# explicitly represent a representable functor

Assume you were given a functor $$F : k\text{-}\mathbf{alg}\to\mathbf{set},$$ with the additional information that it is representable. Is there then a procedure to find an object $A$ that represents $F$? (In other words, an object $A$ such that there is a natural isomorphism $h^A \simeq F$.)

For instance, we could think of the functor $G_m : R \mapsto R^\times$, where $R$ is a $k$-algebra and $R^\times$ is the set of invertible elements in $R$. Then the object $A = k[t,t^{-1}]$ will do, but this already requires a moment of thought. Is there an infallible way to find this?

• My guess is no. Usually in algebraic geometry people prove certain functors are representable and then just name the space representing it something rather than figure out if it is some known space. – Matt Apr 17 '13 at 16:02
• That is not what the functor $\mathbb{G}_m$ does. It sends $R$ to the set of invertible elements in $R$. – Qiaochu Yuan Apr 17 '13 at 18:21
• @Matt: Yes, but in algebraic geometry usually one doesn't ask for affine representability, which makes the issue a lot more complicated than the problem here. – Martin Brandenburg Apr 17 '13 at 23:15
• @Qiaochu Yuan: I must admit I was thinking about fields. Fortunately, people were smart enough to figure out what I meant. – Myself Apr 18 '13 at 7:58
• Maybe the paper "ON THE EQUATIONS DEFINING AFFINE ALGEBRAIC GROUPS" by VLADIMIR L. POPOV could be related. – Watson Nov 25 '17 at 22:36

Zhen Lin's Postscript already answers the general question: If $F$ is isomorphic to a functor of the form $R \mapsto \{a \in R^I : f_j(a)=0 \forall j\}$, where $I$ is a set and $f_j \in k[\{X_i\}_{i \in I}$ are polynomials, then $F$ is represented by $k[\{X_i\}_{i \in I}]/(f_j)$. For example, the functor $R \mapsto \{(a,b,c) \in R^3 : 1+ab=c^2, c \in R^*\}$ is isomorphic to $\{(a,b,c,d) \in R^4 : 1+ab=c^2,cd=1\}$ and therefore represented by $k[x_1,x_2,x_3,x_4]/(1+x_1 x_2-x_3^2,1-x_3 x_4)$.

Let me apply this to a specific situation where you probably don't see a representing object immediately. I hope that this illustrates how one can find a representing object quite systematically.

Let $G$ be a group and fix some $n \in \mathbb{N}$. If $R$ is a $k$-algebra, let $F_G(R)$ be the set of $n$-dimensional representations of $G$ over $R$, i.e. homomorphisms $G \to \mathrm{GL}_n(R)$. The action on morphisms is clear, thus we get a functor $F_G : \mathsf{Alg}(k) \to \mathsf{Set}$. I claim that $F_G$ is representable by some $k$-algebra $R_k(G)$. Of course this is clear by the usual Adjoint Functor Theorem quoted by Zhen Lin. But how does $R_k(G)$ look like?

For $G=\mathbb{Z}$ we have $F_{\mathbb{Z}}(R)=\mathrm{GL}_n(R)$, which is clearly represented by $R_k(\mathbb{Z})=k[\{X_{ij}\}_{1 \leq i,j \leq n},\mathrm{det}(X_{ij})^{-1}]$ (using the universal properties of polynomial algebras and localization).

If $G=G_1 \sqcup G_2$ is a coproduct of two groups, then $F_{G}=F_{G_1} \times F_{G_2}$, thus $R_k(G)=R_k(G_1) \sqcup R_k(G_2)$ as $k$-algebras, where $\sqcup=\otimes_k$ here. The same works for infinitely many groups. So we have found $R_k(G)$ for free groups $G$.

If $G_1 \rightrightarrows G_2 \to G$ is a coequalizer, then $F_G \to F_{G_1} \rightrightarrows F_{G_2}$ is an equalizer, hence $R_{G_1} \rightrightarrows R_{G_2} \to R_G$ is a coequalizer. Since every group $G$ is some coequalizer as above with $G_1,G_2$ free, we have constructed $R_k(G)$.

Note that although the construction of $R_k(G)$ depends on a free presentation of $G$ and is therefore terribly uncanonical, the definition via the universal property $\hom_{\mathsf{Alg}(k)}(R_k(G),R) \cong \hom_{\mathsf{Grp}}(G,\mathrm{GL}_n(R))$ means that $R_k(G)$ is canonically determined.

Here is an example: Let $n=1$ and $G=\mathbb{Z}/5$. We have $R_k(\mathbb{Z})=k[x,x^{-1}]$. Since $\mathbb{Z} \rightrightarrows \mathbb{Z} \to \mathbb{Z}/5$ is a coequalizer, where the two maps are multiplication with $0$ resp. $5$, it follows that (one possible construction for) $R_k(\mathbb{Z}/5)$ is the coequalizer of $k[x,x^{-1}] \rightrightarrows k[x,x^{-1}]$, where the maps send $x \mapsto x^0=1$ resp. $x^5$, i.e. $k[x]/(x^5-1)$.

Actually this example is a special case of a quite general notion of tensor product studied by Freyd in his paper Algebra-valued functors in general and tensor products in particular. If $C,D,E$ are algebraic categories, then right adjoints $C \to D$ and $D \to E$ compose to a right adjoint $C \to E$. But right adjoints $C \to D$ are represented by Co-$C$-algebras in $D$. Thus, if we also have a Co-$D$-algebra in $E$, we may "tensor" them over $D$ to get a Co-$C$-algebra in $E$. It may be described via generators and relations. In the example above, $F_G$ is the composition of the representable functors $\mathrm{GL}_n : \mathsf{Alg}(k) \to \mathsf{Grp}$ and $\hom(G,-):\mathsf{Grp} \to \mathsf{Set}$, hence represented by $A \otimes_{\mathsf{Grp}} G$ where $A$ represents $\mathrm{GL}_n$. Of course, Freyd's tensor products can be applied to lots of other examples.

There are formal results of various kinds. For example, a functor $G : \textbf{Alg}_k \to \textbf{Set}$ is representable if and only if it has a left adjoint $F : \textbf{Set} \to \textbf{Alg}_k$, and the representing object is the $k$-algebra $F 1$. This is because $\textbf{Alg}_k$ is copowered (i.e. tensored) over $\textbf{Set}$.

Here is a slightly better result. Let $\mathcal{C}$ be any locally presentable category (such as $\textbf{Alg}_k$).

Theorem. Let $G : \mathcal{C} \to \textbf{Set}$ be a functor. The following are equivalent:

1. $G$ is representable.

2. $G$ preserves all (small) limits and there exists a regular cardinal $\kappa$ such that $G$ preserves all $\kappa$-filtered colimits.

3. $G$ has a left adjoint.

Proof. 1 ⇒ 2. Suppose $G$ is represented by $A$. Since $\mathcal{C}$ is a locally presentable category, there exists a regular cardinal $\kappa$ such that $A$ is a $\kappa$-presentable object in $\mathcal{C}$; but then $\mathcal{C}(A, -)$ must preserve all limits and all $\kappa$-filtered colimits.

2 ⇒ 3. Apply the accessible adjoint functor theorem.

3 ⇒ 1. Let $F : \textbf{Set} \to \mathcal{C}$ be a left adjoint of $G$. Since $1$ represents $\textrm{id} : \textbf{Set} \to \textbf{Set}$, we have the natural bijections $$G B \cong \textbf{Set}(1, G B) \cong \mathcal{C}(F 1, B)$$ and thus $G \cong \mathcal{C}(F 1, -)$. 　◼

More explicitly, by examining the proof of the accessible adjoint functor theorem, one can extract the following description of $F 1$. First, let $\mathcal{K}$ be the full subcategory of $\kappa$-presentable objects in $\mathcal{C}$, and let $\mathcal{H}$ be the full subcategory of the comma category $(1 \downarrow G)$ spanned by those objects of the form $(B, b)$, where $B$ is in $\mathcal{K}$ and $b \in G B$. Since $1$ is a $\kappa$-presentable object in $\textbf{Set}$, and $G$ preserves $\kappa$-filtered colimits, $\mathcal{H}$ must be a weakly initial family in $(1 \downarrow G)$. However, $\mathcal{H}$ is essentially small, and $G$ preserves small limits, so $(1 \downarrow G)$ has small limits, and thus the limit of the inclusion $\mathcal{H} \hookrightarrow (1 \downarrow G)$ exists. Freyd's initial object theorem says that this is an initial object for $(1 \downarrow G)$, and hence is the representing object $F 1$.

In some sense, however, this is also a purely formal result, since the representing object $F 1$ is itself $\kappa$-presentable and thus is amongst the objects in $\mathcal{H}$...

Postscript. If you have an explicit description of the functor to represent then there is a little bit more that can be said. For example, if $G B = \{ (b_1, \ldots, b_n) \in B^n : \phi [b_1, \ldots, b_n] \}$, where $\phi$ is a (possibly infinite) conjunction of (quantifier-free) equations in $n$ variables, then $G$ is represented by the free algebra on $n$ generators $a_1, \ldots, a_n$ satisfying the condition $\phi [a_1, \ldots, a_n]$. Of course, this is none other than the universal property of $F 1$, where $F$ is a left adjoint of $G$. Your example of $\mathbb{G}_m$ is an instance of this: take $n = 2$ and $\phi(x_1, x_2)$ to be the equation $x_1 x_2 = 1$.

Note that this means is that any moduli functor that can be represented by an affine scheme must be isomorphic to a functor of a very restricted form: each datum for an algebra $B$ can be reduced to a sequence (of fixed but possibly infinite length) of elements of $B$ satisfying a (possibly infinite) conjunction of equations! That is one of the reasons why we have to look at non-affine schemes and more general kinds of algebraic “spaces” to solve moduli problems.

• Either I'm totally missing something from the question/answer or this "answer" has nothing to do with the question. The OP seems to be asking: supposing a functor is representable can one find the representing object? This answer seems to be answering the question: how can we tell if a functor is representable? – Matt Apr 17 '13 at 17:38
• Read the penultimate paragraph. – Zhen Lin Apr 17 '13 at 17:43
• I accepted the other answer, but I would like to stress that I learned a lot from both answers, thank you! – Myself Apr 19 '13 at 8:30