# Standard projective resolution of module of the path algebra

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?

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 $$P(j)\otimes M_i\to \big(P(j)\otimes M_j\big)\oplus\big(P(i)\otimes M_i\big), \quad p\otimes m\mapsto pa\otimes m-p\otimes am.$$

• 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?
– kubo
Oct 28, 2023 at 14:27
• 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
– kubo
Oct 28, 2023 at 15:56
• Also, shouldn't your last time be $ap\otimes m-p\otimes ma$?
– kubo
Oct 28, 2023 at 16:24
• 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 $g$. Oct 28, 2023 at 16:47
• 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. Oct 28, 2023 at 16:54