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I have the quiver $Q$ with two vertices $v_1,v_2$ and two arrows $a_1,a_2$ that go from $v_1$ to $v_2$. Let $kQ$ be the path algebra of my quiver, where $k$ is a field, and let $M=kQ/\langle a_1\rangle$ be a $kQ$-module. I want to find its standard projective resolution (or if you prefer, the standar projective resolution of the representation that corresponds to the $kQ$-module $M$.)

According to Schiffler's Quiver Representations, there is a standard projective resolution of the form $$0\longrightarrow P_1 \longrightarrow P_0\longrightarrow M \longrightarrow 0$$ with $$P_0=\bigoplus_{i\in Q_0}P(i)^{\text{dim }M_i} \quad \mbox{and} \quad P_1=\bigoplus_{\alpha\in Q_1}P(t(\alpha))^{\text{dim } M_{s(\alpha)}}$$ I can construct the $P_0$ and $P_1$, but I have problems when it comes to defining the maps $f:P_1\longrightarrow P_0$ and $g:P_0\longrightarrow M$.

I start with $g$. In my lecture notes it says that I should define $g=(g_j)_j$ with $$g_j:\bigoplus_{i\in Q_0}P(i)_j\otimes M_i\longrightarrow M_j$$ as follows: for $c=\alpha_1\ldots \alpha_r$ a path from $i$ to $j$ in $Q$ and $m_i\in M_i$, we set $g_j(c\otimes m_i)=(\varphi_{\alpha_r}\ldots \varphi_{\alpha_1})(m_i)$ (this is like a coordinate-by-coordinate definition). Now my problem is that I do not know how the $P_0$ relate to the $\bigoplus_{i\in Q_0}P(i)_j\otimes M_i$. I can construct the $g_j$ maps, but then I do not know how to define the $g$ map overall. For example, I have that \begin{align*} g_v(e_v\otimes \lambda e_v)&=\lambda e_v\\ g_w(e_w\otimes (\lambda_1e_w+\lambda_2 b))&=\lambda_1e_w+\lambda_2 b\\ g_w(b\otimes \lambda e_v)&=\lambda b \end{align*} But I do not know how to construct the $g$ from this (the domains of these functions are not $P_0$). Can someone help me?

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1 Answer 1

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Let $M$ be any (left) $kQ$-module. Then for the trivial paths $e_i$ we have that $M_i:=e_iM$ is a $k$-vector space, and for each arrow $a:i\to j$, left multiplication by $a$ induces a $k$-linear map $M(a):M_i\to M_j$. (In this way $M$ determines a $k$-representation of the quiver.)

If we now choose a $k$-basis $m_{ir}$ for $M_i$, then we have an isomorphism of $kQ$-modules $$ P(i)\otimes_kM_i \cong P(i)^{\dim M_i}, \quad \sum_r p_r\otimes m_{ir} \mapsto (p_1,p_2,...) \textrm{ with } p_r\in P(i). $$ As $P(i)=kQe_i$ is a submodule of $kQ$, with basis those paths starting at $i$, we can now define a map $$ P(i)\otimes_kM_i\to M, \quad a_s\cdots a_1\otimes m \mapsto M(a_s)\cdots M(a_1)m, $$ where $a_s\cdots a_1$ is a path starting at $i$. This can then also be written as a map $P(i)^{\dim M_i} \cong P(i)\otimes_kM_i \to M$, and taking the sum over all vertices thus yields a map $P_0\to M$.

Explicitly, for the Kronecker quiver with vertices $1,2$ and arrows $a,b:1\to 2$, a module $M$ is given by two vector spaces $M_1,M_2$, and two linear maps $M(a),M(b):M_1\to M_2$.

The projective module $P(2)$ has $P(2)_1=0$ and $P(2)_2=k$. The projective module $P(1)$ has $P(1)_1=k$ and $P(1)_2=k^2$, with maps $P(1)(a)=\binom10$ and $P(1)(b)=\binom01$.

Then $P(2)\otimes M_2$ is just $M_2$ at vertex 2, whereas $P(1)\otimes M_1$ has $M_1$ at vertex $1$, $M_1^2$ at vertex 2, and the two maps $\binom{\mathrm{id}}0$ and $\binom0{\mathrm{id}}$.

Putting this together, we see that $P_0$ has vector space $M_1$ at vertex 1, $M_1^2\oplus M_2$ at vertex $2$, and linear maps $(\mathrm{id},0,0)^t$ and $(0,\mathrm{id},0)^t$. The map $P_0\to M$ acts as the identity $M_1\to M_1$ at vertex 1, and the map $(M(a),M(b),\mathrm{id})\colon M_1^2\oplus M_2\to M_2$, $(m_1,m_1',m_2)\mapsto M(a)m_1+M(b)m_1'+m_2$ at vertex 2.


Using a bit more theory we know that for any algebra $A$, idempotent $e\in A$, and $A$-module $M$ we have $$ \mathrm{Hom}_A(Ae,M) \cong eM, \quad f\mapsto f(e). $$ This generalises the isomorphism $\mathrm{Hom}_A(A,M)\cong M$, $f\mapsto f(1)$. Choosing a basis $m_r$ for $eM$, we have maps $f_r\colon Ae\to M$, $e\mapsto m_r$, and then a 'universal' map $(f_1,f_2,...):(Ae)^{\dim eM}\to M$, which we can compactly write as $Ae\otimes eM\to M$, $a\otimes m\mapsto am$.

With this more compact notation, we easily get the map $P_1\to P_0$. For, $P_1=\bigoplus_{a:i\to j}P(j)\otimes M_i$. Then $P(j)a\subseteq P(i)$ and $aM_i\subseteq M_j$, so we have a natural map $$ P(j)\otimes M_i\to \big(P(j)\otimes M_j\big)\oplus\big(P(i)\otimes M_i\big), \quad p\otimes m\mapsto pa\otimes m-p\otimes am. $$

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  • $\begingroup$ Thank you very much for your answer. I understand how you form the tensor products of $P(1)$ and $M_1$ (same for $2$), but I do not see why is that the way in which they are formed. Could you please give me a hint on this? $\endgroup$
    – kubo
    Oct 28, 2023 at 14:27
  • $\begingroup$ Also, I wanted to ask how do you define the map $g$. It apperats that you didn't do it the way I said it in my question $\endgroup$
    – kubo
    Oct 28, 2023 at 15:56
  • $\begingroup$ Also, shouldn't your last time be $ap\otimes m-p\otimes ma$? $\endgroup$
    – kubo
    Oct 28, 2023 at 16:24
  • $\begingroup$ The isomorphism from $P(i)\otimes_kM_i$ to $P(i)^{\dim M_i}$ is just choosing a basis. The map $g_i$ from the tensor product as in your notes is exactly the one I wrote down as well, so eg. sending $a\otimes m$ to $M(a)m$ for an arrow $a$ starting at $i$. Choosing bases and taking the direct sum yields the map $g$. $\endgroup$ Oct 28, 2023 at 16:47
  • $\begingroup$ If you do this for the Kronecker $kQ$, with vertices $1,2$ and arrows $a,b:1\to2$, and the $3$-dimensional module $M=kQ/(a)$, we have $M_1=k$ (with basis $e_1$),$M_2=k^2$ (with basis $b,e_2$), and maps $M(a)=0$, $M(b)=(1,0)^t$. The map $g_1$ sends $e_1\otimes m_1$ to $m_1$, $a\otimes m_1$ to 0, and $b\otimes m_1$ to $bm_1$. The map $g_2$ sends $e_2\otimes m_2$ to $m_2$. Using the bases for $M_i$ and taking the direct sum yields $P_0=P(1)\oplus P(2)^2$, with the appropriate map. $\endgroup$ Oct 28, 2023 at 16:54

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