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

Question

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.

Answer 1

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.

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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.

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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?

Answer 2

Your algebra maps onto the free associative algebra of rank 2 (just kill $a_3$), so its growth is exponential.

Answer 3

(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$.