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I am attending a course in foundations in representation theory and I am struggling to grasp the concept of path algebras, more explicitly why paths in a quiver form a $K$-algebra.

Let $Q$ be a quiver and $K$ be a field. Why do paths in a quiver $Q$ form a $K-$algebra, denoted $KQ$?

I understand that the multiplication of paths is associative and that there exists an identity element. Nonetheless no textbook I have studied explains why $KQ$ is a $K$-vector space and how the operation $+: KQ \times KQ \rightarrow KQ$, $(a,b) \mapsto a+b$ of paths is defined. Furthermore they don't explain the scalar multiplication that is required for a $K$-algebra

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To construct the path algebra $KQ$, one first forms the $K$-vector space with basis the set of all paths in $Q$. This then tells you what the addition and scalar multiplication are. In particular, a general element of $KQ$ is a finite $K$-linear combination of paths.

We then need to define the multiplication in $KQ$. Since this must be $K$-linear and distributive, it is enough to declare what the multiplication of two basis elements, i.e. paths, is. One proves that this is associative by checking on paths, and that the trivial paths $e_i$ for vertices $i$ form a complete set of orthogonal idempotents.

In fact, the path algebra $KQ$ is naturally graded by path length, so the trivial paths $e_i$ have degree zero and each arrow has degree 1.

As an example, take the quiver $1\to2\to3$. Then the path algebra is isomorphic to the algebra of lower triangular $3\times3$ matrices over $K$ $$ \begin{pmatrix}\ast&0&0\\\ast&\ast&0\\\ast&\ast&\ast\end{pmatrix}$$ where $e_i$ corresponds to the elementary matrix $E_{ii}$, the arrow $a\colon1\to2$ corresponds to $E_{21}$, the arrow $b\colon2\to3$ to $E_{32}$, and the path $ba\colon1\to3$ to $E_{31}$.

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  • $\begingroup$ Thank you for your answer:) I just don't quite understand how one forms the $K$-vector space with basis the set of all paths in $Q$. $\endgroup$ Nov 16, 2020 at 16:13
  • $\begingroup$ Given any set $X$, the set of all functions $X\to K$ is a $K$-vector space $K^X$ via point-wise addition and scalar multiplication. This has a subspace $K^{(X)}$ consisting of those functions having finite support, so those $f\colon X\to K$ such that $f(x)\neq0$ for only finitely many $x\in X$. This subspace has basis the co-ordinate functions $\delta_x$, where $\delta_x(y)=0$ unless $y=x$, in which case $\delta_x(x)=1$. In this way we have formed a vector space having basis indexed by $X$, also called the free vector space on the set $X$. $\endgroup$ Nov 16, 2020 at 22:09
  • $\begingroup$ @AndrewHubery Hello, I want to join the discussion because I have the same question as mathStudent. I understand what you are saying in your comment, but I do not get how this allows us to form the $K-$vector space with basis the set of all paths in $Q$. Could you please help me? $\endgroup$
    – kubo
    Sep 6, 2023 at 16:53
  • $\begingroup$ Another way to think about this is just to take a formal $K$-linear combination of paths. In the answer there is a quiver with three vertices, two arrows $a$ and $b$, and one path $ba$ of length 2. So our path algebra is six dimensional, containing elements of the form $x_1e_1+x_2e_2+x_3e_3+x_4a+x_5b+x_6ba$ with $x_i\in K$. The multiplication is associative, distributive, $K$-linear, so is completely determined once we know how to multiply paths, and this is exactly the same as matrix multiplication once the coefficients $x_i$ are placed in the appropriate positions in the $3\times3$ matrix. $\endgroup$ Sep 6, 2023 at 22:53
  • $\begingroup$ We put $x_1,x_2,x_3$ on the diagonal, $x_4,x_5$ on the lower diagonal, and $x_6$ in the bottom left corner. $\endgroup$ Sep 6, 2023 at 22:55

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