We know that it is usually much easier to work with highly symmetry objects, the objects that have many automorphisms like the sphere, Lie groups, complete graph,... But is there any advantage of asymmetric objects, the objects that have only trivial automorphisms and what is that advantage? For example, is there a category $C$ and property $P$ that true for all objects of $C$ that it's easier to prove asymmetric objects have property $P$ in that category than for symmetric ones?
Motivation: Give a category $C$, we want to prove all objects of $C$ have property $P$. I have an interesting idea as follow:
For each objects $A$, let $G_A$ be a automorphisms group of $A$. Assume for each group morphism $H\rightarrow G_A$, we have a nice quotient $A/H$, and for some group $H$, it's easy to prove this statement "if $A/H$ has property $P$ then $A$ also has property $P$". For example, we can prove that statement for all cyclic groups, because for each non-trivial group $G$, there exists a non-trivial cyclic group as its subgroup. So we keep quotienting objects $A$, and if we are lucky, we will reach to an asymmetric object in $C$, and now we just need to prove that all asymmetric objects have property $P$.
-
1$\begingroup$ Asymmetric objects in your sense are often called rigid. One nice (almost-)example is algebraic tori, which have no connected algebraic group of automorphisms. $\endgroup$– LSpiceCommented Jul 1, 2022 at 0:17
-
2$\begingroup$ One of the aphorisms I remember from one of my mentors is, "Everything we've ever learned about symmetries we learned by breaking them. Think about it." $\endgroup$– Michael EngelhardtCommented Jul 1, 2022 at 1:13
-
1$\begingroup$ @MichaelEngelhardt, do you have any interesting examples of putting that aphorism into practice? (I suspect my thoughts on it are shallower than those of someone who thinks about it professionally, so I'd like to hear yours!) $\endgroup$– LSpiceCommented Jul 1, 2022 at 1:48
-
4$\begingroup$ @LSpice - At its core, I think all that was meant was that the symmetrized object does not reveal what the symmetry does, being invariant under the symmetry. You have to pick a representative to see it. Simple as that is, it can take interesting guises. E.g., the most direct understanding of the rotating rigid body is in the body-fixed frame; viscerally, you have to get on the merry-go-round to fully experience it. In a gauge theory, physical degrees of freedom only become evident upon fixing a gauge (e.g., the Coulomb gauge in classical electrodynamics, which renders the waves transverse). $\endgroup$– Michael EngelhardtCommented Jul 1, 2022 at 4:21
-
1$\begingroup$ When an object has no automorphism, it is typically quite simple to prove that some other object is isomorphic to it (if it is, of course): there is only one possible choice of isomorphism, so you can't go wrong ! In another direction, bundles of objects with no automorphisms are trivial - for instance any manifold is $\mathbb Z/2$-orientable $\endgroup$– Maxime RamziCommented Jul 1, 2022 at 17:43
3 Answers
Such objects are generally known (in the fields I’m familiar with) as rigid, and yes, working with them can often be very useful.
The advantages I’m familiar with come from the fact that if objects of some type — widgets, say — are rigid, then essentially nothing is lost by working with “widgets up to isomorphism”. A typical example is well-orderings. These are always rigid; their theory is traditionally developed mainly in terms of ordinals, which are canonical representatives of each isomorphism class. The rigidity ensures that every well-ordering is uniquely isomorphic to some ordinal, and that’s simplifies several aspects of the development of the theory.
By contrast, consider some non-rigid objects — say, finite Abelian groups. Even if we have some way choosing of representatives of isomorphism classes (e.g. by the classification theorem), it’s not true that every finite Abelian group is uniquely isomorphic to its chosen representative. In other words, finite Abelian groups can’t be naturally identified with their chosen isomorphic representatives. It turns out that this means working with chosen representatives isn’t nearly so fruitful as in the rigid case.
A way of saying all this in fancy languages is that families of arbitrary objects typically form a (pre-)stack, but families of rigid objects form a (pre-)sheaf — a technically much simpler notion.
The nlab has a page on rigid objects mentioning several more examples.
-
4$\begingroup$ This is completely an aside, but one thing I'm embarrassed to never have properly understood is whether there is some deeper connection between this use of "rigid" and another common use of "rigid" in category theory to mean "objects have duals" (e.g. rigid monoidal category). Is that just a coincidence of terminology? $\endgroup$ Commented Jun 30, 2022 at 17:34
-
6$\begingroup$ @SamHopkins: I’ve wondered the same, and feel equally embarrassed to have no answer… $\endgroup$ Commented Jun 30, 2022 at 17:49
-
3$\begingroup$ @SamHopkins: OK, looking it into a little, the earliest use of “rigid” for “all objects dualisable” I can find is in Saavedra Rivano 1972 Catégories Tanakiennes, LNM 265 (the book, not the short paper of the same title+date) — at least, he doesn’t cite any earlier source for the term/definition. He doesn’t explain the term when he introduces it, and I haven’t read the book fully, but on a quick skim, there are a couple of possible hints towards the motivation: (1) he earlier uses “rigid groupoid” to mean “essentially discrete groupoid” (i.e. all automorphisms are identities); and… [cont’d] $\endgroup$ Commented Jun 30, 2022 at 18:07
-
3$\begingroup$ [cont’d] …and (2) the Théoreme de rigidité (Cor 2.6.2), which seems to maybe kinda-sorta link the two senses — its main hypothesis is polarisability, closely related to existence of duals, and its conclusion is a uniqueness-up-to-unique-equivalence result? So it looks like the “duals” sense is quite possibly based on the older “no non-trivial automorphisms” sense, because they’re somehow connected in the specific setting of Tannakian categories. But the precise connection is still not clear to me, without reading the book or knowing the topic properly! $\endgroup$ Commented Jun 30, 2022 at 18:07
In light of your stated motivation, the following may not be what you had in mind, but objects with no symmetries are often easier to handle when it comes to computation and/or enumeration. For example:
Determining whether two (finite) graphs are isomorphic is usually easy when the graphs have no automorphisms.
There is a nice formula ($n^{n-2}$) for the number of vertex-labeled trees with $n$ vertices, but if the vertices are unlabeled, then you have to account for the automorphisms, and the best you can do is some messy "formula" based on the Redfield–Pólya theorem.
If you're trying to use a SAT solver to prove that an instance of the Boolean satisfiability problem is unsatisfiable, it is usually helpful to add constraints to break symmetries. Otherwise, the solver may end up doing essentially the same (futile) search over and over again before finally concluding that no solutions exist.
-
2$\begingroup$ See also mathoverflow.net/questions/182830/… and mathoverflow.net/questions/232966/… $\endgroup$ Commented Jun 30, 2022 at 23:34
My favourite example is related to some bit of work of my own, so I apologise in advance for a bit of self-promotion. It concerns dealing with symmetric operads (algebraic ones, meaning that the $n$-th component is a representation of $S_n$).
Presence of symmetries, while good in theory, complicates things in practice: it is harder to choose canonical bases that are convenient for calculations etc. The solution is to... forget about symmetries. If one does that naively, the result is very disappointing: for example, if one takes the operad of Lie algebras and considers it as a nonsymmetric operad, it becomes a free operad on infinitely many generators (it is not possible to even describe the Jacobi identity in the realm of nonsymmetric operads, so the dependencies disappear), as proved by Salvatore and Tauraso some 15 years ago. This happens because by considering nonsymmetric operads, we forget some part of the story alongside with the symmetries, some of the structure maps.
How to forget about symmetries in a smart way? The solution is to consider "shuffle operads" (introduced by myself and Khoroshkin): symmetric operads are algebras over the monad of trees, shuffle operads are algebras over the monad of shuffle trees, which just choose a particular way to draw a tree on a plane, so no structural information will be lost. This does exactly what you are talking about: eliminate automorphisms, and the gain is quite noticeable (we did it to define Gröbner bases for operads, and that helped a lot in quite a few questions of algebra and homotopy theory). See our article for more details.
