Minimal genus four manifolds

TL;DR

This paper demonstrates that for any given group, there exists a 4-manifold with that fundamental group whose trisection genus attains the lower bound established by Chu and Tillmann, advancing understanding of 4-manifold topology.

Contribution

It constructs 4-manifolds with prescribed fundamental groups that achieve the minimal possible trisection genus according to Chu and Tillmann's bound.

Findings

Existence of 4-manifolds with any fundamental group and minimal trisection genus

Validation of Chu-Tillmann's lower bound as sharp

Contribution to classification of 4-manifolds by genus and fundamental group

Abstract

In 2018, M. Chu and S. Tillmann gave a lower bound for the trisection genus of a closed 4-manifold in terms of the Euler characteristic of $M$ and the rank of its fundamental group. We show that given a group $G$, there exist a 4-manifold $M$ with fundamental group $G$ with trisection genus achieving Chu-Tillmann's lower bound.

Figures

Equations

$$g\ge \chi (M)-2+3\text{rk}({\pi }_1(M)).$$

$$g\ge \chi (M)-2+3\text{rk}({\pi }_1(M))$$

$$k_i=rk\left({\pi }_1(M)\right)i=1,2,3$$

$$g(M)=\chi (M)-2+3\text{rk}(G)$$

$$t(\hat{L})\le t(L)+|L|=n+|\hat{L}|-1.$$

$$n\le$$

$$\le$$

$$=$$

$$\le$$

Full text

1 Introduction

Let $M$ be a smooth, oriented, closed 4-manifold. D. Gay and R. Kirby showed in [2] that $M$ can be written as the union of three 4-dimensional 1-handlebodiesiiiA 1-handlebody of dimension four is a 4-ball with 1-handles attached along its boundary; i.e. $\natural_{k}S^{1}\times B^{3}$. of genus $k_{1}$, $k_{2}$ and $k_{3}$, respectively; with pairwise intersections 3-dimensional connected handlebodies and triple intersection a connected surface of genus $g$. This is called a $(g;k_{1},k_{2},k_{3})-$trisection of $M$. The trisection genus of a 4-manifold $M$, denoted by $g(M)$, is the smallest $g$ such that $M$ admits a $(g;*,*,*)-$trisection.

Let $\tau$ be a trisection of $M$; i.e. $M=X_{1}\cup X_{2}\cup X_{3}$. Each 1-handlebody $X_{i}$ can be used to span the fundamental group and the first homology of $M$. In [1], M. Chu and S. Tillmann used this to give a lower bound to the trisection genus of a closed 4-manifold. They showed that if $M$ admits a $(g;k_{1},k_{2},k_{3})-$trisection then

[TABLE]

Question 1** (From [1]).**

Given any finitely presented group $G$, is there a smooth, oriented closed 4-manifold $M$ with $\pi_{1}(M)=G$ such that $g(M)=\chi(M)-2+3\text{rk}(G)$?

In this note, we will give a positive answer of Question 1.

Fix a finitely presented group $G$. We divide the answer in two steps as follows: (1) We state an equivalent version of Question 1 in terms of Kirby diagrams of closed 4-manifolds, which translates into a knot theory problem; and then (2) we show the existence of a special type of links.

Along this note, all 4-manifolds will be smooth, oriented and compact. For a link $L$ in a 3-manifold $Y$, $t_{Y}(L)$ will denote the tunnel number of $L$ in $Y$. We will omit the sub-index $Y$ if there is no confusion of the ambient manifold.

2 The proof

Let $M$ be a closed 4-manifold and $\tau$ be a $(g;k_{1},k_{2},k_{3})-$trisection of $M$. In [1], M. Chu and S. Tillmann showed the inequality

[TABLE]

Since $\chi(M)=2+g-(k_{1}+k_{2}+k_{3})$ and $k_{i}\geq\text{rk}(\pi_{1}(M))$ for $i=1,2,3$, equality of Equation (1) is equivalent to

[TABLE]

The following lemma is an application of Lemmas 13 and 14 of [2]. It shows that Question 1 is equivalent to finding links with the correct homotopy class and “small” tunnel number.

Lemma 2.1**.**

Let $G\neq 1$ be a finitely presented group of rank $n$. The following are equivalent:

1. (a)

There is a smooth, oriented closed 4-manifold $M$ with $\pi_{1}(M)=G$ such that

[TABLE] 2. (b)

There is a smooth, oriented closed 4-manifold $M$ with $\pi_{1}(M)=G$ having a handle decomposition with one 0-handle, $n$ 1-handles, $j$ 2-handles, $n$ 3-handles and one 4-handle such that the attaching region of the 2-handles is a link in $\#_{n}S^{1}\times S^{2}$ of tunnel number $n+j-1$. 3. (c)

There is a link $L$ in $\#_{n}S^{1}\times S^{2}$ of tunnel number at most $n+|L|-1$ such that the link $L$, thought of as a set of homotopy classes of loops based at a point, gives relations for a rank $n$ group presentation for $G$.

Proof.

(a) $\Rightarrow$ (b). Let $M$ be a 4-manifold with $\pi_{1}(M)=G$ such that $g(M)=\chi(M)-2+3\text{rk}(G)$. Let $\tau$ be a $(g;k_{1},k_{2},k_{3})-$trisection with $g=g(M)$. In particular, $\tau$ satisfies Equation (2); i.e., $\tau$ is a $\left(g;n,n,n\right)-$trisection of $M$. Lemma 13 of [2] asserts that $M$ admits a handle decomposition with one 0-handle, $n$ 1-handles, $g-n$ 2-handles, $n$ 3-handles and one 4-handle such that the attaching region of the 2-handles is a framed link $L$ contained in the core of one of the handlebodies of a genus $g$ Heegaard splitting for $\#_{n}S^{1}\times S^{2}$. In particular, $t(L)\leq g-1$. The latter must be an equality since Proposition 4.2 of [4] states that $M$ admits a new $(t(L)+1;n,t(L)+1-|L|,n)$-trisection and $t(L)<g-1$ will give us $t(L)+1-|L|<(g-1)+1-(g-n)=n$; contradicting the fact that $k_{2}\geq n$. Hence $t(L)=g-1$ and, (b) holds taking $j=g-n$.

(b) $\Rightarrow$ (a). Let $M$ be a 4-manifold satisfying (b) with the given handlebody decomposition. By taking the tubular neighborhood of the attaching region of the 2-handles and the tunnels, we obtain a Heegaard surface for $\#_{n}S^{1}\times S^{2}$ of genus $g=n-j$ satisfying the assumptions of Lemma 14 of [2]. By the lemma, $M$ admits a $(g;n,n,n)$-trisection. In particular, Equation (2) holds and Equation (1) becomes an equality, thus (a).

(b) $\Rightarrow$ (c). Take $L$ to be the attaching region of the 2-handles.

(c) $\Rightarrow$ (b). Let $L$ be such link and let $x\in\mathbb{Z}^{|L|}$ be any fixed vector. Consider $N$ to be the smooth 4-manifold with a handle decomposition given by one 0-handle, $n$ 1-handles and $|L|$ 2-handles attached along $L$ with framing $x$ with respect to the blackboard. Take $M$ to be the double of $N$. By assumption, $\pi_{1}(N)\cong G$ and so $\pi_{1}(M)\cong G$. Turning the handle decomposition of $N$ upside-down gives $M$ a handle decomposition with one 0-handle, $n$ 1-handles, $2|L|$ 2-handles, $n$ 3-handles and one 4-handle, where half of the 2-handles are attached along $L$ and for each component of $L$ there is a 0-framed unknot linked once along the given component and unlinked from the rest of the diagram.

Let $\widehat{L}=L\cup L^{\prime}$ be the attaching region for the 2-handles of $M$. Notice that

[TABLE]

By a version of Lemma 14 of [2] for unbalanced trisections (see Proposition 4.2 of [4]), $M$ admits a $(t(\widehat{L})+1;n,t(\widehat{L})+1-|\widehat{L}|,n)-$trisection. Using Equation (1) with $k_{2}=t(\widehat{L})+1-|\widehat{L}|$ we obtain

[TABLE]

Thus, $t(\widehat{L})=n+|\widehat{L}|-1$ and $M$ is the desired 4-manifold. Hence (b). ∎

Remark 2.2**.**

We have shown in Lemma 2.1 that to answer Question 1 in the positive, is enough to find a link $L$ in $\#_{n}S^{1}\times S^{2}$ with tunnel number at most $n+|L|-1$ such that the homotopy classes of the components of $L$, together, read a rank $n$ presentation for the given group $G$. The closed 4-manifold answering Question 1 will be the double of a 4-manifold with a Kirby diagram with $n$ 1-handles and 2-handles attached along $L$ with any framing.

The following proposition shows how to build links satisfying (c) in Lemma 2.1.

Proposition 2.3**.**

Let $G$ be a finitely presented group of rank $n$ with a presentation $\langle X|R\rangle$. There exists an $|R|$-component link in $\#_{|X|}S^{1}\times S^{2}$ with tunnel number at most $|X|+|R|-1$ such that the words read by the components of $L$ in $\pi_{1}(\#_{|X|}S^{1}\times S^{2},\star)$ agree with the words in $R$.

Corollary 2.4**.**

Given a finitely presented group $G$, there is a closed 4-manifold with fundamental group isomorphic to $G$ satisfying $g(M)=\chi(M)-2+3\text{rk}(G)$.

Proof of Proposition 2.3.

Take an unknotted graph in $\#_{n}S^{1}\times S^{2}$ made by $n$ loops, denoted by $\Gamma_{0}$, generating $\pi_{1}(\#_{n}S^{1}\times S^{2},\star)$ and $|R|$ unknotted loops $c_{1},\dots,c_{|R|}$ around a neighborhood of $\star$; see Figure 1. For each relation $r_{i}\in R$, take the $i$-th unknotted circle $c_{i}$ and slide one of its ends along the loops on $\Gamma_{0}$ so that $c_{i}$ now reads the word $r_{i}$ as an element of $\pi_{1}(\#_{n}S^{1}\times S^{2},\star)$. Each component of $L$ will be a circle $c_{i}$, and the rest of the graph can be homotoped to be a system of tunnels for $L$.

We have described how to build a link $L$ in $\#_{n}S^{1}\times S^{2}$ with $t(L)\leq n+|L|-1$. ∎

The following figures ilustrate the construction of $L$ for the group $G=\langle x,y,z|x^{3}y^{-2},[y,z]\rangle$.

Remark 2.5**.**

Note that the knot type of the link $L$ doesn’t change the final 4-manifold built in Corollary 2.4. This happens since, when taking double of the 4-manifold, the 0-framed unknots around each component of $L$ allow us to slide the handlebody 2-handles and so to change the crossings of the link. The same reason explains why we only care about the framing of the components of $L$ modulo $2$. In any case, one can consider connected sums of our manifolds with copies of $S^{2}\times S^{2}$ and $\pm CP^{2}$ to build infinitely many 4-manifolds solving Question 1.

One of the two 4-manifolds appearing when running the construction with finite cyclic groups $\langle x|x^{m}=0\rangle$ is the spun lens space $\mathcal{S}_{m}$ trisected by J. Meier in [3]. One can check this by comparing the Kirby diagram we construct with the one drawn by J. Montesinos in [5].