A Homological Action on Sutured Instanton Homology

For the sake of exposition, we write

$$\mathscr{D}=\mathscr{D}_{1}=(Y_{1},R_{1},r_{1},m_{1},\eta_{1},\alpha_{1}),$$

$$\mathscr{D}^{\prime}=\mathscr{D}_{2}=(Y_{2},R_{2},r_{2},m_{2},\eta_{2},\alpha_{2}).$$

Roughly speaking, the map $\underline{\Psi}_{\mathscr{D},\mathscr{D}^{\prime}}$ is defined as a composition

$$\underline{\Psi}_{\mathscr{D},\mathscr{D}^{\prime}}\coloneqq\Theta^{C}_{(Y_{1})_{+}Y_{2}}\circ I_{*}(X_{+}|R_{1})_{(\alpha_{1}+\eta_{1})\times[0,1]}\circ\left(I_{*}(X_{-}|R_{1})_{(\alpha_{1}+\eta_{1})\times[0,1]}\right)^{-1}.$$

Here

• •

  $C$ is a fixed diffeomorphism

  $$C\colon Y_{1}\backslash\operatorname{int}(\operatorname{im}(r_{1}))\to Y_{2}\backslash\operatorname{int}(\operatorname{im}(r_{2})),$$

  which restricts to $m_{2}\circ m_{1}^{-1}$ on $m_{1}(M\backslash N(\gamma))$ for some tubular neighbourhood $N(\gamma)$ and sends $\alpha_{1}\cap(Y_{1}\backslash\operatorname{int}(\operatorname{im}(r_{1})))$ to $\alpha_{2}\cap(Y_{2}\backslash\operatorname{int}(\operatorname{im}(r_{2})))$;

• •

  $X_{-}\colon(Y_{1})_{-}\to Y_{1}$ and $X_{+}\colon(Y_{1})_{-}\to(Y_{1})_{+}$ are certain cobordisms, corresponding to some $\pm 1$-surgeries on $Y_{1}$. The manifold $(Y_{1})+$ is diffeomorphic to $Y_{2}$ via a map that restricts to $C$ on $Y_{1}\backslash\operatorname{int}(\operatorname{im}(r_{1}))$ and to a specific diffeomorphism on a neighbourhood of $r_{1}(R_{1}\times{0})$. This map belongs to a unique isotopy class according to [baldwin2015naturality, Appendix A], and $\Theta^{C}_{(Y_{1})_{+}Y_{2}}$ is the isomorphism on instanton homology associated with this isotopy class.

Hence, it suffices to show that the maps induced by the cobordisms $X_{-}$ and $X_{+}$ and the map $\Theta^{C}_{(Y_{1})_{+}Y_{2}}$ respect the action of $\zeta$.

We can assume that $X_{-}$ and $X_{+}$ are corresponding to a single Dehn twist. In this case, the cobordism $X_{-}$ is constructed by taking the product cobordism $Y\times I$, where $Y=(Y_{1})_{-}$, and attaching a $(-1)$-framed $2$-handle along a curve $r_{1}(a\times{t})\times{1}$ in $Y\times{1}$, where $a$ is a simple closed curve in $R_{1}$ and $t\in(-1,1)$.

Since the Dehn twist is a local operation that does not affect the embedding $m_{1}$, we can consider the tangle closure $\widetilde{T}$, which appears in the definition of the action, as living in both $Y_{1}$ and $(Y_{1})-$. These two instances of $\widetilde{T}$ are related by the product $\widetilde{T}\times I$ contained in $X-$. The action $\mu([\widetilde{T}])$ on $I_{*}(Y_{1})$ (resp. $I_{*}((Y_{1})_{-})$) coincides with the cobordism map $I_{*}(Y_{1}\times I,[\widetilde{T}]\times I)$ (resp. $I_{*}((Y_{1})_{-}\times I,[\widetilde{T}]\times I)$). By the functoriality of instanton Floer homology (cf. [Kronheimer2007MonopolesAT, Theorem 3.4.4] in the context of monopoles),

$$I_{*}(Y_{1}\times I,[\widetilde{T}]\times I)\circ I_{*}(X_{-},[\widetilde{T}]\times I)=I_{*}(X_{-},[\widetilde{T}]\times I)\circ I_{*}((Y_{1})_{-}\times I,[\widetilde{T}]\times I).$$

The action descends onto $I_{*}(Y_{1}|R_{1})$ and $I_{*}((Y_{1})_{-}|R_{1})$ in an equivariant manner, as the operator $\mu([\widetilde{T}])$ commutes with $\mu(R)$ and $\mu(y)$. The same argument holds for $I_{*}(X_{+})$.

We now show the map $\Theta^{C}_{(Y_{1})_{+}Y_{2}}$ also respects the action. Let

$$f\colon(Y_{1})+\to Y_{2}$$

be an orientation-preserving diffeomorphism such that $f$ restricts to $m_{2}\circ m_{1}^{-1}$ on $m_{1}(M\backslash N(\gamma))$ to some certain map on $r_{1}(R\times\{0\})$. We can choose a tangle $T\subset M$ such that it is balanced in both $(Y_{1})_{+}$ and $Y_{2}$ and form the closures $\widetilde{T}_{1}$ and $\widetilde{T}_{2}$ respectively. Recall that $\widetilde{T}_{1}$ is the concatenation of $m_{1}(T)$ and a product tangle $T^{\prime}$ in the image of $r_{1}$. The diffeomorphism $f$ restricts to an orientation-preserving diffeomorphism $f_{1}$ from $R_{1}\times I$ to $R_{2}\times I$ (more explicitly, their images under $r_{1}$ and $r_{2}$), which sends boundaries to boundaries. It induces an isomorphism

$$(f_{1})_{*}\colon H_{1}(R_{1}\times I,\partial(R_{1}\times I))\to H_{1}(R_{2}\times I,\partial(R_{2}\times I)),$$

which must be the identity as $f_{1}$ is orientation-preserving and

$$H_{1}(R_{1}\times I,\partial(R_{1}\times I))\cong\mathbb{Z}.$$

Hence, $f(\widetilde{T}_{1})$ is homologous to $\widetilde{T}_{2}$, and they induce the same action on $I_{*}(Y_{2}|R_{2})$. ∎

In the case where $g(\mathscr{D}^{\prime})=g(\mathscr{D})-1$, the map is defined as the inverse of $\Phi_{\mathscr{D}^{\prime},\mathscr{D}}$, so we only need to prove the result for the case where $g(\mathscr{D}^{\prime})=g(\mathscr{D})+1$. In this situation, the map $\Phi_{\mathscr{D},\mathscr{D}^{\prime}}$ is defined as a composition

$$\Phi_{\mathscr{D}_{3},\mathscr{D}_{4}}\circ\Phi_{\mathscr{D}_{2},\mathscr{D}_{3}}\circ\Phi_{\mathscr{D}_{1},\mathscr{D}_{2}},$$

where $\mathscr{D}_{1}=\mathscr{D}$, $\mathscr{D}_{4}=\mathscr{D}^{\prime}$, $g(\mathscr{D}_{3})-1=g(\mathscr{D}_{2})=g(\mathscr{D}_{1})$. For the sake of exposition, we write

$$\mathscr{D}_{2}=(Y_{2},R_{2},r_{2},m_{2},\eta_{2},\alpha_{2}),$$

$$\mathscr{D}_{3}=(Y_{3},R_{3},r_{3},m_{3},\eta_{3},\alpha_{3}).$$

The closure $\mathscr{d}3$ is a “cut-ready” closure, as defined in [baldwin2015naturality, Section 5.2], modified from $\mathscr{D}_{4}$. Cutting along two tori in $Y_{3}$ and re-gluing yields a two-component $3$-manifold, one of which is diffeomorphic to a mapping torus of the surface $\Sigma_{1,2}$ (a genus one surface with two boundary components), and the other is $Y_{2}$. The maps$\Phi_{\mathscr{D}_{3},\mathscr{D}_{4}}$ and $\Phi_{\mathscr{D}_{1},\mathscr{D}_{2}}$ are well-defined since they are maps between closures of the same genus, and they are equivariant by Proposition 3.1. The map $\Phi_{\mathscr{D}_{2},\mathscr{D}_{3}}$ is induced by a merge-type splicing cobordism corresponding to a torus excision, as described in [kronheimer2010knots, Theorem 7.7]. Since torus excision is a local operation outside $m_{3}(M)$, a similar functoriality argument as in Proposition 3.1 applies here. ∎

Let $\mathscr{D}=\mathscr{D}_{0},\mathscr{D}_{1},\dots,\mathscr{D}_{n}=\mathscr{D}^{\prime}$ be a sequence of marked odd closures of $(M,\gamma)$ such that $|g(\mathscr{D}_{i})-g(\mathscr{D}_{i+1})|\leq 1$. Then the map $\Phi_{\mathscr{D}_{i},\mathscr{D}_{i+1}}$ has been defined previously, and $\Phi_{\mathscr{D},\mathscr{D}^{\prime}}$ is defined as the composition of $\Phi_{\mathscr{D}_{i},\mathscr{D}_{i+1}}$. By Proposition 3.1 and 3.2, it is equivariant. ∎

As explained in [kronheimer2010instanton], one may use disconnected surfaces that satisfy certain conditions as the auxiliary surface. Hence we can form a marked odd closure $\mathscr{D}$ of $(M^{\prime},\gamma^{\prime})$ that also serves as a marked odd closure of $(M,\gamma)$. Hence, the $\mu$-action of the tangle closure $[\widetilde{T}]$ gives the action on $\operatorname{SHI}(M,\gamma)$ and $\operatorname{SHI}(M^{\prime},\gamma^{\prime})$ simultaneously. ∎

Again, we can use disconnected auxiliary surfaces to form closures. The result then follows from the statement in the closed case. ∎

As in [li2020contact, Lemma 4.9], we have

$$(M,\gamma)\cong(M_{1}\sqcup M_{2}\sqcup S^{3}(2),\gamma_{1}\cup\gamma_{2}\cup\delta^{2})\cup h_{1}\cup h_{2}.$$

Here $h_{i}$ is a contact $1$-handle connecting one boundary component of $S^{3}(2)$ to the boundary of $M_{i}$. The first statement then follows from Lemma 4.1 and 4.2. Precisely speaking, let $\zeta_{1}+\zeta_{2}+\zeta^{\prime}$ be the image of $\zeta$ in

$$H_{1}(M_{1},\partial M_{2})\oplus H_{1}(M_{2},\partial M_{2})\oplus H_{1}(S^{3}(2),\partial(S^{3}(2))).$$

Then the action on the right hand side is given by

$$X(a\otimes b\otimes c)=(X_{\zeta_{1}}a)\otimes b\otimes c+a\otimes(X_{\zeta_{2}}b)\otimes c+a\otimes b\otimes(X_{\zeta^{\prime}}c).$$

The remaining task is to calculate the action on $\operatorname{SHI}(S^{3}(2))$. Notice that $S^{1}\times S^{2}(1)$ can be obtained by $S^{3}(1)$ with one contact $1$-handle attached. We have

$$\operatorname{SHI}(S^{3}(2))\cong\operatorname{SHI}(S^{1}\times S^{2}(1))=\operatorname{I}^{\sharp}(S^{1}\times S^{2}).$$

The last group is calculated in [scaduto2015instantons]. As a $\mathbb{C}$-module, it has two $\mathbb{C}$-summands lying in grading $2$ and $3$. Two summands are generated by the relative invariants $[S^{1}\times D^{3}]^{\sharp}$ and $[D^{2}\times S^{2}]^{\sharp}$ (see [scaduto2015instantons, Section 7.2] for the meaning of notations) that are induced by the cobordisms

$$S^{1}\times D^{3}\colon\emptyset\to S^{1}\times S^{2}$$

and

$$D^{2}\times S^{2}\colon\emptyset\to S^{1}\times S^{2}$$

respectively. Let $\gamma$ be an embedded circle $S^{1}\times\{pt\}\subset S^{1}\times S^{2}$. Then $\gamma$ generates $H_{1}(S^{1}\times S^{2})$, and $\mu(\gamma)$ maps $[S^{1}\times D^{3}]^{\sharp}$ to $[D^{2}\times S^{2}]^{\sharp}$ by an adaptation of [donaldson2002floer, Theorem 7.16]. Hence, $\operatorname{I}^{\sharp}(S^{1}\times S^{2})$ is a free $\mathbb{C}[X]/X^{2}$-module of rank $1$ with respect to the action of any nonzero $\eta\in H_{1}(S^{1}\times S^{2})$. The class $\zeta^{\prime}$ is nonzero in $H_{1}(S^{3}(2),\partial(S^{3}(2)))=H_{1}(S^{1}\times S^{2})$ if and only if the algebraic intersection number of $\zeta$ and $S$ is non-zero. The result follows. ∎

We have

$$(M,\gamma)\cong(M_{1}\sqcup Y(1),\gamma_{1}\cup\delta)\cup h.$$

Here $h$ is a contact $1$-handle connecting $M_{1}$ and $Y$. An argument as in Proposition 4.3 applies here, too. ∎

Recall that the proof of [kronheimer2010knots, Proposition 6.9] relies on a reduction to a special case as demonstrated in [kronheimer2010knots, Lemma 6.10]. This reduction only involves the addition of $1$-handles, which preserves the $\mathbb{C}[X]/X^{2}$-module structure by Lemma 4.1.

In this special case, we can assume that $S$ has no closed components, and the oriented boundary of $\partial S$ consists of $n$ simple closed curves $C_{1}^{+},C_{2}^{+},\dots,C_{n}^{+}$ that are linearly independent in $H_{1}(R_{+}(\gamma))$, and $C_{1}^{-},C_{2}^{-},\dots,C_{n}^{-}$ that are linearly independent in $H_{1}(R_{-}(\gamma))$.

Choosing the representative tangle $T$ of $\zeta$ appropriately, we can form a closure $Y=Y(M,\gamma)$ using an auxiliary surface $G$ and a gluing diffeomorphism $h$ as usual with additional requirements that $h$ maps $C_{i}^{+}$ to $C_{i}^{-}$ and $R_{+}(\gamma)\cap T$ to $R_{-}(\gamma)\cap T$. Then $Y$ contains two distinguished closed surfaces: first the usual surface $R$, and second a surface $\overline{S}$ obtained by closing up $S$. The action of $\zeta$ on $\operatorname{SHI}(M,\gamma)$ can be identified with the action of the tangle closure $\mu(\widetilde{T})$ on $I_{*}(Y|R)$.

We now form a closure $Y^{\prime}$ of $(M^{\prime},\gamma^{\prime})$. Cutting along $S$ results in $2n$ new sutures $D_{i}^{\pm}\,(i=1,2,\dots,n)$ corresponding to $C_{i}^{\pm}$. To form the closure, we use an auxiliary surface $G^{\prime}=G\cup G_{1}$, where $G_{1}$ is a collection of $n$ annuli. We glue $G_{1}\times[-1,1]$ to $(M^{\prime},\gamma^{\prime})$ by identifying the sutures $(\partial G_{1})\times\{0\}$ with $D_{i}^{\pm}$. The preclosure has two boundary components

$$\overline{R}^{\prime}_{\pm}=\overline{R}^{\dagger}_{\pm}\cup S_{\pm}\cup G_{1}\times\{\pm 1\},$$

where $\overline{R}^{\dagger}_{\pm}$ are obtained by cutting open $\overline{R}_{\pm}$ along the circles $C_{i}^{\pm}$, and $S_{\pm}$ are copies of $S$. We can then choose a diffeomorphism $h^{\prime}\colon\overline{R}^{\prime}_{+}\to\overline{R}^{\prime}_{-}$ such that it coincides with $h$ on $\overline{R}^{\dagger}_{-}$ and equals to identify on $S_{+}\cup G_{1}\times\{+1\}$. The resulting closure $Y^{\prime}$ contains a distinguished surface $\overline{R}^{\prime}$. Let $T^{\prime}$ be the result of cutting out $T$ along $S$. Then $T^{\prime}$ is a representative of $\iota_{*}(\zeta)$. Let $\widetilde{T}^{\prime}$ be the tangle closure of $T^{\prime}$ in $Y^{\prime}$, and then the action of $\iota_{*}(\zeta)$ can be identified with the action of $\mu(\widetilde{T}^{\prime})$ on $I_{*}(Y^{\prime}|\overline{R}^{\prime})$.