Generalized analogs of the Heisenberg uncertainty inequality

Abstract

We investigate locally compact topological groups for which a generalized analog of the Heisenberg uncertainty inequality hold. In particular, it is shown that this inequality holds for $\mathbb{R}^{n} \times K$ (where K is a separable unimodular locally compact group of type I), Euclidean motion group and several general classes of nilpotent Lie groups which include thread-like nilpotent Lie groups, 2-NPC nilpotent Lie groups and several low-dimensional nilpotent Lie groups.

1 Introduction

In 1927, Heisenberg presented a principle related to the uncertainties in the measurements of position and momentum of microscopic particles. This principle is known as Heisenberg uncertainty principle and can be stated as follows:

It is impossible to know simultaneously the exact position and momentum of a particle. That is, the more exactly the position is determined, the less known the momentum, and vice versa.

In 1933, Wiener gave the following mathematical formulation of the Heisenberg uncertainty principle:

A nonzero function and its Fourier transform cannot both be sharply localized.

Heisenberg’s uncertainty inequality is a precise quantitative formulation of the above principle.

The Fourier transform of $f \in L^{1}(\mathbb{R}^{n})$ is given by

$$ \hat{f}(\xi) =\int_{\mathbb{R}^{n}}{f(x) e^{-2\pi i \langle {x,\xi}\rangle}}\, dx, $$

where $\langle{\cdot,\cdot}\rangle$ denotes the usual inner product on $\mathbb{R}^{n}$. This definition of Fourier transform holds for functions in $L^{1}(\mathbb{R}^{n}) \cap L^{2}(\mathbb{R}^{n})$. Since $L^{1}(\mathbb{R}^{n}) \cap L^{2}(\mathbb{R}^{n})$ is dense in $L^{2}(\mathbb{R}^{n})$, the definition of Fourier transform can be extended to the functions in $L^{2}(\mathbb{R}^{n})$.

The following theorem gives the Heisenberg uncertainty inequality for the Fourier transform on $\mathbb{R}^{n}$. For a proof of the theorem, see [1].

Theorem 1.1

For any $f \in L^{2}(\mathbb{R}^{n})$, we have

$$ \frac{n\|f\|_{2}^{2}}{4\pi} \leq \biggl(\int_{\mathbb {R}^{n}}{\|x\| ^{2} \bigl\vert f(x)\bigr\vert ^{2}}\, dx \biggr)^{1/2} \biggl( \int_{\mathbb{R}^{n}}{\| y\|^{2} \bigl\vert \hat{f}(y)\bigr\vert ^{2}}\, dy \biggr)^{1/2}, $$

(1.1)

where $\|\cdot\|_{2}$ denotes the $L^{2}$-norm and $\|\cdot\|$ denotes the Euclidean norm.

The Heisenberg uncertainty inequality has been established for the Fourier transform on the Heisenberg group by Thangavelu [2]. Further generalizations of the inequality on the Heisenberg group have been established by Sitaram et al. [3] and Xiao and He [4]. For some more details, see [1].

The inequality given below can be proved using Hölder’s inequality and the inequality (1.1).

Theorem 1.2

For any $f \in L^{2}(\mathbb{R}^{n})$ and $a,b \geq1$, we have

$$ \frac{n\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq \biggl(\int_{\mathbb{R}^{n}}{\|x\|^{2a} \bigl\vert f(x)\bigr\vert ^{2}} \, dx \biggr)^{\frac {1}{2a}} \biggl(\int_{\mathbb{R}^{n}}{\|y\|^{2b} \bigl\vert \hat{f}(y) \bigr\vert ^{2}}\, dy \biggr)^{\frac{1}{2b}}, $$

where $\|\cdot\|_{2}$ denotes the $L^{2}$-norm and $\|\cdot\|$ denotes the Euclidean norm.

In Section 2, we shall prove a generalized analog of the Heisenberg uncertainty inequality for $\mathbb{R}^{n} \times K$, where K is a separable unimodular locally compact group of type I. In the next section, a generalized analog of the Heisenberg uncertainty inequality for the Euclidean motion group $M(n)$ is proved. The last section deals with a generalized analog of the Heisenberg uncertainty inequality for several general classes of nilpotent Lie groups for which the Hilbert-Schmidt norm of the group Fourier transform $\pi_{\xi}(f)$ of f attains a particular form. These classes include thread-like nilpotent Lie groups, 2-NPC nilpotent Lie groups and several low-dimensional nilpotent Lie groups.

2 $\mathbb{R}^{n} \times K$, K a locally compact group

Consider $G=\mathbb{R}^{n} \times K$, where K is a separable unimodular locally compact group of type I. The Haar measure of G is $dg=dx \,dk$, where dx is the Lebesgue measure on $\mathbb{R}^{n}$ and dk is the left Haar measure on K. The dual $\widehat{G}$ of G is $\mathbb{R}^{n} \times \widehat{K}$, where $\widehat{K}$ is the dual space of K.

The Fourier transform of $f \in L^{2}(G)$ is given by

$$ \hat{f}(y,\sigma) =\int_{\mathbb{R}^{n}}\int_{K}{f(x,k) e^{-2\pi i \langle{x,y}\rangle} \sigma\bigl(k^{-1}\bigr)} \,dk \,dx, $$

for $(y,\sigma) \in\mathbb{R}^{n} \times\widehat{K}$.

Theorem 2.1

For any $f \in L^{2}(\mathbb{R}^{n} \times K)$ (where K is a separable unimodular locally compact group of type I) and $a,b \geq1$, we have

$$\begin{aligned} \frac{n\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq& \biggl(\int_{\mathbb{R}^{n}}\int _{K}{\|x\|^{2a} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \biggr)^{\frac{1}{2a}} \\ &{}\times \biggl(\int_{\mathbb {R}^{n}} \int_{\widehat{K}}{\|y\|^{2b} \bigl\Vert \hat{f}(y,\sigma) \bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma \biggr)^{\frac{1}{2b}}. \end{aligned}$$

(2.1)

Proof

Without loss of generality, we may assume that both integrals on the right-hand side of (2.1) are finite.

Given that $f \in L^{2}(\mathbb{R}^{n} \times K)$, there exists $A \subseteq K$ of measure zero such that for $k \in K\setminus A=A'$ (say), we have

$$ \int_{\mathbb{R}^{n}}{\bigl\vert f(x,k)\bigr\vert ^{2}} \,dx < \infty. $$

For all $k \in A'$, we define $f_{k}(x)=f(x,k)$, for every $x\in\mathbb{R}^{n}$.

Clearly, for all $k\in A'$, $f_{k} \in L^{2}(\mathbb{R}^{n})$, and for all $y \in\mathbb{R}^{n}$,

$$ \hat{f}_{k}(y) =\int_{\mathbb{R}^{n}}{f(x,k) e^{-2\pi i \langle {x,y}\rangle}} \,dy=\mathscr{F}_{1} f(y,k). $$

By Theorem 1.1, we have

$$ \frac{n}{4\pi}\int_{\mathbb{R}^{n}}{\bigl\vert f(x,k)\bigr\vert ^{2}} \,dx \leq \biggl(\int_{\mathbb{R}^{n}}{\|x \|^{2} \bigl\vert f_{k}(x)\bigr\vert ^{2}} \,dx \biggr)^{1/2} \biggl(\int_{\mathbb{R}^{n}}{\| y\|^{2} \bigl\vert \hat{f}_{k}(y)\bigr\vert ^{2}} \,dy \biggr)^{1/2}. $$

Integrating both sides with respect to dk, we obtain

$$ \frac{n}{4\pi}\int_{A'}\int_{\mathbb{R} ^{n}}{ \bigl\vert f(x,k)\bigr\vert ^{2}} \,dx \,dk \leq\int _{A'} \biggl(\int_{\mathbb{R}^{n}}{\| x \|^{2} \bigl\vert f_{k}(x)\bigr\vert ^{2}} \,dx \biggr)^{1/2} \biggl(\int_{\mathbb{R}^{n}}{\|y\|^{2} \bigl\vert \hat {f}_{k}(y)\bigr\vert ^{2}} \,dy \biggr)^{1/2} \,dk. $$

The integral on the L.H.S. is equal to $\|f\|_{2}^{2}$, so using the Cauchy-Schwarz inequality and Fubini’s theorem, we have

$$ \frac{n\|f\|_{2}^{2}}{4\pi} \leq \biggl(\int_{K}\int _{\mathbb {R}^{n}}{\|x\| ^{2} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dx \,dk \biggr)^{1/2} \biggl(\int_{\mathbb{R}^{n}} \|y\| ^{2}\int_{A'}{\bigl\vert \hat{f}_{k}(y)\bigr\vert ^{2}} \,dk \,dy \biggr)^{1/2}. $$

(2.2)

Now, using Hölder’s inequality, we have

$$\begin{aligned}& \biggl(\int_{\mathbb{R}^{n}}\int_{K}{\|x \|^{2a} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \biggr)^{\frac{1}{a}} \biggl(\int_{\mathbb{R}^{n}}\int _{K}{\bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \biggr)^{1-\frac{1}{a}} \\& \quad \geq\int_{\mathbb{R}^{n}}\int_{K}{\|x \|^{2} \bigl\vert f(x,k)\bigr\vert ^{\frac {2}{a}}\bigl\vert f(x,k)\bigr\vert ^{2 (1-\frac{1}{a} )}} \,dk \,dx \\& \quad = \int_{\mathbb{R}^{n}}\int_{K}{\|x \|^{2} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx, \end{aligned}$$

which implies

$$ \int_{\mathbb{R}^{n}}\int_{K}{\|x\|^{2} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \leq \biggl(\int _{\mathbb{R}^{n}}\int_{K}{\|x\|^{2a} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \biggr)^{\frac{1}{a}} \bigl( \|f\|_{2}^{2} \bigr)^{1-\frac{1}{a}}. $$

(2.3)

Combining (2.2) and (2.3), we obtain

$$\begin{aligned} \frac{n\|f\|_{2}^{2}}{4\pi} \leq& \biggl(\int_{\mathbb{R}^{n}}\int _{K}{\|x\| ^{2a} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \biggr)^{\frac{1}{2a}} \bigl(\|f\| _{2}^{2} \bigr)^{\frac{1}{2}-\frac{1}{2a}} \\ &{}\times \biggl(\int_{\mathbb {R}^{n}}\|y\| ^{2} \int_{A'}{\bigl\vert \hat{f}_{k}(y)\bigr\vert ^{2}} \,dk \,dy \biggr)^{1/2}. \end{aligned}$$

(2.4)

Since

$$ \int_{\mathbb{R}^{n}}\int_{A'}{\bigl\vert \mathscr {F}_{1}f(y,k)\bigr\vert ^{2}} \,dy \,dk =\int _{\mathbb{R}^{n}}\int_{A'}{\bigl\vert f(x,k)\bigr\vert ^{2}} \,dx \,dk =\|f\|_{2}^{2} < \infty, $$

therefore, $\mathscr{F}_{1}f \in L^{2}(\mathbb{R}^{n}\times A')$. Therefore, $\mathscr{F}_{2}\mathscr{F}_{1} f$ is well defined a.e. By approximating $f\in L^{2}(\mathbb{R}^{n} \times A')$ by functions in $L^{1}\cap L^{2}(\mathbb{R}^{n}\times A')$, we have

$$ \mathscr{F}_{2}\mathscr{F}_{1} f =\hat{f}, $$

for all $f \in L^{2}(\mathbb{R}^{n} \times A')$. Applying the Plancherel formula on the locally compact group K, we have

$$ \int_{A'}{\bigl\vert \hat{f}_{k}(y)\bigr\vert ^{2}} \,dk =\int_{\widehat{K}}{\bigl\Vert \hat{f}(y, \sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \,d\sigma. $$

Thus, (2.4) can be written as

$$\begin{aligned} \frac{n\|f\|_{2}^{2}}{4\pi} \leq& \biggl(\int_{\mathbb{R}^{n}}\int _{K}{\|x\| ^{2a} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \biggr)^{\frac{1}{2a}} \bigl(\|f\| _{2}^{2} \bigr)^{\frac{1}{2}-\frac{1}{2a}} \\ &{}\times \biggl(\int_{\mathbb {R}^{n}}\int _{\widehat{K}}{\|y\|^{2}\bigl\Vert \hat{f}(y,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma \biggr)^{1/2}. \end{aligned}$$

(2.5)

Now, again using Hölder’s inequality, we have

$$\begin{aligned}& \biggl(\int_{\mathbb{R}^{n}}\int_{\widehat{K}}{\|y \|^{2b} \bigl\Vert \hat {f}(y,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma \biggr)^{\frac {1}{b}} \biggl(\int_{\mathbb{R}^{n}}\int _{\widehat{K}}{\bigl\Vert \hat {f}(y,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma \biggr)^{1-\frac {1}{b}} \\& \quad \geq\int_{\mathbb{R}^{n}}\int_{\widehat{K}}{\|y \|^{2} \bigl\Vert \hat {f}(y,\sigma)\bigr\Vert _{\mathrm{HS}}^{\frac{2}{b}} \bigl\Vert \hat{f}(y,\sigma )\bigr\Vert _{\mathrm{HS}}^{2 (1-\frac{1}{b} )}} \,dy \,d\sigma \\& \quad = \int_{\mathbb{R}^{n}}\int_{\widehat{K}}{\|y \|^{2} \bigl\Vert \hat {f}(y,\sigma )\bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma, \end{aligned}$$

which implies

$$ \int_{\mathbb{R}^{n}}\int_{\widehat{K}}{\|y\|^{2} \bigl\Vert \hat {f}(y,\sigma )\bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma\leq \biggl(\int_{\mathbb {R}^{n}}\int_{\widehat{K}}{ \|y\|^{2b} \bigl\Vert \hat{f}(y,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma \biggr)^{\frac{1}{b}} \bigl(\|f \|_{2}^{2} \bigr)^{1-\frac{1}{b}}. $$

(2.6)

Combining (2.5) and (2.6), we obtain

$$\begin{aligned} \frac{n\|f\|_{2}^{2}}{4\pi} \leq& \biggl(\int_{\mathbb{R}^{n}}\int _{K}{\|x\| ^{2a} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \biggr)^{\frac{1}{2a}} \bigl(\|f\| _{2}^{2} \bigr)^{\frac{1}{2}-\frac{1}{2a}} \\ &{}\times \biggl(\int_{\mathbb{R}^{n}}\int_{\widehat{K}}{\|y \| ^{2b} \bigl\Vert \hat{f}(y,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma \biggr)^{\frac {1}{2b}} \bigl(\|f\|_{2}^{2} \bigr)^{\frac{1}{2}-\frac{1}{2b}}, \end{aligned}$$

which implies

$$ \frac{n\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq \biggl(\int_{\mathbb{R}^{n}}\int _{K}{\|x\|^{2a} \bigl\vert f(x,k)\bigr\vert ^{2}} \,dk \,dx \biggr)^{\frac{1}{2a}} \biggl(\int_{\mathbb{R}^{n}} \int_{\widehat {K}}{\|y\|^{2b} \bigl\Vert \hat{f}(y,\sigma) \bigr\Vert _{\mathrm{HS}}^{2}} \,dy \,d\sigma \biggr)^{\frac{1}{2b}}. $$

 □

3 Euclidean motion group $M(n)$

Consider $M(n)$ to be the semi-direct product of $\mathbb{R}^{n}$ with $K=\operatorname{SO}(n)$. The group law is given by

$$ (z,k) \bigl(w,k'\bigr) =\bigl(z+k\cdot w, kk'\bigr), $$

for $z,w \in\mathbb{R}^{n}$ and $k,k' \in K$. The group $M(n)$ is called the motion group of the Euclidean plane $\mathbb{R}^{n}$.

As in [5], $M=\operatorname{SO}(n-1)$ can be considered as a subgroup of K leaving the point $e_{1}=(1,0,0,\ldots,0)$ fixed. All the irreducible unitary representations of $M(n)$ relevant for the Plancherel formula are parametrized (up to unitary equivalence) by pairs $(\lambda,\sigma)$, where $\lambda>0$ and $\sigma\in\widehat {M}$, the unitary dual of M.

Given $\sigma\in\widehat{M}$ realized on a Hilbert space $H_{\sigma }$ of dimension $d_{\sigma}$, consider the space,

$$\begin{aligned} L^{2}(K,\sigma) =& \biggl\{ \varphi \Bigm| \varphi: K \rightarrow M_{d_{\sigma}\times d_{\sigma}}, \int{\bigl\Vert \varphi(k)\bigr\Vert ^{2}} \,dk < \infty, \\ & \varphi(uk)=\sigma(u)\varphi(k), \text{for } u \in M \text{ and } k\in K \biggr\} . \end{aligned}$$

Note that $L^{2}(K,\sigma)$ is a Hilbert space under the inner product

$$ \langle{\varphi,\psi}\rangle =\int_{K}{\operatorname{tr} \bigl(\varphi(k)\psi (k)^{\ast}\bigr)} \,dk. $$

For each $\lambda>0$ and $\sigma\in\widehat{M}$, we can define a representation $\pi_{\lambda,\sigma}$ of $M(n)$ on $L^{2}(K,\sigma)$ as follows.

For $\varphi\in L^{2}(K,\sigma)$, $(z,k)\in M(n)$,

$$ \pi_{\lambda,\sigma}(z,k)\varphi(u) =e^{i\lambda\langle {u^{-1}\cdot e_{1},z\rangle}} \varphi(uk), $$

for $u\in K$.

If $\varphi_{j}(k)$ are the column vectors of $\varphi\in L^{2}(K,\sigma )$, then $\varphi_{j}(uk)=\sigma(u)\varphi_{j}(k)$ for all $u \in M$. Therefore, $L^{2}(K,\sigma)$ can be written as the direct sum of $d_{\sigma}$ copies of $H(K,\sigma)$, where

$$\begin{aligned} H(K,\sigma) =& \biggl\{ \varphi \Bigm| \varphi: K \rightarrow\mathbb{C} ^{d_{\sigma}}, \int{\bigl\| \varphi(k)\bigr\| ^{2}} \,dk < \infty, \\ &\varphi (uk)= \sigma(u)\varphi(k), \text{for } u \in M \text{ and } k\in K \biggr\} . \end{aligned}$$

It can be shown that $\pi_{\lambda,\sigma}$ restricted to $H(K,\sigma)$ is an irreducible unitary representation of $M(n)$. Moreover, any irreducible unitary representation of $M(n)$ which is infinite dimensional is unitarily equivalent to one and only one $\pi _{\lambda,\sigma}$.

The Fourier transform of $f \in L^{2}(M(n))$ is given by

$$ \hat{f}(\lambda,\sigma) =\int_{M(n)}{f(z,k) \pi_{\lambda ,\sigma}(z,k)^{\ast}} \,dz \,dk. $$

$\hat{f}(\lambda,\sigma)$ is a Hilbert-Schmidt operator on $H(K,\sigma)$.

A solid harmonic of degree m is a polynomial which is homogeneous of degree m and whose Laplacian is zero. The set of all such polynomials will be denoted by $\mathbb{H}_{m}$ and the restriction of elements of $\mathbb{H}_{m}$ to $S^{n-1}$ is denoted by $S_{m}$. By choosing an orthonormal basis $\{ g_{mj} : j = 1, 2,\ldots,d_{m}\}$ of $S_{m}$ for each $m = 0, 1, 2, \ldots$ , we get an orthonormal basis for $L^{2}(S^{n-1})$.

The Haar measure on $M(n)$ is $dg=dz \,dk$, where dz is Lebesgue measure on $\mathbb{R}^{n}$ and dk is the normalized Haar measure on $\operatorname{SO}(n)$.

The Plancherel formula on $M(n)$ is given as follows (see [6]).

Proposition 3.1

(Plancherel formula)

Let $f \in L^{2}(M(n))$, then

$$ \int_{M(n)}{\bigl\vert f(z_{1},z_{2}, \ldots,z_{n},k)\bigr\vert ^{2}} \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk =c_{n}\int _{0}^{\infty} \biggl(\sum_{\sigma\in \widehat{M}}{d_{\sigma}\bigl\Vert \hat{f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \biggr) \lambda^{n-1}\, d\lambda , $$

where $c_{n}=\frac{2}{2^{n/2} \Gamma (\frac{n}{2} )}$.

We shall now state and prove the following generalized Heisenberg uncertainty inequality for a Fourier transform on $M(n)$.

Theorem 3.2

For any $f \in L^{2}(M(n))$ and $a,b \geq1$, we have

$$\begin{aligned} \frac{\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{2\sqrt {c_{n}}} \leq& \biggl(\int_{K}\int _{\mathbb{R}^{n}}{\|z\|^{2a} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{\frac{1}{2a}} \\ &{}\times \biggl(\int_{0}^{\infty} \sum_{\sigma \in\widehat{M}}{d_{\sigma}\lambda^{2b} \bigl\Vert \hat{f}(\lambda ,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1}\, d\lambda \biggr)^{\frac {1}{2b}}. \end{aligned}$$

(3.1)

Proof

Consider the norm $\|\cdot\|$ on $L^{2}(M(n))$ defined by

$$\begin{aligned} \|f\| : =& \biggl(\int_{\mathbb{R}^{n}}\int_{K}{ \bigl(1+\|z\|^{2a}\bigr) \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{1/2} \\ &{}+ \biggl(\int_{0}^{\infty} \sum_{\sigma\in \widehat{M}}{d_{\sigma}\bigl(1+ \lambda^{2b}\bigr)\bigl\Vert \hat{f}(\lambda,\sigma )\bigr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1} \,d\lambda \biggr)^{1/2}. \end{aligned}$$

This gives us a Banach space $B=\{f\in L^{2}(G):\|f\|<\infty\}$, which is contained in $L^{2}(M(n))$ and the space $\mathcal{S}(M(n))$ of $C^{\infty}$-functions which are rapidly decreasing on $M(n)$ can be shown to be dense in B. It suffices to prove the inequality of Theorem 3.2 for functions in $\mathcal{S}(M(n))$; it is automatically valid for any $f\in B$. If $0\neq f \in L^{2}(M(n))\setminus B$, then the right-hand side of the inequality is always +∞ and the inequality is trivially valid.

Let $f\in\mathcal{S}(M(n))$. Assuming that both integrals on the right-hand side of (3.1) are finite, we have

$$ \int_{\mathbb{R}^{n}}{\bigl\vert f(z,k)\bigr\vert ^{2}} \,dz < \infty, \quad \text {for all } k\in K. $$

For $k \in K$, we define $f_{k}(z)=f(z,k)$, for every $z \in\mathbb {R}^{n}$.

Clearly, $f_{k} \in L^{2}(\mathbb{R}^{n})$, for all $k\in K$.

Take $z=(z_{1},z_{2},\ldots,z_{n})$ and $w=(w_{1},w_{2},\ldots,w_{n})$.

By the Heisenberg inequality on $\mathbb{R}^{n}$, we have

$$\begin{aligned}& \frac{\|f_{k}\|_{2}^{2}}{4\pi} \leq \biggl(\int_{\mathbb {R}^{n}}{|z_{1}|^{2} \bigl\vert f_{k}(z)\bigr\vert ^{2}} \,dz \biggr)^{1/2} \biggl(\int_{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \hat {f}_{k}(w)\bigr\vert ^{2}} \,dw \biggr)^{1/2} \\& \quad \Rightarrow \quad \frac{1}{4\pi}\int_{\mathbb{R}^{n}}{\bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \leq \biggl(\int _{\mathbb{R}^{n}}{|z_{1}|^{2} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \biggr)^{1/2} \biggl(\int _{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \hat{f}_{k}(w)\bigr\vert ^{2}} \,dw \biggr)^{1/2} . \end{aligned}$$

Integrating both sides with respect to dk, we get

$$ \frac{1}{4\pi}\int_{K}\int_{\mathbb{R}^{n}}{ \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \leq \int _{K} \biggl(\int_{\mathbb{R}^{n}}{|z_{1}|^{2} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \biggr)^{1/2} \biggl(\int_{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \hat{f}_{k}(w)\bigr\vert ^{2}} \,dw \biggr)^{1/2} \,dk , $$

which implies

$$\begin{aligned} \frac{\|f\|_{2}^{2}}{4\pi} \leq&\int_{K} \biggl(\int _{\mathbb {R}^{n}}{|z_{1}|^{2} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \biggr)^{1/2} \biggl(\int _{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \hat {f}_{k}(w)\bigr\vert ^{2}} \,dw \biggr)^{1/2} \,dk \\ \leq& \biggl(\int_{K}\int_{\mathbb{R}^{n}}{|z_{1}|^{2} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{1/2} \biggl(\int_{K}\int_{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \hat {f}_{k}(w)\bigr\vert ^{2}} \,dw \,dk \biggr)^{1/2} \\ & (\text{by the Cauchy-Schwarz inequality}) \\ \leq& \biggl(\int_{K}\int_{\mathbb{R}^{n}}{\|z \|^{2} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{1/2} \biggl(\int_{K}\int _{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \hat {f}_{k}(w)\bigr\vert ^{2}} \,dw \,dk \biggr)^{1/2}. \end{aligned}$$

(3.2)

Now,

$$\begin{aligned}& \biggl(\int_{K}\int_{\mathbb{R}^{n}}{\|z \|^{2a} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{\frac{1}{a}} \biggl(\int_{K}\int _{\mathbb{R}^{n}}{\bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{1-\frac{1}{a}} \\& \quad = \biggl(\int_{K}\int_{\mathbb{R}^{n}}{ \bigl(\|z\|^{2} \bigl\vert f(z,k)\bigr\vert ^{\frac {2}{a}} \bigr)^{a}} \,dz \,dk \biggr)^{\frac{1}{a}} \biggl(\int _{K}\int_{\mathbb{R}^{n}}{ \bigl(\bigl\vert f(z,k) \bigr\vert ^{2 (1-\frac {1}{a} )} \bigr)^{\frac{1}{ (1-\frac{1}{a} )}}} \,dz \,dk \biggr)^{1-\frac{1}{a}} \\& \quad \geq\int_{K}\int_{\mathbb{R}^{n}}{\|z \|^{2} \bigl\vert f(z,k)\bigr\vert ^{\frac {2}{a}}\bigl\vert f(z,k)\bigr\vert ^{2 (1-\frac{1}{a} )}} \,dz \,dk \quad \bigl(\text{by H\"{o}lder's inequality}\bigr) \\& \quad = \int_{K}\int_{\mathbb{R}^{n}}{\|z \|^{2} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk . \end{aligned}$$

(3.3)

Combining (3.2) and (3.3), we get

$$\begin{aligned} \frac{\|f\|_{2}^{2}}{4\pi} \leq& \biggl(\int_{K}\int _{\mathbb {R}^{n}}{\|z\| ^{2a} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{\frac{1}{2a}} \bigl(\|f\| _{2}^{2} \bigr)^{\frac{1}{2}-\frac{1}{2a}} \\ &{}\times\biggl(\int_{K}\int _{\mathbb{R} ^{n}}{|w_{1}|^{2} \bigl\vert \hat{f}_{k}(w)\bigr\vert ^{2}} \,dw \,dk \biggr)^{1/2}. \end{aligned}$$

(3.4)

Now, using the Plancherel formula on $\mathbb{R}^{n}$, we have

$$\begin{aligned}& \int_{K}\int_{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \hat{f}_{k}(w)\bigr\vert ^{2}} \,dw \,dk \\& \quad =\int_{K}\int_{\mathbb{R}^{n}}{|w_{1}|^{2} \biggl\vert \int_{\mathbb {R}^{n}}{f(z,k) e^{-2\pi i\langle{z,w}\rangle}} \,dz\biggr\vert ^{2}} \,dw \,dk \\& \quad =\int_{K}\int_{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \mathscr{F}_{1,2,\ldots, n} f(w_{1},w_{2},\ldots ,w_{n},k)\bigr\vert ^{2}} \,dw_{1} \,dw_{2} \cdots \,dw_{n} \,dk \\& \quad =\int_{K}\int_{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \mathscr{F}_{1} f(w_{1},z_{2},\ldots ,z_{n},k)\bigr\vert ^{2}} \,dw_{1} \,dz_{2} \cdots \,dz_{n} \,dk. \end{aligned}$$

(3.5)

Since $\frac{\partial f}{\partial z_{1}}\in\mathcal{S}(M(n))$, we have

$$ \int_{\mathbb{R}}{\biggl\vert \frac{\partial f}{\partial z_{1}}(z_{1},z_{2}, \ldots ,z_{n},k)\biggr\vert ^{2}} \,dz_{1} < \infty, $$

for all $z_{i} \in\mathbb{R}$ and $k \in K$.

Therefore, $w_{1}\mathscr{F}_{1} f(w_{1},z_{2},\ldots,z_{n},k)\in L^{2}(\mathbb {R})$ and

$$ \biggl(\frac{\partial f}{\partial z_{1}}(z_{1},z_{2},\ldots,z_{n},k) \biggr)^{\land}(w_{1}) =2\pi i w_{1} \mathscr{F}_{1} f(w_{1},z_{2},\ldots,z_{n},k), $$

for all $z_{i} \in\mathbb{R}$ and $k \in K$. Then

$$\begin{aligned}& \int_{\mathbb{R}}{|w_{1}|^{2} \bigl\vert \mathscr{F}_{1} f(w_{1},z_{2},\ldots ,z_{n},k)\bigr\vert ^{2}} \,dw_{1} \\& \quad = \frac{1}{4\pi^{2}}\int_{\mathbb{R}}{\biggl\vert \frac{\partial f}{\partial z_{1}}(z_{1},z_{2},\ldots,z_{n},k)\biggr\vert ^{2}} \,dz_{1}, \end{aligned}$$

which implies

$$\begin{aligned}& \int_{K}\int_{\mathbb{R}^{n}}{|w_{1}|^{2} \bigl\vert \mathscr{F}_{1} f(w_{1},z_{2}, \ldots,z_{n},k)\bigr\vert ^{2}} \,dw_{1} \,dz_{2} \cdots \,dz_{n} \,dk \\& \quad =\frac{1}{4\pi^{2}}\int_{K}\int_{\mathbb{R}^{n}}{ \biggl\vert \frac {\partial f}{\partial z_{1}}(z_{1},z_{2}, \ldots,z_{n},k)\biggr\vert ^{2}} \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk. \end{aligned}$$

(3.6)

By Proposition 3.1, we obtain

$$\begin{aligned}& \int_{K}\int_{\mathbb{R}^{n}}{\biggl\vert \frac{\partial f}{\partial z_{1}}(z_{1},z_{2},\ldots,z_{n},k)\biggr\vert ^{2}} \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk \\& \quad =c_{n}\int_{0}^{\infty}\sum _{\sigma\in\widehat{M}}{d_{\sigma}\biggl\Vert \biggl( \frac{\partial f}{\partial z_{1}} \biggr)^{\land}(\lambda,\sigma)\biggr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1} \,d\lambda . \end{aligned}$$

(3.7)

Combining (3.4), (3.5), (3.6), and (3.7), we obtain

$$\begin{aligned} \frac{\|f\|_{2}^{2}}{2\sqrt{c_{n}}} \leq& \biggl(\int_{K}\int _{\mathbb{R} ^{n}}{\|z\|^{2a} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{\frac{1}{2a}} \bigl(\|f\| _{2}^{2} \bigr)^{\frac{1}{2}-\frac{1}{2a}} \\ &{}\times \biggl(\int_{0}^{\infty}\sum _{\sigma\in\widehat {M}}{d_{\sigma}\biggl\Vert \biggl( \frac{\partial f}{\partial z_{1}} \biggr)^{\land}(\lambda,\sigma)\biggr\Vert _{\mathrm{HS}}^{2}} \lambda ^{n-1} \,d\lambda \biggr)^{1/2}. \end{aligned}$$

(3.8)

For each $\lambda>0$ and $\sigma\in\widehat{M}$, consider the representation $\pi_{\lambda,\sigma}(z,k)$ realized on $L^{2}(K,\sigma )$ as

$$ \pi_{\lambda,\sigma}(z,k)g(u) =e^{i\lambda\langle{u^{-1}\cdot e_{1},z}\rangle} g(uk), \quad u \in \operatorname{SO}(n). $$

Denote $u=[u_{ij}]_{n\times n}$; we have

$$ u^{-1}\cdot e_{1} =u^{T}\cdot e_{1}=[ \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} u_{11}& u_{12}& \cdots &u_{1n} \end{array}]^{T}. $$

Therefore, $\langle{u^{-1}\cdot e_{1},z}\rangle=\sum_{i=1}^{n}{u_{1i}z_{i}}$.

Since $f\in\mathcal{S}(M(n))$,

$$\begin{aligned}& \biggl(\frac{\partial f}{\partial z_{1}} \biggr)^{\land}(\lambda,\sigma)g(u) \\& \quad =\int_{\mathbb{R}^{n}}\int_{K}{ \frac{\partial f}{\partial z_{1}}(z_{1},z_{2},\ldots,z_{n},k) \pi_{\lambda,\sigma}(z_{1},z_{2},\ldots ,z_{n},k)^{\ast}g(u)} \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk \\& \quad =\int_{\mathbb{R}^{n}}\int_{K}\lim _{h\rightarrow0}{ \biggl[\frac {f(z_{1}+h,z_{2},\ldots,z_{n},k)-f(z_{1},z_{2},\ldots,z_{n},k)}{h} \biggr]} \\& \qquad {}\times\pi _{\lambda,\sigma}(z_{1},z_{2},\ldots,z_{n},k)^{\ast}g(u)\,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk \\& \quad =\lim_{h\rightarrow0}\frac{1}{h}\biggl[\int _{\mathbb{R}^{n}}\int_{K}{f(z_{1}+h,z_{2}, \ldots,z_{n},k) \pi_{\lambda,\sigma}(z_{1},z_{2}, \ldots ,z_{n},k)^{\ast}g(u)} \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk \\& \qquad {}-\int_{\mathbb{R}^{n}}\int_{K}{f(z_{1},z_{2}, \ldots ,z_{n},k) \pi _{\lambda,\sigma}(z_{1},z_{2}, \ldots,z_{n},k)^{\ast}g(u)} \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk\biggr] \\& \quad =\lim_{h\rightarrow0}\frac{1}{h}\biggl[\int _{\mathbb{R}^{n}}\int_{K}f(z_{1},z_{2}, \ldots,z_{n},k) e^{-i \lambda h u_{11}}\pi_{\lambda ,\sigma}(z_{1},z_{2}, \ldots,z_{n},k)^{\ast}\\& \qquad {}\times g(u) \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk \\& \qquad {}-\int_{\mathbb{R}^{n}}\int_{K}{f(z_{1},z_{2}, \ldots ,z_{n},k) \pi _{\lambda,\sigma}(z_{1},z_{2}, \ldots,z_{n},k)^{\ast}g(u)} \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk\biggr] \\& \quad =\lim_{h\rightarrow0} \biggl[\frac{e^{-i \lambda h u_{11}}-1}{h} \biggr]\int _{\mathbb{R}^{n}}\int_{K}f(z_{1},z_{2}, \ldots ,z_{n},k) \pi_{\lambda,\sigma}(z_{1},z_{2}, \ldots,z_{n},k)^{\ast}\\& \qquad {}\times g(u) \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk \\& \quad =i\lambda u_{11}\int_{\mathbb{R}^{n}}\int _{K}{f(z_{1},z_{2},\ldots ,z_{n},k) \pi _{\lambda,\sigma}(z_{1},z_{2}, \ldots,z_{n},k)^{\ast}g(u)} \,dz_{1} \,dz_{2} \cdots \,dz_{n} \,dk \\& \quad =i\lambda u_{11} \hat{f}(\lambda,\sigma)g(u). \end{aligned}$$

Hence,

$$\begin{aligned} \biggl\Vert \biggl(\frac{\partial f}{\partial z_{1}} \biggr)^{\land}(\lambda,\sigma) \biggr\Vert _{\mathrm{HS}}^{2} =&\sum_{m=0}^{\infty } \sum_{j=1}^{d_{m}}\int_{K}{ \bigl\vert i\lambda u_{11} \hat{f}(\lambda ,\sigma)g_{mj}(u) \bigr\vert ^{2}} \, du \\ \leq&\lambda^{2}\sum_{m=0}^{\infty} \sum_{j=1}^{d_{m}}\int_{K}{ \bigl\vert \hat{f}(\lambda,\sigma)g_{mj}(u)\bigr\vert ^{2}}\, du = \lambda^{2}\bigl\Vert \hat{f}(\lambda,\sigma) \bigr\Vert _{\mathrm{HS}}^{2}. \end{aligned}$$

Therefore, (3.8) can be written as

$$\begin{aligned} \frac{\|f\|_{2}^{2}}{2\sqrt{c_{n}}} \leq& \biggl(\int_{K}\int _{\mathbb {R}^{n}}{\| z\|^{2a} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{\frac{1}{2a}} \bigl(\|f\| _{2}^{2} \bigr)^{\frac{1}{2}-\frac{1}{2a}} \\ &{} \times \biggl(\int_{0}^{\infty}\sum _{\sigma\in\widehat {M}}{d_{\sigma}\lambda^{2}\bigl\Vert \hat{f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1} \,d\lambda \biggr)^{1/2}. \end{aligned}$$

(3.9)

Now, again using Hölder’s inequality, we have

$$\begin{aligned}& \biggl(\int_{0}^{\infty}\sum _{\sigma\in\widehat{M}}{d_{\sigma}\lambda^{2b}\bigl\Vert \hat{f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1} \,d\lambda \biggr)^{\frac{1}{b}} \biggl(\int _{0}^{\infty}\sum_{\sigma\in\widehat{M}}{d_{\sigma}\bigl\Vert \hat {f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1} \,d\lambda \biggr)^{1-\frac{1}{b}} \\& \quad \geq\int_{0}^{\infty}\sum _{\sigma\in\widehat{M}}{d_{\sigma}^{1/b} \lambda^{2} \bigl\Vert \hat{f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{\frac{2}{b}} d_{\sigma}^{ (1-\frac{1}{b} )}\bigl\Vert \hat{f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{2 (1-\frac {1}{b} )}} \lambda^{n-1} \,d\lambda \\& \quad = \int_{0}^{\infty}\sum _{\sigma\in\widehat{M}}{d_{\sigma}\lambda^{2} \bigl\Vert \hat{f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \lambda ^{n-1} \,d\lambda, \end{aligned}$$

which implies

$$\begin{aligned}& \int_{0}^{\infty}\sum_{\sigma\in\widehat{M}}{d_{\sigma}\lambda ^{2} \bigl\Vert \hat{f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1} \,d\lambda \\& \quad \leq \biggl(\int _{0}^{\infty}\sum_{\sigma\in\widehat {M}}{d_{\sigma}\lambda^{2b}\bigl\Vert \hat{f}(\lambda,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1} \,d\lambda \biggr)^{\frac{1}{b}} \bigl(\|f\| _{2}^{2} \bigr)^{1-\frac{1}{b}}. \end{aligned}$$

(3.10)

Combining (3.9) and (3.10), we obtain

$$\begin{aligned} \frac{\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{2\sqrt {c_{n}}} \leq& \biggl(\int_{K}\int _{\mathbb{R}^{n}}{\|z\|^{2a} \bigl\vert f(z,k)\bigr\vert ^{2}} \,dz \,dk \biggr)^{\frac{1}{2a}} \\ &{}\times \biggl(\int_{0}^{\infty} \sum_{\sigma \in\widehat{M}}{d_{\sigma}\lambda^{2b} \bigl\Vert \hat{f}(\lambda ,\sigma)\bigr\Vert _{\mathrm{HS}}^{2}} \lambda^{n-1} \,d\lambda \biggr)^{\frac{1}{2b}}. \end{aligned}$$

 □

4 A class of nilpotent Lie groups

In this section, we shall prove the Heisenberg uncertainty inequality for a class of connected, simply connected nilpotent Lie groups G for which the Hilbert-Schmidt norm of the group Fourier transform $\pi_{\xi}(f)$ of f attains a particular form.

Let $\mathfrak{g}$ be an n-dimensional real nilpotent Lie algebra, and let $G=\exp\mathfrak{g}$ be the associated connected and simply connected nilpotent Lie group [7]. Let $\mathcal{B}=\{ X_{1},X_{2},\ldots,X_{n}\}$ be a strong Malcev basis of $\mathfrak{g}$ through the ascending central series of $\mathfrak{g}$. We introduce a ‘norm function’ on G by setting, for $x=\exp(x_{1}X_{1}+x_{2}X_{2}+\cdots+x_{n}X_{n}) \in G$, $x_{j}\in\mathbb{R}$,

$$ \|x\| =\bigl(x_{1}^{2}+\cdots+x_{n}^{2} \bigr)^{1/2}. $$

The composed map

$$ \mathbb{R}^{n}\rightarrow\mathfrak{g}\rightarrow G, $$

given as

$$ (x_{1},\ldots,x_{n})\rightarrow\sum _{j=1}^{n}{x_{j}X_{j}} \rightarrow \exp \Biggl(\sum_{j=1}^{n}{x_{j}X_{j}} \Biggr), $$

is a diffeomorphism and maps a Lebesgue measure on $\mathbb{R}^{n}$ to a Haar measure on G. In this manner, we shall always identify $\mathfrak {g}$, and sometimes G, as sets with $\mathbb{R}^{n}$. Thus, measurable (integrable) functions on G can be viewed as such functions on $\mathbb{R}^{n}$.

Let $\mathfrak{g}^{\ast}$ denote the vector space dual of $\mathfrak {g}$ and $\{X_{1}^{\ast},\ldots,X_{n}^{\ast}\}$ the basis of $\mathfrak{g}^{\ast}$ which is dual to $\{ X_{1},\ldots,X_{n}\}$. Then $\{X_{1}^{\ast},\ldots,X_{n}^{\ast}\}$ is a Jordan-Hölder basis for the coadjoint action of G on $\mathfrak {g}^{\ast}$. We shall identify $\mathfrak{g}^{\ast}$ with $\mathbb{R}^{n}$ via the map

$$ \xi=(\xi_{1},\ldots,\xi_{n})\rightarrow\sum _{j=1}^{n}{\xi _{j}X_{j}^{\ast}} $$

and on $\mathfrak{g}^{\ast}$ we introduce the Euclidean norm relative to the basis $\{X_{1}^{\ast},\ldots,X_{n}^{\ast}\}$, i.e.

$$ \Biggl\Vert \sum_{j=1}^{n}{ \xi_{j}X_{j}^{\ast}}\Biggr\Vert =\bigl( \xi_{1}^{2}+\cdots +\xi_{n}^{2}\bigr)=\| \xi\|. $$

Let $\mathfrak{g}_{j}=\mathbb{R}\mbox{-} \operatorname{span}\{X_{1},\ldots,X_{n}\}$. For $\xi \in\mathfrak{g}^{\ast}$, $\mathcal{O}_{\xi}$ denotes the coadjoint orbit of ξ. An index $j \in\{ 1,2,\ldots,n\}$ is a jump index for ξ if

$$ \mathfrak{g}(\xi)+\mathfrak{g}_{j} \neq\mathfrak{g}(\xi )+ \mathfrak{g}_{j-1}. $$

We consider

$$ e(\xi)=\{j: j \text{ is a jump index for } \xi\}. $$

This set contains exactly $\dim(\mathcal{O}_{l})$ indices. Also, there are two disjoint sets S and T of indices with $S \cup T =\{1,\ldots,n\}$ and a G-invariant Zariski open set $\mathcal{U}$ of $\mathfrak {g}^{\ast}$ such that $e(\xi)=S$ for all $\xi\in\mathcal{U}$. We define the Pfaffian $\operatorname{Pf} (\xi)$ of the skew-symmetric matrix $M_{S}(\xi)=(\xi([X_{i},X_{j}]))_{i,j\in S}$ as

$$ \bigl\vert \operatorname{Pf}(\xi)\bigr\vert ^{2}= \det{M_{S}(\xi)}. $$

Let $V_{S}=\mathbb{R}\mbox{-} \operatorname{span}\{X_{i}^{\ast}: i \in S\}$, $V_{T}=\mathbb {R}\mbox{-} \operatorname{span}\{X_{i}^{\ast}: i \in T\}$, and dξ be the Lebesgue measure on $V_{T}$ such that the unit cube spanned by $\{X_{i}^{\ast}:i \in T\}$ has volume 1. Then $\mathfrak{g} ^{\ast}=V_{T} \oplus V_{S}$ and $V_{T}$ meets $\mathcal{U}$. Let $\mathcal {W}=\mathcal{U}\cap V_{T}$ be the cross section for the coadjoint orbits through the points in $\mathcal{U}$. If dξ is the Lebesgue measure on $\mathcal{W}$, then $d\mu(\xi )=|\operatorname{Pf} (\xi)| \, d\xi$ is a Plancherel measure for $\widehat{G}$. The Plancherel formula is given by

$$ \|f\|_{2}^{2}=\int_{\mathcal{W}}{\bigl\Vert \pi_{\xi}{(f)}\bigr\Vert _{\mathrm{HS}}^{2}}\, d\mu(\xi ), \quad f \in L^{1}\cap L^{2}(G), $$

where $\|\pi_{\xi}{(f)}\|_{\mathrm{HS}}$ denotes the Hilbert-Schmidt norm of $\pi_{\xi}{(f)}$ and dg is the Haar measure on G.

We shall consider the case in which $\mathcal{W}$ takes the following form:

$$ \mathcal{W} =\bigl\{ \xi=(\xi_{1},\xi_{2},\ldots, \xi_{n})\in\mathfrak {g}^{\ast}: \xi_{j}=0, \text{for }(n-k)\text{ values of }j\text{ with } \bigl\vert \operatorname{Pf}(\xi)\bigr\vert \neq 0\bigr\} . $$

We denote the vanishing variables by $\xi_{j_{1}},\xi_{j_{2}},\ldots,\xi _{j_{n-k}}$.

We consider the class of groups for which for all $\xi\in\mathcal {W}$ and $f \in L^{2}(G)$ the Hilbert-Schmidt norm $\|\pi_{\xi}(f)\|_{\mathrm{HS}}^{2}$ has the following form:

$$ \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2} = \bigl\vert h(\xi)\bigr\vert \int_{\mathbb {R}^{n-k}}{\bigl\vert \mathscr{F}(f\circ\exp) (\xi_{1},\xi_{2}+Q_{2}, \ldots,\xi _{n}+Q_{n} )\bigr\vert ^{2}}\, d \xi_{j_{1}}\, d\xi_{j_{2}} \cdots \, d\xi_{j_{n-k}}, $$

where $\mathscr{F}$ denotes the Fourier transform on $\mathbb {R}^{n-k}$; h is a function from $\mathcal{W}$ to $\mathbb{R}$ which is nonzero on $\mathcal{W}$ and the functions $Q_{m}=Q_{m}(\xi_{1},\xi_{2},\ldots,\xi_{m-1})$ with $2\leq m \leq n$.

We have the following Heisenberg uncertainty inequality for such groups.

Theorem 4.1

For any $f \in L^{1}\cap L^{2}(G)$ and $a,b\geq1$, we have

$$\begin{aligned} \frac{\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq& \biggl(\int_{G}{\|x \|^{2a} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{\frac {1}{2a}} \\ &{}\times\biggl(\int_{\mathcal{W}}{\|\xi \|^{2b} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \frac{1}{\vert h(\xi)\vert ^{b}\vert \operatorname{Pf}(\xi)\vert ^{b-1}}\, d\xi \biggr)^{\frac{1}{2b}}. \end{aligned}$$

(4.1)

Proof

Assuming both integrals on the right-hand side of (4.1) to be finite, we have

$$\begin{aligned}& \biggl(\int_{G}{\|x\|^{2} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{1/2} \biggl(\int _{\mathcal{W}}{\|\xi\|^{2} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \frac {1}{\vert h(\xi)\vert } \,d\xi \biggr)^{1/2} \\& \quad = \Biggl(\int_{\mathbb{R}^{n}}{\sum_{i=1}^{n}{|x_{i}|^{2}} \Biggl\vert (f\circ\exp) \Biggl(\sum_{i=1}^{n}{x_{i} X_{i}} \Biggr)\Biggr\vert ^{2}} \,dx_{1} \cdots \,dx_{n} \Biggr)^{1/2} \\& \qquad {}\times \Biggl(\int_{\mathbb{R}^{k}}\int_{\mathbb{R}^{n-k}}{ \sum_{i=1}^{n}{|\xi _{i}|^{2}} \bigl\vert \mathscr{F}(f\circ\exp) (\xi_{1},\xi _{2}+Q_{2}, \ldots,\xi _{n}+Q_{n} )\bigr\vert ^{2}} \,d \xi_{1} \cdots \,d\xi_{n} \Biggr)^{1/2} \\& \quad \geq \Biggl(\int_{\mathbb{R}^{n}}{|x_{1}|^{2} \Biggl\vert (f\circ\exp) \Biggl(\sum_{i=1}^{n}{x_{i} X_{i}} \Biggr)\Biggr\vert ^{2}} \,dx_{1} \cdots \,dx_{n} \Biggr)^{1/2} \\& \qquad {}\times \biggl(\int_{\mathbb{R}^{k}}\int_{\mathbb{R}^{n-k}}{| \xi _{1}|^{2} \bigl\vert \mathscr{F}(f\circ\exp) ( \xi_{1},\xi_{2}+Q_{2},\ldots,\xi_{n}+Q_{n} )\bigr\vert ^{2}} \,d\xi_{1} \cdots \,d\xi_{n} \biggr)^{1/2} \\& \quad = \biggl(\int_{\mathbb{R}^{n}}{|x_{1}|^{2} \bigl\vert F (x_{1},\ldots ,x_{n} )\bigr\vert ^{2}} \,dx_{1} \cdots \,dx_{n} \biggr)^{1/2} \\& \qquad {}\times \biggl(\int_{\mathbb{R}^{n}}{|\xi_{1}|^{2} \bigl\vert \widehat {F} (\xi _{1},\xi_{2},\ldots, \xi_{n} )\bigr\vert ^{2}} \,d\xi_{1} \,d \xi_{2} \cdots \,d\xi_{n} \biggr)^{1/2}, \end{aligned}$$

(4.2)

where $F(x_{1},\ldots,x_{n})=(f\circ\exp) (\sum_{i=1}^{n}{x_{i} X_{i}} )$ which is in $L^{2}(R^{n})$, $\widehat{F}$ being its Fourier transform.

By the Heisenberg inequality on $\mathbb{R}^{n}$, we have

$$\begin{aligned} \frac{\|F\|_{2}^{2}}{4\pi} \leq& \biggl(\int_{\mathbb{R}^{n}}{|x_{1}|^{2} \bigl\vert F (x_{1},\ldots,x_{n} )\bigr\vert ^{2}} \,dx_{1} \cdots \,dx_{n} \biggr)^{1/2} \\ & {}\times \biggl(\int_{\mathbb{R}^{n}}{|\xi_{1}|^{2} \bigl\vert \widehat {F} (\xi _{1},\xi_{2},\ldots, \xi_{n} )\bigr\vert ^{2}} \,d\xi_{1} \,d \xi_{2} \cdots \,d\xi_{n} \biggr)^{1/2}. \end{aligned}$$

(4.3)

But

$$\begin{aligned} \|F\|_{2}^{2} =&\int_{\mathbb{R}^{n}}{\bigl|F(x_{1}, \ldots,x_{n})\bigr|^{2}} \,dx_{1} \cdots \,dx_{n} \\ =&\int_{\mathbb{R}^{n}}{\Biggl\vert (f\circ\exp) \Biggl(\sum _{i=1}^{n}{x_{i} X_{i}} \Biggr)\Biggr\vert ^{2}} \,dx_{1} \cdots \,dx_{n}= \int_{G}{\bigl\vert f(x)\bigr\vert ^{2}} \,dx =\|f\|_{2}^{2}. \end{aligned}$$

(4.4)

Combining (4.2), (4.3), and (4.4), we get

$$ \frac{\|f\|_{2}^{2}}{4\pi} \leq \biggl(\int_{G}{\|x\|^{2} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{1/2} \biggl( \int_{\mathcal{W}}{\|\xi\|^{2} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \frac{1}{|h(\xi)|} \,d\xi \biggr)^{1/2}. $$

(4.5)

Now, as in the proof of Theorem 3.2, applications of Hölder’s inequality give

$$ \int_{G}{\|x\|^{2} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \leq \biggl(\int_{G}{\|x \|^{2a} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{\frac{1}{a}} \bigl(\|f\|_{2}^{2} \bigr)^{1-\frac {1}{a}} $$

(4.6)

and

$$\begin{aligned}& \int_{\mathcal{W}}{\|\xi\|^{2} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \frac {1}{|h(\xi)|} \,d\xi \\& \quad \leq \biggl(\int_{\mathcal{W}}{\|\xi \|^{2b} \bigl\Vert \pi _{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \frac{1}{|h(\xi)|^{b}|\operatorname {Pf}(\xi)|^{b-1}} \,d\xi \biggr)^{\frac{1}{b}} \bigl(\|f\|_{2}^{2} \bigr)^{1-\frac{1}{b}}. \end{aligned}$$

(4.7)

Combining (4.5), (4.6), and (4.7), we obtain

$$ \frac{\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq \biggl(\int_{G}{\|x \|^{2a} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{\frac {1}{2a}} \biggl(\int_{\mathcal{W}}{\|\xi \|^{2b} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \frac{1}{|h(\xi)|^{b}|\operatorname{Pf}(\xi)|^{b-1}} \,d\xi \biggr)^{\frac{1}{2b}}. $$

 □

Example 4.2

We now list several classes that are included in the above general class.

1. For thread-like nilpotent Lie groups (for details, see [8]), we have $\operatorname{Pf}(\xi)=\xi_{1}$ and

$$ \mathcal{W} =\bigl\{ \xi=(\xi_{1},0,\xi_{3},\ldots, \xi_{n-1},0):\xi_{j} \in\mathbb{R}, \xi_{1}\neq0 \bigr\} . $$

Also, $\|\pi_{\xi}(f)\|_{\mathrm{HS}}$ is given by

$$ \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2} = \frac{1}{|\xi_{1}|}\int_{\mathbb{R} ^{2}}{\bigl\vert \mathscr{F} {(f\circ \exp)} (\xi_{1},t,\xi _{3}+Q_{3},\ldots,\xi _{n-1}+Q_{n-1},s )\bigr\vert ^{2}} \, ds\, dt, $$

where $Q_{j}(\xi_{1},0,\xi_{3},\ldots,\xi_{j-1},t)=\sum_{k=1}^{j-1}{\frac{1}{k!} \frac{t^{k}}{\xi_{1}^{k}} \xi_{j-k}}$, for $3\leq j\leq n-1$.

Thus, for $h(\xi)=\frac{1}{|\xi_{1}|}=\frac{1}{|\operatorname {Pf}(\xi)|}$, one obtains the Heisenberg uncertainty inequality

$$ \frac{\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq \biggl(\int_{G}{\|x \|^{2a} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{\frac {1}{2a}} \biggl(\int_{\mathcal{W}}{\|\xi \|^{2b} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} |\xi_{1}| \,d\xi \biggr)^{\frac{1}{2b}}. $$

2. For 2-NPC nilpotent Lie groups (for details, see [9]), let $\{0\}=\mathfrak{g}_{0} \subset\mathfrak {g}_{1} \subset\cdots \subset\mathfrak{g}_{n} =\mathfrak{g}$ be a Jordan-Hölder sequence in $\mathfrak{g}$ such that $\mathfrak{g}_{m}=\mathfrak{z}(g)$ and $\mathfrak{h}=\mathfrak {g}_{n-2}$. Let us consider the ideal $[\mathfrak{g},\mathfrak{g} _{m+1}]$ of $\mathfrak{g}$ which is one or two dimensional in $\mathfrak{g}$. We discuss the two cases separately:

(a) $\dim{[\mathfrak{g},\mathfrak{g}_{m+1}]}=2$.

In this case, for every basis $\{X_{1},X_{2}\}$ of $\mathfrak{h}$ in $\mathfrak{g}$ and every $Y_{1} \in\mathfrak{g}_{m+1}\setminus\mathfrak{z}(\mathfrak {g})$, the vectors $Z_{1}=[X_{1},Y_{1}]$ and $Z_{2}=[X_{2},Y_{1}]$ are linearly independent and lie in the center of $\mathfrak{g}$. Assume that $\mathfrak{g}_{1}=\mathbb {R}\mbox{-} \operatorname{span}\{Z_{1}\}$, $\mathfrak{g}_{2}=\mathbb{R} \mbox{-} \operatorname{span}\{Z_{1},Z_{2}\}$. Let $Z_{3},\ldots,Z_{m}$ be some vectors such that $\mathfrak{z}(\mathfrak{g})=\mathbb{R}\mbox{-} \operatorname{span}\{Z_{1},\ldots,Z_{m}\}$ and $\mathcal{B}=\{Z_{1},\ldots,Z_{n}\}$ a Jordan-Hölder basis of $\mathfrak{g}$ chosen as follows:

• (i) $\mathfrak{z}(\mathfrak{g})=\mathbb{R}\mbox{-} \operatorname{span}\{Z_{1},\ldots ,Z_{m}\}$;

• (ii) $\mathfrak{h}=\mathbb{R}\mbox{-} \operatorname{span}\{Z_{1},\ldots,Z_{n-2}\}$;

• (iii) $\mathfrak{g}=\mathbb{R}\mbox{-} \operatorname{span}\{Z_{1},\ldots,Z_{n-2}, X_{1}=Z_{n-1},X_{2}=Z_{n}\}$.

For $m_{1}=m+1$ and $m+2\leq m_{2} \leq n-2$, we denote $Z_{m_{1}}=Z_{m+1}=Y_{1}$, $Z_{m_{2}}=Y_{2}$. These vectors can be chosen such that $\xi_{1}=\xi([X_{1},Y_{1}])\neq0$, $\xi_{2,2}=\xi([X_{2},Y_{2}])\neq 0$, for all $\xi\in\mathcal{W}$, where

$$ \mathcal{W} =\bigl\{ \xi=(\xi_{1},\xi_{2},\ldots, \xi_{m},0,0,\xi_{m+3},\xi _{m+4},\ldots, \xi_{n-2},0,0):\xi_{j} \in\mathbb{R}, \bigl\vert \operatorname {Pf}(\xi)\bigr\vert \neq 0\bigr\} . $$

Also, we have $\operatorname{Pf}(\xi)=\xi(Z_{1}) \xi([X_{2},Y_{2}])-\xi ([X_{1},Y_{2}]) \xi(Z_{2})$ and $\|\pi_{\xi}(f)\|_{\mathrm{HS}}$ is given by

$$\begin{aligned} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2} =& \bigl\vert h(\xi)\bigr\vert \int_{\mathbb{R}^{4}} \biggl\vert \mathscr{F} {(f\circ\exp)}\biggl(s_{2},s_{1},P_{n-2} \biggl(\xi,-\frac{t_{1}}{\tilde {\xi}_{1,1}},-\frac{t_{2}}{\tilde{\xi}_{2,2}} \biggr),\ldots, \\ & P_{m+3} \biggl(\xi,-\frac{t_{1}}{\tilde{\xi }_{1,1}},-\frac{t_{2}}{\tilde{\xi}_{2,2}} \biggr),t_{2},t_{1},\xi _{m},\ldots, \xi_{1}\biggr)\biggr\vert ^{2} \, ds_{1}\, ds_{2}\, dt_{1}\, dt_{2}, \end{aligned}$$

where h is the function defined by

$$ h(\xi) =\frac{|\xi_{1}\xi_{2,2}|^{2}}{|\xi_{1}\xi_{2,2}-\xi _{1,2}\xi_{2}|^{2}}, $$

$\xi_{i,j}=\xi([X_{i},Y_{j}])$, $\tilde{\xi}_{i,j}=\xi([X_{i}(\xi ),Y_{j}])$, and $P_{j}(\xi,t)$ is a polynomial function with respect to the variables $t=(t_{1},t_{2})$ and $\xi_{m+1},\ldots,\xi_{j}$ and rational in the variables $\xi_{1},\ldots,\xi_{m}$. Thus, one obtains the Heisenberg uncertainty inequality

$$ \frac{\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq \biggl(\int_{G}{\|x \|^{2a} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{\frac {1}{2a}} \biggl(\int_{\mathcal{W}}{\|\xi \|^{2b} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \frac{1}{|h(\xi)|^{b}|\operatorname{Pf}(\xi)|^{b-1}} \,d\xi \biggr)^{\frac{1}{2b}}. $$

(b) $\dim{[\mathfrak{g},\mathfrak{g}_{m+1}]}=1$.

In this case, we have $\operatorname{Pf}(\xi)=\xi([X_{1},Y_{1}])\cdot \xi([X_{2},Y_{2}])$ and

$$\begin{aligned} \mathcal{W} =&\bigl\{ \xi=(\xi_{1},\xi_{2},\ldots, \xi_{m},0,\xi _{m+2},\ldots ,\xi_{m+d+1},0, \xi_{m+d+3},\ldots,\xi_{n-2},0,0): \\ &\xi_{j} \in\mathbb{R}, \bigl\vert \operatorname {Pf}(\xi)\bigr\vert \neq 0\bigr\} . \end{aligned}$$

Also, $\|\pi_{\xi}(f)\|_{\mathrm{HS}}$ is given by

$$\begin{aligned} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2} =&\frac{1}{|\operatorname{Pf}(\xi)|}\int_{\mathbb{R}^{4}}\biggl\vert \mathscr{F} {(f \circ\exp)}\biggl(s_{2},s_{1},P_{n-2} \biggl(\xi ,- \frac{t_{1}}{\xi_{1}},-\frac{t_{2}+R(-\frac{t_{1}}{\xi_{1}},\xi _{1},\ldots,\xi_{m+d})}{\xi_{2,2}}\biggr), \\ &\ldots,t_{2},\ldots,P_{m+2} \biggl(\xi,-\frac{t_{1}}{\xi_{1}} \biggr),t_{1},\xi_{m},\ldots,\xi_{1}\biggr)\biggr\vert ^{2}\, ds_{1}\, ds_{2}\, dt_{1}\, dt_{2}. \end{aligned}$$

Thus, for $h(\xi)=\frac{1}{|\operatorname{Pf}(\xi)|}$, one obtains the Heisenberg uncertainty inequality,

$$ \frac{\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq \biggl(\int_{G}{\|x \|^{2a} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{\frac {1}{2a}} \biggl(\int_{\mathcal{W}}{\|\xi \|^{2b} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \bigl\vert \operatorname{Pf}(\xi)\bigr\vert \,d\xi \biggr)^{\frac{1}{2b}}. $$

3. For connected, simply connected nilpotent Lie groups $G=\exp {\mathfrak{g} }$ such that $\mathfrak{g}(\xi) \subset[\mathfrak{g},\mathfrak {g}]$ for all $\xi\in\mathcal{U}$ (for details, see [10]), we consider $S=\{ j_{1}<\cdots<j_{d}\}$ and $T=\{t_{1}<\cdots<t_{r}\}$ to be the collection of jump and non-jump indices, respectively, with respect to the basis $\mathcal{B}$. We have $j_{d}=n$ and

$$ \mathcal{W} =\bigl\{ \xi=(\xi_{1},\xi_{2},\ldots, \xi_{n})\in\mathfrak {g}^{\ast}: \xi _{j_{i}}=0, \text{for }j_{i} \in S\text{ with } \bigl\vert \operatorname{Pf}(\xi) \bigr\vert \neq 0\bigr\} . $$

Also, $\|\pi_{\xi}(f)\|_{\mathrm{HS}}$ is given by

$$ \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2} = \frac{|\xi ([X_{j_{1}},X_{n}])|}{|\operatorname{Pf} (\xi)|^{2}}\int_{\mathcal{W}}{\bigl\vert \mathscr{F} {(f\circ \exp )} (\xi,w )\bigr\vert ^{2}} \,dw, $$

where $\xi=(\xi_{t_{i}})_{t_{i}\in T}$ and $w=(w_{j_{i}})_{j_{i}\in S}$. Thus, for $h(\xi)=\frac{|\xi([X_{j_{1}},X_{n}])|}{|\operatorname {Pf}(\xi)|^{2}}$, one obtains the Heisenberg uncertainty inequality

$$ \frac{\|f\|_{2}^{ (\frac{1}{a}+\frac{1}{b} )}}{4\pi} \leq \biggl(\int_{G}{\|x \|^{2a} \bigl\vert f(x)\bigr\vert ^{2}} \,dx \biggr)^{\frac {1}{2a}} \biggl(\int_{\mathcal{W}}{\|\xi \|^{2b} \bigl\Vert \pi_{\xi}(f)\bigr\Vert _{\mathrm{HS}}^{2}} \frac{|\operatorname{Pf}(\xi)|^{b+1}}{|\xi ([X_{j_{1}},X_{n}])|^{b}} \,d\xi \biggr)^{\frac{1}{2b}}. $$

4. For low-dimensional nilpotent Lie groups of dimension less than or equal to 6 (for details, see [11]) except for $G_{6,8}$, $G_{6,12}$, $G_{6,14}$, $G_{6,15}$, $G_{6,17}$, an explicit form of $\|\pi_{\xi}(f)\|_{\mathrm{HS}}$ can be obtained. Thus, an explicit Heisenberg uncertainty inequality can be written down.

5. The classes mentioned above are distinct. For instance, $G_{5,5}$ is thread-like nilpotent Lie group, but it does not belong to the class mentioned in item 3. above. Also, $G_{5,3}$ belongs to the class mentioned in item 3. above, but it is not a thread-like nilpotent Lie group.