Timeline for Nearby homomorphisms from compact Lie groups are conjugate
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Mar 8, 2013 at 4:12 | comment | added | Peter May | I thought I remembered knowing how to do this, but tracing references led backwards. Some details towards this are in LMS (SLN 1213) p. 247, which refers back to tom Dieck "Transformation groups and representation theory" (SLN 766), section 5.6, which refers back to Bredon "Introduction to compact transformation groups" p. 87 and II.5.6 for proofs of the relevant results of Montgomery and Zippen. I don't have Bredon, but you might try looking there. | |
| Mar 8, 2013 at 2:16 | answer | added | Tom Goodwillie | timeline score: 1 | |
| Mar 6, 2013 at 23:34 | answer | added | Johannes Ebert | timeline score: 2 | |
| Mar 6, 2013 at 5:39 | answer | added | Misha | timeline score: 3 | |
| Mar 6, 2013 at 2:52 | comment | added | Tom Goodwillie | It looks like you need the vanishing of a kind of nonabelian continuous cohomology: If $K$ acts continuously on $G$ then any continuous 1-cocycle $K\to G$ (crossed homomorphism, $f(xy)=f(x)^yf(y)$) that is close enough to the trivial one is a coboundary, i.e. is determined by an element $a\in G$ ($f(x)=a^xa^{−1}$). | |
| Mar 5, 2013 at 21:47 | comment | added | YCor | That $H^1(K,\mathgfrak{g})$ is easy: indeed a continuous 1-cocycle induces an affine continuous action of $K$ on the finite-dimensional vector space $\mathfrak{g}$, which has a fixed point iff the 1-cocycle is a 1-coboundary. Now for $K$ compact there is a fixed point (integrate along an orbit, or fix a $K$-invariant Euclidean metric and take the circumcenter of an orbit). | |
| Mar 5, 2013 at 19:47 | comment | added | Charles Rezk | Claudio: it goes in a nice direction, but I don't see that it gets there. The claim is true for $\mathrm{Hom}(U(1),U(1))$, but false for $\mathrm{Hom}(\mathbb{R}, U(1))$, so I don't see how I can prove it purely from Lie algebra considerations. As Misha suggests, I probably need to know something about $H^1(K,\mathfrak{g})$, not just $H^1(\mathfrak{k},\mathfrak{g})$. What I'm missing is probably really easy. | |
| Mar 5, 2013 at 16:11 | comment | added | Claudio Gorodski | Check out the short survey staff.science.uu.nl/~Schat001/survey_Lie_algebras.pdf I think it might go in the direction you want. | |
| Mar 5, 2013 at 16:11 | comment | added | Misha | You could derive this from the fixed point property for affine actions of compact groups, a la the proof of Property T. The point is that lack of local rigidity for representations of a compact group means that $H^1$ of $K$ with coefficients in Lie algebra is nonzero, which is impossible. Note also that geodesic property you are referring to is just the fact that the distance function on a totally normal neighborhood in Riemannian mld is convex. | |
| Mar 5, 2013 at 15:31 | history | asked | Charles Rezk | CC BY-SA 3.0 |