Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I've been trying for a while to get a real concrete handle on the relationship between representations and modules. To frame the question, I'll put here the standard situation I have in mind:

A ring $R$ lives in the category Ab of Abelian groups as an internal monoid $(\mu_R, \eta_R)$. A module is then just an Abelian group $A$ and a map $m : R \otimes A \rightarrow A$ that commutes with the monoid structure in the way you'd expect.

Alternatively, take an Abelian group $A$ and look at its group of endomorphisms $[A,A]$. This has an internal monoid $(\mu_A, \eta_A)$ just taking composition and identity. Then a representation is just a monoid homomorphism $(R, \mu_R, \eta_R) \rightarrow (A, \mu_A, \eta_A)$ in Ab. I.e. a ring homomorphism.

But then, Ab is monoidal closed, so these are the same concept under the iso

$$\hom(R\otimes A, A) \cong \hom(R, [A,A])$$

This idea seems to work for any closed category where one wants to relate a multiplication to composition. So, my question is, since these things are isomorphic in such a general context, why are they taught as two separate concepts? Is it merely pedagogical, or are there useful examples where modules and representations are distinct?

share|improve this question
I have never seen the words "module" and "representation" clearly distinguished. My internal stylist says "module" when $R$ is acting on an abelian group, and "representation" when the action, especially by something not a ring, but rather a group, etc., is acting on a vector space. Eventually you run into trouble with some definitions. For example, it's very hard to say when a map $G\to\operatorname{hom}(V,V)$ is algebraic if $G$ is an algebraic group and $V$ is $\infty$-dimensional. So there the only valid definition of "module" is a map $V\to k[G]\otimes V$. –  Theo Johnson-Freyd Jul 20 '10 at 17:44
Theo, you are correct: "rational representations" of algebraic groups are, in fact, modules. However, the situation in Lie theory is the opposite, i.e. there is a good theory of representations (with some intricate points), but while some modules have been considered, they do not entirely capture it, let alone define it. –  Victor Protsak Jul 20 '10 at 19:30
These and other answers/comments are good reminders of the linguistic subtleties here. There is a related usage issue: I tend to use irreducible and completely reducible for representations, simple and semisimple for modules. But people do mix the terms, and it's hard to say what is "correct". –  Jim Humphreys Jul 20 '10 at 19:55
add comment

4 Answers

up vote 6 down vote accepted

Here is my representation theorist's perspective: the key difference between representations and modules is that representations are "non-linear", whereas modules are "linear". I'll concentrate on the case of groups as the most familiar, but this applies more generally.

As Greg has already mentioned, in the most general sense, a representation is a homomorphism $f:G\to H,$ and usually there is no linear (or additive) structure on $H$, i.e. the set $f(g)$ need not be closed under sums; in fact, if $H$ is a non-abelian group, e.g. the symmetric group, the notion of sum doesn't even make sense (if $H=GL(V)$ then we may view its elements as endomorphisms of $V$ and add them, but this is unnatural since, by definition, $f$ is compatible with multiplicative structure). By contrast, a module involves a linear action $G\times V\to V,$ which is then "completed" by allowing arbitrary linear combinations, leading to certain technical advantages.

Here is an example of a construction that is very useful and makes perfect sense module-theoretically, but not representation-theoretically: change of scalars. Given a module $M$ over a group ring $R[G]$ and a commutative ring homomorphism $R\to S,$ one gets a module $S\otimes_R M$ over the group ring $S[G]$. Common examples involve extensions of scalars (e.g. from $\mathbb{R}$ to $\mathbb{C}$, from a field $K$ of definition to the splitting field, from $\mathbb{Z}$ to $\mathbb{Z}_p$) and, more to the point, reductions (e.g. from $\mathbb{Z}$ or $\mathbb{Z}_p$ to $\mathbb{Z}/p\mathbb{Z}$). The module language is, predictably, also very useful in providing categorical descriptions of various operations on representations, such as functors of induction and restriction,

$$Ind_H^G: H\text{-mod}\to G\text{-mod}\ \text{ and }\ Res_H^G: G\text{-mod}\to H\text{-mod},$$

where $H$ is a subgroup of $G,$ or the monoidal structure on $G$-mod.

Finally, here are two illustrations of the complementary nature of the two approaches besides the group case, in linear algebra. A single linear transformation $T:V\to V$ on a finite-dimensional vector space $V$ over $K$ is most naturally viewed as a representation (no additive structure); in this case, it's a representation of the quiver with a single vertex and a single loop. From this point of view, classification up to isomorphism is a problem about conjugacy classes of linear transformations,

$$T\to gTg^{-1},\ g\in GL(V).$$

By contrast, in the module style description we associate with $T$ a module over the ring $K[x]$ of polynomials in one variable over $K$ and classification problem reduces to the structure of modules over $K[x]$, which is a PID, with all the usual consequences. (Here the module picture is more illuminating.) If we consider a linear operator $S:V\to W$ between two different vector spaces,

$$S\to hSg^{-1},\ g\in GL(V),\ h\in GL(W),$$

and a classification up to isomorphism is accomplished by row and column reduction. The corresponding quiver $\circ\to\circ$ is a single arrow connecting two distinct vertices, but its path algebra is less familiar. (Here the representation theory picture is more illuminating.)

share|improve this answer
add comment

To expand on Tom's answer, the word "representation" is a 19th century word that originally meant "group homomorphism". If $f:G \to H$ is a homomorphism from a group $G$ to a group $H$, then $f(g)$ "represents" the element $g$. $H$ is taken to be a "familiar" or "explicit" group, usually a matrix group but also sometimes a permutation group.

The word "module" is a 20th century word (I think) that means "generalized vector space".

As has been pointed out, a representation of a group $G$ is equivalent to a $k[G]$-module. These days the terms are largely interchangeable; you can also talk about a representation of an algebra instead of a group. Certainly you can add topology to the conditions, for instance by using the group $C^*$-algebra of a locally compact topological group.

To the extent that there is still a useful distinction, there is a difference in emphasis. If a ring $R$ (or a group or whatever) acts on an abelian group $A$, and you consider its action to be a low-level structure, analogous to multiplying a vector by a scalar, then you should call $A$ an $R$-module. On the other hand, if you think of the action as a high-level geometric effect, analogous to a group acting on a manifold, then you should call it a representation. If you don't care, then you can use either term or both and it's all cool AFAIK :-). Possibly the word "module" is slowly supplanting the word "representation", because it's a shorter word as well as more modern and more general.

share|improve this answer
I like the explanation that the key difference in these two terms is emphasis. How far does this analogy go? Would it be fair to say for instance, modules of the enveloping algebra of sl(2) are a bit like vector spaces, but with some pretty beefy scalars? The primary concern here are the symmetries put on the space by the action of sl(2). I would say maybe yes. In plain old vector spaces, the scalar action is an important symmetry, so much so that we like to talk about identities "up to a scalar", etc. –  Aleks Kissinger Jul 20 '10 at 15:59
A purely historical/etymological remark, possibly wrong (maybe someone can confirm, correct, or deny this): I believe that long before anybody talked about $R$-modules (like maybe late 19th century) certain abelian groups were being called modules. Maybe what we now would call a lattice in a real vector space $V$ -- something which has an $\mathbb R$-basis of $V$ as a $\mathbb Z$-basis? Perhaps with $V=\mathbb C$? Is the term "module" in algebra distantly related to the terms "modular group" and "modular form" by this route? –  Tom Goodwillie Jul 20 '10 at 16:24
@Aleks: Sure, that's a valid viewpoint. The sort of thing that blurs the distinction between beefy scalars and puny scalars, even if you are conservative, is a module over (say) a polynomial ring. And even for a simple Lie algebra, the action of the Cartan subalgebra is very module-like, since in the integrable case it amounts to a graded vector space structure. –  Greg Kuperberg Jul 20 '10 at 17:34
@Tom: I looked in early papers in JSTOR and Google Books. In English, a few authors up to the 1930s used the word "module" on an ad hoc basis to sort-of mean an abelian group, for instance E.T. Bell and Harald Bohr. Suddenly in 1938 there are algebra papers by Nakayama that read like papers written today, with left and right modules of algebras, quotient modules, etc. Nakayama had read a German algebra book. It appears that the whole package of definitions came from the Gottingen school. –  Greg Kuperberg Jul 20 '10 at 17:48
Aleks: 1. In the interest of sanity, the word "scalar" should never be applied to non-commuting quantities. 2. The representation-module language use for Lie algebras mirrors the case of groups: a representation of $sl_2$ is a module over its universal enveloping algebra $U(sl_2).$ From the category theory point of view, a better formulation is that there is an equivalence of categories between {representations of $\mathfrak{g}$} and {$U(\mathfrak{g})$-modules}. –  Victor Protsak Jul 20 '10 at 17:55
show 2 more comments

It is certainly true that the category of representations of a group $G$ over a field $k$ is equivalent to the category of modules for the group ring $k[G]$, and it is often productive to rephrase questions about representations about questions about modules. Below, I give some examples of structure which is easier to discuss in terms of representations. But, as I will indicate, it is usually possible to rephrase in terms of modules with enough effort.

Tensor products: If $V$ and $W$ are two representations of $G$, then $V \otimes W$ has a natural structure as a $G$-representation. For $k[G]$ modules, this is not true; the tensor product has to be added as additional structure on the category $k[G]$-rep. Here is an explicit example: Let $G=\mathbb{Z}/4$ and let $H = \mathbb{Z}/2 \times \mathbb{Z}/2$. Then $\mathbb{C}[G]$ and $\mathbb{C}[H]$ are isomrphic rings, but the tensor structures on $\mathbb{C}[G]$-modules and $\mathbb{C}[H]$-modules are inequivalent. The same issue exists with duals. People who like rings better than groups would say that the issue is that I am talking about the algebra structure of $k[G]$ when I should be talking about the Hopf algebra structure.

Topology: Suppose that $G$ is a topological group (maybe a Lie group) and $k$ a toplogical field (maybe $\mathbb{R}$). Then a continuous representation of $G$ is a map $G \times V \to V$ which is a group action, $k$-linear, and continuous. I imagine there is a way to put a topology on $k[G]$ so that a continuous representation is a $k[G]$-module such that $k[G] \times V \to V$ is continuous, but I haven't seen it. And this will get worse with adjectives like smooth, algebraic, ...

share|improve this answer
Concerning your last sentence ('this will get worse with adjectives like ... algebraic'), "algebraic" representations of an affine group scheme $G$ over a field $k$ are actually co-modules for the Hopf algebra of regular functions $k[G]$. This means that the action of $G$ on the representation space $V$ is defined by a mapping $V \to k[G] \otimes V$ satisfying some natural diagrams. When $G$ is smooth and $k$ is alg. closed, one can of course view $V$ as a module for the group algebra of the "abstract" group $G(k)$, but as you point out, it isn't clear this is useful. –  George McNinch Jul 20 '10 at 15:48
TOPOLOGY: Representations of top. groups is probably a good foil for the mod/rep correspondence. For one thing Top fails to be cartesian closed, so the first crack at comparing these things by chasing the iso hom(GxV,V) \cong hom(G,[V,V]) fails unless V and G are nice spaces (compact Hausdorff or something). TENSORS: It seems the natural representation of G on V (x) W already (implicitly) uses the comultiplication from the Hopf algebra structure of k[G], i.e. the induced representation is "(psi (x) phi) o delta" ...where delta is the linear map that copies the basis elements of k[G]. –  Aleks Kissinger Jul 20 '10 at 15:48
Note: positive-dimensional vector spaces over non-discrete locally compact fields are never compact. There are some genuine difficulties in coming up with the correct notion of a group algebra in the topological setting that will faithfully capture representation-theoretic picture (see Kirillov's "Elements of representation theory"). Many of them concern a good category to which it should belong. The reduced $C^*$-algebra of an infinite discrete group is a good illustration. This cannot be fully computed even in easy cases. –  Victor Protsak Jul 20 '10 at 17:30
add comment

I would teach that an $R$-module is an abelian group $A$ plus a map of sets $R\times A\to A$ satisfying certain identities. I would probably also point out that this is the same thing as an abelian group $A$ plus a ring homomorphism $R\to End(A)$. I would also point out that "vector space" is the traditional term for "module" when $R$ is a field.

Similarly, I would teach that an action of a group $G$ on a set $X$ is a map of sets $G\times X\to X$ satisfying certain identities, and I would probably also point out that that is the same as a group homomorphism $G\to Aut(X)$; and if it seemed appropriate for those students I would also say that this second point of view is useful for generalizing the idea so as to make groups act on things other than sets.

An action of a group $G$ on a $k$-module is the same as a module for the group ring $kG$. You can also call this a representation of $G$ over $k$. This is not traditionally called a representation of $kG$.

The fact that there are overlapping definitions is just historical accident. The word "module" was in use in special cases long before there was category theory, even before there was abstract ring theory as we know it. So was the word "representation". The fact that the two terms are both still used is not because somebody decided on a good reason to keep them both; they just survived, as words do.

share|improve this answer
"The word "module" was in use in special cases long before there was category theory, even before there was abstract ring theory as we know it" - do you have a reference in support of that? Note that van der Waerden's Modern algebra used "abelian group with operators" (I personally think that "module" is a great improvement in terminology). I am only aware of unrelated uses of the word "module" (actually, modulus, pl moduli) prior to the modern era. –  Victor Protsak Jul 20 '10 at 17:39
I meant essentially what Greg says in his comment after my comment in the thread of his answer to this question. Namely, "module" was a word sometimes used for an abelian group in some contexts. I don't have any evidence. I wonder how far back it goes. I heard Serre use the expression "group with operators" in the 70s, but he wasn't talking about abelian groups. –  Tom Goodwillie Jul 20 '10 at 18:30
OK, this seems consistent with what I know, but note that the gap between Nakayama 1938 and Eilenberg-MacLane 1945 is less than 10 years. An interesting bit about Serre! French algebra nomenclature diverges from the English one in quite a few places. –  Victor Protsak Jul 20 '10 at 19:38
Serre was lecturing to undergraduates on the Jordan-Hölder theorem, and mentioning a generalization. –  Tom Goodwillie Jul 20 '10 at 20:35
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.