t)A_{1}(t)\dots A_{k-1}(t)]$$ $$=\sum_{k=1}^{n}\operatorname{Tr}[A_{k+1}(t)\dots A_{n}(t)A_{1}(t)\dots A_{k-1}(t)A_{k}'(t)].$$

Corollary: For each $$f\in\mathbb{C}\langle x_{1},\dots,x_{n}\rangle$$, we have $$\operatorname{Tr}(\frac{d}{dt}f(A_{1}(t),\dots,A_{n}(t)))= \operatorname{Tr}[\sum_{k=1}^{n}\phi_{k}(f)(A_{1}(t),\dots,A_{n}(t))A_{k}'(t)].$$

Theorem: Suppose that $$f\in\mathbb{C}\langle x_{1},\dots,x_{n}\rangle$$. Then the following are equivalent

1. $$\operatorname{Tr}(f(X_{1},\dots,X_{n}))=0$$ whenever $$X_{1},\dots,X_{n}$$ are matrices.

2. $$f(0,\dots,0)=0$$ and $$\phi_{k}(f)=0$$ whenever $$1\leq k\leq n$$.

3. $$f\in I$$ where $$I$$ is the sub-vector space of $$\mathbb{C}\langle x_{1},\dots,x_{n}\rangle$$ generated by the vectors of the form $$x_{a_{1}}\dots x_{a_{v}}-x_{a_{2}}\dots x_{a_{v}}x_{a_{1}}$$.

Proof: The direction $$3\rightarrow 1$$ is clear from the cyclicity of the trace. The direction $$3\rightarrow 2$$ is also clear.

$$1\rightarrow 2$$. Suppose that $$\operatorname{Tr}(f(X_{1},\dots,X_{n}))=0$$ whenever $$r\geq 0$$ and $$X_{1},\dots,X_{n}\in M_{n}(K)$$. Then

$$0=\frac{d}{dt}\operatorname{Tr}(f(X_{1}+tY_{1},\dots,X_{n}+tY_{n}))|_{t=0}$$ for each choice of $$X_{1},\dots,X_{n},Y_{1},\dots,Y_{n}$$. However, $$0=\frac{d}{dt}\operatorname{Tr}(f(X_{1}+tY_{1},\dots,X_{n}+tY_{n}))|_{t=0}$$ $$=\operatorname{Tr}(\phi_{1}(f)(X_{1},\dots,X_{n})Y_{1}+\dots+\phi_{n}(f)(X_{1},\dots,X_{n})Y_{n}).$$

Therefore, we conclude using the above lemma that $$\phi_{1}(f)=\dots=\phi_{n}(f)=0$$.

$$2\rightarrow 3$$. Let $$\theta_{i}:\mathbb{C}\langle x_{1},\dots,x_{n}\rangle\rightarrow\mathbb{C}\langle x_{1},\dots,x_{n}\rangle$$ be the linear mapping such that $$\theta_{i}(x_{a_{1}}\dots x_{a_{r}})=\frac{1}{r+1}x_{i}x_{a_{1}}\dots x_{a_{r}}$$ whenever $$r\geq 0,\{a_{1},\dots,a_{r}\}\subseteq\{1,\dots,n\}$$.

Define a linear operator $$L:\mathbb{C}\langle x_{1},\dots,x_{n}\rangle\rightarrow \mathbb{C}\langle x_{1},\dots,x_{n}\rangle$$ by letting $$L(f)=f(0,\dots,0)+\theta_{1}(\phi_{1}(f))+\dots+\theta_{n}(\phi_{n}(f))-f.$$ Then we have $$L(x_{a_{1}}\dots x_{a_{v}})\in I$$ whenever $$v\geq 0$$ and $$L(c)=0\in F$$, so by linearity, we conclude that $$L(f)\in I$$ for each $$f\in\mathbb{C}\langle x_{1},\dots,x_{n}\rangle$$. Therefore, if $$\phi_{1}(f)=0,\dots,\phi_{n}(f)=0,f(0,\dots,0),$$ then $$f=-L(f)\in I$$. Q.E.D.

• Please could you add some detail to the step where you relate differentiation with respect to $t$ to the $\phi_k$ functions? May 2 at 9:58
• I like the lemma, it's a version of non commutative nullstellensatz. May 2 at 12:40
• I have added the relation between differentiation and $\phi_{k}$. It appears that the proof that the proof of the main theorem generalizes just fine as long as the underlying field is infinite or if we restrict ourselves to non-commutative polynomials whose degree is less than the cardinality of the underlying field. May 3 at 16:08