Timeline for The gauge group versus the diffeomorphism group of a manifold
Current License: CC BY-SA 4.0
12 events
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| Nov 2, 2023 at 6:43 | comment | added | Ryan Budney | If you look at the subgroup of $Diff(M)$ that are the diffeomorphisms which are trivial as homotopy-equivalences, then you do get a map to the Gauge group. There's a few cases where it's known there are such diffeomorphisms which produce non-trivial Gauge automorphisms. For example, with diffeomorphisms of the sphere. There is a recent paper by Crowlet, Schick and Steimle to appear in Geometry and Topology. | |
| S Nov 1, 2023 at 19:13 | history | suggested | CommunityBot | CC BY-SA 4.0 |
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| Nov 1, 2023 at 18:26 | review | Suggested edits | |||
| S Nov 1, 2023 at 19:13 | |||||
| Dec 11, 2014 at 21:41 | answer | added | Peter Michor | timeline score: 4 | |
| Sep 20, 2010 at 18:09 | comment | added | Theo Johnson-Freyd | There is a group of "infinitesimal diffeomorphisms"; elements are formal power series starting in degree 1 in a parameter \epsilon with coefficients in Vect(M) (the vector fields on M, i.e. sections of TM). The multiplication is given by the Baker-Campbell-Hausdorff formula. But this group also doesn't map into the group of (formal power series of) sections of GL(TM). The problem is that two vector fields might commute, but if you act on both of them by the same section of your gauge group, then they might not commute anymore. | |
| Sep 20, 2010 at 18:03 | comment | added | Theo Johnson-Freyd | Repeating what Andrew Stacey and Lucas Culler have said in more physics-y language: the Jacobi matrix does not transform as a tensor. So it does not define a section of GL(TM). As a trivial example, let M be the disjoint union of two lines. Pick a coordinate x on one of the lines and a coordinate y on the other one. Then there is a diffeomorphism of the form y(x) = x, x(y) = y. The Jacobi matrix near x=0 is 1 in these coordinates. But under the change of coordinates Y = Y(y), which does not change the x coordinates at all, the Jacobi matrix near x=0 changes to Y'(x). | |
| Sep 20, 2010 at 17:27 | comment | added | Michael Bächtold | Even if you work in coordinates, as you do, observe that your map which associates to a diffeomorphism a gauge transformation is not injective. For example the identity and the shift $x\mapsto x+1$ on $\mathbb{R}$ induce the same gauge transformation. | |
| Sep 20, 2010 at 13:48 | comment | added | E von Tuzzenthaler | The Diff(M) group can be viewed either in an "active" way carrying point x to y or in a passive way changing the coordinate system about a point (the group of coordinate transformations). I use this second picture. | |
| Sep 20, 2010 at 12:10 | history | edited | E von Tuzzenthaler | CC BY-SA 2.5 |
motivation added
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| Sep 20, 2010 at 12:09 | comment | added | Andrew Stacey | Something seems a little odd about your map from Diff(M) to Gauge(M). An element of Diff(M) defines an isomorphism T_xM -> T_yM (where x -> y) but an element of Gauge(M) can only define an isomorphism T_xM -> T_xM. | |
| Sep 20, 2010 at 12:05 | comment | added | Lucas Culler | The diffeomorphism group is not a subgroup of the gauge group, because a diffeomorphism f induces maps $T_x M \to T_{f(x)} M$, rather than from $T_x M$ to itself. In other words, Df is not a map of bundles over $X$. | |
| Sep 20, 2010 at 11:14 | history | asked | E von Tuzzenthaler | CC BY-SA 2.5 |