Let us consider the matrix algebra. $Mat_n(\mathbb{C})$. The Amitsur-Levitzki identity states that for any matrices $X_1, X_2, ..., X_{2n} \in Mat_n(\mathbb{C})$ the sum $\Sigma_{\sigma \in S_{2n}} sgn(\sigma)X_{\sigma(1)}...X_{\sigma(2n)}$ vanishes identically.

Is it true that any identity (noncommutative polynomial, which always vanishes) between the matrices from $Mat_n(\mathbb{C})$ follows from Amitsur-Levitzki in some sense$^*$?

$*$ - I assume the following precise meaning: let us consider the oper of "noncommutative polynomial mappings" with the substitution as the composition of operations. It naturally (also non-linearly) acts on any associative algebra (as we can evaluate an $r$-tuple noncommutative polynomial of $k$ variables on the $k$-tuple of elements of an algebra to obtain an $r$-tuple). Is it true that the polynomials that act trivially on the $Mat_n(\mathbb{C})$ are precisely the ideal, generated by the Amitsur-Levitski polynomial?

  • $\begingroup$ Why this is tagged "commutative algebra"? $\endgroup$ – Anton Klyachko Nov 22 '14 at 5:36
  • $\begingroup$ ... and why "combinatorics"? $\endgroup$ – Anton Klyachko Nov 22 '14 at 6:08
  • 5
    $\begingroup$ Your last paragraph can be expressed succintly using the notion of T-ideals. $\endgroup$ – Mariano Suárez-Álvarez Nov 22 '14 at 8:20

The answer to your question is "no", as explained by Anton Klyachko in his answer. Let me refer you to a remarkable statement of Razmyslov and Procesi that describes all identities. They proved (independently) that in fact all identities of $Mat_n(\mathbb{Q})$ follow, in a sense, from the Cayley--Hamilton theorem. To be more precise:

Let us first define the notion of an algebra with trace as a vector space $A$ over a field $F$ that has a bilinear product, and a linear functional $t\colon A\to F$ satisfying $t(AB)=t(BA)$. Of course we can consider the free algebra with trace generated by a set $X$; it is the free associative algebra generated by $X$ over the ring of polynomials in $t(m)$, where $m$ is a cyclic word in $X$ (a cyclic group orbit on words), and $t$ is extended to this algebra by an obvious rule $t(t(m_1)m_2)=t(m_1)t(m_2)$. This gives us a language to discuss we can talk about identities of algebras with trace. (Alternatively, one can use the language of operads to discuss that; I prefer that latter language but choose to write a more "classical" definition here).

Next, let us define the $n$-th Cayley-Hamilton identity of an algebra with trace as the identity $$ CH_n=\sum_{\sigma\in S_{n+1}}X_{i_1^{(1)}}\cdots X_{i_{k_1}^{(1)}}t(X_{i_1^{(2)}}\cdots X_{i_{k_2}^{(2)}})\cdots t(X_{i_1^{(l)}}\cdots X_{i_{k_l}^{(l)}})=0 $$ where $\sigma$ has the disjoint cycle decomposition $(0,i_1^{(1)},\ldots,i_{k_1}^{(1)})(i_1^{(2)},\ldots,i_{k_2}^{(2)})\cdots(i_1^{(l)},\ldots,i_{k_l}^{(l)})$. (This is a full multilinearisation of the Cayley--Hamilton theorem $\chi_A(A)=0$).

Theorem (Procesi [1], Razmyslov[2]). Every identity of the matrix algebra viewed as an algebra with trace is a consequence of the identity $CH_n=0$.

[1] C.Procesi, The invariant theory of n × n matrices, Advances in Mathematics Volume 19, Issue 3, March 1976, Pages 306–381, http://www.sciencedirect.com/science/article/pii/000187087690027X

[2] Yu. P. Razmyslov, Trace identities of full matrix algebras over a field of characteristic zero, Izv. Akad. Nauk SSSR Ser. Mat., 1974, Volume 38, Issue 4, Pages 723–756, http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=1989&option_lang=eng

Of course, every usual identity is a particular case of a trace identity, so in principle this theorem also classifies all usual identities.

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The answer is No. For $n=2$, there is the Hall identity $[[x,y]^2,z]=0$. Drensky proved in 1981 that these two identities (Hall and standard (=Amitsur-Levitzki)) form a basis of identities of $Mat_2(\Bbb C)$ (i.e. all identities are consequences of these two). For higher $n$, bases of identities are (probably) unknown.

Note that, according to the Kemer theorem, every associative algebra over a field of characteristic zero has a finite basis of identities.

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As mentioned before (for $2\times 2$ matrices), the answer is "no" .

Here I give some reference on the general case.

The identity of algebraicity is defined as:

$$ a_n(x,y_1,\dots, y_n)= \displaystyle \sum _{\sigma \in S_{n+1}} (-1)^\sigma x^{\sigma(0)}y_1 x^{\sigma(1)}y_2 \cdots x^{\sigma(n-1)}y_n x^{\sigma(n)} $$

In boook [1] we have:

Exercise 7.1.12 Show that the identity of algebraicity for $M_k(K)$, $k > 1$, does not follow from the standard identity $St_{2k}$.

In the book there is a hint for the exercise, but the original references for that are preprint [2] and paper [3].

Also in paper [4] there are other polynomial identities for $M_3(K)$ which does not follow from $St_6$ and $a_3$. There you also find a good list of references on the subject.


[1] V. Drensky, Free Algebras and PI-algebras: Graduate Course in Algebra, Springer, Singapore, 1999.

[2] G. M. Bergrnan, Wild automorphisms offree P.I. algebras and some new identities, Preprint, Berkeley (1981). https://math.berkeley.edu/~gbergman/papers/unpub/wild_aut.pdf

[3] V. S. Drensky, A.K. Kasparian, Some polynomial identities of matrix algebras, C. R. Acad. Bulg. Sci. 36 (1983), 565-568.

[4] M. Domokos (1995) New identities for 3 × 3 Matrices, Linear and Multilinear Algebra, 38:3, 207-213, http://dx.doi.org/10.1080/03081089508818356

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