There are four possibilities for $D^{1}$, namely $\db\pm zC$ and $\db \pm \cl{z}C$. We have the following estimates for these operators:

Let $v\in W^{1,2}(\C,\C)$ or $v\in W^{1,2}(\cl{\mathbb{H}},\C,\R)$. Then we have the following estimates:

\begin{equation*}
\begin{aligned}
\norm{\db v}^{2}_{L^{2}}+\norm{zv}^{2}_{L^{2}}&\le \norm{\db v\pm z\cl{v}}_{L^{2}}^{2}+2\norm{v}^{2}_{L^{2}}\\
\norm{\db v}^{2}_{L^{2}}+\norm{zv}^{2}_{L^{2}}&\le \norm{\db v\pm \cl{z}\cl{v}}_{L^{2}}^{2}.
\end{aligned}
\end{equation*}

Using the smooth approximation result Proposition <ref>, we may suppose that $v$ is smooth, compactly supported, and takes real values along the boundary.

To prove the inequalities, we will need to integrate by parts two times. Let us focus on the first estimate. We start by computing:

\begin{equation*}
(-\bd\pm zC)(\db\pm zC)v=-\bd\db v\pm z\cl{\db v}-\pm z\bd \cl{v}-\pm 2\cl{v}+\abs{z}^{2}v.
\end{equation*}

Using the fact that $\cl{\db v}=\bd \cl{v}$ we conclude that two terms cancel and we are left with

\begin{equation}\label{eq:naive-babu}
(-\bd\pm zC)(\db\pm zC)v=-\bd\db v \mp 2\cl{v}+\abs{z}^{2}v.
\end{equation}

The naive idea is to multiply both sides by $\mathrm{Re}\,\mu_{0}(v,-)$, integrate, and use the formal adjoint property for $-\bd\pm zC=(\db\pm zC)^{*}$ and $-\bd=\db^{*}$. This naive argument would require that $\db v$ and $\db v\pm z\cl{v}$ take real values along the boundary, which we do not assume. However, as in the previous section, the fact that we integrate by parts twice will imply that the failures of formal adjointness will cancel out.

Indeed, we compute

\begin{equation*}
\mathrm{Re}\int \mu_{0}(v,-\bd\db v)\d s \d t=\mathrm{Re}\int\mu_{0}(v,-\bd_{s}\db v)\d s\d t+\mathrm{Re}\int \mu_{0}(v,i\bd_{t}\db v) \d s\d t.
\end{equation*}

It is clear that the can integrate by parts with respect to $\bd_{s}$, and conclude

\begin{equation*}
\mathrm{Re}\int \mu_{0}(v,-\bd\db v)\d s \d t=\mathrm{Re}\int\mu_{0}(\bd_{s}v,\db v)\d s\d t+\mathrm{Re}\int \mu_{0}(v,i\bd_{t}\db v) \d s\d t.
\end{equation*}

We can also integrate by parts with respect to $\bd_{t}$, at the expense of a boundary integral term, and (after some simplification) end up with:

\begin{equation*}
\mathrm{Re}\int \mu_{0}(v,-\bd\db v)\d s \d t=\mathrm{Re}\int\mu_{0}(\db v,\db v)\d s\d t-\mathrm{Re}\int_{\R}\mu_{0}(v,i\db v) \d s\d t.
\end{equation*}

We do the same computation with $\db$ replaced by $D=\db\pm zC$, and conclude that

\begin{equation*}
\mathrm{Re}\int \mu_{0}(v,D^{*}Dv)\d s \d t=\mathrm{Re}\int\mu_{0}(D v,D v)\d s\d t-\mathrm{Re}\int_{\R}\mu_{0}(v,iDv)\d s\d t.
\end{equation*}

Finally, we observe that

\begin{equation*}
\begin{aligned}
\mathrm{Re}\int_{\R}\mu_{0}(v,iDv)\d s\d t&=\mathrm{Re}\int_{\R}\mu_{0}(v,i\db v)\d s\d t\pm \mathrm{Re}\int_{\R}\mu_{0}(v,iCz\cl{v})\d s\d t\\ &=\mathrm{Re}\int_{\R}\mu_{0}(v,i\db v)\d s\d t,
\end{aligned}
\end{equation*}

where we have used the fact that $Cz\cl{v}$ takes real values along the boundary. Therefore

\begin{equation*}
\mathrm{Re}\int \mu_{0}(v,D^{*}Dv)\d s \d t-\norm{Dv}_{L^{2}}^{2}=\mathrm{Re}\int \mu_{0}(v,\db^{*} \db v)\d s \d t-\norm{\db v}_{L^{2}}^{2}.
\end{equation*}

Applying $\mathrm{Re}\,\mu_{0}(v,-)$ to (<ref>) and integrating implies

\begin{equation*}
\norm{\db v\pm z\cl{v}}_{L^{2}}^{2}=\norm{\db v}^{2}_{L^{2}}\mp 2\mathrm{Re}\int \mu_{0}(v,\cl{v})+\norm{zv}_{L^{2}}^{2}.
\end{equation*}

We rearrange and estimate to conclude that

\begin{equation*}
\norm{\db v}^{2}_{L^{2}}+\norm{zv}_{L^{2}}^{2} \le \norm{\db v\pm z\cl{v}}_{L^{2}}^{2}+ 2\norm{v}_{L^{2}}^{2},
\end{equation*}

as desired. The second estimate in the statement of the lemma (with $D=\db\pm \cl{z}C$) is proved in the same manner.

§.§ Classifying the kernels of $D^{1}$ (six cases).

The second estimate in Lemma <ref> implies that $\db v\pm \cl{z}\cl{v}=0$ has no non-zero solutions – this takes care of three of the six kinds of operators.

Our next lemma shows that $\db v\pm z\cl{v}=0$ has either a one-dimensional space of solutions or a zero-dimensional space of solutions.

Suppose that $v:\C\to \C$ is in $L^{2}$, then

\begin{equation*}
\begin{aligned}
\db v-z\cl{v}=0&\iff v=ci\exp(-\frac{1}{2}\abs{z}^{2})\text{ for $c\in \R$},\\
\db v+z\cl{v}=0&\iff v=c\exp(-\frac{1}{2}\abs{z}^{2})\text{ for $c\in \R$}.
\end{aligned}
\end{equation*}

On the other hand if $v:\cl{\mathbb{H}}\to \C$ is in $L^{2}$ and takes real values along the boundary, then

\begin{equation*}
\begin{aligned}
\db v-z\cl{v}=0&\iff v=0\\
\db v+z\cl{v}=0&\iff v=c\exp(-\frac{1}{2}\abs{z}^{2})\text{ for $c\in \R$}.
\end{aligned}
\end{equation*}

Morally, this says that positive interior zeros and $(+,+)$ zeros contribute one dimension to the kernel, but $(+,-)$ zeros do not contribute to the kernel.

Observe that if we set $v^{\prime}=iv$, then

\begin{equation*}
\bd_{s}v^{\prime}+i\bd_{t}v^{\prime}-z\cl{v^{\prime}}=i(\bd_{s}v+i\bd_{t}v+z\cl{v}),
\end{equation*}

and hence it suffices to study the equation $\db v-z\cl{v}=0$. Following <cit.>, we prove that the real part of $v$ must vanish identically.

The second estimate from Lemma <ref> implies that $\db{v}\in L^{2}$ and $zv\in L^{2}$ (proof: both $\norm{\db(\rho(z\delta)v)}_{L^{2}}$ and $\norm{z\rho(z\delta)v}_{L^{2}}$ remain bounded as $\delta\to 0$). The $L^{2}$ elliptic estimates then imply that $v\in W^{1,2}$.

Let $y=\mathrm{Re}(v)$. Since $-\Delta v+2\cl{v}+\abs{z}^{2}v=0$, we have

\begin{equation*}
0=-\Delta y+(2+\abs{z}^{2})y.
\end{equation*}

Apply $\mathrm{Re}\,\mu_{0}(\rho(z\delta)y,-)$ to both sides, and integrate by parts to conclude

\begin{equation*}
0=\int \rho(z\delta)\abs{\db y}^{2}\d s\d t+\mathrm{Re}\int \mu_{0}(\db (\rho(z\delta))\cdot y,\db y)\d s\d t+\int \rho(z\delta)(2+\abs{z}^{2})\abs{y}^{2}\d s\d t.
\end{equation*}

When we integrate by parts, we use $\bd_{t}y=0$ (which holds in our case). We can now take the limit $\delta\to 0$, since we have verified that $\db y, zy\in L^{2}$, and conclude that

\begin{equation*}
y}_{L^{2}}^{2}+\norm{(2+\abs{z}^{2})^{1/2}y}_{L^{2}}^{2}\implies y=0,
\end{equation*}

using $\db(\rho(z\delta))=O(\delta)$. It follows that any $L^{2}$ solution of $\db v-z\cl{v}=0$ as in the statement of the lemma is purely imaginary (and hence vanishes along the boundary, if the boundary exists).

We now observe that $v=i\exp(-\frac{1}{2}\abs{z}^{2})$ is certainly in $L^{2}$ and solves $\db v-z\cl{v}=0$. Clearly any other solution $v^{\prime}$ will be $v^{\prime}=gv$ for some holomorphic $g$; moreover, by what we have shown above, $g$ must be real. There are no non-constant holomorphic functions defined on $\mathbb{C}$ or $\mathbb{H}$ which take only real values (the rank of the derivative matrix would be always $0$). Thus $g=c$ must be a real number.

In the case when $v$ is defined on $\cl{\mathbb{H}}$, the only possibility is $c=0$, since otherwise $v$ would take non-zero imaginary values along the boundary.

Finally, we return to the second equation from the statement, $\bd_{s}v+i\bd_{t}v+z\cl{v}=0$. We have shown that this solution is conjugate to the first equation under multiplication by $i$. Therefore all solutions on the disk or half-plane are given by $v=c\exp(-\frac{1}{2}\abs{z}^{2})$ for some real $c$. In this case we can have non-zero $c$ when $v$ is defined on $\cl{\mathbb{H}}$.

This completes the proof.

§.§.§ The formal adjoint near the zeros

Since we chose our metric so that $\abs{\bd_{s}}=1$ in all of the special coordinate charts centered at the zeros of $B$, we can easily compute the coordinate representations of $D^{\sigma,\ast}$:

\begin{equation*}
\begin{aligned}
D^{\sigma}=\db\pm \sigma zC&\implies D^{\sigma,*}(u)=-\bd \pm \sigma zC\\
D^{\sigma}=\db\pm \sigma\cl{z} C&\implies D^{\sigma,*}(u)=-\bd \pm \sigma\cl{z}C.
\end{aligned}
\end{equation*}

Now let $D^{\sigma,\dagger}=-C\circ D^{\sigma,\ast}\circ C$. The above yields:

\begin{equation*}
\begin{aligned}
D^{\sigma}=\db\pm \sigma zC&\implies D^{\sigma,\dagger}(u)=\db \mp \sigma\cl{z}C\\
D^{\sigma}=\db\pm \sigma\cl{z}C&\implies D^{\sigma,\dagger}(u)=\db \mp \sigma zC.
\end{aligned}
\end{equation*}

Thus we can think of $D^{\sigma}\mapsto D^{\sigma,\dagger}$ as defining a “duality involution” on the set of six local model equations. This is illustrated in Figure <ref>.

To explain the labeling scheme used in the figure, we partition the zero set of $B$, denoted $\mathrm{Z}$, into six kinds of zeros:

\begin{equation*}
\mathrm{Z}=\mathrm{Z}^{+}\cup \mathrm{Z}^{-}\cup \mathrm{Z}^{++}\cup \mathrm{Z}^{+-}\cup \mathrm{Z}^{-+}\cup \mathrm{Z}^{--},
\end{equation*}

where $\mathrm{Z}^{\pm}$ are interior positive/negative zeros, and $\mathrm{Z}^{\pm\pm}$ are boundary zeros (let's agree for this notation that the two $\pm$ signs are independent). The convention for assigning labels is via the linearization: the first sign is the linearization of $B$ allowing arbitrary deformations, and the second sign is for the linearization only allowing deformations along the boundary. The local form of $D^{\sigma}$ near a zero $\zeta$ and the corresponding count is summarized in Figure <ref>.

It follows from the construction in Section <ref> that the sum of the counts of all the zeros in $\mathrm{Z}$ is equal to $\mathrm{X}+\mu^{\tau}_{\mathrm{Mas}}$.

[shift=(5.2,0)] (-2.5,-8.5)rectangle+(5,12);
[shift=(10.4,0)] (-2.5,-8.5)rectangle+(10,12);

[black!50!white,fill=black!5!white] (2,0) arc (0:180:2);
(A) at (0,2) [above] $\db+\sigma zC$;

[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (0,2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (45:2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (135:2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);
(B) at (0,-0.2)[below]$\mathrm{count}=+1$;
at (B.south)[below] $\mathrm{Z}^{++}$;

(A) at (0,2) [above] $\db-\sigma zC$;

[black!50!white,fill=black!5!white] (2,0) arc (0:180:2);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (0,2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (45:2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (135:2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);
(B) at (0,-0.2)[below]$\mathrm{count}=0$;
at (B.south)[below] $\mathrm{Z}^{+-}$;

(A) at (3,0.5) [right] $\db-\sigma zC$;
(B) at (A.south) [below] $\mathrm{count}=+1$;
at (B.south)[below]$\mathrm{Z}^{+}$;

[black!50!white,fill=black!5!white] (0,0) circle (2);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (0,2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (0,-2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (45:2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (135:2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (225:2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (315:2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);

(A) at (0,2) [above] $\db-\sigma\cl{z}C$;

[black!50!white,fill=black!5!white] (2,0) arc (0:180:2);

(2,0) arc (0:180:2)–cycle;
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] plot[domain=0.5:2] (,1/);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] plot[domain=0.5:2] (-,1/);

[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (0,2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);
(B) at (0,-0.2)[below]$\mathrm{count}=-1$;
at (B.south)[below] $\mathrm{Z}^{--}$;

(A) at (0,2) [above] $\db+\sigma\cl{z}C$;

[black!50!white,fill=black!5!white] (2,0) arc (0:180:2);

(2,0) arc (0:180:2)–cycle;
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] plot[domain=0.5:2] (,1/);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] plot[domain=0.5:2] (-,1/);

[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (0,2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);
(B) at (0,-0.2)[below]$\mathrm{count}=0$;
at (B.south)[below] $\mathrm{Z}^{-+}$;

(A) at (3,0.5) [right] $\db+\sigma\cl{z}C$;
(B) at (A.south) [below] $\mathrm{count}=-1$;
at (B.south)[below]$\mathrm{Z}^{-}$;
[black!50!white,fill=black!5!white] (0,0) circle (2);

(0,0) circle (2);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] plot[domain=0.5:2] (,1/);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] plot[domain=0.5:2] (,-1/);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] plot[domain=0.5:2] (-,1/);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] plot[domain=0.5:2] (-,-1/);

[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (0,2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (0,-2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);

The six kinds of zeros and the coordinate representation of $D^{\sigma}$ in each chart. Two zeros are in the same box if the operators are dual in the sense defined above.

Applying Lemmas <ref> and <ref> to $D^{1,\dagger}$ yields the following result for $D^{1,\ast}$.

Suppose $v:\C\to \C$ is in $L^{2}$. Then

\begin{equation*}
\begin{aligned}
-\bd v-z\cl{v}=0&\iff v=0\\
-\bd v+\cl{z}\cl{v}=0&\iff v=ic\exp(-\frac{1}{2}\abs{z}^{2})\text{ for some $c\in \R$}.
\end{aligned}
\end{equation*}

Now suppose that $v:\cl{\mathbb{H}}\to \C$ is in $L^{2}$ and takes real values along the boundary. Then

\begin{equation*}
\begin{aligned}
-\bd v\pm z\cl{v}=0&\iff v=0\\
-\bd v+\cl{z}\cl{v}=0&\iff v=0\\
-\bd v-\cl{z}\cl{v}=0&\iff v=c\exp(-\frac{1}{2}\abs{z}^{2})\text{ for some $c\in \R$}.\\
\end{aligned}
\end{equation*}

Heuristically, this says that the zeros with count $-1$ in Figure <ref> contribute a one-dimensional subspace to the kernel of the formal adjoint $D^{\sigma,\ast}$ (and all other zeros contribute nothing).

§.§ Linear compactness and a stabilization of $D^\rho$

In this section we will relate the kernel and cokernel of $D^{\rho}$ to the kernels and cokernels of the local models $D^{1}$. We begin with an explanation of the rescaling scheme we use.

§.§.§ Modified rescaling maps

Suppose that $\zeta$ is a zero and let $z$ be the special coordinate chart centered at $\zeta$. By convention, $z$ is either $\cl{\mathbb{H}}\cap D(1)$ or $D(1)$ valued. Let $\rho$ be a bump function supported in $D(1)$ which is $1$ on $D(1/2)$.

Let $\Phi_{\sigma}:L^{2}(\C,\C)\to L^{2}(\dot\Sigma,\C)$ be the modified rescaling map:

\begin{equation*}
\Phi_{\sigma}(v)=\rho\cdot\sigma^{1/2}v(\sigma^{1/2}z).
\end{equation*}

Observe that $\norm{\Phi_{\sigma}(v)}_{L^{2}}\le \norm{v}_{L^{2}}=\lim_{\sigma\to\infty}\norm{\Phi_{\sigma}(v)}_{L^{2}}$. Dually, we let $\Pi_{\sigma}=\Phi_{\sigma}^{\ast}$ be the adjoint. It is easy to obtain the following explicit formula for $\Pi_{\sigma}$:

\begin{equation*}
\Pi_{\sigma}(u)(z)=\sigma^{-1/2}\rho(\sigma^{-1/2}z)u(\sigma^{-1/2}z).
\end{equation*}

[black!10!white] (0.5,1) circle (0.2);
[fill=black!10!white] (0,-0.4) arc (-90:90:0.2);
(0,-2)–+(0,4) (1,-2)–+(0,4) (0.5,1) circle (0.2) (4,0) circle (2);
(4,0)–+(100:2)coordinate(B) (4,0)–+(220:2)coordinate(C);
[dashed] (0.5,1) +(100:0.2)–(B) (0.5,1) +(240:0.2)–(C);

(-4,-2) arc (-90:90:2)–cycle;
[dashed] (-4,0)+(60:2)–(0,0) (-4,0)+(-60:2)–(0,-.4);

[draw,fill=black,inner sep=1pt,circle] (B) at (4,0) ;
at (B.east)[right]$\zeta\in \mathrm{Z}^{\pm}$;

[draw,fill=black,inner sep=1pt,circle] (B) at (-4,0) ;
at (B.west)[left]$\zeta\in \mathrm{Z}^{\pm\pm}$;

Rescaling sections near the zeros of $B$. The map $\Phi_{\sigma}$ takes a section on the large domain and compresses it to fit inside the small domain (and then cuts it off by $\rho$). The map $\Pi_{\sigma}$ does the opposite, it first cuts off by $\rho$ and then expands the domain of the section. The factors have been chosen so that $\norm{\Phi_{\sigma}(v)}_{L^{2}}=\norm{\rho(\sigma^{-1/2}z)v}_{L^{2}}$.

The relevance of $\Pi_{\sigma},\Phi_{\sigma}$ is how they interact with $D^{\sigma}$.
Suppose that $D^{\sigma}=\db+\sigma \alpha(z)C$ and let $D^{1}=\db+\alpha(z)C$ (where $\alpha=\pm z,\pm \cl{z}$). Then we easily compute

$$D^{\sigma}\circ \Phi_{\sigma}(v)=\sigma^{1/2}\Phi_{\sigma}(D^{1}(v))+(\db \rho)\sigma^{1/2}v(\sigma^{1/2}z).$$
Recall that the $L^{2}$ norm of $\lambda v(\lambda z)$ is constant as function of $\lambda$. A similar computation can be done using $\Pi_{\sigma}$, and we conclude:

\begin{equation}\label{eq:impo-lat-1}
\begin{aligned}
\norm{D^{\sigma}(\Phi_{\sigma}(v))-\sigma^{1/2}\Phi_{\sigma}(D^{1}(v))}_{L^{2}}\le c(\rho)\norm{v}_{L^{2}(D(\sigma)\setminus D(\sigma/2))},\\
\norm{\sigma^{1/2}D^{1}(\Pi_{\sigma}(u))-\Pi_{\sigma}(D^{\sigma}(u))}_{L^{2}}\le c(\rho)\norm{u}_{L^{2}(D(1)\setminus D(1/2))}.
\end{aligned}
\end{equation}

We similarly note the behavior of $D^{\sigma,\ast}$ under $\Phi_{\sigma}$ and $\Pi_{\sigma}$:

\begin{equation}\label{eq:impo-lat-2}
\begin{aligned}
\norm{D^{\sigma,\ast}(\Phi_{\sigma}(v))-\sigma^{1/2}\Phi_{\sigma}(D^{1,\ast}(v))}_{L^{2}}\le c(\rho)\norm{v}_{L^{2}(D(\sigma)\setminus D(\sigma/2))},\\
\norm{\sigma^{1/2}D^{1,\ast}(\Pi_{\sigma}(u))-\Pi_{\sigma}(D^{\sigma,\ast}(u))}_{L^{2}}\le c(\rho)\norm{u}_{L^{2}(D(1)\setminus D(1/2))}.
\end{aligned}
\end{equation}

These estimates will be important later on. They essentially say that a uniform bound on $\norm{D^{\sigma}(u)}_{L^{2}}$ and $\norm{u}_{L^{2}}$ implies that $\norm{D^{1}(v)}_{L^{2}}=O(\sigma^{-1/2})$ where $v=\Pi_{\sigma}(u)$.

§.§.§ A linear compactness result

In this section we will prove a compactness theorem which concerns sequences $\xi_{n}$ with $\norm{D^{\sigma_{n}}(\xi_{n})}<C$ and $\sigma_{n}\to\infty$. To set the stage, let $z_{\zeta}$ be the chosen holomorphic coordinate centered on the zero $\zeta$ (as above), and recall the modified rescaling maps:

\begin{equation*}
\text{$\Phi_{\sigma,\zeta}(v)=\rho\cdot \sigma^{1/2}v(\sigma^{1/2}z_{\zeta})$, and  $\Pi_{\sigma,\zeta}(u)=\sigma^{-1/2}\rho(\sigma^{-1/2}z_{\zeta})u(\sigma^{-1/2}z_{\zeta})$.}
\end{equation*}

Let $\Pi_{\sigma}=\oplus_{\zeta\in \mathrm{Z}}\Pi_{\sigma,\zeta}$ be considered as a map

\begin{equation*}
\Pi_{\sigma}:L^{2}(\dot\Sigma,E)\to \bigoplus_{\zeta\in \mathrm{Z}^{\pm}}L^{2}(\C,\C) \oplus \bigoplus_{\zeta\in \mathrm{Z}^{\pm\pm}}L^{2}(\cl{\mathbb{H}},\mathbb{C})=H.
\end{equation*}

The same formula also defines $\Pi_{\sigma}$ on $L^{2}(\dot\Sigma,\Lambda^{1,0}\otimes E)$. We can think of $H$ as the Hilbert space of $L^{2}$ sections on a disjoint union of finitely many copies of $\C$ and $\cl{\mathbb{H}}$.

We define an operator $D^{1}:H\to H$ (with dense domain) whose restriction to each factor equals the choice of $\db\pm \alpha(z)C$ for $\alpha(z)=z,\cl{z}$ given by Figure <ref>. We similarly define $D^{1,\ast}:H\to H$ where the local form is $-\bd\pm \alpha(z)C$, as appropriate.

The results of Lemmas <ref>, <ref> and Corollary <ref> give a complete classification of the elements in $\ker D^{1}$ and $\ker D^{1,\ast}$. See (<ref>) in the next section for a summary of the kernel of $D^{1}$ and $D^{1,\ast}$.

We let $\mathrm{R}_{\sigma}(\xi)=\xi-\sum_{\zeta\in \mathrm{Z}} \rho(z_{\zeta})\xi$ which we think of as the “remainder” after cutting off. It follows easily from the definitions that

\begin{equation}\label{eq:remainder}
\begin{aligned}
D^{\sigma}(\mathrm{R}_{\sigma}(\xi))&=\db \rho\otimes \xi+\mathrm{R}_{\sigma}(D^{\sigma}(\xi)).\\
\norm{\xi}_{L^{2}}&\le \norm{\mathrm{R}_{\sigma}(\xi)}_{L^{2}}+\norm{\Pi_{\sigma}(\xi)}_{L^{2}}\le 2\norm{\xi}_{L^{2}}.
\end{aligned}
\end{equation}

Let $\xi_{n}\in W^{1,2}(E,F)$ be a sequence so that $\norm{\xi_{n}}_{L^{2}}+\norm{D^{\sigma_{n}}(\xi_{n})}_{L^{2}}$ remains bounded for some sequence $\sigma_{n}\to\infty$. Then

* $\norm{R_{\sigma_{n}}(\xi_{n})}_{L^{2}}\to 0$.

* After passing to a subsequence, $\Pi_{\sigma_{n}}(\xi_{n})\to \mathbf{k}$ in $L^{2}$ for some element $\mathbf{k}\in \ker D^{1}$.

The same holds with $(\xi_{n},E,F,D^{\sigma_{n}},D^{1},\mathbf{k})$ replaced by $(\eta_{n},\Lambda^{0,1}\otimes E,F^{*},D^{\sigma_{n},\ast},D^{1,\ast},\mathbf{c})$.

We will only prove the $\xi_{n}$ case, leaving the $\eta_{n}$ case to the reader. To avoid too much clutter, we suppress some notation and write $\sigma:=\sigma_{n}$, $\xi_{n}:=\xi$. Keep in mind that $\rho$ is a fixed bump function.

Let's begin the proof. Using (<ref>) together with the Bochner-Weitzenböck estimate (<ref>) implies that

\begin{equation*}
\norm{B\mathrm{R}_{\sigma}(\xi)}^{2}_{L^{2}}\le \sigma^{-2}\norm{\db \rho}^{2}_{C^{0}}\norm{\xi}_{L^{2}}^{2}+\sigma^{-2}\norm{D^{\sigma}(\xi)}_{L^{2}}^{2}+C\sigma^{-1}\norm{\xi}_{L^{2}}^{2}.
\end{equation*}

However, $\mathrm{R}_{\sigma}(\xi)$ is supported on $\dot\Sigma\setminus D(\zeta_{1},1/2)\setminus D(\zeta_{2},1/2)\setminus \cdots $ and it follows that $\abs{B}>b>0$ for some fixed constant $b$ on the support of $\mathrm{R}_{\sigma}(\xi)$. Therefore we conclude that

\begin{equation*}
\norm{\mathrm{R}_{\sigma}(\xi)}^{2}_{L^{2}}\le b^{-1}((C\sigma^{-1}+c_{\rho}\sigma^{-2})\norm{\xi}_{L^{2}}^{2}+\sigma^{-2}\norm{D^{\sigma}(\xi)}^{2}_{L^{2}})=O(\sigma^{-1}).
\end{equation*}

This proves part (a).

For part (b), we use (<ref>) to conclude

\begin{equation*}
\norm{D^{1}(\Pi_{\sigma}(\xi))}\le\sigma^{-1/2}(\norm{D^{\sigma}(\xi))}_{L^{2}}+c_{\rho}\norm{\xi}_{L^{2}})=O(\sigma^{-1/2}).
\end{equation*}

Let $v_{n}=\Pi_{\sigma}(\xi)$. Then $\norm{v_{n}}_{L^{2}}$ is bounded and $\norm{D^{1}(v_{n})}=O(\sigma_{n}^{-1/2})$. We will now use the local Bochner Weitzenböck estimates (Lemma <ref>) to conclude that we have

\begin{equation}\label{eq:fancy-man}
\begin{aligned}
\norm{v_{n}}_{L^{2}}+\norm{\db v_{n}}_{L^{2}}+\norm{zv_{n}}_{L^{2}}&=O(1)\\
\norm{D^{1}(v_{n})}_{L^{2}}&=O(\sigma_{n}^{-1/2}).
\end{aligned}
\end{equation}

The first estimate above is actually enough to imply that a subsequence of $v_{n}$ converges to some limit $v_{\infty}$ in $L^{2}$; we will explain this step momentarily. The second estimate will imply that $D^{1}(v_{\infty})=0$. This will complete the proof.

Before we move on, note that the $L^{2}$ elliptic estimates for $\db$ and the first estimate above implies that $v_{n}$ is uniformly bounded in $W^{1,2}$.

We can phrase the next part of our argument rather generally. If we let

\begin{equation*}
W=\set{v\in H\text{ and }\norm{v}_{W^{1,2}}+\norm{zv}_{L^{2}}\le C},
\end{equation*}

(with the obvious induced norm) then the inclusion $W\to H$ is compact; we will prove this below. To see how it applies to our problem, observe that the $L^{2}$ estimates for $\db$ and the first part of (<ref>) imply that $\norm{v_{n}}_{W^{1,2}}+\norm{zv_{n}}_{L^{2}}$ is bounded, and hence $v_{n}$ is bounded in $W$. Therefore, after passing to a subsequence, $v_{n}$ converges to some limit $v_{\infty}$ in $L^{2}$. If $\varphi$ is any test function (taking real values along the boundary) then we have

\begin{equation*}
\ip{D^{1,\ast}\varphi,v_{\infty}}=\lim\ip{D^{1,\ast}\varphi,v_{n}}\to 0,
\end{equation*}

and hence $D^{1}v=0$ weakly. By our elliptic regularity results $v$ is smooth, takes real values along the boundary, and $D^{1}v=0$ holds pointwise, as desired. We can then set $\mathbf{k}=v_{\infty}$ to complete the proof.

It remains to show why $W\to H$ is a compact inclusion. It is well-known that $W^{1,2}(\Omega(r))\subset L^{2}(\Omega(r))$ is a compact inclusion for $\Omega(r)=D(r)$ or $\Omega(r)=D(r)\cap \cl{\mathbb{H}}$. Thus, by a diagonal argument, we can pass to a subsequence $v_{n}$ and that $v_{n}\to v_{\infty}$ for some limit $v\in L^{2}_{\mathrm{loc}}$ (in the $L^{2}_{\mathrm{loc}}$ topology).

We easily estimate

\begin{equation*}
\norm{v_{n}}^{2}_{L^{2}(\Omega(2^{k})\setminus \Omega(2^{k-1}))}\le \frac{1}{4^{k-1}}\norm{zv_{n}}^{2}_{L^{2}(\Omega(2^{k})\setminus \Omega(2^{k-1}))}.
\end{equation*}

Since $\Omega(2r)\setminus \Omega(r)$ is precompact, we must have

\begin{equation*}
\norm{v_{\infty}}^{2}_{L^{2}(\Omega(2^{k})\setminus\Omega(2^{k-1}))}=\lim\norm{v_{n}}^{2}_{L^{2}(\Omega(2^{k})\setminus \Omega(2^{k-1}))}\le \frac{C^{2}}{4^{k-1}}.
\end{equation*}

Since the right hand side is summable, we conclude that $v_{\infty}$ is actually in $L^{2}$. Now for all $k$ we have

\begin{equation*}
\begin{aligned}
\norm{v_{\infty}-v_{n}}_{L^{2}}^{2}&\le \norm{v-v_{n}}_{L^{2}(\Omega(2^{k}))}^{2}+\sum_{\ell>k} \norm{v_{\infty}}_{L^{2}(\Omega(2^{\ell})\setminus\Omega(2^{\ell-1}))}^{2}+\norm{v_{n}}_{L^{2}(\Omega(2^{\ell})\setminus\Omega(2^{\ell-1}))}^{2}.\\
&\le \norm{v_{\infty}-v_{n}}_{L^{2}(\Omega(2^{k}))}^{2}+2C^{2}4^{-k}
\end{aligned}
\end{equation*}

Pick $k$ large enough that the last term is less than $\epsilon$, and then take the limit $n\to\infty$, yielding

\begin{equation*}
\limsup\norm{v_{\infty}-v_{n}}_{L^{2}}^{2}\le \epsilon.
\end{equation*}

This implies that $v_{n}\to v_{\infty}$ in $L^{2}$, completing the proof.

§.§.§ Stabilizing $D^{\sigma}$ and computing its index

In this section we will stabilize $D^{\sigma}$ by adding a cokernel element $\mathbf{c}_{\zeta}$ for each zero $\zeta$ with count $-1$ (Figure <ref>). We will also “co”-stabilize it by adding a kernel element $\mathbf{k}_{\zeta}$ for each $\zeta$ with count $+1$.

We define the following elements of $L^{2}(\C,\C)$ and $L^{2}(\cl{\mathbb{H}},\C)$:

\begin{equation}
\label{eq:kernel-cokernel-elts}
\begin{aligned}
\text{at $\zeta\in \mathrm{Z}^{+}$}&\hspace{1cm}\mathbf{k}_{\zeta}=i\exp(-\frac{1}{2}\abs{z}^{2})\text{ and }\mathbf{c}_{\zeta}=0,\\
\text{at $\zeta\in \mathrm{Z}^{-}$}&\hspace{1cm}\mathbf{k}_{\zeta}=0\text{ and }\mathbf{c}_{\zeta}=i\exp(-\frac{1}{2}\abs{z}^{2}),\\
\text{at $\zeta\in \mathrm{Z}^{++}$}&\hspace{1cm}\mathbf{k}_{\zeta}=\exp(-\frac{1}{2}\abs{z}^{2})\text{ and }\mathbf{c}_{\zeta}=0,\\
\text{at $\zeta\in \mathrm{Z}^{--}$}&\hspace{1cm}\mathbf{k}_{\zeta}=0\text{ and }\mathbf{c}_{\zeta}=\exp(-\frac{1}{2}\abs{z}^{2}),\\
\text{at $\zeta\in \mathrm{Z}^{+-}\cup \mathrm{Z}^{-+}$}&\hspace{1cm}\mathbf{k}_{\zeta}=0\text{ and }\mathbf{c}_{\zeta}=0,
\end{aligned}
\end{equation}

The results of Lemmas <ref>, <ref> and Corollary <ref> show that $\mathrm{span}_{\zeta\in \mathrm{Z}}(\mathbf{k}_{\zeta})=\ker D^{1}\subset H$, and $\mathrm{span}_{\zeta\in \mathrm{Z}}(\mathbf{c}_{\zeta})=\ker D^{1,\ast}\subset H$.

Keeping track of the counts of the various kinds of zeros, we see that

\begin{equation}\label{eq:final-county-boi}
\dim \ker D^{1}-\dim \ker D^{1,\ast}=\mathrm{X}+\mu^{\tau}_{\mathrm{Mas}}.
\end{equation}

Throughout the subsequent arguments, we will use $\mathbf{k}$ and $\mathbf{c}$ to denote linear combinations of the above basic kernel and cokernel elements.

We consider $\Phi_{\sigma}(\mathbf{k})$ and $\Phi_{\sigma}(\mathbf{c})$ as elements of $W^{1,2}(E,F)$ and $W^{1,2}(\Lambda^{0,1}\otimes E,F^{*})$, using the special coordinate charts $z_{\zeta}$ and frames $Y$ defined above.

We define the stabilized operator by the formula:

\begin{equation*}
\begin{aligned}
&D^{\sigma}_{\mathrm{st}}:W^{1,2}(E,F)\oplus \ker D^{1,\ast}\to L^{2}(\Lambda^{0,1}\otimes E)\oplus \ker D^{1}\\
\end{aligned}
\end{equation*}

Note that the second factor is simply an orthogonal projection. The following result will complete the proof of the index formula.

The operator $D^{\sigma}_{\mathrm{st}}$ is an isomorphism for $\sigma$ sufficiently large.

See <cit.> for a similar result.

We summarize the strategy. First prove that $D^{\sigma}_{\mathrm{st}}$ is eventually uniformly injective, in the sense that there are constants $C,\sigma_{0}$ so that

\begin{equation}\label{eq:eventually-inj}
\sigma>\sigma_{0}\implies \norm{(\xi,\mathbf{c})}_{L^{2}}\le C\norm{D^{\sigma}_{\mathrm{st}}(\xi,\mathbf{c})}_{L^{2}}.
\end{equation}

Second, we show that $D^{\sigma}_{\mathrm{st}}$ is eventually surjective. Then $D^{\sigma}_{\mathrm{st}}$ is eventually an isomorphism, as desired.

We prove (<ref>) by contradiction; suppose not and then conclude a sequence $\sigma_{n}\to\infty$ and elements $(\xi_{n},\mathbf{c}_{n})$ so that $\norm{(\xi_{n},\mathbf{c}_{n})}_{L^{2}}=1$ but $\norm{D^{\sigma_{n}}_{\mathrm{st}}(\xi_{n},\mathbf{c}_{n})}_{L^{2}}\to 0$. Let's agree to abbreviate $\sigma=\sigma_{n}$ to avoid excessive subscripts during the course of this argument.

It is clear that

\begin{equation*}
\norm{D^{\sigma}(\xi_{n})}_{L^{2}}\le \norm{D^{\sigma}_{\mathrm{st}}(\xi_{n},\mathbf{c}_{n})}_{L^{2}}+C\norm{\mathbf{c}_{n}}_{L^{2}},
\end{equation*}

for a fixed constant $C$. In particular, we can apply our compactness result to $\xi_{n}$ and conclude that, after passing to a subsequence $\Pi_{\sigma}(\xi_{n})$ converges to $\mathbf{k}$ and $\mathrm{R}_{\sigma}(\xi_{n})$ converges to $0$. However, since $\mathbf{k}_{\zeta}$ form an orthogonal basis for $\ker D^{1}$ we have

\begin{equation*}    \mathbf{k}=\lim_{n\to\infty}\sum_{\zeta}\norm{\mathbf{k}_{\zeta}}^{-2}\ip{\Pi_{\sigma}(\xi_{n}),\mathbf{k}_{\zeta}}\mathbf{k}_{\zeta}.
\end{equation*}

Therefore $D^{\sigma}_{\mathrm{st}}(\xi,\mathbf{c})\to 0$ implies that $\mathbf{k}=0$. Therefore $\Pi_{\sigma}(\xi_{n})$ converges to zero in $L^{2}$, and since we know $\mathrm{R}_{\sigma}(\xi_{n})\to 0$, we conclude $\xi_{n}$ converges to zero in $L^{2}$.

In order to contradict our initial assumption, it suffices to show that the inner product $\ip{\Phi_{\sigma}(\mathbf{c}),D^{\sigma}(\xi)}$ converges to zero (because then $\norm{\mathbf{c}_{n}}^{2}\le \norm{D^{\sigma}_{\mathrm{st}}(\xi_{n},\mathbf{c}_{n})}^{2}+\epsilon$ must hold eventually, by Pythagoras' theorem, for arbitrary $\epsilon$). Using the adjointness property and (<ref>), we have

\begin{equation*}    \ip{\Phi_{\sigma}(\mathbf{c}_{n}),D^{\sigma}(\xi_{n})}=\ip{\mathbf{c}_{n},\Pi_{\sigma}(D^{\sigma(\xi_{n})})}=\sigma^{1/2}\ip{\mathbf{c}_{n},D^{1}(\Pi_{\sigma}(\xi_{n}))}+o(1)=o(1),
\end{equation*}

where we use the fact that $\mathbf{c}_{n}\in \ker D^{1,\ast}$. This completes the proof by contradiction, and hence we have (<ref>).

To prove that $D^{\sigma}_{\mathrm{st}}$ is eventually surjective, we also argue by contradiction. Suppose that it were not. Then by standard properties of Hilbert spaces, we could find a unit norm sequence $\eta_{n},\mathbf{k}_{n}$ (with $\sigma_{n}\to\infty$) so that

\begin{equation*}
\ip{D^{\sigma}(\xi)+\Phi_{\sigma}(\mathbf{c}),\eta_{n}}+\ip{\Pi_{\sigma}(\xi),\mathbf{k}_{n}}=0\text{ for all $n,\xi,\mathbf{c}$},
\end{equation*}

Using $\Pi_{\sigma}^{*}=\Phi_{\sigma}$ and $\mathbf{c}=0$, we conclude that $D^{\sigma,\ast}(\eta_{n})=-\Phi_{\sigma}(\mathbf{k}_{n})$. Since this is bounded in $L^{2}$, we can apply the compactness result to conclude that $\Pi_{\sigma}(\eta_{n})$ converges to a solution of $\ker D^{1,\ast}$. However the assumption that

\begin{equation*}
\ip{\Phi_{\sigma}(\mathbf{c}),\eta_{n}}=0,
\end{equation*}

for all $\mathbf{c}\in \ker D^{1,\ast}$, allows us to conclude that $\Pi_{\sigma}(\eta_{n})$ converges to $0$. It follows that $\eta_{n}$ converges to zero (since we already know $\mathrm{R}_{\sigma}(\eta_{n})$ converges to zero). Now set $\xi_{n}=\Phi_{\sigma}(\mathbf{k}_{n})$ and $\mathbf{c}=0$ to conclude that

\begin{equation*}
\begin{aligned}
\implies &\norm{\Phi_{\sigma}(\mathbf{k}_{n})}=o(1)\implies \norm{\mathbf{k}_{n}}=o(1).
\end{aligned}
\end{equation*}

We have shown that both $\eta_{n},\mathbf{k}_{n}$ converge to zero, which contradicts our assumption that they were unit norm. This completes the proof.

It follows easily from Proposition <ref> that
$$\mathrm{ind}(D^{\sigma})=\dim \ker D^{1}-\dim \ker D^{1,\ast}.$$
To see why, write $D^{\sigma}_{\mathrm{st}}$ in matrix form. Deform the operator by keeping the $1,1$ entry fixed and setting all the other entries to zero. This deformation does not change the Fredholm index. It is easy to compute the Fredholm index after the deformation.

Equation (<ref>) then implies that $\mathrm{ind}(D^{\sigma})=\mathrm{X}+\mu^{\tau}_{\mathrm{Mas}}$, which completes the proof of Lemma <ref>. This in turn completes the proof of Proposition <ref> (the index formula for $\mathrm{ind}(D^{\mathrm{al}})$). Applying our earlier result Proposition <ref> (relating $\mathrm{ind}(D)$ and $\mathrm{ind}(D^{\mathrm{al}})$) completes the proof of our main result, Theorem <ref>.

§ ON THE PARITY OF THE CONLEY-ZEHNDER INDICES

The purpose of this appendix is to explain how the parity of the Conley-Zehnder index of an asymptotic operator $A$ changes with the asymptotic trivialization.

Here is the setup. Suppose that $A_{1}=-i\bd_{t}-S(t)$ is an asymptotic operator. First we suppose that $A_{1}$ is defined on the interval $[0,1]$.

Let $\Omega(t)\in U(n)$ be a path of unitary matrices with the property that $\Omega(0),\Omega(1)$ preserve $\R^{n}$. Referring to Section <ref>, we see that the transition function between any two asymptotic trivializations $\tau_{1},\tau_{2}$ always takes this form.

In this fashion $\Omega(0),\Omega(1)$ can be thought of as elements of $\mathrm{O}(n)$.

Let $A_{2}=\Omega(t)^{-1}A_{1}\Omega(t)$. Then

\begin{equation*}
\begin{aligned} \Omega(0)\Omega(1)\in \mathrm{SO}(n)\implies \mu_{\mathrm{CZ}}(A_{2})-\mu_{\mathrm{CZ}}(A_{1})=0\text{ mod }2,\\
\Omega(0)\Omega(1)\in\mathrm{O}(n)\setminus \mathrm{SO}(n)\implies \mu_{\mathrm{CZ}}(A_{2})-\mu_{\mathrm{CZ}}(A_{1})=1\text{ mod }2.
\end{aligned}
\end{equation*}

To set the stage, consider the Cauchy-Riemann operator on $(\R\times [0,1],\C^{n},\R^{n})$ which equals

\begin{equation*}
\end{equation*}

Here $\beta(s)$ is a cut-off function which vanishes for $s\le 0$ and equals $1$ for $s\ge 1$. We also suppose that $\beta'(s)>0$ on $(0,1)$.

Clearly, the index formula gives

\begin{equation*}    \mathrm{ind}(D)=\mu_{\mathrm{CZ}}(A_{2})-\mu_{\mathrm{CZ}}(A_{1}).
\end{equation*}

Now let $X_{0}$ be the standard frame of $\C^{n}$ and  consider a non-standard frame $X(s,t)$ with the property that

\begin{equation*}
\begin{aligned}
X(s,t)&=\Omega(t)^{-1}X_{0}\text{ for $s\ge 1$}\\
X(s,t)&=X_{0}\text{ for $s\le 0$}.
\end{aligned}
\end{equation*}

A priori, we do not define $X$ on the region $[0,1]\times [0,1]$. Observe that with respect to this non-standard frame both asymptotics of $D$ are equal to $\bd_{s}-A_{1}$. As a consequence, the index formula gives

\begin{equation*}
\mathrm{ind}(D)=\mu_{\mathrm{Mas}}^{\tau},
\end{equation*}

where $\tau=X^{-1}$. Recalling the definition, we conclude $\mu^{\tau}_{\mathrm{Mas}}$ is the signed count of zeros of a generic section of $\det(\C^{n})^{\otimes 2}$ which (a) equals

\begin{equation*}
(X_{1}\wedge \cdots \wedge X_{n})^{\otimes 2}
\end{equation*}

in the ends, and (b) restricts to the canonical generator of $\det(\R^{n})^{\otimes 2}$ along the boundary.

The next step is to prove the result in the case when $\Omega(0),\Omega(1)\in \mathrm{SO}(n)$. Since $\mathrm{SO}(n)$ is connected, we can extend $X$ over the boundary $[0,1]\times \set{0,1}$ so that it is always a frame of $\R^{n}$. Now let

\begin{equation*}
\mathfrak{s}=X_{1}\wedge \cdots \wedge X_{n}=:\mathrm{det}(X)
\end{equation*}

as a section of $\det(\C^{n})$. It is clear that $\mathfrak{s}$ restricts to some generator of $\det(\R^{n})$ along the boundary.

Extend $\mathfrak{s}$ to $[0,1]\times [0,1]$ in two ways, defining $\mathfrak{s}_{1},\mathfrak{s}_{2}$. We require that the zeros of $\mathfrak{s}_{1},\mathfrak{s}_{2}$ are disjoint. Then $\mathfrak{s}_{1}\otimes \mathfrak{s}_{2}$ defines a section of $\det(\C^{n})^{\otimes 2}$ whose zeros count $\mu^{\tau}_{\mathrm{Mas}}$. However, it is easy to show that the signed count of zeros of $\mathfrak{s}_{1}$ and $\mathfrak{s}_{2}$ agree:

\begin{equation*}
\#\mathfrak{s}_{1}^{-1}(0)=\#\mathfrak{s}_{2}^{-1}(0).
\end{equation*}

The argument proving this equality is similar to results in [6].

\begin{equation*}
\mu^{\tau}_{\mathrm{Mas}}=\#\mathfrak{s}_{1}^{-1}(0)+\#\mathfrak{s}_{2}^{-1}(0)=0\text{ mod }2.
\end{equation*}

This proves the theorem in the case $\Omega(0),\Omega(1)\in \mathrm{SO}(n)$.

Next, suppose that $\Omega(t)=\mathrm{diag}(e^{i\pi t},1,\dots,1)$. Then

\begin{equation*}
\det(X)=\left\{
\begin{aligned}
e^{-i\pi t}\det(X_{0})\text{ for $s\ge 1$,}\\
\det(X_{0})\text{ for $s\le 0$}.
\end{aligned}\right.\implies    \det(X)^{\otimes 2}=\left\{
\begin{aligned}
e^{-i2\pi t}\det(X_{0})^{\otimes 2}\text{ for $s\ge 1$,}\\
\det(X_{0})^{\otimes 2}\text{ for $s\le 0$}.
\end{aligned}\right.
\end{equation*}

Therefore we can extend $\mathfrak{s}=\det(X)^{\otimes 2}$ as

\begin{equation*}
\mathfrak{s}=[(1-\beta(s))+\beta(s)e^{-2\pi i t}]\det(X_{0})^{\otimes 2}.
\end{equation*}

It is clear that this restricts to the canonical generator of $\det(\R^{n})^{\otimes 2}$ along the boundary. Our assumption that $\beta'(s)>0$ for $s\in (0,1)$ implies that $\mathfrak{s}$ has a single transverse zero when $\beta(s)=1/2$ and $t=1/2$. As a consequence we conclude that

\begin{equation*}
\mu_{\mathrm{CZ}}(A_{2})-\mu_{\mathrm{CZ}}(A_{1})=\mu^{\tau}_{\mathrm{Mas}}=\#\mathfrak{s}^{-1}(0)=\pm 1.
\end{equation*}

A similar argument works with $\Omega(t)=(e^{i\pi(1-t)},1,\dots,1)$.

Now suppose that $\Omega(0)\in \mathrm{SO}(n)$ but $\Omega(1)\not\in \mathrm{SO}(n)$. Then we can replace

\begin{equation*}
\tilde{\Omega}(t)=\mathrm{diag}(e^{i\pi t},1,\cdots,1)\Omega(t).
\end{equation*}

This modified operator has $\tilde{\Omega}(0),\tilde{\Omega}(1)\in \mathrm{SO}(n)$, and so the first part of our proof applies, and so we conclude

\begin{equation*}
\mu_{\mathrm{CZ}}(\tilde{\Omega}^{-1}A_{1}\tilde{\Omega})-    \mu_{\mathrm{CZ}}(A_{1})=0\text{ mod }2.
\end{equation*}

The second part of our proof implies that

\begin{equation*}
\mu_{\mathrm{CZ}}(\tilde{\Omega}^{-1}A_{1}\tilde{\Omega})-        \mu_{\mathrm{CZ}}(\Omega^{-1}A_{1}\Omega)=1\text{ mod }2.
\end{equation*}

Combining these allows us to conclude the desired result. The other cases are handled similarly.

The $\R/\Z$ case is simpler. If $A_{1}$ is defined on $\R/\Z$ and $\Omega(t)$ is a loop in $U(n)$ then we have:

Let $A_{2}=\Omega(t)^{-1}A_{1}\Omega(t)$. Then

\begin{equation*}    \mu_{\mathrm{CZ}}(A_{2})-\mu_{\mathrm{CZ}}(A_{1})=0\text{ mod }2.
\end{equation*}

The argument is similar to the one given above. The proof is left to the reader. See also <cit.>.

§ ON THE INVARIANCE OF THE EULER CHARACTERISTIC TERM

The analysis in Section <ref> shows that the Euler characteristic term $\mathrm{X}(\Sigma,\Gamma_{\pm})$ is an invariant, as explained in Proposition <ref>. However, it is possible to give an elementary proof of the invariance of $\mathrm{X}(\Sigma,\Gamma_{\pm})$, using techniques similar to those found in [6].

[black!10!white] (2,0) arc (0:180:2)–cycle;
(A) at (0,2) [above] $(+,+)$;

[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (0,2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (45:2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (135:2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);
at (0,-0.2)[below]$\mathrm{count}=+1$;

(A) at (0,2) [above] $(+,-)$;

[black!10!white] (2,0) arc (0:180:2)–cycle;
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (-2,0)–(0,0);
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[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (0,2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (45:2)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (135:2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);
at (0,-0.2)[below]$\mathrm{count}=0$;

(A) at (0,2) [above] $(-,+)$;

[black!10!white] (2,0) arc (0:180:2)–cycle;

(2,0) arc (0:180:2)–cycle;
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] plot[domain=0.5:2] (,1/);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] plot[domain=0.5:2] (-,1/);

[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (0,2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);
at (0,-0.2)[below]$\mathrm{count}=0$;

(A) at (0,2) [above] $(-,-)$;

[black!10!white] (2,0) arc (0:180:2)–cycle;

(2,0) arc (0:180:2)–cycle;
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] plot[domain=0.5:2] (,1/);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] plot[domain=0.5:2] (-,1/);

[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (-2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]>;] (2,0)–(0,0);
[postaction=decorate,decoration=markings,mark=at position 0.5 with [scale=1.5]<;] (0,2)–(0,0);
[draw,circle,inner sep=1pt,fill] at (0,0);
at (0,-0.2)[below]$\mathrm{count}=-1$;

The four models for a boundary zero of $V$. The first sign is from the linearization of $V:\Sigma\to T\Sigma$ and the second sign is from the linearization of the restriction $V:\bd\Sigma\to T\bd\Sigma$.

The argument is as follows: pick two admissible vector fields $V_{0}$ and $V_{1}$ and consider them as defining a partial section $V$ of $\pr^{*}T\dot\Sigma\to \dot\Sigma\times [0,1]$ (lying over $\Sigma\times \set{0,1}$).

Extend $V$ to a section of $\pr^{*}T\bd\dot\Sigma\to \bd\dot\Sigma\times [0,1]$, while keeping it transverse. Note that we only do a partial extension, $V$ is not defined on the interior of $\dot\Sigma\times [0,1]$. The zero set $Z(V)$ is a one-manifold in $\bd\dot\Sigma\times [0,1]$ whose endpoints are the zeros of $V_{0},V_{1}$.

Note that the four kinds of zeros of $V_{0},V_{1}$ can be differentiated by the signs of their linearizations, as in Figure <ref>. Let us agree to say two zeros cancel if they are endpoints of an interval in $Z(V)$ and lie on the same side $\bd\Sigma\times \set{0}$ or $\bd\Sigma\times \set{1}$. We say two zeros match if they are endpoints of an interval whose boundaries lie on opposite sides. Let us also agree to call an interval component of $Z(V)$ a pairing.

The usual argument (see [6]), which only involves the linearization in the boundary direction, then shows that $(\pm,+)$ zeros can only cancel with $(\pm,-)$ zeros, and $(\pm,+)$ zeros can only match with $(\pm,+)$ zeros, etc. Here the various $\pm$ signs are independent (i.e., a $(-,+)$ zero can cancel with a $(+,-)$ zero).

If a $(+,+)$ cancels with $(-,-)$ (along some arc $\gamma$) then we can infinitesimally the extend $V$ into the interior near $\gamma$ so that is non-vanishing away from $\gamma$. This is because the vector fields $V_{0}$ and $V_{1}$ are both pointing upwards along the imaginary axis at the end points of $\gamma$. An explicit formula would be something like $V(s,t)=V(s,0)+t\nu$ where $\nu$ is an inwards pointing vector, and $t$ is distance to the boundary. This can be done compatibly with $V_{0}$ and $V_{1}$. Indeed this formula holds exactly in a standard coordinate chart $z=s+it$ if $V=z\bd_{s}$ or $V=-\cl{z}\bd_{s}$ and $\nu=i\bd_{t}$.

Similarly, we can do this non-vanishing local extension if:

* a $(+,-)$ cancels with a $(-,+)$,

* a $(+,+)$ matches with a $(+,+)$,

* a $(+,-)$ matches with a $(+,-)$,

* a $(-,+)$ matches with a $(-,+)$,

* a $(-,-)$ matches with a $(-,-)$.

Notice that in all cases the count is preserved. The fact that we extend $V$ near these intervals $\gamma$ in a non-vanishing fashion means that these zeros are “protected” from being cancelled by interior zeros once we further extend $V$ to the rest of $\dot\Sigma\times [0,1]$.

Unfortunately, there are other “bad” pairings which can occur, namely:

* a $(+,+)$ cancels with a $(+,-)$,

* a $(-,+)$ cancels with a $(-,-)$,

* a $(+,+)$ matches with a $(-,+)$,

* a $(+,-)$ matches with a $(-,-)$.

In each case, the count of boundary zeros is not preserved. Indeed there is always an imbalance of $\pm 1$. Let $\gamma^{\prime}$ be such a bad pairing.

The next step is to extend $V$ to a neighborhood of $\gamma^{\prime}$ in such a way that a single interior zero enters the neighborhood and hits $\gamma^{\prime}$ (which we consider leaving the surface) – this loss of an interior zero will re-balance the count.

(-3,1)node[draw,circle,inner sep=1pt,fill=black]node[left]$-$to[out=0,in=115](0,0);
[line width=1pt] (-3,0)node[draw,circle,inner sep=1pt,fill=black]node[below,shift=(0,-0.2)]$(+,+)$–(6,0)node[draw,circle,inner sep=1pt,fill=black]node[below,shift=(0,-0.2)]$(-,+)$;
[rotate=25] (-.4,0)–(.4,0)arc (0:180:0.4);
[shift=(-3,0)][rotate=25] (-.4,0)–(.4,0)arc (0:180:0.4);
[shift=(6,0)][rotate=25] (-.4,0)–(.4,0)arc (0:180:0.4);

An interior zero leaves through $\gamma^{\prime}$, re-balancing the count. One should imagine $(\cl{\mathbb{H}}\cap D(1))\times \gamma^{\prime}$ in the figure.

To simplify the construction, let us suppose that $V$ is constant on $\bd\dot\Sigma\times [x_{0},x_{1}]$, and suppose that $p\times [x_{0},x_{1}]$ is contained in $\gamma^{\prime}$. This can be achieved after a small perturbation of $V$. Since $V$ is cut transversally, by assumption, we know that $p$ is a non-degenerate zero of $V_{x}=V|_{\bd\dot\Sigma\times \set{x}}$ for all $x\in [x_{0},x_{1}]$.

Then we can extend $V$ to a non-vanishing section on a neighborhood of the complement
$$\gamma^{\prime}\setminus (p\times (x_{0},x_{1})),$$
by keeping the imaginary part of $V$ always positive or negative. After a perturbation on a fixed boundary coordinate chart of $p$ we may suppose that

\begin{equation*}
V_{x_{0}}=z\text{ and }V_{x_{1}}=\cl{z}\text{ or }V_{x_{0}}=-z\text{ and }V_{x_{1}}=-\cl{z},
\end{equation*}

or vice-versa. This uses the fact that $V_{x_{0}}$ and $V_{x_{1}}$ have non-degenerate zeros. Suppose the chart is valued in $\cl{\mathbb{H}}\cap D(1)$.

On a neighborhood of the interval $\set{p}\times [x_{0},x_{1}]$ we will interpolate between positive imaginary part to negative part by the formula

\begin{equation*}
\end{equation*}

where $\beta$ monotonically increases from $0$ to $1$. Similar formulas work in the other cases. Note that we can explicitly describe the zero set of $V$ near $\gamma^{\prime}$ as

\begin{equation*}
Z(V)=(\set{0}\times [x_{0},x_{1}])\cup i\R\cap \Omega(1)\times \set{\beta^{-1}(1/2)}.
\end{equation*}

i.e., $x_{\ast}=\beta^{-1}(1/2)$ is a singular time where the zero set of $V$ looks like Figure <ref>. Let us call the set $(i\R\times \Omega(1))\times \set{x_{\ast}}$ a sink for $\gamma^{\prime}$. Every “bad pairing” has exactly one sink attached to it.

To simplify the set up, let us suppose that $V$ is $x$-independent for $x\in [0,\epsilon]\cup [1-\epsilon,1]$, and place all of the sinks in these regions. Note that once we do the local extensions of $V$, it will no longer be $x$ independent in this region.

Now extend $V$ to the rest of $\dot\Sigma\times [0,1]$, so that it is a transverse section away from what we have already defined. A straightforward argument shows that every sink is one end of a (compact) interval $I\to Z(V)$; we think of these as pairings between sinks and interior zeros, or pairings between two sinks. The other components of $Z(V)$ are pairings which join interior zeros to interior zeros; these will preserve the counts (by [6]).

Let $I$ be a pairing between an interior zero and a sink. Orient $\dot\Sigma\times [0,1]$ so that $(\bd_{s},\bd_{t},\bd_{x})$ forms an oriented basis. Let $\nu(I)$ denote the normal bundle to $I$. The linearization of the vector field defines a map $\nu(I)\to T\dot\Sigma|_{I}$ which is an isomorphism except at the sink endpoint (i.e., a single endpoint). Near the interior zero $p$, $\nu(I)$ is identified with $T\Sigma_{p}$.

Near the sink, $\nu(I)$ is identified with $\mathrm{span}\set{\bd_{s},\bd_{x}}$, and the linearization depends on what kind of pairing of $Z(V)\cap \bd\Sigma$ it intersects. Let us focus on only the matching parts of $Z(V)\cap \bd \Sigma$ (and not the cancelling parts). Keeping in the local model for the zero set near $\gamma^{\prime}$, we write the linearization near the sinks as

* $\bd_{s}\mapsto +\bd_{s}\text{ and }\bd_{x}\mapsto +\bd_{t}$, if $(-,+)$ matches with $(+,+)$ (net count: $+1$),

* $\bd_{s}\mapsto -\bd_{s}\text{ and }\bd_{x}\mapsto -\bd_{t}$, if $(-,-)$ matches with $(+,-)$ (net count: $+1$),

* $\bd_{s}\mapsto +\bd_{s}\text{ and }\bd_{x}\mapsto -\bd_{t}$, if $(+,+)$ matches with $(-,+)$ (net count: $-1$),

* $\bd_{s}\mapsto -\bd_{s}\text{ and }\bd_{x}\mapsto +\bd_{t}$, if $(+,-)$ matches with $(-,-)$ (net count: $-1$).

Keep in mind that matching is a directed relation. Similar tables hold for cancelling pairs of $V_{0}$ and $V_{1}$.

We orient the interval $I$ so that it ends at a sink. Then $-\bd_{t}$ is identified with the tangent vector near the sink, and since $\set{-\bd_{t},\bd_{s},\bd_{x}}$ has the ambient orientation, we conclude that $\set{\bd_{s},\bd_{x}}$ is an oriented basis of $\nu(I)$ near the sink.

If the initial location of $I$ lies at $x=0$, then $\nu(I)\simeq T\Sigma$ is orientation preserving. If the initial location of $I$ lies at $x=1$ then the identification $\nu(I)\simeq T\Sigma$ is orientation reversing.

In particular, in case (a) or (b), since these linearizations preserve orientation, the initial zero of $I$ must be positive if it lies at $x=0$ or negative if it lies at $x=1$. In either case, the count is re-balanced. In case (c) or (d), the linearization near the sink reverses orientation, and we similarly show that the count is re-balanced.

Let us finally comment on what happens when sinks connect to sinks via an interval $J$. We suppose that both sinks are in one of the cases (a)-(d).

As we have explained, at the terminal end, $\set{\bd_{s},\bd_{x}}$ is an oriented basis of $\nu(J)$. At the initial end, $\set{\bd_{s},\bd_{x}}$ has the wrong orientation of $\nu(J)$. As a consequence, sinks of type (a) or (b) can only cancel with sinks of type (c) or (d) (and vice-versa). This implies that the counts are re-balanced.

For completeness, here are the tables for the linearizations at sinks for cancelling pairs (as opposed to matching pairs).

If a sink lies on a cancelling between two zeros in $V_{0}$, then its linearization is given by:

* $\bd_{s}\mapsto +\bd_{s}$ and $\bd_{x}\mapsto +\bd_{t}$ if $(-,+)$ cancels with $(-,-)$ (net count: $+1$).

* $\bd_{s}\mapsto +\bd_{s}$ and $\bd_{x}\mapsto -\bd_{t}$ if $(+,+)$ cancels with $(+,-)$ (net count: $-1$).

If a sink lies on a cancelling between two zeros in $V_{1}$, then the linearization is given by:

* $\bd_{s}\mapsto +\bd_{s}$ and $\bd_{x}\mapsto +\bd_{t}$ if $(+,-)$ cancels with $(-,-)$ (net count: $-1$).

* $\bd_{s}\mapsto +\bd_{s}$ and $\bd_{x}\mapsto -\bd_{t}$ if $(+,+)$ cancels with $(-,+)$ (net count: $+1$).

The rest of the details are left to the reader.

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††thanks: Co-first authors with equal contribution††thanks: Co-first authors
with equal contribution††thanks: Co-first authors with equal
contribution††thanks: Co-first authors with equal contribution

# Energy-conversion device using a quantum engine with the work medium of two-
atom entanglement

J.-W. Zhang1 B. Wang2,3 W.-F. Yuan2,3 J.-C. Li2,3 J.-T. Bu2,3 G.-Y. Ding2,3
W.-Q. Ding2,3 L. Chen2,1 F. Zhou2,1<EMAIL_ADDRESS>M. Feng1,2,4
<EMAIL_ADDRESS>1Research Center for Quantum Precision Measurement,
Guangzhou Institute of Industry Technology, Guangzhou, 511458, China
2State Key Laboratory of Magnetic Resonance and Atomic and Molecular Physics,
Wuhan Institute of Physics and Mathematics, Innovation Academy of Precision
Measurement Science and Technology, Chinese Academy of Sciences, Wuhan,
430071, China
3School of Physics, University of the Chinese Academy of Sciences, Beijing
100049, China
4Department of Physics, Zhejiang Normal University, Jinhua 321004, China

###### Abstract

Although entanglement is considered as an essential resource for quantum
information processing, whether entanglement helps for energy conversion or
output in the quantum regime is still lack of experimental witness. Here we
report on an energy-conversion device operating as a quantum engine with the
working medium acted by two entangled ions confined in a harmonic potential.
The two ions are entangled by virtually coupling to one of the vibrational
modes shared by the two ions, and the quantum engine couples to a quantum
load, which is another shared vibrational mode. We explore the energy
conversion efficiency of the quantum engine and investigate the useful energy
(i.e., the maximum extractable work) stored in the quantum load by tuning the
two ions in different degrees of entanglement as well as detecting the change
of the phonons in the load. Our observation provides, for the first time,
quantitative evidence that entanglement fuels the useful energy produced by
the quantum engine, but not helpful for the energy conversion efficiency. We
consider that our results may be useful to the study of quantum batteries for
which one of the most indexes is the maximum extractable energy.

Quantum engines (QEs) operated with quantum substances are expected to surpass
their classical counterparts in terms of output power and efficiency by
utilizing quantum features Scully ; Gemmer ; Parrondo ; Chapin ; Robsnagel1 ;
Hardal . The recently growing interest in QEs is additionally motivated by the
experiments of single-spin QEs executed from trapped-ion systems to
optomechanics Klatzow ; Robsnagel2 ; Maslennikov ; Lindenfels ; Zhang2 ;
Chiral ; Quan ; Guthrie ; Ryan ; Peterson ; Koski ; Zhang ; Dechant .

Entanglement Schrodinger  is a unique resource in quantum information
processing, which speeds up computation Kendon , ensures the information
security in communication Ekert ; Bennett1 ; Bennett2  and improves the
signal-to-noise ratio in precision measurement Venzl ; Wineland ; Huelga ;
Giovannetti ; Jozsa . Recently, various ideas with entanglement involved in
QEs have been proposed Hewgill ; Barrios ; Tavakoli ; Josefsson ; Huang ;
Bresque , indicating that entanglement helps improve the efficiency of the
engine over the classical counterparts. Experimentally, enhanced performance
of the QE was demonstrated with linear optics Wang  due to entanglement
between different degrees of freedom in single photons as well as local
measurements. However, a deeper understanding of the QE performance associated
with entanglement needs quantitatively tuning entanglement in the working
medium, which is hard for single-photon experiments.

In the present work, by quantitatively varying the degree of entanglement, we
experimentally investigate, for the first time, the energy conversion from the
QE involving bipartite entanglement to a quantum load explain , in which the
useful energy extracted (also referred to as the work extraction) from the
quantum load is focused. As clarified later, we witness that the entangled
working medium of QE can fuel the work extraction from the load, but not
helpful for the efficiency of the energy conversion from QE to the load.

Figure 1: (a) Level scheme of two 40Ca+ ions entangled via the breath mode
state $|n\rangle$ of the axial vibration. The arrows in red and blue
represent, respectively, the red- and blue-detuning irradiation of the 729-nm
laser for achieving the MS gate, where $\delta$ is the value of the symmetric
detuning. The $\pi$-polarized 854-nm laser beam and the 393-nm spontaneous
emission construct the dissipative channel. (b) A quantum thermodynamic cycle,
where the first stroke absorbs photons by implementing the MS gate; The second
stroke operates with declining detuning; The third stroke is carried out by
red-sideband transition to couple to the load (i.e., the center-of-mass mode),
and the fourth stroke encloses the cycle by enlarging the detuning under
dissipation.

The working medium of our QE is composed of two 40Ca+ ions confined in a
linear Paul trap. We encode the qubits into the pseudo-spin states of each of
the ions, i.e., $|S\rangle=|4^{2}S_{1/2},m_{J}=-1/2\rangle$ and
$|D\rangle=|3^{2}D_{5/2},m_{J}=-3/2\rangle$ with the magnetic quantum number
$m_{J}$. The two-ion system owns two vibrational modes, i.e. the center-of-
mass mode (with frequency of $\omega_{c}/2\pi=0.794$ MHz) and the breath mode
(with frequency of $\omega_{b}/2\pi=1.37$ MHz) along the axial direction. To
reduce the detrimental effect from decoherence, we employ the breath mode to
help entangle the two ions by the M$\phi$lmer-S$\phi$rensen (MS) gate MSgate ,
in which the two qubits are virtually coupled to the breath mode under global
irradiation of 729-nm laser beams. Besides, we detect the output energy by
measuring the variation of the average phonon number in the load (i.e., the
axial center-of-mass mode).

The quantization axis is parallel to the axial direction by a magnetic field
applied with approximately 3.45 Gauss at the trap center. Prior to the
experiment, we have accomplished both the Doppler and sideband cooling, which
reduces the thermal phonons to the average phonon number $n_{\rm h}\approx$
0.03 of the breath mode and $n_{\rm c}\approx$ 0.13 of the center-of-mass
vibrational mode with the Lamb-Dicke parameter of $\eta_{p}=0.0574$. This is
sufficient to avoid thermal phonons yielding offsets of Rabi oscillations, and
ensure correct observation of the output energy from the phonons. Figure 1(a)
sketches the entanglement realization with the MS gate, where the two ions are
driven globally by a red-detuned and a blue-detuned 729-nm laser beams with
the symmetric detuning $\delta$ from the breathe mode. By exactly tuning the
duration of the 729-nm laser irradiation, we may achieve, from the initial
state $|SS\rangle$, entanglement between $|SS\rangle$ and $|DD\rangle$.

Figure 2: Experimental measurement of energy conversion. (a) Time evolution of
the populations, where $|SS\rangle$, $|SD\rangle+|DS\rangle$, and
$|DD\rangle$, are colored, respectively, in blue, black, and red. (b) Time
evolution of the mean number of the phonons. The lines are obtained by
theoretical simulation and dots are experimental results. The error bars
indicating the statistical standard deviation of the experimental data are
obtained by 10 000 measurements for each data point.

For our purpose (e.g., the fourth stroke as plotted in Fig. 1(b)), we must
introduce a dissipative channel for the qubits. To this end, we employ an
extra energy level $|4^{2}P_{3/2},m_{J}=-3/2\rangle$ (labeled as $|P\rangle$),
which couples to $|D\rangle$ by a 854-nm laser (with Rabi frequency
$\Omega_{1}$ and $\pi$ polarization) and dissipates to $|S\rangle$ by
spontaneous emission with the decay rate of $\Gamma_{e}/2\pi=23.1$ MHz (see
Fig. 1(a)). This achieves, under appropriate laser irradiation, an effective
two-level model with the decay $\gamma_{\rm eff}=\Omega^{2}_{1}/\Gamma_{e}$
Zhang1 ; Zhang2 , and described by the Lindblad master equation as
$\dot{\rho}_{m}=-i[H_{s},\rho_{m}]+\sum\limits_{i=1}^{2}\frac{\gamma_{\rm
eff}}{2}(2\sigma_{-}^{i}\rho_{m}\sigma_{+}^{i}-\sigma_{+}^{i}\sigma_{-}^{i}\rho_{m}-\rho_{m}\sigma_{+}^{i}\sigma_{-}^{i})$,
where $i$ is the denotation representing the $ith$ ion, $\rho_{m}$ denotes the
density operator of the working medium, $H_{s}$ is the Hamiltonian of the
working medium in the fourth stroke, and
$\sigma_{-}=(\sigma_{+})^{\dagger}\equiv|g\rangle\langle e|$ is the
annihilation operator of the qubit.

The operations of this QE are depicted in Fig. 1(b), which consists of four
strokes. The first stroke is the heating stroke for absorbing energy, which is
operated by the MS gate and governed by the Hamiltonian

$\begin{split}H_{s}^{\rm
1st}&=\sum\limits_{i=1}^{2}\frac{\eta_{p}\Omega_{r}}{2}({\sigma_{+}^{i}e^{\rm
i\delta_{r}t}a_{h}+\sigma_{-}^{i}e^{\rm-i\delta_{r}t}a^{\dagger}_{h}})\\\
&~{}~{}+\sum\limits_{i=1}^{2}\frac{\eta_{p}\Omega_{b}}{2}({\sigma_{+}^{i}e^{\rm
i\delta_{b}t}a^{\dagger}_{h}+\sigma_{-}^{i}e^{\rm-i\delta_{b}t}a_{h}}),\end{split}$
(1)

where $\Omega_{r}$ ($\Omega_{b}$) and $\delta_{r}$ ($\delta_{b}$) represent
the Rabi frequency and detuning of the red (blue) sideband transition of the
breath vibration mode, respectively, $a_{h}$ and $a^{\dagger}_{h}$ are
phononic operators associated with the axial breath vibration mode.

In our experiment, we may simply consider $\delta_{r}=-\delta_{b}=-\delta$,
$\Omega_{r}=\Omega_{b}=\Omega$. Throughout this work, we fix $\Omega$ to be
$\Omega/2\pi=245.8$ kHz. In this first stroke, we produce different degrees of
entanglement of the two ions by exactly tuning the MS-gating time, which
corresponds to the absorption of different numbers of photons, as clarified
below. Experimentally, the degree of entanglement produced by the MS gate is
quantized by the gating fidelity SM . For the working medium initialized from
$|SS\rangle$, the absorbed optical photon quanta $\Delta n_{t}$ is defined by
$2P_{\rm DD}+P_{\rm SD}+P_{\rm DS}$ Horne , where $P_{\rm DD}$, $P_{\rm SD}$
and $P_{\rm DS}$ represent the populations of $|DD\rangle$, $|SD\rangle$ and
$|DS\rangle$, respectively.

Then we rapidly tune the laser frequency to couple the center-of-mass
vibrational model. In this case, the two ions interact with the globally
irradiated 729-nm laser respectively, given by the Hamiltonian $H_{s}^{\rm
2nd}=\sum\limits_{i=1}^{2}\frac{\eta_{p}\Omega}{2}\sigma_{+}^{i}(e^{-\rm
i\int^{0}_{\omega_{b}-\omega_{c}+\delta}\delta(t)dt}a_{c}+e^{\rm
i\int^{0}_{\omega_{b}-\omega_{c}+\delta}\delta(t)dt}a^{\dagger}_{c})+H.c.$
with $a_{c}$ and $a^{\dagger}_{c}$ representing phononic operators associated
with the axial center-of-mass mode. Due to the rapid change of laser frequency
in a very short duration, the populations in both the qubit states and the
vibrational mode remain unchanged during this stroke. The third stroke aims to
transform the internal energy (represented by the number of the absorbed
optical photons) to the external energy (represented by the number of the
produced phonons in the load), which is accomplished by a Jaynes-Cummings
interaction under the government of the Hamiltonian $H_{s}^{\rm
3rd}=\sum\limits_{i=1}^{2}\frac{\eta_{p}\Omega}{2}(\sigma_{+}^{i}a_{c}+\sigma_{-}^{i}a^{\dagger}_{c})$.
The fourth stroke is the time-reversed process of the second stroke. However,
in this stroke, for enclosing the cycle, we switch on the dissipative channels
in the two ions, i.e., $\gamma_{\rm eff}\neq$ 0\.

Figure 3: Time evolution of the characteristic parameters of the system. (a)
Time evolution of populations and entanglement of the qubits. (b) Time
evolution of the phonons as well as entanglement of the qubits, where the
dashed line $\bar{n}_{1}$ and the solid line $\bar{n}_{3}$ represent,
respectively, the mean phonon numbers in the final states of the first and
third strokes. (c, d) Time evolution of the conversion efficiency $\eta_{c}$
and mechanical efficiency $\eta_{m}$ in comparison with entanglement variation
of the qubits. The vertical gray line represents the time point for maximum
entanglement. In all the panels, $T_{\rm gate}=36$ $\mu$s. Lines are obtained
by simulation and dots are experimental results. The error bars indicating the
statistical standard deviation of the experimental data are obtained by 10 000
measurements for each data point. The relatively large error bars presented in
(c, d) are due to the fact that detecting phonon numbers with ultimate
precision is intricate using current techniques.

Before specifying the experimental details, we elucidate some key points
relevant to the energies and efficiencies of the QE under our consideration.
We evaluate the number of the absorbed optical photons by the net increase of
populations in the qubit levels Horne , and the converted energy to the load
by the the number of the produced phonons in the center-of-mass mode.
Moreover, since the energy of the phonons is tiny compared with that of the
photons, it is not helpful to define the QE efficiency in terms of energies.
To evaluate the effect of the energy conversion from optical photons to
vibrational phonons, we follow the idea in Horne  by considering the
efficiency of conversion of quanta, i.e., by defining the conversion
efficiency of our QE as $\eta_{c}=\Delta n_{t}/\Delta n_{o}$, where $\Delta
n_{t}$ and $\Delta n_{o}$ represent the net increase of the mean phonon number
in the load and the absorbed photon quanta in the working medium,
respectively. Furthermore, for quantitatively figuring out the useful energy
extracted from output phonons as well as the associated efficiency, we employ
the concepts of ergotropy $\mathcal{W}$ ergotropy  and quantum mechanical
efficiency $\eta_{m}=\mathcal{W}/(\Delta n_{o}\hbar\omega_{c})$ where $\Delta
n_{o}\hbar\omega_{c}$ represents the obtained energy of the output phonons.
Ergotropy is defined by the maximum energy extracted using unitary operations,
quantified by the difference in energy between the original state $\rho_{p}$
and the passified state $\widetilde{\rho}_{p}$, i.e.,
$\mathcal{W}=tr[H_{p}\rho_{p}]-tr[H_{p}\widetilde{\rho}_{p}]$, where $H_{p}$
is the Hamiltonian of the phonons, the passified state corresponds to the
considered density matrix being diagonal in the representation of energy
eigenstates with the diagonal terms (i.e., the occupation probabilities)
decreasing for the enlarging energy eigenvalues passive1 ; passive2 ; Francica
.

Experimentally, we first execute the QE with the maximum entanglement produced
in the first stroke, i.e., with the fidelity $F=0.9625$ with respect to the
perfect case SM . This is accomplished by the MS gate for a time duration
$T_{\rm gate}=2\pi/\delta\approx 36$ $\mu$s with $\delta=2\eta_{p}\Omega$.
From Figs. 2(a,b), we observe the time evolution of both the populations in
qubit states and the mean number of phonons, respectively. At the end of this
stroke, we have $P_{\rm DD}\approx 0.4851$ and $P_{\rm SD}+P_{\rm DS}\approx
0.02541$, indicating the absorbed optical photons $\Delta n_{o}=0.99561$SM .
In the second stroke, rapidly reducing the detuning of the center-of-mass mode
from 859 kHz to 0 is accomplished by four successive steps with each of 0.1
$\mu$s. As monitored in Fig. 2(a,b), the populations in qubit states and the
mean number of phonons are almost unchanged during this process which is of
very short time duration. The third stroke is very crucial in our QE cycle,
which outputs the energy of the QE to the load. This is achieved by a red-
sideband transition of the 729-nm laser irradiating globally the two ions.
After well compensating the Stark effect, we have found the optimal duration
of the laser irradiation for the energy conversion, which is 26.5 $\mu$s,
leading to $\Delta n_{t}\approx 0.91-0.13=0.78$ (for excluding the initial
value 0.13). In this stroke, to experimentally evaluate the ergotropy of the
output phonons, we measure the populations $P_{n}$, which are the diagonal
elements of the phononic density matrix $\rho_{p}$, by fitting the
experimental data of the blue-sideband transition of the two ions SM . This is
an approximate treatment based on the assumption that the off-diagonal
elements in $\rho_{p}$ are negligible Horne . Although it is not as good as in
the single-qubit case Horne , this approximate treatment has the nearly
identical variation to the exact result SM . Therefore, due to the initial
thermal state (passified state) of the load, we estimate approximately the
useful energy of the load to be $\mathcal{W}/\hbar\omega_{c}\approx 0.373$.
The fourth stroke enlarges the detuning to the center-of-mass mode from 0 to
859 kHz using five successive steps with each of 1 $\mu$s. Since this is the
last stroke, we have to enclose the QE cycle by switching on the dissipative
channel. We turn on the 854-nm laser, introducing an effective decay with
$\gamma_{\rm eff}=5\Omega$. Therefore, the efficiencies of this QE are
$\eta_{c}=0.7834$ and $\eta_{m}=0.49$, respectively.

To have deeper insight into the role of entanglement in the QE, we need to
quantitatively evaluate the performance of the QEs by tuning the degree of
entanglement of the working medium in the first stroke. This is accomplished
by exactly tuning the gating time $t$ due to the fixed value of $\Omega$.
Experimentally, we choose multiple time intervals around the $T_{\rm gate}$ SM
and monitor variation of the absorbed optical quanta as well as the mean
numbers of the net increased phonons, as illustrated in Fig. 3(a, b). From
these values, we acquire the efficiency $\eta_{c}$ at different degrees of
entanglement, see Fig. 3(c). We also investigate the useful energy stored in
the load by fitting the experimental data of the blue-sideband transition at
the end of third stroke SM , see Fig. 3(d). It is evident that the maximum
value of $\eta_{m}$ appears at the maximum entanglement of the working medium
with values of $\mathcal{W}/\hbar\omega_{c}\approx 0.4242$, $\Delta
n_{t}=0.8108$ and $\eta_{m}=0.523$ SM , while $\eta_{c}$ remains nearly
constant with respect to the change of entanglement. This indicates that the
working medium in entanglement is not helpful for energy conversion, but
strongly supports the mechanical efficiency. Considering the exact solution of
ergotropy that varies from zero to the maximum with the variation of
entanglement, we conjecture that, it is the entanglement in our case that
fuels the useful energy in the load SM .

In summary, we have experimentally executed a QE with the working medium of
tunable bipartite entanglement, working as an energy-conversion device to
output energy to a quantum load. The impact of entanglement of the working
medium on the quantum load has been witnessed quantitatively by measuring the
efficiencies of $\eta_{c}$ and $\eta_{m}$. In particular, the study of the
ergotropy and $\eta_{m}$ associated with the maximum extractable energy of the
load would be helpful for understanding the process of charging the load, if
the vibrational mode of the ions is acted as a quantum battery Yang .
Therefore, our findings would inspire further investigation of microscopic
energy devices, e.g., QEs and quantum batteries, with entanglement involved.

This work was supported by National Natural Science Foundation of China under
Grant Nos. 12304315, U21A20434, 12074346, 12074390, by China Postdoctoral
Science Foundation under Grant Nos. 2022M710881, 2023T160144, by Key Lab of
Guangzhou for Quantum Precision Measurement under Grant No. 202201000010, by
Science and Technology Projects in Guangzhou under Grant Nos. 202201011727 and
2023A04J0050, and by Nansha Senior Leading Talent Team Technology Project
under Grant No. 2021CXTD02.

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