Timeline for Is an invertible biset necessarily a bitorsor?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 27, 2010 at 4:02 | vote | accept | Evan Jenkins | ||
| Mar 27, 2010 at 4:02 | comment | added | Evan Jenkins | 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! | |
| Mar 26, 2010 at 21:55 | comment | added | Tom Church | 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. | |
| Mar 26, 2010 at 21:50 | comment | added | Evan Jenkins | ...the left and right actions be jointly transitive, which is not a very strong condition. | |
| Mar 26, 2010 at 21:50 | comment | added | Evan Jenkins | 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... | |
| Mar 26, 2010 at 20:54 | history | answered | Tom Church | CC BY-SA 2.5 |