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1. • CommentRowNumber1.
   • CommentAuthorziggurism
   • CommentTimeMar 6th 2020
   • PermaLink
   Author: ziggurism
   Format: MarkdownItexA torsor or principal bundle may be characterized as a group action $\rho: P\times G\to P$ such that the associated map $1\times\rho:P\times G \to P\times P$ is an isomorphism (in whatever category, often a comma category $C/X$ especially when using bundle language) A groupoid may be defined as two objects $G_1$ (arrows) and $G_0$ (points) and two maps $s,t\colon G_1\to G_0$, the source and target morphisms (along with some other data and axioms). These two morphisms may be assembled into a product $(s,t)\colon G_1\to G_0\times G_0.$ If the groupoid is discrete (only arrows are identity arrows) then $s$ and $t$ are both the identity morphism and $s\times t$ is the diagonal map. If the groupoid is codiscrete (aka pair groupoid, every pair of points has a unique arrow), then $s$ and $t$ are projections and $s\times t$ is equality. If you consider the action groupoid of the torsor $\rho: P\times G\to P$, and assemble its source and target morphisms into a product as above, you recover the map $1\times\rho:P\times G \to P\times P$ which is stipulated to be an isomorphism. So a codiscrete groupoid has a $s\times t$ map which is equality, whereas a torsor has a $s\times t$ map which is an isomorphism. So would it be correct to characterize a torsor as an "essentially codiscrete" groupoid? As in, it's codiscrete only up to isomorphism, instead of equality? Is there a bigger context to this $P\times G \to P\times P$ map associated to a group action? Is it just a coincidence that it plays an important role both in the definition of the torsor, and the action groupoid?

   A torsor or principal bundle may be characterized as a group action $\rho: P\times G\to P$ such that the associated map $1\times\rho:P\times G \to P\times P$ is an isomorphism (in whatever category, often a comma category $C/X$ especially when using bundle language)

   A groupoid may be defined as two objects $G_1$ (arrows) and $G_0$ (points) and two maps $s,t\colon G_1\to G_0$, the source and target morphisms (along with some other data and axioms). These two morphisms may be assembled into a product $(s,t)\colon G_1\to G_0\times G_0.$ If the groupoid is discrete (only arrows are identity arrows) then $s$ and $t$ are both the identity morphism and $s\times t$ is the diagonal map. If the groupoid is codiscrete (aka pair groupoid, every pair of points has a unique arrow), then $s$ and $t$ are projections and $s\times t$ is equality.

   If you consider the action groupoid of the torsor $\rho: P\times G\to P$, and assemble its source and target morphisms into a product as above, you recover the map $1\times\rho:P\times G \to P\times P$ which is stipulated to be an isomorphism. So a codiscrete groupoid has a $s\times t$ map which is equality, whereas a torsor has a $s\times t$ map which is an isomorphism.

   So would it be correct to characterize a torsor as an “essentially codiscrete” groupoid? As in, it’s codiscrete only up to isomorphism, instead of equality?

   Is there a bigger context to this $P\times G \to P\times P$ map associated to a group action? Is it just a coincidence that it plays an important role both in the definition of the torsor, and the action groupoid?