MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I am reading this paper and have the following question:.

On page 5 you can see different types of $A_n$-modules ($\ A_n=k[x,y]/\langle x^2,y^{n+2},xy^{n+1}\rangle\ $)$\ $ and later in the paper it says that $DA_i^j$ is the dual module of $A_i^j$.

$A_i^j$ is an $A_n$-module with $k$-basis {$x,xy,...,xy^n,\ y^j,...,y^{n+1}$}, $A_n$ acts on $A_i^j$ by multiplication and we have $A_i^j\subseteq A_n$.

Question: Why does the shape of $DA_i^j$ look like what it looks like, that means, why do the pictures of $A_i^j$ and $DA_i^j$ on page 5 not coincide?

In the paper, $k$ need not be algebraically closed, but if it were, then I would think that you could compute $DA_i^j$ by the formula $DA_i^j=$Hom$_k(A_i^j,k)$.

If I do this, the shape of $DA_i^j$ looks like the shape of $A_i^j$, but that isn't the case in the paper.

Is Hom$_k(A_i^j,k)$ isomorphic to the module $DA_i^j$ in the paper, or, in general, in what different way do they compute the shape of $DA_i^j$?

Thanks for the help.

share|cite|improve this question
You're more likely to get people to help if you define the objects you're talking about. What is $A_i^j$? What do you mean by the "shape" of a module? Also, there are multiple possible meanings of "dual module". Often the dual of an $R$-module $M$ is $\mathrm{Hom}(M,R)$. So you might need to replace $k$ with $A_n$ in your definition of dual module. – MTS Oct 11 '12 at 18:32
Also, if these algebras are commutative, you could also try adding the tag ac.commutative-algebra $$\text{ }$$ Besides the dual that MTS suggested, I could also imagine $\text{Hom}(M, \omega_R)$ where $\omega_R$ is a canonical module. – Karl Schwede Oct 11 '12 at 18:55
While the algebras are commutative (they are group algebras of commutative groups) the context is quite not that of commutative algebra, really. – Mariano Suárez-Alvarez Oct 11 '12 at 19:38
Thanks for your comments! I changed my question a little bit and will try to compute Hom$_{A_n}(M,A_n)$ and Hom$(M,{\omega}_{A_n})$ now. @Mariano Suárez-Alvarez: Ok, sorry, I'll delete this tag. – Bernhard Boehmler Oct 11 '12 at 19:46
up vote 1 down vote accepted

In the context of this part of representation theory, $D$ is usually just the standard $k$-duality, so $DM:=Hom_k(M,k)$.

EDIT: But actually in this paper, $DA^j_i$ is not the dual of $A^j_i$. Actually the notion "dual" only appears once in this paper, where it is stated that $DA^0_n$ is dual to $A^0_n$, which is indeed true. For other $i$ and $j$ I think one should interpret the picture of $DA^i_j$ just as its definition (not as some functor $D$ of $A^i_j$).

share|cite|improve this answer
In the pictures (in the paper) of the modules, the vertices coincide with basis vectors and the lines with the action of $A_n$. Therefore, by using your formula, I got as a result that the pictures of $A_i^j$ and $DA_i^j$ should coincide, but that is not the case in the paper, and I wonder, why. – Bernhard Boehmler Oct 11 '12 at 20:04
@Bernhard Boehmler: No, actually applying the duality should turn the picture upside down. – Julian Kuelshammer Oct 20 '12 at 9:08
Yes, you're right. Thank you for the comment. – Bernhard Boehmler Dec 17 '12 at 10:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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