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## Connected objects and a reconstruction theorem

A common theme in mathematics is to replace the study of an object with the study of some category that can be built from that object. For example, we can

• replace the study of a group $G$ with the study of its category $G\text{-Rep}$ of linear representations,
• replace the study of a ring $R$ with the study of its category $R\text{-Mod}$ of $R$-modules,
• replace the study of a topological space $X$ with the study of its category $\text{Sh}(X)$ of sheaves,

and so forth. A general question to ask about this setup is whether or to what extent we can recover the original object from the category. For example, if $G$ is a finite group, then as a category, the only data that can be recovered from $G\text{-Rep}$ is the number of conjugacy classes of $G$, which is not much information about $G$. We get considerably more data if we also have the monoidal structure on $G\text{-Rep}$, which gives us the character table of $G$ (but contains a little more data than that, e.g. in the associators), but this is still not a complete invariant of $G$. It turns out that to recover $G$ we need the symmetric monoidal structure on $G\text{-Rep}$; this is a simple form of Tannaka reconstruction.

Today we will prove an even simpler reconstruction theorem.

Theorem: A group $G$ can be recovered from its category $G\text{-Set}$ of $G$-sets.

Connected objects

The idea of the proof is that we want to find a categorical property that allows us to isolate the subcategory of transitive $G$-sets. The $G$-set $G$ itself can be uniquely identified among transitive $G$-sets as the “largest” one (more precisely, it is the unique weak initial object among transitive $G$-sets), and then $G$ can be recovered as the opposite group of the group of automorphisms of $G$ (as a $G$-set).

The categorical property we want is the following.

Definition: An object $c$ in a category $C$ is connected if the representable functor $\text{Hom}(c, -)$ preserves coproducts.

The idea behind the definition is that if one thinks of coproducts as a disjoint union, then a morphism $c \to \bigsqcup_i d_i$ from a connected object into a disjoint union of objects must land entirely in one of the objects, or else it will be “disconnected” by the fact that it’s spread out over a disjoint union.

Example. In $\text{Set}$, the connected objects are precisely the one-element sets. Note that the empty set is not connected.

Example. In $\text{Graph}$ (which we’ll take to be the category of simple graphs), the connected objects are precisely the connected graphs in the usual sense. Note that the empty graph is not connected.

Example. In $\text{Top}$, the connected objects are precisely the connected topological spaces in the usual sense. Note that the empty space is not connected.

Example. In $\text{Aff}$, any affine scheme $\text{Spec } R$ such that $R$ has nontrivial idempotents (in other words, such that $R$ is not a connected ring) is not connected in the categorical sense. To see this, let $R$ contain a nontrivial idempotent $e$ and consider the map $\text{Spec } R \to \text{Spec } \mathbb{Z} \sqcup \text{Spec } \mathbb{Z}$ induced by the ring homomorphism

$\displaystyle \mathbb{Z} \times \mathbb{Z} \ni (a, b) \mapsto ae + b(1 - e) \in R$.

This map factors through neither of the projections $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$, from which it follows that $\text{Spec } R$ is not connected. (One interpretation of this argument is that $\mathbb{Z} \times \mathbb{Z}$ is the free commutative ring on an idempotent.)

The converse statement – that the spectrum of a connected ring is connected in the categorical sense – is false. For example, let $\prod_{i \in I} \mathbb{F}_2 \to \mathbb{F}_2$ be a homomorphism induced by a non-principal ultrafilter on an infinite set $I$. Then the corresponding morphism $\text{Spec } \mathbb{F}_2 \to \bigsqcup_{i \in I} \text{Spec } \mathbb{F}_2$ does not factor through any of the inclusions, so $\text{Spec } \mathbb{F}_2$ is not connected in the categorical sense. More generally, any ring which can be obtained as a nontrivial ultraproduct does not have connected spectrum in the categorical sense. However, it is true that if $R$ is a connected ring, then $\text{Hom}(\text{Spec } R, -)$ preserves finite coproducts.

Note that the spectrum of the zero ring is not connected. In general, the initial object of a category is never connected; it is too simple to be simple.

Intuitively, a nontrivial coproduct should not be connected. The following result shows that this is true under reasonable hypotheses. These hypotheses don’t hold for $\text{Aff}$ but do hold for $G\text{-Set}$, which is enough; I don’t know how much they can be relaxed.

Proposition: Let $c_1, c_2$ be two objects, neither of which is the initial object, of a concrete category $C$ with finite coproducts such that the forgetful functor $F : C \to \text{Set}$ preserves finite coproducts. Then $c_1 \sqcup c_2$ is not connected.

Proof. We prove the contrapositive, namely that under the hypotheses above, if $c_1 \sqcup c_2$ is connected then one of $c_1, c_2$ is the initial object. Recall that “preserves coproducts” means the following, for binary coproducts. If $c, c_1, c_2$ are objects, the natural inclusions $c_1, c_2 \to c_1 \sqcup c_2$ induce natural maps $\text{Hom}(c, c_1), \text{Hom}(c, c_2) \to \text{Hom}(c, c_1 \sqcup c_2)$ which in turn induces a natural map

$\displaystyle \text{Hom}(c, c_1) \sqcup \text{Hom}(c, c_2) \to \text{Hom}(c, c_1 \sqcup c_2)$

and to say that $\text{Hom}(c, -)$ preserves binary coproducts means that this map is always a bijection for all $c_1, c_2$. In particular, every morphism $c \to c_1 \sqcup c_2$ factors through one of the inclusions $c_1, c_2 \to c_1 \sqcup c_2$.

Now let $c = c_1 \sqcup c_2$. Suppose that $c$ is connected; then WLOG the identity map $\text{id}_c : c_1 \sqcup c_2 \to c_1 \sqcup c_2$ factors through $c_1$. It follows that the inclusion map $c_1 \to c_1 \sqcup c_2$ is a split epimorphism. However, by assumption, the map on underlying sets $F(c_1) \to F(c_1) \sqcup F(c_2)$ is an injection in $\text{Set}$, and since faithful functors reflect monomorphisms, it follows that the inclusion map $c_1 \to c_1 \sqcup c_2$ is a monomorphism, hence an isomorphism. But looking at the corresponding natural isomorphism of representable functors $\text{Hom}(c_1 \sqcup c_2, -) \to \text{Hom}(c_1, -)$, this is possible if and only if $c_2$ is the initial object. $\Box$

(Note that a sufficient condition for $F$ to preserve finite coproducts is that it has a right adjoint. This is true of the forgetful functors from $\text{Graph}, \text{Top}$, and $G\text{-Set}$ to $\text{Set}$.)

Reconstruction from $G\text{-Set}$

Proposition: The connected objects of $G\text{-Set}$ are precisely the transitive $G$-sets (the empty $G$-set is not transitive).

Proof. Every $G$-set can be expressed uniquely as a coproduct of transitive $G$-sets (the orbits of the group action). It follows that a connected object of $G\text{-Set}$ is necessarily transitive. Conversely, if $X$ is a transitive $G$-set, then the image of any homomorphism from $X$ into another $G$-set $Y$ is necessarily also a transitive $G$-set, hence contained in an orbit of $Y$. $\Box$

Proposition: The $G$-set $G$ is the unique (up to isomorphism) transitive $G$-set which admits a morphism to all other transitive $G$-sets.

Proof. Every transitive $G$-set has the form $G/H$ for some subgroup $H$ of $G$, so in particular admits a quotient map $G \to G/H$. On the other hand, if $H$ is not the trivial subgroup, then there exist no morphisms $G/H \to G$, since no map of sets $G/H \to G$ can respect the $G$-action (nontrivial elements of $H$ preserve the identity coset of $G/H$ but can’t preserve its image in $G$). $\Box$

Theorem: A group $G$ can be recovered from its category of $G$-sets.

Proof. We know that from the category of $G$-sets we can recover the subcategory of transitive $G$-sets, and we know that from the transitive $G$-sets we can recover the $G$-set $G$ itself as the unique weak initial object. The automorphism group of $G$, as a $G$-set, is $G^{op}$, from which we can recover $G$ by taking the opposite group. $\Box$

### 7 Responses

1. Nice! On a very slightly related note, I just saw a nice talk on Brauer relations at the British Mathematical Colloquium. The idea is that there’s an obvious functor from $G-\mathrm{Set}$ to $G-\mathrm{Rep}$, but nonisomorphic $G$-sets can get sent to isomorphic representations of $G,$ and we’d like to know precisely how this works. If we assume $G$ is finite and consider only finite $G$-sets and finite-dimensional representations, we get a homomorphism from the Burnside ring of $G$ to its representation ring, and this has a kernel consisting of ‘Brauer relations’. These relations are now, in some sense, completely understood.

2. […] This is from a post Connected objects and a reconstruction theorem: […]

3. To single out the transitive G-sets isn’t it a little easier to say that they are precisely those G-sets which are not nontrivial coproducts of G-sets (so, irreducible instead of connected)?

• Yes, you’re absolutely right. I noticed this after writing the post. The motivation for looking at connected objects now is that it’s analogous to another condition I’ll hopefully later use to prove a more general theorem, but I’m following my usual policy of not promising future posts (because that seems to lead to them not being written).

4. The example with affine schemes is potentially confusing in terms of notation: the inclusion of affine schemes in all schemes does not preserve coproducts. In particular, the object $\sqcup_I \mathrm{Spec}(\mathbf{F_2})$ in the example above is *not* the $I$-fold coproduct of copries of $\mathrm{Spec}(\mathbf{F_2})$; the former is closer to a Stone-Cech compactification of the latter, I think. In any case, the map $\prod_I \mathbf{F_2} \to \mathbf{F_2}$ you construct with an ultrafilter does not come from a map $\mathrm{Spec}(\mathbf{F_2}) \to \sqcup_I \mathrm{Spec}(\mathbf{F_2})$ in the world of schemes. (I was confused by this initially because any such map fo schemes does indeed factor through some factor of the target as $\mathrm{Spec}(\mathbf{F_2})$ admits no non-split covers.)

• Correct. That’s why everything takes place in $\text{Aff}$.