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 $R$. Thus, in some sense, you may view tensor algebras of finitely generated bimodules over a finite dimensional algebra as a generalization of path algebras.

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.

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,

Thus, in some sense, you may view tensor algebras of finitely generated bimodules over a finite dimensional algebra as a generalization of path algebras.

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 $R$. Thus, in some sense, you may view tensor algebras of finitely generated bimodules over a finite dimensional algebra as a generalization of path algebras.