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Hi,

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.

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    $\begingroup$ 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. $\endgroup$
    – MTS
    Oct 11, 2012 at 18:32
  • $\begingroup$ 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. $\endgroup$
    – Karl Schwede
    Oct 11, 2012 at 18:55
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    $\begingroup$ While the algebras are commutative (they are group algebras of commutative groups) the context is quite not that of commutative algebra, really. $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 11, 2012 at 19:38
  • $\begingroup$ 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. $\endgroup$
    – Bernhard Boehmler
    Oct 11, 2012 at 19:46

1 Answer 1

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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 $A^i_j$).

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  • $\begingroup$ 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. $\endgroup$
    – Bernhard Boehmler
    Oct 11, 2012 at 20:04
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    $\begingroup$ @Bernhard Boehmler: No, actually applying the duality should turn the picture upside down. $\endgroup$
    – Julian Kuelshammer
    Oct 20, 2012 at 9:08
  • $\begingroup$ Yes, you're right. Thank you for the comment. $\endgroup$
    – Bernhard Boehmler
    Dec 17, 2012 at 10:11

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