TITLE: Is a cubic hypersurface determined by its Fano variety of lines?
QUESTION [15 upvotes]: Consider a smooth cubic complex hypersurface $X\subset\mathbf{P}^{n+1}$ of dimension $n\geqslant 3$. The associated Fano variety of lines $F(X)$ is a smooth variety of dimension $2n-4$. Can one recover $X$ from $F(X)$? The answer is positive for $n=3$ (due to Clemens-Griffiths, Tjurin) and $n=4$ (due to Voisin, I think). Are there any conjectures for general $n$?

REPLY [5 votes]: For what it's worth, I think I now know a complete proof. I learned of this proof from D. Huybrechts's notes in progress http://www.math.uni-bonn.de/people/huybrech/Notes.pdf, proposition 6.21.
Theorem. Let $X$ be a cubic hypersurface of dimension $d\geqslant 3$, with $d\neq 4$. Then $X$ can be recovered from its Fano variety of lines $F(X)$.
Notations: $F=F(X)$, $G=\mathrm{Gr}(2,d+2)$, $\mathscr{O}_F(1)$ and $\mathscr{O}_G(1)$ the Plücker polarisations of $F$ and $G$, respectively.
For future reference, let me state the following results.
Lemma 1. The canonical map $\mathrm{H}^0(G,\mathscr{O}_G(2))\rightarrow\mathrm{H}^0(F,\mathscr{O}_F(2))$ is injective.
Proof. See J. Starr's answer or Altman-Kleiman, Foundations of the theory of Fano schemes, proposition 1.15 (ii).
Lemma 2. $\omega_{F}=\mathscr{O}_{F}(4-d)$.
Proof. See Altman-Kleiman, proposition 1.8.
Lemma 3. If $V$ is a variety such that $\omega_{V}^{\vee}$ is ample, then $\mathrm{Pic}(V)$ is torsion-free.
Proof. Let $L$ be a torsion line bundle on $V$. Then $\mathrm{H}^i(V,L)=0=\mathrm{H}^i(V,\mathscr{O}_V)$ for $i\geqslant 1$ by Kodaira vanishing. As $\chi(L)=\chi(\mathscr{O}_V)$ by Riemann-Roch, we get $\mathrm{H}^0(V,L)=\mathrm{H}^0(V,\mathscr{O}_V)$, 
which forces $L$ to be trivial.
Proof of the Theorem. Let $X$ and $X'$ be cubic hypersurfaces and $f:F\xrightarrow{\sim} F'$ an isomorphism between their Fano varieties of lines. By lemma 2 we have the equality
$$(4-d)f^\ast(\mathscr{O}_{F'}(1))=f^\ast(\omega_{F'})=\omega_F=(4-d)\mathscr{O}_{F}(1)$$
in $\mathrm{Pic}(F)$. For $d\geqslant 5$ lemma 2 shows that we can use lemma 3 to deduce 
$$f^\ast(\mathscr{O}_{F'}(1))=\mathscr{O}_{F}(1).$$
For $d=3$ this is clear. The theorem then follows from proposition 4 of https://arxiv.org/abs/1209.4509. For completeness, I repeat the proof. Let $\mathbf{P}^{d+1}=\mathbf{P}(V)$ be the ambient projective space containing $X$ and $X'$. By lemma 1,  if $Q$ is a quadric in $\mathbf{P}(\wedge^2 V)$ containing $F$, then $Q$ also contains $G$. In $\mathbf{P}(\wedge^2 V)$ the Grassmannian $G$ is cut out by quadrics, so in particular $G$ is the intersection of quadrics containing $F$. Hence there is an automorphism $g$ of $G$ taking $F$ to $F'$ via $f$, and $g$ also preserves the Plücker polarisation. By a theorem of Chow on automophisms of Grassmannians (Ann. Math. 2), $g$ is induced by an automorphism $h$ of $\mathbf{P}(V)$. The latter automorphism then induces a (polarisation-preserving) isomorphism $X\xrightarrow{\sim} X'$. (If $L$ is a line on $X$ corresponding to $[L]\in F$, then $h(L)$ is a line on $X'$, and $[h(L)]=f([L])$. Of course $X$ is covered by lines.)