# Is there a difference between a model and a representation?

I'm thinking of models in logic here, vs. e.g. group representations.

Is there a difference between a model and a representation?

Could one not explain both at the same time?

A model gives an interpretation, but this might be viewed as a side effect to the work you do. You have some abstract axioms and you model/realize them with certain objects, which are part of another theory (e.g. the ordered pair concept is modeled via sets and $\in$). I don't see a real difference to e.g. a group representation, when you have some abstract multiplication laws and these come to live via a matrix representation of a specific dimension, say.

Also,

What kind of realizations do representable functors deal with?

I mean beyond the realization of groups like above.

Edit, from the comments: Like an example would be to consider integers $\mathbb{Z}$ and build the factor group $\mathbb{Z/2Z}$, this would be representing what is called $\mathbb{Z_2}$ (abstractly defined by the four relations between its two elements). The logic analog would be the sets and the abstract idea of a pair with its characterizing feature.

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Your middle paragraph suggests you are considering group presentations, rather than re presentations. Is it the case? –  M Turgeon Aug 7 '12 at 13:36
@MTurgeon: No, I don't think so. Like an example would be to consider integers $\mathbb{Z}$ and build the factor group $\mathbb{Z/2Z}$, this would be representing what is called $\mathbb{Z_2}$ (abstractly defined by the four relations between its two elements). The logic analog would be the sets and the abstract idea of a pair with its characterizing feature. –  NikolajK Aug 7 '12 at 13:40
I see. Then we don't have the same definition of group (re)presentation. –  M Turgeon Aug 7 '12 at 14:15
@MTurgeon: The example was using quite vague language, but the idea is to take the abstract thing and represent it in term of something which is already there. May that be a map $G\rightarrow GL(V)$ or a realization of some axioms using sets. –  NikolajK Aug 7 '12 at 16:23
@tomasz: I disagree. If you present a group by generators and relations and then try to find representations of it e.g. on sets, you'll end up doing something very similar to what you do when you try to write down models of a set of axioms. See my answer. –  Qiaochu Yuan Aug 7 '12 at 19:02

Well, I mean, they are representations of different kinds of things. But one can think of both as functors out of a given category.

For group representations, a group $G$ may be regarded as a category with one object and morphisms the elements of $G$. A group action of $G$ is then precisely a functor $G \to \text{Set}$; similarly, a linear representation of $G$ is precisely a functor $G \to \text{Vect}$. There are endless variations on this.

For models, a collection of axioms ought to describe category which is roughly speaking the free category containing a model of the axioms. This is easiest to explain for axioms nice enough that they can be described using a Lawvere theory, but I believe this formalism is more general than that. For example, the theory of groups can be described using a Lawvere theory which is described here. Given a Lawvere theory $C$, a model of $C$ is then precisely a product-preserving functor $C \to \text{Set}$. Again there are endless variations on this. See the Wikipedia article on categorical logic.

In the first example, the representable functor gives you the action of $G$ on itself (exercise), which is roughly speaking the free $G$-action on one element. In the second example, the representable functors for the Lawvere theory of groups give you free groups on finite sets (exercise).

Edit: Instead of reducing everything to category theory perhaps I should reduce everything to logic. I claim that representations are a special case of models.

For example, let $G$ be a group. Write down a first-order theory with one unary operation for every element of $G$ and one universally quantified axiom for every entry $g \times h = gh$ in the multiplication table of $G$. Then a model of this theory is precisely a set on which $G$ acts. (A similar but more complicated construction gives linear representations also as a special case of models.) Of course, much of this is redundant: it suffices to specify an operation for every element of a fixed set of generators of $G$ and an axiom for every relation in a fixed presentation relative to the generators.

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I don't know why there was a downvote here on this answer. It seems to answer my question with "there is no fundamental structural difference", but I'm not really sure. The second paragraph is not clear to me, especially the first sentence and which axioms you talk about. Are the axioms objects of a category?? And do you totally replace/reinterpret the model concept from logic with category theory here? –  NikolajK Aug 7 '12 at 14:06
@Nick: in the special case of a Lawvere theory, the objects are a free model $M$ and its finite products. The morphisms are the operations of the theory (e.g. for groups there is a multiplication morphism $m : M \times M \to M$ and an inverse morphism $i : M \to M$). The axioms, in this special case, describe relations between the morphisms (e.g. associativity specifies the relation $m(m(-, -), -) = m(-, m(-, -)) : M \times M \times M \to M$). –  Qiaochu Yuan Aug 7 '12 at 14:09
@QiaochuYuan: More concretely, one could describe the objects of the opposite of a Lawvere theory as the free model and its coproducts, and then the morphisms are just homomorphisms. –  Zhen Lin Aug 7 '12 at 14:38
Okay thanks, the "universal group" blog is interesting. I wonder: Can you abstractly specify e.g. $SU(3)$ without just stating the fundamental represenation from the start. What is the $G$ in $G\longrightarrow GL(\mathbb{C^3})$, if the fundamental representation is already on the right hand side of this expression. If you say "The unitary automorphisms of "$GL(\mathbb{C^3})$ whos elements are the kernel of the $\text{det}$-map, is this it? Is this really more abstract than the fundamental representation? (It feels like it's the same thing.)" –  NikolajK Aug 7 '12 at 20:37
@Nick: it is the same thing. Is this a problem? –  Qiaochu Yuan Aug 7 '12 at 21:37