Let $\mathbb C^n$ be the n-dimensional complex field endowed with a positive definite hermitian form $H(z,w)$. The corresponding symplectic form is $E(z,w)= \Im (H(z,w))$, where $\Im $ denotes the imaginary part of complex numbers.

If we consider the set $H=\mathbb C^n \times \mathbb{U}(1)$, and define on it a law of combination \begin{align} (z,\lambda).(w,\mu) &= \left(z+w, \lambda \mu \, e^{i E(z,w)}\right) \end{align} then $H$ becomes a two step nilpotent group we will call the reduced Heisenberg group.

I will want to know what the expression of the Laplacian $\Delta$ on $H$ ?

Thank you in advance

  • $\begingroup$ Have you seen this reference mathoverflow.net/questions/227019/… ? $\endgroup$ – Jean Marie May 10 '16 at 15:53
  • $\begingroup$ No, this is not the case, I seek the expression of $\Delta$ of $H=\mathbb C^n \times \mathbb{U}(1) $, It is not of $H=\mathbb C^n \times \mathbb R$. $\endgroup$ – Z. Alfata May 10 '16 at 15:56

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