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Let $X$ and $Y$ be two Fano varieties of the same dimension embedded into a same projective space $\mathbb P^N$, assume $Pic X= \mathbb Z\mathcal O_X(1)$ and $Pic Y=\mathbb Z\mathcal O_Y(1)$, where $\mathcal O_X(1)$ means the restriction of $\mathcal O_{\mathbb P^N}(1)$ on $X$.

Then does it hold that $X$ and $Y$ are in fact isomorphic?

Such kinds of problems seem to have been fully understood due to its easy statement, but I do not have an idea on it.

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    $\begingroup$ What exactly is the question here? $\endgroup$
    – pbelmans
    Commented May 24, 2018 at 6:51
  • $\begingroup$ I’m sorry for my carelessness, I post the question now. Thank you! $\endgroup$ Commented May 24, 2018 at 6:56
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    $\begingroup$ Maybe you also want to assume that X and Y have equal index? Otherwise considering hypersurfaces of degree < 5 in P^4 gives counterexamples. $\endgroup$ Commented May 24, 2018 at 7:07

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This is definitely not true; just consider two smooth cubic threefolds in $\mathbb{P}^4$.

Of course many other examples exist, e.g. Fano hypersurfaces in projective space which are not quadrics nor cubic surfaces (here the Picard group is generated by the hyperplane class by the Lefschetz hyperplane section theorem). Also many classes of Fano threefolds of Picard number one.

If what you wanted was true, most of the interesting moduli spaces for Fano varieties would in fact be trivial.

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Even simpler example: Grassmannian Gr(3,6) and $X=Gr(2,7) \cap H$, an hyperplane section of Grassmannian Gr(2,7). They are prime (by Lefschetz), have the same dimension (9), lies in the same space $\mathbb{P}(\bigwedge^3V_6) =\mathbb{P}(\bigwedge^2V_7) \cap H$, they have the same index (6) and the same degree w.r.t. the Pluecker embedding (42). However they are not even diffeomorphic (for example they have different Betti numbers, starting from $b_6$), let alone isomorphic!

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    $\begingroup$ And they are both locally rigid! It's really a good example! $\endgroup$ Commented May 25, 2018 at 10:46

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