Consider the odd-dimensional sphere $S^{2n+1} \subset \mathbb{C}^{n+1}$. One may talk variously about its structure as a contact, CR or Einstein-Sasaki manifold, but I'm looking for some specific down-to-earth detail that is hard to track down, namely charts that are complements of points and which are 'unitary' in the following sense.


The tangent space to $S^{n+1}$ at a point $x$ has a subspace isomorphic to the hermitian complement to the complex span of $x$. These patch together to give a complex vector bundle on the sphere -- and I'm fairly sure this is the holomorphic tangent bundle $HS^{2n+1}$ in the CR setting. Taking for $U$ the standard complement of the south pole, there is an isomorphism (even a CR isomorphism!) $\phi\colon Heis\_n\simeq U$, where $Heis\_n$ is the Heisenberg group, diffeomorphic to $\mathbb{C}^n\times\mathbb{R}$. Hence we can consider the induced isomorphism $H\, Heis\_n \to HU$ and ask whether it is unitary with respect to the standard hermitian structure on the left, and the one induced from $\mathbb{C}^{n+1}$ on the right. This is all controlled by the isomorphism $\phi$, and if I'm not mistaken in my calculations, taking standard stereographic projection rewritten in complex coordinates doesn't do the trick (which is, making the relevant adjustments, only orthogonal).



> 
> What's an explicit chart for an odd-dimensional sphere (considered as embedded in $\mathbb{C}^{n+1}$) satisfying the above condition?
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Asking for such a chart (and analogously over the complement of the north pole) is equivalent to giving a local section of $SU(n+1) \to S^{2n+1}$ over $U$ (and $V$). Note that from a Riemannian point of view I want to consider the sphere with the homogeneous metric from the isomorphism $S^{2n+1} \simeq SU(n+1)/SU(n)$.


I have a hard time believing no one has written this down before.



'Unitary' charts on odd-dimensional spheres
**Part I: The original question:**


Now that the question has been clarified, I can answer it. The answer is 'no', there is no CR-isomorphism $\phi: \mathrm{Heis}\to U$ that is unitary on the holomorphic tangent bundles.


To see this, it's probably better to look at the dual $1$-forms. Let $\alpha:T S^{2n+1}\to\mathbb{R}$ be the (unique) $1$-form whose kernel is the holomorphic tangent bundle $HS^{2n{+}1}\subset TS^{2n+1}$ and such that the $2$-form $\mathrm{d}\alpha$ restricts to $HS^{2n{+}1}$ to be a positive $(1,1)$-form that defines the Hermitian structure on $HS^{2n{+}1}$. Similarly, let $\beta: T \mathrm{Heis}^{2n+1}\to\mathbb{R}$ be the (unique) $1$-form whose kernel is the holomorphic tangent bundle $H\mathrm{Heis}^{2n+1}\subset T\mathrm{Heis}^{2n+1}$ and such that the $2$-form $\mathrm{d}\beta$ restricts to $H\mathrm{Heis}^{2n+1}$ to be a positive $(1,1)$-form that defines the Hermitian structure on $H\mathrm{Heis}^{2n+1}$. 


If $\phi:\mathrm{Heis}\to U$ were a CR-isomorphism that was unitary on the corresponding holomorphic tangent bundles, one would necessarily have $\phi^\*\alpha = \beta$. However, because the integral of $\alpha\wedge(\mathrm{d}\alpha)^n$ over $S^{2n+1}$ is finite, its integral over $U$ is also finite. On the other hand, the integral of $\beta\wedge(\mathrm{d}\beta)^n$ over $\mathrm{Heis}$ is infinite because it is the integral of a left-invariant volume form over a noncompact Lie group. Thus, $\phi$ cannot exist.


By the way, it would not be hard to show, using the method of equivalence, that the inequivalence holds even locally, i.e., there is no open subset of $\mathrm{Heis}^{2n+1}$ that is unitarily CR-diffeomorphic to any open subset of $S^{2n+1}$. However, that would take a little more argument, and I am not sure that the OP would be interested.


**Part II: The modified question:**


Now that the OP has clarified (in the comments below) what is meant by 'comes from' and that it is not actually required that the CR-isomorphism be literally unitary, it's easy to see how to establish the desired CR-isomorphism:


First, it helps to realize the $(2n{+}1)$-sphere as a hypersurface in $\mathbb{CP}^{n+1}$ as the hypersurface 
$$
|X\_0|^2 = |X\_1|^2 + \cdots + |X\_{n+1}|^2,
$$ 
by introducing homogeneous coordinates such that $z\_i = X\_i/X\_0$.
Now one sees that the group $\mathrm{SU}(1,n{+}1)\subset \mathrm{SL}(n{+}2,\mathbb{C})$ acts transitively on this hypersurface (and preserves the CR-structure, of course). If you now 'de-projectivize' this by looking at the part of the hypersurface that lies in the in affine chart that is a complement to the hyperplane $X\_0-X\_{n+1}=0$, one sees that this puts all of the $(2n{+}1)$-sphere except the point $[1,0,0,\ldots,0,1]$ (the 'north pole', if you will) in this affine chart, which is a copy of $\mathbb{C}^{n+1}$. To see this explicitly, make the coordinate change $X\_0 = (Y+W)/2$, $X\_{n+1}=(Y-W)/2$, leaving the others fixed, so that the equation for the hypersurface becomes
$$
\tfrac12(Y\overline{W}+W\overline{Y}) = |X\_1|^2 + \cdots + |X\_n|^2,
$$
and now, on the complement of the hyperplane $W=0$, consider the affine coordinates
$$
u = \frac{Y}{W} = \frac{1+z\_{n+1}}{1-z\_{n+1}} \qquad\text{and}\qquad 
v\_i = \frac{X\_i}{W} = \frac{z\_i}{1-z\_{n+1}} \qquad 1\le i\le n, 
$$
Then the equation for the real hypersurface in these coordinates $(u,v\_1,\ldots,v\_n)$ on $\mathbb{C}^{n+1}$ is
$$
\mathrm{Re}(u) = |v\_1|^2 + \cdots + |v\_n|^2.
$$


This makes it easy to recognize as the Heisenberg model, and the formulae for $u,v\_i$ in terms of the $z\_i$ give the desired mapping.


In fact, the induced CR-isomorphism between the sphere minus a point and the Heisenberg group is 'conformally unitary' since all CR-isomorphisms between strictly pseudoconvex hypersurfaces are conformally unitary in the sense that the induced isomorphism between the corresponding holomorphic tangent spaces is always unitary up to a scalar multiple.


