# Finitely Generated Category Representation

I am studying representation theory of categories, and I am getting stuck on the following definition of finitely generated representations given in Sam & Snowden’s paper on Noetherian Categories.

Let $M$ be a representation of $C$, and define an element of $M$ as $M(x)$ for $x\in C$. Given any set $S$ of elements of $M$, there is a smallest subrepresentation of $M$ containing $S$; we call this the subrepresentation generated by $S$. We say that $M$ is finitely generated if it is generated by a finite set of elements.

I don’t understand what it means to “generate” a representation. For instance, if we consider the poset category $C$ with Ob$(C)=\mathbb{N}$ and Mor$(C)=\{n\to m \text{ if } n\leq m\}$. We could consider the representation $M:C\to Mod_k$ given by $n\mapsto Q^n$ and $(n\to m) \mapsto \Delta^{n\to m}$ where $\Delta^{n\to m}:\mathbb{Q}^n\to\mathbb{Q}^m$ is given by $(x_1,x_2,..,x_n)\mapsto (x_1,..., (m -times),..,x_1)$.

Is it possible to explain these concepts using this example? For instance, is $M$ finitely generated, and why?

Any insight is greatly appreciated, thank you!

• I can't see that your representation $\Delta ^{n\to m}$ is. In $\Bbb Q^n$ there are elements $(x_1,\ldots,x_n)$ where all the $x_i$ are distinct. Where is $\Delta ^{n\to m}$ supposed to send such an element? Mar 14, 2018 at 13:22
• I belive I fixed the problem, sorry for the confusion Mar 14, 2018 at 14:17
• What if $m=n{}$? Mar 14, 2018 at 15:33
• I think the quote from the paper by Sam and Snowden has a minor mistake: an "element of $M$" should be defined to be an element of $M(x)$ for some $x \in C$. Nov 19, 2021 at 20:28

For simplicity, let's first consider what happens if $C$ has only one object $x$, whose endomorphisms form a monoid $A$. A representation of $C$ is then just a $k$-module $M=M(x)$ with an action of the monoid $A$. For each $a\in A$, we have a homomorphism $M(a):M\to M$. We can think of our $M$ as being an algebraic structure which in addition to being a $k$-module also has a unary operation for each element of $A$.
If $S$ is a subset of $M$, then what is the subrepresentation generated by $S$? It's just the subset of $M$ which can be obtained from $S$ using all of the operations we have on $M$. That is, it is the set of all elements of $M$ we can obtain by repeatedly applying the module operations and the maps $M(a)$ for elements $a\in A$.
In the case that $C$ has more than one object the story is similar. You can again think of $M$ as an algebraic structure, consisting of a bunch of sets $M(x)$ for each object $x$, $k$-module structures on each of these sets, and maps between the sets for each morphism in $C$. Given a subset $S(x)\subseteq M(x)$ for each $x$, the subrepresentation they generate then just consists of the subsets of each $M(x)$ which can be obtained by repeatedly applying all these operations.
For a very simple concrete example, let $C$ be the category with two objects $x$ and $y$ and a single morphism $f:x\to y$ (and no other non-identity morphisms). A representation of $C$ then consists of two $k$-modules $M(x)$ and $M(y)$ together with a homomorphism $M(f):M(x)\to M(y)$. Given subsets $S(x)\subseteq M(x)$ and $S(y)\subseteq M(y)$, we can describe the subrepresentation $N$ which they generate as follows. $N(x)$ is the submodule of $M(x)$ generated by $S(x)$. For $N(y)$, we must take into account that we can get elements of $M(y)$ not just from elements of $S(y)$ but also by applying $M(f)$ to elements of $S(x)$. So $N(y)$ is the submodule of $M(y)$ generated by $S(y)\cup M(f)(S(x))$.
• This was extremely helpful. So for a representation M to be finitely generated, it means that there are a finite amount of elements $M(x)$ that generate all elements (modules) of $M$? A finite category will always have finitely generated representations. An infinite category can have finitely generated representations if the above statement holds. Mar 16, 2018 at 8:40
• Right, if $M$ is finitely generated, it means there are finitely many elements from which all others can be generated using all the operations. I don't know what you mean by "have finitely generated representations", though. For any (nonempty) category $C$, some representations are finitely generated and others are not. Mar 16, 2018 at 8:42