3
$\begingroup$

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 $A = \bigoplus_{\tau \in Q_{1}} K\tau$.

Question:

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?

$\endgroup$

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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: http://arxiv.org/abs/0704.0649 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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.