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Are there attempts to classify commutative finite dimensional Frobenius algebras? They appear often in mathematics, such as in algebraic geometry and the famous category equivalence between commutative Frobenius algebras and 2-dimension topolocial quantum field theories. However, I have not yet seen attempts to classify this class of algebras (up to isomorphism of k-algebras). Recall that a commutative Frobenius algebra is a finite dimensional algebra $A$ with $A \cong D(A)$ or equivalently simple socle in case it is local.

Here are two questions related to such a classification (we can assume that commutative Frobenius algebra are connected):

Question 1: Is a commutative Frobenius algebra "field-independent"? This means that in its presentation $KQ/I$ by quiver and relations, there exists such $I$ which only contains the field element 1 or -1 so that a given commutative Frobenius algebra is defined over all fields.

In case question 1 has a positive answer, this would mean that a classification is independet of the field (maybe excluding characteristic 2).

Question 2: For a given integer $d$, are the only finitely many $d$-dimensional commutative Frobenius algebras of vector space dimension $d$?

(here we say that two algebras are isomorphic in case they are isomorphic as $K$-algebras)

A positive answer to question 2 would be surprising, but I think it should be true for $d \leq 5$ at least.

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Answering Question 2, I'll show that there are infinitely many $14$-dimensional commutative Frobenius algebras over an infinite field $k$ so that no two are isomorphic as algebras. Probably $14$ is not minimal.

It was shown by Suprunenko that there are infinitely many pairwise nonisomorphic $7$-dimensional commutative $k$-algebras, and in fact from the proof of this fact by Poonen in

Poonen, Bjorn, Isomorphism types of commutative algebras of finite rank over an algebraic closed field, Lauter, Kristin E. (ed.) et al., Computational arithmetic geometry. AMS special session, San Francisco, CA, USA, April 29–30, 2006. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4320-8/pbk). Contemporary Mathematics 463, 111-120 (2008). ZBL1155.13015. (Slightly updated version available online here.)

it follows that, among the following class of algebras parametrized by $\lambda\in k$, there are only finitely many in each isomorphism class. I'll assume that $\text{char}~k\neq2$, but an easy adjustment deals with the case $\text{char}~k=2$.

$B_\lambda$ will be the algebra with basis $\{1,x_1,x_2,x_3,x_4,s,t\}$, where $x_1^2=s$, $x_2^2=s+t$, $x_3^2=s+2t$ and $x_4^2=s+\lambda t$, and all other products of the last six basis elements are zero. The trivial extension algebras $TB_\lambda = B_\lambda\ltimes DB_\lambda$ are commutative Frobenius algebras, and I'll show, by a rather ad hoc argument, that $TB_\lambda\cong TB_\mu$ only if $B_\lambda\cong B_\mu$. I'm using $D$ here for the vector space dual.

Wakamatsu proved in

Wakamatsu, Takayoshi, Note on trivial extensions of Artin algebras, Commun. Algebra 12, 33-41 (1984). ZBL0537.16008.

that two finite dimensional algebras have isomorphic trivial extension algebras if and only if they are of the form $A\ltimes M$ and $A\ltimes DM$ for some algebra $A$ and $A$-bimodule $M$. Since our algebras are commutative, for us $M$ will just be an $A$-module (or a bimodule with the same action on both sides).

First, note that $\text{rad}^2B_\lambda=\text{soc}B_\lambda$ is two dimensional, spanned by $s$ and $t$.

If $TB_\lambda\cong TB_\mu$ but $B_\lambda\not\cong B_\mu$ then there must be a direct sum decomposition of vector spaces $B_\lambda=A\oplus M$ such that $A$ is a subalgebra, $M$ is a square zero ideal, and $B_\mu\cong A\ltimes DM$. Note that $M\not\cong DM$ or else $B_\lambda\cong B_\mu$. Also, $\text{soc}(A\ltimes DM)$ must be equal to $\text{rad}^2(A\ltimes DM)$ and be two-dimensional. I'll show that these things cannot happen.

Suppose they do.

If $M=S\oplus M'$ has a simple summand $S$, we can replace $A$ by $A\oplus S$ and $M$ by $M'$, so we can assume that $M$, which has radical length two, satisfies $\text{soc}M=\text{rad}M$. Since $\text{soc}(A\ltimes DM)=\text{rad}^2A\oplus\text{soc}DM$ is two dimensional, we must have $\dim\text{rad}M=\dim(M/\text{rad}M)$. It is easy to check that any two dimensional $B_\lambda$-module is self-dual, so $\dim\text{rad}M=\dim(M/\text{rad}M)=2$, and $A$ is three dimensional and $A\cap\text{soc}B_\lambda=0$.

So $\text{soc}A$ is (without loss of generality, as we can freely add multiples of $s$ and $t$ to a basis of $\text{soc}A$) a two dimensional subspace of the span of $\{x_1,x_2,x_3,x_4\}$ with square zero.

But there is no such subspace. If $x=ax_1+bx_2+cx_3+dx_4$ with $x^2=0$, then $a$ and $b$ are each determined up to two possibilities by $c$ and $d$. So in a two dimensional vector space of such elements, all pairs of values of $c$ and $d$ would have to occur, and it is easy to check that no choices of $a$ and $b$ would then give a subspace.

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  • $\begingroup$ Thanks, also the linked article is very interesting. $\endgroup$ – Mare Apr 14 at 17:54
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Question 2 has a negative answer, even in dimension $d = 1$. I'm not sure about question 1, but it's certainly not true that the classification is independent of the field.

For question 2: let $k$ be an infinite field and $\lambda\in k^\times$. Then the bilinear form $\beta(a,b) := \lambda ab$ is nondegenerate, defining a commutative Frobenius algebra structure on $k$. These Frobenius algebras are nonisomorphic for different $\lambda$, exhibiting infinitely many nonisomorphic one-dimensional commutative Frobenius algebras over $k$. These are the only Frobenius algebra structures on $k$: the space of bilinear forms $k\times k\to k$ is one-dimensional, and we've already used everything except the zero form, which doesn't define a Frobenius structure because it's not nondegenerate.

Question 1 is a little strange to me: the quiver-and-relations presentation I know of doesn't tell you a Frobenius form; it only defines an associative algebra. Because the Frobenius form is additional structure, a positive answer wouldn't imply that the classification of commutative Frobenius algebras is independent of $k$, as the possible Frobenius forms could depend on $k$. And indeed, that's what we saw above: fixing $d = 1$, the isomorphism classes of one-dimensional commutative Frobenius $k$-algebras are in bijection with $k^\times$, and this certainly depends on $k$.

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    $\begingroup$ Thanks for your answer but by isomorphic I mean isomorphic as k-algebras. So any 1-dimensional k-algebras are isomorphic. I think you mean isomorphic in a more strict sense (as algebra and coalgebra?). $\endgroup$ – Mare Apr 26 '19 at 8:28
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    $\begingroup$ The Frobenius form can be read of from the socle of the algebra which is determined by the quiver and relations (the socle is the "largest non-zero" path). $\endgroup$ – Mare Apr 26 '19 at 8:31
  • $\begingroup$ @Mare indeed, I meant isomorphic as Frobenius algebras; sorry for the confusion. Since a Frobenius algebra is additional structure, rather than a condition, that seemed more natural to me, but what you're thinking about is still an interesting question. $\endgroup$ – Arun Debray Apr 26 '19 at 14:38
  • $\begingroup$ I didn't realize you could get the Frobenius form from the quiver and relations! That's neat; I'll have to look into it. $\endgroup$ – Arun Debray Apr 26 '19 at 14:39

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