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Let $X \subset \mathbb P(\mathfrak g)$ be an adjoint variety for a simple complex Lie algebra $\mathfrak g$ appearing in the last line of the Freudenthal Magic Square, that is $\mathfrak g \in \{F_4,E_6,E_7,E_8\}$. In particular $\dim X=6m+9$ where $m \in \{1,2,4,8\}$. Consider minimal embedding of the VMRT of $X$ at a point $x$, that is $$ Y^{3m+3} \subset \mathbb P^{6m+7} \subset \mathbb P^{6m+8}=\mathbb P(T_{X,x}). $$ On the other hand we have the contact structure $0 \to F \to T_X \to L \to 0$, and the restriction of such a sequence to $x$ allow us to write $Y \subset \mathbb P(F_x)$.

Now, $F_x$ is endowed with a symplectic form $\omega$ that is the restriction to $F_x$ of the map $\omega=d\theta: F \times F \to L$. In particular $\mathbb P(F_x)$ has a symplectic structure. We blow-up $\mathbb P(F_x)$ along $Y$, obtaining a map $$ b: Bl_Y \mathbb P(F_x) \longrightarrow \mathbb P(F_x). $$ Question: do the map $b$ preserve the symplectic structure? More precisely, can we define a symplectic structure on $Bl_Y \mathbb P(F_x)$?

Any hint of a possible argument for the proof?

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    $\begingroup$ Unless I am missing something, a blow up never has a holomorphic symplectic structure. The canonical bundle of a holomorphic symplectic manifold is trivial. The restriction of the canonical bundle to the exceptional fibre is non-trivial, which may be seen by computing $c_1$ of it. $\endgroup$
    – Nick L
    Commented Nov 8, 2021 at 10:17
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    $\begingroup$ A projective space dos not have a holomorphic symplectic structure, since its canonical bundle is non-trivial, it has a real symplectic structure ( as any Kahler manifold does), but I assumed since you used the "holomorphic symplectic" tag that is what you meant? $\endgroup$
    – Nick L
    Commented Nov 8, 2021 at 10:22
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    $\begingroup$ Ok, are you asking about the existence of a real symplectic structure or a holomorphic symplectic structure? $\endgroup$
    – Nick L
    Commented Nov 8, 2021 at 10:26
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    $\begingroup$ No, I'm asking about the existence of a complex symplectic structure $\endgroup$
    – Bobech
    Commented Nov 8, 2021 at 10:27
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    $\begingroup$ Again: $\Bbb{P}(F_x)$ has no (complex) symplectic structure. $\endgroup$
    – abx
    Commented Nov 8, 2021 at 10:47

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