First take a question as an example:

Let $f:L\to M$ be an irreducible morphism in $\mathrm{mod}-A$ and $X$ be a right $A$-module. Show that $\mathrm{Ext}_A(X,f):\mathrm{Ext}_A(X,L)\to\mathrm{Ext}_A(X,M)$ is a monomorphism,if $\mathrm{Hom}_A(M,X)=0$.

What is the special meaning of $\mathrm{Ext}$ functor in the quiver representation?

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    $\begingroup$ I'm not really sure what you're asking here - Ext means the same thing it always does, it's the derived functor of Hom, and parameterizes extensions. Remember that a quiver representation is the same thing as a module over the path algebra, so Ext groups for representations are the same thing as Ext groups for the corresponding modules. $\endgroup$ – mdp May 7 '12 at 12:32
  • $\begingroup$ @MattPressland Could you detail that in an answer for those of us (like myself) who don't have a good feel for what "it always does"? I would greatly appreciate it! $\endgroup$ – rschwieb May 7 '12 at 15:56
  • $\begingroup$ I am probably not the best person to explain that, but if nobody does it fairly soon I'll have a go! My point at least thus far was that there isn't anything fundamentally different about Ext in the quiver context. $\endgroup$ – mdp May 7 '12 at 16:02
  • $\begingroup$ I know the ext functor in the homological algebra,such as the long exact sequence,but I find it is hard to treat some questions in the quiver representation. $\endgroup$ – Strongart May 8 '12 at 11:50

As Matt has said in the comment, representation of quivers, and modules over the corresponding path algebra are really the same thing. Since you said you understand what it means to take Ext in homological algebra, i.e. taking Ext of modules over an algebra. So what you want for the answer is just $Rep(kQ/I) \simeq A-mod$. Here $A\cong kQ/I$, or in general, can be taken as Morita equivalent to $kQ/I$ (which is a basic algebra, meaning simple modules are all 1 dimensional).

We prove simply $Rep(Q)\simeq kQ-mod$ here.

If you have $M\in kQ-mod$, to get a representation of $Q$, take $M_x = e_x M$ where $e_x$ is the trivial path (i.e. primitive idempotent) corresponding to $x\in Q_0$.

If you have $V=\bigoplus_{x\in Q_0} V_x\in Rep(Q)$, then take the vector space $M=V$, and define the action of path $x_1\to \cdots \to x_k$ on $V$ (this lies in the path algebra $kQ$) as composition $V \twoheadrightarrow V_{x_1}\cdots V_{x_k} \hookrightarrow V$. This then gives you a module over the algebra $kQ$.

Hope this helps. Being able to think of a module in terms of representation of quiver is exactly why quiver is so useful to study representation theory. One should never be afraid of representation of quiver. IMO, I find it much harder to study modules over arbitrary algebra.

  • $\begingroup$ Thank you,but I need some skill to computer.I find the following formula is often used:Ext(M,N)=DHom(M,τN), but how to play the Hom functor and τ,here the module is noted by dimension vector?Assem's book example 6.2.8 is such a case,I can not understand the (T2) step. $\endgroup$ – Strongart Jun 27 '12 at 10:43
  • $\begingroup$ The forumla is Ext($M,N$)$\simeq$DHom($N,\tau M$). $\tau M$ in 6.2.8 is calculated in 4.4.12, which gets you to the last line of (T2), the last equality comes from the usual combinatorial game one plays with quiver representation. $\endgroup$ – Aaron Jun 28 '12 at 16:50
  • $\begingroup$ Note that if $f:M\to N$ is a non-zero map, then the simple modules lying on the top (head) of $M$ must appear in $N$. (or better: quotient of $M$ must appear as submodule of $N$) In 6.2.8 (T2) last line, this means the only Hom we need to look at is the second direct summand in the first term and the first direct summand in the second term (in the last line). The former is in fact $P(4)$ and the later is $S(2)$, so obviously (by combinatorial game, or better by Schur Lemma), there is no non-zero hom $P(4)\to S(2)$, so the whole Hom group is zero. $\endgroup$ – Aaron Jun 28 '12 at 16:55
  • $\begingroup$ Yes, i opposite the M and N, that is a mistake. In 4.4.12,there is an injection to computer τ^(-1), then I think we can get τ from the surjection, in the (T1) part,there are two short sequences, but the first term is just P(2),but the second term is not P(3).Why? $\endgroup$ – Strongart Jun 30 '12 at 10:30
  • $\begingroup$ For the hom functor,how about the nonzero situation? For example, I cannot understand the forth equation in the Ex6.3.11(a). I need more help the treat the quiver. $\endgroup$ – Strongart Jun 30 '12 at 10:35

In answer to your additional question about Example VI.3.11 in [ASS]:

The way too see how the module manifest after tilting is probably not so apparent from the text. The main thing is to note that now the elements in your new algebra $B=End_A(T)$ is given by homomorphisms. The good thing of computation with quiver is you can usually do this rather combinatorially (i.e. looking at the arrows). We look at this example in detail: $3\xleftarrow{\beta}2\xleftarrow{\alpha}1$. Projective indecomposables: $$P(1)=\begin{array}{c}1 \\2\\3\end{array}, P(2)=\begin{array}{c} 2 \\ 3\end{array}, P(3)=\begin{array}{c}3\end{array}$$ On the right hand side of the equal is the way we 'draw' the module according to its Loewy structure (radical+socle series), (we should in fact draw them as $S(1)$, $S(2)$, $S(3)$, by for simplicity, we will always omit $S(-)$). You can 'in some sense' says that for $P(1)$, the $1$ is representing $e_1$, the $2$ representing $\alpha$ and the $3$ representing $\alpha\beta$. A homomophism between modules must map from the top the domain to somewhere in the target module with the same composition factor, the Loewy structure helps us to visualise what kind of map is possible. Now the tilting module: $$T(1) = 1, T(2)=\begin{array}{c}1\\ 2\\ 3\end{array}, T(3) = 3$$ Now there is no hom from $T(1)$ to $T(2)$ because, by above statement, 1 must maps to 1, so we have the following diagram: $$ \begin{array}{ccc} 1 &\to & 1 \\ & & 2 \\& & 3\end{array}$$ which does not make sense because you don't have anything that maps to 2 and 3 in the target. On the other hand, reversing the arrow is a valid homomorphism. Moreover, this map does not factor through $T(3)$, therefore, by denoting this map as $\lambda^o$, on the quiver of $B^{op}$, we have $1\xleftarrow{\lambda^o}2$ (where 1 represents $T(1)$ and so on). $T(3)$ is a simple and is socle of $T(2)$, so we also have a map, which does not factor through $T(1)$, denote by $\mu^o$. After reversing the arrow (taking the opposite ring), this gets us the quiver of $B$, $1\xrightarrow{\lambda}2\xrightarrow{\mu} 3$. The following diagram suffices to say $\mu^o\lambda^o(=\lambda\mu)=0$: $$ \begin{array}{ccccc} 1 &\leftarrow & 1 & & \\ & & 2 & &\\& & 3 & \leftarrow & 3\end{array}$$ Now the projective indecomposables of $B$ are $$ P_B(3) = 3, P_B(2) = \begin{array}{c}2\\ 3\end{array}, P_B(1) = \begin{array}{c}1\\ 2\end{array}$$ Note the primitive idempotents are given by $f_i:T(i)\to T(i)$ (corresponding to the top of these PIMs). The four equations can now be seen:

  1. $Hom_A(T,3)= K-span\{f_3\}\cong S_B(3)$
  2. $Hom_A(T,\begin{array}{c}1\\2\\3\end{array}) = K-span\{f_2,\mu^o\} \cong P_B(2)$
  3. $Hom_A(T,\begin{array}{c} 1 \\ 2\end{array})$: There is only one map, given by canoncial quotienting of $T(3)$ from $T(2)$, which can be visualise as: $$ \begin{array}{ccccc} & & 1 & \xrightarrow{f_2} & 1 \\ & & 2 & \to & 2\\ \hline 3 &\xrightarrow{\mu^o} & 3 & \to & 3 \end{array}$$ Hence this hom group is given by $K$-span$\{ f_2\}$ quotient by $K$-span$\{\mu^o\}$, this corresponds to $P_B(2)$ quotient out by its socle (=$K$-span$\{\mu\}$. Hence giving us $S_B(2)$
  4. $Hom_A(T,1)=K$-span$\{f_1,\lambda^o\}\cong P_B(1)$
  • $\begingroup$ Thanks,I read this answer many times,but also have some questions.Why there is no hom from T(1) to T(2),as we know,if f∈Hom(M,N),f is not must be onto N.Then we note i to T(i) and get new quiver,also have the arrow λ from 1 to 2,does it means the hom functor from T(1) to T(2)? $\endgroup$ – Strongart Jul 7 '12 at 5:20
  • $\begingroup$ why would $f$ be surjective? You can try to write down the map explicity. I will omit K-span from now on, $T(1)=\{\overline{e_1}\}$, $T(2) = P(1) = Ae_1 = \{e_1,\alpha,\alpha\beta\}$. We "guess" the "map" $f:T(1)\to T(2)$ is given by $\overline{e_1}\mapsto e_1$, but this is not an $A$-module map because $f(\overline{e_1})\alpha = \alpha \neq 0 = f(0) = f(\overline{e_1}\alpha)$. I donot quite understand your second point, the arrow $2\xrightarrow{\lambda^o}1$ is a map (element of Hom-group), not a hom functor. When we take opposite ring, this becomes $1\xrightarrow{\lambda}2$. $\endgroup$ – Aaron Jul 7 '12 at 18:19
  • $\begingroup$ Oh,I have not realized the A-module before.For the second part,Does λ∈Hom(T(1),T(2))?Maybe it is also not a A-module map,so λ can be nonzero.Am I right? $\endgroup$ – Strongart Jul 11 '12 at 5:10
  • $\begingroup$ $\lambda$ is $\lambda^o$. When we take endomorphism ring, we take the opposite algebra, so we label the corresponding of $a\in A$ as $a^o\in A^{op}$. The previous comment showed $Hom_A(T(1),T(2))=0$ already. $\endgroup$ – Aaron Jul 11 '12 at 15:21
  • $\begingroup$ Oh,thanks.I have another question about the tilting module:math.stackexchange.com/questions/155187/…, welcome to answer it if it is not too much trouble for you. $\endgroup$ – Strongart Jul 13 '12 at 11:36

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