2
$\begingroup$

A commutative algebra (with unity) over a field gives rise to the covariant functor F: Set_f->Vect from finite sets to vector spaces: F(E) := A^{otimes E}. Is it true that, over complex numbers, a finite dimensional algebra can be reconstructed from the corresponding functor?

(A Gamma-module is a functor from finite pointed sets to vector spaces; so F is not a Gamma-module. I use this term in the title just because I do not know the correct term for F: Set_f->Vect.)


Let me clarify my question. For a commutative algebra $A$ we define a functor $F:\mathrm{Set}_\mathrm{f}\to\mathrm{Vect}$ by

$F(I)=A^{\otimes I}$ for a finite set $I$ and

$F(t):F(I)\to F(J)$, $\bigotimes_{i\in I}a_i\mapsto\bigotimes_{j\in J}\prod_{i\in t^{-1}(j)}a_i$ for a map $t:I\to J$ (exactly as Andreas Blass proposed).

Suppose now that two finite-dimensional algebras $A$ and $B$ over the complex numbers produce isomorphic functors $F$ and $G$. Is it true that then $A$ and $B$ are isomorphic?

The question is not trivial. Let $e:F\to G$ be an isomorphism of functors. Then $e_{\{1\}}:A\to B$ and $e_{\{1,2\}}:A\otimes A\to B\otimes B$ are isomorphisms of vector spaces. If we had $e_{\{1,2\}}=e_{\{1\}}\otimes e_{\{1\}}$, this would imply that $e_{\{1\}}$ is an isomorphism of algebras. The problem is that we have only linear naturality relations between $e_I$.

$\endgroup$
2
  • 1
    $\begingroup$ How is F defined on arrows? Like, what is the morphism in Vect corresponding to the only function {1,2}->{1}? $\endgroup$ Oct 7, 2010 at 18:53
  • 1
    $\begingroup$ I think it would help if (a) you were more explicit about how you define $F$ (b) you explained what you mean by reconstructing the algebra from the functor. I think I can guess the answer to both questions but it is hard to be sure I'm right. $\endgroup$ Oct 7, 2010 at 18:57

2 Answers 2

7
$\begingroup$

I'd guess that the intended functor uses the multiplication operation of $A$ to provide $F(s):A\otimes A\to A$ where $s$ is the surjection $\{1,2\}\to\{1\}$ mentioned by Mattia Talpo, that it uses the unity element of $A$ to provide $F(i):k\to A$ where $i$ is the injection $\emptyset\to\{1\}$ and $k$ is the field of scalars, and that $F$ is to be defined on arbitrary maps between finite sets by a natural generalization and combination of these two examples. The trouble with this guess is that it makes the question trivial, since the algebra structure is contained in $F(s)$ and $F(i)$, so no work is needed to reconstruct the algebra from $F$. Unfortunately, I have no alternative guess as to what $F$ the proposer might have intended.

$\endgroup$
2
  • $\begingroup$ My thoughts exactly :) $\endgroup$ Oct 7, 2010 at 23:59
  • $\begingroup$ @Andreas: I also thought that $F$ has this action on morphisms, but since $ab \otimes c$ and $a \otimes bc$ are not equal in $A^{\otimes 2}$, I doubt that this is functorial. $\endgroup$ Oct 8, 2010 at 7:19
1
$\begingroup$

Now I can prove this, see http://www.pdmi.ras.ru/~ssp/te.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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