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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.

View $A$ as an $R$-bimodule as follows: $(e \cdot f)(a)=e(ha)f(a)$ and $(f \cdot e)(a)=f(a)e(ta))$ for all $a \in Q_{1}$, $e \in R$, $f \in A$

For each $\tau \in Q_{1}$ consider the map $\gamma_{\tau}: Q_{1} \rightarrow K$ by $\gamma_{\tau}(\beta)=0$ if $\beta \neq \tau$ and $1$ if $\beta = \tau$, i.e the characteristic function.

It is clear then that $A = \bigoplus_{\tau \in Q_{1}} K\tau$.


Consider now the tensor product $A \otimes_{R} A$, let $\alpha$,$\tau$ in $Q_{1}$ and suppose we consider an element of this tensor product, say $\gamma_{\alpha} \otimes_{R} \gamma_{\tau}$. Why can this tensor be "identified" with a path of length $2$? Which quiver are we considering, are we constructing a quiver on the tensor product?

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up vote 2 down vote accepted

There is a basis of the tensor product $A^{\otimes n}$ given by $\gamma_1\otimes \cdots \otimes \gamma_n$ where $\gamma_1,\dots, \gamma_n$ are a list of $n$ elements of $Q_1$ that are composable. That is, where concatenating $\gamma_1\cdots \gamma_n$ gives a path of length $n$. So, just as $A$ has a basis given by paths of length 1, $A^{\otimes n}$ has a basis given by paths of length $n$ in the original quiver.

I'm not sure what you mean by "a quiver on the tensor product" but you never use any quiver other than the original one.

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I wanted to comment on Ben Webster's answer but I don't have enough reputation points.

Note that the tensor algebra $T_{R}(A):=\bigoplus_{d=0}^\infty A^{\otimes d}$ (where $A^{\otimes 0} := R$ and all tensor products are over the ring $R$) is isomorphic to the path algebra $KQ$, the $K$-algebra of (finite) linear combinations of formal compositions of (composable) arrows in $Q$ (paths) and where the product of two paths is defined by concatenation is the paths are composable and 0 if they are not.

Conversely, any finite dimensional $R$-$R$ bimodule $B$ is the arrow span of some quiver $Q_B$: Let $(Q_B)_0=Q_0$ and the arrows in $Q_B$ from $i$ to $j$ are in bijective correspondence with a basis of the finite dimensional vector space $e_i B e_j$ where $e_i(j)=\delta_{ij}$ for $i,j\in Q_0$ (note that $1_R=\sum_{i\in Q_0} e_i$). Then $B \cong K^{(Q_B)_1}$ and $KQ_B\cong \bigoplus_{d=0}^\infty B^{\otimes d}=T_R(B)$.

You can find this with more detail in Derksen, Weyman, Zelevinsky's first paper about quivers with potentials: at the beginning of section two.

One important thing of the tensor algebra approach to path algebras is that it is independent of $Q_1$ (of a choice o a basis), you only need the ring $R$. It also shows that, in some sense, you may view tensor algebras of finitely generated bimodules over a finite dimensional algebra as a generalization of path algebras.

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The answer to your first question is that, unless the tail of $\alpha$ equals the head of $\tau$, so that these two arrows form a path of length 2, the tensor product $\gamma_\alpha\otimes\gamma_\tau$ will be zero. To see this, consider any element $e\in R$ such that $e(t\alpha)=1$ (so $\gamma_\alpha\cdot e=\gamma_\alpha$) while $e(h\tau)=0$ (so $e\cdot\gamma_\tau=0$). Then, since the definition of tensor product equates $(\gamma_\alpha\cdot e)\otimes\gamma_\tau$ with $\gamma_\alpha\otimes(e\cdot\gamma_\tau)$, we get that $\gamma_\alpha\otimes\gamma_\tau=0$.

I don't understand your second question.

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