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Question 1. Can a reflexive Delzant polytope of some dimension contain a $2$-face with more than $11$ edges?

Motivation. I would like more generally to get an answer to the following question:

Question 2. Suppose $X$ is a smooth Fano variety with a $\mathbb C^*$-action. Let $Y\subset X$ be the connected component of $X^{\mathbb C^*}$ on which all the weights of the action are positive. Is it true that variety $Y$ can be deformed to a Fano variety?

A positive answer to the first question gives a negative to the second one.

Explanation. Indeed, a reflexive Delzant polytope by definition corresponds to a toric Fano. For any face of a Delzant polytope we can find a rational linear support function. This will give us a $\mathbb C^*$ action as in Question 2. A surface that deforms to a Fano surface has $b_2\le 9$. Finally, a toric surface corresponding to a polygon with $n$ sides has $b_2=n-2$.

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  • $\begingroup$ May I suggest that you remove the term "Obviously" ? There is nothing obvious here (and few obvious things in mathematics in general). $\endgroup$
    – Nicolast
    Commented Oct 16, 2022 at 17:51
  • $\begingroup$ Thanks for your comment Nicolas. Indeed, I wrote "Obviously" without thinking twice. Question 2 was the reason for Question 1, I was thinking of how to get a counter-example. Anyway, I replaced "Obviously" by an explanation $\endgroup$
    – aglearner
    Commented Oct 23, 2022 at 11:28

1 Answer 1

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Haase and Melnikov have proved here that every lattice polytope can be realized as a face of some reflexive polytope. They do this by an iterative procedure that increases the dimension by one until the polytope becomes reflexive. At each step the new polytope is a wedge over a facet of the previous one, and one can show that this operation takes Delzant polytopes to Delzant polytopes. So if you start with a smooth polygon with more than 11 edges will produce a positive answer to your first question.

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