2
$\begingroup$

I'm looking at a simple example module over an k-algebra, and trying to calculate the nakayama functor applied to that module. I can't seem to get it right/understand how to do it.

Let k-algebra $A = k[1 \rightarrow2]$

Let left A-module $P = 0 \rightarrow k$

I know P is projective indecomposable (corresponding to vertex 2 in the quiver), so applying the nakayama functor I know I should get the injective indecomposable corresponding to vertex 2:

$I = k \xrightarrow{1} k$

However another definition of the nakayama functor is also:

$\nu(-)=D(Hom_A(-,A))=Hom_k(Hom_A(-,A), k)$

Where A is seen as a left module over itself. Applying this to P, I'm struggling a bit understanding how to calculate $Hom_A(P,A)$, and how it should end up as I after applying D.

My attempt:

I want to look at the space of maps from P to A (as left modules). We can represent the left module A as:

$[k \xrightarrow{1} k] \oplus [0 \rightarrow k] = k \xrightarrow{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} k^2 $

I want to figure out how left A-module homomorphisms from P to A look like.

Let f: $[0 \rightarrow k] \rightarrow [k\xrightarrow{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} k^2] $

Need to determine what f does with each component in P. The first component in P is 0, so there it's always the zero map (into the first component of A). In the second component of P, we need to specify a map $k \rightarrow k^2$, so here there are 2 dimensions of freedom. So atleast we know the space of maps from P to A is 2-dimensional as a k-vector space, the same as I. This should be promising, but at this point is where I'm kinda stuck. If I try to represent $Hom_A(P,A)$ as a $A^{op}$ module, I need to put 2 copies of k somewhere. What seems most reasonable is to do like this:

$0 \leftarrow k^2$,

since both dimensions of maps came from this second component. (Also, is it correct to turn the arrow around here, because $Hom_A(-,A)$ is contravariant, and hence $Hom_A(P,A)$ is a $A^{op}$ module?)

Afterwards I want to apply duality $D(-)=Hom_k(-,k)$, which if i understand correctly, amounts to just turning the arrows again. Thus i end up with

$0 \rightarrow k^2$

Which I know is incorrect, since I should end up with the injective I.

Some thoughts:

A important part for a morphism to be part of $Hom_A(P,A)$, is that it has to respect the action of the arrows. However in my example, since the arrow is just the zero map, it has no effect on what the maps from the other component can be. Does this change when the arrow changes direction?

I have an idea of how it could make sense to put one of the copies of k in the first vertex, but I can't quite seem to formalize it. Will try to describe it here. In the representation for A, one dimension of the second vertex "comes from" the first vertex through the arrow. This same dimension is also responsible for one of the degrees of freedom when looking at maps from P to A. Then I want to argue that we pull this dimension back to the first vertex in P, since this is the common "parent" for both these maps. But I have no idea if this makes any sense.

To summarize my questions:

I know the dimension of $Hom_A(P,A)$ as a k-vector space, how do I figure out the rest, where to put those copies of k, and what the maps between them should be?

When I apply duality, is it as simple as just turning the arrows?

Is there something I seem to not understand correctly? Are my definitions/assumptions correct?

$\endgroup$

1 Answer 1

2
$\begingroup$

There are several parts to computing this, since one should first understand both constructions $\mathrm{Hom}_A(M,A)$ and $DN$, which are best thought of in terms of $A$-modules, and then we want to turn $D\mathrm{Hom}_A(M,A)$ back into a quiver representation. I will go through these constructions in general, before answering your specific problem at the end.

If $M$ is a left $A$-module, then the vector space $\mathrm{Hom}_A(M,A)$ becomes a right $A$-module via $$ fa \colon M\to A, \quad m\mapsto f(m)a. $$ Let $e_i$ be the idempotent corresponding to vertex $i$. Then $Ae_i=P_i$ is the indecomposable projective $A$-module corresponding to vertex $i$, and $$ \mathrm{Hom}_A(M,A)e_i = \mathrm{Hom}_A(M,Ae_i) = \mathrm{Hom}_A(M,P_i). $$ If $\alpha\colon i\to j$ is an arrow, then this yields an $A$-module map $$ \alpha^\ast \colon P_j \to P_i, \quad a \mapsto a\alpha \quad\textrm{for $a$ in }P_j=Ae_j. $$ We therefore get an induced linear map $$ \alpha^\ast \colon \mathrm{Hom}_A(M,P_j) \to \mathrm{Hom}_A(M,P_i), \quad f\mapsto\alpha^\ast f, \textrm{ which sends $m$ to }f(m)\alpha. $$

We next consider the vector space duality $D$. If $N$ is a right $A$-module, then $DN$ is a left $A$-module via $$ a\xi\colon N\to k, \quad n\mapsto\xi(na). $$ In particular, $e_iDN=D(Ne_i)$. For an arrow $\alpha\colon i\to j$ we have the linear map $Ne_j\to Ne_i$, $n\mapsto n\alpha$, and then the induced linear map $$ \alpha \colon D(Ne_i) \to D(Ne_j), \quad \xi\mapsto\alpha\xi, \textrm{ which sends $n\in Ne_j$ to $\xi(n\alpha)$.} $$ (Note that this is where the arrows get reversed: $\alpha\colon i\to j$ yields a map $Ne_j\to Ne_i$ for a right $A$-module $N$.)

We now put this together. For a left $A$-module $M$, the quiver representation corresponding to $D\mathrm{Hom}_A(M,A)$ has the vector space $D\mathrm{Hom}_A(M,P_i)$ at vertex $i$, and for an arrow $\alpha\colon i\to j$, we have the linear map $$ \alpha\colon D\mathrm{Hom}_A(M,P_i) \to D\mathrm{Hom}_A(M,P_j), \quad \xi\mapsto\alpha\xi, \textrm{ which sends $g\in\mathrm{Hom}_A(M,P_j)$ to $\xi(\alpha^\ast g)$}. $$

OK. So what does this do in your special case, where $M=P_2$? We have $\mathrm{Hom}_A(M,P_2)=k\,\mathrm{id}$, and $\mathrm{Hom}_A(M,P_1)=k\,\alpha^\ast$, where $\mathrm{id}$ is the identity map and $\alpha\colon 1\to 2$ is the arrow.

Let $\xi$ and $\eta$ be the respective dual bases, so that $D\mathrm{Hom}_A(M,A)$ is one dimensional at each vertex, with basis $\eta$ at vertex 1 and $\xi$ at vertex 2.

Finally, $\alpha\eta(\mathrm{id})=\eta(\alpha^\ast)=1=\xi(\mathrm{id})$, so $\alpha$ sends $\eta$ to $\xi$, and we have the quiver representation $k \xrightarrow{1} k$ as expected.

$\endgroup$
3
  • $\begingroup$ Thank you very much for the answer. One question: You say Hom(M,A) is best thought of as an A-module. As far as I understand, quiver representations and A-modules are essentially the same (equivalent categories). So if I want to find a representation of HomA(M,A), will it then be a representation of the opposite quiver? Will I get any useful information out of this approach? Or is there a specific reason you prefer to look at this from the perspective as A-modules? $\endgroup$
    – Hodge
    Commented Mar 16, 2022 at 0:40
  • $\begingroup$ The categories are equivalent, but forming a quiver representation requires decomposing the underlying vector space using a complete set of orthogonal idempotents. It’s a bit like always choosing a basis for a vector space, so is sometimes useful, but not always. $\endgroup$ Commented Mar 16, 2022 at 6:43
  • $\begingroup$ A right $kQ$ module is the same as a left module over the opposite algebra, which is just the path algebra for the opposite quiver $kQ^{\mathrm{op}}$. In my answer this shows up when we look at the duality and the action of an arrow. $\endgroup$ Commented Mar 16, 2022 at 6:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .