## Empty sets

December 3, 2010 by Qiaochu Yuan

Here’s something I had never really thought about before. Any group has a unique action on the one-element set , since there is a unique morphism . Abstractly, this is because is a terminal object in the category of sets. Now, any group also has a unique action on the **empty** set , since there is also a unique morphism (the “empty function”). Abstractly, this is because is an initial object in the category of sets.

This action is quite strange. There are two definitions (that I’m aware of) of what it means for an action of a group on a set to be transitive, and they **disagree** for this action:

- An action of on is transitive if, for every , there exists such that . This is vacuously true when is empty.
- An action of on is transitive if it has one orbit. This is false when is empty; there are
**zero** orbits.

Wikipedia takes the stance that a transitive -set must be nonempty, which I suppose corresponds to supporting the second definition. There is a great reason for doing this: you want transitive -sets to correspond to conjugacy classes of subgroups of , and this only works if you can take stabilizers. And you can only take stabilizers if the -set is non-empty. Instead of using stabilizers you can use the kernel of the action, but then there are two actions – the unique action on , and the unique action on – that both correspond to the entire group .

**Edit:** Perhaps an even better reason to declare that the empty -set is not transitive is that otherwise, the decomposition of a -set into a coproduct of transitive -sets is not unique. (As I mention in the comments, the empty -set itself is the empty coproduct.)

These aren’t entirely idle thoughts; there are some major theorems identifying certain categories with the category of certain types of actions of a group , and as far as I can tell these theorems are wrong as stated in the literature because they don’t take into account the empty case.

It’s likely I’ve said several things on this blog which are false because I didn’t take into account the empty case. Hopefully I’ll do this less in the future.

**Exercise:** Define the topology on the empty topological space . Is it Hausdorff? Compact? Metrizable? Connected? Path-connected? Simply connected? What is its fundamental group? What does the algebra of continuous functions look like?

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on July 30, 2012 at 1:03 am |Chris HuntShouldn’t the empty topology be considered connected? To allow it to be disconnected disagrees with the definition of disconnected as “a union of disjoint nonempty open sets”; and to drop the “nonempty” would make any space disconnected. The solution I found more elegant (from the Wikipedia page) is to take connected components to be “maximal connected subsets”, which maintains unique decomposition into components, and the simple definition of connectedness still applies to all topologies.

on July 30, 2012 at 5:00 am |Qiaochu YuanThe standard definition is wrong for the empty space; see the nLab page too simple to be simple. A better definition is that a connected space is a space with exactly one connected component (the empty space has zero).

on June 22, 2012 at 8:51 pm |anonIf you take the graph induced by G on X, you can reduce the question: Is the empty graph a connected graph? I am voting “no” because of a theorem, similar to fundamental theorem of arithmetic:

Every graph can be uniquely represented as a disjoint sum of connected graphs.

on June 23, 2012 at 1:05 pm |Qiaochu YuanI completely agree; this is also why I think the empty graph is not connected.

on December 9, 2010 at 3:25 am |Peter CameronA similar problem arises at the next stage. Does a primitive action have to be on a set with more than one element? (An action is primitive if the only two G-invariant congruences are equality and the “all equivalent” relation. But do we want these to be distinct or not? For much the same reasons as in your initial post, we probably do. This is the same as forbidding 1 to be a prime, or forbidding the trivial group to be simple.

But there is another problem. Does a primitive group have to be transitive? The trivial group acting on a set of size 2 is primitive according to the above definition, since there are only two equivalence relations on a set of size 2. Does this matter? It turns out that it has some implications at a still deeper level. In some metaphysical sense, it is this fact that allows a primitive permutation group to have two minimal normal subgroups.

on December 6, 2010 at 10:29 am |Todd TrimbleIt works nicely in Aff precisely because Aff is an extensive (even infinitary lextensive) category! By the way, I recommend the nLab’s article “extensive category”.

on December 6, 2010 at 9:33 am |ka7thFascinating.

on December 6, 2010 at 5:58 am |Todd TrimbleBenoit’s comment is very much to the point. As a plug for the nLab, notice that it gets the definitions of torsor and connected space “right” (by saying that the objects should be inhabited). In fact, if we define an object X in (preferably an extensive) category to be connected precisely when preserves finite coproducts, then the connected objects in G-Set are exactly the transitive G-sets under the second definition, and are exactly the (nonempty) connected spaces in Top.

on December 6, 2010 at 9:40 am |Qiaochu YuanI like this definition! Looks like it even works nicely in Aff.

on December 4, 2010 at 7:48 pm |BenoitI agree with your “edit”. Similarly, the uniqueness of the decomposition into connected components or prime factors answers the questions wether {} is connected or 1 is prime.

on December 3, 2010 at 7:39 pm |TheoI tend to think that “quotient $G$-set of $G$ acting on itself” is a very important notion, and so I’m happy having a single word for this.

The only reason I can think of for wanting $\emptyset$ to be a transitive G-set is that I can imagine wanting to consider “Galois $G$-sets”, i.e. $G$-torsors, which are $G$-sets $X$ such that the map $G \times X \to X \times X$ given by $(g,x) \mapsto (gx,x)$ is an iso. This certainly is an iso for $X = \emptyset$. But I don’t really think that this is a $G$-torsor at all, so I think I want to include a condition making it not be. So I think I want the word “$G$-torsor” to mean “$G$-set $X$ satisfying the above condition and such that $X \to 1$ is a cover”. This is the definition that works well in algebraic geometry land.

on December 4, 2010 at 4:36 am |Qiaochu YuanIf you agree with me that the empty -set shouldn’t be transitive, then you can just require that -torsors are transitive.

on December 3, 2010 at 2:30 pm |PeterAh, my bad. Sorry about that

on December 3, 2010 at 2:05 pm |PeterWhy not modify the second definition to “all orbits are equal”? This seems to capture the intended meaning just as well and poses no trouble.

Besides, http://en.wikipedia.org/wiki/Group_action seems to define it by “$Gx=X$ for all $x\in X$” which is fine. The article mentions the ‘one orbit’ bit following ‘all elements are equivalent’ which again is ok — so at most this is a lapse on wikipedia’s part for not saying ‘at most one orbit’.

on December 3, 2010 at 2:14 pm |Qiaochu YuanI’m

agreeingwith the second definition; I want transitive G-sets to correspond exactly to conjugacy classes of subgroups of G, and this theorem is false if the empty G-set is transitive. I also want the decomposition of a G-set into a coproduct of transitive G-sets to be unique, and this is false if the empty G-set is transitive. (The empty G-set itself is the empty coproduct.)