The n-Category Café

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 $T\otimes _{A} S\equiv \widehat{T\otimes S}$.