1 Introduction
Let
$\mathbb D^2\subseteq \mathbb C^2$
be the unit polydisc,
$H^2(\mathbb D^2)$
be the Hardy space over
$\mathbb {D}^2,$
and
$H^\infty (\mathbb D^2)$
be the space of bounded holomorphic functions. We will study the Fredholmness and Fredholm index of Toeplitz pair
$(T_{f_1},T_{f_2})$
on
$H^2(\mathbb D^2)$
with
$f_i\in H^\infty (\mathbb D^2)$
.
The Fredholmness of Toeplitz pair
$(T_{f_1},T_{f_2})$
with bounded holomorphic symbols on the Bergman space
$L^2_a(\mathbb B_n)$
over the unit ball
$\mathbb B_n$
was studied by Putinar [Reference Putinar10], which was generalized to the case of strongly pseudoconvex domains with
$C^3$
boundary by Andersson and Sandberg [Reference Andersson and Sandberg1]. For Toeplitz tuple
$(T_{\varphi _1},\cdots ,T_{\varphi _m})$
on the Bergman space
$L^2_a(\Omega )$
on a strongly pseudoconvex domains
$\Omega \subseteq \mathbb C^n$
with symbols
$\varphi _i\in C(\overline {\Omega })$
, the Fredholm index was studied by Guo [Reference Guo5], in which the essential commutativity shown in [Reference Salinas, Sheu and Upmeier12] of the Toepliz algebra played the key role.
In [Reference Guo and Wang6], Guo and the first author studied the Fredholm index of Toeplitz pairs on
$H^2(\mathbb D^2)$
with rational inner symbols, in which the Fredholmness can be deduced by Spectral Mapping Theorem directly. However, for Toeplitz pair
$(T_{f_{1}},T_{f_{2}})$
with bounded holomorphic symbols, the Spectral Mapping Theorem does not work. In the present note, the Fredholmness will be obtained by using the idea in the proof for existence of
$L^p$
solution of Corona Problem over
$\mathbb {D}^2$
[Reference Lin8, Reference Lin9]. In [Reference Douglas and Sarkar3], Douglas and Sarkar highlighted the relationship between the Corona Theorem and Fredholmness of Toeplitz tuples over the ball and the disk, and this connection was also emphasized by [Reference Andersson and Sandberg1]. To state the result, for
$0<r<1$
, set
$$ \begin{align*}\mathbb{D}_r=\{z,|z|<r\},\quad\mathbb{D}_r^n=\mathbb{D}_r\times\cdots \times\mathbb{D}_r,\quad \text{and}\quad \mathbb{U}_r^n=\mathbb{D}^n\,{\backslash}\, \mathbb{D}_r^n,\end{align*} $$
and we have the following result.
Theorem 1.1 For
$f_{1}, f_{2}\in H^\infty (\mathbb D^2)$
, the Toeplitz pair
$\left (T_{f_{1}}, T_{f_{2}}\right )$
on
$H^{2}\left (\mathbb {D}^{2}\right )$
is Fredholm if and only if there is
$0<r<1$
and
$c>0$
such that
$\left |f_{1}(z)\right |{}^{2}+\left |f_{2}(z)\right |{}^{2} \geqslant c$
on
${\mathbb U}_r^2$
, and in this case, the Fredholm index of the Toeplitz pair is given by
$$ \begin{align*}\operatorname{Ind} \left(T_{f_{1}}, T_{f_{2}}\right)=-\operatorname{dim} H^{2}(\mathbb{D}^{2}) /( f_{1} H^{2}\left(\mathbb{D}^{2}\right)+f_{2} H^{2}(\mathbb{D}^{2})). \end{align*} $$
Such a result can be seen as a polydisc analogue to the result of Putinar [Reference Putinar10]. The difficulty to prove the above result is to characterize when
$f_1 H^2(\mathbb D^2)+f_2 H^2(\mathbb D^2)$
is of finite codimension, which was solved by using the characteristic space theory of polynomials in [Reference Guo and Wang6] for the case that
$f_1$
and
$f_2$
are rational inner functions.
For Toeplitz tuple
$(T_{p_1},\cdots ,T_{p_n})$
with symbols in ploynomial ring
$\mathbb {C}[z_1,\cdots ,z_n],$
the Fredholm index was given in Kaad and Nest [Reference Kaad and Nest7, Corollary 9.4]. If
${\partial \mathbb D^n \cap (\bigcap \limits _{i=1}^n Z(p_i)) = \emptyset} $
, then
$(T_{p_1},\cdots ,T_{p_n})$
is Fredholm, and
$$ \begin{align} \operatorname{Ind}(T_{p_{1}},\cdots,T_{p_{n}})=-\operatorname{deg} (0,P_n,\mathbb{D}^n), \end{align} $$
where
$P_n=(p_1,\cdots ,p_n):\mathbb D^n\rightarrow \mathbb C^n$
is the polynomial map.
Remark 1.2 Let
$f_i\in A(\mathbb D^2)$
. Suppose that
$Z(f_1)\cap Z(f_2)\cap \partial \mathbb D^2=\emptyset $
. Then,
$(T_{f_1}, T_{f_2})$
is Fredholm. Moreover, for
$i=1,2$
we can find a sequence of polynomials
$p_{1,n}$
and
$p_{2,n}$
, such that
$p_{i,n}$
converge to
$f_i$
uniformly. Hence, for n large enough,
$$ \begin{align*}Z(p_{1,n})\cap Z(p_{2,n})\cap\partial\mathbb D^2=\emptyset.\end{align*} $$
Set the map
$P_n=(p_{1,n},p_{2,n})$
and
$F=(f_1, f_2)$
. Since both the Fredholm index and
$deg(0,F,\mathbb D^2)$
are topological invariants, combining with (1.1), we have
$$ \begin{align*}\mathrm{Ind}(T_{f_1}, T_{f_2})=-\deg(0, F, \mathbb D^2). \end{align*} $$
The paper is arranged as follows. In Section 2, we will characterize the Fredholmness of Toeplitz pairs with bounded holomorphic symbols. In Section 3, the case that the symbols are in
$L^\infty $
will be considered.
2 The Fredholm index of Toeplitz pairs with bounded holomorphic symbols
The present section is devoted to the Fredholmness and the Fredholm index of Toeplitz pairs on
$H^2(\mathbb D^2)$
with bounded holomorphic symbols. First, we recall the notation of the Koszul complex for a commuting pair. Let
$\left (T_{1}, T_{2}\right )$
be a commuting pair of operators on H; the Koszul complex associated with
$\left (T_{1}, T_{2}\right )$
was introduced in [Reference Taylor13],
$$ \begin{align*}0 \rightarrow H \stackrel{d_{1}}{\rightarrow} H \oplus H \stackrel{d_{2}}{\rightarrow} H \rightarrow 0, \end{align*} $$
where the boundary operators
$d_{1}, d_{2}$
are given by
$$ \begin{align*}d_{1}(\xi)=\left(-T_{2} \xi, T_{1} \xi\right), d_{2}\left(\xi_{1}, \xi_{2}\right)=T_{1} \xi_{1}+T_{2} \xi_{2}, \quad \text { for } \xi, \xi_{1}, \xi_{2} \in H. \end{align*} $$
Obviously,
$d_{2} d_{1}=0$
. The commuting pair
$\left (T_{1}, T_{2}\right )$
is called Fredholm [Reference Curto2] if
$$ \begin{align} \mathcal{H}_{0}=\operatorname{ker}\left(d_{1}\right), \mathcal{H}_{1}=\left.\operatorname{ker}\left(d_{2}\right)\right/\operatorname{Ran}\left(d_{1}\right) \text{ and } \mathcal{H}_{2}=\left.H\right/\operatorname{Ran}\left(d_{2}\right) \end{align} $$
are all of finite dimension, and in this case, the Fredholm index of T is defined by
$$ \begin{align*}\operatorname{Ind}T=-\operatorname{dim} \mathcal{H}_{0}+\operatorname{dim} \mathcal{H}_{1}-\operatorname{dim} \mathcal{H}_{2}.\end{align*} $$
The following easy lemma may be well-known before, and the proof is left as an exercise for the readers.
Lemma 2.1 Let
$T_{1}, T_{2},\cdots ,T_{n}$
be bounded operators on a Hilbert space H. Then,
${T_{1}H+\cdots +T_{n}H}$
is closed if and only if the operator
$T_{1}T_{1}^{*}+\cdots +T_{n}T_{n}^{*}$
has the closed range, and in this case,
$$ \begin{align*}\operatorname{Ran}\left(T_{1}T_{1}^{*}+\cdots+T_{n}T_{n}^{*}\right)=T_{1}H+\cdots+T_{n}H.\end{align*} $$
To continue, we need the following lemma; the proof is inspired by the proof of the
$L^p$
-solution of the corona problem [Reference Lin8, Reference Lin9].
Lemma 2.2 Suppose
$f_{1}, f_{2} \in H^{\infty }\left (\mathbb {D}^{2}\right )$
. Then,
$$ \begin{align*}\operatorname{dim} H^{2}\left(\mathbb{D}^{2}\right) \left/ (f_{1}H^{2}\left(\mathbb{D}^{2}\right)+f_2 H^2(\mathbb D^2))<\infty\right.\end{align*} $$
if and only if there is
$0<r<1$
and
$\delta>0$
such that
$|f_1|^2+|f_2|^2>\delta $
on
$\mathbb {U}_r^2.$
Proof
Necessity. Since
$f_{1}H^{2}\left (\mathbb {D}^{2}\right )+f_2 H^2(\mathbb D^2)$
is of finite codimension, by Lemma 2.1, we have
$$ \begin{align*}Ran\left(T_{f_1}T_{f_1}^*+T_{f_2}T_{f_2}^*\right)=f_{1}H^{2}\left(\mathbb{D}^{2}\right)+f_2 H^2(\mathbb D^2). \end{align*} $$
Then,
$T_{f_1}T_{f_1}^*+T_{f_2}T_{f_2}^*$
is a positive Fredholm operator. Hence, there exists a positive invertible operator X and a compact operator K such that
$$ \begin{align*}T_{f_1}T_{f_1}^*+T_{f_2}T_{f_2}^*=X+K.\end{align*} $$
For
$\lambda =(\lambda ^{(1)},\lambda ^{(2)})\in \mathbb {D}^2,$
let
$k_\lambda =\frac {\sqrt {1-|\lambda ^{(1)}|^2}\sqrt {1-|\lambda ^{(2)}|^2}}{(1-\overline {\lambda ^{(1)}}z_1)(1-\overline {\lambda ^{(2)}}z_2)}$
be the normalized reproducing kernel of
$H^2(\mathbb D^2)$
which converges to
$0$
weakly as
$\lambda \rightarrow \partial \mathbb {D}^{2}.$
It follows that there is positive constant
$c>0$
such that
$$ \begin{align*}\lim\limits _{\lambda\rightarrow \partial \mathbb{D}^{2}}\langle K k_{\lambda}, k_{\lambda}\rangle \rightarrow 0 , ~\text{and}~\liminf\limits _{\lambda\rightarrow \partial \mathbb{D}^{2}} \left\langle X k_{\lambda}, k_{\lambda}\right\rangle =c.\end{align*} $$
Therefore, for
$\lambda _0 \in \partial \mathbb {D}^{2},$
$$ \begin{align*}\begin{aligned} \liminf_{\lambda\rightarrow\lambda_0}|f_1(\lambda)|^2+|f_2(\lambda)|^2&= \liminf_{\lambda\rightarrow\lambda_0} ||T_{f_{1}}^{*} k_{\lambda}||^{2}+||T_{f_{2}}^{*} k_{\lambda}||^{2}\\&=\liminf_{\lambda\rightarrow\lambda_0}(\langle X k_{\lambda}, k_{\lambda}\rangle+\langle K k_{\lambda}, k_{\lambda}\rangle)\geq c. \end{aligned} \end{align*} $$
Thus, there is
$0<r<1$
and
$\delta>0$
such that
$|f_1(\lambda )|^2+|f_2(\lambda )|^2\geq \delta $
for
$\lambda \in \mathbb {U}_r^2.$
Sufficiency. Notice that
$ V=Z(f_1)\cap Z(f_2)$
is a compact zero subvariety of
$\mathbb D^2$
. Then, by [Reference Rudin11, Theorem 14.3.1], it is a finite set. Let
$\mathcal {O}$
be the sheaf of germs of analytic functions in
$\mathbb {D}^2$
and
$(f_1,f_2)\mathcal {O}$
be the ideal generated by
$\{f_1,f_2\}.$
Since the support of the analytic sheaf
$\mathcal {O} /\left (f_1, f_2\right ) \mathcal {O}$
coincides with the set
$V,$
the space
$\mathcal {O}(\mathbb D^2) /\left (f_1, f_2\right ) \mathcal {O}(\mathbb D^2)$
is finite-dimensional, where
$\mathcal {O}(\mathbb D^2)$
is the space of holomorphic functions over
$\mathbb D^2.$
Let
$$ \begin{align*}M=\{f \in H^{2}\left(\mathbb{D}^{2}\right): f=f_1 h_1+f_2h_2 ~\text{for some} ~h_1, h_2 \in \mathcal{O}(\mathbb D^2) \}. \end{align*} $$
To show that the subspace
$f_1 H^2(\mathbb D^2)+f_2H^2(\mathbb D^2)$
is closed and of finite codimension, it suffices to show that
$$ \begin{align} M\subset f_1 H^2(\mathbb D^2)+f_2 H^2(\mathbb D^2). \end{align} $$
In fact, (2.2) implies that
$$ \begin{align*}A: H^2(\mathbb D^2)\left/ \sum \limits_{i=1}^2 f_iH^2(\mathbb D^2)\right. \longrightarrow \mathcal{O}(\mathbb D^2) \left/ \left(f_1, f_2\right) \mathcal{O}(\mathbb D^2)\right.\end{align*} $$
is injective. Fix
$g \in M$
. Then,
$g=f_1 h_1+f_2h_2$
for some
$h_1, h_2 \in \mathcal {O}(\mathbb D^2).$
Let
$\phi :[0,1]\times [0,1] \rightarrow [0,1]$
be a smooth function with
$$ \begin{align*}\phi(t_1, t_2)=1 ~\text{for}~ (t_1,t_2)\in [0,r+\frac{1-r} {3}] \times [0,r+\frac{1-r} {3}],\end{align*} $$
and
$$ \begin{align*}\phi(t_1, t_2)=0 ~\text{for}~ t_1 \geq r+\frac{2(1-r)}{3}~ \text{or}~ t_2 \geq r+\frac{2(1-r)}{3}.\end{align*} $$
Set
$$ \begin{align*}\chi(z_1,z_2)=\phi(|z_1|,|z_2|),\end{align*} $$
and
$$ \begin{align} \varphi_{j}= \begin{cases} \frac{g \bar{f_{j}}}{|f_{1}|^{2}+|f_{2}|^{2}},& |f_{1}|^{2}+|f_{2}|^{2} \neq 0, \\ 0 ,& |f_{1}|^{2}+|f_{2}|^{2} = 0. \end{cases} \end{align} $$
It is easy to see that the functions
$l_j=\chi h_j+(1-\chi ) \varphi _j$
are smooth and satisfy the identity
$$ \begin{align} f_1 l_1+f_2 l_2=g. \end{align} $$
In general,
$l_{1}$
and
$l_{2}$
are not holomorphic. As in [Reference Lin8], we will use the technique on normal family of holomorphic functions. Set
$$ \begin{align*}G_i=\frac{1}{g}\left[l_{1} \frac{\partial l_{2}}{\partial \bar{z}_{i}}-l_{2} \frac{\partial l_{1}}{\partial \bar{z}_{i}}\right],\end{align*} $$
i = 1, 2. By straightforward calculations,
$G_i\in C^\infty ({\mathbb {D}^2}),i=1,2,$
$$ \begin{align*}\frac{\partial G_{1}}{\partial \bar{z}_{2}}=\frac{\partial G_{2}}{\partial \bar{z}_{1}} \end{align*} $$
in
$\mathbb {D}^2,$
and
$$ \begin{align*}G_1=\frac{1}{g}\left[\varphi_{1} \frac{\partial \varphi_{2}}{\partial \bar{z}_{1}}-\varphi_{2} \frac{\partial \varphi_{1}}{\partial \bar{z}_{1}}\right]=\frac{g\left(\bar{f_{1}} \frac{\partial \bar{f_{2}}}{\partial \bar{z}_{1}}-\bar{f_{2}} \frac{\partial \bar{f_{1}}}{\partial \bar{z}_{1}}\right)}{\left(| f_{1}|^{2}+|f_{2}|^{2}\right)^{2}}, \end{align*} $$
$$ \begin{align*}G_2=\frac{1}{g}\left[\varphi_{1} \frac{\partial \varphi_{2}}{\partial \bar{z}_{2}}-\varphi_{2} \frac{\partial \varphi_{1}}{\partial \bar{z}_{2}}\right]=\frac{g\left(\bar{f_{1}} \frac{\partial \bar{f_{2}}}{\partial \bar{z}_{2}}-\bar{f_{2}} \frac{\partial \bar{f_{1}}}{\partial \bar{z}_{2}}\right)}{\left(| f_{1}|^{2}+|f_{2}|^{2}\right)^{2}}\end{align*} $$
in
$\mathbb {U}_{r+\frac {2(1-r)}{3}}^2.$
Notice that
$$ \begin{align*}\frac{\partial G_{1}}{\partial \bar{z}_{2}}=\frac{\partial G_{2}}{\partial \bar{z}_{1}} \end{align*} $$
implies that the system
$$ \begin{align} \left\{ \begin{aligned} \frac{\partial b}{\partial \bar{z}_{1}}&=G_1, \\ \frac{\partial b}{\partial \bar{z}_{2}}&=G_2 \end{aligned} \right. \end{align} $$
is
$\bar {\partial }-$
closed. Then, by the proof in [Reference Lin8], the equations (2.5) admit a solution
$b\in L^2(\mathbb T^2).$
Put
$$ \begin{align} g_{1}=l_{1}+b f_{2}, \quad g_{2}=l_{2}-b f_{1}.\end{align} $$
Then,
$g_i\in H^2(\mathbb D^2)$
and
$$ \begin{align*}f_{1}g_{1}+f_{2}g_{2}=g.\end{align*} $$
It follows that
$M\subset f_1H^2(\mathbb D^2)+f_2H^2(\mathbb D^2)$
.
For
$\{f_1,\cdots ,f_m\}\subset H^\infty (\mathbb D^n)$
, using the same reasoning as the proof of Lemma 2.2, we have that
$$ \begin{align*}\operatorname{dim} H^{2}\left(\mathbb{D}^{n}\right) \left/ \sum\limits_{i=1}^m f_i H^2(\mathbb D^n)<\infty\right.\end{align*} $$
if and only if there is a constant
$\delta>0$
and
$0<r<1$
such that
$$ \begin{align*}\sum\limits_{i=1}^m |f_i(z)|^2\geq \delta \quad\text{for}\quad z\in \mathbb{U}_r^n.\end{align*} $$
For example, for
$m=2, n=3,$
we only prove the sufficiency. Similarly,
$ V=Z(f_1)\cap Z(f_2)$
is compact and hence is a finite set. Then, the space
$\mathcal {O}(\mathbb D^3) /\left (f_1, f_2\right ) \mathcal {O}(\mathbb D^3)$
is finite-dimensional. Let
$$ \begin{align*}M=\{f \in H^{2}\left(\mathbb{D}^{3}\right): f=f_1 h_1+f_2h_2 ~\text{for some} ~h_1, h_2 \in \mathcal{O}(\mathbb D^3) \}. \end{align*} $$
Fix
$g \in M$
. Then,
$g=f_1 h_1+f_2h_2$
for some
$h_1, h_2 \in \mathcal {O}(\mathbb D^3).$
Let
$\phi :[0,1]^3 \rightarrow [0,1]$
be a smooth function with
$$ \begin{align*}\phi(t_1, t_2,t_3)=1 ~\text{for}~ (t_1,t_2,t_3)\in [0,r+\frac{1-r} {3}] \times [0,r+\frac{1-r} {3}] \times [0,r+\frac{1-r} {3}],\end{align*} $$
and
$$ \begin{align*}\phi(t_1,t_2,t_3)=0 ~\text{for}~ t_1 \geq r+\frac{2(1-r)}{3}~ \text{or}~ t_2 \geq r+\frac{2(1-r)}{3}~\text{or}~ t_3 \geq r+\frac{2(1-r)}{3}.\end{align*} $$
Set
$$ \begin{align*}\chi(z_1,z_2,z_3)=\phi(|z_1|,|z_2|,|z_3|),\end{align*} $$
and
$$ \begin{align} \varphi_{j}= \begin{cases} \frac{g \bar{f_{j}}}{|f_{1}|^{2}+|f_{2}|^{2}},& |f_{1}|^{2}+|f_{2}|^{2} \neq 0, \\ 0 ,& |f_{1}|^{2}+|f_{2}|^{2} = 0. \end{cases} \end{align} $$
It is easy to see that the functions
$l_j=\chi h_j+(1-\chi ) \varphi _j$
are smooth and satisfy the identity
$$ \begin{align*}f_1 l_1+f_2 l_2=g.\end{align*} $$
Set
$$ \begin{align*}G_i=\frac{1}{g}\left[l_{1} \frac{\partial l_{2}}{\partial \bar{z}_{i}}-l_{2} \frac{\partial l_{1}}{\partial \bar{z}_{i}}\right],\end{align*} $$
i = 1, 2, 3. Considering the equation
$$ \begin{align} \left\{ \begin{aligned} \frac{\partial b}{\partial \bar{z}_{1}}&=G_1, \\ \frac{\partial b}{\partial \bar{z}_{2}}&=G_2,\\ \frac{\partial b}{\partial \bar{z}_{3}}&=G_3, \end{aligned} \right. \end{align} $$
by [Reference Lin9, Section 3], there exists a solution
$b\in L^2(\mathbb T^3).$
Put
$$ \begin{align} g_{1}=l_{1}+b f_{2}, \quad g_{2}=l_{2}-b f_{1}.\end{align} $$
Then,
$g_i\in H^2(\mathbb D^3)$
and
$$ \begin{align*}f_{1}g_{1}+f_{2}g_{2}=g.\end{align*} $$
It follows that
$M\subset f_1H^2(\mathbb D^3)+f_2H^2(\mathbb D^3)$
.
Now, we can prove the main result Theorem 1.1.
The proof of Theorem 1.1
The necessity is an easy application of Lemma 2.2. In fact, suppose that
$\left (T_{f_{1}}, T_{f_{2}}\right )$
is Fredholm. By the definition, the subspace
$f_1 H^2(\mathbb D^2)+f_2 H^2(\mathbb D^2)$
is closed and of finite codimension. By Lemma 2.2, there is
$0<r<1$
and
$c>0$
such that
$$ \begin{align*}\left|f_{1}(z)\right|{}^{2}+\left|f_{2}(z)\right|{}^{2} \geqslant c,\,\text{ for } z\in \mathbb{U}_r^2.\end{align*} $$
For the other direction, let
$\mathcal {H}_{0}, \mathcal {H}_{1}, \mathcal {H}_{2}$
be defined as in (2.1) for
$\left (T_{f_{1}}, T_{f_{2}}\right )$
, and suppose there is
$0<r<1$
and
$c>0$
such that
$\left |f_{1}(z)\right |{}^{2}+\left |f_{2}(z)\right |{}^{2} \geqslant c$
on
$\mathbb {U}_r^2$
. Then, by Lemma 2.2, the subspace
$f_{1} H^{2}\left ( \mathbb {D}^{2}\right )+f_{2} H^{2}\left ( \mathbb {D}^{2}\right )$
is closed and
$$ \begin{align*}\dim{\cal H}_2=\operatorname{dim} H^{2}\left( \mathbb{D}^{2}\right) / (f_{1} H^{2}\left( \mathbb{D}^{2}\right)+f_{2} H^{2}\left( \mathbb{D}^{2}\right))<\infty.\end{align*} $$
Next, for any
$\zeta \in H^2(\mathbb D^2)$
,
$$ \begin{align*} \|d_1\zeta\|^2&=\left\|-T_{f_{2}} \zeta\right\|^{2}+\left\|T_{f_{1}} \zeta\right\|^{2}\\ &=\sup _{0 \leq r <1} \int_{\mathbb{T}^{2}}\left|f_{2}\zeta (r \xi)\right|{}^{2} d m+\sup _{0 \leq r <1} \int_{\mathbb{T}^{2}}\left|f_{1}\zeta (r \xi)\right|{}^{2} d m\\ &\geq \sup _{0 \leq r<1} \int_{\mathbb{T}^{2}}(\left|f_{2}(r \xi)\right|{}^{2}+\left|f_{1}(r \xi)\right|{}^{2})|\zeta (r \xi)|^{2} d m \\ & \geqslant c \sup _{0 \leq r <1} \int_{\mathbb{T}^{2}}\left|\zeta (r \xi)\right|{}^{2} d m =c\|\zeta\|^{2}, \end{align*} $$
which implies that the boundary operator
$d_{1}$
is injective and has closed range, and hence,
${\cal H}_0$
is trivial. Moreover, it is easy to see that
$(\zeta _1,\zeta _2)\in \mathcal {H}_{1}$
if and only if
$(\zeta _1,\zeta _2)$
solves the following equations:
$$ \begin{align*}\left\{\begin{array}{@{}c@{}l} & T_{f_{1}} \zeta_{1}+T_{f_{2}} \zeta_{2}=0 , \\ &-T_{f_{2}}^{*} \zeta_{1}+T_{f_{1}}^{*} \zeta_{2}=0. \end{array}\right. \end{align*} $$
Since
$T_{f_{1}} \zeta _{1}+T_{f_{2}} \zeta _{2}=0$
, then
$\frac {-\zeta _{2}}{f_{1}}=\frac {\zeta _{1}}{f_{2}}$
on
$\mathbb {D}^{2} \,{\backslash}\, \left ( Z\left (f_{1}\right ) \cup Z\left (f_{2}\right )\right )$
. Set
$$ \begin{align*}\phi=\frac{-\zeta_{2}}{f_{1}}=\frac{\zeta_{1}}{f_{2}},\end{align*} $$
which
$\phi $
can be holomorphically extended to
$\mathbb {D}^{2} \,{\backslash}\, \left ( Z\left (f_{1}\right ) \bigcap Z\left (f_{2}\right )\right )$
naturally. Notice that
$Z\left (f_{1}\right ) \cap Z\left (f_{2}\right ) \cap \mathbb {D}^{2}$
is a compact holomorphic subvariety of
$\mathbb D^2$
, and hence,
$Z\left (f_{1}\right ) \cap Z\left (f_{2}\right ) \cap \mathbb {D}^{2}$
is a finite set. By Hartogs’ Theorem,
$\phi $
can be holomorphically extended to
$\mathbb {D}^{2}$
. Furthermore, it is easy to see that there exists
$0<s<1$
and
$\varepsilon>0$
such that for any
$1>r>s$
,
$$ \begin{align*}\left|f_{1}(r \xi)\right|{}^{2}+\left|f_{2}(r \xi)\right|{}^{2}>\varepsilon, \quad \text{for all } \xi\in\mathbb T^2. \end{align*} $$
Therefore, for all
$\xi \in \mathbb {T}^{2},$
$$ \begin{align*}|\phi(r \xi)|^{2}=\frac{\left|-\zeta_{2}(r \xi)\right|{}^{2}}{\left|f_{1}(r \xi)\right|{}^{2}}=\frac{\left|\zeta_{1}(r \xi)\right|{}^{2}}{\left|f_{2}(r \xi)\right|{}^{2}} =\frac{\left|\zeta_{2}(r \xi)\right|{}^{2}+\left|\zeta_{1}(r \xi)\right|{}^{2}}{\left|f_{1}(r \xi)\right|{}^{2}+\left|f_{2}(r \xi)\right|{}^{2}}<\frac{\left|\zeta_{2}(r \xi)\right|{}^{2}+\left|\zeta_{1}(r \xi)\right|{}^{2}}{\varepsilon}. \end{align*} $$
It follows that
$\phi \in H^{2}\left (\mathbb {D}^{2}\right )$
, and hence,
$$ \begin{align*}\zeta_{2}=-f_{1} \phi, ~\zeta_{1}=f_{2} \phi. \end{align*} $$
Combining with that
$-T_{f_{2}}^{*} \zeta _{1}+T_{f_{1}}^{*} \zeta _{2}=0$
, we have
$ (T_{f_{1}}^{*} T_{f_{1}}+T_{f_{2}}^{*} T_{f_{2}}) \phi =0; $
thus,
$$ \begin{align*}\|\zeta_1\|^2+\|\zeta_2\|^2=\|f_1\phi\|^2+\|f_2\phi\|^2=\left\langle(T_{f_{1}}^{*} T_{f_{1}}+T_{f_{2}}^{*} T_{f_{2}}) \phi, \phi\right\rangle=0,\end{align*} $$
and hence,
$(\zeta _1,\zeta _2)=0$
. This implies that
$\mathcal {H}_{1}=0$
. Thus,
$\left (T_{f_{1}}, T_{f_{2}}\right )$
is Fredholm. It follows that
$$ \begin{align*}\operatorname{Ind}\left(T_{f_{1}}, T_{f_{2}}\right)=-\operatorname{dim} \mathcal{H}_{2}=-\operatorname{codim}\left(f_{1} H^{2}\left(\mathbb{D}^{2}\right)+f_{2} H^{2}\left(\mathbb{D}^{2}\right)\right). \end{align*} $$
As an easy application, we have the following corollary.
Corollary 2.3 Let
$f_{i}\in H^{\infty }\left (\mathbb {D}^{2}\right )$
,
$i=1,2$
, and
$F=(f_1,f_2):\mathbb D^2 \to \mathbb C^2$
be the corresponding map. Then, the essential spectrum
$$ \begin{align*}\sigma_{e}\left(T_{f_{1}}, T_{f_{2}}\right)=\bigcap \limits_{0<r<1}\overline{F\left(\mathbb{U}_{r}^2\right)}.\end{align*} $$
Proof It follows from Theorem 1.1 and the definition of
$\sigma _{e}$
that
$(\lambda _1,\lambda _2)\not \in \sigma _{e}(T_{f_1},T_{f_2})$
if and only if there is a
$\delta>0$
and
$0<r<1$
such that
$$ \begin{align*}|\lambda_1-f_1(z)|+|\lambda_2-f_2(z)|>\delta,\quad\text{for}\quad z \in \mathbb{U}_{r}^2, \end{align*} $$
which is equivalent to saying that
$(\lambda _1,\lambda _2)\not \in \overline {F(\mathbb {U}_{r}^2)}$
for some
$0<r<1$
.
3 Fredholmness of Toeplitz tuples with symbols in
$L^\infty $
In the present section, we will consider the Toeplitz tuple
$(T_{f_1},T_{f_2},\cdots ,T_{f_n})$
with symbols in
$L^\infty .$
It is easy to see that, in general,
$(T_{f_1},T_{f_2},\cdots ,T_{f_n})$
is not essentially commuting. To get the commutativity of the Toeplitz tuple, let
$L_{z_i}^\infty (\mathbb T)$
be the space of essentially bounded functions over
$\mathbb {T}$
on the variable
$z_i,$
and the symbols
${f_i\in L_{z_i}^\infty (\mathbb T).}$
To avoid the confusion, denote by
$T_{f_i}^{(i)}(resp. ~T_{f_i})$
the Toeplitz operator on
$H^2_{z_i}(\mathbb T)(resp. ~H^2(\mathbb D^n)).$
We have the following Proposition.
Proposition 3.1 For
$f_i\in L_{z_i}^\infty (\mathbb T),$
assume that all
$T_{f_i}^{(i)}$
are not invertible. Then, the Toeplitz tuple
$(T_{f_1},T_{f_2},\cdots ,T_{f_n})$
on
$H^2(\mathbb D^n)$
is Fredholm if and only if all
$T_{f_i^{(i)}}$
are Fredholm, and in this case,
$$ \begin{align*}\operatorname{Ind}(T_{f_1},T_{f_2},\cdots,T_{f_n}) =(-)^{n+1}\prod\limits_{i=1}^n\operatorname{Ind} T^{(i)}_{f_i}. \end{align*} $$
The proof of Proposition 3.1 comes from the following lemma easily, which is a generalization of [Reference Curto2, Proposition 15.4]. The sufficient part was proved by [Reference Kaad and Nest7, Lemma 7.3], by using cohomology, and we will give an elementary proof.
Lemma 3.2 Let
$\{H_i\}_{i=1}^n$
be a set of Hilbert spaces of infinite dimension and set
$$ \begin{align*}H=H_1\otimes H_2\otimes\cdots\otimes H_n.\end{align*} $$
Suppose all
$T_i\in L(H_i)$
are not invertible. Set
$$ \begin{align*}{\widetilde{T_i}}=I_{H_1}\otimes \cdots \otimes I_{H_{i-1}}\otimes T_{i}\otimes I_{H_{i+1}}\otimes\cdots\otimes I_{H_{n}}, \qquad i\leq n.\end{align*} $$
Then, the tuple
$\mathcal {T}=({\widetilde {T}_1},\cdots ,{\widetilde {T}_n})$
is a Fredholm tuple on H if and only if all
$T_i$
are Fredholm, and in this case,
$$ \begin{align*}\operatorname{Ind}\mathcal{T}=(-1)^{n+1}\prod\limits_{i=1}^n\operatorname{Ind}{T}_{i}.\end{align*} $$
Proof At first, suppose that
$\mathcal {T}$
is Fredholm. Notice that all
$T_j$
are not invertible. Then,
$$ \begin{align*}H_j/\operatorname{ran} T_j\not=0,\quad\text{or}\quad \operatorname{ker} T_j\not=0. \end{align*} $$
Let
$\Lambda _1=\{i, \ker T_i\not =0\}$
and
$\Lambda _2=\{j,\ker T_j=0\},$
and set
$S_i=T_{i}^*$
for
$i\in \Lambda _1$
and
${S_i=T_i}$
for
$i\in \Lambda _2$
. Since
$\mathcal {T}$
is doubly commutative, that is,
$$ \begin{align*}[\widetilde{T}_i,\widetilde{T}_j]=[\widetilde{T}_i,\widetilde{T}_j^*]=0, \quad 1\leq i,j\leq n. \end{align*} $$
By [Reference Curto2, Corollary 3.7], the tuple
$(\widetilde {S}_1,\cdots ,\widetilde {S}_n)$
is Fredholm, where
$\widetilde {S}_i$
is defined as same as
$\widetilde {T}_i$
for every i. If follows that as a linear space,
$$ \begin{align*}H\left/\sum\limits_{j=1}^n\widetilde{S}_i H\right.=\left(H_1/\operatorname{ran} S_1\right)\otimes\cdots \otimes \left(H_n/\operatorname{ran} S_n\right), \end{align*} $$
which is of finite dimension. Since all
$H_i/\operatorname {ran} S_i$
are not trivial,
$$ \begin{align*}\dim H_i/\operatorname{ran} S_i<\infty. \end{align*} $$
It follows that all
$\operatorname {ran} S_i$
are closed. Next, we will show that all
$\ker S_j$
are of finite dimension. It suffices to show that
$\dim \ker S_j<\infty $
for
$j\in \Lambda _1$
. For
$i\in \Lambda _2$
, since
$S_i$
is not invertible, we have that
$\ker S_i^*\not =0$
, and hence, for any
$1\leq i\leq n$
,
$\ker S_i^*\not =0$
. For
$i\in \Lambda _1$
, by [Reference Curto2, Proposition 3.7] again, the tuple
$(\widetilde {S}_1^*,\cdots , \widetilde {S}_{i-1}^*, \widetilde {S}_{i},\widetilde {S}_{i+1}^*,\cdots ,\widetilde {S}_n^*)$
is Fredholm. It can be verified that
$$ \begin{align*}\left(\bigcap\limits_{j\not=i} \ker \widetilde{S}^*_{j}\right)\cap \ker \widetilde{S}_{i}=\ker S_1^*\otimes\cdots\otimes\ker S_{i-1}^*\otimes \ker S_{i}\otimes \ker S_{i+1}^*\otimes\cdots\otimes \ker S_n^* \end{align*} $$
is of finite dimension. It follows that all
$\ker S_{i}$
are of finite dimension. Therefore, all
$S_i$
are Fredholm, and hence, all
$T_i$
are also Fredholm.
Next, assume that all
$T_i$
are Fredholm. We will prove that the tuple
$\mathcal {T}$
is Fredholm. By [Reference Curto2, Corollary 3.7], the tuple
$({\widetilde {T}_1},\cdots ,{\widetilde {T}_n})$
is Fredholm if and only if
$\sum \limits _{i=1}^n {}^f {\widetilde {T_i}}$
is Fredholm for every function
$f:\{1, \cdots , n\} \rightarrow \{0,1\},$
where
$$ \begin{align*} {}^f {\widetilde{T_i}}= \begin{cases}{\widetilde{T_i}}^* {\widetilde{T_i}}, & f(i)=0, \\ {\widetilde{T_i}} {\widetilde{T_i}}^*, & f(i)=1.\end{cases} \end{align*} $$
Now, we will show that
$\widetilde {T}_1\widetilde {T}_1^*+\cdots +\widetilde {T}_n\widetilde {T}_n^*$
is Fredholm. In fact,
$$ \begin{align*}\widetilde{T}_1H+\cdots+\widetilde{T}_n H&=T_1H_1\otimes H_2\otimes\cdots\otimes H_n\\&\quad+H_1\otimes T_2 H_2\otimes H_3\otimes\cdots\otimes H_n+\cdots+ H_1\otimes\cdots\otimes H_{n-1}\otimes T_n H_n. \end{align*} $$
Since
$T_iH$
is closed,
$\widetilde {T}_1H+\cdots +\widetilde {T}_n H$
is closed. By Lemma 2.1,
$$ \begin{align*}\operatorname{ran}\left(\sum\limits_{i=1}^n \widetilde{T}_i\widetilde{T}_i^*\right)=\widetilde{T}_1H+\cdots+\widetilde{T}_n H\end{align*} $$
is closed. Moreover,
$$ \begin{align*}\begin{aligned} \ker\left( \sum\limits_{i=1}^n{\widetilde{T}}_i{\widetilde{T}_i^*}\right)&=\bigcap\limits_{i=1}^n\ker \widetilde{T}_i^*=\ker T_1^* \otimes\cdots \otimes \ker T_n^* \end{aligned} \end{align*} $$
is of finite dimension. This implies that
$\sum \limits _{i=1}^n{\widetilde {T}}_i{\widetilde {T}_i^*}$
is Fredholm. The same discussions as above show that for every function
$f:\{1, \cdots , n\} \rightarrow \{0,1\}$
,
$\sum \limits _{i=1}^n {}^f {\widetilde {T_i}}$
is Fredholm.
By [Reference Kaad and Nest7, Lemma 7.3], if
$\mathcal {T}=({\widetilde {T}_1},\cdots ,{\widetilde {T}_n})$
is Fredholm, then
$$ \begin{align*}\operatorname{Ind}\mathcal{T}=(-1)^{n+1}\prod\limits_{i=1}^n\operatorname{Ind}{T}_{i}.\\[-42pt]\end{align*} $$
For the remainder of the section, we will consider the Toeplitz tuple
$(T_{f_1},\cdots ,T_{f_n})$
on
$H^2(\mathbb {D}).$
By Andersson and Sandberg [Reference Andersson and Sandberg1], for
$\{f_{i}\}_{i=1}^n\subset H^\infty (\mathbb D)$
,
$\left ( T_{f_{1}},T_{f_{2}},\cdots ,T_{f_{n}}\right ),$
acting on
$H^{2}\left (\mathbb {D}\right ),$
is Fredholm if and only if there exists
$0<s<1$
and
$\delta>0$
such that
$$ \begin{align*}\left|f_{1}(z)\right|{}^{2}+\left|f_{2}(z)\right|{}^{2}+\cdots+\left|f_{n}(z)\right|{}^{2}>\delta,\quad \text{for}\ |z|>s.\end{align*} $$
Proposition 3.3 If
$\left (T_{f_{1}},T_{f_{2}},\cdots ,T_{f_{n}}\right )$
is Fredholm, then
$$ \begin{align*}\operatorname{Ind}(T_{f_{1}},\cdots,T_{f_{n}})=0.\end{align*} $$
Before giving the proof of Proposition 3.3, we should point out that in Yang [Reference Yang14], it was proved that if
$[T_{\phi }^{*},T_{\psi }]$
is compact, then
$\operatorname {Ind}\left (T_{\phi },T_{\psi }\right )=0.$
The proof of Proposition 3.3
Since
$\left (T_{f_{1}},T_{f_{2}},\cdots ,T_{f_{n}}\right )$
is Fredholm, there are
$\delta>0$
and
$0<s<1$
such that
$$ \begin{align*}|f_1(z)|^2+|f_2(z)|^2+\cdots+|f_n(z)|^2\geq \delta, \quad \text{for}\ |z|>s. \end{align*} $$
It follows that there is a finite Blaschke product B such that
$$ \begin{align*}f_i=B \tilde{f_{i}}\quad{\text{and}}\quad\sum\limits_{i=1}^n|\tilde{f_{i}}|^2\geq \frac{\delta}{|B|^2}\geq \delta \ \text{on}\ \mathbb D \,{\backslash}\, Z(B).\end{align*} $$
Then,
$$ \begin{align*}\sum\limits_{i=1}^n|\tilde{f_{i}}|^2\geq \frac{\delta}{|B|^2}\geq \delta \ \text{on}\ \mathbb D.\end{align*} $$
By Corona Theorem, we have
$(T_{\tilde {f_{1}}},\cdots ,T_{\tilde {f_{n}}})$
is invertible, and hence,
$$ \begin{align*}\operatorname{Ind}(T_{\tilde{f_{1}}},\cdots,T_{\tilde{f_{n}}})=0.\end{align*} $$
Moreover, since
$T_B$
is Fredholm, by [Reference Curto2, Proposition 11.1] for any bounded analytic function
$\varphi _i$
, the tuple of Toeplitz operators
$(T_{\varphi _1},\cdots , T_{\varphi _i},T_{B}, T_{\varphi _{i+1}},\cdots , T_{\varphi _n})$
is Fredholm and
$$ \begin{align*}\operatorname{Ind}(T_{\varphi_1},\cdots, T_{\varphi_i},T_{B}, T_{\varphi_{i+1}},\cdots, T_{\varphi_n})=0.\end{align*} $$
By [Reference Fang4, Proposition 1],
$$ \begin{align*}\begin{aligned} \operatorname{Ind}(T_{f_1},\cdots,T_{f_n})&=\operatorname{Ind}(T_B,T_{f_2},\cdots,T_{f_n}) +\operatorname{Ind}(T_{\tilde{f_1}},T_{f_2},\cdots,T_{f_n})\\&= \operatorname{Ind}(T_{\tilde{f_1}},T_{f_2},\cdots,T_{f_n}). \end{aligned} \end{align*} $$
By induction,
$$ \begin{align*}\operatorname{Ind}(T_{{f_1}},T_{f_2},\cdots,T_{f_n})=\operatorname{Ind} (T_{\tilde{f_1}},T_{\tilde{f}_2},\cdots,T_{\tilde{f}_n})=0.\end{align*} $$
Acknowledgements
We would like to thank K. Guo (Fudan University) for valuable discussions on this topic. The authors thank the referee for helpful suggestions, which make this paper more readable.
