The $C^{*}$-algebras of connected real two-step nilpotent Lie groups

Abstract

Using the operator valued Fourier transform, the $C^{*}$-algebras of connected real two-step nilpotent Lie groups are characterized as algebras of operator fields defined over their spectra. In particular, it is shown by explicit computations, that the Fourier transform of such $C^{*}$-algebras fulfills the norm controlled dual limit property.

Appendix

Lemma 1

Let V be a finite-dimensional euclidean vector space and S an invertible, skew-symmetric endomorphism. Then V can be decomposed into an orthogonal direct sum of two-dimensional S-invariant subspaces.

Proof

S extends to a complex endomorphism $S_{\mathbb {C}}$ on the complexification $V_{\mathbb {C}}$ of V, which has purely imaginary eigenvalues.

If $i\lambda \in i{\mathbb {R}}$ is an eigenvalue, then also $-i\lambda$ is a spectral element. Denote by $E_{i\lambda }$ the corresponding eigenspace. These eigenspaces are orthogonal to each other with respect to the Hilbert space structure of $V_{\mathbb {C}}$ coming from the euclidean scalar product $\langle \cdot ,\cdot \rangle$ on V.

Let for $i\lambda$ in the spectrum of $S_{\mathbb {C}}$

$$\begin{aligned} V^{\lambda }:=(E_{i\lambda }+E_{-i\lambda })\cap V. \end{aligned}$$

If $\lambda \ne 0$, $dim (V^{\lambda } )$ is even and $V^{\lambda }$ is S -invariant and orthogonal to $V^{\lambda '}$, whenever $|\lambda |\ne |\lambda '|$:

Indeed, one then has for $x\in V^{\lambda }, x'\in V^{\lambda '}$ that

$$\begin{aligned} x+iy\in & {} E_{i\lambda } \quad \text {and} \quad x-iy\in E_{-i\lambda }\quad \text { for some }y\in V \quad \text {as well as}\\ x'+iy'\in & {} E_{i\lambda '} \quad \text {and} \quad x'-iy'\in E_{-i\lambda '} \quad \text { for some }y'\in V. \end{aligned}$$

Therefore,

$$\begin{aligned} \langle {x+iy},{x'+iy'}\rangle =0 \quad ~\text {and} \quad \langle {x-iy},{x'+iy'}\rangle =0. \end{aligned}$$

Thus, one has

$$\begin{aligned} \langle {x},{x'+iy'}\rangle =0 \quad \text {and hence} \quad \langle {x},{x'}\rangle =0. \end{aligned}$$

Suppose that $dim (V^{\lambda } )>2$, choose a vector $x\in V^{\lambda }$ of length 1 and let $y=S(x)$. Since $S_{\mathbb {C}}^2=-\lambda ^2 \text {Id}$, both on $E_{i\lambda }$ and on $E_{-i\lambda }$,

$$\begin{aligned} S(y)=S^2(x)=-\lambda ^2 x. \end{aligned}$$

This shows that $W_1^{\lambda }:= \text {span} \{x,y\}$ is an S-invariant subspace of $V^{\lambda }$. If $V_1^{\lambda }$ denotes the orthogonal complement of $W_1^{\lambda }$ in $V^{\lambda }$, then $V_1^{\lambda }$ is S -invariant, since $S^t=-S$.

In this way one can find a decomposition of $V^{\lambda }$ into an orthogonal direct sum of two-dimensional S -invariant subspaces $W_j^{\lambda }$ and by summing up over the eigenvalues, one obtains the required decomposition of V.