Rank of a co-doubly commuting submodule is 2

TL;DR

This paper proves that non-trivial co-doubly commuting submodules of the Hardy space over the bidisk have rank 2, providing a precise characterization and answering a question by Douglas and Yang.

Contribution

It establishes the exact rank of co-doubly commuting submodules and characterizes rank-one submodules as those generated by one-variable inner functions.

Findings

Rank of non-trivial co-doubly commuting submodules is 2.

Rank-one co-doubly commuting submodules are generated by one-variable inner functions.

Answers a question posed by Douglas and Yang regarding submodule ranks.

Abstract

We prove that the rank of a non-trivial co-doubly commuting submodule is $2$. More precisely, let $\varphi, \psi \in H^\infty(\mathbb{D})$ be two inner functions. If $\mathcal{Q}_{\varphi} = H^2(\mathbb{D})/ \varphi H^2(\mathbb{D})$ and $\mathcal{Q}_{\psi} = H^2(\mathbb{D})/ \psi H^2(\mathbb{D})$, then \[ \mbox{rank~}(\mathcal{Q}_{\varphi} \otimes \mathcal{Q}_{\psi})^\perp = 2. \] An immediate consequence is the following: Let $\mathcal{S}$ be a co-doubly commuting submodule of $H^2(\mathbb{D}^2)$. Then $\mbox{rank~} \mathcal{S} = 1$ if and only if $\mathcal{S} = \Phi H^2(\mathbb{D}^2)$ for some one variable inner function $\Phi \in H^\infty(\mathbb{D}^2)$. This answers a question posed by R. G. Douglas and R. Yang.

Equations

$$\text{mbox}rank({\mathcal{Q}}_{\phi }\otimes {\mathcal{Q}}_{\psi }{)}^{\perp }=2.$$

$$\text{mbox}rank(T)=\min \{\#E:[E{]}_T=\mathcal{H},E\subseteq \mathcal{H}\}.$$

$$\text{mbox}rank\mathcal{S}=\text{mbox}rank(M_{z_1}{|}_{\mathcal{S}},\dots ,M_{z_n}{|}_{\mathcal{S}}).$$

$$M_z{M}_w=M_w{M}_z,M_z^{\ast }{M}_z=M_w^{\ast }{M}_w=I_{H^2({\mathbb{D}}^2)}.$$

$$\mathcal{S}={\mathcal{S}}_{\phi }:=\phi H^2(\mathbb{D}),$$

$$\mathcal{Q}={\mathcal{Q}}_{\phi }:=({\mathcal{S}}_{\phi }{)}^{\perp }=H^2(\mathbb{D})/\phi H^2(\mathbb{D}).$$

$$\text{mbox}rank(M_z{|}_{{\mathcal{S}}_{\phi }})=\text{mbox}rank(P_{{\mathcal{Q}}_{\phi }}{M}_z{|}_{{\mathcal{Q}}_{\phi }})=1.$$

$$\mathcal{Q}={\mathcal{Q}}_1\otimes {\mathcal{Q}}_2,$$

$$\text{mbox}rank\mathcal{S}=1.$$

$$\text{mbox}rank({\mathcal{Q}}_{\phi }\otimes {\mathcal{Q}}_{\psi }{)}^{\perp }=2.$$

$$\text{mbox}rank(P_{\mathcal{S}}{T}_1{|}_{\mathcal{S}},\dots ,P_{\mathcal{S}}{T}_n{|}_{\mathcal{S}})\le \text{mbox}rank(T_1{|}_{{\mathcal{S}}_1},\dots ,T_n{|}_{{\mathcal{S}}_1}).$$

$$(P_{\mathcal{S}}{T}_i{P}_{\mathcal{S}})(P_{\mathcal{S}}{T}_j{P}_{\mathcal{S}})=P_{\mathcal{S}}(T_i{T}_j){|}_{{\mathcal{S}}_1}(i,j=1,\dots ,n).$$

$$\mathcal{S}={\left({\mathcal{Q}}_{\phi }\otimes {\mathcal{Q}}_{\psi }\right)}^{\perp },$$

$$\mathcal{S}=\text{mbox}ranX,$$

$$X=((M_{\phi }{M}_{\phi }^{\ast })\otimes (I_{H^2(\mathbb{D})}-M_{\psi }{M}_{\psi }^{\ast }))\oplus (I_{H^2(\mathbb{D})}\otimes M_{\psi }{M}_{\psi }^{\ast }),$$

$$\mathcal{S}=\left({\mathcal{S}}_{\phi }\otimes {\mathcal{Q}}_{\psi }\right)\oplus \left(H^2(\mathbb{D})\otimes {\mathcal{S}}_{\psi }\right).$$

$$\mathcal{E}=\mathcal{S}\ominus \left({\mathcal{S}}_{\phi }\otimes {\mathcal{S}}_{\psi }\right).$$

$$\mathcal{E}=\left({\mathcal{S}}_{\phi }\otimes {\mathcal{Q}}_{\psi }\right)\oplus \left({\mathcal{Q}}_{\phi }\otimes {\mathcal{S}}_{\psi }\right).$$

$$\text{rank}(P_{\mathcal{E}}{M}_z{|}_{\mathcal{E}},P_{\mathcal{E}}{M}_w{|}_{\mathcal{E}})\le \text{rank}(M_z{|}_{\mathcal{S}},M_w{|}_{\mathcal{S}})=\text{rank}(\mathcal{S}).$$

$$P_{\mathcal{E}}=(P_{{\mathcal{S}}_{\phi }}\otimes P_{{\mathcal{Q}}_{\psi }})\oplus (P_{{\mathcal{Q}}_{\phi }}\otimes P_{{\mathcal{S}}_{\psi }}).$$

$$P_{\mathcal{E}}{M}_z{|}_{\mathcal{E}}=(M_z{|}_{{\mathcal{S}}_{\phi }}\otimes P_{{\mathcal{Q}}_{\psi }})\oplus (P_{{\mathcal{Q}}_{\phi }}{M}_z{|}_{{\mathcal{Q}}_{\phi }}\otimes P_{{\mathcal{S}}_{\psi }}),$$

$$P_{\mathcal{E}}{M}_w{|}_{\mathcal{E}}=(P_{{\mathcal{S}}_{\phi }}\otimes P_{{\mathcal{Q}}_{\psi }}{M}_w{|}_{{\mathcal{Q}}_{\psi }})\oplus (P_{{\mathcal{Q}}_{\phi }}\otimes M_w{|}_{{\mathcal{S}}_{\psi }}).$$

$$\tilde{\mathcal{E}}=\mathcal{E}\ominus (({\mathcal{S}}_{{\phi }^2}\otimes {\mathcal{Q}}_{\psi })\oplus ({\mathcal{Q}}_{\phi }\otimes {\mathcal{S}}_{{\psi }^2})).$$

$${\mathcal{S}}_{\theta }\ominus {\mathcal{S}}_{{\theta }^2}=\theta {\mathcal{Q}}_{\theta }.$$

$$\begin{array}{rl}\tilde{\mathcal{E}}& =(({\mathcal{S}}_{\phi }\otimes {\mathcal{Q}}_{\psi })\oplus ({\mathcal{Q}}_{\phi }\otimes {\mathcal{S}}_{\psi }))\ominus (({\mathcal{S}}_{{\phi }^2}\otimes {\mathcal{Q}}_{\psi })\oplus ({\mathcal{Q}}_{\phi }\otimes {\mathcal{S}}_{{\psi }^2}))\\ & =(\phi {\mathcal{Q}}_{\phi }\otimes {\mathcal{Q}}_{\psi })\oplus ({\mathcal{Q}}_{\phi }\otimes \psi {\mathcal{Q}}_{\psi }).\end{array}$$

$$\text{rank}(P_{\tilde{\mathcal{E}}}{M}_z{|}_{\tilde{\mathcal{E}}},P_{\tilde{\mathcal{E}}}{M}_w{|}_{\tilde{\mathcal{E}}})\le \text{rank}(P_{\mathcal{E}}{M}_z{|}_{\mathcal{E}},P_{\mathcal{E}}{M}_w{|}_{\mathcal{E}})\le \text{rank}(\mathcal{S})\le 2.$$

$$\text{rank}(P_{\tilde{\mathcal{E}}}{M}_z{|}_{\tilde{\mathcal{E}}},P_{\tilde{\mathcal{E}}}{M}_w{|}_{\tilde{\mathcal{E}}})>1.$$

$$⟨(z^p\otimes w^q)\xi ,{\eta }_{\xi }⟩=0(p,q\in \mathbb{N}).$$

$$\xi =(\sum _{k,l}{a}_{kl}\phi f_k\otimes g_l)\oplus (\sum _{k,l}{b}_{kl}{f}_k\otimes \psi g_l),$$

$$\sum _{k,l}|a_{kl}{|}^2,\sum _{k,l}|b_{kl}{|}^2<\infty .$$

Full text

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An immediate consequence is the following: Let $\mathcal{S}$ be a co-doubly commuting submodule of $H^{2}(\mathbb{D}^{2})$. Then $\mbox{rank~{}}\mathcal{S}=1$ if and only if $\mathcal{S}=\Phi H^{2}(\mathbb{D}^{2})$ for some one variable inner function $\Phi\in H^{\infty}(\mathbb{D}^{2})$. This answers a question posed by R. G. Douglas and R. Yang [4].

1. Introduction

Let $T=(T_{1},\ldots,T_{n})$ be an $n$-tuple of commuting bounded linear operators on a Hilbert space $\mathcal{H}$. For a subset $E\subseteq\mathcal{H}$ we denote $[E]_{T}$ by the close subspace $\overline{\mbox{span}}\{T_{1}^{k_{1}}\cdots T_{n}^{k_{n}}E:k_{j}\in\mathbb{N},j=1,\ldots,n\}$ of $\mathcal{H}$. Then the rank of $T$ [3] is the unique number

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A closed subspace $\mathcal{S}$ of $H^{2}(\mathbb{D}^{n})$, the Hardy space over the unit polydisc $\mathbb{D}^{n}$, is said to be shift invariant if $M_{z_{i}}(\mathcal{S})\subseteq\mathcal{S}$ for $i=1,2,\ldots,n$, where $M_{z_{i}}$ is the co-ordinate multiplication operator on $H^{2}(\mathbb{D}^{n})$. The rank of a shift invariant subspace $\mathcal{S}$ of $H^{2}(\mathbb{D}^{n})$ is the rank of the corresponding $n$-tuple of restricted co-ordinate shift operators, that is

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The rank of a bounded linear operator (or, of a commuting tuple of bounded linear operators) on a Hilbert space is an important numerical invariant. Very briefly, the rank of a bounded linear operator is the cardinality of a minimal generating set (see the definition below). One of the most intriguing and important problems in operator theory and function theory is the existence of a finite generating set for a commuting tuple of operators. Alternatively, one may ask when the rank of a commuting tuple of operators is finite.

Prototype examples of rank one operators are the co-ordinate multiplication operator tuple $(M_{z_{1}},\ldots,M_{z_{n}})$ on the Hardy space, the (weighted) Bergman space over the unit ball and the polydisc in $\mathbb{C}^{n}$, $n\geq 1$, and the Drury-Arveson space over the unit ball in $\mathbb{C}^{n}$. Moreover, a particular version of the celebrated invariant subspace theorem of Beurling says: A shift invariant (or, shift co-invariant) subspace of the one variable Hardy space is of rank one.

Computation of ranks of shift invariant as well as shift co-invariant subspaces beyond the case of the one variable Hardy space is an excruciatingly difficult problem, even if one considers only shift invariant (as well as co-invariant) subspaces of the Hardy space over the unit polydisc in $\mathbb{C}^{n}$, $n>1$ (see however [2, 6, 7, 8, 14]).

The purpose of this paper is to compute the rank of a tractable class of shift invariant subspaces of the two variables Hardy space, $H^{2}(\mathbb{D}^{2})$, over the bidisc $\mathbb{D}^{2}$ in $\mathbb{C}^{2}$. In order to state the precise contribution of this paper, we need to introduce first some definitions and notations.

We denote the open unit disc of $\mathbb{C}$ by $\mathbb{D}$, and the unit circle by $\mathbb{T}$. The Hardy space over the unit disc $\mathbb{D}$ (bidisc $\mathbb{D}^{2}$), denoted by $H^{2}(\mathbb{D})$ ($H^{2}(\mathbb{D}^{2})$), is the Hilbert space of all square summable holomorphic functions on $\mathbb{D}$ (on $\mathbb{D}^{2}$). Also we will denote by $M_{z}$ and $M_{w}$ the multiplication operators on $H^{2}(\mathbb{D}^{2})$ by the coordinate functions $z$ and $w$, respectively. It is easy to see that $(M_{z},M_{w})$ is a pair of commuting isometries, that is,

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Identifying $H^{2}(\mathbb{D}^{2})$ with the $2$-fold Hilbert space tensor product $H^{2}(\mathbb{D})\otimes H^{2}(\mathbb{D})$, one can represent $(M_{z},M_{w})$ as $(M_{z}\otimes I_{H^{2}(\mathbb{D})},I_{H^{2}(\mathbb{D})}\otimes M_{w})$.

Let $\mathcal{S}$ and $\mathcal{Q}$ be closed subspaces of $H^{2}(\mathbb{D}^{2})$. Then $\mathcal{S}$ is said to be a submodule if $M_{z}(\mathcal{S})\subseteq\mathcal{S}$ and $M_{w}(\mathcal{S})\subseteq\mathcal{S}$. We say that $\mathcal{Q}$ is a quotient module if $\mathcal{Q}^{\perp}$ is a submodule.

A well-known result due to Beurling states that if $\mathcal{S}$ is a submodule of $H^{2}(\mathbb{D})$ (that is, $\mathcal{S}$ is closed subspace of $H^{2}(\mathbb{D})$ and $M_{z}\mathcal{S}\subseteq\mathcal{S}$), then $\mathcal{S}$ can be represented as

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where $\varphi\in H^{\infty}(\mathbb{D})$ is an inner function (that is, $\varphi$ is a bounded holomorphic function on $\mathbb{D}$ and $|\varphi|=1$ a.e. on $\mathbb{T}$). Consequently, a quotient module $\mathcal{Q}$ (that is, $\mathcal{Q}$ is a closed subspace of $H^{2}(\mathbb{D})$ and $M_{z}^{*}\mathcal{Q}\subseteq\mathcal{Q}$) of $H^{2}(\mathbb{D})$ can be represented as

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It readily follows that

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Rudin [10], however, pointed out that there exists a submodule $\mathcal{S}$ of $H^{2}(\mathbb{D}^{2})$ such that the rank of $\mathcal{S}$ is not finite (see also [7], [12] and [13]).

A quotient module $\mathcal{Q}$ of $H^{2}(\mathbb{D}^{2})$ is doubly commuting if $C_{z}C_{w}^{*}=C_{w}^{*}C_{z}$, where $C_{z}=P_{\mathcal{Q}}M_{z}|_{\mathcal{Q}}$ and $C_{w}=P_{\mathcal{Q}}M_{w}|_{\mathcal{Q}}$. A submodule $\mathcal{S}$ of $H^{2}(\mathbb{D}^{2})$ is co-doubly commuting if the quotient module $\mathcal{S}^{\perp}(\cong H^{2}(\mathbb{D}^{2})/\mathcal{S})$ is doubly commuting.

The following useful characterization of co-doubly commuting submodules is essential for our study (see [9, 11]): If $\mathcal{Q}$ is a quotient module of $H^{2}(\mathbb{D}^{2})$, then $\mathcal{Q}$ is a doubly commuting quotient module if and only if

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for some quotient modules $\mathcal{Q}_{1}$ and $\mathcal{Q}_{2}$ of $H^{2}(\mathbb{D})$.

Let $\mathcal{S}=(\mathcal{Q}_{1}\otimes\mathcal{Q}_{2})^{\perp}$ be a non-zero co-doubly commuting submodule. If $\mathcal{Q}_{j}=H^{2}(\mathbb{D})$, for some $j=1,2$, then it is easy to see that

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Now let both $\mathcal{Q}_{1}$ and $\mathcal{Q}_{2}$ be non-trivial quotient modules of $H^{2}(\mathbb{D})$, that is, $\mathcal{Q}_{j}\neq\{0\},H^{2}(\mathbb{D})$, $j=1,2$. Then there exist inner functions $\varphi,\psi\in H^{\infty}(\mathbb{D})$ such that $\mathcal{Q}_{1}=\mathcal{Q}_{\varphi}$ and $\mathcal{Q}_{2}=\mathcal{Q}_{\psi}$. The main purpose of the present paper is to prove that (see Theorem 2.1)

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As a consequence of this, we give a complete and affirmative answer to a conjecture of Douglas and Yang (see page 220 [4]): If $\mathcal{S}$ is a rank one co-doubly commuting submodule, then $\mathcal{S}=\Phi H^{2}(\mathbb{D}^{2})$ for some one variable inner function $\Phi\in H^{\infty}(\mathbb{D})$.

2. Proof of the main result

We begin with a simple but crucial observation on the rank of a joint semi-invariant subspace of a commuting tuple of operators.

Lemma 2.1*.*

Let $T=(T_{1},\ldots,T_{n})$ be an $n$-tuple of commuting operators on a Hilbert space $\mathcal{H}$. Let $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ be two joint $T$-invariant subspaces of $\mathcal{H}$ and $\mathcal{S}_{2}\subseteq\mathcal{S}_{1}$. If $\mathcal{S}=\mathcal{S}_{1}\ominus\mathcal{S}_{2}$, then

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Proof.

Let $m\in\mathbb{N}$ be the right side of the above inequality. Let $\{f_{j}\}_{j=1}^{m}\subseteq\mathcal{S}_{1}$ be a generating set for $(T_{1}|_{\mathcal{S}_{1}},\ldots,T_{n}|_{\mathcal{S}_{1}})$. Clearly, $P_{\mathcal{S}}T_{j}P_{\mathcal{S}}=P_{\mathcal{S}}T_{j}|_{\mathcal{S}_{1}}$ for all $j=1,\ldots,n$. This yields

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It hence follows that $\{P_{\mathcal{S}}f_{j}\}_{j=1}^{m}$ is a generating set for $(P_{\mathcal{S}}T_{1}|_{\mathcal{S}},\ldots,P_{\mathcal{S}}T_{n}|_{\mathcal{S}})$. This completes the proof. ∎

We now prove the main result of this paper.

Theorem 2.1**.**

Let $\varphi,{\psi}\in H^{\infty}(\mathbb{D})$ be two inner functions. If

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then $\mbox{rank~{}}\mathcal{S}=2$.

Proof.

Let $X=I_{H^{2}(\mathbb{D}^{2})}-(I_{H^{2}(\mathbb{D}^{2})}-M_{\varphi}M_{\varphi}^{*}\otimes I_{H^{2}(\mathbb{D})})(I_{H^{2}(\mathbb{D}^{2})}-I_{H^{2}(\mathbb{D})}\otimes M_{\psi}M_{\psi}^{*})$. Since

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and

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it follows that

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Since by Theorem 6.2 of [1], $\mbox{rank~{}}\mathcal{S}\leq 2$, we only need to show that $\mbox{rank~{}}\mathcal{S}\geq 2$. Set

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It follows that

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Since $\mathcal{S}_{\varphi}\otimes\mathcal{S}_{{\psi}}\subseteq\mathcal{S}$ is a submodule of $H^{2}(\mathbb{D}^{2})$, by Lemma 2.1, it follows that

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Note that

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and hence, an easy calculation yields

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and

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Therefore it follows from the above equalities that $(\mathcal{S}_{\varphi^{2}}\otimes\mathcal{Q}_{\psi})\oplus(\mathcal{Q}_{\varphi}\otimes\mathcal{S}_{{\psi}^{2}})$ is a joint $(P_{\mathcal{E}}M_{z}|_{\mathcal{E}},P_{\mathcal{E}}M_{w}|_{\mathcal{E}})$ invariant subspace of $\mathcal{E}$. Set

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Notice that for any inner function $\theta\in H^{\infty}(\mathbb{D})$, we have

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From this and the representation of $\mathcal{E}=(\mathcal{S}_{\varphi}\otimes\mathcal{Q}_{{\psi}})\oplus(\mathcal{Q}_{\varphi}\otimes\mathcal{S}_{{\psi}})$ it follows that

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Then Lemma 2.1 and (2.1) implies that

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To finish the proof of the theorem it is now enough to prove the following:

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Equivalently, it is enough to prove that the set $\{\xi\}$, for any $\xi\in\tilde{\mathcal{E}}$, is not a generating set corresponding to $(P_{\tilde{\mathcal{E}}}M_{z}|_{\tilde{\mathcal{E}}},P_{\tilde{\mathcal{E}}}M_{w}|_{\tilde{\mathcal{E}}})$. Equivalently, given $\xi\in\tilde{\mathcal{E}}$, we show that there exists $\eta_{\xi}(\neq 0)\in\tilde{\mathcal{E}}$ such that

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To this end, let $\{f_{i}\}$ and $\{g_{j}\}$ be orthonormal bases of $\mathcal{Q}_{\varphi}$ and $\mathcal{Q}_{\psi}$, respectively, and let $\xi\in\tilde{\mathcal{E}}$ where

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$\{a_{kl}\},\{b_{kl}\}\subseteq\mathbb{C}$, and

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Again we observe that for any inner function $\theta\in H^{\infty}(\mathbb{D})$ and $f=\sum_{m\geq 0}c_{m}z^{m}\in\mathcal{Q}_{\theta}$ we have

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where $\bar{f}=\sum_{m\geq 0}\bar{c}_{m}e^{-imt}\in L^{2}(\mathbb{T})$. This follows from the fact that $\theta$ is a bounded holomorphic function on $\mathbb{D}$ and $M_{z}^{*}(\theta\bar{f})\perp z^{m}$ for all $m<0$ (which gives that $M_{z}^{*}(\theta\bar{f})\in H^{2}(\mathbb{D})$), and then $M_{z}^{*}(\theta\bar{f})\perp\theta z^{m}$ in $L^{2}(\mathbb{T})$ for all $m\geq 0$ (which gives that $M_{z}^{*}(\theta\bar{f})\in\mathcal{Q}_{\theta}$). It should be noted that $M_{z}^{*}(\theta\bar{f})=\theta\overline{zf}=C_{\theta}(f)$, where the conjugation map $C_{\theta}:\mathcal{Q}_{\theta}\to\mathcal{Q}_{\theta}$, $f\mapsto M_{z}^{*}(\theta\bar{f})$, is called a $C$-symmetry and it is used extensively in the study of Toeplitz operators on model spaces (for more details see [5]).

Coming back to our context, this immediately yields that

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and hence $s_{0}\otimes s_{1},t_{0}\otimes t_{1}\in\mathcal{Q}_{\varphi}\otimes\mathcal{Q}_{\psi}$, where

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and

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Set

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Then $\eta_{\xi}\neq 0$ and for every $p,q\in\mathbb{N}$ we have

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We have thus shown that $\{\xi\}$ is not a minimal generating subset of $\tilde{\mathcal{E}}$ with respect to $(P_{\tilde{\mathcal{E}}}M_{z}|_{\tilde{\mathcal{E}}},P_{\tilde{\mathcal{E}}}M_{w}|_{\tilde{\mathcal{E}}})$ as desired. ∎

As a consequence of the above theorem we have the following corollary which provides an affirmative answer of the question raised by Douglas and Yang [4].

Corollary 2.2*.*

Let $\mathcal{S}$ be a co-doubly commuting submodule of $H^{2}(\mathbb{D}^{2})$. Then rank $(\mathcal{S})=1$ if and only if $\mathcal{S}=\Theta H^{2}(\mathbb{D}^{2})$ for some one variable inner function $\Theta\in H^{\infty}(\mathbb{D})$.

Proof.

If $\mathcal{S}=\Theta H^{2}(\mathbb{D}^{2})$ for some one variable inner function $\Theta\in H^{\infty}(\mathbb{D})$, then $\mathcal{S}\cong H^{2}(\mathbb{D}^{2})$ and hence rank $\mathcal{S}=1$. To prove the the sufficient part let $\mathcal{S}$ be a rank one co-doubly commuting submodule of $H^{2}(\mathbb{D}^{2})$. Then there exist quotient modules $\mathcal{Q}_{1}$ and $\mathcal{Q}_{2}$ of $H^{2}(\mathbb{D})$ such that (see [9, 11])

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Since rank $(\mathcal{S})=1$, it follows from Theorem 2.1 that $\mathcal{Q}_{j}=H^{2}(\mathbb{D})$, for some $j=1,2$. This shows that

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for some inner functions $\varphi,\psi\in H^{\infty}(\mathbb{D})$. This concludes the proof of the corollary. ∎

There is now the following interesting and natural question: Let $m\geq 2$ and let $\{\varphi_{j}\}_{j=1}^{m}\subseteq H^{\infty}(\mathbb{D})$ be inner functions. Is then

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Our present approach does not seem to work for $m>2$ case.

Acknowledgement: The first named author acknowledge Fulbright-Nehru Postdoctoral Research Fellowship (Award No. 2164/FNPDR/2016) and University of New Mexico for warm hospitality. The second author’s research work is supported by DST-INSPIRE Faculty Fellowship No. DST/INSPIRE/04/2015/001094. The research of the third author is supported in part by NBHM (National Board of Higher Mathematics, India) Research Grant NBHM/R.P.64/2014.