Torsors and Their Classification III

Interlude on constructions with torsors

Let $P$ be a right torsor for some group object $G$ . Twisting with $P$ will be a functor $\objl{G}\to \ca C$ from objects with a left $G$ –action to objects in $\ca C$ .

Let $X$ be an object with a left $G$ –action. The twist $P\times^G X$ is defined as the quotient of $P\times X$ by the diagonal action of $G$ , that is, it is the coequalizer $\begin{tikzcd} G\times P\times X \arrow[r,shift left=.5ex, "\act"] \arrow[r, shift right=.5ex, "\proj"'] & P\times X \arrow[onto,r] & P\times^G X \end{tikzcd}$ where $\act(g, p, x) = (p g^{-1}, gx)$ on generalized elements. If $X$ additionally carries a right action by some group object $H$ , then evidently the twist $P\times^G X$ also carries a right $H$ –action; similarly, if the torsor $P$ carries a left action by some group object $H'$ , then the twist also carries a left $H'$ –action.

Let’s introduce some notation for the action of the epimorphism $P\times X\onto P\times^G X$ on generalized elements. Namely, given a generalized element $(p,x)\colon T\to P\times X$ , I will write $[p,x]$ for the composition $T\to P\times X\to P\times^G X$ . Because $P\times X\onto P\times^G X$ is an epimorphism, to check some relation between generalized elements downstairs it will be enough to check it on generalized elements of the form $[p,x]$ . Furthermore, if $(p,x)$ actually comes from a generalized element $(g,p,x)\colon T\to G\times P\times X$ via projection, then by the coequalizer property we find $[p,x] = [p g^{-1}, g x]$ .

The process of twisting has some exactness properties:

The functor $P\times^G{-}\colon\objl{G}\to\ca C$ preserves colimits and finite products.

Regarding preservation of colimits, we can in fact construct a right adjoint to $P\times^G{-}$ . It is given by the functor $({-})^P$ which sends $T$ to the function object $T^P$ whose $G$ –action is given by precomposing with the action on $P$ . This won’t be particularly important in what follows, so I’ll skip the detailed calculation.

Showing that binary products are preserved is a little more difficult. Let’s first observe that the functor $G\times^G{-}$ is naturally isomorphic to the identity functor: Consider the commutative diagram $\begin{tikzcd} G\times X & X \arrow[l, "e\times{\id}"'] \\ G\times G\times X \arrow[u, shift left=.5ex, "\proj"] \arrow[u, shift right=.5ex, "\act"'] & X \arrow[l, "e\times e\times{\id}"] \ar[u, shift left=.5ex, "{\id}"] \ar[u, shift right=.5ex, "{\id}"'] \end{tikzcd}$ This induces a natural transformation ${\id}\to G\times^G{-}$ which is easily checked to be an isomorphism on generalized elements.

Now, choose a trivializing cover $U\onto *$ of the torsor $P$ . Because pullback in a topos commutes with colimits and limits, it will then be enough to check that, for any $X$ and $Y$ , the natural morphism $P\times^G (X\times Y)\to (P\times^G X) \times (P\times^G Y)$ after pulling back to the corresponding natural morphism in $\ca C/U$ becomes an isomorphism. Since there is a $G\times U$ –equivariant isomorphism $P\times U\isom G\times U$ , the functors $P\times U\times^{G\times U}{-}$ and $G\times U\times^{G\times U}{-} \isom \id$ are naturally isomorphic; in particular $P\times U\times^{G\times U}{-}$ preserves products, which is exactly what we needed to show.

Preservation of finite products in particular implies that $P\times^G{-}$ sends group objects in $\objl{G}$ to group objects in $\ca C$ . This allows the following construction.

When endowed with the conjugation action of $G$ on itself, $G$ becomes a group object in $\objl{G}$ . In particular, ${}^P G\coloneqq P\times^G G$ is naturally a group object in $\ca C$ .

It is enough to check that the structure maps $e\colon *\to G$ , $m\colon G\times G\to G$ and $i\colon G\to G$ are equivariant with respect to conjugation. This can be checked on generalized elements where it reduces to the corresponding statement about genuine groups. There, it is a simple calculation.

More generally, if $G$ acts on itself via automorphisms, then the twist $P\times^G G$ with respect to this action becomes a new group object. Sometimes, this is called a form of $G$ . This construction let’s us refine the process of twisting a little bit. It turns out that the twisted object $P\times^G X$ comes naturally equipped with a left group action by ${}^P G$ . In fact, $P$ itself carries a natural left action by ${}^P G$ . Recall that, since $P$ is a righ $G$ –torsor, the map $P\times G\to P\times P$ given by $(p,g)\mapsto (p, pg)$ on generalized elements is an isomorphism. Let’s denote the inverse map by $(p, q) \mapsto (p, \tdiv{p}{q})$ on generalized elements. Then we can construct a commutative diagram $\begin{tikzcd} P\times G \times P\arrow[r, "\varphi"] & P \arrow[d, equal] \\ G\times P\times G\times P\arrow[u, shift left=.5ex]\arrow[u, shift right=.5ex] \arrow[r, "\psi"'] & P \end{tikzcd}$ where, on generalized elements, $\varphi(p,g,q) = p\cdot g \tdiv{p}{q}$ and $\psi(h,p,g,q) = \varphi(p,g,q)$ . The commutativity follows from the following calculation: $\varphi(ph^{-1}, hgh^{-1}, q) = ph^{-1}\cdot hgh^{-1} \tdiv{ph^{-1}}{q} = p\cdot g h^{-1} h \tdiv{p}{q} = \varphi(p,g,q).$ Hence, because colimits commute with products in toposes, we have a morphism ${}^P G\times P\to P$ where the axioms for a left action can be checked by direct calculation.

This action makes $P$ into a left ${}^P G$ –torsor.

Take an epimorphism $U\onto *$ which trivializes $P$ . Then it will be enough to check that there is a ${}^P G$ –equivariant isomorphism $P\times U\iso {}^P G\times U$ . Since everything in sight commutes with pullbacks, we might as well assume that $P$ is isomorphic to the trivial right $G$ –torsor and check that then it is ${}^P G$ –isomorphic to the trivial left ${}^P G$ –torsor.

We first need to produce an isomorphism $P\iso {}^P G$ given a $G$ –isomorphism $\varphi\colon P\iso G$ . I claim that the composition $G \to[g\mapsto(e,g)] G\times G\onto G\times^G G$ is an isomorphism. The inverse map is defined by the commutative¹ diagram $\begin{tikzcd} G\times G \arrow[r, "\conj"] & G \arrow[d, equal] \\ G\times G\times G \arrow[u, shift left=.5ex]\arrow[u, shift right=.5ex] \arrow[r, "\conj\circ\proj"'] & G \end{tikzcd}$ where $\conj$ denotes the conjugation action. Hence, we get a chain of isomorphisms² $\begin{tikzcd} P \arrow[r,"\varphi"] & G \arrow[r, "{g\mapsto(e,g)}"] & G\times^G G \arrow[r, "\varphi^{-1}_*"] & P\times^G G = {}^P G. \end{tikzcd}$ The last thing to check is that this isomorphism $P\iso {}^P G$ is in fact ${}^P G$ –equivariant, that is, we need to check that the diagram $\begin{tikzcd} {}^P G\times P \arrow[r, "\act"] \arrow[d] & P \arrow[d] \\ {}^P G\times G \arrow[d] & G \arrow[d] \\ {}^P G\times (G\times^G G) \arrow[d] & G\times^G G \arrow[d] \\ {}^P G\times {}^P G \arrow[r, "m"'] & {}^P G \end{tikzcd}$ commutes. On generalized elements, the top composition gives $[p, g], q \mapsto [\varphi^{-1}(e), \varphi(p) g \varphi(p)^{-1}\varphi(q)]$ while the bottom composition is $[p,q], q\mapsto [p, g\cdot\tdiv{\varphi^{-1}(e)}{p}\varphi(q)\tdiv{p}{\varphi^{-1}(e)}] = [p, g\varphi(p)^{-1}\varphi(q)\varphi(p)]$ But this is equal to $[\varphi^{-1}(e), \varphi(p) g\varphi(p)^{-1}\varphi(q)\varphi(p)\varphi(p)^{-1}]$ which proves the required commutativity.

A consequence of this theorem is that the twisting functor $P\times^G{-}$ may be considered as a functor $\objl{G}\to\objl{{}^P G}$ . Furthermore, if $Q$ is a left $G$ –torsor, then we can choose an epimorphism $U\onto *$ which trivializes both $Q$ and $P$ . Then, because twisting commutes with products, we find $(P\times^G Q)\times U\isom G\times U \times^{G\times U} G\times U\isom G\times U\isom {}^P G\times U$ similarly to the proof of the theorem. Hence, twisting restricts to a functor $P\times^G{-}\colon \Tors[G]\to\Tors[{}^P G]$ from the category of left $G$ torsors to the category of left ${}^P G$ torsors. Furthermore, if $Q$ is a left $K$ –torsor and a right $H$ –torsor, $P$ a left $H$ –torsor and a right $G$ –torsor then there is a natural isomorphism $Q\times^H (P\times^G{-})\isom (Q\times^H P)\times^G{-}$ which follows from the fact that colimits commute with colimits and with binary products in toposes.

If $P$ is a right $G$ –torsor and hence a left ${}^P G$ –torsor, let $\op{P}$ denote the right ${}^P G$ –torsor with action given by $p\cdot g = g^{-1}\cdot p$ on generalized elements. Then $\op{P}\times^{{}^P G} P$ is a trivial right $G$ –torsor. Furthermore, ${}^{\op{P}}({}^P G)\isom G$ as group objects and along this identification $P\times^G \op{P}$ is a trivial right ${}^P G$ –torsor. It follows that $P\times^G{-}\colon \Tors[G]\to\Tors[{}^P G]$ is an equivalence of categories with weak inverse $\op{P}\times^{{}^P G}{-}\colon\Tors[{}^P G]\to\Tors[G].$

Let’s first produce an isomorphism ${}^{\op{P}}({}^P G)\iso G$ . Let $\varphi\colon P\times P\times G\to G$ be such that $\varphi(p,q,g) = \tdiv{q}{p} \cdot g \cdot \tdiv{p}{q}$ on generalized elements. It is straightforward, albeit tedious, to check that $\varphi$ descends to a group homomorphism $\op{P}\times^{{}^P G} (P\times^G G) = {}^{\op{P}}({}^P G)\to G.$ I claim that this is an isomorphism. To verify this claim, it will be enough to pull back along an epimorphism $U\onto *$ which trivializes both the right $G$ –torsor $P$ and the right ${}^P G$ –torsor $\op{P}$ , that is, we can assume that $P$ is trivial.³ So let $\psi\colon G\iso P$ be a $G$ –equivariant isomorphism. We can then set $\widetilde\varphi\colon G\to {}^{\op{P}}({}^P G)$ to be the morphism such that $\widetilde\varphi(g) = [\psi(e), [\psi(e), g]]$ on generalized elements. Computing the compositions $\varphi\circ\widetilde\varphi$ and $\widetilde\varphi\circ\varphi$ on generalized elements, we find $\varphi(\widetilde\varphi(g)) = \varphi([\psi(e), [\psi(e), g]]) = g$ and $\widetilde\varphi(\varphi([p, [q, g]])) = [\psi(e), [\psi(e), \tdiv{q}{p}\cdot g\cdot \tdiv{p}{q}]].$ Noticing that $[p, [q, g]] = [p_0, [p_0, \tdiv{p_0}{p}]\, [p_0, \tdiv{q}{p_0}\cdot g\cdot \tdiv{p_0}{q}]\, [p_0, \tdiv{p}{p_0}]]$ where $p_0 = \psi(e)$ finishes the proof of ${}^{\op{P}}({}^P G)\isom G$ .

Next, let’s check that $\op{P}\times^{{}^P G} P$ is a trivial right $G$ –torsor. Let $P\to P\times P\onto \op{P}\times^{{}^P G} P$ be the diagonal composed with the canonical projection. I claim that this map descends to the quotient $P/G\isom *$ which will provide a global section of $\op{P}\times^{{}^P G} P$ . This will then imply the claim. The calculation $[pg, pg] = [p, [p, g^{-1}]\cdot(pg)] = [p, p g g^{-1}] = [p,p]$ shows that the diagonal is $P\to \op{P}\times^{{}^P G} P$ is invariant under the right action of $G$ which proves the claim. The last assertion of the theorem is proven in a completely analogous way.

In summary, we have found that twisting with a right torsor $P$ is an invertible operation up to isomorphism. This will become essential in the computation of the homotopy groups of $BG$ at any basepoint.

Remember that in $G\times^G G$ the $G$ on the left is the trivial right torsor and the $G$ on the right is taken with the conjugation action.↩
Note that $\varphi^{-1}_*$ denotes the morphism induced by the $G$ –equivariant isomorphism $\varphi^{-1}\colon G\to P$ , using the fact that twisting is functorial with respect to $G$ –equivariant morphisms in both factors.↩
By the proof of the previous theorem, this also implies that $\op{P}$ is trivial.↩