The Weyl transform of a measure

Abstract

(1) Suppose $\mu$ is a smooth measure on a smooth hypersurface of positive Gaussian curvature in ${\mathbb {R}}^{2n}$. If $n\ge 2$, then $W(\mu )$, the Weyl transform of $\mu$ is a compact operator, and if $p>n\ge 6$, then $W(\mu )$ belongs to the p-Schatten class. (2) There exist Schatten class operators with linearly dependent quantum translates.

Appendix

We prove Lemma 2.2 here. We need another technical lemma.

Lemma A.1

Let M be a compact connected smooth hypersurface in ${\mathbb {R}}^m$ with positive Gaussian curvature. Let H be an affine subspace of ${\mathbb {R}}^m$ and suppose H intersects M transversally. Let $N=M\cap H$. Then N is a smooth hypersurface in H, and for all $y\in N$ and $X, Y \in T_yN$, we have

$$\begin{aligned} K_{(N \subseteq H)}(X,Y) \ge K_{(M \subseteq {\mathbb {R}}^m)}(X,Y), \end{aligned}$$

where $K_{(N \subseteq H)}$ and $K_{(M \subseteq {\mathbb {R}}^m)}$ are the second fundamental forms with respect to the inward pointing normal on the smooth hypersurfaces N and M respectively.

Proof

By the Jordan–Brower separation theorem, $M=\partial \Omega$ where $\Omega$ is a bounded open set in ${\mathbb {R}}^m$. Let $\vec {n}_M$ denote the unit normal to M which points into $\Omega$. Let $N=M\cap H$. It is a standard fact from differential topology that N is a smooth hypersurface in H (see [5, p. 30]). Let $\vec {n}_N$ denote the unit normal vector to N which points into $\Omega \cap H$. Observe that $\vec {n}_N\cdot \vec {n}_M>0$.

If W is a submanifold of the Riemannian manifold Z, let $\vec {K}_{(W \subseteq Z)}$ denote the vector-valued second fundamental form of W in Z, i.e.,

$$\begin{aligned} \vec {K}_{(W \subseteq Z)}(X,Y)={\text {Proj}}_{TZ \ominus TW} \nabla _X Y, \end{aligned}$$

where $\nabla$ is the Levi–Civita connection on Z and $TZ \ominus TW$ denotes the orthogonal complement of the tangent bundle of W in the tangent bundle of Z. Moreover, if $\vec n$ is a unit normal vector field along W, we may define the second fundamental form of W with respect to $\vec n$ by

$$\begin{aligned} K_{(W\subseteq Z)}=\vec {K}_{(W\subseteq Z)}\cdot \vec {n}. \end{aligned}$$

Let $x\in N$ and let $X, Y \in T_xN$. We have orthogonal decompositions

$$\begin{aligned} \begin{aligned} \vec {K}_{(N \subseteq {\mathbb {R}}^m)}(X,Y) =&\; \vec {K}_{(N \subseteq H)}(X,Y) + \vec {K}_{(H \subseteq {\mathbb {R}}^m)}(X,Y), \qquad \text {and}\\ \vec {K}_{(N \subseteq {\mathbb {R}}^m)}(X,Y) =&\; \vec {K}_{(N \subseteq M)}(X,Y) + \vec {K}_{(M \subseteq {\mathbb {R}}^m)}(X,Y). \end{aligned} \end{aligned}$$

Since $\vec {K}_{(H \subseteq {\mathbb {R}}^m)}(X,Y)=0$, it follows that

$$\begin{aligned} \vec {K}_{(N \subseteq H)}(X,Y) = \vec {K}_{(N \subseteq M)}(X,Y) + \vec {K}_{(M \subseteq {\mathbb {R}}^m)}(X,Y) \end{aligned}$$

Since $\vec {K}_{(N \subseteq M)}(X,Y)$ is tangent to M, it follows that

$$\begin{aligned} \begin{aligned} K_{(N \subseteq H)}(X,Y) \vec {n}_N \cdot \vec {n}_M =&\; \vec {K}_{(N \subseteq H)}(X,Y) \cdot \vec {n}_M\\ =&\; \vec {K}_{(M \subseteq {\mathbb {R}}^m)}(X,Y) \cdot \vec {n}_M\\ =&\; K_{(M \subseteq {\mathbb {R}}^m)}(X,Y). \end{aligned} \end{aligned}$$

Therefore

$$\begin{aligned} K_{(N \subseteq H)}(X,Y) = \frac{K_{(M \subseteq {\mathbb {R}}^m)}(X,Y)}{ \vec {n}_M\cdot \vec {n}_N} \ge K_{(M \subseteq {\mathbb {R}}^m)}(X,Y). \end{aligned}$$

Taking $H=\{(x,y) \,|\,y \in {\mathbb {R}}^n\}$ in Lemma A.1, we see that $S_x$ is a smooth hypersurface in H with $K_{(S_x \subseteq H)}(X,Y) \ge K_{(S \subseteq {\mathbb {R}}^{2n})}(X,Y)$. Since $\Pi _2$ is an isometry between H and ${\mathbb {R}}^n$, it follows that $\Pi _2(S_x)$ is a smooth hypersurface in ${\mathbb {R}}^n$ with $K_{(\Pi _2(S_x) \subseteq {\mathbb {R}}^n)}(X,Y) \ge K_{(S \subseteq {\mathbb {R}}^{2n})}(X,Y)$. Therefore the smallest principal curvature of $\Pi _2(S_x)$ is greater than or equal to the smallest principal curvature of S. Therefore the Gaussian curvature of $S_x$ is bounded below by the $(n-1)$-th power of the smallest principal curvature of S.