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Question

Let $G$ be a group, and let $X$ be a $G$-biset that is (weakly) invertible with respect to the contracted product. Is $X$ necessarily a bitorsor?

Background

By $G$-biset, I mean a set equipped with commuting left and right $G$-actions. There is a standard tensor product on the category of $G$-bisets called the contracted product; it is defined by $X \times_G Y = X \times Y / (x \cdot g, y) \sim (x, g \cdot y)$, where $G$ acts on the left by its left action on $X$, and on the right by its right action on $Y$. The unit object is the group $G$, where $G$ acts by left and right multiplication on itself.

A left $G$-torsor is a left $G$-set $X$ such that the map $G \times X \to X \times X$, $(g, x) \mapsto (g \cdot x, x)$ is a bijection. A right $G$-torsor is defined analogously. A $G$-bitorsor is a $G$-biset that is both a left and a right $G$-torsor. A $G$-bitorsor $X$ is necessarily invertible with respect to the contracted product; its inverse is the opposite $G$-bitorsor $X^{\operatorname{op}}$. This bitorsor has the same objects as $X$, but $g \in G$ acts on the left (resp. the right) by the right (resp. left) action of $g^{-1}$.

It follows from a simple counting argument that when $G$ is a finite group, any invertible $G$-biset is a $G$-bitorsor. Is this true for arbitrary groups (and more generally, in an arbitrary topos)? What about if we replace "invertible" by "right- (or left-) invertible"?

I can show, at least in the punctual topos (and I think it's true in general), that if $X^{\operatorname{op}}$ is an inverse to $X$, then $X$ must be a $G$-bitorsor. So the question is whether a $G$-biset can have an inverse not of this form.

The reason I'm interested in this question is that I want to understand how to generalize bitorsors to higher categorical settings. A possible generalization would be an invertible profunctor, but this is only a good definition if the answer to my question is affirmative.

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    $\begingroup$ +1 for formatting, giving definitions, etc. I mean, such things should be completely standard, and questions without them closed without comment. But alas, this is not always the case. $\endgroup$
    – Theo Johnson-Freyd
    Commented Mar 26, 2010 at 23:31

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A torsor is a faithful transitive $G$-set. If the left $G$-action on $X$ is not faithful, the left $G$-action on $X\times_G Y$ will not be faithful. If the left $G$-action on $Y$ is not transitive, the left $G$-action on $X\times_G Y$ will not be transitive. By symmetry, it follows that a $G$-biset with a left and right inverse is a $G$-bitorsor.

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  • $\begingroup$ I don't see why any of the actions need to be transitive. The left action on $X \times_G Y$ is given by the left action of $G$ on $X$. (I forgot to mention the action in the question; I'll fix this.) Even if the action is not transitive on $X$ or $Y$, that does not a priori mean that it's not transitive on $X \times_G Y$. We could get from $(x, y)$ to $(x', y')$, for instance, by taking $x' = x \cdot g$ for some $g \in G$, whence $(x, y) = (x, g \cdot y)$, and then taking some $g'$ such that $(g \cdot y) \cdot g' = y'$. So it seems that the only (obvious) requirement is that... $\endgroup$
    – Evan Jenkins
    Commented Mar 26, 2010 at 21:50
  • $\begingroup$ ...the left and right actions be jointly transitive, which is not a very strong condition. $\endgroup$
    – Evan Jenkins
    Commented Mar 26, 2010 at 21:50
  • $\begingroup$ That the left action on $X\times_G Y$ is transitive means that for any $(x,y)$ and $(x',y')$, there is some $g\in G$ such that $g\cdot(x,y)\sim (x',y')$. This means there exists $h\in G$ such that $(g\cdot x\cdot h^{-1},h\cdot y)=(x',y')$. In particular, $h\cdot y=y'$, so the left action on $Y$ is transitive. $\endgroup$
    – Tom Church
    Commented Mar 26, 2010 at 21:55
  • $\begingroup$ Aha, true! I don't know how I missed that; I made the exact same argument when I proved that $Y \cong X^{\operatorname{op}}$ implies that it's a bitorsor, but my brain somehow failed to make the connection. This still leaves open the question of whether having just a right inverse is sufficient, and I'm also not sure how well this proof generalizes (I'm not exactly sure how well "simply transitive" translates to an arbitrary topos), but it certainly answers my question as posed. Thanks! $\endgroup$
    – Evan Jenkins
    Commented Mar 27, 2010 at 4:02

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