October 16, 2025
In this paper we study non-standard holomorphic structures on line bundles over the quantum projective line \({\mathbb{C}}P^1_q\). We show that there exist infinitely many non-gauge equivalent holomorphic structures on those line bundles. This gives a negative answer to a question raised by Khalkhali, Landi, and Van Suijlekom in 2011.
Over the past three decades, noncommutative differential geometry has witnessed substantial progress [1]. By contrast, the corresponding theory of noncommutative complex geometry is still at a relatively early stage of development. An important step in this direction was made by Khalkhali, Landi, and Van Suijlekom in [2], where the authors introduced, for a deformation parameter \(0<q<1\), the quantum projective line \({\mathbb{C}}P^1_q\). This space provides a rich and instructive example of a noncommutative complex manifold. They further demonstrated that many of the classical features of the complex projective line \({\mathbb{C}}P^1\) continue to hold in the quantum setting. In particular, for each \(n\in\mathbb{Z}\) they constructed holomorphic line bundles \(\mathcal{L}_n\) on \({\mathbb{C}}P^1_q\), which may be regarded as noncommutative analogs of the classical line bundles \(\mathcal{O}(n)\) on \({\mathbb{C}}P^1\).
A fundamental property of the classical line bundle \(\mathcal{O}(n)\) over \({\mathbb{C}}P^1\) is that its holomorphic structure is unique up to gauge equivalence. More concretely, a holomorphic structure on a complex vector bundle \(E\) over a complex manifold \(X\) is given by a flat \(\bar{\partial}\)-connection \[\overline{\nabla}: E \longrightarrow \Omega^{0,1}(X)\otimes E,\] and two such structures \(\overline{\nabla}_1\) and \(\overline{\nabla}_2\) are said to be gauge equivalent if there exists an invertible bundle map \(g\in \operatorname{Aut}(E)\) such that \[g\circ \overline{\nabla}_1\circ g^{-1} = \overline{\nabla}_2.\] It is a classical fact that any holomorphic structure on \(\mathcal{O}(n)\) is gauge equivalent to the standard one. Motivated by this, the authors of [2] asked whether the same statement remains true for the quantum line bundles \(\mathcal{L}_n\) on \({\mathbb{C}}P^1_q\).
The purpose of this paper is to provide a negative answer to this question.
Theorem 1 (See Theorem 4 below). For \(0<q<1\), each quantum line bundle \(\mathcal{L}_n\) over \({\mathbb{C}}P^1_q\) admits a flat \(\bar{\partial}\)-connection \(\overline{\nabla}\) which is not gauge equivalent to the standard \(\bar{\partial}\)-connection.
We refer to such \(\bar{\partial}\)-connections as non-standard holomorphic structures. Our work further investigates the nature of these non-standard structures and their mutual gauge equivalence. A key observation is that the dimension of spaces of holomorphic sections of non-standard holomorphic structures can grow indefinitely.
Proposition 1 (See Corollary 6 below). For \(0<q<1\), on \(\mathcal{L}_0\) over \({\mathbb{C}}P^1_q\), and for every \(N\in\mathbb{N}\), there exists a \(\bar{\partial}\)-connection \(\overline{\nabla}\) such that \[\dim \ker \overline{\nabla} \geq N.\]
The above theorem, combined with the general fact that \(\dim \ker \overline{\nabla}\) is finite, implies the following theorem, which is the main result of this paper:
Theorem 2 (See Theorem 5 below). For \(0<q<1\), the line bundle \(\mathcal{L}_n\) over \({\mathbb{C}}P^1_q\) carries infinitely many holomorphic structures, no two of which are gauge equivalent.
Organization of the paper. In Section 2, we review the construction of the quantum projective line and recall the definition of holomorphic structures in this setting. Section 3 focuses on a distinguished sub-\(C^*\)-algebra of the quantum projective line, where we explicitly construct non-standard holomorphic structures and analyze their gauge equivalence classes. In Section 4, we study the holomorphic sections associated with these structures and establish the existence of classes with arbitrarily large finite dimension. Finally in Section 5 we briefly discuss some possible future work.
The authors would like to thank Jonathan Block for his encouragement and valuable suggestions. They are also grateful to David Gao and Réamonn Ó Buachalla for kindly answering questions related to this work, and to Ryszard Szwarc for directing them to the reference [3].
This work was carried out during the TADM-REU program in Summer 2025, supported by the National Science Foundation under Grant No DMS-2243991. The authors would like to thank East Texas A&M University for providing an excellent research environment throughout the program. They also want to thank Padmapani Seneviratne and Mehmet Celik for their organization and support during the program.
Z.W.’s research is partially supported by the AMS-Simons Research Enhancement Grants for Primarily Undergraduate Institution (PUI) Faculty.
In this section we mainly follow [2].
The algebra \(\mathcal{A}(SU_q(2))\) is the unital Hopf \(*\)-algebra with defining matrix \[\begin{bmatrix}a & -qc^* \\ c & a^*\end{bmatrix}\] that is, the unital Hopf \(*\)-algebra defined by the relations \[\label{eq:32relations32in32A40SUq4024141} \begin{align} &ac=qca,~ac^*=qc^*a,~cc^*=c^*c,\\ & a^*a+c^*c=aa^*+q^2cc^*=1. \end{align}\tag{1}\]
It can be shown that \(\mathcal{A}(SU_q(2))\) is a compact quantum group.
The algebra \(\mathcal{A}(S_q^2)\to B(\ell^2)\) is a \(*\)-subalgebra of \(\mathcal{A}(SU_q(2))\) which is given by the generators \[\label{eq:32generators32of32A40CP141} B_- =ac^*,~ B_+=ca^*,~ B_0=cc^*,\tag{2}\] one can calculate that these generators must obey the basic relations
\[\begin{align} &B_-B_0=q^2B_0B_{-},~ B_{+}B_0=q^{-2}B_0B_{+},\\ &B_-B_+=q^2B_0(1-q^2B_0),~B_+B_-=B_0(1-B_0)\\ &B_0^*=B_0,~ B_+^*=B_-. \end{align}\]
Remark 2. In the classical case where \(q = 1,\) and we consider \(S^2\) as the standard sphere, we see that \(B_-, B_+,\) and \(B_0\) correspond to \(\frac{x-iy}{2}, \frac{x+iy}{2},\) and \(\frac{1-z}{2}\) in \(\mathcal{A}(S^2).\)
Line bundles \(\mathcal{L}_n\) can be defined on \(\mathcal{A}(S_q^2)\) as the \(\mathcal{A}(S_q^2)\)-sub-bimodules of \(\mathcal{A}(SU_q(2)).\) These are generated by
\[\begin{align} &\{(c^*)^m(a^*)^{n-m},~m=0,\ldots n\} \text{ for }n\geq 0;\\ &\{c^ma^{-n-m},~m=0,\ldots -n\} \text{ for }n\leq 0 \end{align}\] Note that \(\mathcal{L}_n\)’s are projective \(\mathcal{A}(S_q^2)\)-modules.
One can observe that \[\label{eq:32line32bundle32product} \mathcal{L}_0=\mathcal{A}(S_q^2),~\mathcal{L}_n^*=\mathcal{L}_{-n},~ \mathcal{L}_m\otimes_{\mathcal{A}(S_q^2)}\mathcal{L}_n\cong \mathcal{L}_{m+n}.\tag{3}\] If we denote by \({\mathrm {End}}_{\mathcal{A}(S_q^2)}(\mathcal{L}_n)\) the ring endomoprhisms of \(\mathcal{L}_n\) as left \(\mathcal{A}(S_q^2)\)-modules, then we have \[\label{eq:32end32of32line32bundles} {\mathrm {End}}_{\mathcal{A}(S_q^2)}(\mathcal{L}_n)=\mathcal{A}(S_q^2),\tag{4}\] where the right hand side means right multiplication by \(\mathcal{A}(S_q^2)\).
In [2] the authors introduced the \(1\)-form \[\omega_-:=c^* da^*-qa^*dc^*\] which satisfies ( [2]) \[\label{eq:32omega-32commuting} \begin{align} &\omega_-a=q^{-1}a\omega_-,~\omega_-c=q^{-1}c\omega_-,\\ &\omega_-a^*=qa^*\omega_-,~ \omega_-c^*=qc^*\omega_- \end{align}\tag{5}\] It is clear from 5 that \(\omega_-\) commutes with elements in \(\mathcal{A}(S_q^2)\).
Let \(\Omega^{0,1}({\mathbb{C}}P^1_q)\) be \(\mathcal{L}_{-2}\omega_-\) as an \(\mathcal{A}(S_q^2)\)-bimodule. We can define the \(\bar{\partial}\) operator on \(\mathcal{A}(S_q^2)\) as the map generated by the actions:
\[\label{eq:32pb32on32generators} \overline{\partial}(B_0)=-q^{-\frac{1}{2}}ca\omega_-,~\overline{\partial}(B_+)=q^{\frac{1}{2}}c^2\omega_-,~\overline{\partial}(B_-)=-q^{-\frac{1}{2}}a^2\omega_-.\tag{6}\] This makes \((\mathcal{A}(S_q^2), \overline{\partial})\) an algebra with complex structure in the sense of [2]. From now on we denote \(\mathcal{A}(S_q^2)\) by \(\mathcal{A}({\mathbb{C}}P^1_q)\).
Lemma 1. We have the following relations \[\label{eq:32B032pb32B-32B4332commutes} \begin{align} &(\bar{\partial}B_0 )B_0=q^2 B_0 \bar{\partial}B_0, ~ (\bar{\partial}B_- )B_-=q^2 B_- \bar{\partial}B_-, ~(\bar{\partial}B_+ )B_+=q^2 B_+ \bar{\partial}B_+,\\ & (\bar{\partial}B_0 )B_-=B_-\bar{\partial}B_0=q^2 B_0 \bar{\partial}B_-,~(\bar{\partial}B_0) B_+ = -\bar{\partial}B_+ + q^2 B_0 \bar{\partial}B_+. \end{align}\qquad{(1)}\]
Lemma 2. [[2] Lemma 3.6] For any integer n, there is a twisted flip isomorphism \[\label{eq:32twist32flip32isomorphism} \Phi_{(n)}: \mathcal{L}_n\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)} \Omega^{0,1}({\mathbb{C}}P^1_q)\overset{\sim}{\to}\Omega^{0,1}({\mathbb{C}}P^1_q) \otimes_{\mathcal{A}({\mathbb{C}}P^1_q)} \mathcal{L}_n\qquad{(2)}\] as \(\mathcal{A}({\mathbb{C}}P^1_q)\)-bimodules.
Lemma 3. Under the isomorphism \(\mathcal{L}_m\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)}\mathcal{L}_n\cong \mathcal{L}_{m+n}\) in 3 , we have the identity \[\label{eq:32associatitity32of32Phi} \Phi_{(m+n)} =(\Phi_{(m)}\otimes {\text{id}})\circ({\text{id}}\otimes \Phi_{(n)})\qquad{(3)}\] as isomorphisms from \(\mathcal{L}_m\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)}\mathcal{L}_n\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)} \Omega^{0,1}({\mathbb{C}}P^1_q)\) to \(\Omega^{0,1}({\mathbb{C}}P^1_q) \otimes_{\mathcal{A}({\mathbb{C}}P^1_q)} \mathcal{L}_m\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)}\mathcal{L}_n\).
Proof. Since \(\Omega^{0,1}({\mathbb{C}}P^1_q)= \mathcal{L}_{-2}\omega_-\), ?? follows from 5 and the associativity of tensor products. ◻
Definition 3. Let \(\mathcal{E}\) be a left \(\mathcal{A}({\mathbb{C}}P^1_q)\)-module. A left \(\bar{\partial}\)-connection on \(\mathcal{E}\) is a linear map \(\overline{\nabla}: \mathcal{E}\to \Omega^{0,1}({\mathbb{C}}P^1_q)\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)}\mathcal{E}\) satisfying the left Leibniz rule \[\overline{\nabla}(fe)=\bar{\partial}(f)\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)}e+f\overline{\nabla}(e) \text{ for any } f\in \mathcal{A}({\mathbb{C}}P^1_q), ~e\in \mathcal{E}.\]
We can define right \(\bar{\partial}\)-connections on right \(\mathcal{A}({\mathbb{C}}P^1_q)\)-modules in the same way.
We will need following definition in later construction:
Definition 4. [[2] Definition 2.11] Let \(\mathcal{E}\) be an \(\mathcal{A}({\mathbb{C}}P^1_q)\)-bimodule. A left \(\bar{\partial}\)-connection \(\overline{\nabla}\) on \(\mathcal{E}\) is called a bimodule \(\bar{\partial}\)-connection if there exists an \(\mathcal{A}({\mathbb{C}}P^1_q)\)-bimodule isomorphism \[\sigma(\overline{\nabla}): \mathcal{E}\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)}\Omega^{0,1}({\mathbb{C}}P^1_q)\overset{\sim}{\to}\Omega^{0,1}({\mathbb{C}}P^1_q)\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)}\mathcal{E}\] such that for any \(f\in \mathcal{A}({\mathbb{C}}P^1_q)\) and \(s\in\mathcal{E}\), the following twisted right Leibniz rule holds \[\overline{\nabla}(sf)=\overline{\nabla}(s)f+\sigma(\overline{\nabla})(s\bar{\partial}(f)).\]
Definition 5. Let \(\mathcal{E}\) be a left \(\mathcal{A}({\mathbb{C}}P^1_q)\)-module. A left holomorphic structure on \(\mathcal{E}\) is a flat left \(\bar{\partial}\)-connection on \(\mathcal{E}\), i.e. a left \(\bar{\partial}\)-connection \(\overline{\nabla}\) on \(\mathcal{E}\) such that \(\overline{\nabla}\circ \overline{\nabla}=0\).
We can define right holomorphic structures on right \(\mathcal{A}({\mathbb{C}}P^1_q)\)-modules in the same way.
Remark 6. For \(\mathcal{A}({\mathbb{C}}P^1_q)\)-modules, the condition \(\overline{\nabla}\circ \overline{\nabla}=0\) is automatically satisfied by dimension reason.
We can define the standard \(\bar{\partial}\)-connection \(\overline{\nabla}^{(n)}: \mathcal{L}_n \to \Omega^{0,1}({\mathbb{C}}P^1_q)\otimes_{\mathcal{A}({\mathbb{C}}P^1_q)}\mathcal{L}_n\). In particular on \(\mathcal{L}_0\) the \(\bar{\partial}\)-connection \(\overline{\nabla}^{(0)}\) coincides with \(\bar{\partial}\).
Observe that for \(f\in \mathcal{A}({\mathbb{C}}P^1_q)\) and \(s\in \mathcal{L}_n,\) we satisfy the Leibniz Rule \[\overline{\nabla}^{(n)}(fs)=\bar{\partial}(f)s+f\overline{\nabla}^{(n)}(s).\]
According to [2], the standard \(\bar{\partial}\)-connection \(\overline{\nabla}^{(n)}\) also satisfies the Leibniz rule with respect to the right multiplication \[\label{eq:32right32Leibniz32rule} \overline{\nabla}^{(n)}(sf)=\overline{\nabla}^{(n)}(s)f+\Phi_{(n)}(s \otimes\bar{\partial}(f)),\tag{7}\] where \(\Phi_{(n)}\) is the twist flip isomorphism in ?? . In other words on each \(\mathcal{L}_n\) the standard \(\bar{\partial}\)-connection \(\overline{\nabla}^{(n)}\) is a bimodule \(\bar{\partial}\)-connection in the sense of Definition 4.
The standard \(\bar{\partial}\)-connection induces the following cochain complex: \[0 \xrightarrow{0} \mathcal{L}_n \xrightarrow{\overline{\nabla}^{(n)}} \Omega^{0,1}(S_q^2)\otimes \mathcal{L}_n \xrightarrow{0} 0\]
Proposition 7. [[2] Theorem 4.4] With the standard \(\bar{\partial}\)-connection, the \((0,0)\)-cohomologies on \(\mathcal{L}_n\) is given by \[H^{0,0}_{\overline{\nabla}^{(n)}}(\mathcal{L}_n)=\begin{cases}0, ~n>0\\ {\mathbb{C}}^{|n|+1},~ n\leq 0\end{cases}\]
In particular \(H^{0,0}_{\overline{\nabla}^{(0)}}(\mathcal{L}_0)={\mathbb{C}}\).
We also have the following result on \(H^{0,1}\)
Proposition 8. [[4] Proposition 7.2] With the standard \(\bar{\partial}\)-connection, we have \[H^{0,1}_{\overline{\nabla}^{(0)}}(\mathcal{L}_0)=0.\]
Since the difference of any two \(\bar{\partial}\)-connections is \(\mathcal{A}({\mathbb{C}}P^1_q)\)-linear, any \(\bar{\partial}\)-connection \(\mathcal{L}_n\) is expressible as \(\overline{\nabla}^{(n)} + D\) where \(\overline{\nabla}^{(n)}\) is the standard \(\bar{\partial}\)-connection on \(\mathcal{L}_n\) and \[D \in Hom_{\mathcal{A}({\mathbb{C}}P^1_q)}(\mathcal{L}_n, \Omega^{0,1}({\mathbb{C}}P^1_q) \otimes_{\mathcal{A}({\mathbb{C}}P^1_q)} \mathcal{L}_n).\]
Moreover by Lemma 2 and 3 such \(D\) is realizable through right multiplication by a \((0,1)\) form, so we choose to express holomorphic structures of \(\mathcal{L}_n\) on \({\mathbb{C}}P^1_q\) as \[\label{eq:32pb-connections32omega} \overline{\nabla}_\theta^{(n)}(s) := \overline{\nabla}^{(n)} s - \Phi_{(n)}(s\theta).\tag{8}\]
Remark 9. \(\overline{\nabla}_\theta^{(n)}\) is a left \(\bar{\partial}\)-connection on \(\mathcal{L}_n\), but unlike the standard \(\bar{\partial}\)-connection \(\overline{\nabla}^{(n)}\), \(\overline{\nabla}_\theta^{(n)}\) is not a bimodule \(\bar{\partial}\)-connection on \(\mathcal{L}_n\) in general.
The following result generalized [2].
Definition-Proposition 10. Let \(\overline{\nabla}^{(m)}\) be the standard \(\bar{\partial}\)-connection on \(\mathcal{L}_m\) and \(\overline{\nabla}_\theta^{(n)}\) be a left \(\bar{\partial}\)-connection on \(\mathcal{L}_n\), then we define the tensor product \(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)}\) as \[\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)}:=\overline{\nabla}^{(m)}\otimes {\text{id}}+ (\Phi_{(m)}\otimes {\text{id}})\circ ({\text{id}}\otimes \overline{\nabla}_\theta^{(n)}).\] \(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)}\) is a left \(\bar{\partial}\)-connection on \(\mathcal{L}_{m+n}\). Moreover we have \[\label{eq:32tensor32product32connections32equal} \overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)}=\overline{\nabla}_\theta^{(m+n)}.\qquad{(4)}\]
Proof. First we check \(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)}\) is well defined. For \(s\in \mathcal{L}_m\), \(t\in \mathcal{L}_n\) and \(f\in \mathcal{A}({\mathbb{C}}P^1_q)\), we have \[(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(sf\otimes t)=\overline{\nabla}^{(m)}(sf)\otimes t+(\Phi_{(m)}\otimes {\text{id}}) (sf \otimes \overline{\nabla}_\theta^{(n)}(t)).\]
By 7 we know \(\overline{\nabla}^{(m)}(sf)=\overline{\nabla}^{(m)}(s)f+\Phi_{(m)}(s\otimes \bar{\partial}(f))\) hence \[\begin{align} (\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(sf\otimes t)=&\overline{\nabla}^{(m)}(s)f\otimes t+\Phi_{(m)}(s\otimes \bar{\partial}(f))\otimes t+(\Phi_{(m)}\otimes {\text{id}}) (sf \otimes \overline{\nabla}_\theta^{(n)}(t))\\ =&\overline{\nabla}^{(m)}(s)\otimes ft+(\Phi_{(m)}\otimes {\text{id}}) (s\otimes \bar{\partial}(f)\otimes t+s \otimes f\overline{\nabla}_\theta^{(n)}(t)). \end{align}\] On the other hand \[\begin{align} (\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(s\otimes ft)=&\overline{\nabla}^{(m)}(s)\otimes ft+ (\Phi_{(m)}\otimes {\text{id}})(s\otimes \overline{\nabla}_\theta^{(n)}(ft))\\ =&\overline{\nabla}^{(m)}(s)\otimes ft+(\Phi_{(m)}\otimes {\text{id}})(s\otimes \bar{\partial}(f)\otimes t+s\otimes f\overline{\nabla}_\theta^{(n)}t). \end{align}\] Therefore \((\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(sf\otimes t)=(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(s\otimes ft)\).
Next we check that \(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)}\) is a left \(\bar{\partial}\)-connection. For \(s\in \mathcal{L}_m\), \(t\in \mathcal{L}_n\) and \(f\in \mathcal{A}({\mathbb{C}}P^1_q)\), we have \[\begin{align} (\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(fs\otimes t)=&\overline{\nabla}^{(m)}(fs)\otimes t+(\Phi_{(m)}\otimes {\text{id}}) (fs \otimes \overline{\nabla}_\theta^{(n)}(t))\\ =& \bar{\partial}(f)s\otimes t+f\overline{\nabla}^{(m)}(s)\otimes t+(\Phi_{(m)}\otimes {\text{id}}) (fs \otimes \overline{\nabla}_\theta^{(n)}(t)). \end{align}\] Since \(\Phi_{(m)}\) is an \(\mathcal{A}({\mathbb{C}}P^1_q)\)-bimodule map, we have \[(\Phi_{(m)}\otimes {\text{id}}) (fs \otimes \overline{\nabla}_\theta^{(n)}(t))=f(\Phi_{(m)}\otimes {\text{id}}) (s \otimes \overline{\nabla}_\theta^{(n)}(t))\] hence \[(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(fs\otimes t)=\bar{\partial}(f)s\otimes t+f(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(s\otimes t).\]
Lastly we check ?? . For \(s\in \mathcal{L}_m\) and \(t\in \mathcal{L}_n\) we have \[\label{eq:32tensor32connection32expansion} \begin{align}(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)})(s\otimes t)=&\overline{\nabla}^{(m)}(s)\otimes t+(\Phi_{(m)}\otimes {\text{id}}) (s \otimes \overline{\nabla}_\theta^{(n)}(t))\\ =&\overline{\nabla}^{(m)}(s)\otimes t+(\Phi_{(m)}\otimes {\text{id}}) (s \otimes \overline{\nabla}^{(n)}(t)-s\otimes\Phi_{(n)}(t\otimes \theta))\\ =&\overline{\nabla}^{(m)}(s)\otimes t+(\Phi_{(m)}\otimes {\text{id}}) (s \otimes \overline{\nabla}^{(n)}(t))-(\Phi_{(m)}\otimes {\text{id}})(s\otimes\Phi_{(n)}(t\otimes \theta)).\end{align}\tag{9}\] By [2] we know that \(\overline{\nabla}^{(m)}\otimes {\text{id}}+ (\Phi_{(m)}\otimes {\text{id}})\circ ({\text{id}}\otimes \overline{\nabla}^{(n)})=\overline{\nabla}^{(m+n)}\). The equality in ?? then follows from Lemma 3 and 9 . ◻
Remark 11. \(\overline{\nabla}^{(m)}\otimes \overline{\nabla}_\theta^{(n)}\) is flat by dimension reason.
Remark 12. In general we cannot define the tensor production connection \(\overline{\nabla}_{\theta_1}^{(m)}\otimes \overline{\nabla}_{\theta_2}^{(n)}\) of two left \(\bar{\partial}\)-connections \(\overline{\nabla}_{\theta_1}^{(m)}\) and \(\overline{\nabla}_{\theta_2}^{(n)}\).
Remark 13. In the sequel we will usually omit the superscript "\((n)\)" in the notation of \(\bar{\partial}\)-connections and simply denote it by \(\overline{\nabla}_\theta\).
As in [2], we denote by \(\mathcal{C}(SU_q(2))\) the \(C^*\)-completion of \(\mathcal{A}(SU_q(2))\). By definition \(\mathcal{C}(SU_q(2))\) is the universal \(C^*\)-algebra generated by \(a\) and \(c\) subject to relations in 1 . Moreover, the exists a unique left invariant Haar state \(h\) on \(\mathcal{C}(SU_q(2))\) such that \(h(1)=1\). As a result we can consider \(L^2(SU_q(2))\) via the GNS-construction on \(\mathcal{C}(SU_q(2))\).
We can also define the \(C^*\)-subalgebra of \(\mathcal{C}(SU_q(2))\) generated by \(B_0\), \(B_+\), and \(B_-\), which we denote by \(\mathcal{C}({\mathbb{C}}P^1_q)\); and its \(L^2\)-completion \(L^2({\mathbb{C}}P^1_q)\). We consider \(\mathcal{C}({\mathbb{C}}P^1_q)\) and \(L^2({\mathbb{C}}P^1_q)\) the algebras of continuous and \(L^2\)-functions on \({\mathbb{C}}P^1_q\), respectively.
We shall introduce a \(C^*\)-representation of \(\mathcal{C}({\mathbb{C}}P^1_q)\). Let \(\ell^2\) be the standard separable Hilbert space with orthonormal basis \(\{e_n\}_{n\geq 0}\), and \(B(\ell^2)\) be the \(C^*\)-algebra of bounded operators on \(\ell^2\).
Proposition 14. [[5] Proposition 4, [6] Proposition 4.1] There exists a faithful representation \(\pi: \mathcal{C}({\mathbb{C}}P^1_q)\to B(\ell^2)\) such that \[\label{eq:32standar32representation} \begin{align} \pi(B_-)(e_n)&=q^{n}\sqrt{1-q^{2n}}e_{n-1};\\ \pi(B_0)(e_n)&=q^{2n}e_n;\\ \pi(B_+)(e_n)&=q^{n+1}\sqrt{1-q^{2n+2}}e_{n+1}. \end{align}\qquad{(5)}\]
In particular, the spectrum of \(B_0\) is \(\{0\} \cup \{q^{2n} | n \in {\mathbb{Z}}_{\geq 0} \}\).
Remark 15. We can further show that \(\mathcal{C}({\mathbb{C}}P^1_q)\) is \(*\)-isomorphic to the \(C^*\)-subalgebra of \(B(\ell^2)\) generated by \(1\) and all compact operators. Nevertheless we do not need this fact in our paper.
As in [2], we can define \(\Gamma(\mathcal{L}_n)\) and \(L^2(\mathcal{L}_n)\) as spaces of continuous and \(L^2\)-sections of the line bundle \(\mathcal{L}_n\), respectively. It is clear that \(\Gamma(\mathcal{L}_n)\) is a \(\mathcal{C}({\mathbb{C}}P^1_q)\)-bimodule. Similar to 3 we have \[{\mathrm {End}}_{\mathcal{C}({\mathbb{C}}P^1_q)}(\Gamma(\mathcal{L}_n))=\mathcal{C}({\mathbb{C}}P^1_q).\]
In particular we can consider the Hilbert space \(L^2(\Omega^{0,1}({\mathbb{C}}P^1_q))\). According to [4], the map \(\bar{\partial}: \mathcal{A}({\mathbb{C}}P^1_q)\to \Omega^{0,1}({\mathbb{C}}P^1_q)\) has a Hermitian conjugate \[\bar{\partial}^{\dagger}: \Omega^{0,1}({\mathbb{C}}P^1_q)\to \mathcal{A}({\mathbb{C}}P^1_q).\] Moreover we have the following theorem:
Theorem 3. [[7] Theorem 6.21] \((\mathcal{A}({\mathbb{C}}P^1_q), L^2(\Omega^{0,\bullet}({\mathbb{C}}P^1_q)), D_{\bar{\partial}})\) forms a spectral triple in the sense of [1], where \(D_{\bar{\partial}}:=\bar{\partial}+\bar{\partial}^{\dagger}\).
Remark 16. In particular, the map \(\bar{\partial}: \mathcal{A}({\mathbb{C}}P^1_q)\to \Omega^{0,1}({\mathbb{C}}P^1_q)\) extends to a closed map \[\bar{\partial}: L^2({\mathbb{C}}P^1_q)\to L^2(\Omega^{0,1}({\mathbb{C}}P^1_q)).\]
We will need the following result
Proposition 17. [[2]] There are no nontrivial holomorphic functions in \({\mathrm {Dom}}(\bar{\partial})\cap \mathcal{C}({\mathbb{C}}P^1_q)\), i.e. we have \(\ker\bar{\partial}\cap \mathcal{C}({\mathbb{C}}P^1_q)={\mathbb{C}}\).
For any \(\bar{\partial}\)-connection \(\overline{\nabla}_{\theta}\) on \(\mathcal{L}_0\), we can define its Hermitian conjugate \(\overline{\nabla}_{\theta}^{\dagger}\) and the Dirac operator \(D_{\overline{\nabla}_{\theta}}\) in the same way. We have the following corollary:
Corollary 1. For any \(\bar{\partial}\)-connection \(\overline{\nabla}_{\theta}\) on \(\mathcal{L}_0\), \((\mathcal{A}({\mathbb{C}}P^1_q), L^2(\Omega^{0,\bullet}({\mathbb{C}}P^1_q)), D_{\overline{\nabla}_{\theta}})\) forms a spectral triple. In particular \(\ker \overline{\nabla}_{\theta}\) is finite dimensional for any \(\bar{\partial}\)-connection \(\overline{\nabla}_{\theta}\) on \(\mathcal{L}_0\).
Proof. Since \(\theta\wedge(-)\) is a bounded operator, the difference between \(D_{\bar{\partial}}\) and \(D_{\overline{\nabla}_{\theta}}\) is a bounded self-adjoint operator. As spectral triples are preserved by bounded perturbations, \((\mathcal{A}({\mathbb{C}}P^1_q), L^2(\Omega^{0,\bullet}({\mathbb{C}}P^1_q)), D_{\overline{\nabla}_{\theta}})\) is still a spectral triple.
The finite dimensionality of \(\ker \overline{\nabla}_{\theta}\) follows from the fact that \(D_{\overline{\nabla}_{\theta}}\) has compact resolvent. ◻
Khalkhali et. al extend the notion of gauge equivalence in the noncommutative case in [2]. Two \(\bar{\partial}\)-connections \(\overline{\nabla}_{\theta_1}, \overline{\nabla}_{\theta_2}\) on \(\mathcal{L}_n\) are said to be gauge equivalent if there exists an invertible element \(g \in {\mathrm {End}}_{\mathcal{A}({\mathbb{C}}P^1_q)}(\mathcal{L}_n)\) such that \[\overline{\nabla}_{\theta_1} = g^{-1} \circ \overline{\nabla}_{\theta_2}\circ g.\]
Lemma 4. Two \(\bar{\partial}\)-connections \(\overline{\nabla}_{\theta_1}, \overline{\nabla}_{\theta_2}\) on \(\mathcal{L}_n\) are gauge equivalent if and only if there exists an invertible element \(g\in \mathcal{A}({\mathbb{C}}P^1_q)\) such that \[\theta_1=g\theta_2 g^{-1}-\bar{\partial}(g)g^{-1}.\]
In particular, a \(\bar{\partial}\)-connection \(\overline{\nabla}_{\theta}\) is gauge equivalent to the standard \(\bar{\partial}\)-connection \(\overline{\nabla}\) if and only if there exists an invertible element \(g\in \mathcal{A}({\mathbb{C}}P^1_q)\) such that \(\bar{\partial}(g)=g\theta\).
Proof. By 4 we know that \({\mathrm {End}}_{\mathcal{A}({\mathbb{C}}P^1_q)}(\mathcal{L}_n)=\mathcal{A}({\mathbb{C}}P^1_q)\) where the right hand side means right multiplication by elements in \(\mathcal{A}({\mathbb{C}}P^1_q)\). The result then follows from 7 and 8 . ◻
However, the condition that \(g\in \mathcal{A}({\mathbb{C}}P^1_q)\) in Lemma 4 is too restrictive as shown in the following example.
Example 1. Let \(\theta=\bar{\partial}(B_0)\). Using ?? and induction we can get \[\label{eq:32pb32of32B95094n} \bar{\partial}(B_0^n)=\sum_{k=0}^{n-1}q^{2k}B_0^{n-1}\bar{\partial}(B_0)=q^{n-1}[n]_{q^2}B_0^{n-1}\bar{\partial}(B_0),\qquad{(6)}\] where \([n]_{q^2}=\frac{q^{2n}-q^{-2n}}{q^2-q^{-2}}\) is the \(q^2\)-integer as in [2]. We define \([n]_{q^2}!:=\prod_{k=1}^n [k]_{q^2}\) and \([0]_{q^2}!:=1\). Then we can check that \[\label{eq:32g32for32B950} g:=\sum_{n=0}^{\infty}\frac{B_0^n}{q^{\frac{n(n-1)}{2}}[n]_{q^2}!}\qquad{(7)}\] satisfies \(\bar{\partial}(g)=g\bar{\partial}(B_0)\). It is clear that \(g\in \mathcal{C}({\mathbb{C}}P^1_q)\) is invertible and \(g\in {\mathrm {Dom}}(\bar{\partial})\) but \(g\notin \mathcal{A}({\mathbb{C}}P^1_q)\).
Inspired by Example 1 we have the following modified definition.
Definition 18. We call two \(\bar{\partial}\)-connections \(\overline{\nabla}_{\theta_1}, \overline{\nabla}_{\theta_2}\) on \(\mathcal{L}_n\) gauge equivalent if there exists an invertible element \(g\in \mathcal{C}({\mathbb{C}}P^1_q)^{\times}\cap {\mathrm {Dom}}(\bar{\partial})\) such that \[\theta_1=g\theta_2 g^{-1}-\bar{\partial}(g)g^{-1}.\] hence \[\label{eq:32gauge32equivalence32general} \bar{\partial}(g)= g\theta_2 -\theta_1 g.\qquad{(8)}\]
In particular, a \(\bar{\partial}\)-connection \(\overline{\nabla}_{\theta}\) on \(\mathcal{L}_n\) is gauge equivalent to the standard \(\bar{\partial}\)-connection if there exists an invertible element \(g\in \mathcal{C}({\mathbb{C}}P^1_q)^{\times}\cap {\mathrm {Dom}}(\bar{\partial})\) such that \[\label{eq:32gauge32equivalence32to32standard32general} \bar{\partial}(g)=g\theta.\qquad{(9)}\]
By Proposition 8 we know that \(H^{0,1}_{\overline{\nabla}}(\mathcal{L}_0)=0.\) As a result for any \(\theta\in \Omega^{0,1}({\mathbb{C}}P^1_q)\) there exists an \(f\in \mathcal{A}({\mathbb{C}}P^1_q)\) such that \(\theta=\bar{\partial}(f)\). Therefore a \(\bar{\partial}\)-connection \(\overline{\nabla}_{\theta}=\overline{\nabla}_{\bar{\partial}(f)}\) on \(\mathcal{L}_n\) is gauge equivalent to the standard \(\bar{\partial}\)-connection if there exists an invertible element \(g\in \mathcal{C}({\mathbb{C}}P^1_q)^{\times}\cap {\mathrm {Dom}}(\bar{\partial})\) such that \[\label{eq:32exponential32equation} \bar{\partial}(g)=g\bar{\partial}(f),\tag{10}\] which is a noncommutative analogue of the exponential equation.
Remark 19. If the algebra was commutative, then \(g=\exp(f)\) would give a solution to 10 . Hence we call 10 the noncommutative exponential equation.
The following lemma plays a key role in the contruction of non-standard holomorphic structures:
Lemma 5. If there exists a non-zero non-invertible \(h \in \mathcal{C}({\mathbb{C}}P^1_q)\cap {\mathrm {Dom}}(\bar{\partial})\) such that \(\bar{\partial}(h) = h \bar{\partial}(f)\), then there cannot exist an invertible \(g\) such that \(\bar{\partial}(g) = g \bar{\partial}(f)\).
Proof. For the sake of contradiction, let such a \(g\) exist. Since \(g\) is invertible, we can write \(h = ag\) for \(a = h g^{-1}\). We have \[\bar{\partial}(h)=\bar{\partial}(ag)=\bar{\partial}(a)g+a\bar{\partial}(g)=\bar{\partial}(a)g+ag\bar{\partial}(f)=\bar{\partial}(a)g+h\bar{\partial}(f).\] Since \(\bar{\partial}(h) = h \bar{\partial}(f)\),we have \[\bar{\partial}(a) g = 0.\] Since \(g\) is invertible, we then have \(\bar{\partial}(a) = 0\). By Proposition 17, this means that \(a\in {\mathbb{C}}\) is a constant. However, this implies that either \(h = 0\) or that \(h\) is invertible, a contradiction. ◻
Proposition 21. There is a one-to-one correspondence between sets of gauge equivalence classes of holomorphic structures on \(\mathcal{L}_m\) and \(\mathcal{L}_n\) for any \(m\) and \(n\).
Proof. By Proposition 10, there exists a one-to-one correspondence between holomorphic structures on \(\mathcal{L}_m\) and \(\mathcal{L}_n\). The compatibility with gauge equivalences follows from ?? and ?? . ◻
Let \(C^*(1, B_0)\) denote the unital \(C^*\)-subalgebra of \(\mathcal{C}({\mathbb{C}}P^1_q)\) generated by \(B_0\). Since \(B_0\) is self-adjoint hence normal, by continuous functional calculus we have a \(*\)-isomorphism \[\label{eq:32functional32calculus32B0} \Psi: C^*(1, B_0)\overset{\sim}{\to}C(sp(B_0)),\tag{11}\] where \(C(sp(B_0))\) is the \(C^*\)-algebra of continuous functions on \(sp(B_0)\) the spectrum of \(B_0\). Recall Proposition 14 tells us \[\label{eq:32sp32B0} sp(B_0)=\{0\} \cup \{q^{2n} | n \in {\mathbb{Z}}_{\geq 0} \}.\tag{12}\] Note that since \(sp(B_0)\) only has a single limit point at zero, continuity of a function \(f\) on \(sp(B_0)\) is equivalent to \(\lim_{n \to \infty} f(q^{2n}) = f(0).\) Additionally, an element \(f \in C(sp(B_0))\) is invertible iff it never vanishes on the spectrum, as this is the necessary and sufficient condition for \(1/f\) being well-defined.
We want to study the restriction of \(\bar{\partial}\) to \(C^*(1, B_0)\) in more details. First we introduce the following operator.
Definition 22. Let \({\mathbb{C}}[x]\) denote the algebra of polynomials. We define the linear map \(\bar{\delta}: {\mathbb{C}}[x] \to {\mathbb{C}}[x]\) as \[\label{eq:32def32of32db} \bar{\delta}(f)(x):=\frac{f(x)-f(q^2x)}{x-q^2x}.\qquad{(10)}\]
Remark 23. The same formula as ?? appeared in [3]. Nevertheless analytic properties of \(\bar{\delta}\) like Proposition 26 below have not been covered in [3] .
Remark 24. The operator \(\bar{\delta}\) is not a derivation on \({\mathbb{C}}[x]\) in the usual sense. Actually we can show that \(\bar{\delta}\) satisfies a twisted Leibniz rule as in [3], but we do not need this fact in our paper.
If we define the dilation operator \(m_{c}\) for \(c\in \mathbb{R}\) on \({\mathbb{C}}[x]\) by \[\label{eq:32mc} m_c(f)(x):=f(cx),\tag{13}\] then [def:32db] can be rewritten as \[\bar{\delta}(f)(x)=\frac{f(x)-m_{q^2}(f)(x)}{x-q^2x}.\] We can consider \({\mathbb{C}}[x]\) as a subspace of \(C(sp(B_0))\) by restricting \(f(x)\) to \(sp(B_0)\). If \(0<c<1\), then we can also extend \(m_c\) to a bounded operator on \(C(sp(B_0))\).
Lemma 6. The map \(\bar{\delta}\) corresponds to \(\bar{\partial}\) under the functional calculus isomorphism 11 . In more details, for any \(f\in {\mathbb{C}}[x]\subset C(sp(B_0))\), we have \[\label{eq:32db32and32pb32correspond} \bar{\partial}(\Psi^{-1}(f))=(\Psi^{-1}(\bar{\delta}f))\bar{\partial}B_0.\qquad{(11)}\]
Proof. The definition ?? gives \[\bar{\delta}(x^n) = \frac{1 - q^{2n}}{1 - q^2} x^{n-1},\] and \(\bar{\delta}(1) = 0\). On the other hand ?? gives \[\bar{\partial}(B_0^n) =\sum_{k=0}^{n-1}q^{2k}B_0^{n-1}\bar{\partial}(B_0)= \frac{1 - q^{2n}}{1 - q^2} B_0^{n-1} \bar{\partial}B_0.\] Since \(\Psi(B_0)=x\), the lemma then follows by linearity. ◻
By the Stone-Weierstrass Theorem, \({\mathbb{C}}[x]\) is a dense subset of \(C(sp(B_0))\). However the operator \(\bar{\delta}\) is not bounded, for example for \(f_n(x)=(1-x)^n\) we have \(\|f_n\|=1\) when we take the maximal norm as elements in \(C(sp(B_0))\). On the other hand \(\|\bar{\delta}(f_n)\|\geq |\bar{\delta}(f_n)(0)|=n\).
Therefore we cannot extend \(\bar{\delta}\) to an operator on \(C(sp(B_0))\). To further study the analytic properties of \(\bar{\delta}\), we introduce the \(\bar{I}\) operator inspired by [3].
Definition 25. The linear map \(\bar{I}: {\mathbb{C}}[x]\to {\mathbb{C}}[x]\) is defined by \[\label{eq:32def32of32Ibar} \bar{I}(x^n)=\frac{1-q^2}{1-q^{2n+2}}x^{n+1}\qquad{(12)}\] and \(\bar{I}(0)=0\).
Lemma 7. For all \(f \in {\mathbb{C}}[x]\) we have \[\label{eq:32I32db32inverse} \bar{\delta}(\bar{I}(f)) = f \text{ and }\bar{I}(\bar{\delta}(f)) = f-f(0).\qquad{(13)}\]
Proof. It By direct computation we can check that ?? holds for any \(f(x)=x^n\). The general case then follows by linearity. ◻
Lemma 8. Given a \(f \in {\mathbb{C}}[x]\) we have \[\label{eq:32expansion32of32I} \bar{I}(f)(x) = (1-q^2)x \sum_{n=0}^{\infty} q^{2n}(m_{q^{2n}}f)(x),\qquad{(14)}\] where \(m_{q^{2n}}\) is defined in 13 .
Proof. For \(f(x)=x^k\), ?? gives \[\begin{align} \bar{I}(f)(x)=&(1-q^2)x\frac{x^k}{1-q^{2k+2}}\\ =&(1-q^2)x\sum_{n=0}^{\infty}q^{(2k+2)n}x^k\\ =&(1-q^2)x\sum_{n=0}^{\infty} q^{2n}(q^{2n}x)^k\\ =&(1-q^2)x\sum_{n=0}^{\infty}q^{2n}(m_{q^{2n}}f)(x). \end{align}\] The general case then follows by linearity. ◻
Lemma 9. Given a \(f \in {\mathbb{C}}[x]\), we have \(\|\bar{I}(f)\| \leq \|f\|.\)
Proof. For \(f\in {\mathbb{C}}[x]\), by Lemma 8 we have \[\|\bar{I}(f)\| = \|(1-q^2) x \sum_{n=0}^\infty q^{2n}(n_{q^{2n}} f) \| \leq (1-q^2) \| x \|\cdot \| \sum_{n=0}^\infty q^{2n}(m_{q^{2n}} f) \|\] Since \(sp(B_0)=\{0\} \cup \{q^{2n} | n \in {\mathbb{Z}}_{\geq 0} \}\subset [0,1]\), we have \(\|x\| = 1\). Hence \[\|\bar{I}(f)\| \leq (1-q^2) \sum_{n=0}^\infty \| q^{2n}(m_{q^{2n}} f) \| = (1-q^2) \sum_{n=0}^\infty q^{2n} \| (m_{q^{2n}} f) \|\] Now, since \(q^{2n}\leq 1\), the dilation \(m_{q^{2n}}\) does not increase the norm of \(f,\) we have \[\|\bar{I}(f)\| \leq (1-q^2) \sum_{n=0}^\infty q^{2n} \| f \| = \|f\| \cdot \Big( (1-q^2) \sum_{n=0}^\infty q^{2n} \Big) = \|f\|.\] ◻
Lemma 10. \(\bar{I}\) extends to a bounded map \(\bar{I}: C(sp(B_0)) \to C(sp(B_0))\). Moreover, ?? holds for any \(f\in C(sp(B_0))\).
Proof. By the Stone-Weierstrass theorem, \({\mathbb{C}}[x]\) is dense in \(C(sp(B_0))\). The result then follows from Lemma 9. ◻
Lemma 11. The map \(\bar{I}: C(sp(B_0)) \to C(sp(B_0))\) is injective.
Proof. Let \(f\in C(sp(B_0))\) be in the kernel of \(\bar{I}\). Since \(sp(B_0)=\{0\} \cup \{q^{2n} | n \in {\mathbb{Z}}_{\geq 0} \}\), for any \(k\geq 0\) we have \(\bar{I}(f)(q^{2k})=0\). By ?? we have \[\bar{I}(f)(q^{2k})=(1-q^2)q^{2k} \sum_{n=0}^{\infty} q^{2n}(m_{q^{2n}}f)(q^{2k})=(1-q^2)q^{2k} \sum_{n=0}^{\infty} q^{2n}f(q^{2n+2k}).\] Therefore \(\bar{I}(f)(q^{2k})=0\) implies \[\label{eq:32sum32is32zero1} \sum_{n=0}^{\infty} q^{2n}f(q^{2n+2k})=0.\tag{14}\]
Notice that we also have \(\bar{I}(f)(q^{2(k+1)})=0\) hence \[\label{eq:32sum32is32zero2}\sum_{n=0}^{\infty} q^{2n}f(q^{2n+2(k+1)})=\sum_{n=0}^{\infty} q^{2n}f(q^{2n+2+2k})=0.\tag{15}\] Compare 14 and 15 we get \[f(q^{2k})=0 \text{ for any }k\geq 0.\] Since \(f\) is continuous, we also get \[f(0)=\lim_{k\to \infty}f(q^{2k})=0.\] Hence \(f\equiv 0\). ◻
Proposition 26. \(\bar{\delta}\) is a closable operator on \(C(sp(B_0))\).
Proof. Recall that \(\bar{\delta}\) being closable means that for all \(\{f_n\} \in {\mathbb{C}}[x]\) such that \(f_n \to 0\) and \(\bar{\delta}f_n \to g,\) for some \(g \in C(sp(B_0)),\) then \(g = 0.\)
Now, by Lemma 10 we know \(\bar{I}\) is bounded hence \(\bar{\delta}f_n \to g\) implies \(\bar{I}\bar{\delta}f_n \to \bar{I}g\). By Lemma 7, \(\bar{I}\bar{\delta}f_n=f_n - f_n(0)\) hence we have \(f_n - f_n(0) \to \bar{I}g.\) However, since \(f_n \to 0,\) we also have \(f_n - f_n(0) \to 0,\) which means that \(\bar{I}g = 0.\) The injectivity of \(\bar{I}\) as in Lemma 11 then implies \(g=0\). ◻
Proposition 26 tells us that we can extend \(\bar{\delta}\) to a closed operator on \(C(sp(B_0))\).
Remark 27. We can deduce that \(\bar{\delta}\) is closable from the fact that \(\bar{\partial}\) is a closed operator and the relation ?? . We give a direct proof here because the operator \(\bar{I}\) which is introduced in the proof is important in the proof of Proposition 28 below.
Proposition 28. \(f\in C(sp(B_0))\) is in the domain of \(\bar{\delta}\) if and only if \[\frac{f(x)-f(q^2 x)}{x-q^2 x}\] is a continuous function on \(sp(B_0),\) i.e. \[\lim_{k\to \infty}\frac{f(q^{2k})-f(q^{2k+2})}{q^{2k}-q^{2k+2}} \text{ exists.}\] In this case we have \[(\bar{\delta}f)(q^{2k})=\frac{f(q^{2k})-f(q^{2k+2})}{q^{2k}-q^{2k+2}} \text{ and }(\bar{\delta}f)(0)= \lim_{k\to \infty}\frac{f(q^{2k})-f(q^{2k+2})}{q^{2k}-q^{2k+2}}.\]
Proof. Recall that the domain of \(\bar{\delta}\) consists of all functions \(f\in C(sp(B_0))\) such that there exists a sequence \(f_n\in {\mathbb{C}}[x]\) such that \(\lim_{n \to \infty} f_n = f\) and \(\lim_{n \to \infty} \bar{\delta}f_n\) converges.
Now, if \(f\in {\mathrm {Dom}}(\bar{\delta})\), let \(f_n\in {\mathbb{C}}[x]\) be a sequence such that \(f_n\to f\) and \(\bar{\delta}f_n\) converges with limit \(\bar{\delta}f\). We know that \[(\bar{\delta}f_n)(x)=\frac{f_n(x)-f_n(q^2 x)}{x-q^2 x}.\] Since \(f_n\to f\) we get \[\frac{f(x)-f(q^2 x)}{x-q^2 x} = \lim_{n \to \infty} \frac{f_n(x)-f_n(q^2 x)}{x-q^2 x}.\] for any \(x\neq 0\). Since \(\bar{\delta}f_n\) converges we know \(\frac{f(x)-f(q^2 x)}{x-q^2 x}\) is continuous on \(sp(B_0)\) and \[(\bar{\delta}f)(x)=\frac{f(x)-f(q^2 x)}{x-q^2 x}.\]
On the other hand, if \(\frac{f(x)-f(q^2 x)}{x-q^2 x}\) is a continuous function on \(sp(B_0)\). Since \({\mathbb{C}}[x]\) is dense in \(C(sp(B_0))\), there exists a sequence \(g_n\in {\mathbb{C}}[x]\) such that \[g_n\to \frac{f(x)-f(q^2 x)}{x-q^2 x}\] Since \(\bar{I}\) is a bounded operator on \(C(sp(B_0))\) we get \[\bar{I}g_n\to\bar{I}\big(\frac{f(x)-f(q^2 x)}{x-q^2 x}\big).\] By ?? we can check \[\bar{I}\big(\frac{f(x)-f(q^2 x)}{x-q^2 x}\big)=f(x)-f(0)\] therefore \(\bar{I}g_n\to f-f(0)\). We then define \[f_n=\bar{I}g_n+f(0).\] It is then clear that \(f_n\to f\) and \(\bar{\delta}f_n=\bar{\delta}\bar{I}g_n=g_n\) converges with limit \(\bar{\delta}f\). ◻
Corollary 2. For any \(f\in {\mathrm {Dom}}(\bar{\partial})\cap C^*(1, B_0)\) we have \(\Psi(f)\in {\mathrm {Dom}}(\bar{\delta})\). Moreover we have \[\bar{\partial}f=\big (\Psi^{-1}(\bar{\delta}(\Psi f))\big )\bar{\partial}B_0.\] Sometimes we abuse the notation and simply write it as \[\label{eq:32db32and32pb32correspond32general} \bar{\partial}f=(\bar{\delta}( f))\bar{\partial}B_0.\qquad{(15)}\]
We can now tackle our problem of finding an invertible \(g\) such that \(\bar{\partial}g = g \bar{\partial}f,\) in the case of restricting both \(g\) and \(f\) to \(C^*(1, B_0).\)
Proposition 29. Let \(f\in C(sp(B_0))\) be a function contained in \({\mathrm {Dom}}(\bar{\delta}).\) Then, a solution to \(\bar{\delta}g = g \bar{\delta}f\) is given by \[\label{eq:32def32of32g} g(q^{2n}) = g(1) \prod_{k = 1}^n (1 - f(q^{2k - 2}) + f(q^{2k})) \text{ for any }n\geq 1,\qquad{(16)}\] and \[\label{eq:32def32of32g0} g(0) = g(1)\prod_{k = 1}^\infty (1 - f(q^{2k - 2}) + f(q^{2k}))\qquad{(17)}\] The \(g\) defined above is always in \({\mathrm {Dom}}(\bar{\delta})\).
Proof. First, recall from Proposition 28 that \(f \in Dom(\bar{\delta})\) if and only if \(\frac{f(x) - f(q^2x)}{1 - q^2x}\) is a continuous function on \(sp(B_0),\) and if so, \(\bar{\delta}f = \frac{f(x) - f(q^2x)}{1 - q^2x}.\) Then, given such an \(f,\) a solution \(g\) such that \(\bar{\delta}g = g \bar{\delta}f\) is equivalent to a \(g\) such that \[\label{eq:32solve32for32g321} \frac{g(x) - g(q^2x)}{1 - q^2x} = g(x) \frac{f(x) - f(q^2x)}{1 - q^2x}\tag{16}\] Now, since we are only considering \(x \in [0,1],\) \(1 - q^2x\) is never zero, so we may reduce 16 to \[g(x) - g(q^2x) = g(x) (f(x) - f(q^2x)),\] which gives \[\label{eq:32solve32for32g322} g(q^2x) = g(x)(1 - f(x) + f(q^2x)).\tag{17}\] Now, if we let \(x = q^{2n - 2},\) this gives us the recursive formula \[g(q^{2n}) = g(q^{2n - 2})(1 - f(q^{2n - 2}) + f(q^{2n})).\] Thus, if we write \(g(1) = c\) for any \(c \in {\mathbb{C}},\) we obtain \[\label{eq:32def32of32g32in32proof} g(q^{2n}) = c \prod_{k = 1}^n (1 - f(q^{2n - 2k}) + f(q^{2n - 2k + 2})) = c \prod_{k = 1}^n (1 - f(q^{2k - 2}) + f(q^{2k}))\tag{18}\]
We know \(g\) is continuous if and only if \(\lim_{n \to \infty} g(q^{2n}) = g(0).\) Since the \(g\) is defined pointwise in 18 , we need only set \(g(0)\) as the limit to the above expression as \(n \to \infty.\) Then, \(g\) is continuous if the product \[g(0) := c \prod_{k = 1}^\infty (1 - f(q^{2k - 2}) + f(q^{2k}))\] converges. By taking logarithm, it is easy to see that the above infinite product converges if \[\sum_{j=1}^\infty\big( f(q^{2n})- f(q^{2n-2})\big)\] converges absolutely. Since \(f\in {\mathrm {Dom}}(\bar{\delta}),\) we have \[|f(q^{2n})- f(q^{2n-2})| = |(q^{2n} - q^{2n - 2}) \bar{\delta}f(q^{2n - 2}) |\] where \(\bar{\delta}f\) is a continuous function on \(sp(B_0)\), hence bounded. As a result there exists a number \(M\) such that \[|f(q^{2n})- f(q^{2n-2})|\leq M(1-q^2)q^{2n-2}\] for all \(n\). Hence \(\sum_{j=1}^\infty\big( f(q^{2n})- f(q^{2n-2})\big)\) converges absolutely hence \(g\) is continuous.
Now \(g\) is continuous and satisfies 16 for any \(x=q^{2n}\). By Proposition 28 the right hand side of 16 is continuous, hence the left hand side, which is \(\frac{g(x) - g(q^2x)}{1 - q^2x}\), is also continuous. Again by Proposition 28 we know that \(g\in {\mathrm {Dom}}(\bar{\delta})\). ◻
Corollary 3. The solution \(g\) in Proposition 29 is invertible if and only if \(g(1)\neq 0\) and \[\label{eq:32defective32spot32of32f} f(q^{2n}) - f(q^{2n - 2}) \neq 1\text{ for all }n \in {\mathbb{N}}.\qquad{(18)}\]
Proof. We know that \(g\) is invertible if and only if \(g(q^{2n})\neq 0\) for each \(n\geq 0\) and \(g(0)=\lim_{n\to \infty}g(q^{2n})\neq 0\). If \(g(1)=0\) then \(g\) is clearly non-invertible. So now we assume \(g(1)\neq 0\).
By ?? , we know that \(g(q^{2n}) = 0\) for some \(n\) if and only if there exists some \(k \leq n\) such that \(f(q^{2k}) - f(q^{2k - 2}) = 1.\)
Also, if \((1 - f(q^{2k - 2}) + f(q^{2k})) \neq 0\) for each \(k\), then since \[\sum_{k=1}^{\infty}\log((1 - f(q^{2k - 2}) + f(q^{2k})))\] does not go to \(-\infty\) as in the proof of Proposition 29, we know that the infinite product \[g(0)= g(1)\prod_{k = 1}^\infty (1 - f(q^{2k - 2}) + f(q^{2k}))\] is also not zero. We finished the proof. ◻
Inspired by Corollary 3 we have the following definition.
Definition 30. We say that \(f \in C^*(1, B_0)\) has a defective spot* at \(n \in {\mathbb{N}}\) if \[(\Psi f)(q^{2n}) - (\Psi f)(q^{2n - 2}) = 1,\] where \(\Psi: C^*(1, B_0)\overset{\sim}{\to}C(sp(B_0))\) is the functional calculus isomorphism as in 11 .*
We denote the set of defective spots of \(f\) by \(S_f\).
Remark 31. We know that \(S_f\) must be a finite subset of \({\mathbb{N}}\) as \(\Psi f\in C(sp(B_0))\) is continuous at \(0\).
Note that \(\frac{B_0}{q^{2n-2} - q^{2n}}\) is a function which has a defective spot at \(n.\)
Theorem 4. Given \(f \in \mathcal{A}({\mathbb{C}}P^1_q)\cap C^*(1, B_0),\) there exists an invertible \(g\in\mathcal{C}({\mathbb{C}}P^1_q)^{\times}\cap {\mathrm {Dom}}(\bar{\partial})\) such that \(\bar{\partial}g = g \bar{\partial}f\) if and only if \(f\) has no defective spot.
In other words, for \(f \in \mathcal{A}({\mathbb{C}}P^1_q)\cap C^*(1, B_0),\) the \(\bar{\partial}\)-connection \(\overline{\nabla}_\theta\) on \(\mathcal{L}_n\) with \(\theta=\bar{\partial}f\) is gauge equivalent to the standard \(\bar{\partial}\)-connection \(\overline{\nabla}\) if and only if \(f\) has no defective spot.
Proof. By Corollary 3, if \(f\) has no defective spot. an invertible solution \(g\) to \(\bar{\partial}g = g \bar{\partial}f\) exists.
On the other hand, if \(f\) has a defective spot, then by Corollary 3, \(\bar{\partial}g = g \bar{\partial}f\) has a not-invertible, nonzero solution. By Lemma 5, \(\bar{\partial}g = g \bar{\partial}f\) cannot have any invertible solution. ◻
Example 2. We notice that the element \(B_0\) has no defect spot. Actually in Example 1 we found explicitly an invertible element \(g\) such that \(\bar{\partial}g=g \bar{\partial}B_0\).
On the other hand we consider \(f=\frac{B_0}{1-q^2}\in \mathcal{A}({\mathbb{C}}P^1_q)\). It is clear that the defective spot \(S_f=\{1\}\). By ?? we can get \[\bar{\partial}(B_0^{\infty})=B_0^{\infty}\bar{\partial}(f)\] where \[B_0^{\infty}=\lim_{n\to\infty}B_0^n\in C^*(1, B_0).\] Since \(\Psi(B_0)=x\) we have \[\Psi(B_0^{\infty})(q^{2n})=\begin{cases}1 &n=0\\0 & n\geq 1 \end{cases}\] which is a continuous function on \(sp(B_0)=\{0\} \cup \{q^{2n} | n \in {\mathbb{Z}}_{\geq 0} \}\). In particular \(B_0^{\infty}\) is not invertible. Therefore \(\overline{\nabla}_{\bar{\partial}(\frac{B_0}{1-q^2})}\) is not gauge equivalent to the standard \(\bar{\partial}\)-connection \(\overline{\nabla}\), which gives a concrete example of non-standard holomorphic structure on \(\mathcal{L}_n\).
On the other hand, we have the following affirmative result for \(\bar{\partial}\)-connections which are gauge equivalent to the standard ones.
Corollary 4. For any \(f\in \mathcal{A}({\mathbb{C}}P^1_q)\cap C^*(1, B_0)\) with \(\|f\|<\frac{1}{2}\), the \(\bar{\partial}\)-connection \(\overline{\nabla}_{\bar{\partial}f}\) on \(\mathcal{L}_n\) is gauge equivalent to the standard \(\bar{\partial}\)-connection \(\overline{\nabla}\).
Proof. Since \(\|f\|<\frac{1}{2}\), we know that \[|f(q^{2k})-f(q^{2k-2})|<1 \text{ for any }k,\] hence \(f\) cannot have defective spot. ◻
We now turn to the question that when \(\overline{\nabla}_{\bar{\partial}f}\) and \(\overline{\nabla}_{\bar{\partial}h}\) are gauge equivalent for \(f, h \in \mathcal{A}({\mathbb{C}}P^1_q)\cap C^*(1, B_0)\).
This means that the existence of a non-invertible \(g\) such that \(\bar{\partial}g = g \bar{\partial}f + \bar{\partial}h \cdot g\) does not mean that \(f\) and \(h\) must lie in different gauge equivalency classes. However, the existence of such an invertible \(g\) still implies that \(f\) and \(h\) are gauge equivalent.
Lemma 12. For \(g, h \in {\mathrm {Dom}}(\bar{\partial})\cap C^*(1, B_0)\) we have \[\label{eq:32commuting32partials32in32CB0} \bar{\partial}h \cdot g = (m_{q^2}g) \cdot \bar{\partial}h,\qquad{(19)}\] where \(m_{q^2}\) is the dilation map in 13 extended to \(C^*(1, B_0)\) via the functional calculus isomorphism.
Proof. By ?? we get \[\label{eq:32commuting32partial32B0321} \bar{\partial}(B_0^{m+n})=\frac{1-q^{2m+2n}}{1-q^2}B_0^{m+n-1}\bar{\partial}B_0.\tag{19}\] On the other hand we have \[\label{eq:32commuting32partial32B0322}\bar{\partial}(B_0^{m+n})=\bar{\partial}(B_0^m)B_0^n+B_0^m\bar{\partial}(B_0^n)\tag{20}\] where \[B_0^m\bar{\partial}(B_0^n)=B_0^m\frac{1-q^{2n}}{1-q^2}B_0^{n-1}\bar{\partial}B_0=\frac{1-q^{2n}}{1-q^{2m+2n}}\bar{\partial}(B_0^{m+n}).\] Therefore 20 becomes \[\frac{1-q^{2m+2n}}{1-q^{2n}}B_0^m\bar{\partial}(B_0^n)=\bar{\partial}(B_0^m)B_0^n+B_0^m\bar{\partial}(B_0^n)\] hence \[\label{eq:32commuting32partial32B0323} \bar{\partial}(B_0^m)B_0^n=q^{2m}B_0^m\bar{\partial}(B_0^n)=(m_{q^2}B_0^m)\bar{\partial}(B_0^n).\tag{21}\] We proved that ?? holds for any monomials hence for any polynomials. The general case now follows from the fact that \(\bar{\partial}\) is a closed operator and multiplication and \(m_{q^2}\) are bounded operators. ◻
Proposition 32. Let \(f, h \in \mathcal{A}({\mathbb{C}}P^1_q)\cap C^*(1, B_0),\) and write \(S_f, S_h \subset {\mathbb{N}}\) for the sets of defective spots of \(f, h\) respectively. Then, there exists an invertible \(g \in C^*(1, B_0)^{\times}\cap {\mathrm {Dom}}(\bar{\delta})\) such that \[\label{eq:32gauge32equivalence32in32B0} \bar{\partial}g = g \bar{\partial}f -\bar{\partial}h \cdot g\qquad{(20)}\] if and only if \(S_f = S_h.\)
In particular if \(S_f = S_h\), then the two \(\bar{\partial}\)-connections \(\overline{\nabla}_{\bar{\partial}f}\) and \(\overline{\nabla}_{\bar{\partial}h}\) are gauge equivalent.
Proof. The second assertion follows from the first one and Definition 18.
We again use the functional calculus isomorphism \(\Psi\) to identify \(C^*(1, B_0)\) and \(C(sp(B_0))\). Equation ?? then becomes \[\bar{\delta}g = g \bar{\delta}f - (\bar{\delta}h) g.\] By Lemma 12 it becomes \[\label{eq:32gauge32equivalent32dilated} \bar{\delta}g = g \bar{\delta}f - (m_{q^2} g) \bar{\delta}h,\tag{22}\] By Proposition 28 we can write 22 as \[\frac{g(x) - g(q^2x)}{x - q^2x} = g(x) \frac{f(x) - f(q^2x)}{x - q^2x} - g(q^2x) \frac{h(x) - h(q^2x)}{x - q^2x}\] Since \(x - q^2x\) is never zero for \(x \in (0,1],\) this becomes \[g(x) - g(q^2x) = g(x) (f(x) - f(q^2x)) - g(q^2x) (h(x) - h(q^2x))\] hence \[\label{eq:32recursive32eq32for32g} g(q^2x)[1 - h(x) + h(q^2x)] = g(x)[1 - f(x) + f(q^2x)]\tag{23}\] For \(x=q^{2n}\), 23 becomes \[\label{eq:32recursive32eq32for32g322} g(q^{2n+2})[1 - h(q^{2n}) + h(q^{2n+2})] = g(q^{2n})[1 - f(q^{2n}) + f(q^{2n+2})]\tag{24}\]
If \(S_f\neq S_h\), then there must exist an \(n\) such that one of \(1 - f(q^{2n}) + f(q^{2n+2})\) and \(1 - h(q^{2n}) + h(q^{2n+2})\) is zero and the other is non-zero, hence one of \(g(q^{2n})\) and \(g(q^{2n+2})\) must be zero. Therefore \(g\) cannot be invertible.
If \(S_f= S_h\), then \(1 - f(q^{2n}) + f(q^{2n+2})\) and \(1 - h(q^{2n}) + h(q^{2n+2})\) are both zero or both nonzero. If both are nonzero, then we have \[g(q^{2n+2})=g(q^{2n})\frac{1 - f(q^{2n}) + f(q^{2n+2})}{1 - h(q^{2n}) + h(q^{2n+2})}.\] If both are zero, then 24 implies that \(g(q^{2n+2})\) can be any number. We can therefore define \(g\) inductively at any \(q^{2n}\) so that \(g(q^{2n})\neq 0\).
Moreover since \(S_f=S_h\) is a finite set, let \[N= \text{the maximum of }S_f.\] Then for any \(n>N\), \(1 - f(q^{2n}) + f(q^{2n+2})\) and \(1 - h(q^{2n}) + h(q^{2n+2})\) are both nonzero hence \(g(q^{2n})\) is uniquely determined by \(g(q^{2N})\) by \[g(q^{2n}) = g(q^{2N})\prod_{k=N}^{n-1} \frac{1 - f(q^{2k}) + f(q^{2k+2})}{1 - h(q^{2k}) + h(q^{2k+2})} .\]
Therefore \[\label{eq:32infinity32product32quotient} \lim_{n \to \infty} g(q^{2n}) =\prod_{k = N }^{\infty} \frac{1 - f(q^{2k}) + f(q^{2k + 2})}{1 - h(q^{2k}) + h(q^{2k + 2})}\tag{25}\]
Since \(f,h\in {\mathrm {Dom}}(\bar{\delta})\), by the same argument as in the proof of Proposition 29 and Corollary 3, the infinite product on the right hand side of 25 converges with a nonzero limit. Thus, \(g(0)\) exists, and is nonzero. Hence \(g\) is a continuous function which is invertible.
It remains to show that \(g\in {\mathrm {Dom}}(\bar{\delta})\). But this follows from \[\bar{\delta}g = g \bar{\delta}f - \bar{\delta}h \cdot g,\] and the fact that \(\bar{\delta}f, \bar{\delta}h,\) and \(g\) are all continuous. ◻
Remark 33. Notice that Proposition 32 gives a sufficient but not necessary condition: if \(S_f\neq S_h\), we do not know if \(\overline{\nabla}_{\bar{\partial}f}\) and \(\overline{\nabla}_{\bar{\partial}h}\) are gauge equivalent or not. The main reason is that we do not have a generalization of Lemma 5 to solutions of \[\bar{\partial}g=g \bar{\partial}f-\bar{\partial}h g.\]
We will study non-gauge equivalent \(\bar{\partial}\)-connections using a different method in Section 4.
By Corollary 1, for any \(\bar{\partial}\)-connection \(\overline{\nabla}_\theta\) on \(\mathcal{L}_0\) the space of holomorphic sections \(\ker (\overline{\nabla}_\theta)\) is finite dimensional. In this section we look for elements in \(\ker(\overline{\nabla}_\theta)\subset \mathcal{L}_0=\mathcal{A}({\mathbb{C}}P^1_q)\) of the form \(fB_-^n\) for some \(n \in {\mathbb{Z}}_{\geq 0}\), where \(f\in C^*(1, B_0)\).
We first prove the following results:
Lemma 13. For any \(f\in C^*(1,B_0)\) and \(n\in {\mathbb{N}}\) we have \[\label{eq:32commuting32partials32in32CB032and32B-} B_-^n ~f=(m_{q^{2n}}f)B_-^n,\qquad{(21)}\] where \(m_{q^{2n}}\) represents a dilation operator as in 13 .
Proof. Similar to the proof of Lemma 12, by ?? we can check that ?? holds when \(f\) is a polynomial. The general case follows by continuity of multiplications and \(m_{q^{2n}}\). ◻
Lemma 14. For \(f\in {\mathrm {Dom}}(\bar{\partial})\cap C^*(1, B_0)\), \(h\in \mathcal{A}({\mathbb{C}}P^1_q)\cap C^*(1, B_0)\), and \(\theta=\bar{\partial}h\), we have \[\overline{\nabla}_\theta(f B_-^n) = \Big( q^{2n} B_0 \bar{\delta}f + \big(\frac{1-q^{2n}}{1-q^2} - q^{2n} (m_{q^{2n}} \bar{\delta}h) B_0\big) f\Big) B_-^{n-1} \bar{\partial}B_-,\] Here we abuse the notation and denote \(\Psi^{-1}(\bar{\delta}(\Psi f))\) simply by \(\bar{\delta}f\).
Proof. By the definition of \(\overline{\nabla}_\theta\) we get \[\overline{\nabla}_\theta(f B_-^n) = \bar{\partial}(f B_-^n) - (f B_-^n)\theta= \bar{\partial}(f B_-^n) - (f B_-^n)\bar{\partial}h.\] Notice that we are working with \(\mathcal{L}_0\) hence there is no need of \(\Phi_{(n)}\) as in 8 .
By ?? and ?? we further get \[\label{eq:32nb32f32B-321} \begin{align} \overline{\nabla}_\theta(f B_-^n) & = (\bar{\delta}f) \bar{\partial}B_0 B_-^n + f \bar{\partial}(B_-^n) - f B_-^n \bar{\partial}h\\ =& q^2 (\bar{\delta}f) B_0 (\bar{\partial}B_-) B_-^{n-1} + \frac{1-q^{2n}}{1-q^2} f B_-^{n-1} \bar{\partial}B_- - f B_-^n \bar{\partial}h\\ =& q^{2n} (\bar{\delta}f) B_0 B_-^{n-1} \bar{\partial}B_- + \frac{1-q^{2n}}{1-q^2} f B_-^{n-1} \bar{\partial}B_- - f B_-^n \bar{\partial}h. \end{align}\tag{26}\]
We write \(\bar{\partial}h = \bar{\delta}h \bar{\partial}B_0.\) Then we have \[\label{eq:32B-n32pb32h} B_-^n \bar{\partial}h = B_-^n \bar{\delta}h \bar{\partial}B_0 = (m_{q^{2n}} \bar{\delta}h) B_-^n \bar{\partial}B_0.\tag{27}\] By ?? , \(\bar{\partial}B_0\) commutes with \(B_-,\) so the right hand side of 27 becomes \[\label{eq:32mq32db32h} (m_{q^{2n}} \bar{\delta}h) (\bar{\partial}B_0) B_-^n = q^2 (m_{q^{2n}} \bar{\delta}h) B_0 (\bar{\partial}B_-) B_-^{n-1} = q^{2n} (m_{q^{2n}} \bar{\delta}h) B_0 B_-^{n-1} \bar{\partial}B_-,\tag{28}\] 28 together with 26 give \[\label{eq:32nb32f32B-322} \begin{align} \overline{\nabla}_\theta(f B_-^n) =& q^{2n} (\bar{\delta}f) B_0 B_-^{n-1} \bar{\partial}B_- + \frac{1-q^{2n}}{1-q^2} f B_-^{n-1} \bar{\partial}B_- - q^{2n} f (m_{q^{2n}} \bar{\delta}h) B_0 B_-^{n-1} \bar{\partial}B_-\\ =& \Big( q^{2n} (\bar{\delta}f) B_0 + \frac{1-q^{2n}}{1-q^2}f - q^{2n} f (m_{q^{2n}} \bar{\delta}h) B_0 \Big) B_-^{n-1} \bar{\partial}B_- \end{align}\tag{29}\]
Since everything in these large parentheses is in the commutative \(C^*\)-algebra \(C^*(1, B_0),\) we can rewrite 29 as \[\overline{\nabla}_\theta(f B_-^n)=\Big( q^{2n} B_0 \bar{\delta}f + \big( \frac{1-q^{2n}}{1-q^2} - q^{2n} (m_{q^{2n}} \bar{\delta}h) B_0\big) f \Big) B_-^{n-1} \bar{\partial}B_-\] ◻
Corollary 5. Let \(h\in \mathcal{A}({\mathbb{C}}P^1_q)\cap C^*(1, B_0)\) and \(\theta = \bar{\delta}h\). Consider the \(\bar{\partial}\)-connection \(\overline{\nabla}_\theta\) on \(\mathcal{L}_0\). Suppose the defective spot \(S_h\neq \emptyset\). Then for any \(0\leq n< {\mathrm {max}}{S_h}\), there exists an element \(f\in {\mathrm {Dom}}(\bar{\partial})\cap C^*(1, B_0)\) such that \[(\Psi f)(1)\neq 0, \text{ and } f B_-^n\in \ker(\overline{\nabla}_\theta),\] where \(\Psi: C^*(1, B_0)\overset{\sim}{\to}C(sp(B_0))\) is the functional calculus isomorphism
Proof. Again we use \(\Psi\) to identify \(C^*(1, B_0)\) and \(C(sp(B_0))\). By Lemma 14, to find \(f\) such that \(f B_-^n\in \ker(\overline{\nabla}_\theta)\), it is sufficient to find an \(f\in {\mathrm {Dom}}(\bar{\delta})\cap C(sp(B_0))\) such that \[q^{2n} B_0 \bar{\delta}f + \big( \frac{1-q^{2n}}{1-q^2} - q^{2n} (m_{q^{2n}} \bar{\delta}h) B_0\big) f=0,\] i.e. for any \(q^{2k}\in sp(B_0)\), \(k\geq 0\) we have \[\label{uosdrcbq} q^{2k+2n} \frac{f(q^{2k}) - f(q^{2k+2})}{q^{2k} - q^{2k+2}} + \frac{1-q^{2n}}{1-q^2}f(q^{2k}) - q^{2n} \frac{h(q^{2k+2n}) - h(q^{2k+2n+2})}{q^{2k+2n} - q^{2k+2n+2}} q^{2k} f(q^{2k}) = 0\tag{30}\]
From 26 we get \[f(q^{2k+2})=\frac{1-h(q^{2k+2n})+h(q^{2k+2n+2})}{q^{2n}}f(q^{2k}).\] Therefore for any \(m\geq 1\) we have \[\label{eq:32nb32fB-32recursive} f(q^{2m})= f(1) \prod_{k = 1}^{m} \frac{1 - h(q^{2n + 2k-2}) + h(q^{2n+2k })}{q^{2n}} .\tag{31}\]
Since \(n<{\mathrm {max}}{S_h}\), there exists \(m_0> 0\) such that \(n + m_0 \in S_h\). By 31 we have \[f(q^{2m})=0, \text{ for any }m \geq m_0.\] Therefore we can choose \(f(1)\neq 0\) and the function \(f\) defined by 31 is continuous and belongs to \({\mathrm {Dom}}(\bar{\delta})\). Moreover it satisfies \(\overline{\nabla}_\theta(fB_-^n)=0\). ◻
Remark 34. If \(n\geq {\mathrm {max}}{S_n}\), then \[1 - h(q^{2n + 2k-2}) + h(q^{2n+2k })\neq 0 \text{ for any } k>0,\] and \[\lim_{k\to \infty}\big(1 - h(q^{2n + 2k-2}) + h(q^{2n+2k })\big)=1.\] Since \(0<q^{2n}<1\), the \(f(q^{2m})\) defined by 31 diverges unless \(f(1)=0\).
Lemma 15. Let \(n_1,\ldots, n_k\) be distinct nonnegative integers. Then for any \(f_1,\ldots f_k\in C^*(1,B_0)\) such that \((\Psi f_i)(1)\neq 0\) for each \(i\), the elements \(f_1B_-^{n_1},\ldots, f_kB_-^{n_k}\) are linearly independent over \({\mathbb{C}}\).
Proof. Suppose we have \(c_1,\ldots, c_k\in {\mathbb{C}}\) which are not all zeros. Let \(n_s\) be the smallest \(n_i\) such that \(c_i\neq 0\). Recall the faithfull representation \(\pi: \mathcal{C}({\mathbb{C}}P^1_q)\to B(\ell^2)\) in Proposition 14. It is clear that\((\Psi f_s)(1)\neq 0\) implies \(\pi(f_s)(e_0)\neq 0\). Moreover ?? implies \[\pi(B_-^{n_s})\pi(B_+^{n_s})(e_0)=\lambda e_0, \text{ for some }\lambda\neq 0,\] and \[\pi(B_-^{n})\pi(B_+^{n_s})(e_0)=0, \text{ for any }n> n_s.\]
We apply \(\pi(\sum_{i=1}^kc_if_iB_-^{n_i})\) to the vector \(\pi(B_+^{n_s})(e_0)\in \ell^2\) and get \[\begin{align} &\pi(\sum_{i=1}^kc_if_iB_-^{n_i})\pi(B_+^{n_s})(e_0)=\\ =& c_s\lambda (\pi (f_s))(e_0)+\sum_{n_i>n_s}c_i(\pi (f_i)) \pi(B_-^{n_i})\pi(B_+^{n_s})(e_0)\\ =& c_s \lambda (\pi (f_s))(e_0)+0=c_s \lambda (\pi (f_s))(e_0)\neq 0. \end{align}\] So we have \(\sum_{i=1}^kc_if_iB_-^{n_i}\neq 0\). ◻
Corollary 6. Let \(h\in \mathcal{A}({\mathbb{C}}P^1_q)\cap C^*(1, B_0)\) and \(N\) be the maximal element in \(S_h\). Then for \(\theta = \bar{\partial}h\) and the \(\bar{\partial}\)-connection \(\overline{\nabla}_\theta\) on \(\mathcal{L}_0\), we have \[\dim(\ker(\overline{\nabla}_\theta)) \geq N.\]
Remark 35. If we want to extend the result to \(B_+\), then we notice that we have an analogue of Lemma 15 for \(B_+^{n_1}f_1,\ldots, B_+^{n_k}f_k\) instead of \(f_1B_+^{n_1},\ldots, f_kB_+^{n_k}\).
However, a careful computation shows \[\overline{\nabla}_\theta(B_+^n f) = B_+^{n-1} \bar{\partial}B_+ \Big( \frac{1-q^{2n}}{1-q^2} f + (q^{-4} B_0 - q^{-2}) \big((m_{q^{-2}} \bar{\delta}f) - (m_{q^{-2}} f)(m_{q^{-2}} \bar{\delta}h)\big) \Big).\] Since \(q^{-2} > 1,\) this dilation operator \(m_{q^{-2}}\) is unbounded, so we cannot use functional calculus to solve this equation.
The following theorem is the main result of this paper:
Theorem 5. There exist infinitely many gauge equivalent classes of holomorphic structures on \(\mathcal{L}_0\), hence on \(\mathcal{L}_n\).
Proof. We know that the element \[h=\frac{B_0}{q^{2N-2}-q^{2N}}\] has \(S_h=\{N\}\). Therefore by Corollary 6, for any \(N\), we can find a \(\bar{\partial}\)-connection \(\overline{\nabla}_\theta\) on \(\mathcal{L}_0\) such that \[\dim(\ker(\overline{\nabla}_\theta)) \geq N.\]
On the other hand, by Corollary 1, \(\ker(\overline{\nabla}_\theta)\) is finite dimensional for any \(\bar{\partial}\)-connection \(\overline{\nabla}_\theta\) on \(\mathcal{L}_0\). Since the dimension of \(\ker(\overline{\nabla}_\theta)\) is invariant under gauge equivalence, there exist infinitely many gauge equivalent classes of holomorphic structures on \(\mathcal{L}_0\).
The \(\mathcal{L}_n\) case follows from the \(\mathcal{L}_0\) case and Proposition 21. ◻
Remark 36. It is a classical result that on commutative \({\mathbb{C}}P^1\), there exists a unique holomorphic structure up to gauge equivalence on each \(\mathcal{O}(n)\). Therefore the existence of infinitely many holomorphic structures in Theorem 5 is a new phenomenon in noncommutative geometry which has no counterpart in the commutative world.
Note that Theorem 5 does not provide a classification of the gauge equivalence classes of holomorphic structures on \(\mathcal{L}_n\) over \({\mathbb{C}}P^1_q\). It would be interesting to classify and parametrize all such gauge equivalence classes, that is, to determine the Picard group of the quantum projective line \({\mathbb{C}}P^1_q\).
We also notice that higher dimensional quantum projective spaces \({\mathbb{C}}P^l_q\) and line bundles over them were introduced and studied in [9]. It is interesting to study non-standard holomorphic structures on line bundles over \({\mathbb{C}}P^l_q\). The analysis will be more involved in higher dimensional case as the flatness condition \(\overline{\nabla}_\theta\circ \overline{\nabla}_\theta=0\) does not hold automatically on \({\mathbb{C}}P^l_q\) for \(l\geq 2\).