## What is growth of ass. algebra  with 3 generators and relation a1a2a3 + a2a3a1 +a3a1a2 - a1a3a2 - a2a1a3 -a3a2a1 ?

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Consider ass. algebra with 3 generators a1 a2 a3 and relation: a1a2a3 + a2a3a1 +a3a1a2 - a1a3a2 - a2a1a3 -a3a2a1 = 0.

i.e. $$\sum_{ s \in S_3} (-1)^{sgn (s)} a_{s(1)} a_{s(2)} a_{s(3 )} = 0.$$

Informal questions: how far this algebra is from commutative polynomial algebra k[a1 a2 a3] ? What is known about this algebra ?

Formal question: what is the Hilbert series of this algebra ?

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Some reformulations of the defining condition:

Consider 3 Grassman variables $\psi_i$ i = 1,2,3. Define $$\psi = a_1 \psi_1 + a_2 \psi_2 + a_3 \psi_3$$

The condition above is equivalent to $$\psi ^3 =0$$.

For commutative algebras we clearly have $\psi ^2=0$. So this algebra extends the commutative ones in certain sense. So I wonder how far it is away from commutative one ?

Yet another way to reformulate the defining relation is the following:denote by "M" 3* matrix:

M =

a1 a1 a1

a2 a2 a2

a3 a3 a3

The condition above is the same as $det^{column} M =0$ where column-determinant is used i.e. first take elements from the first column, second from the second and so forth.

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Some example.

Consider $E_{ij}$ - "elementary matrices" i.e. n*n matrix with zeros everywhere except position (i,j) where we put 1.

Take for example $E_{11} , E_{21} , E_{31}$,

Observation: they satisfy the above relation.

More generally one can take $E_{i p} , E_{j p} , E_{k p}$ (important the second index is the same).

This means that the algebra above admits a homomorphism to universal enveloping of $E_{11} , E_{21} , E_{31}$. Universal enveloping algebras are very close to commuttive (at least their size is the same). So it suggests that in general such algebra is close to commutative, but probably this is wrong...

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It seems Roland Berger discusses similar alegbras at section 3 of

http://arxiv.org/abs/0801.3383

as far as I can understand he proves that such algebra is N-Koszul (i.e. generalization of Koszul duality to non-quadratic algebras). But I cannot get far in his theory.

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## 3 Answers

Put a term order on your (noncommutative) monomials such that $a_i a_j > a_j a_i$ for $i \lt j$. So the leading term of your equation is $a_1 a_2 a_3$. A basis for your ring is (noncommutative) monomials not divisible by $a_1 a_2 a_3$. In other words, a basis for the degree $n$ part of your algebra is length $n$ sequences of $1$'s, $2$'s and $3$'s which don't contain the sequence $123$. The rest of this post is the combinatorial task of counting the number of such sequences.

Let $A_n$ be the number of such sequences ending in $1$. Let $B_n$ be the number of such sequences ending in $12$. Let $C_n$ be the number of such sequences not in the other two classes (including the empty sequence). Then,

$$A_n = A_{n-1}+ B_{n-1} + C_{n-1}$$ $$B_n = A_{n-1}$$ $$C_n = A_{n-1} + B_{n-1} + 2 C_{n-1} + [n=0]$$

Where $[n=0]$ is $1$ if $n=0$ and is $0$ otherwise. So ```$$\begin{pmatrix} A_n \\ B_n \\ C_n \\ \end{pmatrix}
= \begin{pmatrix} 
1 & 1& 1 \\
1 & 0 & 0 \\
1 & 1 & 2 \\
\end{pmatrix}^n
\begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix}$$``` The total number of terms of degree $n$ is $A_n+B_n+C_n$, so ```$$\begin{pmatrix} 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} 
1 & 1& 1 \\
1 & 0 & 0 \\
1 & 1 & 2 \\
\end{pmatrix}^n
\begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix}$$```

The spectral radius of this matrix is $\approx 2.87939$, so your Hilbert series grow exponentially. That is very different from a commutative ring, whose Hilbert series will grow polynomially.

With a little hacking around with Mathematica, I get that the Hilbert series is $\frac{1}{1-3x+x^3} = 1+3x+9x^2+26x^3+75 x^4 + 216 x^5 + 622 x^6 + \cdots$. Does that match your data?

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1
How do you show that the images of these monomials are linearly independent in this algebra? – mt Nov 20 2011 at 16:02

1
Let $\Delta$ be the defining cubic. Suppose for contradiction that the standard monomials are not linearly independent in degree $n$. So we have $\sum b_i w_i = \sum c_i u_i \Delta v_i$, where $w_i$ are standard monomials, $b_i$ and $c_i$ are nonzero constants and the $u_i$ and $v_i$ are monomials. Continued... – David Speyer Nov 20 2011 at 16:25

2
This is the "standard" noncommutative Groebner basis argument. As always, I am terrible for references. If you are good at references, please post one! – David Speyer Nov 20 2011 at 16:27

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Say $u_1=123$ (I'll drop the $a$s). Then if $v_1=123123$ we have in $u_1\Delta v_1$ the monomial $u_1 123 v_1 = 123\cdot 123 \cdot 123123$, and you say this can't appear elsewhere. But if $u_2=123123$, bigger than $u_1$ in the lex order, and $v_2=123$, then a term in $u_2\Delta v_2$ is $123123 \cdot 123 \cdot 123$. Or am I misunderstanding? – mt Nov 20 2011 at 16:35

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Certainly it can. Take each of those $123$'s and replace them by $213+132-231-312+321$, then expand out to get $25$ terms. – David Speyer Nov 20 2011 at 16:54
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Your algebra maps onto the free associative algebra of rank 2 (just kill $a_3$), so its growth is exponential.

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(edited a bit to cover a few questions about the notation)

A slight simplification of David Speyer's argument: his argument using Groebner bases explains that is we degenerate the relation into $a_1a_2a_3=0$, the resulting algebra has the same Hilbert series. Now, the latter algebra $B$ has a very economic resolution of the trivial module by free right modules: $$0\to span_k(a_1a_2a_3)\otimes_k B\to span_k(a_1,a_2,a_3)\otimes_kB\to B\to k\to 0$$ (the leftmost differential maps $a_1a_2a_3\otimes 1$ to $a_1\otimes a_2a_3$, the next one maps $a_i\otimes 1$ to $a_i$). Computing the Euler characteristics, we get H_B(t)(1-3t+t^3)=1\$.

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@Volodya Thank You ! But I not quite get: span - means what ? Algebra or module ? spanned from the left or from the both sides over B or over all algebra ? Tensor product - as algebras over C or as C vector spaces ? Why this is right modules than ? This probably works for any number of a_i - does it have some name ? where is it used ? – Alexander Chervov Nov 21 2011 at 11:55

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Spans are over the ground field $k$, as well as tensor products (because of this we can compute Euler characteristics). The right $B$-module on $V\otimes B$ is via the action on $B$ alone: $(v\otimes b)b':=v\otimes(bb')$. Of course it works for any number of $a_i$'s, and the construction I use is a very special case of a construction due to David Anick (ams.org/mathscinet-getitem?mr=846601). – Vladimir Dotsenko Nov 21 2011 at 12:08

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Oh, and for your last question: it is used in an awful lot of places, from deformation theory to theoretical computer science. Since the publication of Anick's paper in 1986, quite a few people rediscovered his results in various contexts! – Vladimir Dotsenko Nov 21 2011 at 12:39

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Thanks for the reference! Anick's paper looks very definitive. – David Speyer Nov 21 2011 at 14:46
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