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Jun 5 at 15:08 answer added Nick L timeline score: 2
Jun 4 at 19:56 comment added Nick L Ok. I will write the argument. Sorry, it does not give the associated polytopes.
Jun 4 at 19:18 comment added Yromed @NickL Yes, I would be interested in seeing such a proof, especially if one can work out the associated polytopes in a systematic way.
Jun 4 at 19:05 comment added Nick L I can prove that a smooth complete intersection in $\prod \mathbb{P}^{a_i}$ of multidegree $(1,\ldots,1)$ is toric only if all but one of the $a_i$ is one. It follows pretty simply from combining some known results. Is it interesting for you?
Dec 16, 2023 at 17:49 comment added Yromed Thank you everyone for your comments !
Dec 16, 2023 at 17:48 vote accept Yromed
Dec 15, 2023 at 16:29 answer added Igor Makhlin timeline score: 3
Dec 10, 2023 at 21:37 comment added Igor Makhlin @Sasha Or, in other words, the toric hypersurface of degree $(1,1)$ in $\mathbb P^2\times\mathbb P^2$ is not smooth =) It is the zero set of $x_1y_1-x_2y_2$ which has 7 torus-fixed points of which one is singular. It is indeed given by the Minkowski sum of two triangles.
Dec 6, 2023 at 15:18 comment added Sasha A smooth hypersurface of degree $(1,1)$ in $\mathbb{P}^2 \times \mathbb{P}^2$ is not toric.
Dec 6, 2023 at 15:15 comment added pinaki At the first glance it looks you might want to consider the sums, not unions, of polytopes: the toric variety of the sum of two polytopes is isomorphic to the closure of the "diagonal image" of the torus in the product of the toric varieties of the original polytopes. It is easy to show; I worked out the details in Proposition VI.6 in "How many zeroes?" (arxiv.org/abs/1806.05346).
Dec 6, 2023 at 15:02 history asked Yromed CC BY-SA 4.0