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I am trying to understand the following theorem about symplectomorphisms of projective bundles. Theorem 1.5 of "Characteristic Classes in Symplectic Topology" A.G. Reznikov. Selecta Mathematica, volume 3, pages 601–642(1997).

Theorem: Let $E_i \rightarrow M_i$ be Hermitian vector bundles, $i = 1, 2$. Let $ \mathbb{P}(E_i)$ be the projectivization of $E_i$. Let $f : \mathbb{P}( E_1) → \mathbb{P} (E_2)$ be a fiber-like symplectomorphism, covering a map $\varphi : M_1 \rightarrow M_2$. Then $\varphi^{*}(c_k(E_2)) = c_k(E_1)$ for $k \geq 2$.

Here is where I get confused: taking a simple case where $M_1=M_2=\,\mathbb{CP}^2$. Suppose that $E_{1} = \mathcal{O} \oplus \mathcal{O}$. $E_{2} = \mathcal{O}(1) \oplus \mathcal{O}(1)$. Then, if I am not mistaken, there should be a fibre-wise symplectomorphism $\mathbb{P}(E_1) \rightarrow \mathbb{P}(E_2)$ since they are both isomorphic to the $3$-fold $\mathbb{CP}^2 \times \mathbb{CP}^1$. But there is no diffeomorphism of $\mathbb{CP}^2$ mapping $c_{2}(\mathcal{O} \oplus \mathcal{O}) = 0$ to $c_{2}(\mathcal{O}(1) \oplus \mathcal{O}(1)) = H^2$, where $H$ is the hyperplane class (it is not hard to see that a fibre preserving symplectomorphism must induce a diffeomorphism of the base).

I am thinking possibly that this theorem is for a specific choice of symplectic form?

I would be grateful if somebody could clear up my confusion. I am by no means questioning this theorem, I just want to understand the statement correctly. thanks.

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    $\begingroup$ Read carefully the statement of the Theorem. The symplectomorphism is between $\mathbb{E}_1$ and $\mathbb{E}_2$, not $\mathbb{P}(\mathbb{E_1})$ and $\mathbb{P}(\mathbb{E_2})$. $\endgroup$
    – abx
    Commented Mar 9, 2021 at 4:50
  • $\begingroup$ Sorry, I misquoted the theorem. The symplectomorphism in the paper is between the projective bundles. Edited. $\endgroup$
    – Nick L
    Commented Mar 9, 2021 at 5:10

1 Answer 1

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I think you spotted an imprecision in Reznikov's paper. Clearly, the statement has a problem because it is not robust under tensoring with a line bundle.

It seems that by "$c_k(E_i)$ for $k\geq 2$" Reznikov really means the characteristic classes of the $PU(n)$ principal bundle obtained by quotienting the $U(n)$ bundle underlying $E_i$ by its center $U(1)$. This is the algebra generated by classes $\tilde{c}_k, 2\leq k \leq n$ of degree $2k$ which are sort of ``projections of $c_k$ orthogonally to $\mathbb Z[c_1]$''.

More concretely, I think the theorem works if you replace $c_k$ by $\tilde c_k$, where $$1 + \sum_{k=2}^n \tilde c_k(E) t^k = \left [\mathrm{det}\left( \mathrm I_n + \frac{i t}{2\pi} \left( \Omega - \frac{1}{n} \mathrm{Trace}(\Omega) \mathrm{I_n}\right) \right)\right]~,$$ where $\Omega$ is the curvature form of some Hermitian connection.

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