Representable functors play a large role in algebraic geometry when developed through the 'functor of points' approach. One finds schemes represent Zariski sheaves and this gives access to the great power of sheaf theory and topos theory.

My problem is I don't really understand representability, especially geometrically. Formally speaking, knowing some object $X$ represents a functor $F$ says that $F$ "probes" $X$ by giving at every object $Y$ the $Y$-points of $X$. But I just can't appreciate the (especially geometric) significance behind this.

What are some instructive geometric examples of representability?

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    $\begingroup$ In order for representability to have a geometric meaning, you must first give geometric meaning to non-representable sheaves. Otherwise representability is simply the property of having geometric meaning. $\endgroup$
    – Zhen Lin
    Commented Jan 12, 2016 at 23:40
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    $\begingroup$ If a functor is representable, then you can try doing geometry to the representing object. Otherwise it can be unclear how to do geometry to a functor (although you can do this too). $\endgroup$ Commented Jan 13, 2016 at 2:06

1 Answer 1


Although I agree with Zhen Lin's and Qiaochu's comments, I thought it might be useful to give some classical examples where you can write down a functor first, and then ask whether it's representable. (Of course, you could always write down something that is already representable to begin with, but I doubt you're interested in that.)

Example. Here are some down-to-earth examples of representable functors:

  • The functor $X \mapsto \Gamma(X,\mathcal O_X)$ is represented by $\mathbb A^1$ (also denoted $\mathbb G_a$ in this setting).
  • The functor $X \mapsto \Gamma(X,\mathcal O_X)^\times$ is represented by $\mathbb A^1 \setminus \{0\}$ (also denoted by $\mathbb G_m$).
  • The functor $X \mapsto \operatorname{GL}_n(\Gamma(X,\mathcal O_X))$ is represented by an open subscheme of $\mathbb A^{n \times n}$: the nonvanishing locus of the determinant. It is denoted simply by $\operatorname{GL}_n$. The case $n = 1$ gives $\mathbb G_m$.

Arguably, the cleanest approach to linear algebraic groups, and especially if you want to consider more general group schemes, is by considering the fppf sheaves they define. For example, a sequence of algebraic groups $$1 \to G_1 \to G_2 \to G_3 \to 1$$ is exact if it is so as fppf sheaves. Giving a definition in more geometric terms is awkward to say the least.

Example. To give some more interesting geometric examples of representable functors:

  • The Picard scheme $\operatorname{\mathbf{Pic}}_{X/k}$ represents a suitable¹ Picard functor. This generalises the notion of the Jacobian of a curve: for any smooth projective variety, the Picard group now has a continuous part ($\operatorname{\mathbf{Pic}}_{X/k}^0$; the Jacobian of $X$) and a discrete part ($\operatorname{Pic}(X)/\operatorname{Pic}^0(X)$, the Néron–Severi group of $X$; for higher-dimensional varieties this need not be just $\mathbb Z$).
  • The Hilbert scheme represents a functor that, loosely speaking, associates to a variety its family of subvarieties. Sometimes, you want to add some numerical conditions, e.g. one can consider Hilbert schemes of $n$-tuples of points in $X$ (including fat points, counted with multiplicity), which is birational to $\operatorname{Sym}^n X$: they are isomorphic on the part where the $n$ points are distinct.
  • Any moduli problem is a functor, and one can ask if it is representable. This is often not the case, until you pull the French trick of defining a larger class of objects (algebraic spaces or algebraic stacks) where this is true. For example, you can ask for the moduli space of curves of genus $3$. The functor assigns to each scheme $X$ the set of isomorphism classes of families $\mathscr C \to X$ whose fibres are smooth projective curves of genus $3$.

Remark. Finally, observe that representability of functors is by no means a quality that's reserved for algebraic geometry! In any category, you can ask whether a functor on it is representable. It's a good exercise to keep your eyes open for any representable functors around, especially when you're dealing with easy categories (like abelian groups, $R$-modules, sets, or other categories that are relatively easy to describe).

Example. The forgetful functor $\operatorname{\underline{Ab}} \to \operatorname{\underline{Set}}$ is represented by $\mathbb Z$.

Example. The forgetful functor $\operatorname{\underline{Ring}} \to \operatorname{\underline{Set}}$ is represented by $\mathbb Z[x]$. (Compare with the very first example I gave above).

Example. The dualisation functor $\operatorname{\underline{Vect}}_k^{\operatorname{op}} \to \operatorname{\underline{Vect}}_k$ is represented by $k$. This is a slight abuse of language, since representable functors technically have to go to $\operatorname{\underline{Set}}$.

Most dualities are given by representable functors, often by definition. For a less trivial example, see Hartshorne's definition of Serre duality: the functor is $H^n(X, (-))^*$, and the representing object is $\omega_X^\circ$.

Exercise. One of my favourite examples is the functor $\operatorname{\underline{Top}}^{\operatorname{op}} \to \operatorname{\underline{Set}}$ that associates to a topological space $(X, \mathcal T)$ the topology $\mathcal T$, and to a continuous map $X \to Y$ the inverse image map $\mathcal T_Y \to \mathcal T_X$. Try to write down a topological space that represents this functor (it exists!).

(I think Zhen Lin might have told me this example when I was learning about representable functors.)

¹Defining the correct functor is not so obvious, and there are multiple different things people might mean by the Picard scheme. The weakest notion is the representability of the fppf sheafification of the presheaf $U \mapsto \operatorname{Pic}(U)$.

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    $\begingroup$ Is there an accepted or acceptable name for the functor in your last example? $\endgroup$
    – magma
    Commented Jan 13, 2016 at 20:12
  • $\begingroup$ None that I'm aware of, but I would be interested to know if there is. $\endgroup$
    – Remy
    Commented Jan 13, 2016 at 22:16
  • $\begingroup$ Is it the two point space {0, 1} with {1} the only non-trivial open set? $\endgroup$
    – Xipan Xiao
    Commented Dec 11, 2023 at 10:10
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    $\begingroup$ @XipanXiao that's right! Good old Sierpiński space. $\endgroup$
    – Remy
    Commented Dec 11, 2023 at 10:33

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