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This is a very simple question, but for some reason I'm having trouble with it. Here is the set up:

Let $k$ be an alg. closed field, let $Q$ be a quiver, and let $A = kQ$ be the path algebra of $Q$ over $k$. Then we have two functors, the $k$-dual (also called 'standard duality'), which is the contravariant Hom functor $D = Hom_k(-,k)$, and the $A$-dual, which is the contravariant functor $(-)^* = Hom_A(-,A)$.

My question is: how are these two functors related? How are they the same, and how are they different? I'm specifically thinking of quiver representations.

But, I'd prefer an answer in general: how does a/the "vector space duality" differ from a/the "algebra duality"? (Is that not the right way to pose the question?) Perhaps, how does the k-linear duality differ from the A-linear duality, conceptually?

[Edit, re: my comment, it was pointed out to me I originally messed up this example, so I am rewriting it to get the arrows correct] For concreteness: Let $Q = 1 \rightarrow 2 \rightarrow 3$ be a quiver, let $A=kQ$ be its path algebra. Take a representation, say, $M = k \rightarrow k \rightarrow 0$. Then $DM = Hom(k,k)\leftarrow Hom(k,k)\leftarrow Hom(0,k)$ and this is isomorphic to, say, $N = k \leftarrow k \leftarrow 0$. Then $N^*= Hom_A(A,k)\rightarrow Hom_A(A,k)\rightarrow Hom_A(A,0)$. Does this "just" end up being isomorphic to $M = k \rightarrow k \rightarrow 0$ again?

That being said, was my example too simple as to be trivial and not shed light on the situation? Or was I too fast and loose with my calculations, and perhaps said something very silly?

For the very generous: Can someone hold my hand through a more enlightening calculation, or otherwise point out where my lack of insight might be coming from?

Otherwise, just the answers to the highlighted questions would be appreciated!

Thanks!

For the record, my references for these questions are:

  • Auslander, Reiten, Smalø - Representations Theory of Artin Algebras

  • Assem, Simson, Skowronski - Elements of the Representation Theory of Associative Algebras

  • Schiffler - Quiver Representations

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    $\begingroup$ Oops! It looks like I reversed the position of the objects AND the orientation of the arrows, instead of just choosing one. $\endgroup$ – Samantha Y Dec 19 '16 at 16:21
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One great difference between the functors $D$ and $(-)^*$ is the following: $D$ is a duality (meaning that as a functor, it admits a quasi-inverse), while $(-)^*$ is not. (The name $A$-dual for $(-)^*$ is perhaps slightly misleading in this respect).

For instance, you can convince yourself that in your example with $Q = 1\to 2\to 3$, the functor $(-)^*$ sends all non-projective indecomposable representations to zero.

That being said, I'm not sure what can be said about comparing the two functors. Their composition, $\nu:= D(-)^*$, is called the Nakayama functor, and has many great properties (for instance, it restricts to an equivalence from the subcategory of projective representations to the subcategory of injective representations).

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  • $\begingroup$ Your comment on the misleading name seems like it might have been a place where I got confused. It fails to be a genuine duality (that is, a contravariant equivalence of categories), and for instance, you say $(-)^*$ sends a non-projective indecomposable representation to 0. Thus, take $S(1)$ to be the simple (non-projective) module $k\rightarrow 0 \rightarrow 0$, then $(S(1))^*=Hom_A(k, A) \leftarrow Hom_A(0, A) \leftarrow Hom_A(0,A)$... But this should be the 0 representation? I suppose my bigger issue is a lack of understanding how to compute $Hom_A(-, A)$. Could you explain? $\endgroup$ – Samantha Y Dec 19 '16 at 16:43
  • $\begingroup$ @SamanthaY It seems that there may be a slight misunderstanding in the way one computes $Hom_A(M,A)$. The exact place where it fails is when one writes $Hom_A(k,A)$; this does not make sense, as $k$ is not an $A$-module. Computing $Hom_A(S(1),A)$ will be harder, since the $A$-module structure of both $S(1)$ and $A$ have to be taken into account. However, your method for computing $D(M)$ works fine. $\endgroup$ – Pierre-Guy Plamondon Dec 19 '16 at 16:51
  • $\begingroup$ Ah, so I was 'playing it fast and loose' with my computation. Then I suppose my question once more boils down to something else: Where might I find a walkthrough of calculations involving this functor? I have trouble keeping track of 'reality' when I begin calculating. It seems that it is reasonable to say that $D$ "just turns around the arrows", but $(-)^*$ is much trickier! In particular, I don't quite know how to make sense of "since the $A$-module structure of both $S(1)$ and $A$ have to be taken into account." Earlier I had wanted to ask, how does $kQ$ "look" as an $A$-module? $\endgroup$ – Samantha Y Dec 19 '16 at 17:32
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    $\begingroup$ @SamanthaY You are right in saying that $(-)^*$ is much trickier to compute than $D$. Perhaps one way of computing $M^*$ would be the following: let $P_1\to P_0\to M\to 0$ be a projective presentation of $M$. Then the sequence $0\to M^* \to P_0^*\to P_1^*$ is exact, and $P_0^*$ and $P_1^*$ are projective $A^{op}$ modules. So $M^*$ is the kernel of the map $P_0^*\to P_1^*$. Moreover, $(-)^*$ induces a duality between the category of projective $A$-modules and that of projective $A^{op}$-modules. Hence the map $P_0^* \to P_1^*$ should not be too hard to deduce from $P_1\to P_0$. $\endgroup$ – Pierre-Guy Plamondon Dec 19 '16 at 17:56

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