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Jul 6, 2023 at 21:15 comment added Tom Goodwillie How do you rule out the possibility of there being no non-zero lattice points at all in the interior of $K$? I think an additional argument is needed. For example by perturbing the norm slightly without introducing more lattice points to $K$, you can arrange for one of the boundary lattice points to be now in the interior, in such a way that the new norm never increases and therefore the new dual norm never decreases and therefore $\lambda_1$ does not decrease.
Jun 27, 2023 at 13:58 comment added Mikhail Katz I sent you Averkov's explanation via email.
Jun 27, 2023 at 13:50 comment added David E Speyer Okay, I guess I'm wrong, I'm not claiming to have a proof for the $\lambda_1(L) \lambda_1(L^{\ast})$ optimum. I'd be curious to see the example that achieves $\sqrt{2}$.
Jun 27, 2023 at 13:39 comment added Mikhail Katz Wait a minute, but Mahler seems to say otherwise: $\lambda_1(L)\lambda_1(L^\ast)$ reaches $\sqrt2$. A colleague sent me an explanation of how this follows from Mahler's '48 paper.
Jun 27, 2023 at 13:35 comment added David E Speyer For this particular lattice, $\lambda_1(L) = 2/3$ (because shrinking $P$ by $2/3$ still includes $\pm (1,0)$, so $\lambda_1(L) \lambda_1(L^{\ast}) = 1$. I suspect that the optimal value of $\lambda_1(L) \lambda_1(L^{\ast})$ is $4/3$, using the hexagon I posted as a comment earlier.
Jun 27, 2023 at 8:31 comment added Mikhail Katz Awsome! Here your convention is to work with $\lambda_2(L)\lambda_1(L^\ast)$. How big is the difference between $\lambda_1(L)$ and $\lambda_2(L)$ for this extremal lattice?
Jun 26, 2023 at 14:50 vote accept Mikhail Katz
Jun 26, 2023 at 14:50 history bounty ended Mikhail Katz
Jun 25, 2023 at 8:38 history edited Mikhail Katz CC BY-SA 4.0
typo
Jun 24, 2023 at 16:18 history edited David E Speyer CC BY-SA 4.0
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Jun 24, 2023 at 15:27 history answered David E Speyer CC BY-SA 4.0