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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!

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  • $\begingroup$ 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? $\endgroup$ Mar 14, 2018 at 13:22
  • $\begingroup$ I belive I fixed the problem, sorry for the confusion $\endgroup$ Mar 14, 2018 at 14:17
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    $\begingroup$ What if $m=n{}$? $\endgroup$ Mar 14, 2018 at 15:33
  • $\begingroup$ 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 $x \in C$. $\endgroup$ Nov 19, 2021 at 20:28

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The example you have asked about does not make sense so I will not address it, but this notion of "generation" really is quite simple and is no different from the familiar sense of "generation" for any other algebraic structure.

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 $S(y)\cup M(f)(S(x))$.

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  • $\begingroup$ 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. $\endgroup$ Mar 16, 2018 at 8:40
  • $\begingroup$ 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. $\endgroup$ Mar 16, 2018 at 8:42

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