Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Elementary Torsors

B. Wilson (Jan 22 2025 at 08:37):

Set the stage with a torsor $\underbar T = \left(T, \alpha: G\times T\to T \right)$ and trivialization $\rho: G \to T$. Naïvely, I'd think this would induce a division like $\hat d: T\times T\to G : (u,v) \mapsto \left[\rho^{-1}(u)\right]^{-1}\cdot\rho^{-1}(v)$.

However, Remark 2.13 on nlab's torsor page suggests something completely different and I don't see what it's trying to point at.

I'm guessing that by "the canonical map", it means that $p_1$ is the left projection, and as far as I can tell $D = \rho\circ d$ which has type $D: T\times T\to T$, but that doesn't immediately match up with the statement that $D(g,g) = \rho$.

If, instead, we define $D(g, h)$ to be the trivialization such that $g^{-1} h\mapsto\rho(e)$, then it kind of works out. That said, we still need to invert one of the elements like in my $\hat d$, which is nowhere in the 2.13 remark.

Am I just missing something obvious here?

Matt Earnshaw (Jan 22 2025 at 09:33):

seems wrong to me as well... not even $d(x,x) = e$ holds for the map there. I would encourage you to change it

John Baez (Jan 22 2025 at 16:03):

Yes, $p_1$ typically means the projection onto the first component of a cartesian product, and that makes sense here. But then the formula for division in Remark 2.13 looks terribly wrong. It takes an element $(t,t') \in T \times T$, uses the inverse of the trivialization $\rho : G \xrightarrow{\sim} T$ to convert $t$ into a group element $\rho^{-1}(t)$, and then throws out $t'$ altogether, leaving us with $\rho^{-1}(t)$. This is nothing like dividing $t$ by $t'$.

John Baez (Jan 22 2025 at 16:10):

Also, the definition of "trivialisation" in Definition 2.11 is misleading: we don't want just an isomorphism of sets, a trivialization of $T$ should be a $G$-torsor isomorphism $G \to T$. I'm going to fix that one now.

B. Wilson (Jan 23 2025 at 02:53):

Thank you for the sanity checks! I went ahead and munged the remark.

John Baez (Jan 23 2025 at 20:00):

Thanks! Demunged, I hope.

John Baez (Jan 23 2025 at 20:05):

Here's one thing I'm not satisfied with. If we're trying to show how a trivialization $\rho: G \to T$ equips $T$ with a group structure, we don't need to introduce "division". We can just use the isomorphism $\rho$ to transfer the group structure from $G$ to $T$.

On the other hand, if we're interested in "division", we don't need to bring in a trivialization. Any $G$-torsor has a canonical division map

$T \times T \to G$

which we can define without use of a trivialization. This division map answers the question "given two elements of $T$, which group element maps the second to the first?"

John Baez (Jan 23 2025 at 20:05):

So, someday I may try to work on this article some more. I am a big fan of torsors.

B. Wilson (Jan 24 2025 at 05:08):

Hah! The action's regularity gives a canonical division map. That's way better, and now I have even less an idea what the original Remark was trying to get at.

B. Wilson (Jan 24 2025 at 05:13):

Very nice Torsor article! Thanks for sharing. These past two night have actually had me pondering on generalizing to R-torsors and more, which you hint at right at the end. This seems like the natural structure for Energy and your other examples before you give them units.

B. Wilson (Jan 24 2025 at 05:42):

The algebraist in me wants to start with a "ring action" $R \to {\rm Aut}(X)$, but that requires $X$ to be an abelian group to work.

Taking a more nuts 'n bolts approach, at first blush we want an action like $R \times R \times X \to X$ where

$$r + s \cdot \left(s' + r' \cdot x\right) = \left(r + sr'\right) + \left(ss'\right)\cdot x$$

holds. This suggests a group with multiplication $(r, s)(r', s') \mapsto (r + sr', ss')$. Checking associativity and invertability, it's clear this does turn $G \coloneqq R^+ \times \left( R^\times \setminus \left\{0\right\}\right)$ into a group as long as $R$ is a division ring.

B. Wilson (Jan 27 2025 at 22:54):

I think a slightly bigger picture clicked with me last night in bed!

The above multiplication corresponds to the group $R^+\rtimes\left(R^\times\setminus\left\{0\right\}\right)$ with action $\phi :\left(R^\times\setminus\left\{0\right\}\right)\times R^+\to R^+ : \left(r, s\right)\mapsto sr$, i.e. simple scalar multiplication.

But that's just a nice decomposition of affine symmetries into scaling and translation. As whole it's essentially a subgroup of ring automorphisms ${\rm Aut}\!\left(R\right)$. So the slightly wider view is that the group in our $G$-torsor is just some subgroup of symmetries of ${\rm Aut}\!\left(X\right)$ that we're interested in!

B. Wilson (Jan 27 2025 at 22:57):

Pretty sure all this is well-known and spelled out somewhere, but I couldn't find the right keywords and had to scrounge around and learn about semi-direct products before things started to fall into place. Am I on the right track here?

John Baez (Jan 27 2025 at 23:09):

I'm confused about why you're calling scaling and translation "ring automorphisms". The usual concept of ring automorphism is: a map from a ring to itself that preserves 0, 1, addition and multiplication.

B. Wilson (Jan 27 2025 at 23:19):

Ooohh, okay. I'm actually thinking of $R$ as an $R$-vector space it seems. So the ${\rm Aut}\!\left(R\right)$ we should be thinking of is ${\rm GL}\!\left(R\right)$?

John Baez (Jan 27 2025 at 23:38):

Translations aren't in $\text{GL}(R)$.

I agree that for defining 'ring-torsors' we are interested in the group of transformations of $R$ of the form

$x \mapsto a x + b$

where $a,b \in R$ and $a$ is invertible. This group is the semidirect product of the groups $GL(R)$ (the group of transformations $x \mapsto a x$ with $a$ invertible) and $(R, +)$ (the group of transformations $x \mapsto x + b$, also called the additive group of $R$).

Maybe this group has some name like $\text{IGL}(R)$, the 'inhomogeneous' version of $\text{GL}(R)$. I haven't actually seen anyone call it that, but people use $\text{ISO}(\mathbb{R}^n)$ to mean the semidirect product of the rotation group $\text{SO}(\mathbb{R}^n)$ and the translation group of $\mathbb{R}^n$.

John Baez (Jan 27 2025 at 23:41):

Anyway, whatever it's called, this group is just a subgroup of the group of all automorphisms of the underlying set of $R$. I think a 'ring-torsor' of $R$ should be a $\text{IGL}(R)$-set (i.e. a set equipped with an action of this group) which is isomorphic to $R$ as an $\text{IGL}(R)$-set.