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Robert Bryant
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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 $$ |X_0|^2 = |X_1|^2 + \cdots + |X_{n+1}|^2. $$$$ |X_0|^2 = |X_1|^2 + \cdots + |X_{n+1}|^2, $$ by introducing homogeneous coordinates such that $z_i = X_i/X_0$. Now it's clearone 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_1=0$$X_0-X_{n+1}=0$, one sees that this puts all of the $(2n{+}1)$-sphere except the point $[1,1,0,\ldots,0]$$[1,0,0,\ldots,0,1]$ (the 'north pole', if you will) in this affine chart, which is a copy of $\mathbb{C}^{n+1}$. One can now show that there are complex coordinates on $\mathbb{C}^{n+1}$ To see this explicitly, saymake the coordinate change $X_0 = (Y+W)/2$, $(u,v_1,\ldots,v_n)$$X_{n+1}=(Y-W)/2$, leaving the others fixed, so that the imageequation for the hypersurface is described bybecomes $$ \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{Im}(u) = |v_1|^2 + \cdots + |v_n|^2. $$$$ \mathrm{Re}(u) = |v_1|^2 + \cdots + |v_n|^2. $$

This makes it easy to recognize as the Heisenberg model and, 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.

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 $$ |X_0|^2 = |X_1|^2 + \cdots + |X_{n+1}|^2. $$ Now it's clear 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 $X_0-X_1=0$, one sees that this puts all of the $(2n{+}1)$-sphere except the point $[1,1,0,\ldots,0]$ (the 'north pole', if you will) in this affine chart, which is a copy of $\mathbb{C}^{n+1}$. One can now show that there are complex coordinates on $\mathbb{C}^{n+1}$, say, $(u,v_1,\ldots,v_n)$ so that the image hypersurface is described by the equation $$ \mathrm{Im}(u) = |v_1|^2 + \cdots + |v_n|^2. $$

This makes it easy to recognize as the Heisenberg model and, in fact, the induced CR-isomorphism between the sphere minus a point and the Heisenberg group is 'conformally unitary'.

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.

added 1397 characters in body
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Robert Bryant
  • 108.4k
  • 8
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  • 453

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 $$ |X_0|^2 = |X_1|^2 + \cdots + |X_{n+1}|^2. $$ Now it's clear 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 $X_0-X_1=0$, one sees that this puts all of the $(2n{+}1)$-sphere except the point $[1,1,0,\ldots,0]$ (the 'north pole', if you will) in this affine chart, which is a copy of $\mathbb{C}^{n+1}$. One can now show that there are complex coordinates on $\mathbb{C}^{n+1}$, say, $(u,v_1,\ldots,v_n)$ so that the image hypersurface is described by the equation $$ \mathrm{Im}(u) = |v_1|^2 + \cdots + |v_n|^2. $$

This makes it easy to recognize as the Heisenberg model and, in fact, the induced CR-isomorphism between the sphere minus a point and the Heisenberg group is 'conformally unitary'.

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 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 $$ |X_0|^2 = |X_1|^2 + \cdots + |X_{n+1}|^2. $$ Now it's clear 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 $X_0-X_1=0$, one sees that this puts all of the $(2n{+}1)$-sphere except the point $[1,1,0,\ldots,0]$ (the 'north pole', if you will) in this affine chart, which is a copy of $\mathbb{C}^{n+1}$. One can now show that there are complex coordinates on $\mathbb{C}^{n+1}$, say, $(u,v_1,\ldots,v_n)$ so that the image hypersurface is described by the equation $$ \mathrm{Im}(u) = |v_1|^2 + \cdots + |v_n|^2. $$

This makes it easy to recognize as the Heisenberg model and, in fact, the induced CR-isomorphism between the sphere minus a point and the Heisenberg group is 'conformally unitary'.

Source Link
Robert Bryant
  • 108.4k
  • 8
  • 342
  • 453

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.