In this section we will talk about structures with three operations. These are called algebras. We will start by defining an algebra over a field, which is a vector space with a bilinear vector product. After giving some examples, we will then move to a discussion of quivers and their path algebras. == Algebras over a Field == Definition 1: Let F {\displaystyle F} be a field, and let A {\displaystyle A} be an F {\displaystyle F} -vector space on which we define the vector product ⋅ : A × A → A {\displaystyle \cdot \,:\,A\times A\rightarrow A} . Then A {\displaystyle A} is called an algebra over F {\displaystyle F} provided that ( A , + , ⋅ ) {\displaystyle (A,+,\cdot )} is a ring, where + {\displaystyle +} is the vector space addition, and if for all a , b , c ∈ A {\displaystyle a,b,c\in A} and α ∈ F {\displaystyle \alpha \in F} , a ( b c ) = ( a b ) c {\displaystyle a(bc)=(ab)c} , a ( b + c ) = a b + a c {\displaystyle a(b+c)=ab+ac} and ( a + b ) c = a c + b c {\displaystyle (a+b)c=ac+bc} , α ( a b ) = ( α a ) b = a ( α b ) {\displaystyle \alpha (ab)=(\alpha a)b=a(\alpha b)} .The dimension of an algebra is the dimension of A {\displaystyle A} as a vector space. Remark 2: The appropriate definition of a subalgebra is clear from Definition 1. We leave its formal statement to the reader. Definition 2: If ( A , + , ⋅ ) {\displaystyle (A,+,\cdot )} is a commutative ring, A {\displaystyle A} is called a commutative algebra. If it is a division ring, A {\displaystyle A} is called a division algebra. We reserve the terms real and complex algebra for algebras over R {\displaystyle \mathbb {R} } and C {\displaystyle \mathbb {C} } , respectively. The reader is invited to check that the following examples really are examples of algebras. Example 3: Let F {\displaystyle F} be a field. The vector space F n {\displaystyle F^{n}} forms a commutative F {\displaystyle F} -algebra under componentwise multiplication. Example 4: The quaternions H {\displaystyle \mathbb {H} } is a 4-dimensional real algebra. We leave it to the reader to show that it is not a 2-dimensional complex algebra. Example 5: Given a field F {\displaystyle F} , the vector space of polynomials F [ x ] {\displaystyle F[x]} is a commutative F {\displaystyle F} -algebra in a natural way. Example 6: Let F {\displaystyle F} be a field. Then any matrix ring over F {\displaystyle F} , for example ( F 0 F F ) {\displaystyle \left({\begin{array}{cc}F&0\\F&F\end{array}}\right)} , gives rise to an F {\displaystyle F} -algebra in a natural way. == Quivers and Path Algebras == Naively, a quiver can be understood as a directed graph where we allow loops and parallell edges. Formally, we have the following. Definition 7: A quiver is a collection of four pieces of data, Q = ( Q 0 , Q 1 , s , t ) {\displaystyle Q=(Q_{0},Q_{1},s,t)} , Q 0 {\displaystyle Q_{0}} is the set of vertices of the quiver, Q 1 {\displaystyle Q_{1}} is the set of edges, and s , t : Q 1 → Q 0 {\displaystyle s,t\,:\,Q_{1}\rightarrow Q_{0}} are functions associating with each edge a source vertex and a target vertex, respectively.We will always assume that Q 0 {\displaystyle Q_{0}} is nonempty and that Q 0 {\displaystyle Q_{0}} and Q 1 {\displaystyle Q_{1}} are finite sets. Example 8: The following are the simplest examples of quivers: The quiver with one point and no edges, represented by 1 {\displaystyle 1} . The quiver with n {\displaystyle n} point and no edges, 1 2 . . . n {\displaystyle 1\quad 2\quad ...\quad n} . The linear quiver with n {\displaystyle n} points, 1 ⟶ a 1 2 ⟶ a 2 . . . → a n − 1 n {\displaystyle 1\,{\stackrel {a_{1}}{\longrightarrow }}\,2\,{\stackrel {a_{2}}{\longrightarrow }}\,...\,{\xrightarrow {a_{n-1}}}\,n} . The simplest quiver with a nontrivial loop, 1 ⇆ b a 2 {\displaystyle 1{\underset {a}{\stackrel {b}{\leftrightarrows }}}2} .Definition 9: Let Q {\displaystyle Q} be a quiver. A path in Q {\displaystyle Q} is a sequence of edges a = a m a m − 1 . . . a 1 {\displaystyle a=a_{m}a_{m-1}...a_{1}} where s ( a i ) = t ( a i − 1 ) {\displaystyle s(a_{i})=t(a_{i-1})} for all i = 2 , . . . , m {\displaystyle i=2,...,m} . We extend the domains of s {\displaystyle s} and t {\displaystyle t} and define s ( a ) ≡ s ( a 0 ) {\displaystyle s(a)\equiv s(a_{0})} and t ( a ) ≡ t ( a m ) {\displaystyle t(a)\equiv t(a_{m})} . We define the length of the path to be the number of edges it contains and write l ( a ) = m {\displaystyle l(a)=m} . With each vertex i {\displaystyle i} of a quiver we associate the trivial path e i {\displaystyle e_{i}} with s ( e i ) = t ( e i ) = i {\displaystyle s(e_{i})=t(e_{i})=i} and l ( e i ) = 0 {\displaystyle l(e_{i})=0} . A nontrivial path a {\displaystyle a} with s ( a ) = t ( a ) = i {\displaystyle s(a)=t(a)=i} is called an oriented loop at i {\displaystyle i} . The reason quivers are interesting for us is that they provide a concrete way of constructing a certain family of algebras, called path algebras. Definition 10: Let Q {\displaystyle Q} be a quiver and F {\displaystyle F} a field. Let F Q {\displaystyle FQ} denote the free vector space generated by all the paths of Q {\displaystyle Q} . On this vector space, we define a vector product in the obvious way: if u = u m . . . u 1 {\displaystyle u=u_{m}...u_{1}} and v = v n . . . v 1 {\displaystyle v=v_{n}...v_{1}} are paths with s ( v ) = t ( u ) {\displaystyle s(v)=t(u)} , define their product v u {\displaystyle vu} by concatenation: v u = v n . . . v 1 u m . . . u 1 {\displaystyle vu=v_{n}...v_{1}u_{m}...u_{1}} . If s ( v ) ≠ t ( u ) {\displaystyle s(v)\neq t(u)} , define their product to be v u = 0 {\displaystyle vu=0} . This product turns F Q {\displaystyle FQ} into an F {\displaystyle F} -algebra, called the path algebra of Q {\displaystyle Q} . Lemma 11: Let Q {\displaystyle Q} be a quiver and F {\displaystyle F} field. If Q {\displaystyle Q} contains a path of length | Q 0 | {\displaystyle |Q_{0}|} , then F Q {\displaystyle FQ} is infinite dimensional. Proof: By a counting argument such a path must contain an oriented loop, a {\displaystyle a} , say. Evidently { a n } n ∈ N {\displaystyle \{a^{n}\}_{n\in \mathbb {N} }} is a linearly independent set, such that F Q {\displaystyle FQ} is infinite dimensional. Lemma 12: Let Q {\displaystyle Q} be a quiver and F {\displaystyle F} a field. Then F Q {\displaystyle FQ} is infinite dimensional if and only if Q {\displaystyle Q} contains an oriented loop. Proof: Let a {\displaystyle a} be an oriented loop in Q {\displaystyle Q} . Then F Q {\displaystyle FQ} is infinite dimensional by the above argument. Conversely, assume Q {\displaystyle Q} has no loops. Then the vertices of the quiver can be ordered such that edges always go from a lower to a higher vertex, and since the length of any given path is bounded above by | Q 0 | − 1 {\displaystyle |Q_{0}|-1} , there dimension of F Q {\displaystyle FQ} is bounded above by d i m F Q ≤ | Q 0 | 2 − | Q 0 | < ∞ {\displaystyle \mathrm {dim} \,FQ\leq |Q_{0}|^{2}-|Q_{0}|<\infty } . Lemma 13: Let Q {\displaystyle Q} be a quiver and F {\displaystyle F} a field. Then the trivial edges e i {\displaystyle e_{i}} form an orthogonal idempotent set. Proof: This is immediate from the definitions: e i e j = 0 {\displaystyle e_{i}e_{j}=0} if i ≠ j {\displaystyle i\neq j} and e i 2 = e i {\displaystyle e_{i}^{2}=e_{i}} . Corollary 14: The element ∑ i ∈ Q 0 e i {\displaystyle \sum _{i\in Q_{0}}e_{i}} is the identity element in F Q {\displaystyle FQ} . Proof: It sufficed to show this on the generators of F Q {\displaystyle FQ} . Let a {\displaystyle a} be a path in Q {\displaystyle Q} with s ( a ) = j {\displaystyle s(a)=j} and t ( a ) = k {\displaystyle t(a)=k} . Then ( ∑ i ∈ Q 0 e i ) a = ∑ i ∈ Q 0 e i a = e j a = a {\displaystyle \left(\sum _{i\in Q_{0}}e_{i}\right)a=\sum _{i\in Q_{0}}e_{i}a=e_{j}a=a} . Similarily, a ( ∑ i ∈ Q 0 e i ) = a {\displaystyle a\left(\sum _{i\in Q_{0}}e_{i}\right)=a} . To be covered: - General R-algebras