Timeline for Can a manifold have a curvature-free connection that is not torsion-free?
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| Apr 17, 2020 at 8:12 | comment | added | Robert Bryant | @EthanDlugie: No. The assumptions you need for global parallel coframe are flatness and simple-connectivity. The closedness of the $\theta_i$ is then equivalent to the torsion-free condition: If $X_i$ are the vector fields dual to the $\theta_i$, then they will parallel as well, so $T(X_i,X_j) = -[X_i,X_j]$. Torsion-free is then $[X_i,X_j]=0$, and then the definition of exterior derivative says $\mathrm{d}\theta_i = 0$. For the converse, read backwards. | |
| Apr 17, 2020 at 2:46 | comment | added | Ethan Dlugie | Thanks for answering Robert. Two questions still: is torsion-free necessary to find the global parallel coframe? I would think that's just a consequence of flatness. And how to see closedness of the $\theta_i$? | |
| Apr 16, 2020 at 18:08 | comment | added | Robert Bryant | @EthanDlugie: Yes. The point is that, if $S^3$ admitted a torsion-free flat connection, then, because it is simply connected, it would have a global parallel coframe field, $(\theta_1,\theta_2,\theta_3)$ and the parallel condition together with the torsion-free condition would imply that the $1$-forms $\theta_i$ are closed. Since $S^3$ is simply-connected, these would have to be exact, i.e., $\theta_i = \mathrm{d}x_i$ for some functions $x_i$ on $S^3$. But these functions could not have any critical points because their differentials never vanish. Since $S^3$ is compact, this is not possible. | |
| Apr 16, 2020 at 16:31 | comment | added | Ethan Dlugie | Can someone explain this last step, the contradiction about a connection on $S^3$? | |
| Nov 21, 2013 at 21:17 | vote | accept | pavpanchekha | ||
| Nov 21, 2013 at 4:02 | history | answered | Robert Bryant | CC BY-SA 3.0 |