3
$\begingroup$

I'm reading the nlab entry on torsors and am struggling to show the two definitions below are equivalent using only universal properties. All torsors in question are over a point.

Definition 1. A $G$-torsor is a $G$-action $\varphi:G\curvearrowright X$ which is isomorphic as an action to multiplication $m:G\curvearrowright G$.

Definition 2. A $G$-torsor is a $G$-action $\varphi:G\curvearrowright X$ such that $X$ admits a global point $x:\mathbf 1\to X$ and the arrow $(\varphi,\pi_2):G\times X\to X\times X$ is an isomorphism.

Given the first definition I thought of showing that $(m,\pi_2):G\times G\to G\times G$ is an isomorphism iff $(\varphi,\pi_2):G\times X\to X\times X$ is an isomorphism, since the latter is always an iso, but I don't know how to show this.

For the converse, for instance here, the nlab assumes the second definition and says the following square is a pullback, but I don't understand why. The "elements" of the pullback are triples $(g,x,y)$ such that $gx=y$ and saying this is the same as giving $G$ seems to assume we're already dealing with a $G$-torsor. $$\require{AMScd} \begin{CD} G\times \mathbf 1 @>{1\times x}>> G\times X\\ @VVV @VV{(\varphi,\pi_2)}V\\ X\times \mathbf 1 @>>{1\times x}> X\times X \end{CD}$$

  • 1
    what do the bent arrows mean? – 2017-01-17
  • 1
    A group action. – 2017-01-17
  • 0
    never seen anyone use that. – 2017-01-17
  • 0
    Are you sure you gave the correct definitions? Both definitions describe _trivial_ $G$-torsors only. In Def. 2, for example, I think you rather want $X\to 1$ be an epi instead of requiring a global point. – 2017-01-17
  • 0
    @Hanno I'm only interested in the definition of torsors over a point, so I thought there's no issues of local triviality. Regarding the definition, I definitely may have misunderstood it, but I don't understand why we want $X\to \bf 1$ to be an epi instead. Could you explain the picture? – 2017-01-17
  • 0
    @Hanno I think you do want a global point for a trivial torsor. Over a base $B$, I'll define torsors are locally trivial torsors in the sense of admitting an effective descent morphism pulling back to a trivial torsor. – 2017-01-17
  • 0
    @Arrow I meant that there's no need for multiple definitions over a point or over a general base. And for a not necessarily trivial torsor over a point the condition should read $X\to 1$ epi instead of the existence of global point. Then the trivial torsors are singled out by the existence of a section/global point. – 2017-01-18
  • 0
    @Arrow: Ok let's clean this up here – 2017-01-18

1 Answers 1

3

I did not check everything properly, so I hope nothing is wrong in the following. I will head back here tonight if time permits to be a little more precise.

Defn 1 $\implies$ Defn 2. Call $f$ the iso $G \to X$; show that indeed $(m,\pi_2)$ is an iso and that there is a global element in $G$ (namely its unit); post/pre-compose by $f$ and $f^{-1}$ to get back the same properties on $X$ and $\varphi$.

Defn 2 $\implies$ Defn 1. Try defining $f : G \to X$ as the composite $$ G \simeq G \times 1 \stackrel{{\rm id}\times x} \to G\times X \stackrel \varphi \to X $$ and show it has the following as inverse: $$ X \simeq X \times 1 \stackrel{{\rm id}\times x} \to X\times X \stackrel {(\varphi,\pi_2)^{-1}} \to G \times X \stackrel {\pi_1} \to G $$

  • 0
    Thanks for the answer! To prove $(m,\pi_2)$ is an iso I thought of just constructing an arrow which divides the left coordinate by the right one, so the composite $(m\times 1)\circ (1\times (-)^{-1}\times 1)\circ (1\times \Delta)$. Does this make sense? – 2017-01-17
  • 0
    @Arrow Yes, that was the map I had in mind. – 2017-01-18