## June 3, 2013

### Torsors and Enriched Categories

#### Posted by Simon Willerton

In another instalment of my occasional series on ‘Things you didn’t realise were enriched categories (unless you’re an expert!)’ I want to talk about torsors and how they can be considered enriched categories where the enriching category is a group, considered as a discrete monoidal category.

I’ll start off by telling you what a torsor is and then explain how it can be thought of as having something like a ‘group-valued distance’ and how this relates to enriched category theory.

### What is a torsor?

“Torsor” is a word that used to strike a dread fear into my heart. It is a word that was seemingly only used by high-brow mathematicians who wished to intimidate more lowly souls. That was until I read Dan Freed’s paper Higher Algebraic Structures and Quantization which made me realise that torsor is just a fancy name for a simple concept. For a group $G$ a torsor is something that looks like $G$ but isn’t actually a group. That might sound like an overly cryptic description but hopefully it will make things clear in a minute.

On my desk I have a coffee cup — we are in a café, after all.

The rim $R$ of my coffee cup is a circle: it looks like the group of unit complex numbers $\mathbb{T}$, but I don’t know how to multiply two points on the coffee cup rim, $R$, so the coffee-cup rim $R$ is not canonically a group. However, the group of unit complex number $\mathbb{T}$ acts on the rim $R$: if you give me a unit complex number $e^{i\theta }$, I can rotate my coffee cup through the angle $\theta$.

This action is transitive in that I can move any point on the rim to any other point by a rotation and the action is free in that precisely one rotation will move a specified point to another specified point. A set with a $G$ action which is free and transitive is known as a principal homogeneous $G$-set, or simply a $G$-torsor. So the rim of my coffee cup is a $\mathbb{T}$-torsor.

If I pick any point $p$ on the rim $R$ then I get an bijection between the points of $R$ and the points in the group $\mathbb{T}$ by identifying the point $p$ with the identity in $\mathbb{T}$. So there are as many ways to identify the rim with the circle group $\mathbb{T}$ as there are points in the rim but none of them is canonical.

Just as an affine space can be thought of as a vector space without an origin, so a torsor can be thought of as a group without an identity: a torsor is to a group as an affine space is to a vector space.

John has written a nice piece explaining how torsors, particularly in physics, are more prevalent than you might first think, because our gut reaction is to think in terms of groups. We see a line, so we want to put an origin on it and identify it with $\mathbb{R}$ rather than accept that it doesn’t naturally have an origin, and should be thought of as an $\mathbb{R}$-torsor.

### Another view of torsors

Torsors can be thought of in a different fashion and thinking in this fashion this will lead us enriched categories. Between each pair of points in a torsor there is something like a ‘group-valued distance’. More precisely, associated to each ordered pair $(t_{1},t_{2})$ of elements of a $G$-torsor $T$ there is a unique element of the group $G$ which sends $t_{1}$ to $t_{2}$, we can write this as $g(t_{2},t_{1})$, or as $g_{T}(t_{2},t_{1})$ if we want to emphasise the torsor $T$. Symbolically, this is the unique element of the group $G$ that satisfies $g(t_{2},t_{1})\cdot t_{1} = t_{2}$. This is the ‘distance’ from $t_{1}$ to $t_{2}$.

[I have chosen to write $g(t_{2},t_{1})$ rather than the more usual category theoretic convention $g(t_{1},t_{2})$, because the actions are on the left and the chosen conventions make the formulas look a bit nicer.]

Of course, in the coffee cup example, the ‘distance’ from one point to another is the rotation required to move the first point to the second.

Because we have a $G$-action on $T$ we know that $(g h)\cdot t_{1}=g\cdot (h\cdot t_{1})$ and this implies $g(t_{3},t_{2})g(t_{2},t_{1})=g(t_{3},t_{1})$ for all $t_{1},t_{2},t_{3}$.; for the same reason we also have $g(t_{1},t_{1})=e$. Moreover, we can recover the $G$-action from this data because given $g\cdot t_{1}$ is the unique element of $T$ such that $g(g\cdot t_{1}, t_{1})=g$.

In summary, we can think of a $G$-torsor as a set $T$ such that

1. for each pair $t_{1},t_{2}\in T$ there is a group element $g(t_{2},t_{1})\in G$;

2. for each triple $t_{1},t_{2},t_{3}\in T$ there is an equality $g(t_{3},t_{2})g(t_{2},t_{1})=g(t_{3},t_{1})$;

3. for each element $t_{1}\in T$ there is the equality $g(t_{1}, t_{1})=e$;

4. for each $t_{1}\in T$, $g\in G$ there is a unique $t_{2}$ such that $g(t_{2},t_{1})=g$.

You might have noticed that the first three of the above conditions form precisely the definition of a certain kind of enriched category, namely a category enriched over the monoidal category $\mathcal{V}_{G}$ which is the group $G$ considered as a discrete monoidal category. This means that $\mathcal{V}_{G}$ has the elements of $G$ as its objects, has only identity morphisms and has the group multiplication as the monoidal multiplication: $g\otimes h \coloneqq g h$.

We have shown that a $G$-torsor is a $\mathcal{V}_{G}$-category satisfying an extra condition. A reasonable question to ask at this point is “What’s the good in that?” One answer is that it gives us another layer of intuition; it allows us to make analogies with categories, with metric spaces, with posets, amongst other things, and it allows us to use the tools from enriched category theory.

We should now ask what a $\mathcal{V}_{G}$-functor $\phi \colon T\to S$ between $\mathcal{V}_{G}$-torsors is. This is a function $\phi \colon T\to S$ such that $g(t_{1},t_{2})=g(\phi (t_{1}),\phi (t_{2})).$ This is quite a rigid condition, saying that the right notion of map is some kind of isometry. Unsurprisingly, if $T$ and $S$ correspond to $G$-torsors then this is precisely the condition that $\phi$ is an equivariant map: $\phi (g\cdot t)=g\cdot \phi (t)\quad \text{for all}\,\, g\in G, t \in T.$

### New torsors from old

When we look at torsors over abelian groups there are ways of combining torsors. We will see below how these relate to standard enriched category theory constructions.

Hom torsor: If $A$ is an abelian group and both $T$ and $S$ are $A$-torsors then we can form the set $\text{Hom}_{A}(T,S)$ of equivariant maps from $T$ to $S$. I learnt many years ago from Freed’s paper that this set of maps $\text{Hom}_{A}(T,S)$ is itself an $A$-torsor. We can define the action of $A$ on an equivariant map $\phi$ by $(a\cdot \phi )(t):=a\cdot (\phi (t))$. You can check that $a\cdot \phi$ is an equivariant map, but you will see that it is necessary that $A$ is abelian for this to work.

Tensor torsor: If $A$ is an abelian group and both $T$ and $S$ are $A$-torsors then we can also define the tensor product $T\otimes _{A} S$ as $T\times S/ \sim$ where $\sim$ is the relation defined by $(a\cdot t,s)\sim (t,a\cdot s)\quad \text{for all}\,\, a\in A, t \in T, s\in S.$ We find that $T\otimes _{A} S$ is also an $A$-torsor if we define the action by $a\cdot (t,s)\coloneqq (a\cdot t,s)$. Again, if you check, you will see that you need $A$ to be abelian for this to be well defined.

### Properties of $\mathcal{V}_{G}$

Things get more interesting with enriched categories when we enrich over categories which are closed, braided and bicomplete. Let’s consider each of these conditions for $\mathcal{V}_{G}$.

Closedness: A monoidal category $\mathcal{V}$ is left-closed if for each object $v$ the functor $v\otimes -\colon \mathcal{V}\to \mathcal{V}$ has a right adjoint $[v,-]_{\text{left}}\colon \mathcal{V}\to \mathcal{V}$. Unwrapping this definition for $\mathcal{V}_{G}$ we find that for each $g\in G$ we need a function of sets $[g,-]_{\text{left}}:G\to G$ such that for every $h\in G$ we have $g h=x\quad \text{if and only if} \quad h=[g,x]_{\text{left}}.$ Clearly, we can take $[g,x]_{\text{left}}:=g^{-1}x$. So for any group $G$ the monoidal category $\mathcal{V}_{G}$ is left-closed.

Similarly, a monoidal category $\mathcal{V}$ is right-closed if for each object $v$ the functor $-\otimes v\colon \mathcal{V}\to \mathcal{V}$ has a right adjoint $[v,-]_{\text{right}}\colon \mathcal{V}\to \mathcal{V}$. For $\mathcal{V}_{G}$, we can take $[g,x]_{\text{right}}:=x g^{-1}$. So for any group $G$ the monoidal category $\mathcal{V}_{G}$ is both left- and right-closed.

Braidedness: For a monoidal category to be braided we need isomorphisms $v\otimes w \cong w\otimes v$ for all objects $v$ and $w$. For the category $\mathcal{V}_{G}$ that would mean we need $g h=h g$ for all $g,h\in G$, in other words, we need that $G$ is abelian. In that case we have that $\mathcal{V}_{G}$ is in fact symmetric.

Bicompleteness: A category is bicomplete if it has all limits and colimits. We have no hope of any non-trivial group $G$ having a bicomplete category $\mathcal{V}_{G}$. This is because this category is discrete: to form a product $g\prod h$ we would need projections to $g$ and $h$ but, because the only morphisms are identities, that would mean $g\prod h=g$ and $g\prod h =h$. You can check that the self products are actually defined though: $g\prod g=g$. This looks like it might spoil our fun, but it will transpire that the only limits that we need will be the self-products.

Summary: In summary then, if $G$ is abelian then $\mathcal{V}_{G}$ is a closed symmetric monoidal category; if $G$ is non-abelian then $\mathcal{V}_{G}$ is a closed monoidal category which is not braided.

As the definition of the tensor product of $\mathcal{V}$-categories requires that $\mathcal{V}$ is braided, we will restrict ourselves to the case that $G$ is an abelian group, and we will emphasize this fact by renaming it $A$.

### Functor category and tensor product

We can now look at how the hom torsor and tensor torsor from above arise in enriched category theory. We will work with an abelian group $A$ so that $\mathcal{V}_{A}$ is closed symmetric monoidal.

Functor categories and hom torsors: Suppose $\mathcal{V}$ is a closed symmetric monoidal category and that $\mathcal{C}$ and $\mathcal{D}$ are $\mathcal{V}$-categories then (provided that $\mathcal{V}$ has sufficiently many limits) there is $\mathcal{V}$-category $[[\mathcal{C},\mathcal{D}]]$ of $\mathcal{V}$-morphisms where the objects are the $\mathcal{V}$-functors and the hom object $[[\mathcal{C},\mathcal{D}]](\phi ,\theta )$ is given by the equalizer of the following diagram. $\prod _{c \in \mathcal{C}} \mathcal{D}(\phi (c),\theta (c)) \rightrightarrows \prod _{c',c'' \in \mathcal{C}} [\mathcal{C}(c',c''),\mathcal{D} (\phi (c'), \theta (c''))]$ We don’t need to worry here about what the two maps are as, in the case of interest when $\mathcal{V}=\mathcal{V}_{A}$, they are both going to be equalities.

If $T$ and $S$ are $A$-torsors, thought of as $\mathcal{V}_{A}$-categories, then we try to construct the functor category $[[T,S]]$. This has $\mathcal{V}_{A}$-functors, i.e. equivariant maps as morphisms. The hom object $g(\phi ,\theta )$, between equivariant maps $\phi ,\theta \colon T\to S$ is given by the equalizer of the above diagram, but as mentioned, the maps are equalities, so the equalizer is just the left-hand-side term, so $g_{[[T,S]]}(\phi ,\theta )=\prod _{t \in T} g_{S}(\phi (t),\theta (t)).$ As mentioned in the previous section, $\mathcal{V}_{A}$ does not have many limits, but fortunately it does have this product as all the terms in the product are the same: a calculation shows that $g_{S}(\phi (t_{1}),\theta (t_{1}))=g_{S}(\phi (t_{2}),\theta (t_{2}))\quad \text{for all} \,\,t_{1},t_{2}\in T.$ Hence we find that the “distance” between two functors is the distance between the two images of any point: $g_{[[T,S]]}(\phi ,\theta )= g_{S}(\phi (t),\theta (t))\quad \text{for all} \,\,t\in T.$ This means that the functor $\mathcal{V}_{A}$-category $[[T,S]]$ exists and is actually a torsor. You can easily check that the $A$-action is exactly that of the hom torsor, so $[[T,S]]=\text{Hom}_{A}(T,S).$

Tensor product categories and tensor torsors: If $\mathcal{V}$ is a braided monoidal category then you can define the tensor product $\mathcal{C}\otimes \mathcal{D}$ of two $\mathcal{V}$-categories $\mathcal{C}$ and $\mathcal{D}$. An object of $\mathcal{C}\otimes \mathcal{D}$ is a pair $(c,d)\in \text{ob}\mathcal{C}\times \text{ob}\mathcal{D}$ and the hom objects are defined by $(\mathcal{C}\otimes \mathcal{D})((c,d),(c',d'))\coloneqq \mathcal{C}(c,c')\otimes \mathcal{D}(d,d').$ The braiding of $\mathcal{V}$ is needed to define the composition morphisms.

If $T$ and $S$ are $A$-torsors, thought of as $\mathcal{V}_{A}$-categories, then we can construct the tensor product $\mathcal{V}_{A}$-category $T\otimes S$. You might expect that this is actually a torsor equal to the tensor torsor $T\otimes _{A} S$ but that’s not quite true!

In the torsor $T\otimes _{A} S$ we have made the identification $(a\cdott,s)=(t,a\cdot s)$ however in the $\mathcal{V}_{A}$-category $T\otimes S$ these two points are distinct despite having no distance’ between them: you can check that $g_{T\otimes S}((a\cdot t,s),(t,a\cdot s))=e$.

This means that the $\mathcal{V}_{A}$-category $T\otimes S$ is not actually a torsor but it is equivalent to the torsor $T\otimes _{A} S$: the quotient map on objects gives an equivalence of $\mathcal{V}_{A}$-categories $T\otimes S\to T\otimes _{A} S$. You can think of $T\otimes S$ as a fattened-up version of $T\otimes _{A} S$; it is like using a stack instead of a quotient space. This is what you might expect from a categorical approach, in that you don’t violently quotient out but rather encode the equivalence relation via isomorphisms.

### Final words

I’ll just finish by mentioning the Yoneda map. For any $\mathcal{V}_{A}$-category $T$, the presheaf $\mathcal{V}_{A}$-category $\widehat{T}\coloneqq [[T^{\text{op}},\mathcal{V}_{A}]]$ is a torsor.

If $T$ is actually a torsor then the Yoneda map $T\to \widehat{T}$ is an isomorphism, which is not something you would expect from the Yoneda map!

If $T$ is not a torsor then the Yoneda map $T\to \widehat{T}$ is a torsorization of $T$. For instance, in the tensor example above we have $T\otimes _{A} S\equiv \widehat{T\otimes S}$.

Posted at June 3, 2013 12:10 AM UTC

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### Re: Torsors and enriched categories

I read this immediately after watching this video (don’t ask me why I watched it, just a random thing - if you don’t want to watch it, is a bit like Vi Hart in pace and style), and in my head it was in that narrator’s voice and at that speed. o.O Might be an idea to make a video explaining a cute piece of category theory in all seriousness with the energy and speed of a Vi Hart video (not you necessarily, Simon, but someone).

OK, more seriously, a torsor is a principal G-bundle over a point, and the map $T\times T \to G$ you give is the cocycle for the cover $T\to \ast$. Here in Adelaide we call it the difference map (as it exists for any principal $G$-bundle). What confused me for a while when I first hear the term is that algebraic geometers think of torsors in the category of schemes over a base, hence something like a principal $G$-bundle without asking for local triviality. Except sometimes they do.

Thinking of $T$ as the objects of a category enriched over $\mathcal{V}_G$ is interesting, because the usual view (or at least, my usual view) of this is that we take the coarse (=banal=indiscrete=chaotic=etc) groupoid with objects $T$ and a unique arrow between any two objects, and then a $G$-torsor structure on this is a fibration to the one-object groupoid with arrows $G$.

The issue with needing abelianness is remedied by taking bitorsors - sets with left and right commuting $G$-action, such that both of them are torsors. Can this be seen in the light of this post, apart from saying there are two compatible enrichments over $\mathcal{V}_G$?

(Still racing from that video, need to slow down)

Posted by: David Roberts on June 3, 2013 3:02 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

David R said

Might be an idea to make a video explaining a cute piece of category theory in all seriousness with the energy and speed of a Vi Hart video (not you necessarily, Simon, but someone).

Could that someone not be you David? I keep meaning to make more Catster videos, but don’t seem to be able to decide on a topic.

On your more serious points, I was going to say something about torsors appearing as the fibres in principal fibre bundles, but that didn’t make the final edit. I was vaguely wondering if you could think of a principal bundle as a category enriched over $\mathcal{V}_G \cup {\infty}$ where $\infty \otimes g = \infty$ for all $g\in G$, so that points in different fibres have $\infty$ distance between them. (Do I need $\infty$ to be initial?) But I didn’t follow this far enough to see if it worked.

The issue with needing abelianness is remedied by taking bitorsors

Well it all depends what you mean by “the issue” and “remedied”. Certainly, if you consider bitorsors over a non-abelian group then you can define a tensor product of such bitorsors. But, no, I don’t see immediately how that fits in with this. Does it help to define an internal hom? I’m not sure. Abelian torsors are, without doubt, much easier to handle which is why abelian gerbes are so much easier to define.

Posted by: Simon Willerton on June 3, 2013 5:24 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

Simon wrote:

Could that someone not be you David?

It may very well be. I like that sort of public outreach.

Well it all depends what you mean by “the issue” and “remedied”. Certainly, if you consider bitorsors over a non-abelian group then you can define a tensor product of such bitorsors. But, no, I don’t see immediately how that fits in with this.

Yes, it is a bit like this. Just thought I’d throw it in the ring, though.

Posted by: David Roberts on June 4, 2013 3:24 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

Wow, those videos of Vi Hart are absolutely spectacular.

Posted by: Bruce Bartlett on June 6, 2013 12:51 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

Thanks, Simon. By the way, your “extra condition” says that the enriched category is tensored (or copowered, in modern language).

Posted by: Steve Lack on June 3, 2013 3:19 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

I think the extra condition is also equivalent to saying that the enriched category is Cauchy complete.

Posted by: Mike Shulman on June 3, 2013 2:13 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

Is there an easy way to see that Mike?

Posted by: Simon Willerton on June 3, 2013 5:00 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

Well, every object of a group is invertible, hence dualizable, and copowers with a dualizable object are absolute colimits. To show that you don’t need anything else, I think you can look at adjoint profunctors from the unit category and show that they consist of giving a consistent ‘distance’ to every object, which in a torsor will be represented.

This makes me want to generalize from groups to monoids.

Posted by: Mike Shulman on June 4, 2013 6:33 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

Mike said:

This makes me want to generalize from groups to monoids.

If you mean what I think you mean, then I don’t think you get much else.

Given a monoid $M$ you can look at categories enriched the discrete monoidal categroy $\mathcal{V}_M$. Composition implies that

$\mathcal{C}(c,d)\mathcal{C}(d,c)=\mathcal{C}(c,c)=e$

which means that $\mathcal{C}(c,d)$ is invertible in $M$ and it suffices to consider the group of invertible elements in $M$.

Posted by: Simon Willerton on June 5, 2013 3:04 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

)-:

Posted by: Mike Shulman on June 6, 2013 5:03 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

Hi Steve. I’d not thought about the condition in terms of copowers. I think that’s only half the story — the other half has to do with being skeletal and I had thought about that. Let’s have a look at these conditions, and see how I’ve mucked up my conventions!

For $\mathcal{V}$ closed symmetric monoidal, a $\mathcal{V}$-category $\mathcal{C}$ is copowered if for every $v\in \text{ob}(V)$ there is a functor $v\odot -\colon \mathcal{C}\to \mathcal{C}$ such that for all

$\mathcal{C}(v\odot c,d)\cong [v,\mathcal{C}(c,d)].$

This implies (exercise) that $(v\otimes w)\odot c\cong v\odot (w\odot c)$. So for $A$ an abelian group, a $\mathcal{V}_{A}$-category $T$ is copowered if for every $h\in G$ we have a $\mathcal{V}_{A}$-functor $h\odot - \colon T\to T$ such that (bearing in mind my conventions)

$g(t_{2},h\odot t_{1}) = g(t_{2},t_{1})h^{-1}.$

Functoriality means that $g(h\odot t_{2},h\odot t_{1})=g(t_{2},t_{1})$. Putting this all together means that $g(h\odot t,t)=g(h^{-1} \odot h \odot t,h^{-1}\odot t)= g(t, h^{-1}\odot t)=g(t,t)h=h.$ This means that for $h\in G$ and $t\in T$ we have a $t_{2}$ such that $g(t_{2},t_{1})=h$, namely $t_{2}=h\odot t_{1}$. However — unless I’ve missed something — there’s no uniqueness here.

For the uniqueness it is necessary and sufficient that $g(t_{2},t_{1})=e$ implies $t_{2}=t_{1}$. This is a skeletal condition. The underlying category $T_{0}$ of $T$ has hom-sets given by

$T_{0}(t_{2},t_{1})= \begin{cases} \{ \ast \} &\text{if }g(t_{2},t_{1})=e,\\ \emptyset &\text{otherwise} \end{cases}$

So

\begin{aligned} T_{0} \,\,\text{is skeletal} &\Longleftrightarrow T_{0}\,\, \text{is discrete}\\ &\Longleftrightarrow g(t_{2},t_{1})=e\,\,\text{implies}\,\, t_{2}=t_{1}\\ &\Longleftrightarrow \text{if the uniqueness condition holds.} \end{aligned}

This means that when we are dealing with abelian torsors it looks like the extra condition is copowered plus the underlying category is skeletal.

In the non-abelian case, we need to talk about copowers over a non-braided enriching category. I suspect in that case we’d have to talk about left and right copowers, with only one of them being relevant, because we just need a left action.

Posted by: Simon Willerton on June 3, 2013 4:58 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

I would guess that Steve, like me, was thinking ‘like a category theorist’ with uniqueness always being up to isomorphism. (Insert shameless plug for homotopy type theory.)

IIRC, there is really only one ‘handedness’ of copowers (and more generally colimits) that ‘makes sense’ when enriching over a nonsymmetric base. It might or might not be possible to define both of them, but I think there’s one of them that’s clearly ‘correct’.

Posted by: Mike Shulman on June 4, 2013 6:27 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

I have no experience with enrichments over non-symmetric categories so my intuition may be way off but I find this strange. For any monoidal category $(\mathcal{V}, \otimes)$ you can consider its “opposite” $(\mathcal{V}, \otimes^\mathrm{op})$ and there is a correspondence between enrichments in $(\mathcal{V}, \otimes)$ and enrichments in $(\mathcal{V}, \otimes^\mathrm{op})$ and “handedness” of colimits is inverted under this correspondence. But it seems to me that there is a perfect symmetry between these two situations while you seem to be saying that one is preferred over the other. So what’s going on here?

Posted by: Karol Szumiło on June 5, 2013 6:13 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

Having studied enriched categories over not necessarily symmetric bases some time ago, I suspect that an enrichment over $(\mathcal{V},\otimes)$ cannot be turned into an enrichment over $(\mathcal{V},\otimes^{op})$ in the sense that you might have in mind. The issue is that there really are two different notions of enriched category: one with a composition like $C(a,b)\otimes C(b,c) \longrightarrow C(a,c)$ and one with a composition like $C(b,c) \otimes C(a,b) \longrightarrow C(a,c).$ Taking the oppposite of the base corresponds to exchanging these two notions of enrichment.

Posted by: Tobias Fritz on June 5, 2013 8:14 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

Yeah, it’s definitely not the case that a $\mathcal{V}$-enrichment can be turned into a $\mathcal{V}^{rev}$-enrichment. For instance, one of my favorite non-symmetric monoidal categories is $2Cat$ with the lax Gray tensor product. When enriched over itself, its hom-2-categories consist of 2-functors, lax natural transformations, and modifications, and the laxness of the tensor product captures the fact that given lax natural transformations $\alpha:f\to g$ and $\beta:h\to k$, the interchange law holds only laxly in that there is a transformation $(k.\alpha)(\beta.f) \to (\beta.g)(h.\alpha)$ which is one of the lax 2-cell components of $\beta$. Reversing the enrichment would swap the roles of $\alpha$ and $\beta$, which would make the morphism go the other way, which would require the transformations to be oplax instead.

Posted by: Mike Shulman on June 6, 2013 5:10 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

You are both right of course. Can you explain the difference between “left colimits” and “right colimits” in some particular example? For example in categories enriched with respect to the Gray tensor product?

Posted by: Karol Szumiło on June 6, 2013 8:09 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

Presumably all this works in a smooth setting. Is there, then, an enriched story to tell about torsoroids?

Posted by: David Corfield on June 3, 2013 11:03 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

Is there, then, an enriched story to tell about torsoroids?

Interesting idea. Let’s see, thinking aloud here. Take Simon’s 4 axioms for a $G$-torsor from an enriched category perspective, and apply it to the case where everything is smooth.

In axioms 1 and 3, if we keep $t_1$ fixed and let $t_2 = t$ vary, then differentiating gives us a linear map $\phi : Lie(G) \rightarrow Vect(T)$.

I think if we differentiate axiom 2, we get that $\phi$ respects Lie brackets. Not sure about that.

So it seems that to get an infinitesimal $G$-torsor, we don’t need axiom 4.

If we want to capture the concept of a Lie torsoroid, we’ll want to drop the condition that $\phi$ is a homomorphism in some directions… i.e. modify axiom 2.

Mmm, I’m not getting anywhere but it is an interesting idea.

Posted by: Bruce Bartlett on June 6, 2013 12:36 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

OK, having now got to the last of your original post, and since Mike has inserted his “shameless plug”, I’ll just remark:

1. under Univalence, the type of all $G$-torsors is (if you can write it down) a fine $\mathbf{B} G$, and even has a fine basepoint, $G$ itself

2. this $\widehat{\cdot}$ construction of $\otimes_A$ seems to give a nice something structure on $\mathbf{B} A$ without seeming to mention quotients, (though of course there may be a similar question of can you write it down). Dare we hope it be nearly the right abelian group structure?

I mention (2) because the main point in using (1) is that it doesn’t seem to mention quotients, unless that’s your favourite way to spell “$A_n$”. There’s another way that uses only products and a big nearly-cubic diagram, but actually writing it in, say, coq` is a pain… Anyways, I don’t know if that’s useful to anyone else, or if they can see ways to fill-in the gaps, but it seemed worth mentioning.

Posted by: Jesse C. McKeown on June 5, 2013 12:29 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

One minor problem with the type of $G$-torsors as a $\mathbf{B}G$ is that it raises the universe level.

Posted by: Mike Shulman on June 5, 2013 2:14 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

Yes, that is irksome… there ought to be something like a Downward Löwenheim+Skolem theorem for this kind of situation: here is a “big” model of $\mathbf{B}G$, but the condition that makes something a $\mathbf{B}G$ is not much “larger” than $G$ itself, nor very complicated…

Posted by: Jesse C. McKeown on June 5, 2013 5:01 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

At least we can say that in general $\mathbf{B}$(Nerve$(\mathcal{V}_G)) \subseteq$ Nerve($\mathcal{V}_G$-Cat). If we restrict to the subcategory of $\mathcal{V}_G$-categories tensored over $\mathcal{V}_G$ then that subset symbol becomes an equality. I don’t know if this helps the universe situation. (It’s also been a while since I thought about this. See the date on this shameless plug: page 11 of a talk from long ago.)

Posted by: stefan on June 5, 2013 8:08 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

Voevodsky has proposed various “resizing axioms” for type theory, one of which I think would make this $B G$ small. Of course, you can also construct a version of $B G$ as a HIT, which works just as well and is small. I believe the latter is also more in line with the history of type theory — or at least the calculus of constructions — where inductive definitions were first noticed in an “impredicative” form constructed out of the universe (but raising the universe level), and then later introduced as basic type forming operations not requiring any impredicativity or resizing. The definition as a basic type former also has the advantage of a dependent elimination principle, which is harder to get with impredicative definitions; I suspect that the torsory definition of $B G$ may be harder to prove things about than the HIT version.

Posted by: Mike Shulman on June 6, 2013 5:31 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

When you defined a torsor, you forgot to specify that the underlying set must be nonempty. This is important because while all of the other conditions can naturally be stated algebraicly, the natural way to make nonemptiness an algebraic condition is to specify an element in the underlying set, which ruins the symmetry.

Posted by: Itai Bar-Natan on June 5, 2013 1:17 PM | Permalink | Reply to this

### Re: Torsors and enriched categories

Itai wrote:

the natural way to make nonemptiness an algebraic condition is to specify an element in the underlying set,

Here one could use the HoTT expression “merely exists an element”, rather than “exists an element”. The former naturally captures the notion that there is an element, but we don’t know what it is, whereas the latter necessitates specifying an element.

Posted by: David Roberts on June 6, 2013 12:37 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

merely

Was that the adverb that was settled on? There was a long, long terminological discussion not so long ago here, where lots of adverbs were tried on for size. I didn’t think it had reached a firm conclusion.

Posted by: Todd Trimble on June 6, 2013 2:04 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

I think it was settled on somewhere, because the book that the Univalent foundations program has produced has the term throughout.

Posted by: David Roberts on June 6, 2013 3:45 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

In fact a better link to the book is this, once the book is put there :-) (that page promises “soon”).

Posted by: David Roberts on June 6, 2013 3:47 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

Thanks! But then how did you get hold of the book? I didn’t see how to do it myself, maybe for lack of trying hard to figure it out. Do I need to talk to Mike (who will no doubt see this anyway)? :-)

Posted by: Todd Trimble on June 6, 2013 5:15 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

I followed the instructions and downloaded the LaTeX source from the GitHub link provided, and compiled it myself. No doubt my copy is out of date already, but merely in insignificant ways.

Posted by: David Roberts on June 6, 2013 8:44 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

I have purposely not been posting any links to the book yet publically, nor are there compiled versions available anywhere, because it’s not finished yet. You can, of course, as David says, download the source from github and compile it yourself. Personally, I would suggest that no compiled versions be distributed at this time (although you can technically do whatever you want, since the book has a CC BY-SA licence), just so that anyone obtaining a copy has to go through the extra step ensuring they understand that it’s not the official release version. It is true that at this point we expect to be making only very minor changes before release.

Posted by: Mike Shulman on June 6, 2013 11:28 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

“Merely” was indeed settled on for purposes of The HoTT Book. (Not all the discussion happened on this blog. For which you should all probably be thankful…) It remains to be seen whether it will catch on more widely. (FWIW, the book also states clearly that the choice of “there exists” and “there merely exists” may not be the best one in all cases. Sometimes one may want “mere existence” to be the default, in which case one can say instead “there purely exists” (or perhaps “there constructively exists”) and “there exists”.) My current guess for availability of the book is by the end of June.

For the benefit of any bystander who may be confused, let me point out that David’s comment applies only when working in HoTT. In traditional foundations (which I assume Simon is implicitly using), there is no need to say “merely” in front of “there exists”; it’s only in propositions-as-types logic that saying “there exists” necessitates specifying an element.

Posted by: Mike Shulman on June 6, 2013 5:23 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

Mike wrote: “…applies only when working in HoTT”

though it is surely also conceivable when working in the algebraic setting, as Itai wrote. To specify that “the object $a$ is inhabited” when working with (say) a Lawvere theory - and its natural internal logic - the only way we can say that is to give an element of $a$, namely a 0-ary operation in the Lawvere theory.

Think of this as a very truncated and restricted-in-logic form of the HoTT version.

Posted by: David Roberts on June 6, 2013 8:42 AM | Permalink | Reply to this

### Re: Torsors and enriched categories

But mere existence is not an algebraic condition. It’s true that the terminology could be used in any type theory which admits propositional truncation, but that’s not the case for the simple type theory that corresponds to finite-product logic (Lawvere theories) or any other notion that I’ve heard called “algebraic”. Moreover, that sort of type theory is not generally used as a foundational system — being rather a way of specifying a kind of structure inside some other system — so I would think that the conventions of English-to-mathematics should be inherited from the ambient system, to avoid serious confusion.

Posted by: Mike Shulman on June 6, 2013 10:58 AM | Permalink | Reply to this

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