14
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
17
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
20
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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.

References:

[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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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