Timeline for Is a Lagrangian subgroup of a metric group isomorphic to its quotient?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 28 at 14:04 | history | edited | LSpice | CC BY-SA 4.0 |
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| Jun 28 at 14:00 | history | edited | Andi Bauer | CC BY-SA 4.0 |
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| Jun 28 at 13:59 | comment | added | Andi Bauer | @SeanEberhard Thanks, I had also just noticed that $\rho_a=M(\bullet, a)$ is exactly the $\mathbb{R}/\mathbb{Z}$ representation I was looking for. And yes, I forgot to say $M$ is non-degenerate. | |
| Jun 28 at 13:30 | review | Close votes | |||
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| Jun 28 at 13:11 | comment | added | Sean Eberhard | Each element of $G$ induces a map $H \to \mathbb R / \mathbb Z$, and this map $G \to H^*$ factors through $G/H$ since the bilinear form is zero on $H$. Assuming the bilinear form is nondegenerate, the image of $G/H \to H^*$ separates the points of $H$, so it is a surjective homomorphism and it follows that $G/H$ is canonically isomorphic to $H^*$ since $|G/H| = |H| = |H^*|$. | |
| Jun 28 at 13:08 | comment | added | Sean Eberhard | Do you assume the bilinear form is nondegenerate? If not, take $q = 0$ and you get easy counterexamples. | |
| Jun 28 at 11:38 | history | edited | YCor | CC BY-SA 4.0 |
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| Jun 28 at 10:36 | history | asked | Andi Bauer | CC BY-SA 4.0 |