2. A dimension inequality

A vertical tangle in a balanced sutured manifold $(M,\gamma )$ is a properly embedded 1-manifold

\[ T=T_1\cup \cdots \cup T_n \]

in $M$, with boundary in $R(\gamma )$, whose components $T_i$ satisfy

\[ \partial T_i \cap R_+(\gamma) \neq \emptyset \neq \partial T_i \cap R_-(\gamma), \]

and are oriented from $R_+(\gamma )$ to $R_-(\gamma )$ (in particular, $T$ has no closed components). One forms an associated balanced sutured manifold $(M_T,\gamma _T)$ by removing tubular neighborhoods of the components $T_i$, and adding positively-oriented meridians of these components to the suture $\gamma$, as in [Reference Li and YeLY22, § 3] and depicted in Figure 7. In this section, we prove Theorem 1.9, which states that

\[ \dim_\mathbb{C} \mathit{SHI}(M,\gamma)\leq \dim_\mathbb{C}\mathit{SHI}(M_T,\gamma_T) \]

when $[T]=0$ in $H_1(M,\partial M;\mathbb {Q})$. The rough idea is to turn $T$ into a related tangle $T'$ whose components are rationally nullhomologous, and apply Theorem 1.8.

Proof of Theorem 1.9 Let $T_1,\ldots,T_n$ be the components of $T$. For $i=1,\ldots,n$, let

\[ \partial T_i = q_i-p_i, \]

where $p_i\in R_+(\gamma )$ and $q_i \in R_-(\gamma )$. We may assume that $\gamma$ is connected, because we can achieve this by adding contact 1-handles to $(M,\gamma )$, an operation which does not change $\mathit {SHI}(M,\gamma )$ (equivalently, $\mathit {SHI}$ is invariant under product disk decomposition [Reference Kronheimer and MrowkaKM10, Reference Baldwin and SivekBS16]). Then we can find a sequence of pairwise disjoint arcs

\[ \xi_1,\ldots,\xi_n\subset\partial M \]

such that, for each $i=1,\ldots,n$, we have:

• • $\partial \xi _i = p_{i+1}-q_i$ (where $p_{n+1}:=p_1$); and

• • $\xi _i$ intersects $\gamma$ in exactly one point.

For every $i\in \{2,\ldots,n\}$, choose an arc $t_{i} \subset \partial M$ in a neighborhood of the unique intersection point $\xi _{i-1}\cap \gamma$, as depicted in Figures 1 and 2. Push the interior of $t_{i}$ into the interior of $M$ to turn this arc into a vertical tangle $T_{i}'$, and let

\begin{align*} T'' = T_2' \cup\cdots\cup T_n'. \end{align*}

Then

\begin{align*} (M_{T''},\gamma_{T''}) = (M-N(T''),\gamma\cup \mu_2'\cup\cdots\cup\mu_n'), \end{align*}

where $\mu _i'$ is a positively oriented meridian of $T_i'$. Each component $T_{i}'$ cobounds a disk in $M$ with the arc $t_{i}$. These disks then restrict to properly embedded disks

\begin{align*} D_2,\ldots, D_n\subset M_{T''} \end{align*}

with $|D_i\cap \gamma |=1$ and $|D_i\cap \mu _j'|=\delta _{ij}$, so that

\begin{align*} |D_i\cap \gamma_{T''}|=2. \end{align*}

Thus, each $D_i$ is a product disk.

Figure 1. The point of view is from the interior of $M$, looking at $\partial M$.

Figure 2. The tangle $T = T_1\cup \cdots \cup T_n$ in $M$ and the arcs $\xi _i$ and $t_i$ in $\partial M$, together with the suture $\gamma$.

Next, consider the arc

\[ t_1 = T_1\cup\xi_1\cup T_2\cup\xi_2\cup\cdots\cup T_{n-1}\cup \xi_{n-1}\cup T_n \subset M_{T''}. \]

Push its interior into the interior of $M_{T''}$ to form a vertical tangle $T_1'$ with $\partial T_1' = q_n-p_1,$ as in Figure 3. Let $T'$ be the tangle in $M$ given by

\[ T' = T_1'\cup T'' = T_1'\cup \cdots\cup T_n'. \]

We refer to a tangle $T'$ formed in this way as a mixed tangle for $T$. Note that

\[ (M_{T'},\gamma_{T'}) = ((M_{T''})_{T_1'}, (\gamma_{T''})_{T_1'}) = (M-N(T'),\gamma\cup \mu_1'\cup\cdots\cup\mu_n'), \]

where $\mu _1'$ is a positively oriented meridian of $T_1'$. Observe that the disks $D_2,\ldots, D_n\subset M_{T''}$ restrict to properly embedded annuli

\[ A_2,\ldots,A_n\subset M_{T'} \]

because $T_1'$ intersects each disk in exactly one point, as shown in Figure 4.

Figure 3. The tangle $T' = T_1' \cup \cdots \cup T_n'$ in $M$.

Figure 4. The annulus $A_i$ in $M_{T'}$ near the boundaries of the tubular neighborhoods of the components $T_1'$ and $T_i'$, as seen from inside $M_{T'}$.

The endpoints of the arc $\xi _n\subset \partial M_{T''}$ agree with $\partial T_1'$, and $|\xi _n\cap \gamma |=1$. We can, thus, use $\xi _n$ together with $\gamma$ and $\mu _1'$ to define a sequence of sutures $\Gamma _m\subset \partial M_{T'}$ for $m\in \mathbb {N}$, as in [Reference Li and YeLY22, § 3.2], which one should regard as ‘longitudinal’ sutures for $T_1'$; see Figure 5. By the construction of $T_1'$ and the assumption that $T$ is rationally nullhomologous in $(M,\partial M)$, we have

\[ [T_1'] = [T] =0 \in H_1(M_{T''},\partial M_{T''};\mathbb{Q}). \]

Therefore, by [Reference Li and YeLY22, Lemmas 3.21 and 3.22], we have the following.

Lemma 2.1 There is an exact triangle

coming from the surgery exact triangle associated to surgeries on the meridian $\mu _1'$ of $T_1'$. Furthermore, $G_m\equiv 0$ for $m$ sufficiently large.

Figure 5. (a) The suture $-\Gamma _{m+1}$ and the bypass arc $\eta _-$ shown in bold. (b) The suture $-\gamma _{T'}$ resulting from the bypass attachment along $\eta _-$, and the negatively stabilized annulus $A_i^-\subset (-M_{T'},-\gamma _{T'})$.

For each $i=2,\ldots,n$ and every $m\in \mathbb {N}$, we have that

\[ |A_i\cap \Gamma_m|=4. \]

Let us orient each $A_i$ so that the induced orientation on $\partial A_i$ is opposite the orientation of $\partial D_i$ coming from that of $T_i'$, as in Figures 4 and 5. By [Reference Ghosh and LiGL19], the disks $D_2,\ldots,D_n$ induce a $\mathbb {Z}^{n-1}$-grading on

\begin{align*} \mathit{SHI}(-M_{T''},-\gamma_{T''}). \end{align*}

Similarly, the annuli $A_2,\ldots,A_n$ induce a $\mathbb {Z}^{n-1}$-grading on

\[ \mathit{SHI}(-M_{T'},-\Gamma_m) \]

for each $m\in \mathbb {N}$, and we have the following graded version of the triangle in Lemma 2.1.

Lemma 2.2 The exact triangle of Lemma 2.1 restricts to the exact triangle

Proof of Lemma 2.2 We prove that the map $F_{m+1}$ preserves the gradings. The arguments for the other two maps are similar.

Let us first recall the definition of $F_{m+1}$ from [Reference Li and YeLY22, § 3]. Pick a closure

\[ (Y_{m+1},R_{m+1},\omega_{m+1}) \]

for $(-M_{T'},-\Gamma _{m+1})$ so that each annulus $A_i$ extends to a closed surface $\bar A_i\subset Y_{m+1}$, as in [Reference LiLi21, § 3]. By the construction therein,

\begin{align*} g(\bar A_i)=2 \end{align*}

for $i=2,\ldots,n$, because each component of $\partial A_i$ intersects $\Gamma _{m+1}$ in two points. The sutured instanton homology of $(-M_{T'},-\Gamma _{m+1})$ is defined as a certain direct summand

\[ \mathit{SHI}(-M_{T'},-\Gamma_{m+1}) = I_*(Y_{m+1}|R_{m+1})_{\omega_{m+1}} \]

of the instanton Floer homology $I_*(Y_{m+1})_{\omega _{m+1}}$ as in [Reference Kronheimer and MrowkaKM10, § 7]. The summand

\[ \mathit{SHI}(-M_{T'},-\Gamma_{m+1},(A_2,\ldots,A_n), (i_2,\ldots,i_n)) \]

is defined as the simultaneous (generalized) $(2i_2,\ldots,2i_n)$-eigenspace of commuting operators

\[ \mu(\bar A_2),\ldots,\mu(\bar A_n):\mathit{SHI}(-M_{T'},-\Gamma_{m+1}) \to \mathit{SHI}(-M_{T'},-\Gamma_{m+1}) \]

associated to these surfaces.

The meridian $\mu _1'$ of $T_1'$ can be thought of as an embedded circle in $Y_{m+1}$. Let $Y$ be the manifold obtained from $Y_{m+1}$ via $0$-surgery on $\mu _1'$, with respect to the framing of $\mu _1'$ induced by $\partial M_{T'}$. Since $\mu _1'$ is disjoint from $R_{m+1}$ and the $\bar A_i$, these surfaces survive in $Y$. By [Reference Baldwin and SivekBS16, § 3.3], $(Y,R_{m+1},\omega _{m+1})$ is a closure of $(-M_{T''}, -\gamma _{T''})$. The map $F_{m+1}$ is then induced by the cobordism given by the trace of $0$-surgery on $\mu _1'$. Since $\bar A_i\subset Y_{m+1}$ is homologous to $\bar A_i\subset Y$ in this cobordism, $F_{m+1}$ respects the eigenspaces of $\mu (\bar A_i)$. Thus, $F_{m+1}$ maps

\[ \mathit{SHI}(-M_{T'},-\Gamma_{m+1},(A_2,\ldots,A_n), (i_2,\ldots,i_n)) \]

into

\[ \mathit{SHI}(-M_{T''},-\gamma_{T''},(A_2,\ldots,A_n), (i_2,\ldots,i_n)). \]

Now, the $0$-surgery on $\mu _1'$ makes each $\bar A_i$ compressible in $Y$; in particular, each $\bar A_i\subset Y$ is homologous to the disjoint union of two tori

\[ T_i^1\cup T_i^2\subset Y. \]

One of these tori, say $T_i^1$, is the extension $\bar D_i\subset Y$ of $D_i\subset -M_{T''}$ that is used to define the grading on $\mathit {SHI}(-M_{T''},-\gamma _{T''})$ associated to $D_i$. Since

\[ \bar A_i=\bar D_i+T_i^2 \]

in $H_2(Y)$, the $k$-eigenspace of $\mu (\bar A_i)$ agrees with the $k$-eigenspace of $\mu (\bar D_i)$ for every $k$, by [Reference Baldwin and SivekBS22, Corollary 2.9]. Thus, we have that

\begin{align*} &\mathit{SHI}(-M_{T''},-\gamma_{T''},(A_2,\ldots,A_n), (i_2,\ldots,i_n))\\ &\quad =\mathit{SHI}(-M_{T''},-\gamma_{T''},(D_2,\ldots,D_n), (i_2,\ldots,i_n)). \end{align*}

Putting these arguments together, we see that $F_{m+1}$ preserves the $\mathbb {Z}^{n-1}$-gradings as claimed in the lemma.

Note that decomposing $(-M_{T''},-\gamma _{T''})$ along $D_2\cup \cdots \cup D_n$ yields $(-M,-\gamma )$. By [Reference LiLi21, Lemma 4.2], we therefore have

\[ \mathit{SHI}(-M_{T''},-\gamma_{T''},(D_2,\ldots,D_n), (0,\ldots,0))\cong \mathit{SHI}(-M,-\gamma). \]

Hence, for $m$ sufficiently large, Lemmas 2.1 and 2.2 imply that

(2.1)\begin{align} \dim_\mathbb{C}\mathit{SHI}(-M,-\gamma) &=\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\Gamma_{m+1},(A_2,\ldots,A_n), (0,\ldots,0))\nonumber\\ &\quad -\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\Gamma_{m},(A_2,\ldots,A_n), (0,\ldots,0)), \end{align}

since $G_m\equiv 0$.

Next, we consider attaching a bypass to $(-M_{T'},-\Gamma _{m+1})$ along the arc $\eta _-$ in Figure 5. By [Reference Baldwin and SivekBS22, § 4], this attachment gives rise to a bypass exact triangle. As discussed in [Reference Li and YeLY22, § 3], the other two sutures involved in the triangle are $-\Gamma _m$ and $-\gamma _{T'}$. It is straightforward to check that the bypass attachment along $\eta _-$ creates a negative stabilization

\[ A_i^-\subset (-M_{T'},-\gamma_{T'}) \]

of $A_i$, for each $i=2,\ldots,n$, in the sense of [Reference LiLi21, Definition 3.1]. Hence, as in the proof of [Reference LiLi21, Proposition 5.5], we have the following graded version of the bypass exact triangle of [Reference Baldwin and SivekBS22, Theorem 1.20]:

which implies that

(2.2)\begin{align} &\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\gamma_{T'},(A_2^-,\ldots,A_n^-), (0,\ldots,0))\nonumber\\ &\quad \geq\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\Gamma_{m+1},(A_2,\ldots,A_n), (0,\ldots,0))\nonumber\\ &\qquad -\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\Gamma_{m},(A_2,\ldots,A_n), (0,\ldots,0)). \end{align}

From the grading shifting property [Reference LiLi21, Theorem 1.12] and [Reference WangWan20, Proposition 4.1], we have

(2.3)\begin{align} &\mathit{SHI}(-M_{T'},-\gamma_{T'},(A_2^-,\ldots,A_n^-), (0,\ldots,0)) \nonumber\\ &\quad = \mathit{SHI}(-M_{T'},-\gamma_{T'},(A_2^+,\ldots,A_n^+), (-1,\ldots,-1)), \end{align}

where $A_i^+$ is a positive stabilization of $A_i$. Moreover, from the construction of the gradings and stabilizations in [Reference LiLi21, § 3], we have

(2.4)\begin{align} &\mathit{SHI}(-M_{T'},-\gamma_{T'},(A_2^+,\ldots,A_n^+), (-1,\ldots,-1)) \nonumber\\ &\quad = \mathit{SHI}(-M_{T'},-\gamma_{T'},(-(A_2^+),\ldots,-(A_n^+)), (1,\ldots,1))\nonumber\\ &\quad =\mathit{SHI}(-M_{T'},-\gamma_{T'},((-A_2)^-,\ldots,(-A_n)^-), (1,\ldots,1)). \end{align}

By [Reference LiLi21, Lemma 4.2], this last group is isomorphic to the sutured instanton homology of the manifold obtained from $(-M_{T'},-\gamma _{T'})$ by decomposing along $(-A_2)^-\cup \cdots \cup (-A_n)^-$. By [Reference LiLi21, Lemma 3.2], this is the same as the manifold obtained by decomposing along $-A_2\cup \cdots \cup -A_n$, which is, after reversing orientation, the manifold obtained from $(M_{T'},\gamma _{T'})$ by decomposing along $A_2\cup \cdots \cup A_n$. It is straightforward to check that the latter manifold is simply $(M_T,\gamma _T)$, as indicated in Figure 6 in the case $n=2$. Thus,

(2.5)\begin{equation} \mathit{SHI}(-M_{T'},-\gamma_{T'},((-A_2)^-,\ldots,(-A_n)^-), (1,\ldots,1)) \cong \mathit{SHI}(-M_T,-\gamma_T). \end{equation}

Figure 6. The result of decomposing $(M_{T'},\gamma _{T'})$ along $A_2\cup \cdots \cup A_n$ is simply $(M_T,\gamma _T)$. This is illustrated above in the case $n=2$.

Finally, combining (2.1)–(2.5), we have that

\begin{align*} \dim_\mathbb{C}\mathit{SHI}(-M_T,-\gamma_T)&=\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\gamma_{T'},((-A_2)^-,\ldots,(-A_n)^-), (1,\ldots,1))\\ &=\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\gamma_{T'},(A_2^-,\ldots,A_n^-), (0,\ldots,0)) \\ &\geq\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\Gamma_{m+1},(A_2,\ldots,A_n), (0,\ldots,0))\\ &\quad-\dim_\mathbb{C}\mathit{SHI}(-M_{T'},-\Gamma_{m},(A_2,\ldots,A_n), (0,\ldots,0))\\ &=\dim_\mathbb{C}\mathit{SHI}(-M,-\gamma). \end{align*}

Given the symmetry of $\mathit {SHI}$ under orientation reversal, this proves Theorem 1.9.

3. Proof of Theorem 1.1

3.1 Full tangles

Let $(M,\gamma )$ be a balanced sutured manifold. Let

\[ \mathcal{H} = \big(\Sigma,\alpha = \{\alpha_1,\ldots,\alpha_k\},\beta = \{\beta_1,\ldots,\beta_k\}\big) \]

be any (not necessarily admissible) sutured Heegaard diagram for $(M,\gamma )$. This means that $M$ is obtained from $\Sigma \times [-1,1]$ by attaching 3-dimensional 2-handles

\begin{align*} \mathbb{D}_{\alpha_i} &=D_{\alpha_i}^2\times I,\\ \mathbb{D}_{\beta_i} &=D_{\beta_i}^2\times I, \end{align*}

along $A_i\times \{-1\}$ and $B_i\times \{+1\}$, where $A_i$ and $B_i$ are annular neighborhoods of $\alpha _i$ and $\beta _i$, respectively, for $i=1,\ldots,k$. The suture $\gamma$ is given by

\[ \gamma = \partial \Sigma\times \{0\}. \]

We next define a special class of vertical tangles in $(M,\gamma )$ associated to $\mathcal {H}$.

Let $R_1,\ldots,R_n$ be the regions of $\Sigma$–$\alpha$–$\beta$ disjoint from $\partial \Sigma$. For each $i=1,\ldots,n$, let $p_{i1},\ldots,p_{ia_i}$ be $a_i$ distinct points in $R_i$, for some integer $a_i\geq 1$. Let

\[ T_{ij} = p_{ij}\times[-1,1]\subset \Sigma\times[-1,1], \]

and let

\[ T = \bigcup_{i=1}^n \bigcup_{j=1}^{a_i} T_{ij}. \]

Then $T$ is a vertical tangle in $(M,\gamma )$, oriented from $R_+(\gamma )$ to $R_-(\gamma )$. Let $D_{ij}^2$ be a tubular neighborhood of the point $p_{ij}\in R_i$, let

\[ N_{ij} =D^2_{ij}\times [-1,1] \]

be a tubular neighborhood of the component $T_{ij}$ in $M$, and let

\[ \gamma_{ij} = \partial D^2_{ij}\times \{0\}\subset \partial N_{ij} \]

be a positively oriented meridian of this component; see Figure 7. Let $(M_T, \gamma _T)$ be the balanced sutured manifold obtained from $M$ by removing these tubular neighborhoods,

\[ M_T = M-\bigcup_{i=1}^n\bigcup_{j=1}^{a_i}N_{ij}, \]

where $\gamma _T$ is the union of $\gamma$ with meridians of the $T_{ij}$,

\[ \gamma_T = \gamma\cup \bigcup_{i=1}^n\bigcup_{j=1}^{a_i}\gamma_{ij}, \]

as in § 2. We refer to any tangle obtained in this way as a full tangle for $\mathcal {H}$.

Figure 7. (a) The component $T_{ij}=p_{ij}\times [-1,1]\subset M$. (b) The complement of $N_{ij}$ with the meridian $\gamma _{ij}$.

The main result of this section is the following.

Proposition 3.1 If $T$ is a full tangle for $\mathcal {H}$, then $\dim _\mathbb {C}\mathit {SHI}(M_T,\gamma _T) = |\mathfrak {S}(\mathcal {H})|.$

Remark 3.2 The analogue of this proposition for $\mathit {SFH}$ is immediate. One forms a sutured Heegaard diagram $\mathcal {H}_T$ for $(M_T,\gamma _T)$ from $\mathcal {H}$ by removing neighborhoods of the $p_{ij}\in \Sigma$. Since there is at least one such point in every region of $\Sigma$–$\alpha$–$\beta$ not intersecting $\partial \Sigma$, this ensures that $\mathcal {H}_T$ is admissible and that the differential on $\mathit {SFC}(\mathcal {H}_T)$ is zero, so that

\[ \textrm{rk}_\mathbb{Z}\mathit{SFH}(M_T,\gamma_T) = \textrm{rk}_\mathbb{Z}\mathit{SFC}(\mathcal{H}_T) = |\mathfrak{S}(\mathcal{H}_T)| = |\mathfrak{S}(\mathcal{H})|. \]

This was the inspiration for our result above.

We need the following for the proof of Proposition 3.1; see [Reference Ghosh and LiGL19, Corollary 4.3].

Proposition 3.3 Suppose $(M,\gamma )$ is an irreducible balanced sutured manifold and $D\subset M$ is a properly embedded disk which intersects $\gamma$ in four points. Then

\[ \mathit{SHI}(M,\gamma) \cong \mathit{SHI}(M',\gamma')\oplus \mathit{SHI}(M'',\gamma''), \]

where $(M',\gamma ')$ and $(M'',\gamma '')$ are the decompositions of $(M,\gamma )$ along $D$ and $-D$, respectively.

Before proving this proposition, we record the following lemma. This lemma is well-known (Kronheimer and Mrowka prove the much harder converse in [Reference Kronheimer and MrowkaKM10, Theorem 7.12]), but since we could not find a concrete reference in the literature we provide a short proof here.

Lemma 3.4 Suppose $(M,\gamma )$ is an irreducible balanced sutured manifold. If $(M,\gamma )$ is not taut, then $\mathit {SHI}(M,\gamma )= 0$.

Proof. As mentioned in the proof of Lemma 2.2, $\mathit {SHI}(M,\gamma )$ is defined in [Reference Kronheimer and MrowkaKM10, § 7] in terms of a closure $(Y,R,\omega )$ of $(M,\gamma )$, where $Y$ is a closed oriented $3$-manifold, $R\subset Y$ is a connected closed oriented surface which we can take to have genus $g(R)>1$, and $\omega \subset Y$ is a simple closed curve with $|\omega \cap R|=1$. Specifically, $\mathit {SHI}(M,\gamma )$ is the generalized $(2g(R){-}2)$-eigenspace of the operator $\mu (R)$ on $I_*(Y)_\omega$.

Now suppose that $(M,\gamma )$ is irreducible but not taut. Then, by the definition of tautness (e.g. [Reference JuhászJuh06, Definition 9.18]), either one of $R_{\pm }(\gamma )$ is compressible or $R_{\pm }(\gamma )$ do not minimize the Thurston-norm in their homology classes. In either case, there exists a surface $R'\subset Y$ such that $[R']=[R]\in H_2(Y)$ but $1\leq g(R')< g(R)$. This implies that $\mathit {SHI}(M,\gamma )=0$ by [Reference Kronheimer and MrowkaKM10, Proposition 7.5].

Proof of Proposition 3.3 Ghosh and Li prove this in [Reference Ghosh and LiGL19, Corollary 4.3] under the additional assumption that $(M,\gamma )$ is taut and at least one of $(M',\gamma ')$ and $(M'',\gamma '')$ is taut. We show that this additional assumption is unnecessary, by showing that the proposition still holds when the assumption is not true.

First, suppose $(M,\gamma )$ is not taut. Then neither $(M',\gamma ')$ nor $(M'',\gamma '')$ is taut, by [Reference GabaiGab87, Lemma 0.4]. Note that the irreducibility of $M$ implies the irreducibility of any manifold obtained by cutting $M$ open along a properly embedded disk; in particular, $M'$ and $M''$ are irreducible. Therefore,

\begin{align*} \mathit{SHI}(M,\gamma) \cong \mathit{SHI}(M',\gamma')\cong \mathit{SHI}(M'',\gamma'') =0, \end{align*}

by Lemma 3.4, and the proposition holds.

Next, suppose neither $(M',\gamma ')$ nor $(M'',\gamma '')$ is taut. Consider the bypass exact triangle of [Reference Baldwin and SivekBS22]:

determined by an initial bypass attachment to $(M,\gamma )$ along an arc in $\partial D$ as shown in Figure 8. The other manifolds $(M,\gamma '_1)$ and $(M,\gamma ''_1)$ in the triangle product disk decompose (along a copy of $D$ which intersects the new sutures in two points) to $(M',\gamma ')$ and $(M'',\gamma '')$, respectively. Since $\mathit {SHI}$ is invariant under product disk decomposition, we therefore have

\begin{align*} \mathit{SHI}(M,\gamma'_1)&\cong \mathit{SHI}(M',\gamma')\cong 0,\\ \mathit{SHI}(M,\gamma''_1)&\cong \mathit{SHI}(M'',\gamma'')\cong 0, \end{align*}

by Lemma 3.4. It then follows from the bypass triangle, and the symmetry of $\mathit {SHI}$ under orientation reversal, that $\mathit {SHI}(M,\gamma )=0$ as well, so the proposition holds.

Proof of Proposition 3.1 Recall that $D_{ij}^2$ denotes a tubular neighborhood of $p_{ij}\in R_i$. There exists a (possibly empty) set of disjoint, properly embedded arcs

\begin{align*} d_1,\ldots,d_m\subset \Sigma-\bigcup_{i=1}^{n}\bigcup_{j=1}^{a_i} D_{ij} \end{align*}

which satisfy the following three conditions:

1. (1) for every $e=1,\ldots,m$, the arc $d_e$ is contained in some region of $\Sigma$–$\alpha$–$\beta$;

2. (2) for every $e$, either both endpoints of $d_e$ are on $\partial \Sigma$, or each is on some $\partial D_{ij}^2$; and

3. (3) $\Sigma -\bigcup _{i=1}^{n}\bigcup _{j=1}^{a_i} D_{ij} - d_1-\ldots - d_m$ deformation retracts onto $\alpha \cup \beta$.

Now consider the disk

\[ \delta_e = d_e\times[-1,1]\subset \biggl(\Sigma-\bigcup_{i=1}^{n}\bigcup_{j=1}^{a_i} D_{ij} \biggr)\times [-1,1] \subset \partial M_T, \]

for $e=1,\ldots,m$. The boundary of each $\delta _e$ intersects $\gamma _T$ in two points; hence, $\delta _e$ is a product disk. Since $\mathit {SHI}$ is invariant under product disk decomposition, let $(M_T,\gamma _T)$ henceforth refer to the balanced sutured manifold obtained after decomposing along $\delta _1,\ldots,\delta _m$.

Figure 8. (a) A neighborhood of the disk $D\subset M$ whose boundary intersects the suture $\gamma$ in four points. (b–d) The arc of attachment for the initial bypass in the triangle.

Then $M_T$ admits the following description. Let $q_1,\ldots,q_t$ denote the intersection points between the $\alpha$ and $\beta$ curves. If $q_\ell$ is an intersection point between $\alpha _i$ and $\beta _j$, let $r_\ell \subset \Sigma$ denote the rectangular component of $A_i\cap B_j$ which contains $q_\ell$. Then $M_T$ is the given by the union

\[ M_T = \mathbb{D}_{\alpha_1}\cup \cdots \cup \mathbb{D}_{\alpha_k} \cup \mathbb{D}_{\beta_1}\cup \cdots \cup \mathbb{D}_{\beta_k} \cup \tau_1 \cup\cdots \cup \tau_t \]

of the usual 2-handles with the tubes

\[ \tau_\ell = r_\ell\times[-1,1], \]

as shown in Figure 9. Let $c_\ell$ denote the union of the four corners of the rectangle $r_\ell$. Then the suture $\gamma _T$ is given by

\[ \gamma_T = \bigcup_{i=1}^k (\partial D_{\alpha_i}^2 \times \partial I)\cup \bigcup_{i=1}^k (\partial D_{\beta_i}^2 \times \partial I) \cup \bigcup_{\ell=1}^t (c_\ell \times[-1,1]) - \bigcup_{\ell=1}^t (\partial r_\ell\times\{-1,1\}) \]

as shown and oriented in the figure near a tube $\tau _\ell$. Let

\[ m_\ell = r_\ell\times\{0\}\subset \tau_\ell \]

denote the meridional disk of $\tau _\ell$, oriented as in Figure 9, for $\ell =1,\ldots,t$. Note that the boundary of each $m_\ell$ intersects the suture $\gamma _T$ in four points.

Figure 9. (a) An intersection point $q_\ell \in \alpha _i\cap \beta _j$. The rectangular component $r_\ell$ of $A_i\cap B_j$ containing $q_\ell$ is shown in darker gray. (b) The 2-handles $\mathbb {D}_{\alpha _i}$ and $\mathbb {D}_{\beta _j}$ glued together by the tube $\tau _\ell$. The meridional disk $m_\ell$ is shown in gray; it intersects the suture $\gamma _T$ in four points, and its oriented normal points upwards. (c) The result of decomposing along $-m_\ell$. (d) The result of decomposing along $m_\ell$.

Note that $M_T$ is a handlebody and therefore irreducible. We may thus apply Proposition 3.3 to it, as well as to the manifolds obtained by decomposing $M_T$ along any collection of the meridional disks $m_\ell$, as these manifolds are simply disjoint unions of handlebodies and, hence, also irreducible. For each $t$-tuple of signs

\[ I = (\epsilon_1,\ldots,\epsilon_t)\in \{+,-\}^t, \]

let $(M_T^I,\gamma _T^I)$ be the sutured manifold obtained by decomposing $(M_T,\gamma _T)$ along the disks

\[ \epsilon_1m_1 \cup \cdots \cup \epsilon_tm_t. \]

Then

\[ \mathit{SHI}(M_T,\gamma_T)\cong \bigoplus_{I \in \{+,-\}^t} \mathit{SHI}(M_T^I,\gamma_T^I), \]

by Proposition 3.3. Each $(M_T^I,\gamma _T^I)$ is simply a union of 3-balls

\[ M_T^I\cong \mathbb{D}_{\alpha_1}\cup \cdots \cup \mathbb{D}_{\alpha_k}\cup\mathbb{D}_{\beta_1}\cup \cdots \cup \mathbb{D}_{\beta_k}, \]

which means that $\mathit {SHI}(M_T^I,\gamma _T^I)$ is either $\mathbb {C}$ or trivial, according as whether $\gamma _T^I$ has exactly one component on the boundary of each of these 3-balls or not. We claim that the nonzero summands $\mathit {SHI}(M_T^I,\gamma _T^I)$ are in one-to-one correspondence with the elements of $\mathfrak {S}(\mathcal {H})$, which will then complete the proof.

For this claim, we consider the restriction of $\gamma _T^I$ to the ball $\mathbb {D}_{\alpha _i}$. Let

\[ q_{\ell_1},\ldots,q_{\ell_p}\in\{q_1,\ldots,q_t\} \]

denote the intersection points between $\alpha _i$ and $\beta$. Then $\gamma _T^I$ restricts to exactly one component on the boundary of $\mathbb {D}_{\alpha _i}$ if and only if exactly one of $\epsilon _{\ell _1},\ldots \epsilon _{\ell _p}$ is $-$ and the rest are $+$. The analogous statement holds for the restriction of $\gamma _T^I$ to $\mathbb {D}_{\beta _j}$. Thus, if we let

\[ q(I) = \{q_{\ell}\mid \epsilon_\ell = -\}, \]

then $\gamma _T^I$ restricts to exactly one component on each 3-ball in $M_T^I$ if and only if $q(I)\in \mathfrak {S}(\mathcal {H})$.

Remark 3.5 Proposition 3.1 also follows from the fact that $\mathit {SHI}$ and $\mathit {SFH}$ obey the same decomposition laws (Proposition 3.3, and the invariance under product disk decomposition), agree in rank for sutured 3-balls, and

\[ \textrm{rk}_\mathbb{Z}\mathit{SFH}(M_T,\gamma_T) = |\mathfrak{S}(\mathcal{H})|, \]

per Remark 3.2. Our original proof has the advantages that it does not rely on the definition of the differential in $\mathit {SFC}$, and it establishes a very concrete bijection between the nonzero summands $\mathit {SHI}(M_T^I,\gamma _T^I)$ and elements of $\mathfrak {S}(\mathcal {H})$.

3.2 The proof

Recall that a sutured Heegaard diagram $\mathcal {H}$ for a balanced sutured manifold $(M,\gamma )$ is admissible if and only if every nontrivial periodic domain has both positive and negative multiplicities [Reference JuhászJuh06]. This is automatically true of any $\mathcal {H}$ when $H_1(M,\partial M;\mathbb {Q})=0$ (there are no nontrivial periodic domains in this case), though every balanced sutured manifold admits an admissible diagram.

Proof of Theorem 1.1 Let $\mathcal {H} = (\Sigma,\alpha,\beta )$ be an admissible sutured Heegaard diagram for $(M,\gamma )$. Then we can assign a positive integer area $a_i$ to each region $R_i$ of $\Sigma$–$\alpha$–$\beta$ disjoint from $\partial \Sigma$, so that the signed area of every periodic domain is zero; see [Reference Ozsváth and SzabóOS04b, Lemma 4.12].Footnote ⁴ Fix $a_i$ distinct points $p_{i1},\ldots,p_{ia_i}\in R_i$ for each $i$, and let

\[ T = \bigcup_{i=1}^n\bigcup_{j=1}^{a_i} T_{ij} \subset M \]

be the corresponding full tangle for $\mathcal {H}$, as in § 3.1.

We claim that $[T]=0$ in $H_1(M,\partial M;\mathbb {Q})$. To see this, note that for every periodic domain $P$ of $\mathcal {H}$, the intersection number of $T$ with the 2-cycle in $M$ represented by $P$ is negative the signed area of $P$, which is zero. Since the homology classes represented by periodic domains span $H_2(M)$, the claim follows. Theorem 1.9 therefore implies that

\[ \dim_\mathbb{C}\mathit{SHI}(M,\gamma)\leq \dim_\mathbb{C}\mathit{SHI}(M_{T},\gamma_{T}). \]

Theorem 1.1 then follows from the fact that

\[ \dim_\mathbb{C}\mathit{SHI}(M_{T},\gamma_{T})=|\mathfrak{S}(\mathcal{H})|, \]

by Proposition 3.1.

Remark 3.6 It is not true that the inequality

\[ \dim_\mathbb{C}\mathit{SHI}(M,\gamma)\leq |\mathfrak{S}(\mathcal{H})| \]

holds for any sutured Heegaard diagram $\mathcal {H}$ for an arbitrary balanced sutured $(M,\gamma )$. For example, consider the diagram $\mathcal {H} = (T^2-D^2,\alpha _1,\beta _1)$ for

\[ (M,\gamma) = ((S^1\times S^2)(1),\delta) \]

in which $\alpha _1$ and $\beta _1$ are disjoint curves on the punctured torus. In this case, we know that

\[ \dim_\mathbb{C}\mathit{SHI}(M,\gamma) = 2, \]

while $|\mathfrak {S}(\mathcal {H})|=0$. The issue here is that $\mathcal {H}$ is not admissible.

4. Further directions

Let $\mathcal {H}$ be an admissible sutured Heegaard diagram for a balanced sutured manifold $(M,\gamma )$. Let $T$ be a full tangle for $\mathcal {H}$, as defined in § 3. In ongoing work, we prove that

\[ \mathit{SHI}(-M_T,-\gamma_T)\cong\mathbb{C}^{|\mathfrak{S}(\mathcal{H})|} \]

has a basis given by the contact invariants of the tight contact structures on $(M_T,\gamma _T)$. In particular, this sutured instanton homology group is naturally graded by homotopy classes of 2-plane fields. We discuss potential applications of this fact in the following.

Let $T'$ be a mixed tangle for $T$, as defined in § 2. Let $V_{T}$ and $V_m$ be the groups

\begin{align*} V_T&=\mathit{SHI}(-M_{T'},-\gamma_{T'},(A_2^-,\ldots,A_n^-), (0,\ldots,0)),\\ V_m&=\mathit{SHI}(-M_{T'},-\Gamma_m, (A_2,\ldots,A_n), (0,\ldots,0)) \end{align*}

from § 2, for $m\in \mathbb {N}$. To prove Theorem 1.1, we proved that $V_T\cong \mathit {SHI}(-M_T,-\gamma _T)$ and

\[ \dim_\mathbb{C} \mathit{SHI}(-M,-\gamma) = \dim_\mathbb{C} V_{m+1} - \dim_\mathbb{C} V_m\leq \dim_\mathbb{C} V_T \]

for sufficiently large $m$. However, in fact, this inequality can be viewed as coming from a spectral sequence similar to that in [Reference Li and YeLY22, § 4]. This spectral sequence can also be described as follows. From the bypass exact triangle used in § 2, we have that

\[ V_T\cong H_*(\textrm{Cone}(\psi_-:V_m\to V_{m+1})), \]

where $\psi _-$ is the map associated to the bypass attachment along the arc $\eta _-$. One can prove, on the other hand, that

\[ \mathit{SHI}(-M,-\gamma)\cong H_*(\textrm{Cone}(\psi_--\psi_+:V_m\to V_{m+1})) \]

for $m$ large, where $\psi _+$ is a related bypass attachment map. The groups $V_m$ and $V_{m+1}$ can be graded using the rational Seifert surface for $T'$, as in [Reference LiLi21]. After adjusting this grading by an overall shift, the map $\psi _-$ is grading-preserving while $\psi _+$ decreases the grading by $1$, for $m$ large. The complex $\textrm {Cone}(\psi _--\psi _+)$ is then filtered, and the $E_1$ page of the associated spectral sequence is $H_*(\textrm {Cone}(\psi _-))$. In sum, we have a spectral sequence

(4.1)\begin{align} \mathbb{C}^{|\mathfrak{S}(\mathcal{H})|}\cong V_T\cong H_*(\textrm{Cone}(\psi_-)) \implies H_*(\textrm{Cone}(\psi_--\psi_+))\cong \mathit{SHI}(-M,-\gamma). \end{align}

The first potential application of these ideas involves defining a grading on $\mathit {SHI}(-M,-\gamma )$ by homotopy classes of plane fields. Indeed, $\mathit {SHI}(-M_T,-\gamma _T)$ has such a grading, as mentioned previously, as it is generated by contact invariants of contact structures. The manifold $(M_{T'},\gamma _{T'})$ is obtained by gluing $(M_T,\gamma _T)$ along annuli, as in § 2, and we believe that the tight contact structures on the latter glue to give tight contact structures on the former whose invariants form a basis for $V_T$. Thus, there should be a natural grading by homotopy classes of 2-plane fields on $V_T$ as well. The bypass maps $\psi _-,\psi _+$ are natural from a contact-geometric standpoint, and should therefore shift plane field gradings in a sensible way. We expect that one can then use the relation between $V_T$ and $\textrm {Cone}(\psi _-)$ and the structure of the latter to define a plane field grading on $\textrm {Cone}(\psi _-)$, and then on $\textrm {Cone}(\psi _--\psi _+)$.

A grading by homotopy classes of 2-plane fields on $\mathit {SHI}$ would enable one to define $\textrm {Spin}^c$ decompositions of these groups, as well as an analogue of the Maslov grading in Heegaard Floer homology (see [Reference Li and YeLY22, § 4] for another approach to such a decomposition). The current lack of such structure makes it difficult to translate arguments from the Heegaard Floer setting to the instanton Floer setting.

A related second application is towards proving the isomorphism (1.1). Indeed, there is some hope that one could understand the spectral sequence (4.1) purely in terms of contact geometry, and thereby obtain a more axiomatic proof that

(4.2)\begin{equation} \mathit{SHI}(M,\gamma) \cong \mathit{SFH}(M,\gamma)\otimes \mathbb{C}\cong \mathit{SHM}(M,\gamma)\otimes \mathbb{C}, \end{equation}

since the analogous spectral sequences can be defined in the Heegaard Floer and monopole Floer settings by the same contact-geometric means.

A more pedestrian approach to (4.2) is the following: first, prove that one can understand the spectral sequence

\begin{align*} \mathbb{C}^{|\mathfrak{S}(\mathcal{H})|}\implies \mathit{SHI}(-M,-\gamma) \end{align*}

as coming from a differential

\[ \partial=\partial_1 + \partial_2 + \ldots:\mathbb{C}^{|\mathfrak{S}(\mathcal{H})|}\to \mathbb{C}^{|\mathfrak{S}(\mathcal{H})|}, \]

where $\partial _k$ shifts a grading (coming from homotopy classes of 2-plane fields) on $\mathbb {C}^{|\mathfrak {S}(\mathcal {H})|}\cong V_T$ by $k$, such that

\[ \mathit{SHI}(-M,-\gamma)\cong H_*(\mathbb{C}^{|\mathfrak{S}(\mathcal{H})|},\partial). \]

Then, for generators $x,y\in \mathfrak {S}(\mathcal {H})$ and the corresponding basis elements $e_x,e_y\in \mathbb {C}^{|\mathfrak {S}(\mathcal {H})|}$, perhaps one could use the 2-plane field gradings to show that the coefficient

(4.3)\begin{equation} \langle \partial e_x,e_y\rangle \end{equation}

is nonzero only if there is a homotopy class of Whitney disks

(4.4)\begin{equation} \varphi\in\pi_2(x,y) \end{equation}

with positive domain in $\mathcal {H}$ and Maslov index one. If even this were true, then one could prove, for example, that the inequality in Corollary 1.7 is an equality,

\[ \dim_\mathbb{C}\mathit{KHI}(L(p,q),K)\leq \textrm{rk}_\mathbb{Z}\widehat{\mathit{HFK}}(L(p,q),K) \]

for $(1,1)$-knots $K\subset L(p,q)$.

More generally, the hope would be that for a nice diagram $\mathcal {H}$ (one in which the regions of $\Sigma$–$\alpha$–$\beta$ disjoint from $\partial \Sigma$ are bigons or rectangles), one could show that the coefficient (4.3) is nonzero if and only if there is a class as in (4.4) with positive domain and Maslov index one (the domain of such a class is necessarily an empty embedded bigon or rectangle in this case). These are precisely the domains counted in the differential on $\mathit {SFC}(\mathcal {H})$ in this case, by [Reference Sarkar and WangSW10]. If one could further show that these domains are counted with the same nonzero complex coefficients in both the sutured instanton and Heegaard Floer settings, then this would prove (1.1) and then (4.2) by the same methods. If one could work with coefficients in $\mathbb {F}=\mathbb {Z}/2\mathbb {Z}$, then the last step would be unnecessary, as all nonzero elements of this field are equal. Unfortunately, $\mathit {SHI}$ is not defined over $\mathbb {F}$. On the other hand, $\mathit {SHM}$ is, and therefore the strategy outlined above minus the last step would be sufficient to give an alternative, more axiomatic proof that

\[ \mathit{SFH}(M,\gamma;\mathbb{F})\cong \mathit{SHM}(M,\gamma;\mathbb{F}). \]