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I study the quiver just some weeks and I cannot understand the projective module in the quiver representation well.Here are some questions:

Suppose $Q$ is a quiver, $a\in Q_{0}$.

1)Show that the projective $KQ$-module $P(a)$ is simple iff $a$ is a sink.

2)If $Q$ is acyclic, $P(a)=(P(a)_{b},\phi_{\beta})$ be the indecomposable projective. Show that for each arrow $\beta$, the map $\phi_{\beta}$ is injective.

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    Q is a finite quiver.2012-04-07

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I'll assume for now that your quiver is unbound, i.e. you aren't taking a quotient by an admissible ideal.

1) I'm assuming from the phrasing of question 2) that you know what the indecomposable projectives are and the simples are. Then if $a$ is a sink, there is only the constant path $a\to a$, so $P(a)_a=K$, and there are no paths $a\to b$ for any $b\ne a$, so $P(a)_b=0$ when $b\ne a$. But this is exactly the description of the simple module $S(a)$.

Conversely, if $a$ is not a sink, then $\dim{P(a)_a}\geq1$, as there is at least the constant path at $a$. Assume first that $Q$ is acyclic. Then as $a$ is not a sink, there must be a vertex $b\ne a$ and an arrow $a\to b$, so $\dim P(a)_b\geq1$, and again $P(a)$ cannot be simple, as simples of an acyclic quiver have non-zero vector spaces at only a single vertex.

Now if $Q$ is cyclic, but there are no paths from $a$ to a vertex involved in an cycle, then all of the vector spaces in $P(a)$ associated to vertices on the cycle are zero (as there are no paths). So we induce a representation of a quiver obtained by deleting edges in the cycles (which have associated map the zero map between zero vector spaces) to obtain an acyclic subquiver $\widetilde{Q}$. The representation $P(a)$ on $Q$ restricts to the representation $P(a)$ on $\widetilde{Q}$, so we can apply the argument from the acyclic case to find a proper submodule. Adding back the deleted edges, and their associated zero maps, we obtain a proper submodule of $P(a)$ on $Q$.

Finally, if $Q$ is cyclic and there is a path from $a$ to a vertex on a cycle $C$, then the dimension of $P(a)_b$ is infinite for any $b$ such that there is a path to $b$ from a vertex of $C$. Take $m$ to be the length of the shortest path from $a$ to $c$, and $n$ to be the number of edges in $C$. We get a subrepresentation $R$ by choosing $R_v=0$ if there is no path to $v$ from a vertex of $C$, and otherwise taking $R_v$ to be the subspace of the infinite dimensional space $P(a)_v$ spanned by paths $a\to v$ of length at least $m+n$ (this is to ensure the subspace is proper - we have excluded the direct path from $a\to v$ that does not include a full lap of the cycle). The restriction of the maps in $P(a)$ to $R$ are either zero, or increase the length of paths, so we have a proper subrepresentation and therefore $P(a)$ is not simple.

(Note: The acyclic case is more complicated than I was expecting, and you should check carefully for any errors. If anybody can think of a simpler argument I'd be interested to see it.)

2) Note that $P(a)_b$ is the $K$-vector space spanned by paths $w$ from $a\to b$. For each arrow $\beta:a\to c$, the map $\phi_\beta$ takes $w$ to $w\beta$, a path from $a$ to $c$. So let $v=(a|v_1,v_2,\ldots,v_k|b)$ and $w=(a|w_1,w_2,\ldots,w_l|b)$ be paths $a$ to $b$. Then if $v\beta=w\beta$, the sequence $(w_1,\ldots,w_l,\beta)$ is equal to the sequence $(v_1,\ldots,v_k,\beta)$, and so $(w_1,\ldots,w_l)=(v_1,\ldots,v_k)$ and $v=w$. So the map is injective.

I think the reason that acyclic-ness is introduced here is that if the quiver has a directed cycle, the zero ideal is not admissible, so you're forced to take a non-trivial quotient of the path algebra by an admissible ideal $\mathcal{I}$, which means the vector space $P(a)_b$ is no longer generated by paths $a\to b$, but by equivalence classes of paths modulo this ideal. In this case the result in 2) may no longer apply, as although $w\beta$ and $v\beta$ are distinct paths, it is possible that $v\beta-w\beta\in\mathcal{I}$.

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    Maybe I do not understand the sink well,because of the dual claim"the injective KQ-module I(a) is simple iff a is a sourse",I just guess it is a target.But from your answer,maybe a is a sink means no paths from a→b for any b≠a.Am I right?2012-04-07
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    Yes, that would be the usual definition of a sink. A source only has outgoing arrows, and a sink only has incoming arrows.2012-04-07
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    In fact,I am reading the book "Elements of the Representation Theory of Associative Algebras(1)" by Assem,it do not define the sink,but define the source as a start point,dual with the target.2012-04-09
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    In the book p76,there is lemma2.1 says "The set {S(a);a∈Q$_0$} is complete set of representative of the isomorphism classes of the simple A-module."Here S(a)=(S(a)$_b$,0),S(a)$_b$=0,if b≠a,so we can do it directly.2012-04-09
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    For the question 2,the arrow in the answer maybe do not match well and I need a reason to cancel β.2012-04-09
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    Ah, right, Assem refers to the source of the arrow for what I would call the tail. Source and sink as I defined above are (I believe) standard terms in graph theory. Did you get the questions from the book? I can't seem to find them. The reason I was a little wary in my answers is that Lemma 2.1 depends on the quiver being bound, so it must either be acyclic or come with a non-trivial ideal. This is why I tried to show directly that there would be a proper subrepresentation, rather than show that $P(a)$ wasn't one of the $S(a)$s - if $Q$ is unbound then there might be other simples.2012-04-09
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    The idea in 2 is not so much to cancel $\beta$, but to use the definition of the path algebra to observe that the only way two paths can be equal after you add $\beta$ to the end of them is if they were already equal before. (This definitely need not be true if you're working with $KQ/\mathcal{I}$ for some admissible ideal $\mathcal{I}$, even if the quiver is acyclic).2012-04-09
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    Yes,my question is from Assem's book,Ch3,Ex9a and 16a.In fact,I do not understand the projective module in quiver representations well.I am ashame to ask too such,so just choose two questions as some examples2012-04-12
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    Now I can understand the acyclic situation.but for the cyclic,I do not know the meaning of "P(a) is simple",how does it correspond to the quiver picture? For the case 2,I want to know where to use the condition "indecomposable".2012-04-12
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    You use the indecomposable condition in 2) because it's what gives you a concrete description of the map $\phi_\beta$. (There is also a general philosophy of only needing to understand indecomposable objects because of Krull-Schmidt). In the cyclic case, simples can take all sorts of strange forms, which is why my argument reduces to a more abstract representation theoretic one rather than using the quiver. I don't have a good picture of what simples can look like in cyclic quivers.2012-04-12
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    An example I tend to keep in mind when thinking about the projectives is the linear $A_n$ case (so an $A_n$ quiver with all arrows pointing the same direction). Then the projectives (and indeed injectives) have a particular nice form in terms of their dimension vectors.2012-04-12
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    Thank you very much.Maybe I ask too much,I will try to digest these answers.At last,maybe Assem's book is a little hard for the beginner,does any easier book?2012-04-15
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    That is the most useful book on quiver representations that I know, but there may be other good ones. What kind of background do you have in ring theory and module theory in general? He seems to assume quite a lot of that.2012-04-16
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    I have finished Lam's book:A First Course in Noncommutative Rings and Lectures on Modules and Rings(Some chapters in the second book are not understand well).I think my problem is how to translate the ring theory into quiver.maybe I need a table to show the corresponding relationship.2012-04-17
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    I think really it would be good to understand the equivalence of the categories of $KQ$-modules and of representations of $Q$. This equivalence tells you how to translate between $KQ$-modules and $Q$ representations. Assem also explains when the category of $A$-modules for some general associative algebra $A$ is equivalent to a category of $KQ$-modules for some $Q$ (I think this was true for all $A$, with $Q=Q_B$, where $B$ is the basicization of $A$, but maybe this statement is slightly too general).2012-04-17
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    Oh,I remember that I have read a note which says that the equivalence of the categories of KQ -modules and of representations of Q is similar with the group representations.2012-04-19
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    I think it again and find a simple model to explain it.Take the group representations as a example,G is a group,V is representation space,the equivalence is just"(GV)V=G(vv)",here VV induce the matrix in the representations.2012-04-19