Timeline for Moduli space of flat connection over homology 3-sphere
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 15, 2022 at 9:39 | comment | added | Steven Sivek | I don't claim that one always exists, but that this construction applies when it does. E.g. if you splice together two copies of the same knot complement $E_K$ in this way, then it suffices to find an irreducible representation $\rho: \pi_1(E_K) \to SU(2)$ with $\rho(\mu)=\rho(\lambda)$ and use this $\rho$ on each copy of $E_K$. But this is the same as an irreducible representation of the (-1)-surgery group $\pi_1(S^3_{-1}(K)) \cong \pi_1(E_K)/\langle\mu\lambda^{-1}\rangle$, and Kronheimer and Mrowka's proof of property P showed that for $\pm1$-surgery, such representations always exist. | |
| Apr 15, 2022 at 9:22 | comment | added | Nicolast | Nice! One needs to be a bit careful however : not every representation of $pi_1(T)$ extends to $pi_1(Y_i)$ and we need a representation that extends on both sides. Do you have an argument for that? | |
| Apr 15, 2022 at 9:06 | comment | added | Steven Sivek | Rigidity can fail even for an irreducible homology sphere $Y$. Say $Y$ splits along an incompressible torus $T$ as $Y=Y_1 \cup_T Y_2$ (example: glue two nontrivial knot complements, meridian to longitude and vice versa), and $\rho:\pi_1(Y)\to SU(2)$ is irreducible on each $Y_i$ separately. Then $\rho$ is reducible on $T$, with image in a $U(1)$ subgroup, and you can "bend" it by the same trick as for connected sums: keep $\rho|_{\pi_1(Y_1)}$ fixed and replace $\rho|_{\pi_1(Y_2)}$ with $g\rho g^{-1}$ for elements $g$ in that subgroup. This gives a $U(1)$ family of non-conjugate representations. | |
| Apr 15, 2022 at 7:16 | history | edited | Nicolast | CC BY-SA 4.0 |
added 2 characters in body
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| Apr 15, 2022 at 7:16 | comment | added | Nicolast | For SU(2)-connections, it does (the orthogonal of an invariant subundle is invariant) | |
| Apr 15, 2022 at 7:01 | comment | added | ThiKu | Why can a reducible connection not take values (up to conjugation) in the upper triangle matrices? Reducible need not be completely reducible. | |
| Apr 15, 2022 at 6:53 | history | answered | Nicolast | CC BY-SA 4.0 |