M}\|{\sigma}_{0}(P)(x)\|$

###### Proof.

By Proposition 5.9, for any $f\in\mathcal{D}(T_{x}M)\otimes A_{x}$ non zero,

$\frac{\|P_{k}\phi_{k}^{f}\|}{\|\phi_{k}^{f}\|}=\frac{\|{\sigma}_{0}(P)(x)f\|}{\|f\|}+\mathcal{O}(k^{-\frac{1}{2}}).$

So
$\liminf_{k\rightarrow\infty}\|P_{k}\|\geqslant\|{\sigma}_{0}(P)(x)f\|/\|f\|$
. ∎

##  6 Derivatives

The class $\mathcal{L}(A,B)$ has been defined without any control on the
derivatives of the Schwartz kernels. The reason was merely to simplify the
exposition but in the applications it is natural and necessary to understand
the composition of operators of $\mathcal{L}(A,B)$ with covariant derivatives.
We start from general considerations, then we define a subclass
$\mathcal{L}^{\infty}(A,B)$ where the asymptotic expansion of the Schwartz
kernels hold with respect to a convenient $C^{\infty}$ topology. Finaly we
apply this to complete the proofs of the theorems stated in the introduction.

###  6.1 The class $\mathcal{O}_{\infty}(k^{-N})$

Consider as before a Hermitian line bundle $L\rightarrow M$ and an auxiliary
Hermitian vector bundle $A\rightarrow M$. Let $\mathcal{F}$ be the space of
families

$s=(s_{k}\in{\mathcal{C}}^{\infty}(M,L^{k}\otimes A),\,k\in{\mathbb{N}}).$

Recall that $s\in\mathcal{O}(k^{-N})$ if for any $x\in M$,
$|s_{k}(x)|=\mathcal{O}(k^{-N})$ with a $\mathcal{O}$ uniform on any compact
subsets of $M$. Here we do not assume that $M$ is compact.

The definition of $\mathcal{O}_{\infty}$ involves the derivatives. If $L$ and
$A$ are trivial bundles so that $(s_{k})$ is a sequence of
${\mathcal{C}}^{\infty}(M,{\mathbb{C}}^{r})$ with $r$ the rank of $A$, then we
say that $(s_{k})\in\mathcal{O}_{\infty}(k^{-N})$ if for any
$m\in{\mathbb{N}}$, the derivatives of order $m$ of $(s_{k})$ are in
$\mathcal{O}(k^{-N+m})$. More precisely, for any vector fields $X_{1}$, …,
$X_{m}$ of $M$, we require that

$\displaystyle X_{1}\ldots X_{m}s_{k}=\mathcal{O}(k^{-N+m}).$ (77)

So we loose one power of $k$ for each derivative. Because of this, the class
$\mathcal{O}_{\infty}(k^{-N})$ is invariant by multiplication by $e^{ikh}$,
where $h$ is any real-valued function of $M$.

For actual vector bundles $L$ and $A$, we introduce unitary frames $u$ and
$(v_{j})_{j=1}^{r}$ of $L$ and $A$ over the same open set $U$ of $M$ and write
$s_{k}=\sum f_{k,j}u^{k}\otimes v_{j}$ with
$f_{k}\in{\mathcal{C}}^{\infty}(U,{\mathbb{C}}^{r})$. Then we say that
$(s_{k})$ belongs to $\mathcal{O}_{\infty}(k^{-N})$ if for all choices of
unitary frames of $L$ and $A$, the corresponding local representative sequence
$(f_{k})$ is in $\mathcal{O}_{\infty}(k^{-N})$. Observe that changing the
frame $u$ of $L$ amounts to multiply $f_{k}$ by $e^{ikh}$, so the condition
that $f_{k}\in\mathcal{O}_{\infty}(k^{-N})$ does not depend on the frame
choice when these frames are defined on the same open set.

The typical example of a family in $\mathcal{O}_{\infty}(k^{-N})$ is an
oscillating sequence

$s_{k}(x)=k^{-N}e^{-k\varphi(x)}a(x)$

with $\varphi\in{\mathcal{C}}^{\infty}(M)$ having a non negative real part and
$a\in{\mathcal{C}}^{\infty}(M,{\mathbb{C}}^{r})$. More generally, for actual
bundles, we can set

$s_{k}(x)=k^{-N}E^{k}(x)a(x)$

where $E\in{\mathcal{C}}^{\infty}(M,L)$ is such that $|E|\leqslant 1$ and
$a\in{\mathcal{C}}^{\infty}(M,A)$.

Obviously, if $N^{\prime}\geqslant N$,
$\mathcal{O}_{\infty}(k^{-N^{\prime}})\subset\mathcal{O}_{\infty}(k^{-N})$.
Define
$\mathcal{O}_{\infty}(k^{-\infty}):=\cap_{N}\mathcal{O}_{\infty}(k^{-N})$. We
will need the following result.

######  Lemma 6.1.

Let $(s_{\ell})$ be a sequence of $\mathcal{F}$ such that for any $\ell$,
$s_{\ell}\in\mathcal{O}_{\infty}(k^{-p(\ell)})$ where $(p(\ell))$ is an
increasing real sequence, and $p(\ell)\rightarrow\infty$ as
$\ell\rightarrow\infty$. Then

  1. 1.

There exists $s\in\mathcal{F}\cap\mathcal{O}_{\infty}(k^{-p(0)})$ unique
modulo $\mathcal{O}_{\infty}(k^{-\infty})$ such that

$\displaystyle
s_{k}=\sum_{\ell=0}^{N-1}s_{\ell,k}+\mathcal{O}_{\infty}(k^{-p(N)}),\qquad\forall
N$ (78)

  2. 2.

Let $s\in\mathcal{F}$ such that $s\in\mathcal{O}_{\infty}(k^{p})$ for some $p$
and $s_{k}=\sum_{\ell=0}^{N-1}s_{\ell,k}+\mathcal{O}(k^{-p(N)})$ for any $N$.
Then $s\in\mathcal{O}_{\infty}(k^{-p(0)})$ and (78) holds.

The first part is a variation of Borel Lemma, the second part follows from
interpolation inequalities, cf. as instance [25, Lemma 32.]

###  6.2 Application to $\mathcal{L}(A,B)$

Choose a section $E$ as in section 5.1 and let
$b\in{\mathcal{C}}^{\infty}(M^{2},B\boxtimes\overline{A})$ vanishing to order
$m$ along the diagonal. Then by the same proof as Lemma 5.1, the family
$(E^{k}b)$ is in $\mathcal{O}_{\infty}(k^{-\frac{m}{2}})$. Actually, we even
have a better result if instead of using any derivatives, we only consider
covariant derivatives for the connection of
$(L\boxtimes\overline{L})^{k}\otimes(B\boxtimes\overline{A})$ induced by the
connection of $L$ and any connections of $A$ and $B$.

######  Lemma 6.2.

For any $\ell\in{\mathbb{N}}$, any vector fields $X_{1}$, …, $X_{\ell}$ of
$M^{2}$, we have $\nabla_{X_{1}}\ldots\nabla_{X_{\ell}}(E^{k}b)$ is in
$\mathcal{O}(k^{-\frac{1}{2}(m-\ell)})$.

The improvement is that we only loose a half power of $k$ for each derivative.

###### Proof.

The main observation is that $\nabla E$ vanishes on the diagonal. Indeed,
$\nabla_{X}E=0$ on the diagonal when $X$ is tangent to the first factor
because of the second equation in (54), but also when $X$ is tangent to the
diagonal by the first equation in (54). So on a neighborhood of the diagonal,
we have $\nabla_{X}E=fE$ with $f\in\mathcal{O}(1)$. By Leibniz rule,
$\nabla_{X}(E^{k}b)=E^{k}(kfb+\nabla_{X}b)$. Using this repeatedly, we obtain

$\nabla_{X_{1}}\ldots\nabla_{X_{\ell}}(E^{k}b)=E^{k}(k^{\ell}b_{\ell}+k^{\ell-1}b_{\ell-1}+\ldots+b_{0})$

where $b_{\ell}=\mathcal{O}(m+\ell)$, $b_{\ell-1}\in\mathcal{O}(m+\ell-2)$, …,
$b_{0}\in\mathcal{O}(m-\ell)$. And we conclude as in the proof of Lemma 5.1. ∎

Recall that the Schwartz kernel family of an operator $P\in\mathcal{L}(A,B)$
has by definition an expansion of the form

$\displaystyle
P_{k}(x,y)=\Bigl{(}\frac{k}{2\pi}\Bigr{)}^{n}E^{k}(x,y)\sum_{\ell+m(\ell)\leqslant
N}k^{-\frac{\ell}{2}}b_{\ell}(x,y)+R_{N,k}(x,y)$ (79)

with $R_{N,k}\in\mathcal{O}(k^{n-\frac{N+1}{2}})$. Define the classes

$\mathcal{L}^{\infty}(A,B):=\mathcal{L}(A,B)\cap\mathcal{O}_{\infty}(k^{n}),\quad\mathcal{L}_{q}^{\infty}(A,B):=\mathcal{L}_{q}(A,B)\cap\mathcal{O}_{\infty}(k^{n})$

By the following proposition, these new classes have the same properties than
the $\mathcal{L}(A,B)$ and this follows directly from Lemma 6.1.

######  Proposition 6.3.

  1. 1.

If $P\in\mathcal{L}^{\infty}(A,B)$ and the expansion (79) holds with
$R_{N,k}\in\mathcal{O}(k^{n-\frac{N+1}{2}})$ for any $N$, then
$R_{N,k}\in\mathcal{O}_{\infty}(k^{n-\frac{N+1}{2}})$ for any $N$.

  2. 2.

For any $P\in\mathcal{L}(A,B)$ there exist $Q\in\mathcal{L}^{\infty}(A,B)$
unique modulo $\mathcal{O}_{\infty}(k^{-\infty})$ such that
$Q=P+\mathcal{O}(k^{-\infty})$.

  3. 3.

For any $P\in\mathcal{L}^{\infty}(A,B)$,

    1. (a)

the adjoint of $P$ belongs to $\mathcal{L}^{\infty}(B,A)$,

    2. (b)

$(P_{k}Q_{k})\in\mathcal{L}^{\infty}(A,C)$ for any
$Q\in\mathcal{L}^{\infty}(B,C)$,

    3. (c)

$(f\circ P_{k})$ belongs to $\mathcal{L}^{\infty}(A,C)$ for any
$f\in{\mathcal{C}}^{\infty}(M,\operatorname{Hom}(B,C))$; $(P_{k}\circ g)$
belongs to $\mathcal{L}^{\infty}(C,B)$ for any
$g\in{\mathcal{C}}^{\infty}(M,\operatorname{Hom}(C,A))$

    4. (d)

for any vector field $X$ of $M$ and connections on $A$ and $B$,
$(k^{-\frac{1}{2}}P_{k}\circ\nabla^{L^{k}\otimes A}_{X})$ and
$(k^{-\frac{1}{2}}\nabla^{L^{k}\otimes B}_{X}\circ P_{k})$ belong to
$\mathcal{L}^{\infty}(A,B)$.

Furthermore, if $(P_{k})$ is even (resp. odd), these two operators are odd
(resp. even).

  4. 4.

$\mathcal{L}_{q}^{\infty}(A,B)=\mathcal{L}(A,B)\cap\mathcal{O}_{\infty}(k^{n-\frac{q}{2}})$,
the restriction of ${\sigma}_{q}$ to $\mathcal{L}_{q}^{\infty}(A,B)$ is onto
and has kernel $\mathcal{L}_{q+1}^{\infty}(A,B)$.

###### Proof.

Assertion 1 follows the preliminary observation on $E^{k}b$ and the second
part of Lemma 6.1. Assertion 2 follows from the first part of Lemma 6.1. Claim
3b follows from the fact that the composition of two kernels in
$\mathcal{O}_{\infty}(k^{n})$ is in $\mathcal{O}_{\infty}(k^{2n})$, and
$\mathcal{O}_{\infty}(k^{2n})\cap\mathcal{L}(A,C)=\mathcal{L}^{\infty}(A,C)$
by the second part of Lemma 6.1. Claims 3a and 3c are straightforward. Ones
proves claim 3d by arguing as in the proof of Lemma 6.2. Part 4 follows from
the second part of Lemma 6.1. ∎

######  Remark 6.4.

We can adapt Theorems 3.1 and 3.4 to the spaces $\mathcal{L}^{\infty}$:

  1. 1.

in Theorem 3.1, if we start with $P\in\mathcal{L}^{\infty}(A)$, then
$\chi(P)\in\mathcal{L}^{\infty}(A)$.

  2. 2.

in Theorem 3.4, if $\Pi$ and $\Pi^{\prime}$ are in $\mathcal{L}^{\infty}(A)$
and $\mathcal{L}^{\infty}(B)$ respectively, then we can choose
$U\in\mathcal{L}^{\infty}(A,B)$.

In both cases, the only change in the proof is the fact that for any families
of operators $Q_{k},Q^{\prime}_{k}:{\mathcal{C}}^{\infty}(M,L^{k}\otimes
A)\rightarrow{\mathcal{C}}^{\infty}(M,L^{k}\otimes A)$ and
$Q^{\prime\prime}_{k}:L^{2}(M,L^{k}\otimes A)\rightarrow L^{2}(M,L^{k}\otimes
A)$, by [9, Section 4.3], if the Schwartz kernel families of $(Q_{k})$ and
$(Q^{\prime}_{k})$ are respectively in $\mathcal{O}_{\infty}(k^{-N})$ and
$\mathcal{O}_{\infty}(k^{-N^{\prime}})$, and the operator norms of
$Q_{k}^{{}^{\prime\prime}}$ are in $\mathcal{O}(1)$, then the Schwartz kernel
family of $Q_{k}Q^{\prime\prime}_{k}Q_{k}^{\prime}$ is in
$\mathcal{O}_{\infty}(k^{-(N+N^{\prime})})$. ∎

By Theorem 2.2, we already know how to compute the symbols of $P^{*}$, $PQ$,
$fP$ or $gP$ in terms of the symbols of $P$ and $Q$. To complete this, we
compute the symbol of the compositions of $P$ with the covariant derivatives
$\nabla_{X}^{L^{k}\otimes A}$ and $\nabla_{X}^{L^{k}\otimes B}$. Recall that
for any $Y\in T_{x}M$, we defined in the introduction some endomorphisms
$\rho(Y)\in\operatorname{End}(\mathcal{D}(T_{x}M))$ in (10).

######  Lemma 6.5.

For any $P\in\mathcal{L}^{\infty}(A,B)$ and vector field $X$ of $M$, we have

$\displaystyle{\sigma}_{0}(k^{-\frac{1}{2}}P_{k}\circ\nabla^{L^{k}\otimes
A}_{X})(x)={\sigma}_{0}(P_{k})(x)\circ\rho(X(x))$
$\displaystyle{\sigma}_{0}(k^{-\frac{1}{2}}P_{k}\circ\nabla^{L^{k}\otimes
A}_{X})(x)=\rho(X(x))\circ{\sigma}_{0}(P_{k})(x).$

###### Proof.

We deduce one formula from the other by taking adjoint. To prove the first
one, it suffices by Proposition 5.9 to show that if $(\Phi_{k}^{f})$ is the
peaked section associated to $f\in\mathcal{D}(T_{x}M)\otimes A_{x}$, then

$k^{-\frac{1}{2}}\nabla_{X}\Phi_{k}^{f}=\Phi_{k}^{g}+\mathcal{O}(k^{-\frac{1}{2}})$

where $g=\rho(X(x))f$. This is easily checked if we use the normal coordinates
as in (58) centered at $p_{0}=x$. We have to derivate (76). We have first that
$k^{-\frac{1}{2}}\nabla_{X}(E^{k})=k^{\frac{1}{2}}E^{k}(\nabla_{X}E)E^{-1}$
and by (59),

$(\nabla_{\partial_{i}}E)E^{-1}=-\overline{z}_{i}+\mathcal{O}(2),\qquad(\nabla_{\overline{\partial}_{i}}E)E^{-1}=\mathcal{O}(2).$

Second we have $k^{-\frac{1}{2}}\partial_{i}f(k^{\frac{1}{2}}\xi)=0$ since
$f\in{\mathbb{C}}[\overline{z}_{1},\ldots,\overline{z}_{n}]$ and
$k^{-\frac{1}{2}}\overline{\partial}_{i}f(k^{\frac{1}{2}}\xi)=(\partial
f/\partial\overline{z}_{i})(k^{\frac{1}{2}}\xi)$. To conclude recall that
$\rho(\partial_{i})$ is the multiplication by $-\overline{z}_{i}$ whereas
$\rho(\overline{\partial}_{i})$ is the derivation with respect to
$\overline{z}_{i}$. ∎

###  6.3 Proofs of Theorems 1.1, 1.3 and 1.4

In this last section, we complete the proof of the theorems stated in the
introduction, a generalization actually since we will consider more general
projectors.

Let $\Pi\in\mathcal{L}^{\infty}(A)\cap\mathcal{L}^{+}(A)$ be a self-adjoint
projector with symbol $\pi=\pi_{m}\otimes\operatorname{id}_{A}$, where
$\pi_{m}$ is the projector of $\mathcal{D}(TM)$ onto $\mathcal{D}_{m}(TM)$.
Such an operator exists by Theorem 3.1 and Remark 6.4. Alternatively, the
projector $\Pi=(\Pi_{m,k})$ onto the $m$-th Landau level defined in (40) has
the expected properties [10, Theorems 5.2, 5.3].

By Theorem 3.6, the dimension of $\mathcal{H}_{k}=\operatorname{Im}(\Pi_{k})$
is

$\operatorname{dim}\mathcal{H}_{k}=\int_{M}\operatorname{ch}(L^{k}\otimes
A\otimes\mathcal{D}_{m}(TM))\;\operatorname{Td}(M)$

when $k$ is sufficiently large, which implies Theorem 1.1.

Define the Toeplitz algebra

$\mathcal{T}^{\infty}=\\{P\in\mathcal{L}^{\infty}(A)\cap\mathcal{L}^{+}(A)/\;\Pi
P\Pi=P\\}.$

Clearly $\mathcal{T}^{\infty}$ is contained in the Toeplitz algebra
$\mathcal{T}$ defined in (45), and by assertion 2 of Proposition 6.3, the
difference is rather small: for every $P\in\mathcal{T}$, there exists
$P^{\prime}\in\mathcal{T}^{\infty}$ unique modulo
$\mathcal{O}_{\infty}(k^{-\infty})$ such that
$P^{\prime}=P+\mathcal{O}(k^{-\infty})$. Recall the symbol map $\tau_{0}$
introduced in Theorem 3.2 and denote by $\tau$ its restriction to
$\mathcal{T}^{\infty}$

$\tau:\mathcal{T}^{\infty}\rightarrow{\mathcal{C}}^{\infty}(M,\operatorname{End}(\mathcal{D}_{m}(TM)\otimes
A)).$

We could as well consider the maps $\tau_{q}$ with $q\geqslant 1$ but we will
limit ourselves to $\tau_{0}$. By Theorem 3.2 and Assertions 2, 4 of
Proposition 6.3, $\tau$ is onto and its kernel is
$k^{-1}\mathcal{T}^{\infty}$. It follows as well from Theorem 3.2 that for any
$P,Q\in\mathcal{T}^{\infty}$

$\displaystyle\tau(PQ)=\tau(P)\tau(Q),\quad\|P_{k}\|=\sup_{x\in
M}\|\tau(P)_{x}\|+\mathcal{O}(k^{-1})$ $\displaystyle
P_{k}(x,x)=\Bigl{(}\frac{k}{2\pi}\Bigr{)}^{n}\operatorname{tr}(\tau(P)_{x})+\mathcal{O}(k^{-1})$

Choose any connection on $A$. By Proposition 6.4, for any
$f\in{\mathcal{C}}^{\infty}(M,\operatorname{End}A)$, $p\in{\mathbb{N}}$ and
vector fields $X_{1}$, …, $X_{2p}$ of $M$, the operator

$\displaystyle
T_{k}(f,X_{1},\ldots,X_{2p})=k^{-p}\Pi_{k}f\nabla_{X_{1}}^{L^{k}\otimes
A}\ldots\nabla_{X_{2p}}^{L^{k}\otimes A}\Pi_{k}$ (80)

belong to $\mathcal{T}^{\infty}$ and its $\tau$-symbol is
$(\pi_{m}\rho(X_{1})\ldots\rho(X_{2p})\pi_{m})\otimes f$. Since these symbols
generates
${\mathcal{C}}^{\infty}(M,\operatorname{End}(\mathcal{D}_{m}(TM)\otimes A))$
as a vector space, we deduce that any $P\in\mathcal{T}^{\infty}$ is of the
form

$P_{k}=\sum_{\ell=0}^{N}k^{-\ell}P_{\ell,k}+\mathcal{O}(k^{-(N+1)}),\qquad\forall
N\in{\mathbb{N}}$

where for any $\ell$, $(P_{\ell,k})_{k}$ is a finite sum of operators such as
(80). So in the case where the auxiliary bundle $A$ is trivial,
$\mathcal{T}^{\infty}$ is the space $\mathcal{T}^{\operatorname{sc}}_{m}$
defined in the introduction. To end the proof of Theorem 1.3, we will show
that for any $f,g\in{\mathcal{C}}^{\infty}(M)$,

$\displaystyle
T_{k}(f)T_{k}(g)=T_{k}(fg)+k^{-1}T_{k}(1,X,Y)+\mathcal{O}(k^{-2})$ (81)

where $X$ and $Y$ are the Hamiltonian vector fields of $f$ and $g$
respectively. Our sign convention is ${\omega}(X,\cdot)+df=0$. The proof is
based on the following result, interesting by itself, which involves the
Kostant-Souriau operator.

######  Lemma 6.6.

For any $f\in{\mathcal{C}}^{\infty}(M,{\mathbb{R}})$ with Hamiltonian vector
field $X$, for any $P\in\mathcal{L}^{\infty}_{0}(A)$ we have

  * •

$[f,P]$ belongs to $\mathcal{L}_{1}^{\infty}(A)$

  * •

$[f,P]\equiv(ik)^{-1}[\nabla_{X}^{L^{k}\otimes A},P]$ modulo
$\mathcal{L}_{2}^{\infty}(A)$.

So the commutator $[f+\frac{i}{k}\nabla_{X}^{L^{k}\otimes A},P]$ belongs to
$\mathcal{L}_{2}^{\infty}(A)$.

###### Proof.

The Schwartz kernel of $[f,P]$ is the product of $g(x,y)=f(x)-f(y)$ by the
Schwartz kernel of $P$. Since $g$ vanishes along the diagonal, this implies
that $[f,P]\in\mathcal{L}_{1}(A)$. Furthermore, it follows from the definition
(69) of the symbol that if
$\widetilde{{\sigma}}_{0}(P)(x)=\operatorname{Op}(b)$ with
$b\in\mathcal{P}(T_{x}M)$, then
$\widetilde{{\sigma}}_{1}(P)(x)=\operatorname{Op}(\ell b)$ where
$\ell=d_{x}f\in T_{x}^{*}M$. Now a computation from (52) shows that

$[\operatorname{Op}(b),a_{i}]=\operatorname{Op}(z_{i}b),\qquad[\operatorname{Op}(b),a_{i}^{*}]=\operatorname{Op}(\overline{z}_{i}b)$

where as in Section 4 we use the annihilition and creation operators
$a_{i}=\partial_{\overline{z}_{i}}$,
$a_{i}^{*}=\overline{z}_{i}-\partial_{z_{i}}$. Since
$\rho(\overline{\partial}_{i})$ and $\rho(\partial_{i})$ are the restrictions
of $a_{i}$ and $-a_{i}^{*}$ to
${\mathbb{C}}[\overline{z}_{1},\ldots,\overline{z}_{n}]$ respectively, we
deduce from Lemma 6.5 that

${\sigma}_{1}([f,P])=\tfrac{1}{i}{\sigma}_{0}(\bigl{[}k^{-\frac{1}{2}}\nabla_{X},P])$

and the result follows. ∎

###### Proof of (81).

By a straightforward computation, we have

$\Pi f\Pi g\Pi=\Pi fg\Pi+\Pi[f,\Pi][g,\Pi]\Pi$

By lemma 6.6, $\Pi[f,\Pi][g,\Pi]\Pi$ belongs to $\mathcal{L}_{2}(A)$ and its
symbol is

$\displaystyle{\sigma}_{2}(\Pi[f,\Pi][g,\Pi]\Pi)$
$\displaystyle=-\pi_{m}[\rho(X),\pi_{m}][\rho(Y),\pi_{m}]\pi_{m}$
$\displaystyle=\pi_{m}\rho(X)\rho(Y)\pi_{m}\qquad$ since
$\pi_{m}\rho(X)\pi_{m}=0$ and $\pi_{m}\rho(Y)\pi_{m}=0$
$\displaystyle={\sigma}_{2}(k^{-1}\Pi\nabla_{X}^{L^{k}\otimes
A}\nabla_{Y}^{L^{k}\otimes A}\Pi)$

and the result follows. ∎

Let us prove now Theorem 1.4. Introduce a quantization $(\mathcal{H}_{F,k})$
of $(M,L)$ twisted by $F=\mathcal{D}_{m}(TM)\otimes A$. We can adapt the
definition (24) of $W_{k}$ with the auxiliary bundle $A$ by setting

$\displaystyle\begin{split}W_{k}:{\mathcal{C}}^{\infty}(M,L^{k}\otimes
A)\rightarrow{\mathcal{C}}^{\infty}(M,L^{k}\otimes F),\qquad
k\in{\mathbb{N}}\\\ W_{k}=R_{m}D_{G^{\otimes(m-1)}\otimes A,k}\circ
D_{G^{\otimes(m-2)}\otimes A,k}\circ\ldots\circ D_{G\otimes A,k}\circ
D_{A,k}\end{split}$ (82)

######  Lemma 6.7.

The operator $(V_{k}=\frac{1}{m!}k^{-\frac{m}{2}}\Pi_{F,k}W_{k}\Pi_{k}$,
$k\in{\mathbb{N}}$) belongs to $\mathcal{L}^{\infty}(A,F)$, has the same
parity as $m$ and its symbol ${\sigma}_{0}(V)$ viewed as a morphism from
$\mathcal{D}(TM)\otimes A$ to $\mathcal{D}(TM)\otimes F$ is given by

$\forall\,f\in\mathcal{D}_{p}(TM),\;\forall\,a\in A,\;{\sigma}_{0}(V)(f\otimes
a)=\begin{cases}1\otimes f\otimes a\text{ if $p=m$}\\\ 0\text{
otherwise}\end{cases}$

###### Proof.

By Proposition 6.3, for any even (resp. odd) operator
$P\in\mathcal{L}^{\infty}(B,A)$, $(k^{-\frac{1}{2}}D_{A,k}\circ P)$ belongs to
$\mathcal{L}^{\infty}(B,A\otimes G)$, is odd (resp. even) and its symbol is
$\varphi_{A}\circ{\sigma}_{0}(P)$ where

$\varphi_{A}=\sum_{i}a_{i}\otimes\overline{z}_{i}\otimes\operatorname{id}_{A}\in\mathcal{S}(TM)\otimes
G\otimes\operatorname{End}A.$

Here we have introduced an orthonormal frame $(\partial_{i})$ of $T^{1,0}M$,
$(z_{i})$ is the dual frame of $(T^{1,0}M)^{*}$ and
$a_{i}=\partial_{\overline{z}_{i}}$ is the annihilation operator.

Consequently, $k^{-\frac{m}{2}}R_{m}D_{A\otimes G^{m-1}}\circ\ldots\circ
D_{A,k}\circ\Pi_{k}$ belong to $\mathcal{L}(A,F)$ with symbol
$\varphi_{A}^{m}\circ\pi_{m}$ where $\varphi_{A}^{m}$ is the morphism from
$\mathcal{D}(TM)\otimes A$ to
$\mathcal{D}(TM)\otimes\mathcal{D}_{m}(TM)\otimes A$ given by

$\varphi_{A}^{m}=\sum_{i_{1},\ldots,i_{m}=1}^{n}a_{i_{1}}\ldots
a_{i_{m}}\otimes(\overline{z}_{i_{1}}\ldots\overline{z}_{i_{m}})\otimes\operatorname{id}_{A}$

Computing $\varphi_{A}^{m}(\overline{z}^{{\beta}})$ for any multi-index
${\beta}$, we show that for any $f\in\mathcal{D}_{p}(TM)$ and $a\in A$,
$\varphi_{A}^{m}(f\otimes a)=m!(1\otimes f\otimes a)$ when $p=m$ and $0$
otherwise. ∎

The symbol $\sigma_{0}(V)$ is exactly the symbol $\rho$ introduced in Lemma
3.3. Now Theorem 1.4 and the remark on the unitarization of $V_{k}$ follows by
the same proof as Theorem 3.4.

##  7 Appendix

In this appendix we discuss the known results for surfaces with constant
curvature. Two important features appear for the negatively curved surfaces:
only the lower part of the spectrum consists of Landau levels, and moreover,
there is an isomorphism between the $m$-th Landau level and the first level of
a Laplacian twisted by the $m$-th power of the complex determinant bundle.

These results appeared in the physics literature, cf. in particular [15] for
the case with surface with genus $\geqslant 2$. A more recent mathematical
reference is [26].

#### The plane [18]

Consider a quantum particle confined in a two dimensional plane $(x,y)$ and
subject to a constant magnetic field perpendicular to this plane. Its
Hamiltonian is the operator

$\displaystyle H=-\tfrac{1}{2}(\nabla_{x}^{2}+\nabla_{y}^{2})\quad\text{ with
}\quad\nabla_{x}=\tfrac{\partial}{\partial
x}+\tfrac{i}{2}By,\;\nabla_{y}=\tfrac{\partial}{\partial y}-\tfrac{i}{2}Bx.$
(83)

$B$ is a positive constant representing the strength of the magnetic field.
The spectrum of $H$ is $B(\frac{1}{2}+{\mathbb{N}})$ and the Landau levels
$\mathcal{H}_{m}=\operatorname{ker}(H-B(\frac{1}{2}+m))$ are given in terms of
the ladder operators $\nabla_{z}=\nabla_{x}-i\nabla_{y}$,
$\nabla_{\overline{z}}=\nabla_{x}+i\nabla_{y}$ by

$\displaystyle\mathcal{H}_{0}=\operatorname{ker}(\nabla_{\overline{z}}),\qquad\mathcal{H}_{m}=(\nabla_{z})^{m}\mathcal{H}_{0},\quad
m\geqslant 1.$ (84)

#### Surface with constant curvature, [15], [26]

Let $M$ be a compact orientable surface with a Riemannian metric having a
constant Gauss curvature $S$. Introduce a Hermitian line bundle $L\rightarrow
M$ with a connection
$\nabla:{\mathcal{C}}^{\infty}(M,L)\rightarrow\Omega^{1}(M,L)$. Assume that
the curvature satisfies

$\displaystyle i\operatorname{curv}(\nabla)=B\operatorname{vol}_{g}$ (85)

where $B$ is a non-zero constant and $\operatorname{vol}_{g}$ is the
Riemannian volume. Choosing the convenient orientation for $M$, we can assume
that $B$ is positive. The quantum Hamiltonian is the Laplacian
$\Delta:=\frac{1}{2}\nabla^{*}\nabla$ acting on sections of $L$. Then denoting
its eigenvalue by $0\leqslant{\lambda}_{0}<{\lambda}_{1}<\ldots$, it is known
that

$\displaystyle{\lambda}_{m}=B(\tfrac{1}{2}+m)+S\tfrac{m(m+1)}{2}\qquad\text{
if }B+mS>0$ (86)

For a sphere or a torus, $S\geqslant 0$, and these formulas describe the whole
spectrum. If the genus of $M$ is larger than $2$, then $S<0$ and the condition
$B+mS>0$ is satisfied only for a finite number of $m$. In this case, it is not
reasonable to expect an explicit formula for the other eigenvalues. Indeed, if
$S=-1$ and $L=K^{r}$ with $K$ the canonical bundle of $M$ and $r$ a positive
integer, then (86) gives the first $(r+1)$ eigenvalues, and for any
$n\in{\mathbb{N}}$, ${\lambda}_{r+n}={\lambda}_{r}+\tfrac{1}{2}\mu_{n}$, where
$\\{\mu_{n},\;n\in{\mathbb{N}}\\}$ is the spectrum of the Laplace-Beltrami
operator of $M$. This latter spectrum depends in an essential way on the
metric. Indeed, by Huber’s theorem [6, Theorem 9.2.9], $\\{\mu_{n}\\}$
determines the length spectrum of $M$.

The multiplicity of the first eigenvalues is equal to:

$\displaystyle\operatorname{mult}({\lambda}_{m})=B\frac{\operatorname{vol}_{g}(M)}{2\pi}+(\tfrac{1}{2}+m)\chi(M)\qquad\text{
if }B+(m+1)S>0.$ (87)

Here $\chi(M)$ is the Euler characteristic of $M$ and observe that
$B\operatorname{vol}_{g}(M)/(2\pi)$ is the degree of $L$, so it is an integer.
(87) follows from a description of the corresponding eigenspace
$\mathcal{H}_{m}=\ker(\Delta-{\lambda}_{m})$ similar to (84), that we explain
below.

#### Proof of formulas (86) and (87)

To start with, we do not assume that the Gauss curvature $S$ and the function
$B$ defined in (85) are constant. We choose any orientation on $M$. Let $j$ be
the complex structure of $M$ compatible with $g$, i.e. $g(jX,jY)=g(X,Y)$ for
any tangent vectors $X$, $Y\in T_{p}M$ and $X\wedge jX>0$ if $X\neq 0$. Since
we are in real dimension 2, $j$ is integrable. Furthermore the associated
volume form $\operatorname{vol}_{g}$ is the symplectic form
${\omega}(X,Y)=g(jX,Y)$.

$L$ has a natural holomorphic structure such that its
$\overline{\partial}$-operator is $\nabla^{0,1}$. We denote it by
$\overline{\partial}_{L}:{\mathcal{C}}^{\infty}(L)\rightarrow{\mathcal{C}}^{\infty}(L\otimes\overline{K})$
with $K=(T^{*}M)^{1,0}$ the canonical bundle. The curvature of $\nabla$ has
the form $\frac{1}{i}B{\omega}$ with
$B\in{\mathcal{C}}^{\infty}(M,{\mathbb{R}})$. The canonical bundle has a
natural metric induced by $g$. Its Chern connection, that is its connection
compatible with both its metric and holomorphic structure, has curvature
$iS{\omega}$ where $S$ is the Gauss curvature.

######  Theorem 7.1.

The following identities holds:

  1. 1.

Weitzenböck formula:
$\Delta_{L}=\overline{\partial}^{*}_{L}\overline{\partial}_{L}+\frac{1}{2}B$.

  2. 2.

Bosonic commutation relation:
$\overline{\partial}_{L}\overline{\partial}_{L}^{*}=\overline{\partial}_{L\otimes
K^{-1}}^{*}\overline{\partial}_{L\otimes K^{-1}}+(B+S)$.

The Weitzenböck formula is a classical relation, it holds more generally on
Kähler manifolds. We call the second formula the bosonic commutation relation
because it replaces the canonical commutation relation satisfied by the
creation/annihilation operators $[a,a^{*}]=1$. In this formula, we identify
$\overline{K}$ with $K^{-1}$ through the metric so that the operators
$\overline{\partial}_{L}\overline{\partial}_{L}^{*}$ and
$\overline{\partial}_{L\otimes K^{-1}}^{*}\overline{\partial}_{L\otimes
K^{-1}}$ act on the same space
${\mathcal{C}}^{\infty}(L\otimes\overline{K})={\mathcal{C}}^{\infty}(L\otimes
K^{-1})$. A similar formula were obtained in [26, Proposition 9] for the same
purpose of computing the spectrum of $\Delta_{L}$.

###### Proof of the bosonic identity.

Introduce a local holomorphic frame $s$ of $L$ and a complex coordinate $z$ on
$M$. We have first $\overline{\partial}_{L}(fs)=f_{\overline{z}}\;s\otimes
d\overline{z}$. To compute the adjoint, recall that the scalar products of
${\mathcal{C}}^{\infty}(L)$ and ${\mathcal{C}}^{\infty}(L\otimes\overline{K})$
are defined by integrating the pointwise scalar products against the volume
form. Write ${\omega}=ihdz\wedge d\overline{z}$ and $|s|^{2}=e^{-\varphi}$
with $h$ and $\varphi$ real valued functions. Then
$|d\overline{z}|^{2}=h^{-1}$ and a direct computation leads to

$\overline{\partial}_{L}^{*}(fs\otimes
d\overline{z})=h^{-1}(-f_{z}+f\varphi_{z})s.$

With the identification $\overline{K}\simeq K^{-1}$, we have
$d\overline{z}=h^{-1}(dz)^{-1}$. We deduce that

$\overline{\partial}_{L}\overline{\partial}_{L}^{*}(fs\otimes(dz)^{-1})=h^{-1}\bigl{(}-f_{z\overline{z}}+f_{\overline{z}}(\varphi_{z}-\tfrac{h_{z}}{h})+f(\varphi_{z\overline{z}}-\partial_{\overline{z}}(\tfrac{h_{z}}{h}))\bigr{)}s\otimes(dz)^{-1}$

Similar computations by using $|s\otimes dz^{-1}|^{2}=he^{-\varphi}$ leads to

$\overline{\partial}_{L\otimes K^{-1}}^{*}\overline{\partial}_{L\otimes
K^{-1}}(fs\otimes(dz)^{-1})=h^{-1}\bigl{(}-f_{z\overline{z}}+f_{\overline{z}}(\varphi_{z}-\tfrac{h_{z}}{h})\bigr{)}s\otimes(dz)^{-1}.$

To conclude, observe that $B=h^{-1}\varphi_{z\overline{z}}$ and
$S=-h^{-1}\partial_{z}\partial_{\overline{z}}\,\ln h$. ∎

From now on, we will assume that $B$ and $S$ are constant. With the
Weitzenböck formula, we pass directly from the spectrum of $\Delta_{L}$ to the
one of $\overline{\partial}^{*}_{L}\overline{\partial}_{L}$. We can use the
Bosonic relation exactly as it is usually done with the Landau Hamiltonian,
cf. proof of Proposition 4.1. We deduce that for any ${\lambda}\neq 0$,
${\lambda}$ is an eigenvalue of
$\overline{\partial}^{*}_{L}\overline{\partial}_{L}$ if and only if
${\lambda}-(B+S)$ is an eigenvalue of $\overline{\partial}^{*}_{L\otimes
K^{-1}}\overline{\partial}_{L\otimes K^{-1}}$. Moreover the eigenspaces have
the same dimension. Indeed $\overline{\partial}_{L}$ restricts to an
isomorphism

$\operatorname{Ker}({\lambda}-\overline{\partial}^{*}_{L}\overline{\partial}_{L})\rightarrow\operatorname{ker}({\lambda}-(B+S)-\overline{\partial}^{*}_{L\otimes
K^{-1}}\overline{\partial}_{L\otimes K^{-1}})$

with inverse the restriction of ${\lambda}^{-1}\overline{\partial}_{L}^{*}$.
Besides this, $\ker(\overline{\partial}^{*}_{L}\overline{\partial}_{L})$ is
the space $H^{0}(L)$ of holomorphic sections of $L$. By Riemann-Roch theorem,
$H^{0}(L)$ has dimension $d+\frac{1}{2}\chi(M)$ if the degree
$d=B\operatorname{Vol}(M)/(2\pi)$ of $L$ is larger that $-\chi(M)$.

To summarize, when $B$ is sufficiently large, $0$ is an eigenvalue of
$\overline{\partial}^{*}_{L}\overline{\partial}_{L}$ with multiplicity equal
to $B\operatorname{Vol}(M)/(2\pi)+\frac{1}{2}\chi(M)$, and the remainder of
the spectrum is identical with the spectrum of
$(B+S)+\overline{\partial}^{*}_{L\otimes K^{-1}}\overline{\partial}_{L\otimes
K^{-1}}$, multiplicities included. We can iterate this argument and deduce by
induction the formulas (86), (87) giving the first eigenvalues of $\Delta_{L}$
with their multiplicity. Since $\operatorname{deg}(L\otimes
K^{-1})=\operatorname{deg}(L)+\chi(M)$, we can repeat ad infinitum this
argument when $\chi(M)\geqslant 0$ and obtain the whole spectrum of
$\Delta_{L}$; whereas for $\chi(M)<0$, only a finite number of iterations is
possible.

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Laurent Charles
Sorbonne Université, Université de Paris, CNRS
Institut de Mathématiques de Jussieu-Paris Rive Gauche
F-75005 Paris, France
E-mail<EMAIL_ADDRESS>