Let $(M,\omega)$ be a symplectic manifold. One can define its minimal Chern number $N_M$ as: $$ N_M := \text{inf} \lbrace k > 0 \ |\ \exists A \in H_2(M; \mathbb{Z}), \langle c_1, A \rangle = k \rbrace, $$ or alternatively, as the positive generator of $\langle c_1, H_2(M; \mathbb{Z}) \rangle$, where $c_1 \in H^2(M; \mathbb{Z})$ is the first Chern class of the symplectic manifold $(M, \omega)$, and $\langle ., . \rangle$ is the natural pairing between cohomology and homology groups.

On the other hand, one has the cuplength $cl(M)$ of $M$, defined as the minimal positive integer $k$ such that any cup-product $a_1 \cup ... \cup a_k$ of cohomology classes $a_j \in H^*(M; \mathbb{Z})$ of degree greater or equal to $1$ vanishes.

I have seen in several papers (for instance in the introduction of the paper "a fixed point theorem for toric manifolds", by A. Givental), that $$ N_M \leq cl(M), $$ with equality only if $M = \mathbb{C} P^n$ endowed with the Fubini-Study form, in which case both quantities equal $n+1$.

However, I have never seen a proof of this statement. Does someone have an idea of how to prove this fact?

Thank you in advance!

| cite | improve this question | | | | |
  • $\begingroup$ Dear Brian, I am curious about this statement $N_M\le cl(M)$. I looked briefly in the introduction to Givental's paper and I have not found it. Is it really there? Maybe you have a different version of the paper? Also, what are the other papers where such a statement is claimed? $\endgroup$ – Dmitri Panov Dec 22 '19 at 12:41

I had a look at the paper of Givental, https://math.berkeley.edu/~giventh/papers/tor.pdf and don't see this statement... If this statement were true, Conjecture 6.1 of Eliashberg from 2015 would be wrong.

Conjecture, Eliashberg. On any manifold $M$ of dimension $n = 2k > 4$ with a cohomology class $\eta \in H^2(M)$ with $\eta^k\ne 0$ and a non-degenerate 2-form $\omega_0$, there exists a symplectic form $ω$ homotopic to $\omega_0$ through non-degenerate forms, whose cohomology class $[\omega]$ can be deformed to $\eta$ keeping its $k$-th power non-vanishing.

Here is the paper of Eliashberg: https://www.ams.org/journals/bull/2015-52-01/S0273-0979-2014-01470-3/S0273-0979-2014-01470-3.pdf

So, why the inequality $N_M\le cl(M)$ contradicts the conjecture of Eliashberg? This is because on $\mathbb CP^3$ there exist almost complex structures with $c_1$ as large as you want. However $cl(\mathbb CP^3)=4$. Now, for $n>4$ take an almost complex structure $J$ on $\mathbb CP^3$ with $c_1(J)=n\in \mathbb Z=H^2(\mathbb CP^3,\mathbb Z)$, choose a (non-closed) two-form $\omega_0$ defining the same $J$. Applying then the conjecture of Eliashberg, we get $\omega$, contradicting the inequality $N_{\mathbb CP^3}\le 4$.

So, my guess is that the statement $N_M\le cl(M)$ was never proven for the class of all symplectic manifolds (otherwise Eliashberg would not make the conjecture)

However, something analogous is known for smooth complex projective varieties. Namely, if you have a smooth Fano variety $X$ with $c_1$ divisible by $\dim X+1$, then $X$ is $\mathbb CP^n$. And $c_1$ can not be divisible by $\dim X+k$ for $k\ge 2$.

| cite | improve this answer | | | | |
  • $\begingroup$ Dear Dmitri Panov. Please forgive me, my statement was wrong. It seems that this is true when the symplectic manifold is monotone and toric. $\endgroup$ – BrianT Dec 22 '19 at 14:35
  • $\begingroup$ I have looked again at Givental's paper, and you are completely right. The question then is: in the case of a toric manifold, is it always possible to compare the cuplength and the minimal Chern number? What is the cuplength? $\endgroup$ – BrianT Dec 22 '19 at 18:04
  • $\begingroup$ Brian, reading the definition of the cup length that you give, I get that for a symplectic 2n manifold the cup length is $ n+1$, indeed by definition the nth power of symplectic form is non zero. If this is the correct definition, then yes, the only toric variety for which $N_M=cl(M)$ is $CP^n$, otherwise $N_M<cl(M)$ $\endgroup$ – Dmitri Panov Dec 22 '19 at 18:54
  • $\begingroup$ Oh yes, of course. Thank you very much. $\endgroup$ – BrianT Dec 22 '19 at 19:59
  • $\begingroup$ How can one prove that $N_M < cl(M)$ when $M \neq \mathbb{C} P^n$? And isn’t your example a contradiction with what was just said? $\endgroup$ – BrianT Dec 22 '19 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.