Let $Q=(Q_{0},Q_{1},h,t)$ be a finite quiver where $Q_{0}$ are the vertices, $Q_{1}$ the arrows and we have two maps $h: Q_{1} \rightarrow Q_{0}$ (head) and $t: Q_{1} \rightarrow Q_{0}$ (tail). Fix a field $K$ and associative to $Q$ two vector spaces $R=K^{Q_{0}}$ and $A=K^{Q_{1}}$ i.e vector spaces consisting of $K$-valued functions on $Q_{0}$ and $Q_{1}$ respectively. Then the path algebra of $Q$ is defined as the tensor algebra $$R\langle A\rangle=\bigoplus_{d=0}^{\infty}A^d$$ You can see more details in Derksen, Weyman, Zelevinsky's first paper about quivers with potentials: http://arxiv.org/abs/0704.0649 at the beginning of section two.

But I know the path algebra $KQ$ of $Q$ is the K-algebra whose underlying K-vector space has as its basis the set of all paths and multiplication given by concatenation of paths.

My question is how to understand the two def. are equivalent? Whether Derksen, Weyman, Zelevinsky define the path algebra as the tensor algebra firstly? If not ,who can give me some related papers?


1 Answer 1


Both definitions are equivalent. To see this, consider the following decomposition of the path algebra $KQ$. Let us call $P(Q)$ the set of all paths of $Q$ and $P_n(Q)$ the set of all paths of length $n$, that is, compositions of $n$ arrows. We think of elements of $Q_0$ as paths of length $0$. It should be clear that

$$ P(Q) = \coprod_{n=0}^{\infty} P_n(Q) $$

where $\amalg$ denotes the disjoint union of sets. Since $KQ$ is the $K$-vector space with basis $P(Q)$, we have

$$ KQ \cong \bigoplus_{n=0}^{\infty} K P_n(Q) $$

where $KP_n(Q)$ is the $K$-vector space with basis $P_n(Q)$. Now let us see that $KP_n(Q)$ corresponds to $A^n$ in your notation.

First of all, you have to realize that $A^n$ is short for the tensor product of $n$ copies of $A$ over $R$, that is:

$$ A \otimes_R \ldots \otimes_R A $$

When $n=0$, this is $R$ by convention. Note that the structure of $A$ as a left and right $R$-module is described in that paper. Now consider the map of $K$-vector spaces defined on the basis

$$ KP_n(Q) \to A^n $$

$$ (f_1,\ldots,f_n) \mapsto f_n^* \otimes f_{n-1}^* \otimes \ldots \otimes f_1^* $$

where $f_i^* \colon Q_1 \to K$ takes the value $1$ in $f_i$ and $0$ in the rest of elements of $Q_1$. My notation $(f_1,\ldots,f_n)$ here means an $n$-tuple of elements of $Q_1$ with $h(f_i)=t(f_{i+1})$. It is not hard to check that this map is well defined and surjective. Then you can check that both $K$-vector spaces have the same dimension and so it is an isomorphism of $K$-vector spaces (or check injectivity by hand). The case $n=0$ is a bit different, just send an element of $Q_0$ to its dual.

All of these isomorphisms together give you an isomorphism $KQ \to R\langle A \rangle$ of $K$-vector spaces. You would also have to check that it is a map of algebras, but it is not hard. If you have problems with that, let me know and I can write more details.

  • $\begingroup$ Thank you very much. Good idea, beautiful proof. It is easy to understand. If we check that two vector spaces satisfy the compatible with the multiplication, whether it can guarantee the isomorphism is a map of algebras? $\endgroup$
    – Daisy
    May 27, 2017 at 2:18
  • $\begingroup$ can you show me why it is a well definied? $\endgroup$
    – Daisy
    May 27, 2017 at 3:54
  • $\begingroup$ Yes, if you have a map of $K$-algebras which is an isomorphism as a map of $K$-vector spaces, then it is an isomorphism of $K$-algebras. That is, the inverse is automatically a map of $K$-algebras. $\endgroup$
    – Goa'uld
    May 27, 2017 at 15:11
  • $\begingroup$ As for why it is well defined, to define a $K$-linear map on a $K$-vector space it suffices to define it on a basis. And the image of $(f_1,\ldots,f_n)$ is an element of $A^n$, so that's it. $\endgroup$
    – Goa'uld
    May 27, 2017 at 15:14

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