# Comptes Rendus
Mathématique

Yunhyung Cho

Remark on the Betti numbers for Hamiltonian circle actions

Volume 359, issue 2 (2021), p. 113-117

Published online: 17 March 2021

https://doi.org/10.5802/crmath.127

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Symplectic geometry / Géométrie symplectique

# Remark on the Betti numbers for Hamiltonian circle actions

Yunhyung Cho<sup>a</sup>

<sup>a</sup> Department of Mathematics Education, Sungkyunkwan University, Seoul, Republic of Korea.

E-mail: yunhyung@skku.edu

**Abstract.** In this paper, we establish a certain inequality in terms of Betti numbers of a closed Hamiltonian $S^1$-manifold with isolated fixed points.

**Résumé.** Dans cet article, nous établissons une certaine inégalité en termes de nombres de Betti d'une $S^1$-variété hamiltonienne avec des points fixes isolés.

**2020 Mathematics Subject Classification.** 53D20, 53D05.

**Funding.** This work is supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP; Ministry of Science, ICT & Future Planning) (NRF-2020R1C1C1A01010972).

Manuscript received 12 March 2020, revised 2 September 2020 and 5 October 2020, accepted 2 October 2020.

## 1. Introduction

Let $(M, \omega)$ be a $2n$-dimensional closed symplectic manifold admitting a Hamiltonian torus action with only isolated fixed points. It has been a long-standing open problem whether $M$ admits a Kähler metric or not. Historically, Delzant [9] proved that if $M$ admits a Hamiltonian $T^n$-action, where the fixed point set is automatically discrete, then $M$ admits a $T^n$-invariant Kähler metric. Restricting to an $S^1$-action case, several results on the existence of a Kähler metric were provided in some special cases. For instance, Karshon [13] proved that every closed symplectic four manifold admitting a Hamiltonian circle action admits a Kähler metric. (In fact, the $S^1$-action is induced from a toric action when the fixed points are isolated.) Also if $\dim M = 6$ with $b_2(M) = 1$, then it turned out that $M$ admits a Kähler metric, which was proved by Tolman [18] and McDuff [15]. Recently, the author has shown that any 6-dimensional monotone closed semifree Hamiltonian $S^1$-manifold admits a Kähler metric, see [5-7].

As a counterpart, there were “candidates” of closed Hamiltonian $T$-manifolds (with isolated fixed points) which possibly fail to admit Kähler metrics. Tolman [17] and Woodward [19] constructed a six-dimensional closed Hamiltonian $T^2$-manifold with only isolated fixed points and with no $T^2$-invariant Kähler metric. Surprisingly Goertsches-Kostantis-Zoller [10] have recently shown that the examples of Tolman and Woodward indeed admit Kähler metrics that are not
$T^2$-invariant. Thus their result provides a positive evidence for the conjecture of the existence of Kähler metrics.

On the other hand, it seems reasonable to ask whether $(M, \omega)$ enjoys Kählerian properties, such as the hard Lefschetz property of the symplectic form $\omega$ or the unimodality of even Betti numbers. Recall that every closed Kähler manifold $(M, \omega, J)$ satisfies the *hard Lefschetz property*, that is,

$$[\omega]^{n-k} : H^k(M; \mathbb{R}) \to H^{2n-k}(M; \mathbb{R}) \\ \alpha \mapsto \alpha \cup [\omega]^{n-k}$$

is an isomorphism for every $k = 0, 1, \dots, n$. This implies that

$$[\omega] : H^k(M; \mathbb{R}) \to H^{k+2}(M; \mathbb{R})$$

is injective for every $k$ with $0 \le k < n$, and therefore the sequence of even (as well as odd) Betti numbers of $M$ is unimodal. In other words,

$$b_k \le b_{k+2}, \quad k = 0, 1, \dots, n-1$$

where $b_i$ denotes the $i$-th Betti number of $M$. In this paper we deal with the following conjecture.

**Conjecture 1 ([12]).** Let $(M, \omega)$ be a $2n$-dimensional closed symplectic manifold equipped with a Hamiltonian $S^1$-action with only isolated fixed points. Then the sequence of even Betti numbers is unimodal, i.e.,

$$b_{2i} \le b_{2i+2} \quad \text{for every} \quad 0 \le i < \lfloor \frac{n}{2} \rfloor.$$

It is worth mentioning that every odd Betti number of $M$ vanishes by Frankel's theorem which states that a moment map is a Morse function whose critical points are of even indices. (See [2, Theorem IV.2.3].) Therefore we only need to care about even Betti numbers of $M$.

In [8], the author and Kim proved Conjecture 1 when $\dim M = 8$. The main goal of this article is to improve the result of [8] and prove the following inequality, which is automatically satisfied when Conjecture 1 is true.

**Theorem 2.** Let $(M, \omega)$ be a closed symplectic manifold admitting a Hamiltonian circle action with only isolated fixed points where $\dim M = 8n$ or $8n+4$. Then

$$b_2 + \dots + b_{2+4(n-1)} \le b_4 + \dots + b_{4+4(n-1)}.$$

In particular when $\dim M = 8$ or $12$, we have

$$b_2 \le b_4.$$

## 2. Proof of the main Theorem 2

The main technique for proving Theorem 2 is the ABBV-localization due to Atiyah-Bott and Berline-Vergne. Recall that for an $S^1$-manifold $M$, the *equivariant cohomology* is defined by $H_{S^1}^*(M) := H^*(M \times_{S^1} ES^1)$ where $ES^1$ is a contractible space on which $S^1$ acts freely. Then $H_{S^1}^*(M)$ inherits an $H^*(BS^1)$-module structure induced from the projection

$$\pi : M \times_{S^1} ES^1 \to BS^1 := ES^1/S^1.$$

Note that $H^*(BS^1; \mathbb{R}) \cong H^*(CP^\infty; \mathbb{R}) = \mathbb{R}[u]$. Moreover, for the inclusion map $i : M^{S^1} \hookrightarrow M$, we have an induced ring homomorphism

$$i^* : H_{S^1}^*(M; \mathbb{R}) \to H_{S^1}^*(M^{S^1}; \mathbb{R}) \cong H^*(BS^1; \mathbb{R}) \otimes H^*(M^{S^1}; \mathbb{R}).$$
When $M^{S^1} = \{p_1, \cdots, p_m\}$ is discrete, we may express as

$$H^*(BS^1; \mathbb{R}) \otimes H^*(M^{S^1}; \mathbb{R}) \cong \bigoplus_{i=1}^{m} H^*(BS^1; \mathbb{R})$$

and so

$$i^*(\alpha) = (f_1, \cdots, f_m), \quad f_i \in \mathbb{R}[u]$$

for $\alpha \in H_{S^1}^*(M; \mathbb{R})$. We denote by $\alpha|_{p_i} := f_i$ and call it the restriction of $\alpha$ to $p_i$. By the Kirwan's injectivity theorem [14], the map $i^*$ is injective and hence $H_{S^1}^*(M; \mathbb{R})$ is a free $H^*(BS^1; \mathbb{R})$-module.

**Theorem 3 (ABBV Localization Theorem [1,3]).** Let $M$ be a closed $S^1$-manifold with only isolated fixed points and $\alpha \in H_{S^1}^*(M; \mathbb{R})$. Then we have

$$\int_M \alpha = \sum_{p \in M^{S^1}} \frac{\alpha|_p}{(\prod_{i=1}^n w_i(p)) u^n}.$$

where $w_1(p), \cdots, w_n(p)$ denote the weights of the tangential $S^1$-representation at $p$.

To obtain Theorem 2, we will apply Theorem 3 to canonical classes which form a basis of $H_{S^1}^*(M; \mathbb{R})$ as an $H^*(BS^1; \mathbb{R})$-module.

**Theorem 4 (16, Lemma 1.13)<fnref n="1" />).** Let $(M, \omega)$ be a $2n$-dimensional closed Hamiltonian $S^1$-manifold with only isolated fixed points. For each fixed point $p \in M^{S^1}$ of index $2k$, there exists a unique class $\alpha_p \in H_{S^1}^{2k}(M; \mathbb{Z})$ such that

* $\alpha_p|_q = 0$ for every $q(\neq p) \in M^{S^1}$ with either $H(q) \le H(p)$ or $\mathrm{ind}(q) \le 2k$,

* $\alpha_p|_p = \prod_{i=1}^k \lambda_i u$, where $\lambda_1, \cdots, \lambda_k$ are negative weights of the $S^1$-action at $p$.

Moreover, the set $\{\alpha_p | p \in M^{S^1}\}$ is a basis of $H_{S^1}^*(M; \mathbb{R})$ as an $H^*(BS^1; \mathbb{R})$-module.

Now we are ready to prove Theorem 2.

**Proof of Theorem 2.** We first consider the case $\dim M = 8n$. Suppose that

$$b_2 + \cdots + b_{2+4(n-1)} > b_4 + \cdots + b_{4+4(n-1)}. \quad (1)$$

Since $H_{S^1}^*(M)$ is a free module over $H^*(BS^1)$, we have

$$H_{S^1}^{4n-2}(M) \cong u^0 \otimes H^{4n-2}(M) \oplus u^1 \otimes H^{4n-4}(M) \oplus \cdots \oplus u^{(2n-1)} \otimes H^0(M)$$

which implies that

* $\dim_{\mathbb{R}} H_{S^1}^{4n-2}(M; \mathbb{R}) \cong b_0 + b_2 + \cdots + b_{4n-2}$, and

* $\{\alpha_p \cdot u^{2n-1-\frac{1}{2}\mathrm{ind}(p)} | p \in M^{S^1}, \mathrm{ind}(p) \le 4n-2\}$ is a basis of $H_{S^1}^{4n-2}(M; \mathbb{R})$ (as an $\mathbb{R}$-vector space) by Theorem 4.

Now, consider the following map

$$\Phi: H_{S^1}^{4n-2}(M; \mathbb{R}) \to (\mathbb{R}^{b_0} \oplus \mathbb{R}^{b_4} \oplus \cdots \oplus \mathbb{R}^{b_{4(n-1)}}) \oplus (\mathbb{R}^{b_{4n}} \oplus \cdots \oplus \mathbb{R}^{b_{8n-4}}) \\ \alpha \mapsto (\alpha_0, \cdots, \alpha_{4n-4}, \alpha_{4n}, \cdots, \alpha_{8n-4})$$

with the identification

$$\mathbb{R}^{b_{4i}} = \bigoplus_{\mathrm{ind}(p)=4i} \mathbb{R} \cdot u^{2n-1} \quad \text{and} \quad \alpha_{4i} := (\alpha|_p)_{\mathrm{ind}(p)=4i} \in \bigoplus_{\mathrm{ind}(p)=4i} \mathbb{R} \cdot u^{2n-1} \quad (2)$$

for each $i = 1, \cdots, n$. Since the dimension of the range of the map $\Phi$ satisfies

$$\dim \mathrm{Im} \Phi \le b_0 + \cdots + b_{4n-4} + (b_{4n} + b_{4n+4} + \cdots + b_{8n-4}) < b_0 + \cdots + b_{4n-4} + (b_{4n-2} + \cdots + b_2)$$

<footnote n="1">¹See also [18, Proposition 2.2] and [11, Lemma 2.10]</footnote>
by (1) and Poincaré duality, the map $\Phi$ has a non-trivial kernel. In other words, there exists a nonzero element $\alpha \in H_{S^1}^{4n-2}(M; \mathbb{R})$ such that

$$ \alpha|_p = 0 $$

for every fixed point $p \in M^{S^1}$ of index $0, 4, \dots, 8n-4$.

Now fix a moment map $H$ for the $S^1$-action on $(M, \omega)$ such that $H$ attains the maximum value 0. Denote by $p_{\max}$ the maximal fixed point and so $\mathrm{ind}(p_{\max}) = 8n$. The equivariant extension $[\omega_H] \in H_{S^1}^2(M; \mathbb{R})$ of $\omega$ with respect to the moment map $H$ satisfies

$$ [\omega_H]|_p = -H(p)u \in \mathbb{R}[u] $$

for every $p \in M^{S^1}$, see [4, Proposition 2.6]. Since $H(p) < 0$ for every $p \neq p_{\max}$ by the choice of $H$, we obtain

$$ 0 = \int_M \alpha^2 \cdot [\omega_H] = \sum_{p \in M^{S^1}} \frac{-\alpha^2|_p \cdot H(p)u}{\left( \prod_{i=1}^{4n} w_i(p) \right) u^{4n}} = \sum_{\mathrm{ind}(p) \equiv 2 \pmod 4} \frac{-\alpha^2|_p \cdot H(p)u}{\left( \prod_{i=1}^{4n} w_i(p) \right) u^{4n}} \quad (3) $$

by Theorem 3 and the fact $[\omega_H]|_{p_{\max}} = -H(p_{\max})u = 0$. Moreover, there exists at least one fixed point $p \in M^{S^1}$ such that

$$ \alpha|_p \neq 0 \quad \text{and} \quad \mathrm{ind}(p) < 8n $$

because

* $\alpha|_p \neq 0$ for some $p \in M^{S^1}$ by the Kirwan's Injectivity Theorem [14], and

* if $\alpha|_p = 0$ for every $p \in M^{S^1}$ with $p \neq p_{\max}$, then $\alpha|_{p_{\max}} \neq 0$ and it violates Theorem 3

$$ 0 = \int_M \alpha = \frac{\alpha|_{p_{\max}}}{\left( \prod_{i=1}^{4n} w_i(p) \right) u^{4n}} \neq 0. $$

Consequently, each summand of the rightmost equation of (3) has non-positive coefficient (of $\frac{1}{u}$) and at least one of those should be negative. Therefore it leads to a contradiction.

Now it remains to consider the case of $\dim M = 8n + 4$. Under the same assumption (1), we similarly define

$$ \Phi: H_{S^1}^{4n}(M; \mathbb{R}) \to (\mathbb{R}^{b_0} \oplus \mathbb{R}^{b_4} \oplus \cdots \oplus \mathbb{R}^{b_{4n}}) \oplus (\mathbb{R}^{b_{4n+4}} \oplus \cdots \oplus \mathbb{R}^{b_{8n}}) \\ \alpha \mapsto (\alpha_0, \cdots, \alpha_{4n}, \alpha_{4n+4}, \cdots, \alpha_{8n}) $$

with the same identification as in (2). Note that $\dim_{\mathbb{R}} H_{S^1}^{4n}(M; \mathbb{R}) = b_0 + b_2 + \cdots + b_{4n-2} + b_{4n}$ and

$$ \dim \mathrm{Im} \Phi \le b_0 + \cdots + b_{4n} + (b_{4n+4} + \cdots + b_{8n}) = b_0 + \cdots + b_{4n} + (b_{4n} + \cdots + b_4) $$

$$ < b_0 + \cdots + b_{4n} + (b_{4n-2} + \cdots + b_2) = \dim_{\mathbb{R}} H_{S^1}^{4n}(M; \mathbb{R}) $$

by (1) and Poincaré duality. Thus $\Phi$ has a non-trivial kernel $\alpha \in H_{S^1}^{4n}(M; \mathbb{R})$ and so there exists a nonzero $\alpha \in H_{S^1}^{4n}(M; \mathbb{R})$ such that $\alpha|_p = 0$ for every fixed point $p$ of index $0, 4, \dots, 4n$. Therefore, we obtain

$$ 0 = \int_M \alpha^2 \cdot [\omega_H] = \sum_{\mathrm{ind}(p) \equiv 2 \pmod 4} \frac{-\alpha^2|_p \cdot H(p)u}{\left( \prod_{i=1}^{4n+2} w_i(p) \right) u^{4n+2}} \neq 0 $$

which leads to a contradiction. This completes the proof of Theorem 2. $\square$
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