(e-1-\epsilon) md -(1+\epsilon)k  < \gen{c_1(\spincv), [P_\gamma]} &<  (e+1+\epsilon)md + (1+\epsilon)k \\
%(-1-\epsilon)(k+md) + k+md < \frac{\nu}{m}\gen{c_1(\spincv), [R]} < (1+\epsilon)(k+md) + k+md
(e-1-\epsilon)(k+md) < \frac{\nu}{m} \gen{c_1(\spincv), [R]} &< (e+1+\epsilon)(k+md).
\end{align*}

Comparing with (<ref>) and (<ref>) respectively, for sufficiently large $m$, this implies that $p,q \in \{0,e\}$. Also through (<ref>)and (<ref>), we have

\[
2(q-p-1)md + (2q-1)k < ke < 2(q-p+1)md + (2q+1)k.
\]

It then follows that $p=q$, as required.

§.§.§ The map $h_1^t$

We recall the definition of the truncated map $h^t_{1}$ from Hedden-Levine's argument.

Let $\tilde h^t_1$ denote the sum of all terms $h^t_{1,\spincv}$ for which $\spincv$ satisfies

\begin{equation} \label{eq: h1-trunc-Q-bound}
\gen{c_1(\spincv), [Q]} = \pm m.
\end{equation}

Similar to the previous case, the truncated map is enough to fulfill the required condition in Proposition <ref>, as indicated by the following lemma.

Fix $t \in \N$. For all $m$ sufficiently large, $\tilde h_1^t$ is a null-homotopy of $f_2^t \circ f_1^t$.

An extra spin$^c$ constraint is needed in the proof, given by the Maslov grading bound.

Fix $t \in \N$. For all $m$ sufficiently large, if $\spincv$ is any spin$^c$ structure with $\abs{\gen{c_1(\spincv), [Q]}} = m$, and $h^t_{1,\spincv} \ne 0$, then $\abs{\gen{c_1(\spincv), [P_\gamma]}} < md$.

We are ready to prove the following proposition, parallel to <cit.>.

Fix $t \in \N$. For all $m$ sufficiently large, the map $\tilde h_1^t$ is filtered with respect to the filtrations $\JJ_{\alpha\gamma}$ and $\JJ_{\alpha\beta}$.

Suppose $\spincv$ is a spin$^c$ structure for which $h^t_{1,\spincv} \ne 0$, and assume $\gen{c_1(\spincv), [Q]} = em$, where $e=\pm 1.$  Let $\q \in \T_\alpha \cap \T_\gamma$ and $\x \in \T_\alpha \cap \T_\beta$ and  be generators such that   $\spincs_z(\q)=\spincv|_{Y_\lambda}$ and $\spincs_z(\x)=\spincv|_Y$.

Similar as before, note that $\theta \circ h^\circ_{1,\spincv}$  lands in the summand $[\x,i] \otimes T^r \GR$, where $r\in \Q$ is given by

\[
\frac{-k-md}{2d} \le r < \frac{-k+md}{2d} \quad \text{and} \quad r \equiv -\frac{1}{2d}( \gen{c_1(\spincv), [P_\delta]} +k+md) \pmod m,
\]

so for some $p\in \Z$, we can write

\[
\gen{c_1(\spincv), [P_\delta]} = -2dr-k + (2p-1) md,
\]

which implies that

\begin{equation} \label{eq: h1-filt-p-bounds}
(2p-2)md < \gen{c_1(\spincv), [P_\delta]} \le (2p)md.
\end{equation}

By Lemma <ref>, we may assume that $\abs{\gen{c_1(\spincv), [P_\gamma]}} < md$. Recall that $[P_\gamma] = [P_\delta] + d[Q]$, and therefore

\[
\gen{c_1(\spincv), [P_\delta]} = \gen{c_1(\spincv), [P_\gamma]} - emd.
\]

If $e = 1$, this implies $p=0$; if $e = -1$, this implies $p=1$. Either way, it holds that
By (<ref>), we have

\begin{align*}
(2p-2)nm & =  -\gen{c_1(\spincv), n[Q]} -nm\\
&=-2m \left( -n_z(\rho) + n_{z_n}(\rho) -  \sum_{j=1}^n\left( n_{z_j}(\rho) - n_{u_j}(\rho) \right)  \right) +nm -nm\\
&=2m n_z(\rho) - 2m n_{z_n}(\rho) + 2m \sum_{j=1}^n\left( n_{z_j}(\rho) - n_{u_j}(\rho) \right).
\end{align*}

Now, we compute:

\begin{align*}
&\JJ_{\alpha\gamma}([\q,i]) - \JJ_{\alpha\beta}([\x,i-n_z(\rho)] \otimes T^r) \\
&= \AlNorm_{w,z_n}(\q) +  \frac{2nd r +nk+n^2 d}{2k} +n_z(\rho) \\
&= \frac{1}{2k} \left( 2\Al(\q) + 2ndr + nk + n^2 d \right) +n_z(\rho) \\
&= \frac{1}{2k} \left( 2\Al(\q)  - n\gen{c_1(\spincv),[P_\delta]}  + (2p-1)nmd + n^2 d \right) +n_z(\rho) \\
&= \frac{1}{2k} \left( 2\Al(\q) -2\Al(\x) + 2d n_w(\rho) - 2dn_z(\rho) +  (2p-1)nmd -n(k+md) + n^2 d \right) +n_z(\rho) \\
&= \frac{1}{2k} \Big( 2\MultComb(\rho) +  2d n_w(\rho) - 2dn_z(\rho) + (2p-2)nmd \Big) +n_z(\rho) \\
&= \frac{1}{k} \Big( -nd n_w(\rho) - (k+md -nd) n_z(\rho) + (k+md)n_{z_n}(\rho) -md\sum_{j=1}^n\left( n_{z_j}(\rho) - n_{u_j}(\rho) \right)\\
&\qquad +  d n_w(\rho) - dn_z(\rho) +  md n_z(\rho) -md n_{z_n}(\rho) +md  \sum_{j=1}^n\left( n_{z_j}(\rho) - n_{u_j}(\rho) \right)   \Big) +n_z(\rho) \\
&= n_{z_n}(\rho) \ge 0.
\end{align*}

§.§.§ The map $h_2^t$

Let us start by recalling the definition of the truncated map $\tilde h^t_2$ from [14], which is similar to the previous case.
For any $\spincv \in \Spin^c_0(X_{\alpha\delta\beta\gamma})$,  let $\tilde h^t_2$ denote the sum of all terms $h^t_{2,\spincv}$ such that

\begin{equation} \label{eq: h2-trunc-Q-bound}
\gen{c_1(\spincv), [Q]} = \pm m.
\end{equation}

As before, the next two lemmas show that $\tilde h^t_2$ satisfies the first condition in Proposition <ref> and give an extra spin$^c$ evaluation bound that we will use in the proof, respectively.

Fix $t \in \N$. For all $m$ sufficiently large, $\tilde h_2^t$ is a null-homotopy of $f_0^t \circ f_2^t$.

Fix $t \in \N$ and $\epsilon > 0$. For all $m$ sufficiently large, if $\spincv$ is any spin$^c$ structure with $\abs{\gen{c_1(\spincv), [Q]}} = m$ (where $e$ is an odd integer), and $h^t_{1,\spincv} \ne 0$, then

\begin{equation} \label{eq: h2-trunc-R-bound}
\abs{\gen{c_1(\spincv), [R]}} < \frac{m(k+md)}{\nu}. %\qed
\end{equation}

The following proposition is parallel to <cit.>.

Fix $t \in \N$. For all $m$ sufficiently large, the map $\tilde h_2^t$ is filtered with respect to the filtrations $\JJ_{\alpha\delta}$ and $\JJ_{\alpha\gamma}$.

Let $\a \in \T_\alpha \in \T_\delta$ and $\q \in \T_\alpha \cap \T_\gamma$ be generators such that $\spincs_z(\q)=\spincv|_{Y_\lambda}$ and $\spincs_z(\a)=\spincv|_{Y_{\lambda+m\mu}}=\spincu$, and suppose $\rho \in \pi_2(\a, \Theta_{\delta\beta}, \Theta_{\beta\gamma}, \q)$ is a rectangle that contributes to $h^t_{2,\spincv}([\a,i])$. Associated with $\spincu$ is a
number $s_\spincu$, which by its definition (<ref>) satisfies

\[
-(k+md) < 2d s_\spincu \le k+md \quad \text{and} \quad  2nd s_\spincu \equiv  \gen{c_1(\spincv),[P_{n,\delta}]} + n (k+md)   \pmod {2n(k+md)}.
\]

Therefore, for some $q\in \Z$ we have

\[
\gen{c_1(\spincs_z(\rho)), [P_{n,\delta}]} = 2nd s_\spincu + (2q-1)n(k+md),
\]

so that

\[
(2q-2)(k+md) < \gen{c_1(\spincs_z(\rho)), [P_\delta]} \le 2q(k+md).
\]

Next we will assume that $\gen{c_1(\spincv), [Q]} = em$, where $e=\pm 1$, and $h^t_{2,\spincv} \ne 0$, so that $\spincv$ satisfies (<ref>). We have

\begin{gather*}
\gen{c_1(\spincv), [P_\delta]} = \frac{\nu}{m} \gen{c_1(\spincv), [R]} - \frac{k+md}{m} \gen{c_1(\spincv), [Q]}.
\end{gather*}

Using (<ref>), compare the range of both sides of the equation. If $e=1$, this implies $q=0$; if $e=-1$, this implies $q=1$. In either case, it holds that $2q-1 = -e$.

\begin{align*}
-(2q-1)m&= em=\gen{c_1(\spincs_z(\rho)), [Q]}\\
&=2n_w(\rho) - 2n_z(\rho) + m.
\end{align*}

We now compute:

\begin{align*}
&\JJ_{\alpha\delta}([\a,i]) - \JJ_{\alpha\gamma}([\q,i-n_w(\rho)] ) \\
&= \frac{\Al(\a)}{k+md}  + \frac{nd^2m(2s_\spincu-n)}{2k(k+md)} - \frac{\Al(\q)}{k}  + n_z(\rho) \\
&= \frac{2k \Al(\a) - 2(k+md)\Al(\q) + nd^2m(2s_\spincu-n)}{2k(k+md)}  + n_z(\rho) \\
&= \frac{2(k+md)( \Al(\a) - \Al(\q)) - 2md \Al(\a) + nd^2m(2s_\spincu-n)}{2k(k+md)}  + n_z(\rho) \\
&= \frac{1}{k} \left( \Al(\a) - \Al(\q) \right) + \frac{nd^2m(2s_\spincu-n)}{2k(k+md)} + n_z(\rho)  - \frac{md\Al(\a)}{2k(k+md)}\\
&= \frac{1}{k} \left( \MultComb(\rho) - \frac{nmd}{2} \right) + \frac{md\gen{c_1(\spincs_z(\rho)), [P_{n,\delta}]} -(2q-1)m\cdot nd(k+md) -n^2 d^2 m}{2k(k+md)}  + n_z(\rho) \\
& - \frac{ md\left( \gen{c_1(\spincs_z(\rho)), [P_{n,\delta}]} -2(k+md)\left( n_z(\rho) - n_{z_n}(\rho) + \sum_{j=1}^n\left( n_{z_j}(\rho) - n_{u_j}(\rho) \right) \right) - n^2 d \right) } {2k(k+md)} \\
&= \frac{1}{k} \left( \MultComb(\rho) - \frac{nmd}{2} \right)  + \frac{nd(k+md)(2n_w(\rho) - 2n_z(\rho) + m )}{2k(k+md)}   + n_z(\rho) \\
&+ \frac{ 2md(k+md)\left( n_z(\rho) - n_{z_n}(\rho) + \sum_{j=1}^n\left( n_{z_j}(\rho) - n_{u_j}(\rho) \right) \right)}{2k(k+md)}\\
&= \frac{1}{k} \left( -nd n_w(\rho) - (k+md -nd) n_z(\rho) + (k+md) n_{z_n}(\rho) -md\sum_{j=1}^n\left( n_{z_j}(\rho) - n_{u_j}(\rho) \right)\right)      \\
& + \frac{ nd\left(n_w(\rho) - n_z(\rho)\right) + md \left( n_z(\rho) - n_{z_n}(\rho) + \sum_{j=1}^n\left( n_{z_j}(\rho) - n_{u_j}(\rho) \right)    \right) } {k} + n_z(\rho) \\
&= \frac{1}{k}\Big( -nd n_w(\rho) - (k -nd) n_z(\rho) + k n_{z_n}(\rho)  +  nd\left(n_w(\rho) - n_z(\rho)\right) \Big)       + n_z(\rho)\\
&= n_{z_n}(\rho) \\
&\ge 0.
\end{align*}

§.§ Pentagon maps

In this section we aim to prove the second part of Proposition <ref>: $\tilde h^t_{j+1} \circ f^t_j +  f^t_{j+2} \circ \tilde h^t_j$ (where $j \in \Z/3$) are filtered quasi-isomorphisms. We will only focus on the case of $j=0$, which is the most technically difficult because of the twisted coefficients. The arguments for $j=1$ and $j=2$ are similar.
Again, our argument here is completely parallel to the argument presented in <cit.>, and we will leave out some of the technical details and focus on the adjustment needed due to a different setting. For a more detailed read, see <cit.>.

To begin, let $\tilde{\bm\beta} = (\tilde \beta_1, \dots, \tilde\beta_g)$ denote a small Hamiltonian isotopy of $\bm\beta$, such that each $\tilde \beta_i$ meets $\beta_i$ in a pair of points. We further require that $v$ is an extra reference point such that $v$ is in the same region of $\Sigma \minus (\bm\alpha \cup \bm\beta)$ as $z$ and in the same region of $\Sigma \minus (\bm\alpha \cup \tilde{\bm\beta})$ as $w$.  Finally, let $\Theta_{\beta\tilde\beta} \in \T_\beta \cap \T_{\tilde\beta}$ denote the canonical top-dimensional generator. [This is the same setting as depicted in <cit.>, although in their text description  the order of $w$ and $z$ is switched by mistake.]

\[
\tilde \Psi^t_0  = \tilde h^t_{1} \circ f^t_0 +  f^t_{2} \circ \tilde h^t_0.
\]

The fact that $\tilde \Psi^t_0$ is filtered follows immediately  from the previous sections, since it is a sum of filtered maps. In order to show it is a quasi-isomorphism, we relate it to the map

\begin{equation} \label{eq: Phi0}
\Phi_0^t \co \ul\CF^t(\Sigma, \bm\alpha, \bm\beta, z; \GR) \to \ul\CF^t(\Sigma, \bm\alpha, \tilde{\bm\beta}, z; \GR)
\end{equation}

given by

\begin{equation} \label{eq: Phi0-def}
\Phi_0^t(T^s \cdot [\x, i]) =
\sum_{\tilde\y \in \T_\alpha \cap \T_{\tilde\beta}} \sum_{\substack{\psi \in \pi_2(\x, \Theta_{\beta\tilde\beta}, \tilde\y) \\ \mu(\psi)=0 }} \#\MM(\psi) \, T^{s + n_w(\psi)-n_z(\psi)} \cdot [\tilde\y, i-n_z(\psi) ].
\end{equation}

By the work of [17], $\Phi_0^t$ is a chain isomorphism. Moreover, for any $\psi \in \pi_2(\x, \Theta_{\beta\tilde\beta}, \tilde\y)$, we have $\AlNorm(\x) - \AlNorm(\tilde\y) = n_z(\psi) - n_w(\psi)$. It follows that $\Phi_0^\circ$ is a filtered isomorphism with respect to $\JJ_{\alpha\beta}$ and $\JJ_{\alpha\tilde\beta}$.

The aim is then to show  $\tilde \Psi^t_0$ is filtered homotopy equivalent to  $\Phi_0^t$.   The argument hinges on the following pentagon-counting map, defined in [14]:

\begin{equation} \label{eq: g0-Phi0}
g_0^\circ \co \ul\CF^\circ(\Sigma, \bm\alpha, \bm\beta, z; \GR) \to \ul\CF^\circ(\Sigma, \bm\alpha, \tilde{\bm\beta}, z; \GR)
\end{equation}

given by

\begin{equation}
\label{eq: g0-def}
g_0^\circ(T^s \cdot [\x,i]) = \sum_{\y \in \T_\alpha \cap \T_{{\tilde \beta}}}  \sum_{\substack{\sigma \in \pi_2(\x, \Theta_{\beta\gamma}, \Theta_{\gamma\delta}, \Theta_{\delta{\tilde \beta}}, \y) \\ \mu(\sigma)=-2 \\ \mathclap{s + n_w(\sigma) - n_v(\sigma) \equiv 0 \pmod m}}}   \#\MM(\sigma) \, T^{n_v(\sigma) - n_z(\sigma)} [\y, i-n_z(\sigma)].
\end{equation}

Hedden and Levine defined a truncated version of $g_0^t$, again by throwing away bad terms, and they proceed to show the truncated map $\tilde g^t_0$ establishes the required filtered homotopy equivalence between $\tilde \Psi^t_0$ and $\Phi_0^t$. This argument works for our case as well, except that the grading shift on $\tilde g^t_0$ needs to be recalculated.

Before we discuss the filtration shifts, let us state analogues of the results of Section <ref> for pentagons. To begin, let $V$ be the $(\beta,\tilde\beta)$ periodic domain with $\partial V = \beta_g - \tilde\beta_g$. In other word, $V$ is the thin domain between  $\beta_g $ and $\tilde\beta_g$. We have  $n_v(V) = 1$, and $n_w(V) = n_z(V) = n_{z_j}(V) = n_{u_j}(V) = 0$ for $j=1,\cdots, n$. Let $\tilde P_\gamma$, $\tilde P_\delta$, and $\tilde Q$ be the analogues of $P_\gamma$, $P_\delta$, and $Q$ with $\beta$ circles replaced by ${\tilde \beta}$ circles: up to thin domains, we have

\[
[\tilde P_\gamma] = [P_\gamma] + k[V], \quad [\tilde P_\delta] = [P_\delta]+(k+md)[V], \quad \text{and} \quad
[\tilde Q] = [Q]-m[V].
\]

The Heegaard diagram determines a $4$-manifold $X_{\alpha\beta\gamma\delta\tilde\beta}$, which admits various decompositions into the pieces described in Section <ref>; for instance, we have

\[
X_{\alpha\beta\gamma\delta\tilde\beta} = X_{\alpha\beta\gamma\delta} \cup_{Y_{\alpha\delta}} X_{\alpha\delta\tilde\beta} = X_{\alpha\beta\gamma} \cup_{Y_{\alpha\gamma}} X_{\alpha\gamma\delta\tilde\beta} .
\]

In the intersection pairing form on $H_2(X_{\alpha\beta\gamma\delta\tilde\beta})$, we have

\[
[V] \cdot [V] = [V] \cdot [Q] = [V] \cdot [\tilde Q] = 0,
\]

and all other intersection numbers can be deduced accordingly. Compare the following with <cit.>.

For any $\x \in \T_\alpha \cap \T_\beta$, $\tilde\y \in \T_\alpha \cap \T_{\tilde\beta}$, and $\sigma \in \pi_2(\x, \Theta_{\beta\gamma}, \Theta_{\gamma\delta}, \Theta_{\delta\tilde\beta}, \tilde \y)$, we have:

\begin{align}
\label{eq: abgdb-alex}
n\Al(\x) - n\Al(\tilde \y) &= -nd n_w(\sigma) - (k+md - nd) n_z(\sigma) + (k+md) n_{z_n}(\sigma) \\
\nonumber & \qquad \qquad - md \sum_{j=1}^n\left( n_{z_j}(\sigma) - n_{u_j}(\sigma) \right)   \\
\label{eq: abgdb-c1-Pg}
\qquad \qquad \gen{c_1(\spincs_z(\sigma)), [P_\gamma]} &= 2 \Al(\x) + 2dn_w(\sigma) - 2d n_v(\sigma) + k \\
\label{eq: abgdb-c1-Pdt}
\qquad \qquad \gen{c_1(\spincs_z(\sigma)), [\tilde P_\delta]} &= 2 \Al(\tilde\y) + 2dn_z(\sigma) - 2d n_v(\sigma) + (k+md) \\
\label{eq: abgdb-c1-Q}
\qquad \qquad \gen{c_1(\spincs_z(\sigma)), n[Q]} &= -2m \sum_{j=1}^n\left( n_{z_j}(\sigma) - n_{u_j}(\sigma) \right) -nm \\
\label{eq: abgdb-c1-Qt}
\qquad \qquad \gen{c_1(\spincs_z(\sigma)), n[\tilde Q]} &= 2m\Big( -n_z(\sigma) + n_{z_n}(\sigma) - \sum_{j=1}^n\left( n_{z_j}(\sigma) - n_{u_j}(\sigma) \right) \Big) -nm \\
% \label{eq: abgdb-c1-V-single}
% \qquad \qquad \gen{c_1(\spincs_z(\sigma)), [V]} &= 2n_z(\sigma) - 2n_{z_1}(\sigma)=2n_{z_1}(\sigma) - 2n_{z_2}(\sigma)\\
% \nonumber &= \quad \cdots \quad = 2n_{z_{n-1}}(\sigma) - 2n_{z_n}(\sigma)\\
\label{eq: abgdb-c1-V}
\qquad \qquad \gen{c_1(\spincs_z(\sigma)), n[V]} &= 2n_z(\sigma) - 2n_{z_n}(\sigma)
\end{align}

This proof  resembles the one for Proposition <ref>, building upon the calculations made in Propositions <ref> and <ref>.

For any $\x \in \T_\alpha \cap \T_\beta$, $\tilde\y \in \T_\alpha \cap \T_{\tilde\beta}$, and $\sigma \in \pi_2(\x, \Theta_{\beta\gamma}, \Theta_{\gamma\delta}, \Theta_{\delta\tilde\beta}, \tilde \y)$, choose $\a \in \T_\alpha \cap \T_\delta$, $\rho \in \pi_2(\x, \Theta_{\beta\gamma}, \Theta_{\gamma\delta},\a)$ and $\psi \in \pi_2(\a,\Theta_{\delta\tilde\beta}, \tilde \y)$ such that $\spincs_z(\rho) = \spincs_z(\sigma)|_{X_{\alpha\beta\gamma\delta}}$ and $\spincs_z(\psi) = \spincs_z(\sigma)|_{X_{\alpha\delta{\tilde\beta}}}$. Up to adding copies of $\Sigma,$ which does not change the spin$^c$ structure condition, we may assume that $n_z(\sigma) = n_z(\rho) + n_z(\psi)$. Hence, $S = \DD(\sigma) - \DD(\rho * \psi)$ is a quintuply periodic domain with $n_z(S)=0$. Since the function $\MultComb$ vanishes on all periodic domains, we have:

\begin{align*}
n\Al(\x) - n\Al(\tilde \y) &= (n\Al(\x) - \Al(\a)) + (\Al(\a) - n\Al(\tilde \y)) \\
&= \MultComb(\rho) - \frac{n(k+md)-n^2 d}{2} + \MultComb(\psi) + \frac{n(k+md)-n^2 d}{2}\\
&= \MultComb(\sigma)
\end{align*}

which proves (<ref>).

Next, up to thin domains, we have $S = x P_\gamma + y R$, similar as before, we can solve

\[
x =  \frac{n_w(S)}{k}.
\]

Using (<ref>), (<ref>), and (<ref>), we compute:

\begin{align*}
\gen{c_1(\spincs_z(\sigma)), [P_\gamma]}
&= \gen{c_1(\spincs_z(\rho)), [P_\gamma]} + 2x [P_\gamma]^2  \\
&= 2 \Al(\x) + 2d(n_w(\rho)-n_z(\rho)) + k + 2d( n_w(\sigma) - n_w(\rho) - n_w(\psi)) \\
&= 2 \Al(\x) + 2d(n_w(\sigma) - n_z(\rho) -n_w(\psi)) +k  \\
&= 2 \Al(\x) + 2d(n_w(\sigma) - n_v(\rho) -n_v(\psi)) +k  \\
&= 2 \Al(\x) + 2d(n_w(\sigma) -n_v(\sigma)) +k,
\end{align*}

which proves (<ref>).
The computations for the rest of the items are similar and left for the reader.

We can adopt the perspective in Remark <ref>: for a fixed $n$, the relations in Lemma <ref> hold for all $n'$ with $1\leq n' \leq n$. In particular, taking $n'=1,$ then (<ref>) implies that $\gen{c_1(\spincs_z(\sigma)), [V]}$ is always an even integer, and (<ref>) implies that $\gen{c_1(\spincs_z(\sigma)), [\tilde Q]}=em$ for some odd integer $e.$

Moreover, by varying $n'$, we obtain
$n_z(\sigma) - n_{z_1}(\sigma)=n_{z_j}(\sigma) - n_{z_{j+1}}(\sigma)$ for $j=1,\cdots,n-1.$ This is true because the pentagon class $\sigma$ has no $(\alpha,\gamma)$ or $(\alpha,\delta)$ endpoints.

Fix $0 < \epsilon < \frac13$. Let $\tilde g^t_0$ denote the sum of all terms $g^t_{0,\spincv}$ for which $\spincv$  satisfies either both

\begin{align}
\label{eq: g0-trunc-Pg} \abs{\gen{c_1(\spincv), [P_\gamma]}} &< \epsilon md  \\
\label{eq: g0-trunc-Qt} \gen{c_1(\spincv), [\tilde Q]} &= \pm m, \\
\intertext{or it satisfies both}
\label{eq: g0-trunc-Pdt} \abs{\gen{c_1(\spincv), [\tilde P_\delta]}} &< (1+\epsilon)(k+md)  \\
\label{eq: g0-trunc-Q} \gen{c_1(\spincv), [Q]} &= \pm m,  \\
\intertext{or it satisfies}
\label{eq: g0-trunc-V} \gen{c_1(\spincv), [V]} &= 0.
\end{align}

The following lemma puts further constraints on the spin$^c$ evaluations. It is stated and proved in [14].

Fix $t \in \N$ and $0<\epsilon < \epsilon' < 1$. For all $m$ sufficiently large, if $\spincv$ is any spin$^c$ structure for which $g^t_{0,\spincv} \ne 0$, then the following implications hold:

If $\abs{\gen{c_1(\spincv), [P_\gamma]}} < \epsilon md$ and $\gen{c_1(\spincv), [\tilde Q] } = \pm m$, then

\begin{equation} \label{eq: g0-trunc-Pgt}
\abs{\gen{c_1(\spincv), [\tilde P_\gamma]}} < \epsilon' md.
\end{equation}

If $\gen{c_1(\spincv), [Q]} = \pm m$ and $\abs{\gen{c_1(\spincv), [\tilde P_\delta]}} < (1+\epsilon)(k+md)$, then
\begin{equation} \label{eq: g0-trunc-Pd}
\abs{\gen{c_1(\spincv), [P_\delta]}} < (1+\epsilon')(k+md).
\end{equation}

If $\gen{c_1(\spincv), [V]}=0$, then

\[
\gen{c_1(\spincv), [Q]} = \gen{c_1(\spincv), [\tilde Q]}  = \pm m.
\]

We are ready to compute the filtration shifts. The following lemma is parallel to <cit.>.

Fix $t \in \N$ and $\epsilon>0$. For all $m$ sufficiently large, the map $\tilde g^t_0$ is filtered with respect to the filtrations $\JJ_{\alpha\beta}$ and $\JJ_{\alpha\tilde\beta}$.

For any $\x \in \T_\alpha \cap \T_\beta$, we have

\begin{align*}
(\theta_{\alpha\tilde\beta} &\circ  g^t_0  \circ \theta_{\alpha\beta}^{-1}) ([\x,i] \otimes T^r) \\
\sum_{\tilde\y \in \T_\alpha \cap \T_{\tilde\beta}}  \sum_{\substack{\sigma \in \pi_2(\x, \Theta_{\beta\gamma}, \Theta_{\gamma\delta}, \Theta_{\delta\tilde\beta}, \tilde\y) \\ \mu(\sigma)=-2 \\ \mathclap{r + k/d +\AlNorm_{w,z}(\x) + n_z(\sigma) - n_v(\sigma) \equiv 0 \pmod m}}}   \#\MM(\sigma) \, [\tilde\y, i-n_z(\sigma)] \otimes T^{-k/d -\AlNorm_{w,z} (\tilde\y) + n_v(\sigma) - n_z(\sigma)}
\end{align*}

So as before, the composition $(\theta_{\alpha\tilde\beta} \circ  g^t_{0,\spincv}  \circ \theta_{\alpha\beta}^{-1})$ is only nonzero on the summand $\CF^t(\bm\alpha, \bm\beta, \spincs, z) \otimes T^r \GR $ and lands in $\CF^t(\bm\alpha, \bm\beta, \tilde \spincs, z) \otimes T^s \GR $, where $\spincs = \spincv |_{Y_{\alpha\beta}}$ , $\tilde\spincs = \spincv |_{Y_{\alpha\tilde\beta}}$ and $r,s \in \Q$ are given by

\begin{equation} \label{eq: g0-r-s-def}
\begin{aligned}
\frac{-k-md}{2d} &\le r < \frac{-k+md}{2d} & r &\equiv -\frac{1}{2d}( \gen{c_1(\spincv), [P_\gamma]} +k) \pmod m \\
\frac{-k-md}{2d} &\le s < \frac{-k+md}{2d} & s &\equiv -\frac{1}{2d}( \gen{c_1(\spincv), [\tilde P_\delta]} +k+md) \pmod m.
\end{aligned}
\end{equation}

Therefore, for some $p,q\in \Z$ we can write

\begin{align*}
\gen{c_1(\spincv), [P_\gamma]} &= -2dr -k + 2pmd \\
\gen{c_1(\spincv), [\tilde P_\delta]} &= -2ds -k + (2q-1) md,
\end{align*}

so that

\begin{align}
\label{eq: g0-filt-Pg-bounds} (2p-1)md &< \gen{c_1(\spincv), [P_\gamma]} \le (2p+1)md \\
\label{eq: g0-filt-Pdt-bounds} (2q-2)md &< \gen{c_1(\spincv), [\tilde P_\delta]} \le 2q md.
\end{align}

Next, assume that $\gen{c_1(\spincv), [\tilde Q]} = em$, where $e$ is an odd integer.     For any $\spincv$ that appears in the definition of $\tilde g^t_0$,      we now claim

\[
\]

Following the same argument in [14], we will first establish that $\gen{c_1(\spincv), [V]}$ is small compared to $m$. To be precise, we will show $\abs{k \gen{c_1(\spincv), [V]}} < 2md$ in all three cases in Definition <ref>.

* If $\spincv$ satisfies (<ref>) and (<ref>), hence (<ref>) by Lemma <ref>(<ref>),where we take $\epsilon' = 2\epsilon$. Then

\begin{align*}
\abs{k \gen{c_1(\spincv), [V]}} &= \abs{ \gen{c_1(\spincv), [P_\gamma] - [\tilde P_\gamma]}} \\
&\le \abs{\gen{c_1(\spincv), [P_\gamma]}} + \abs{\gen{c_1(\spincv), [\tilde P_\gamma]}}  \\
&< 3\epsilon md \\
&< 2md
\end{align*}

as required.

* Suppose that $ \spincv$ satisfies (<ref>) and (<ref>) and hence (<ref>) by Lemma <ref>(<ref>), again taking $\epsilon'=2\epsilon$.

\begin{align*}
(k+md) \abs{\gen{c_1(\spincv), [V]}} = \abs{\gen{c_1(\spincv), [\tilde P_\delta] - [P_\delta]}} &< (2+3\epsilon)(k+md).
\end{align*}

For $m$ sufficiently large, we again obtain $\abs{k \gen{c_1(\spincv), [V]}} <2md$, as required.

* If $ \spincv$ satisfies (<ref>), then the requirement is met immediately.

We proceed by observing since

\[
[P_\gamma] - [P_\delta] = d[Q], \quad [\tilde P_\delta] = [P_\delta] + (k+md) [V] \quad \text{and} \quad[\tilde Q] = [Q] -m[V],
\]

it follows

\begin{equation}
\gen{c_1(\spincv),[\tilde P_\delta ] - [P_\gamma] + d [\tilde Q]} = \gen{c_1(\spincv),k[V]}.
\end{equation}

Comparing the range of both sides of the equation, using (<ref>) and (<ref>), we obtain

\begin{equation} \label{eq: g0-filt-V-bounds}
2q-2p-3 + e <     \frac{k\gen{c_1(\spincv), [V]}}{md} < 2q-2p+1 + e.
\end{equation}

When $\abs{\gen{c_1(\spincv), [V]}}$ is sufficiently small, this implies $e=2p-2q+1,$ proving the claim.

For any $\sigma \in \pi_2(\x, \Theta_{\beta\gamma}, \Theta_{\gamma\delta}, \Theta_{\delta\tilde\beta}, \tilde \y)$ contributing to $\tilde g^t_0$, we compute:

\begin{align*}
\JJ_{\alpha\beta}([\x,i] \otimes T^r) &- \JJ_{\alpha\tilde\beta}([\tilde\y,i-n_z(\sigma)] \otimes T^s) \\
&= i - \frac{2nd r +nk+n^2 d}{2k} - (i-n_z(\sigma)) + \frac{2nd s +nk+n^2 d}{2k} \\
&= \frac{nd(s-r)}{k} + n_z(\sigma) \\
&= \frac{\gen{c_1(\spincv), n[P_\gamma]-n[\tilde P_\delta]} + (2q-2p-1)nmd }{2k} + n_z(\sigma) \\
&= \frac{\gen{c_1(\spincv), n[P_\delta] + nd[Q] -n[\tilde P_\delta]} -enmd }{2k} + n_z(\sigma) \\
&= \frac{\gen{c_1(\spincv), -(k+md)n[V] + nd[Q]} - \gen{c_1(\spincv),   nd[\tilde Q]} }{2k} + n_z(\sigma) \\
&= \frac{\gen{c_1(\spincv), -(k+md)n[V] + nmd[V]} }{2k} + n_z(\sigma) \\
&= \frac{-k\gen{c_1(\spincv), n[V]} }{2k} + n_z(\sigma) \\
&= n_{z_n}(\sigma) - n_z(\sigma) + n_z(\sigma)\\
&= n_{z_n}(\sigma)\\
&\geq 0.
\end{align*}

Fix $t \in \N$. For all $m$ sufficiently large, the map $\tilde g^t_0$ is a (filtered) chain homotopy between $\tilde \Psi^t_0$ and $\Phi^t_0$.

The proof in [14] can proceed without modulation, except that in the last item (P-$5$) where $\sigma$ has the decomposition $\sigma = \rho * \psi$, where $\rho \in \pi_2(\Theta_{\beta\gamma}, \Theta_{\gamma\delta}, \Theta_{\delta\tilde\beta}, \tilde \Theta)$ and $\psi \in \pi_2(\x, \Theta, \tilde\y)$ with $\tilde \Theta \in \T_\beta \cap \T_{\tilde\beta}$, now $z$ and $z_n$ are in the same region of both $(\Sigma, \bm\beta, \bm\gamma, \bm\delta, \bm{\tilde \beta})$ and $(\Sigma, \bm\alpha, \bm\beta, \bm{\tilde \beta})$. As a result, $n_z(\sigma) = n_{z_n}(\sigma)$.  By (<ref>), $\gen{c_1(\spincs_z(\sigma)), [V]}=0$, so $\spincv$ satisfies the condition (<ref>) and must appear in the definition of $\tilde g^t_0$.

This concludes the proof of Proposition <ref>, and hence of Theorem <ref>.

§ PROOF OF THE FILTERED MAPPING CONE FORMULA

We are ready to prove the filtered mapping cone formula. According to Lemma <ref>, it suffices to prove the mapping cone formula for  $\CF^t$. The proof follows closely the recipe from <cit.>, and we will write down the steps for completeness.

We will only prove the case when $k>0$. Given $t\in \N$, by Theorem <ref> , for a diagram $(\Sigma, \bm\alpha, \bm\beta, \bm\delta, w, z, z_n)$ and large enough $m$, we have doubly-filtered homotopy equivalence

\[
\begin{pmatrix} f_1^t \\ h_1^t \end{pmatrix}  \co \CF^t(\bm\alpha, \bm\gamma, z) \to \operatorname{Cone}(f_2^t).
\]

Fixing a spin$^c$ structure $\spinct \in \Spin^c(Y_\lambda(K))$, consider all its extension in $\Spin^c(W_\lambda).$  As discussed at the end of Section <ref>, these spin$^c$ structures form an orbit with the action of $\PD[C]$, where $C$ denotes the $2$-handle attached to $Y$. Their image under the bijection $E_{Y,\lambda,K}$ form an orbit in $\ul\Spin^c(Y,K)$ with the action of $\PD[K];$ call them $\xi_l$. The Alexander gradings  $A_{Y,K}(\xi_l)$ with $l \in \Z$ are precisely the arithmetic sequence $s_l$ with step length $k/d$, as defined  in Section <ref>, where the index is fixed by

\[
\frac{(2l-1)k}{2d}  < s_l \le \frac{(2l+1)k}{2d}.
\]

Under the bijection $E_{Y_\lambda, K_{n,\lambda}}\circ  (E_{Y,\lambda,K})^{-1} \co \ul\Spin^c(Y,K) \to \ul\Spin^c(Y_\lambda,K_{n,\lambda})$,  the set $\{\xi_l\}$ is identified with a sequence of spin$^c$ structures in  $\ul\Spin^c(Y_\lambda,K_{n,\lambda})$ (also with the action of $\PD[K]$). Their  $A_{Y_\lambda,K_{n,\lambda}}$ values form a $\Q/n\Z$ coset, which we denote by $A_{Y_\lambda,K_{n,\lambda}}(\spinct).$ By (<ref>), we have

\[
nds_l \equiv kA_{Y_\lambda,K_{n,\lambda}}(\spinct) + \frac{-nk+n^2 d}{2}  \pmod{\Z}
\]

Next, define $\spincv_l \in \Spin^c_0(\bar X_{ \alpha\gamma\delta})$ to be  the spin$^c$ structure extending $\spinct$ with

\[
\frac{(2l-1) mk}{\nu} < \gen{c_1(\spincv_l), [R]} \le \frac{(2l+1)mk}{\nu},
\]

and let  $\spincu_l = \spincv_l | _{Y_{\alpha\delta}}$. We want to show $s_{\spincu_l}=s_l$. (For the definition of $s_{\spincu_l}$ see (<ref>). By (<ref>), we have

\begin{align*}
\frac{n\nu}{2m}\gen{c_1(\spincv_l),[R]}&=\Al(\q) - k \sum_{j=1}^n \left( n_{z_j}(\psi) - n_{u_j}(\psi) \right) +\frac{-nk + n^2 d}{2}\\
&\equiv kA_{Y_\lambda,K_{n,\lambda}}(\spinct) +\frac{-nk + n^2 d}{2}   \pmod{k}\\
&\equiv nd s_l.  \pmod{k}
\end{align*}

\[
\gen{c_1(\spincv_l),[R]}= \frac{2mds_l}{\nu}.
\]

On the other hand, according to Lemma <ref>,

\[
\gen{c_1(\spincv_l),[R]}= \frac{2mds_{\spincu_l}}{\nu},
\]

which shows $s_{\spincu_l}=s_l$, as required. Lemma <ref> further shows that $f^t_{1,\spincv_l}$ is only non zero when $-L\leq l \leq L$ for some $L$.  In other words, the  image of $f^t_1$ restricted to $\CF^t(\bm\alpha,\bm\gamma,\spinct)$  is contained in the direct sum of $\CF^t(\bm\alpha,\bm\delta,\spincu_l)$ for $-L\leq l \leq L$.

Recall that spin$^c$ structures $\spincx_{\spincu_l}, \spincy_{\spincu_l}$ on $W_{\alpha\delta\beta}$, both restricting to $\spincu_l$ on $Y_{\alpha\delta}$,  are characterized by

\begin{align*}
\gen{c_1(\spincx_{\spincu_l}), [P_\delta]} &= 2ds_l - k-md \\
\gen{c_1(\spincy_{\spincu_l}), [P_\delta]} &= 2ds_l + k+md.
\end{align*}

We will write $\spincx_l = \spincx_{\spincu_l}$ and $\spincy_l = \spincy_{\spincu_l}$ to simplify the notation. The proof of Proposition <ref> shows that $\spincx_{l}$ and  $\spincy_{l}$ are the only two spin$^c$ structures that may contribute to the map $f^t_{2}$ on $\CF^t(\bm\alpha,\bm\delta, \spincu_l)$.
Observe $\spincx_{l}$ and $\spincy_{l-1}$ restrict to the same spin$^c$ structure  $\spincs_{\spincu_l}$ on $Y$. We will write  $\spincs_l =\spincs_{\spincu_l}.$

Moreover,  the images of the maps $\theta \circ f^t_{2, \spincx_{l}}$ and $\theta \circ f^t_{2, \spincy_{l-1}}$ both lie in the summand $\CF^t(\bm\alpha, \bm\beta, \spincs_l) \otimes T^{-s_l}$.

Therefore, the complex $\CF^t(\Sigma, \bm\alpha, \bm\gamma, w, \spinct)$ equipped with filtrations $\II_{\alpha\gamma}$ and $\JJ_{\alpha\gamma}$, is doubly-filtered quasi-isomorphic to the doubly-filtered complex

\begin{equation} \label{eq: mapping-cone-CFt}
<EMAIL_ADDRESS>\cdots \ar[dr] &\CF^t(\bm\alpha, \bm\delta, \spincu_{l-1}) \ar[d]|{F^t_{W'_m, \spincx_{l-1}}} \ar[dr]|{F^t_{W'_m, \spincy_{l-1}}} & \CF^t(\bm\alpha, \bm\delta, \spincu_{l}) \ar[d]|{F^t_{W'_m, \spincx_{l}}} \ar[dr]|{F^t_{W'_m, \spincy_{l}}} & \cdots  \ar[d]  \\
&\CF^t(\bm\alpha, \bm\beta, \spincs_{l-1}) \otimes T^{-s_{l-1}} & \CF^t(\bm\alpha, \bm\beta, \spincs_{l}) \otimes T^{-s_{l}} & \cdots
\end{equation}

where the filtrations are inherited from those on $\CF^t(\bm\alpha, \bm\delta, z)$ and $\ul\CF^t(\bm\alpha, \bm\beta, z; \GR)$, respectively.

By Theorem <ref>, there are doubly-filtered quasi-isomorphisms

\[
\Lambda^t_{\spincu_l}\co \CF^t(\Sigma, \bm\alpha, \bm\delta, \spincu_l) \to A^t_{\spincs_l, s_l}
\]

where the Alexander filtration on $\Lambda^t_{\spincu_l}$ is identified with the filtration $\JJ_{\spincu_l}$ from (<ref>). Whereas each $\CF^t(\bm\alpha, \bm\beta, \spincs_{l}) \otimes T^{-s_{l}}$ can be identified with $B^t_{\spincs_l} = C_{\spincs_l}\{0 \le i \le t\}$, so that the complex in (<ref>) is quasi-isomorphic to

\begin{equation} \label{eq: mapping-cone-AtBt}
\xymatrix<EMAIL_ADDRESS>\cdots \ar[dr] & A^t_{\xi_{l-1}} \ar[d]|{v^t_{\xi_{l-1}}} \ar[dr]|{h^t_{\xi_{l-1}}} & A^t_{\xi_{l}} \ar[d]|{v^t_{\xi_{l}}} \ar[d]|{v^t_{\xi_{l}}} \ar[dr]|{h^t_{\xi_{l}}}  &\cdots  \ar[d] \\
& B^t_{\spincs_{l-1}}   & B^t_{\spincs_{l}}& \cdots
\end{equation}

This is  the complex $X^t_{\lambda,\spinct,n,-L,L}$ from Section <ref> by definition.

Disregarding the second filtration, the underlying mapping cone is isomorphic to the one defined in [14], and with the mapping cone in [34], for this purpose. The maps induced by cobordisms are independent of the choice of basepoints. Thus the absolute grading shift on our mapping cone matches with Hedden-Levine's. Placing the $z_n$ basepoint differently amounts to imposing a different $\JJ$ filtration to the mapping cone. To complete the proof, we just need to check the $\JJ$ filtration agrees with the descriptions (<ref>) and (<ref>) in Section <ref>.

* On each summand $\CF^t(\bm\alpha, \bm\delta, \spincu_l)$, $\JJ_{\alpha\delta}$ is defined in (<ref>) as the Alexander filtration plus $\frac{nd^2m(2s_{\spincu_l}-n)}{2k(k+md)}$, thus $\JJ$ on $A^t_{\xi_l}$ would be $\JJ_{\spincu_l}$ plus the same shift:

\begin{align*}
\JJ([\x, i, j]) &= \JJ_{\spincu_l}([\x,i,j]) + \frac{nd^2m(2s_{\spincu_l}-n)}{2k(k+md)} \\
&= \max\{i-n, j-s_l\} + \frac{nd(2s_l - n)}{2(k+md)}  + \frac12 + \frac{nd^2m(2s_l-n)}{2k(k+md)} \\
%&= \max\{i-1, j-s_l\} + \frac{d(2s_l - 1)}{2k}  + \frac12 \\
&= \max\{i-n, j-s_l\} + \frac{nds_l +nk-n^2 d}{2k}
\end{align*}

as required.

* On each summand $\CF^t(\bm\alpha, \bm\beta, \spincs_l) \otimes T^{-s_l}$, according to (<ref>) we have

\begin{align*}
\JJ([\x, i, j]) &= \JJ_{\alpha\beta}([\x, i] \otimes T^{-s_l}) \\
&= i - \frac{2nd(-s_l) + nk + n^2 d}{2k} \\
&= i-n + \frac{2nds_l + nk - n^2 d}{2k}
\end{align*}

as required.

We have proved that $\CFK^t(Y_\lambda, K_{n,\lambda}, \spinct)$ is doubly-filtered quasi-isomorphic to  $X^t_{\lambda,\spinct,n,-L,L}$. According to Lemma <ref>, this implies the doubly-filtered quasi-isomorphism between   $\CFKm(Y_\lambda, K_{n,\lambda}, \spinct)$ and $X^-_{\lambda,\spinct,n,-L,L}$. To pass to the infinity version, tensor both complexes by $\F[U,U^{-1}]$, and we conclude that $\CFKi(Y_\lambda, K_{n,\lambda}, \spinct)$ is doubly-filtered quasi-isomorphic to $X^\infty_{\lambda,\spinct,n,-L,L}$, as required.

§ RATIONAL SURGERY

Following the same approach spelled out in <cit.>, our formula also supports a generalization to the rational surgery. For the simplicity, in this section we only demonstrate the process for $1/p$ surgery on a null-homologous knot $K$ inside a rational homology sphere $Y$, where $p$ is a positive integer. But this can be generalized to the case of arbitrary rational surgeries on any knot $K$ in a rational homology sphere without too much trouble.

Let $K_{n,1/p}$ denote the knot in $Y_{1/p}(K)$ obtained from $(n,1)$–cable of a left-handed meridian of $K$. Note that the left-handed meridian is no longer isotopic to the surgery solid torus as in the case of integral surgery.  Following Hedden and Levine's notation, let $Y'$ denote $Y\conn -L(p,1)$ and $K'=K\conn O_p$, where $O_p \subset -L(p,1)$ is obtained from the Hopf link by performing a $-p$ surgery on one of the unknot components. Then   $Y_{1/p}(K)$ is obtained from $Y'$ via a $2$-handle cobordism along $K';$ let $W$ denote this cobordism.

For $q\in \{0, \cdots, p-1 \}$, the complex $\HFKa(-L(p,1), O_p, \spincu_q)$ is generated by a single generator in Alexander grading $-\frac{p-2q-1}{2p}$. For any $\spincs \in \Spin^c(Y)$, let  $\spincs_{q} = \spincs \conn \spincu_q \in \Spin^c(Y \conn -L(p,1))$. The Künneth principle for connected sums (see <cit.>) implies that $\CFKi(Y',K', \spincs_q)$ is isomorphic to $\CFKi(Y,K)$, with the Alexander grading shifted by $-\frac{p-2q-1}{2p}$. Moreover, all the  $\spincs_{q}$ are cobordant to the same spin$^c$ structure on $Y_{1/p}(K)$ through $W$; denoted this spin$^c$ structure by $\spinct \in \Spin^c(Y_{1/p}(K))$.

In order to compute $\CFKi(Y_{1/p}(K),K_{n,1/p},\spinct),$ we apply the filtered mapping cone formula to $(Y',K')$. Using the notation from Section <ref>, we take $d=p$ and the framing on $K'$ corresponds to $k=1$. Consider all the spin$^c$ structures on $W$ that extends $\spinct$. Through the bijection between $\Spin^c(W)$ and $\ul\Spin^c(Y',K'),$ they form a sequence $\xi_l \in \ul\Spin^c(Y',K')$ for $l\in \Z$, where

\[
\frac{2l-1}{2p} < \AlNorm_{Y',K'}(\xi_l) \le \frac{2l+1}{2p}.
\]

We set $s_l = \AlNorm_{Y',K'}(\xi_l)$. Note that each $\AlNorm_{Y',K'}(\xi_l)$ satisfies

\[
\AlNorm_{Y',K'}(\xi_l) \equiv \frac{-p+2q+1}{2p} \pmod \Z,
\]

for some $q=0,1,\cdots,p-1.$ So we can write

\[
s_l = \frac{-p+2q+1}{2p} + r
\]

for some $r \in \Z.$ The above arithmetic constraint is enough to pin down $s_l$ and $r$, as demonstrated by the following computation from [14]: Observe $2l-1 < (2r-1)p + 2q+1 \le 2l+1$, so we have $(2r-1)p \le 2(l-q) < (2r-1)p + 2$. There are two cases:

* If $p$ is even, we deduce that $2(l-q) = (2r-1)p$, which implies

\[
s_l = \frac{2l+1}{2p};
\]

* if $p$ is odd, then $2(l-q) = (2r-1)p+1$, and therefore

\[
s_l= \frac {l}{p}.
\]

In both cases  we have

\[
r=\floor {\frac{2l+p}{2p}}.
\]

In the mapping cone, the complexes $A^\infty_{\xi_l}$ and $B^\infty_{\xi_l}$ are each copies of $\CFKi(Y',K', \spincs_q)$ by definition. As noted above, each of these complexes is in turn isomorphic to $\CFKi(Y,K,\spincs)$, but with the $j$ coordinate shifted by $-\frac{p-2q-1}{2p}$. Now we can directly compute the filtrations on $A^\infty_{\xi_l}$ and $B^\infty_{\xi_l}$ using the formulas (<ref>), (<ref>), (<ref>), and (<ref>). We collect the results as follows.

\begin{align}
\intertext{On $A^\infty_{\xi_l}$,}
\label{eq: It-ratl-A} \II_\spinct([\x,i,j]) &= \max\{i,j-r\} \\
\label{eq: Jt-ratl-A} \JJ_\spinct([\x,i,j]) &= \max\{i-n,j-r\} + nl -
\begin{cases}
\frac{n}{2}(np-1)  \quad p \text{ odd},  \\
\frac{n}{2}(np-2)  \quad p \text{ even},  \\
\end{cases}
%\label{eq: grt-ratl-A} \gr_\spinct([\x,i,j]) &= \absgr([\x,i,j]) + \frac{(2ds_l - k)^2 }{4dk} + \frac{2-3\sign(k)}{4} \\
\intertext{On $B^\infty_{\xi_l}$,}
\label{eq: It-ratl-B} \II_\spinct([\x,i,j]) &= i \\
\label{eq: Jt-ratl-B} \JJ_\spinct([\x,i,j]) &= i-n + nl -
\begin{cases}
\frac{n}{2}(np-1)  \quad p \text{ odd},  \\
\frac{n}{2}(np-2)  \quad p \text{ even}.  \\
\end{cases}%\\
%\label{eq: grt-ratl-B} \gr_\spinct([\x,i,j]) &= \absgr([\x,i,j]) + \frac{(2ds_l - k)^2 }{4dk} + \frac{-2-3\sign(k)}{4}
\end{align}

In particular, the above formula takes the same form as the mapping cone defined by Ozsváth and Szabó, namely each $A_s$ and $B_s$ complex appears $p$ times in the mapping cone (though in our case the $\JJ_t$ filtration of each copy is shifted accordingly).

§ EXAMPLES AND APPLICATIONS

In this section, we compute a series of examples that lead us to the proof of Proposition <ref> and other applications. For the reader's convenience, we restate the mapping cone formula for any knot  $K\subset S^3$ under the $+1$-surgery, where the $(n,1)$–cable of the meridian is denoted by $K_{n,1}$. Take $k=d=1$, and $s_l=s \in \Z$ in this case, since the Alexander grading for any relative spin$^c$ structure of $K$ is an integer.

According to the main theorem, the knot floer complex  $\CFKi(S^3_{1}(K),K_{n,1})$ is given by $X^\infty_{1,n,a,b}(K)$ for some $a\ll 0$ and $b\gg 0$. Specifically, let us consider the  mapping cone $X^\infty_{1,n,-g+1,g+n-1}(K)$ (we will see this gives suitable values for $a$ and $b$ later):

\begin{align}
\bigoplus^{g+n-1}_{s=-g+1}A^\infty_s \xrightarrow{v^\infty_s+h^\infty_s} \bigoplus^{g+n-1}_{s=-g+2}B^\infty_s,
\end{align}

where each $A^\infty_s$ and $B^\infty_s$ is isomorphic to $\CFKi(S^3,K)$, the map $v^\infty_s\co A^\infty_s \to B^\infty_s $ is the identity and $h^\infty_s\co A^\infty_s \to B^\infty_{s+1} $ is the reflection map precomposed with $U^s$. The $\II$ and  $\JJ$  filtrations are given by

\begin{align}
\intertext{For $[\x,i,j] \in A^\infty_{s}$,}
\label{eq: filtration_s3_1}
\II([\x,i,j]) &= \max\{i,j-s\} \\
\label{eq: filtration_s3_2}
\JJ([\x,i,j]) &= \max\{i-n,j-s\} + ns - \frac{n(n-1)}{2} \\
\intertext{For $[\x,i,j] \in B^\infty_{s}$,}
\label{eq: filtration_s3_3}
\II([\x,i,j]) &= i \\
\label{eq: filtration_s3_4}
\JJ([\x,i,j]) &= i-n + ns - \frac{n(n-1) }{2}.
\end{align}

It is straightforward to check that for $s<-g+1$, the map $h^\infty_s$ induces an isomorphism on the homology; for $s>g+n-1,$
the map $v^\infty_s(K)$ induces an isomorphism on the homology. Thus the filtered quasi-isomorphism type of $X^\infty_{1,n,a,b}(K)$ does not depend on $a,b$ if $a\leq -g+1$ and $b\geq g+n-1$. We will write

\[
X_n^\infty(K) = X^\infty_{1,n,-g+1,g+n-1}(K).
\]

For the purpose of computations,   looking at the associated graded complex is usually helpful. We will describe the  associated graded complex with $\II=0$, and the associated graded complex with other  $\II$ values can be obtained from translations. The complex $X_n^\infty (K)\{i=0\}$ is given by

\begin{align}
\bigoplus^{g+n-1}_{s=-g+1}A^\infty_s \{\text{max}(i,j-s)=0 \} \xrightarrow{v^\infty_s+h^\infty_s} \bigoplus^{g+n-1}_{s=-g+2}B^\infty_s\{i=0\},
\end{align}

where each $A^\infty_s \{\text{max}(i,j-s)=0 \}$ admits exactly $n+1$ filtration levels with respect to $\JJ$, namely $\JJ= ns+\frac{n(n-1)}{2}, 1+ns+\frac{n(n-1)}{2},\cdots, n+ns+\frac{n(n-1)}{2}, $ while   $B^\infty_s\{i=0\}$ corresponds to the filtration level $-n+ns+\frac{n(n-1)}{2}.$ See Figure <ref>.

$ns+n(n-1)/2$  at 100 290
$-1+ns+n(n-1)/2$  at 193 268
$\cdot$  at 143 249
$\cdot$  at 143 245
$\cdot$  at 143 241
$-n+ns+n(n-1)/2$  at 193 215

$A_{s-1}$  at 110 314

$h_{s-1}$  at 200 164
$v_s$  at 292 164

$n+ns+n(n-1)/2$  at 280 288
$n-1+ns+n(n-1)/2$  at 382 268
$\cdot$  at 330 248
$\cdot$  at 330 244
$\cdot$  at 330 240
$ns+n(n-1)/2$  at 389 215

$A_s$  at 290 314

$ns+n(n-1)/2$  at 200 100
$B_s$  at 290 110

A portion of the complex $X_n^\infty(K) \{i=0\}$, where the vertical bar of each complex is at $i=0$ and the horizontal bar of $A_{s}$ (resp.  $A_{s-1}$) is at $j=s$ (resp.  $j=s-1$).

The general strategy for computation involves finding a reduced basis for $X^\infty_n(K),$ where every term in the differential strictly lowers at least one of the filtrations. This can be achieved through a cancellation process (see for example [23]) as follows: suppose $\partial x_i = y_i$ $+ $ lower filtration terms, where the double filtration of $y_i$ is the same as $x_i$, then the subcomplex of  $X^\infty_n(K)$ generated by all such $\{x_i,  \partial x_i\}$ is acyclic, and $X^\infty_n(K)$ quotient by this complex is reduced.

In the rest of the section, we wish to shed a light on the richness of examples coming from the new filtered mapping cone formula. But at first we have to go on a detour for some preliminaries of Heegaard Floer concordance invariants. The experts can skip the next subsection.

§.§ Concordance invariants and other rings

The concordance invariants in our interests are defined using Heegaard Floer homology over other rings. Pointing the reader to the sources, we would not go in depth about the detailed construction. Instead, we will review some results that are useful to us and use an example to explain some of the intuition.

To start, Ian Zemke provided a reinterpretation of Heegaard Floer homology over the ring $\F[U,V]$ (see for example [39]).   In the knot Floer complex over $\F[U,V]$,  there is a bigrading $\gr=(\gr_U,\gr_V)$ assigned to each generator, where $\gr(U)=(-2,0)$ and $\gr(V)=(-2,0)$.  Two knot Floer complexes are locally equivalent over $\F[U,V]$ if the two knots are (smoothly) concordant (<cit.>). The local equivalence takes on slightly different meanings when defined over different rings, but is defined to always respect the knot concordance. We are generally interested in, within different contexts,  the morphism between the monoid of knots over concordance and the set of knot Floer complexes over the local equivalence relation.

No matter over which ring it is defined, the knot Floer complex always possesses certain symmetric property. For example, over the ring  $\F[U,V]$ this entails that after exchanging the role of $U$ and $V$, the resulting complex is homotopy equivalent to the original knot Floer complex.

In [3],  Dai-Hom-Stoffregen-Truong studied  knot Floer complexes over the ring $\F[U,V]/(UV).$ The results were initially for knots in $S^3,$ but easily generalizable to knots in integer homology sphere $L$–spaces. According to <cit.>, a standard complex over $\F[U,V]/(UV)$ is freely generated by $\{x_0, x_1, \cdots, x_{2l} \}$ for some $l\in \N,$ where each pair of generators $x_{2i}$ and $x_{2i+1}$ is connected by some “U-arrow”  and each pair of generators $x_{2i+1}$ and $x_{2i+2}$ is connected by some “V-arrow”  for $i=0,\cdots, l-1$. All information of a standard complex can be encoded using a sequence of signed integers describing  the length and the direction of the arrows in order. Thus we shall use this sequence to indicate a standard complex. It turns out that the complexes over the ring   $\F[U,V]/(UV)$ are quite nice. Indeed, we have
[Theorem 6.1 in [3]]
The knot Floer complex of any knot in an integer homology sphere $L$–space is locally equivalent to a unique standard complex over $\F[U,V]/(UV).$

Therefore, the sequence of signed integers is a concordance invariant through this identification. In <cit.>, the invariant $\varphi_i$ is defined to be a signed count of the times number $i$ appear in the sequence in odd position, and  subsequently  in <cit.>, is shown to be a homomorphism from the knot concordance group $\mathcal{C}$ to $\Z$. Equivalently we can define $\varphi_i$ by counting only in even position (inverting the sign for each integer since we want the two definitions to be consistent); either way only half of the sequence is needed due to the symmetry. The two definitions correspond respectively to  counting all the “U-arrows” or  “V-arrows” of a certain length.

Setting $U=0$ in the ring $\F[U,V]/(UV)$ makes the knot Floer complex a module over $\F[V]$; call the homology of the resulting complex the vertical homology. Since $\F[V]$ is a PID,   the vertical homology of any knot Floer complex  has the decomposition

\begin{align*}
\F[V] \oplus \big( \bigoplus_j \F[V]/(V^j)  \big).
\end{align*}

A basis on the knot Floer complex that realizes this decomposition is called a vertical basis; namely, a vertical basis consists of generators that pairwise generate the  $V$–torsions and a standalone generator (i.e. in the kernel but not the image of $\partial$) that generates the single copy of $\F[V]$. Furthermore by setting $V=1,$ the resulting complex computes $\HFa(Y) \cong \F,$ where $Y$ is the ambient manifold. Therefore the  standalone generator in a vertical basis has $\gr_V=d(Y),$ or we say the $d$–invariant is supported in said generator.

In [2], the above construction is generalized by the same group of authors. They devised the following ring

\begin{align*}
\X=\frac{\F[U_B,\{W_{B,i}\}_{i\in \Z},V_T,\{W_{T,i}\}_{i\in \Z}]}{(U_B V_T, U_B W_{B,i}-W_{B,i+1},V_T W_{T,i}-W_{T,i+1})}
\end{align*}

with a bigrading $(\gr_1,\gr_2) \in \Z \times \Z,$ where

\begin{gather*}
\begin{aligned}
\hspace{8em} \gr(U_B)=(-2,0) &\hspace{2em} \text{and} \hspace{2em}&  \gr(W_{B,i})=(-2i,-2) &\hspace{3em}
\shortintertext{and}
\gr(V_T)=(0,-2) &\hspace{2em} \text{and} \hspace{2em}&  \gr(W_{T,i})=(-2,-2i) &\hspace{3em}.
\end{aligned}
\end{gather*}

Inside $\X$ there are two subrings $\mathcal{R}_U$ and  $\mathcal{R}_V$  given by

\begin{align*}
\mathcal{R}_U &= \F[U_B,\{W_{B,i}\}_{i\in \Z}]/(U_B W_{B,i}-W_{B,i+1})
\intertext{and}
\mathcal{R}_V &= \F[V_T,\{W_{T,i}\}_{i\in \Z}]/(V_T W_{T,i}-W_{T,i+1}).
\end{align*}

Note that elements in $\mathcal{R}_U$ can be uniquely written down as $U^i_B W^j_{B,0},$ where

\begin{align*}
(i,j)\in (\Z \times \Z^{\geq 0}) - (\Z^{<0} \times \{0\}).
\end{align*}

Namely, $i$ (the power of $U_B$) can be negative when $j$ (the power of $W_{B,0}$) is positive, but is required to be non-negative when $j=0.$  This is analogous to the ring $\F[U,V]$ localized at the maximal ideal $(U)$, denoted by $\F[U,U^{-1},V]$. Here $U_B$ plays the role of $U$ and $W_{B,0}$ plays the role of $V$, such that we are allow to invert $U$ but not $V$. However, a difference is that in $\mathcal{R}_U$  when there is no $V$ power (when $j=0$) we “remember” the power of $U$ ($i$ needs to be non-negative).

Similarly, the elements in $\mathcal{R}_V$ can be written down  uniquely as well. Whenever talking about the elements in $\mathcal{R}_U$ and $\mathcal{R}_V$, we canonically use this expression.

The complexes over $\F[U,V]$ can be translated to $\X$ using the map

\begin{align*}
U&\xmapsto[\hspace{2em}]{} U_B + W_{T,0}\\
V&\xmapsto[\hspace{2em}]{} V_T + W_{B,0},
\end{align*}

so this allows us to talk about knot Floer complexes over the ring $\X.$

Similar to the case of $\F[U,V]/(UV)$, there is the notion of the standard complex over $\X$ (<cit.>). A standard complex is freely generated by $\{x_0, x_1, \cdots, x_{2l} \}$ for some $l\in \N,$ where each pair of generators $x_{2i}$ and $x_{2i+1}$ is connected by a differential using elements in $\mathcal{R}_U$  and each pair of generators $x_{2i+1}$ and $x_{2i+2}$ is connected by a differential using elements in $\mathcal{R}_V$,  for $i=0,\cdots, l-1$. Again, quite remarkably, we have the following.
[Theorem 7.1 in [2]]
Every knot Floer complex over the ring $\X$ is locally equivalent to a unique standard complex.

Therefore one can  define the concordance invariants $\varphi_{i,j}$ by either counting  $\mathcal{R}_U$  edges or $\mathcal{R}_V$  edges in the standard complex that is locally equivalent to the given complex (<cit.>). For $(i,j)\in (\Z \times \Z^{\geq 0}) - (\Z^{<0} \times \{0\})$, $\varphi_{i,j}$ are indeed homomorphisms according to <cit.>. When $j=0$, the map $\varphi_{i,0}$ agrees with the previously defined homomorphism $\varphi_i$, and when $j \neq 0$, all $\varphi_{i,j}$ vanish for any knot in an integer homology sphere $L$–space.

Note that for a knot Floer complex over $\X$, suppose we set $V_{T}=1$. This forces $\mathcal{R}_U=0$ due to the relation $U_B V_T=0$. The resulting complex is a graded module over $\F[W_{T,0}]$, and once again we obtain that the $d$–invariant is supported in the non-torsion element in this complex.

We hope that the following example provides some intuition in dealing with complexes over the ring $\X.$

[The complex $C$ in Example <ref>. ]

[thin, black!20!white]
ıin -2,...,3
(ı-0.5, 3) – (ı-0.5, -3);
(3, ı-0.5) – (-3, ı-0.5);

(2, 0) circle (2pt) node[] (x1);
(2, 1) circle (2pt) node[] (y2);
(2, 2) circle (2pt) node[] (z);
(1, 2) circle (2pt) node[] (x2);
(0, 2) circle (2pt) node[] (y1);
(-1, -2) circle (2pt) node[] (w1);
(-2, -1) circle (2pt) node[] (w2);

(2,2) – (2,1);
(2,2) – (1,2);
(2,1) – (-2,-1);
(2,1) – (-1,-2);
(1,2) – (-1,-2);
(1,2) – (-2,-1);
(0,2) – (-2,-1);
(2,0) – (-1,-2);

[right] at (x1) $x_1$;
[right] at (z) $z$;
[right] at (y2) $y_2$;
[above] at (y1) $y_1$;
[above] at (x2) $x_2$;
[below] at (w1) $w_1$;
[below] at (w2) $w_2$;

[The  complex $C$ after a change of basis in $\F[U,V,V^{-1}]$.]
[thin, black!0!white]
(5, 0) – (-5,0);

[thin, black!20!white]
ıin -2,...,3
(ı-0.5, 3) – (ı-0.5, -3);
(3, ı-0.5) – (-3, ı-0.5);

(2, 0) circle (2pt) node[] (x1);
(2, 1) circle (2pt) node[] (y2);
(2, 2) circle (2pt) node[] (z);
(1, 2) circle (2pt) node[] (x2);
(0, 2) circle (2pt) node[] (y1);
(-1, -2) circle (2pt) node[] (w1);
(-2, -1) circle (2pt) node[] (w2);

(2,2) – (2,1);

(1,2) – (-1,-2);

(0,2) – (-2,-1);

[right] at (z) $z$;
[right] at (y2) $y_2+UV^{-1}x_2$;
[left] at (y1) $y_1$;
[above] at (x2) $x_2+Uy_1$;
[below] at (w1) $w_1$;
[below] at (w2) $w_2$;
at (2.1, -0.25) $x_1+UV^{-1}x_2+U^2V^{-2}y_1$;

Consider the following complex $C$ over $\F[U,V]$ generated by $x_1,x_2,y_1,y_2,$ $w_1,w_2$ and $z$ with differentials as follows (see Figure <ref>)

\begin{gather*}
\begin{aligned}
\partial x_1 &= U^3V^2w_1  \hspace{4em}   &\partial x_2 = U^2V^4w_1 + U^3V^3w_2 \\
\partial y_1 &= U^2V^3w_2              &\partial y_2 = U^3V^3w_1 + U^4V^2w_2\\
\partial z &= Ux_2 + Vy_2. &
\end{aligned}
\end{gather*}

In fact, $C$ is locally equivalent to the complex $CFK(S^3_1(T_{2,11}),(T_{2,11})_{2,1}).$ (This follows from Lemma <ref> and the proof of Proposition <ref>.) We aim to compute $\varphi_{i,j}(C)$ for some suitable $i$ and $j,$ which necessitates studying $C$ over the ring $\X.$ To provide some intuition, we would like to first examine $C$ over $\F[U,V]$.

Inverting $V$, in the localized ring  $\F[U,V,V^{-1}]$, we can perform a change of basis by replacing $y_2$ with $y_2+UV^{-1}x_2.$ This simplifies the differentials:

\begin{gather}\label{eq: example_middlestep_changebasis}
\begin{aligned}
\partial x_1 &= U^3V^2w_1  \hspace{4em} &\partial x_2 = U^2V^4w_1 + U^3V^3w_2 \\
\partial y_1 &= U^2V^3w_2      &\partial z = V(y_2 + UV^{-1}x_2 ).  \hspace{0.8em}
\end{aligned}
\end{gather}

Observing that $x_1, x_2$ and $y_1$  and their differentials make up  “a connected series of line segments", we further perform the following change of basis:

\begin{align*}
x_1 &\xmapsto[\hspace{1em}]{} x_1 + UV^{-2}x_2 + U^2V^{-2}y_1\\
x_2 &\xmapsto[\hspace{1em}]{} x_2 + Uy_1.
\end{align*}

Under this change of basis, (the image of) $y_1$ and $w_2$, $x_2$ and $w_1$, $z$ and $y_2$ pairwise generate the torsion in the homology while the standalone generator $x_1$  generates the single copy of the base ring, realizing the decomposition

\[
H_*(C\otimes \F[U,V,V^{-1}]) \cong \F[U,V,V^{-1}] \oplus \big(\bigoplus_j \F[U,V,V^{-1}]/(U^j)\big).
\]

If we instead invert $U$, then one could also write down a similar change of basis over the ring $\F[U,U^{-1},V],$ such that there is one standalone generator while all other generators appear in pairs forming the torsion in the homology. Indeed,

\begin{gather*}
\begin{aligned}
y_1 &\xmapsto[\hspace{1em}]{} y_1 + U^{-2}Vy_2 + U^{-2}V^2x_1,  & \hspace{2em}
y_2 \xmapsto[\hspace{1em}]{} y_2 + Vy_1, \\
x_2 &\xmapsto[\hspace{1em}]{} x_2 + U^{-1}Vy_2  &
\end{aligned}
\end{gather*}

accomplishes the job.

Next, denote by $C_\X$ the complex $C$ over the ring $\X.$ Translating using the language of ring $\X$, the differentials become

\begin{align*}
\partial x_1 &= (U_B^3 W_{B,0}^2 + V_T^2W_{T,0}^3)w_1\\
\partial x_2 &= (U_B^2 W_{B,0}^4 + V_T^4 W_{T,0}^2 )w_1 + (U_B^3W_{B,0}^3 + V_T^3 W_{T,0}^3  )w_2 \\
\partial y_1 &= (U_B^2 W_{B,0}^3 + V_T^3 W_{T,0}^2  )w_2\\
\partial y_2 &= (U_B^3 W_{B,0}^3 + V_T^3 W_{T,0}^3 )w_1 + (U_B^4W_{B,0}^2 + V_T^2 W_{T,0}^4) w_2\\
\partial z &= (U_B + W_{T,0}) x_2 + (W_{B,0} + V_T ) y_2.
\end{align*}

We wish to formulate a change of basis, under which $C_\X$ is a standard complex. To motivate such a change of basis, consider the following. Inside the differential of each generator, there are terms in the subring $\mathcal{R}_V$ and terms in the subring $\mathcal{R}_U.$ As we have seen, the ring $\mathcal{R}_V$ is the analogue of inverting $V$ inside $\F[U,V]$ and the ring $\mathcal{R}_U$ is the analog of inverting $U$ inside $\F[U,V]$. If we ignore all the terms in $\mathcal{R}_U$ and focus on the terms in $\mathcal{R}_V$, there clearly exists a set of basis analogous to the case of $\F[U,V,V^{-1}]$. Similar for $\mathcal{R}_U$. Thus the intuition would be to perform the two sets of the change of basis simultaneously.

First, if we set

\begin{gather*}
\begin{aligned}
\hspace{2.5em} \widetilde{x}_2 &= x_2 + U^{-1}_B W_{B,0} y_2    \hspace{5.5em} &\widetilde{y}_2 &= y_2 + V^{-1}_T W_{T,0} x_2,
\end{aligned}
\end{gather*}

the differentials can be simplified to

\begin{gather*}
\begin{aligned}
\hspace{4em}    \partial x_1 &= (U_B^3 W_{B,0}^2 + V_T^2W_{T,0}^3)w_1 \hspace{2em}    &\partial \widetilde{x}_2 &=  V_T^4 W_{T,0}^2 w_1 +  V_T^3 W_{T,0}^3  w_2 \hspace{0.3em}\\
\partial y_1 &= (U_B^2 W_{B,0}^3 + V_T^3 W_{T,0}^2  )w_2      &\partial \widetilde{y}_2 &= U_B^3 W_{B,0}^3 w_1 + U_B^4W_{B,0}^2  w_2\\
\partial z &= U_B \widetilde{x}_2 + V_T  \widetilde{y}_2.
\end{aligned}
\end{gather*}

Compare the $\mathcal{R}_V$ terms in the differential of each generator with the differentials in (<ref>) and notice the similarity.   With this observation in mind, we further perform the following change of basis

\begin{gather*}
\begin{aligned}
\hspace{6em}\widehat{x}_1 &= x_1 + V_T^{-2}W_{T,0} \widetilde{x}_2 + V_T^{-2}W^{2}_{T,0} y_1 \hspace{1em}
&\widehat{x}_2 &= \widetilde{x}_2 + W_{T,0} y_1 \hspace{0.3em}\\
\widehat{y}_1 &= y_1 + U_B^{-2}W_{B,0} \widetilde{y}_2 + U_B^{-2}W^{2}_{B,0} x_1
&\widehat{y}_2 &= \widetilde{y}_2 + W_{B,0} x_1,
\shortintertext{making the differentials become}
\partial \widehat{x}_1 &= U_B^3 W_{B,0}^2 w_1  &\partial \widehat{x}_2 &=  V_T^4 W_{T,0}^2 w_1  \\
\partial  \widehat{y}_1 &=  V_T^3 W_{T,0}^2 w_2    &\partial \widetilde{y}_2 &=  U_B^4W_{B,0}^2  w_2\\
\partial z &= U_B \widehat{x}_2 + V_T  \widehat{y}_2.
\end{aligned}
\end{gather*}

We conclude that $C_\X$ under this final basis is a standard complex given by

\[
\]

In particular, we have

\begin{align*}
\varphi_{i,j}(C)=\begin{cases} -1 \quad &\text{if} \hspace{0.5em} (i,j)=(3,2) \text{ or }(4,2)\\
1 &\text{if} \hspace{0.5em} (i,j)=(1,0)\\
0 &\text{otherwise.}
\end{cases}
\end{align*}

In the following subsection, we will perform most of the computations (especially when we are invoking the filtered mapping cone formula) over the original ring $\F[U,U^{-1}].$ Only after collecting enough information and ready to compute the concordance invariants, we then move on to either the ring $\F[U,V]/(UV)$ or $\X$ depending on the case at hand. We will always specify the ring we are working on whenever the base ring is not  $\F[U,U^{-1}]$ or $\F[U]$. It is our hope that the current approach causes minimal confusion.

§.§ Examples and applications

We begin by describing a family of concrete examples by applying the filtered mapping formula to the $+1$-surgery on the torus knot $T_{2,3}$  with varying $n$.

Using the notion of the standard complex from either [2] or [3], we have

The complex $\CFKi(S^3_1(T_{2,3}),(T_{2,3})_{n,1})$ is locally equivalent to the standard complex $(-1,n,1,-1,-2,\cdots,n-i+1,1,-1,-i-1,\cdots,2,1,-1,-n,1)$ for some fixed $n\geq 1$, where $i=1,\cdots, n-1.$

The standard complex described in the proposition is simply the concatenation of $(n-i+1,1,-1,-i-1)$ for $i=1,\cdots,n-1$ in that order, with $-1$ and $1$ in the front and back.

at (-2,6) $A_s$;

[]() at (0,4)

[step=1, black!80!white, thin] (0, 0) grid (1, 3);

(0.5, 2.5) circle (2pt) node[] (b);
(0.5, 1.5) circle (2pt) node[] (a);
(0.5, 0.5) circle (2pt) node[] (c);

[thick, ->] (a) – (c);

at (-0.3,1.5) $\alpha_s$;
at (-0.3,2.5) $x_s$;
at (-0.3,0.5) $y_s$;

at (3.3,1.5) $-\frac{1}{2}n^2  + \frac{1}{2}n + ns -s$;
at (3.3,2.5) $-\frac{1}{2}n^2  + \frac{1}{2}n + ns -(s-1)$;
at (3.3,0.5) $-\frac{1}{2}n^2  + \frac{1}{2}n + ns -(s+1)$;

at (-5,-1) $B_s$;
[]() at (-3,0)

(0, 0) circle (2pt) node[] ;
at (-0.4,0.6) $\beta_s$;
at (2.5,0) $-\frac{1}{2}n^2  + \frac{1}{2}n + n(s -1)$;

at (2.5,-1) $B_{s+1}$;
[]() at (4,0)

(0, 0) circle (2pt) node[] ;
at (0.3,0.7) $\beta_{s+1}$;
at (2.5,0) $-\frac{1}{2}n^2  + \frac{1}{2}n + ns$;

[thick, ->] (-2.5,2) – (-4,0.6);
[thick, ->] (-1.3,2) – (1,0.4);

at (-3.6,1.5) $v_s$;
at (0.3,1.4) $h_s$;

A part of the complex $X^\infty_n(T_{2,3})\{i=0\}$ under the chosen basis.

The case of $n=1$ follows from a simple calculation. Now consider when $n>1.$

Let us denote the generators of $\CFKi(S^3,T_{2,3})$ by $a,b$ and $c$ with coordinates $(0,0),(0,1)$ and $(0,-1)$ respectively, where $b$ has Maslov grading $0$. They satisfy

\[
\partial a=Ub + c.
\]

Via the isomorphism with $\CFKi(S^3,T_{2,3})$, we denote the generators in $A_s$ (resp.  $B_s$) by $a_s,b_s$ and $c_s$  (resp.  $a'_s,b'_s$ and $c'_s$) for $0\leq s \leq n$. We will abuse the notation and use $\partial$ for the differential in $X^\infty_n(T_{2,3})$ from now on. For each $s=1,\cdots, n-1$, We have

\begin{align*}
\partial a_s &= Ub_s + c_s + a'_i + U^s a'_{s+1} \qquad \qquad \null \\
\partial b_s &= b'_s + U^{s-1} c'_{s+1} \\
\partial c_s &= c'_s + U^{s+1} b'_{s+1}\\
\partial a'_s &= Ub'_s +  c'_s.
\end{align*}

We will establish a set of basis for the reduced complex of $X^\infty_n(T_{2,3})$.   First we can let $c'_s + Ub'_s$ replace $c'_s$ and quotienting out the acyclic summand generated by $a'_s$ and $c'_s + Ub'_s$. Then considering the first three relations, it is natural to perform the following change of basis (for all $0\leq s\leq n$):

\begin{gather*}
\begin{aligned}
\alpha_s &= a_s    \qquad \qquad   &   x_s &= b_s + U^{s-1} a'_{s+1}\\
\beta_s &=c'_s    \qquad  \qquad  &       y_s &= c'_s + a_s,
\end{aligned}
\end{gather*}

such that under the new basis, the relations are simplified.  See Figure <ref>. For each $s=1,\cdots, n-1$,

\begin{align*}
\partial \alpha_s &= Ux_s + y_s \qquad \qquad \qquad \qquad \qquad \null \\
\partial x_s &= \beta_s + U^s \beta_{s+1} \\
\partial  y_s &= U\beta_s + U^{s+1} \beta_{s+1}.
\end{align*}

When considered as a generator of the complex $B_s\{ i =0\},$  $\beta_s$ is the unique generator of homology under the chosen basis. The $(\II,\JJ)$ filtrations of the generators can be computed using (<ref>),(<ref>),(<ref>) and (<ref>). All $\alpha_s, \beta_s, x_s$ and $y_s$ have $\II=0$ and their $\JJ$ filtrations are as follows:

\begin{align*}
\JJ(\alpha_s)&=-\frac{1}{2}n^2  + \frac{1}{2}n + ns -s\\
\JJ(\beta_s)&=-\frac{1}{2}n^2  + \frac{1}{2}n + n(s -1)\\
\JJ(x_s)&=-\frac{1}{2}n^2  + \frac{1}{2}n + ns -(s-1)\\
\JJ(y_s)&=-\frac{1}{2}n^2  + \frac{1}{2}n + ns -(s+1).
\end{align*}

The differential of $\alpha_s$ consists of $Ux_s$ and $y_s$, which differ by $(1,-1)$ in  the $(\II,\JJ)$ grading. Next    consider the differentials of $x_s$ and $y_s$. The $(\II,\JJ)$ grading shift is equal to $(0,n-s+1)$  from $x_s$ to $\beta_s$ and equal to $(s,1)$  from $x_s$ to $U^s\beta_{s+1}$. Note that $ \partial  y_s = \partial Ux_s$, from this immediately follows that  the $(\II,\JJ)$ grading shifts in the differential of $y_s$ are $(1,n-s)$ and $(s+1,0)$ respectively.

We then consider the generators in $A_0$. After performing a change of basis

\begin{align*}
x_0 &\xmapsto[]{}  x_0 + U^{-1}y_0\\
\alpha_0 &\xmapsto[]{} \alpha_0,
\end{align*}

and quotienting out the resulting acyclic summand, the only remaining generator in $A_0$ is $y_0$, with the differential given by

\[
\partial y_0 = U\beta_1.
\]

The $\JJ$ filtrations of $y_0 $ and $ U\beta_1$ are equal.

With the above computations ready,  the rest of the proof  assesses the complex  over the ring $\F[U,V]/(UV)$,  using the language of [3]. The sub-quotient complex generated by $\alpha_s, \beta_s x_s$ and $y_s$  corresponds to the sequence $(n-s+1,1,-1,-s-1)$, starting from $\beta_s$. See Figure <ref>.  Next observe that each $\beta_s$ is in the image of $x_s$ and  $y_{s+1}$ (with some appropriate $U,V$ decoration), thus as $s$ ranges from $1$ to $n-1$,  the sub-quoitent complex generated by all $\alpha_s, \beta_s, x_s$ and $y_s$  corresponds to the  concatenation of $(n-s+1,1,-1,-s-1)$.

Finally, it is easy to see that by setting $U=0$, each pair of generators $\alpha_s$ and $y_s$,  $\beta_s$ and  $x_s$ for $n\geq s\geq 1$ makes up a torsion summand in the vertical homology while $y_0$ generates the single copy of $\F[V]$. Therefore by further setting $V=1$, $\HFa(S^3_1(T_{2,3}))$ (and thus the $d$-invariant) is supported in $y_0$ as required.  The differential of $y_0$ gives the $-1$ in the beginning of the sequence. Following from the symmetry, we obtain the $1$ at the end of the sequence.

In fact, the proof shows that over the ring $\F[U,V],$ $\CFK(S^3,T_{2,3})/(UV)$ is isomorphic to the standard complex $(-1,\cdots,n-i+1,1,-1,-i-1,\cdots,1)$, with $i=1,\cdots, n-1.$ Furthermore, this information completely determines $\CFKi(S^3,T_{2,3})$ (via the condition $\partial^2 = 0$).

[thick] (2, 2) – (1, 2);
[thick] (2, 2) – (2, 1);
[thick] (1, 2) – (1, -2);
[thick] (2, 1) – (-2, 1);

[] at (-2.5,1.3) $\beta_{s+1}$;
[] at (2.4,0.7) $y_s$;
[] at (2.4,2.3) $\alpha_s$;
[] at (0.7,2.3) $x_s$;
[] at (1.3,-2) $\beta_{s}$;

[] at (1.5,2.2) $1$;
[] at (2.2,1.6) $1$;

[] at (-0.5,1.3) $s+1$;
[] at (0,-0.5) $n-s+1$;

The portion of $X^\infty_n (T_{2,3})$ that corresponds to the sequence $(n-s+1,1,-1,-s-1)$ in a standard complex using vertical basis. The generators are denoted abstractly, without their $U,V$ decoration.

We are ready to prove the computational result Proposition <ref> which leads to Theorem <ref>.
Recall that torus knots $T_{2,4k+3}$ are $L$–space knots with genus $g=2k + 1$, where $k\in \Z_{\geq 0}$. (We have so far reserved the letter $k$ for the surgery coefficient. Since in this section the surgery is mostly fixed, hopefully this new assignment of letter $k$ does not cause confusion.)  The knot Floer complex $\CFKi(S^3,T_{2,4k+3})$ is generated by $a_i$ for $i=1,\cdots,g$ with coordinate $(0,-g+2i-1)$ and $b_i$ for $i=1,\cdots,g+1$ with coordinate $(0,-g+2i-2)$. The Maslov grading is supported in $b_{g+1}$ and the differentials are given by

\[
\partial a_i = Ub_{i+1} + b_i, \quad \text{for} \hspace{0.5em} i=1,\cdots,g.
\]

As before, let $(T_{2,4k+3})_{n,1}$ denote the $(n,1)$–cable of the meridian inside $+1$-surgery on $T_{2,4k+3}$ and let $J_{n,k}$ denote $(T_{2,4k+3})_{n,1}$ connected sum with the unknot in $-S^3_1(T_{2,4k+3})$. The ambient manifold  $S^3_1(T_{2,4k+3}) \conn -S^3_1(T_{2,4k+3})$ is homology cobordant to $S^3$.

This immediately implies Theorem <ref>. In fact, we can prove a stronger version.

For any integers $i$ and $j$ such that $i>j\geq 0,$ and a given sequence of arbitrary  integers $(h_{j+1},\cdots, h_i)$,
there exists some knot $K\in \CZhat$ such that $\varphi_{t,j}(K)=h_t$ for $t=j+1,\cdots, i,$ and $\varphi_{t,j}(K)=0$ for $t>i.$

The concordance invariant $\varphi_{i,j}$  is additive under the group action of $\CZhat$. By Proposition <ref>, we can first choose a knot  with $\varphi_{i,j}=h_i$. Assuming $i-j>1,$ then by adding copies of $J_{i-1,j}$ to the chosen knot, we can make it such that $\varphi_{i-1,j}=h_{i-1}$ while the value of $\varphi_{i,j}$ remains unchanged. Repeat this process until  a knot with desired property is obtained.

[thin, black!50!white]
[<->] (-6, 0) – (6, 0);
[<->] (0, -6) – (0, 6);

(-6,4) circle (2pt) node (b_6) ;
[above] at (b_6) $U^3b_{6}$;
ıin -2,...,2

(-2*ı,2*ı) circle (2pt) node (a_ı) ;
(-2*ı,-2+2*ı) circle (2pt) node (b_ı) ;

[very thick] (-2*ı,2*ı)–(-2*ı-2,2*ı);
[very thick] (-2*ı,2*ı)–(-2*ı,-2+2*ı);

[below] at (b_2) $U^2b_{5}$;
[below] at (b_1) $Ub_{4}$;
[below] at (b_0) $b_{3}$;
[below] at (b_-1) $U^{-1}b_{2}$;
[below] at (b_-2) $U^{-2}b_{1}$;

[right] at (a_2) $U^2a_{5}$;
[right] at (a_1) $Ua_{4}$;
[right] at (a_0) $a_{3}$;
[right] at (a_-1) $U^{-1}a_{2}$;
[right] at (a_-2) $U^{-2}a_{1}$;

The knot Floer complex $\CFKi(S^3,T_{2,11})$. The torus knot $T_{2,11}$ is an $L$–space knot with genus equal to $5$.

Before moving on to the proof of Proposition <ref>, let us study the complex $X^\infty_n (T_{2,4k+3})$ closely. The case when $k=0$ has been studied, so we assume $k>0$ from now on. Via the isomorphism with $\CFKi(S^3,T_{2,4k+3})$, denote the generators in  $A_s$  by $(a_i)_s$ for $i=1,\cdots, g$ and $(b_i)_s$ for $i=1,\cdots, g+1$ and the generators in $B_s$ by $(a_i)'_s$ and $(b_i)'_s$, where $s=-g+1,\cdots, g+n-1.$ The differential on the mapping cone is given by

\begin{align*}
\partial (a_i)_s &= (b_i)_s + U(b_{i+1})_s + (a_i)'_s + U^{s+g+1-2i} (a_{g-i+1})'_{s+1} \\
\partial (b_i)_s &= (b_i)'_s + U^{s+g+2-2i} (b_{g-i+2})'_{s+1} \\
\partial (a_i)'_s &= (b_i)'_s + U(b_{i+1})'_s.
\end{align*}

We first aim to choose a reduced basis for the complex of $X^\infty_n (T_{2,4k+3})$. Note that $(a_i)'_s $ and $ (b_i)'_s \in \partial  (a_i)'_s$ are in the same $(\II,\JJ)$ filtration, so we may quotient out the acyclic complex generated by $\{(a_i)'_s, \partial  (a_i)'_s \}$. Under this identification, we have

\[
(b_{g+1})'_s = U(b_{g})'_s = \cdots = U^g(b_{1})'_s. \quad \null
\]

\begin{align*}
(\alpha_i)_s &=(a_i)_s   \qquad \qquad \qquad \qquad \qquad \null\\
(x_i)_s &=(b_i)_s       \\
\beta_s &=(b_{g+1})'_s,
\end{align*}

the differentials are now simplified considerably:

\begin{align}
\label{eq: diff_a}  \partial (\alpha_i)_s &= (x_i)_s + U(x_{i+1})_s \\
\label{eq: diff_x}  \partial (x_i)_s &= U^{g+1-i}\beta_s + U^{s+g+1-i} \beta_{s+1}.
\end{align}

Next, notice for certain values of $i$ and $s$,  at least one term in $\partial (\alpha_i)_s$ preserve the $(\II,\JJ)$ filtration, and we shall quotient out the acyclic summands generated by these $\{(a_i)_s, \partial  (a_i)_s \}$  as well. See Figure <ref>. When $-g\leq s\leq g$, we claim for $i \in \{1,\cdots, g\}$  the above condition is satisfied if and only

\begin{gather*}
\begin{aligned}
i\geq \frac{s+g+1}{2} & \quad \text{or} \quad i\leq \frac{s+g-1}{2}-\floor{\frac{n-1}{2}}  & \text{ if } s \text{ is even;} \\
i\geq \frac{s+g+2}{2} & \quad  \text{or} \quad  i\leq \frac{s+g}{2}-\ceil{\frac{n-1}{2}}  & \text{ if } s \text{ is odd},
\end{aligned}
\end{gather*}

and when $s\geq g$, if and only if

\[
i\leq \frac{g+1}{2} - \ceil{\frac{n-s}{2}}.
\]

We will show the claim when  $-g\leq s\leq g$ and $s$ is even. The other two cases are similar and left for the reader to verify. The first half of the claim is clear, and we focus on the second half.

[The complex $A_s \{\text{max}(i,j-s)=0 \}$ when $-g \leq s \leq g$ and $s$ is even. ]
[thin, black!0!white]
(-10, 0) – (12, 0);

[thin, black!40!white]
(0, 0) – (0, -10);
(2, 2) – (2, -10);
(-4, 0) – (0, 0);
(-4, 2) – (2, 2);
(0, 0) – (2, 0);
(0, -2) – (2, -2);
(0, -4) – (2, -4);

(-1,1) circle (2pt) node () ;

ıin 0,...,2

(1,1-4*ı) circle (2pt) node () ;
(1,-1-4*ı) circle (2pt) node () ;

[very thick] (1,1-4*ı)–(1,-1-4*ı);

[very thick] (-1,1)–(1,1);

[] at (-2.1,0.8) $x_{\frac{s+g+3}{2}}$;
[right] at (1.2,1) $\alpha_{\frac{s+g+1}{2}}$;
[right] at (1.2,-1) $x_{\frac{s+g+1}{2}}$;
[right] at (1.2,-3) $\alpha_{\frac{s+g-1}{2}}$;

[right] at (2,-5) $\cdot$;
[right] at (2,-5.5) $\cdot$;
[right] at (2,-4.5) $\cdot$;

[right] at (1.2,-7) $\alpha_i$;
[right] at (1.2,-9) $x_i$;

[The complex $A_s \{\text{max}(i,j-s)=0 \}$ when $-g \leq s \leq g$ and $s$ is odd. ]
[thin, black!0!white]
(-10, 0) – (8, 0);

[thin, black!40!white]
(0, 0) – (0, -10);
(2, 2) – (2, -10);
(-4, 0) – (0, 0);
(-4, 2) – (2, 2);
(0, 0) – (2, 0);
(0, -2) – (2, -2);
(0, -4) – (2, -4);

(-1,1) circle (2pt) node () ;
(-3,1) circle (2pt) node () ;
[very thick] (-1,1)–(-3,1);

ıin 0,...,1

(1,-1-4*ı) circle (2pt) node () ;
(1,-3-4*ı) circle (2pt) node () ;

[very thick] (1,-1-4*ı)–(1,-3-4*ı);

(1,1) circle (2pt) node () ;

[right] at (1.2,1) $x_{\frac{s+g+2}{2}}$;
[right] at (1.2,-1) $\alpha_{\frac{s+g}{2}}$;
[right] at (1.2,-3) $x_{\frac{s+g}{2}}$;

[right] at (2,-5) $\cdot$;
[right] at (2,-5.5) $\cdot$;
[right] at (2,-6) $\cdot$;

[The complex $A_s \{\text{max}(i,j-s)=0 \}$ when $ s \geq g$.]
[thin, black!0!white]
(-10, 0) – (8, 0);

[thin, black!40!white]
(0, 6) – (0, -4);
(2, 6) – (2, -4);
ıin 0,...,2

(0, 2*ı) – (2, 2*ı);

(1,5) circle (2pt) node () ;
(1,3) circle (2pt) node () ;
(1,1) circle (2pt) node () ;
(1,-1) circle (2pt) node () ;
(1,-3) circle (2pt) node () ;

[very thick] (1,3)–(1,1);
[very thick] (1,-3)–(1,-1);

[right] at (1.2,5) $x_{g+1}$;
[right] at (1.2,3) $\alpha_g$;
[right] at (1.2,1) $x_g$;

[right] at (1.5,-2) $\cdot$;
[right] at (1.5,-1.5) $\cdot$;
[right] at (1.5,-1) $\cdot$;

Part of the associated graded complex $X^\infty_n(T_{2,4k+3})\{i=0\}$,  the complex $A_s \{\text{max}(i,j-s)=0 \}$ with the $\JJ$ filtration level depicted.

In order to simplify the notation when talking about the filtration,   define

\begin{equation}\label{eq: def_f}
f(n,s)=-\frac{(n-1)n}{2} + ns,
\end{equation}

so that the $\JJ$ filtration on the complex $A_s \{\text{max}(i,j-s)=0 \}$ ranges between $f(n,s)$ and $f(n,s)-n.$ When $s$ is even, for any $l\geq 0$  the only generator supported in the coordinate $(0,s-l)$  is $ x_{(s+g+2-l)/2}$ if $l$ is odd and $\alpha_{(s+g+1-l)/2}$ if $l$ is even. Therefore $\JJ(\alpha_i)=f(n,s)-n$ if and only if

\begin{align*}
\begin{cases}
i\leq (s+g-n)/2   & n     \text{ is odd}\\
i\leq (s+g+1-n)/2    &  n    \text{ is even}.
\end{cases}
\end{align*}

In either case, the second half of the claim is verified. Thus we have obtained a reduced basis for the complex $X^\infty_n (T_{2,4k+3})$ as follows. The generators in $A^\infty_s$ are given by $(\alpha_i)_s, (x_i)_s$  for $s=-g+1,\cdots, g+n-1$, such that when $s\leq g,$

\begin{gather*}
\begin{aligned}
\frac{s+g+1}{2} \geq i \geq \text{max}\{1,\frac{s+g+1}{2}-\floor{\frac{n-1}{2}}\}  & \text{ if } s \text{ is even,} \\
\frac{s+g+2}{2} \geq i \geq \text{max}\{1,\frac{s+g+2}{2}-\ceil{\frac{n-1}{2}}\}  & \text{ if } s \text{ is odd},
\end{aligned}
\end{gather*}

when $s\geq g$,

\[
g+1 \geq i\geq \text{max} \{1,\frac{g+3}{2} - \ceil{\frac{n-s}{2}}\};
\]

the generators in $B^\infty_s$ are $\beta_s$ for $s=-g+2,\cdots, g+n-1$ and the differentials are given by (<ref>) and (<ref>).

Before moving on, let us introduce some notational shorthands:
denote by $x_s$ the “top” generator in each $A_s$ complex, i.e. the one with the largest $i$ index; for example, when $-g+1 \leq s \leq g$ and $s$ is even, $x_s=(x_{(s+g+1)/2})_s$. Let $i^{(t)}_{n,s}$ be this “top” $i$ index. Similarly, let  $i^{(b)}_{n,s}$  be the smallest integer that $i$ index can take in each case. Here letters $t$ and $b$ are chosen to indicate “top” and “bottom” respectively.

We want to take a look at the $(\II,\JJ)$ filtration shifts next. Clearly, the filtration shifts in the differential of each $(\alpha_i)_s$ are $(1,0)$ and $(0,1).$ So we focus on the filtration shifts in the differential of each $(x_i)_s$.  Suppose $U^c \beta_{s'}$ is a nontrivial term in $\partial (x_i)_s$, where $s'=s$ or $s+1.$ Define

\begin{equation}
\Delta_{\II,\JJ}((x_i)_s,\beta_{s'}) = (\II,\JJ)((x_i)_s) - (\II,\JJ)(U^c \beta_{s'})
\end{equation}

and similarly define $\Delta_\II$ and $\Delta_\JJ$. Note that we have

\begin{equation}
\Delta_{\II,\JJ}((x_{i-1})_s,\beta_{s'})=\Delta_{\II,\JJ}((x_i)_s,\beta_{s'}) + (1,-1)
\end{equation}

for $ i^{(t)}_{n,s} \geq i\geq i^{(b)}_{n,s}+1$, therefore  to completely understand the filtration shifts of all the generators in each complex $A_s$, we need only compute the filtration shift of the “top” generator $x_s$.

When $-g+1\leq s \leq g,$ if $s$ is even,

\begin{align}
\label{eq: del_sleqg_even_s} \Delta_{\II,\JJ}(x_s,\beta_s)&=\Big(\frac{-s+g+1}{2},n+\frac{-s+g-1}{2}\Big)\\
\label{eq: del_sleqg_even_sp} \Delta_{\II,\JJ}(x_s,\beta_{s+1})&=\Big(\frac{s+g+1}{2},\frac{s+g-1}{2}\Big);\\
\intertext{if $s$ is odd,}
\label{eq: del_sleqg_odd_s} \Delta_{\II,\JJ}(x_s,\beta_s)&=\Big(\frac{-s+g}{2},n+\frac{-s+g}{2}\Big)\\
\label{eq: del_sleqg_odd_sp} \Delta_{\II,\JJ}(x_s,\beta_{s+1})&=\Big(\frac{s+g}{2},\frac{s+g}{2}\Big);\\
\intertext{when $g\leq s \leq g+n-1$,}
\label{eq: del_sgeqg_s} \Delta_{\II,\JJ}(x_s,\beta_s)&=(0,n+g-s)\\
\label{eq: del_sgeqg_sp} \Delta_{\II,\JJ}(x_s,\beta_{s+1})&=(s,g).
\end{align}

For each case, $\JJ(\beta_s)=f(n,s)-n$ and $\JJ(\beta_{s+1})=f(n,s)$ by (<ref>). When
$-g+1\leq s \leq g,$ and $s$ is even,

\begin{align*}
\partial x_s &= U^{(-s+g+1)/2} \beta_s + U^{(s+g+1)/2} \beta_{s+1},\\
\JJ(x_s)&=f(n,s)-1.
\end{align*}

\begin{align*}
\Delta_{\II,\JJ}(x_s,\beta_s)&= \Big(\frac{-s+g+1}{2},\frac{-s+g+1}{2}\Big) + (0,n-1)\\
\Delta_{\II,\JJ}(x_s,\beta_{s+1})&= \Big(\frac{s+g+1}{2},\frac{s+g+1}{2}\Big) + (0,-1)\\
\end{align*}

When $-g+1\leq s \leq g,$ and $s$ is odd, the computation is parallel. Suppose now $g\leq s \leq g+n-1$, then we have

\begin{align*}
\partial x_s &= \beta_s + U^s \beta_{s+1},\\
\JJ(x_s) &= f(n,s) - (s-g).
\end{align*}

So we may compute

\begin{align*}
\Delta_{\II,\JJ}(x_s,\beta_s)&= (0,n+g-s)\\
\Delta_{\II,\JJ}(x_s,\beta_{s+1})&= (s,s) + (0,g-s)\\
\end{align*}

Next we claim that for the purpose of computing concordance invariants, we need only  look at a portion of the mapping cone.
Define  $X^\infty_n (K) \langle l \rangle =  X^\infty_{1,n,-l+n,l} (K)$ for $l\in \Z$, which is the filtered mapping cone

\begin{align*}
\bigoplus^{l}_{s=-l+n}A^\infty_s \xrightarrow{v^\infty_s+h^\infty_s} \bigoplus^{l}_{s=-l+n+1}B^\infty_s.
\end{align*}

Note that under this notation  $X^\infty_n (K) = X^\infty_n (K) \langle g+n-1 \rangle$.

For any $k \geq 1$ and $n\geq 1$, the filtered complex $X^\infty_n (T_{2,4k+3}) \langle g+n-1 \rangle$  is isomorphic to $ X^\infty_n (T_{2,4k+3}) \langle n \rangle \oplus D $ up to a change of basis, where $H_*(D)=0$.

It suffices to show for each $l=n+1,\cdots, g+n-1,$ the complex $X^\infty_n (T_{2,4k+3}) \langle l \rangle$ is isomorphic to $X^\infty_n (T_{2,4s+3}) \langle l-1 \rangle \oplus D$ up to a change of basis, where $H_*(D)=0$. For every such $l$, we will demonstrate a filtered change of basis such that the complex given by

\begin{equation}\label{eq: summand1}
A^\infty_{-l+n} \xrightarrow[]{h_{-l+n}}B^\infty_{-l+n+1}
\end{equation}

becomes a summand under the new basis. Thus following from the symmetry, there is also a filtered change of basis such that

\begin{equation}\label{eq: summand2}
A^\infty_{l} \xrightarrow[]{\hspace{0.3em} v_{l}\hspace{0.3em}}B^\infty_{l}
\end{equation}

becomes a summand under the new basis as required.

Let $s=-l+n+1$ in the following proof, and we have $s\leq 0.$ There are two cases to consider depending on the parity of $s.$

* When $s$ is even, perform the change of basis

\begin{align*}
(x_i)_{s} \xmapsto[\qquad\null]{}  &(x_i)_{s} + U^{\frac{-s+g+3}{2} -i} x_{{s}-1},
\quad \text{for} \hspace{0.5em}   i^{(t)}_{n,s} \geq i \geq i^{(b)}_{n,s}.
\end{align*}

By (<ref>) and (<ref>) it is straightforward to check that the complex given by (<ref>) is a summand under this basis. (Note that in complex $X^\infty_n (T_{2,4k+3}) \langle l \rangle$ the map $v_{-l+n}=v_{s-1}$ is the zero map.)  It remains to show that this change of basis is filtered. This in fact amounts to showing for $i^{(t)}_{n,s} \geq i \geq i^{(b)}_{n,s}$,

\begin{align*}
\Delta_{\II,\JJ}((x_i)_s, \beta_s ) \geq  \Delta_{\II,\JJ}(x_{s-1}, \beta_s).
\end{align*}

Thus we compute

\begin{align*}
\Delta_{\II,\JJ}&((x_i)_s, \beta_s ) - \Delta_{\II,\JJ}(x_{s-1}, \beta_s)\\
&= \Delta_{\II,\JJ}(x_s, \beta_s ) +
\left( \frac{s+g+1}{2} - i\right)(1,-1) - \Delta_{\II,\JJ}(x_{s-1}, \beta_s) \\
&= \left( \frac{-s+g+1}{2} ,  n +  \frac{-s+g-1}{2} \right) + \left( \frac{s+g+1}{2} - i ,  -\frac{s+g+1}{2} + i \right) \\
& \hspace{2em}- \left( \frac{s-1+g}{2} ,   \frac{s-1+g}{2} \right)\\
&= \left( \frac{-s+g+3}{2} -i ,   n-1 + \frac{-3s-g+1}{2} + i \right).
\end{align*}

The $\II$ filtration shift is clearly positive for all $i$ since

\begin{align*}
\frac{-s+g+3}{2} -i &\geq \frac{-s+g+3}{2} - \frac{s+g+1}{2}\\
&> 0,
\end{align*}

while for the  $\JJ$ filtration shift there are two possible cases depending on the value of $i^{(b)}_{n,s}$:

* Suppose

\[
1\geq \frac{s+g+1}{2} - \floor{\frac{n-1}{2}},
\]

then we have

\begin{align*}
n-1 + \frac{-3s-g+1}{2} + i &\geq  n + \frac{-3s-g+1}{2} \\
& \geq 2\floor{\frac{n-1}{2}} +1 + \frac{-3s-g+1}{2}\\
&\geq s+g + \frac{-3s-g+1}{2}\\
&\geq \frac{-s+g+1}{2}\\
\end{align*}

* If instead

\[
1\leq \frac{s+g+1}{2} - \floor{\frac{n-1}{2}},
\]

this is when $n$ is relatively small, and we have

\begin{align*}
n-1 + \frac{-3s-g+1}{2} + i &\geq  n-1 + \frac{-3s-g+1}{2} + \frac{s+g+1}{2} - \floor{\frac{n-1}{2}} \\
& \geq -s+1 - \floor{\frac{n-1}{2}} +n -1 \\
&\geq -s+1 + \frac{n-1}{2} \\
\end{align*}

* When $s$ is odd, perform the change of basis

\begin{align*}
(x_i)_s \xmapsto[\qquad\null]{}  &(x_i)_s + U^{\frac{-s+g+2}{2} -i} x_{s-1},  \quad \text{for} \hspace{0.5em}  i^{(t)}_{n,s} \geq i \geq i^{(b)}_{n,s}.
\end{align*}

Again one can verify under this basis the complex given by (<ref>) is a summand.  The difference in $(\II,\JJ)$ filtration is

\begin{align*}
\Delta_{\II,\JJ}&((x_i)_s, \beta_s ) - \Delta_{\II,\JJ}(x_{s-1}, \beta_s)\\
&= \Delta_{\II,\JJ}(x_s, \beta_s ) +
\left( \frac{s+g+2}{2} - i\right)(1,-1) - \Delta_{\II,\JJ}(x_{s-1}, \beta_s) \\
&= \left( \frac{-s+g}{2} ,  n +  \frac{-s+g}{2} \right) + \left( \frac{s+g+2}{2} - i ,  -\frac{s+g+2}{2} + i \right) \\
& \hspace{2em}- \left( \frac{s+g}{2} ,   \frac{s+g-2}{2} \right)\\
&= \left( \frac{-s+g+2}{2} -i ,   n + \frac{-3s-g}{2} + i \right).
\end{align*}

The $\II$ filtration shift is clearly positive since

\begin{align*}
\frac{-s+g+2}{2} -i &\geq \frac{-s+g+2}{2} - \frac{s+g+2}{2}\\
&> 0,
\end{align*}

while the $\JJ$ filtration has two possible cases depending on the value of $i^{(b)}_{n,s}$.

* Suppose that

\[
1\geq \frac{s+g+2}{2} - \ceil{\frac{n-1}{2}},
\]

then we have

\begin{align*}
n + \frac{-3s-g}{2} + i &\geq  n + \frac{-3s-g}{2} + 1 \\
& \geq 2\ceil{\frac{n-1}{2}} +  \frac{-3s-g}{2} + 1\\
&\geq s+g + \frac{-3s-g}{2} + 1\\
&\geq \frac{-s+g}{2} + 1\\
\end{align*}

* If instead

\[
1\leq \frac{s+g+1}{2} - \floor{\frac{n-1}{2}},
\]

in this case we have

\begin{align*}
n + \frac{-3s-g}{2} + i &\geq  n    + \frac{-3s-g}{2} + \frac{s+g+2}{2} - \ceil{\frac{n-1}{2}} \\
& \geq -s + 1 + \ceil{\frac{n-1}{2}}  \\
\end{align*}

as required.

By Lemma <ref>, in order to compute concordance invariants  it suffices to consider the complex $X^\infty_n(T_{2,4k+3})\langle n \rangle$. We claim that over the ring $\X$ discussed in the previous subsection <ref>, after a change of basis $X^\infty_n(T_{2,4k+3})\langle n \rangle$ is a standard complex.

Translating to the ring $\X$, the differential on the complex is given by

\begin{align*}
\partial  (\alpha_i)_s &= (U_B + W_{T,0}) (x_{i+1})_s + (W_{B,0} + V_T) (x_i)_s  \\
\partial {{(x_i)}}_0 &= \big( U^{\Delta_\II({(x_i)}_0, \beta_{1})}_B W^{\Delta_\JJ({(x_i)}_0, \beta_{1})}_{B,0} + V^{\Delta_\JJ({(x_i)}_0, \beta_{1})}_T W^{\Delta_\II({(x_i)}_0, \beta_{s+1})}_{T,0}     \big)\beta_1\\
\partial {{(x_i)}}_n &= \big( U^{\Delta_\II({(x_i)}_n, \beta_n)}_B W^{\Delta_\JJ({(x_i)}_n, \beta_n)}_{B,0} + V^{\Delta_\JJ({(x_i)}_n, \beta_n)}_T W^{\Delta_\II({(x_i)}_n, \beta_n)}_{T,0}     \big)\beta_n\\
\intertext{\hspace{1em}and if $n>s>0,$ then}
\partial {{(x_i)}}_s &= \big( U^{\Delta_\II({(x_i)}_s, \beta_s)}_B W^{\Delta_\JJ({(x_i)}_s, \beta_s)}_{B,0} + V^{\Delta_\JJ({(x_i)}_s, \beta_s)}_T W^{\Delta_\II({(x_i)}_s, \beta_s)}_{T,0}     \big)\beta_s \\
+\big( &U^{\Delta_\II({(x_i)}_s, \beta_{s+1})}_B W^{\Delta_\JJ({(x_i)}_s, \beta_{s+1})}_{B,0} + V^{\Delta_\JJ({(x_i)}_s, \beta_{s+1})}_T W^{\Delta_\II({(x_i)}_s, \beta_{s+1})}_{T,0}     \big)\beta_{s+1}.
\end{align*}

We first perform the change of basis

\begin{align*}
(\widetilde{x}_i)_s &= \begin{cases}
(x_i)_s + U^{-1}_B W_{B,0} (x_{i-1})_s  &\quad \text{if} \hspace{0.5em} i=i^{(t)}_{n,s}\\
(x_i)_s + U^{-1}_B W_{B,0} (x_{i-1})_s + V^{-1}_T W_{T,0} (x_{i+1})_s &\quad \text{if} \hspace{0.5em}  i^{(b)}_{n,s}<i<i^{(t)}_{n,s}\\
(x_i)_s + V^{-1}_T W_{T,0} (x_{i+1})_s  &\quad \text{if} \hspace{0.5em} i=i^{(b)}_{n,s},
\end{cases}
\intertext{which simplifies the differentials considerably:}
\partial  (\alpha_i)_s &= U_B  (\widetilde{x}_{i+1})_s +  V_T (\widetilde{x}_i)_s  \\
\partial {(\widetilde{x}_i)}_0 &=  \begin{cases}
V^{\Delta_\JJ({(x_i)}_0, \beta_{1})}_T W^{\Delta_\II({(x_i)}_0, \beta_{1})}_{T,0} \beta_{1}, \hspace{0.5em} i=i^{(t)}_{n,0} \\
U^{\Delta_\II({(x_i)}_0, \beta_{1})}_B W^{\Delta_\JJ({(x_i)}_0, \beta_{1})}_{B,0}\beta_{1}, \hspace{0.5em} i=i^{(b)}_{n,0}\\
0,   \hspace{10em}\text{ otherwise}
\end{cases}\\
\partial {(\widetilde{x}_i)}_n &=\begin{cases}
V^{\Delta_\JJ({(x_i)}_n, \beta_n)}_T W^{\Delta_\II({(x_i)}_n, \beta_n)}_{T,0} \beta_n, \hspace{0.5em} i=i^{(t)}_{n,n} \\
U^{\Delta_\II({(x_i)}_n, \beta_n)}_B W^{\Delta_\JJ({(x_i)}_n, \beta_n)}_{B,0}\beta_n, \hspace{0.5em} i=i^{(b)}_{n,n}\\
0,   \hspace{10.4em}\text{ otherwise}
\end{cases}
\end{align*}

and for $n>s>0,$

\begin{align*}
\partial {(\widetilde{x}_i)}_s = &\begin{cases}
V^{\Delta_\JJ({(x_i)}_s, \beta_s)}_T W^{\Delta_\II({(x_i)}_s, \beta_s)}_{T,0} \beta_s
+ V^{\Delta_\JJ({(x_i)}_s, \beta_{s+1})}_T W^{\Delta_\II({(x_i)}_s, \beta_{s+1})}_{T,0} \beta_{s+1}, \hspace{0.5em} i=i^{(t)}_{n,s} \\
U^{\Delta_\II({(x_i)}_s, \beta_s)}_B W^{\Delta_\JJ({(x_i)}_s, \beta_s)}_{B,0}\beta_s
+ U^{\Delta_\II({(x_i)}_s, \beta_{s+1})}_B W^{\Delta_\JJ({(x_i)}_s, \beta_{s+1})}_{B,0}\beta_{s+1}, \hspace{0.5em} i=i^{(b)}_{n,s}\\
0,   \hspace{24.5em}\text{ otherwise.}
\end{cases}
\end{align*}

Next observe that

\begin{align*}
\Delta_\II(x_s,\beta_{s+1}) -  \Delta_\II(x_{s+1},\beta_{s+1}) = \begin{cases}
s+1 \quad &\text{if} \hspace{0.5em} 0\leq s \leq g\\
s &\text{otherwise,}
\end{cases}
\end{align*}

by (<ref>),   (<ref>), (<ref>),   (<ref>), (<ref>) and (<ref>). In particular, for any $s$ with $0\leq s \leq n-1,$ we have  $\Delta_\II(x_s,\beta_{s+1}) -  \Delta_\II(x_{s+1},\beta_{s+1})>0.$ Similarly, one can show $\Delta_\JJ((x_{i^{(b)}_{n,s}})_s,\beta_s) -  \Delta_\JJ((x_{i^{(b)}_{n,s-1}})_{s-1},\beta_s)>0$  for $n\geq s \geq 1$.

Therefore we may further perform a change of basis

\begin{align}
\label{eq: changebasis1}
(\widetilde{x}_{i^{(t)}_{n,s}})_s &\xmapsto[\hspace{2em}]{} \sum_{n\geq s'\geq s}V^{\lambda_{s'}}_T W^{\mu_{s'}}_{T,0} (\widetilde{x}_{i^{(t)}_{n,s'}})_{s'} \quad \text{for all} \hspace{0.5em} n > s \geq 0,\\
\label{eq: changebasis2}
(\widetilde{x}_{i^{(b)}_{n,s}})_s &\xmapsto[\hspace{2em}]{} \sum_{0\leq s'\leq s} U^{\theta_{s'}}_B W^{\eta_{s'}}_{B,0}(\widetilde{x}_{i^{(b)}_{n,s'}})_{s'} \quad \text{for all} \hspace{0.5em} 0 < s \leq n,
\end{align}

\begin{align*}
\lambda_{s'} &= \sum_{s'-1 \geq t\geq s} \big(  \Delta_\JJ(x_t,\beta_{t+1}) -  \Delta_\JJ(x_{t+1},\beta_{t+1}) \big)\\
\mu_{s'} &= \sum_{s'-1 \geq t\geq s} \big(  \Delta_\II(x_t,\beta_{t+1}) -  \Delta_\II(x_{t+1},\beta_{t+1}) \big)  \hspace{5em},
\end{align*}

and $\theta_{s'}, \eta_{s'}$  can be defined similarly. Note that the definition implies $\lambda_s=\mu_s = 0.$ When $s' > s$ (resp. $s' < s$), $\mu_{s'}$ (resp.  $\eta_{s'}$) is positive, while $\lambda_{s'}$ (resp.  $\theta_{s'}$) needs not to be (in fact they are negative).  This change of basis results in a standard complex as required.

Let $(\widetilde{x})_s$ denote the image of $(\widetilde{x}_{i^{(t)}_{n,s}})_s$ under the above change of basis. Then we have $(\widetilde{x})_0 \in \operatorname{ker} \partial$, and starting from $(\widetilde{x})_0$ every odd number of edge is marked by elements in $\mathcal{R}_U$ and every even number of edge is marked by elements in $\mathcal{R}_V$. In order to compute $\varphi_{i,j},$ we need only consider the set of (say) $\mathcal{R}_V$ edges. Each $\mathcal{R}_V$ edge that is not  equal to $V_T$ is marked by the differential of $(\widetilde{x})_s$ for some $s=1,\cdots,n.$ For $1\leq s\leq n,$

\begin{align*}
\partial(\widetilde{x})_s = V^{\Delta_\JJ(x_s,\beta_s)}_T W^{\Delta_\II(x_s,\beta_s)}_{T,0}\beta_s
\end{align*}

thus we simply need to look at the values of $\Delta_{\II,\JJ}(x_s,\beta_s).$

For the knot $T_{2,4k+3}$, $g=2k+1.$ By (<ref>), (<ref>) and (<ref>),  when $n\geq g$, the sequence of $\Delta_{\II,\JJ}(x_s,\beta_s)$ for $s=1,\cdots,n$ is given by

\begin{align*}
\end{align*}

when $n\leq g$, the above sequence terminates at $(k+1-n/2,n/2+k)$ if $n$ is even and at $(k-(n-1)/2,(n+1)/2+k)$ if $n$ is odd.

Therefore we conclude that for the interested complex, the concordance invariant $\varphi_{i,j}=-1$ if and only if $(i,j)=(n+k,k),(n+k-1,k),\cdots$ and $\varphi_{i,j}=0$ if $(i,j)$ is not among those values and also $(i,j)\neq (1,0).$

We now prove Proposition <ref> with a more precise restatement. The result is stated for knots in $S^3$, but a similar result should hold for knots in rational homology spheres as well.

For a knot $K\subset S^3,$ when $n\geq 2g,$ as a quotient complex of the filtered mapping cone (for any surgery), $A_{\ceil{n/2}}$ is filtered homotopy equivalent to $\CFKi(S^3,K).$

When $n\geq 2g,$ $\ceil{n/2}\geq g.$ Note that all the generators of $\CFKi(S^3,K)$ are within the region of $-g\leq i+j \leq g.$ It is then straightforward to check using (<ref>) and (<ref>) that this region is filtered.

Finally let us analyze the behavior of the complex $X^\infty_n(K)$ when $n\geq 2g.$ Comparing $X^\infty_n(K)$ and $X^\infty_{n+1}(K)$, we can identify $A_s$ in both complex for $s\leq \floor{n/2}.$  Under this identification, one can check that $\JJ(y_s) - \JJ(h_s(y_s))$ is constant while  $\JJ(y_s) - \JJ(v_s(y_s))$ increases as $n$ increases, and the $\II$ filtration shift is constant. At the same time, we can identify $A_s \subset X^\infty_n(K)$ with $A_{s+1} \subset X^\infty_{n+1}(K)$ for $s\geq \floor{n/2}+1 > g$, and under this identification, similarly the filtration shifts from the $v$ map is constant while the filtration shifts from the $h$ map increases as $n$ increases. Furthermore, by Proposition <ref>, the “middle” complex $A_{\ceil{(n+1)/2}}$ is simply a copy of $\CFKi(S^3,K)$ (with some differentials pointing to the rest of the complex).

Therefore, when $n\geq 2g$ and as $n$ increases by $1$, we conclude that there are two things happen to $X^\infty_n(K)$:

* certain edges are extended;

* a copy of $\CFKi(S^3,K)$ is added to the “middle” of the complex.

Aside from these two changes, the complex remains the same “shape” as $n$ increases.  We name this phenomenon stabilization. For an example, note that the whole family described by Proposition <ref> can be seen as the result of continued stabilizations. In general, with slightly more careful analysis, one should be able to pin down the exact behavior of a given complex during the stabilization.

[1]
Kenneth L. Baker, J. Elisenda Grigsby, and Matthew Hedden, Grid diagrams
for lens spaces and combinatorial knot Floer homology, Int. Math. Res.
Not. IMRN (2008), no. 10, Art. ID rnm024, 39. 2429242

[2]
Irving Dai, Jennifer Hom, Matthew Stoffregen, and Linh Truong, Homology
concordance and knot floer homology, arXiv preprint arXiv:2110.14803 (2021).

[3]
, More concordance homomorphisms from knot Floer homology,
Geom. Topol. 25 (2021), no. 1, 275–338. 4226231

[4]
Eaman Eftekhary, Heegaard floer homology and morse surgery, arXiv
preprint math/0603171 (2006).

[5]
, Bordered Floer homology and existence of incompressible tori
in homology spheres, Compos. Math. 154 (2018), no. 6, 1222–1268.

[6]
Michael H. Freedman and Frank Quinn, Topology of 4-manifolds, Princeton
Mathematical Series, vol. 39, Princeton University Press, Princeton, NJ,
1990. 1201584

[7]
Joshua Evan Greene, L-space surgeries, genus bounds, and the cabling
conjecture, J. Differential Geom. 100 (2015), no. 3, 491–506.

[8]
Matthew Hedden, On Floer homology and the Berge conjecture on knots
admitting lens space surgeries, Trans. Amer. Math. Soc. 363 (2011),
no. 2, 949–968. 2728591

[9]
Kristen Hendricks, Jennifer Hom, Matthew Stoffregen, and Ian Zemke,
Surgery exact triangles in involutive heegaard floer homology, arXiv
preprint arXiv:2011.00113 (2020).

[10]
, An involutive dual knot surgery formula, arXiv preprint
arXiv:2205.12798 (2022).

[11]
Matthew Hedden, Se-Goo Kim, and Charles Livingston, Topologically slice
knots of smooth concordance order two, J. Differential Geom. 102
(2016), no. 3, 353–393. 3466802

[12]
Jennifer Hom, ÇağrıKarakurt, and Tye Lidman, Surgery
obstructions and Heegaard Floer homology, Geom. Topol. 20
(2016), no. 4, 2219–2251. 3548466

[13]
Jennifer Hom and Tye Lidman, A note on surgery obstructions and
hyperbolic integer homology spheres, Proc. Amer. Math. Soc. 146
(2018), no. 3, 1363–1365. 3750247

[14]
Matthew Hedden and Adam Simon Levine, A surgery formula for knot floer
homology, arXiv preprint arXiv:1901.02488 (2019).

[15]
Jennifer Hom, Adam Simon Levine, and Tye Lidman, Knot concordance in
homology cobordisms, arXiv preprint arXiv:1801.07770 (2018).

[16]
Jennifer Hom, Tye Lidman, and Nicholas Zufelt, Reducible surgeries and
Heegaard Floer homology, Math. Res. Lett. 22 (2015), no. 3,
763–788. 3350104

[17]
Matthew Hedden and Thomas E. Mark, Floer homology and fractional Dehn
twists, Adv. Math. 324 (2018), 1–39. 3733880

[18]
Jennifer Hom, A survey on Heegaard Floer homology and concordance,
J. Knot Theory Ramifications 26 (2017), no. 2, 1740015, 24.

[19]
Matthew Hedden and Olga Plamenevskaya, Dehn surgery, rational open books
and knot Floer homology, Algebr. Geom. Topol. 13 (2013), no. 3,
1815–1856. 3071144

[20]
Matthew PF Jackson, The invariance of knot lattice homology, arXiv
preprint arXiv:2111.05229 (2021).

[21]
András Juhász, Dylan Thurston, and Ian Zemke, Naturality and
mapping class groups in Heegard Floer homology, Mem. Amer. Math. Soc.
273 (2021), no. 1338, v+174. 4337438

[22]
Eileen Li and Yi Ni, Half-integral finite surgeries on knots in $S^3$,
Ann. Fac. Sci. Toulouse Math. (6) 24 (2015), no. 5, 1157–1178.

[23]
Robert Lipshitz, Peter S. Ozsvath, and Dylan P. Thurston, Bordered
Heegaard Floer homology, Mem. Amer. Math. Soc. 254 (2018),
no. 1216, viii+279. 3827056

[24]
Ciprian Manolescu and Peter Ozsváth, Heegaard floer homology and
integer surgeries on links, arXiv preprint arXiv:1011.1317 (2010).

[25]
Seppo Niemi-Colvin, Invariance of knot lattice homology and homotopy,
Ph.D. thesis, Duke University, 2022.

[26]
Yi Ni and Zhongtao Wu, Cosmetic surgeries on knots in $S^3$, J. Reine
Angew. Math. 706 (2015), 1–17. 3393360

[27]
Peter Ozsváth and Zoltán Szabó, Holomorphic disks and knot
invariants, Adv. Math. 186 (2004), no. 1, 58–116. 2065507

[28]
, Holomorphic disks and three-manifold invariants: properties and
applications, Ann. of Math. (2) 159 (2004), no. 3, 1159–1245.

[29]
, Holomorphic disks and topological invariants for closed
three-manifolds, Ann. of Math. (2) 159 (2004), no. 3, 1027–1158.

[30]
, On the Heegaard Floer homology of branched double-covers,
Adv. Math. 194 (2005), no. 1, 1–33. 2141852

[31]
, Holomorphic triangles and invariants for smooth four-manifolds,
Adv. Math. 202 (2006), no. 2, 326–400. 2222356

[32]
, An introduction to Heegaard Floer homology, Floer homology,
gauge theory, and low-dimensional topology, Clay Math. Proc., vol. 5, Amer.
Math. Soc., Providence, RI, 2006, pp. 3–27. 2249247

[33]
Peter S. Ozsváth and Zoltán Szabó, Knot Floer homology and
integer surgeries, Algebr. Geom. Topol. 8 (2008), no. 1, 101–153.

[34]
, Knot Floer homology and rational surgeries, Algebr. Geom.
Topol. 11 (2011), no. 1, 1–68. 2764036

[35]
Jacob Rasmussen, Lens space surgeries and l-space homology spheres,
arXiv preprint arXiv:0710.2531 (2007).

[36]
Linh Truong, A refinement of the Ozsváth-Szabó large integer
surgery formula and knot concordance, Proc. Amer. Math. Soc. 149
(2021), no. 4, 1757–1771. 4242330

[37]
Ian Zemke, Graph cobordisms and heegaard floer homology, arXiv preprint
arXiv:1512.01184 (2015).

[38]
, Connected sums and involutive knot Floer homology, Proc.
Lond. Math. Soc. (3) 119 (2019), no. 1, 214–265. 3957835

[39]
, Link cobordisms and absolute gradings on link Floer homology,
Quantum Topol. 10 (2019), no. 2, 207–323. 3950650

[40]
Hugo Zhou, Homology concordance and an infinite rank free subgroup, J.
Topol. 14 (2021), no. 4, 1369–1395. 4406694