Hi,


I am reading [this](http://math.bnu.edu.cn/~huwei/paper/Holm-Hu-1.pdf) 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.



Question about the shape of the dual module of a module
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 $A^i\_j$).


