Timeline for Orbits of some action of SL2 on Pontryagin dual of the field of formal Laurent series
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2015 at 22:37 | comment | added | YCor | I see, so the action on the dual should be computed by hand. | |
| Dec 5, 2015 at 14:28 | comment | added | m07kl | Maybe the following fact is not true on K-rational points? If a unipotent group acts on an affine variety, all its orbits are closed. BTW, the action of SL(2,K) on K^3 is not given by K-linear endomorphism, because in general $\lambda^2 \neq \lambda$ for $\lambda \in K$. | |
| Dec 4, 2015 at 17:14 | comment | added | YCor | No it's not true in general that orbits are locally closed in the Zariski topology. It's true for intersections of the orbits of the group over an algebraic closure with $K^3$, but in general the orbits can be smaller. I'd expect anyway that orbits are easy to describe and that most of them are closed and remaining ones are singletons or complements of singletons in closed subsets, but since the characteristic is 2 I'm not sure. | |
| Dec 4, 2015 at 17:04 | comment | added | m07kl | Thanks. So we are done, because the algebraic group SL(2,K) acts algebraically on the affine varety K^3. So all orbits are locally closed in Zariski topology, hence in particualr, in locally compact topology. | |
| Dec 4, 2015 at 16:04 | comment | added | YCor | Yes it's topological, but I already wrote it. | |
| Dec 4, 2015 at 16:02 | comment | added | m07kl | Dear YCor: Thanks very much for your answer. Is the identification of its Pontryagin dual orbit with its ordinary orbits topological? because I would like to know whether all orbits are locally closed. | |
| Dec 4, 2015 at 14:46 | comment | added | YCor | (...) by restriction we have a right group action of $\mathrm{GL}(V)$ (group of $K$-linear isomorphism, which composed with the inversion map yields a left action. So if $H$ acts on $K^d$ through a map $j:H\to\mathrm{GL}_d(K)$, then the Pontryagin dual of $K^d$ is $H$-equivariantly isomorphic to $K^d$ with the action of $H$ given by $h\mapsto (j(h)^t)^{-1}$, if I'm correct. In particular when $H\subset\mathrm{GL}_d(K)$ is stable under transposition, then its Pontryagin dual orbit can be identified to its ordinary orbits. | |
| Dec 4, 2015 at 14:44 | comment | added | YCor | If $V$ is a finite-dimensional $K$-vector space, then $\hat{V}$ is isomorphic (not canonically) to $V$ as a topological group. View $V$ as an $R$-module, where $R$ is the ring of $K$-linear endomorphism. Then $V$ is naturally a left $R$-module, which makes $\hat{V}$ naturally a (continuous) right-module, which is the commutant in the full endomorphism group (as topological group) of the multiplication by $t$. With a choice of basis, the action is just given by transpose matrices. (...) | |
| Dec 4, 2015 at 13:00 | history | edited | m07kl | CC BY-SA 3.0 |
added 136 characters in body
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| Dec 4, 2015 at 12:51 | history | asked | m07kl | CC BY-SA 3.0 |