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Let $V$ be a representation of the path algebra $P_{Q}$. From this representation, we can construct a representation of $Q$ as follows: let $V_i =p_i V$, and for any edge $h$, let $x_h = a_h|_{p_h'} v : p_{h'} V \to p_{h'' } V$ be the operator corresponding to the one edge path $h$

My brain has some what exploded trying to understand thsi definition. I have tried to capture the three main doubts I have here:

  1. Before, I learned representation of an associative algebra, we consider a representation as a homomorphism from the algebra to the endomorphism set of a vector space. So, by extension, which vector space are we taking the endomorphism set of when dealing with quivers?

  2. And, secondly, what on earth is $x_h$ supposed to be here? I don't get the $a_h|_{p_{h}'}$ thing. What is a $p_{h}'$ doing under the $A$, and why are we mapping between the two $V_is$?

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There are two ways of defining a representation of a quiver:

First way: Construct the path algebra of the quiver, which is the associative (non-commutative) algebra over a field whose elements are linear combinations of paths in the quiver. Multiplication is defined by concatenating two paths whenever the first one ends where the other starts, setting the product to be zero if the two paths aren't compatible, and extending the multiplication linearly to linear combinations.

(Your source prefers to distinguish between the path in the quiver and the element in the algebra, which is where the $x_h$ comes from. If $h: a \rightarrow b$ is an edge of the quiver, then $x_h$ is the element of the algebra such that for any other path $P$:

  • the product $x_hx_P$ is given by adding the edge $h$ to the end of $P$ if $P$ ends at $a$, and $0$ otherwise.
  • the product $x_Px_h$ is given by adding the edge $h$ to the start of $P$ if $P$ starts at $b$, and $0$ otherwise.

The source does the same with the constant paths: if $i$ is a vertex of the quiver then $p_i$ is an idempotent element of the algebra that doesn't change paths that start (resp. end) at $i$, and sends other paths to zero.)

With this path algebra construct, define a representation of the quiver to be a representation of the algebra in the manner you already understand: a vector space $V$ and endomorphisms on that vector space for each algebra element.

Second way: Construct a new object called a representation, directly on top of the diagram that is the quiver. (In a manner reminiscent to a functor between categories, if that helps.) For each vertex $i$ pick a vector space $V_i$. For each edge $h:a \rightarrow b$, pick a linear transformation $x_h: V_a \rightarrow V_b$. This results in a nice little graph of vector spaces and maps drawn on top of the underlying quiver.


The point of your quoted statement is that these two definitions of representation are actually the same. For each vertex $i$ the element $p_i$ of the algebra is idempotent, so in the representation it is a projection that picks out a particular subspace of $V$. The sum of all the vertices is the identity of the algebra, so $V$ is a direct sum of the projected subspaces. This gives the vector spaces $V_i$ for each vertex.

Then for each edge $h:a \rightarrow b$, compatibility with the vertex elements in the algebra means that it maps one of these direct summands $V_a$ into another $V_b$, and is zero on the other summands. This means we can look at $x_h$ restricted to that summand $V_a$, which gives a linear map $V_a \rightarrow V_b$ for each $h$.

The combination of these gives a representation in the second sense from a representation in the first sense.

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  • $\begingroup$ Goddamn you're good at explainin $\endgroup$ Apr 1, 2023 at 12:17
  • $\begingroup$ So the points in the grap project sub spaces, and the edges moves between the projections $\endgroup$ Apr 1, 2023 at 12:28

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