##SU(2, \mathbb C)## parametrization using Euler angles

[cianfa72]

TL;DR Summary

About the $SU(2, \mathbb C)$ parametrization using Euler angles.

Hi,
I found on some lectures the following parametrization of $SU(2, \mathbb C)$ group elements

\begin{pmatrix}
e^{i(\psi+\phi)/2}\cos{\frac{\theta}{2}}\ \ ie^{i(\psi-\phi)/2}\sin{\frac{\theta}{2}}\\
ie^{-i(\psi-\phi)/2}\sin{\frac{\theta}{2}}\ \ e^{-i(\psi+\phi)/2}\cos{\frac{\theta}{2}}
\end{pmatrix}
where $\theta, \psi$ and $\phi$ are the three Euler angles. In particular $\theta$ runs in the closed interval $[0, \pi]$ whereas $\psi$ and $\phi$ in the range $[0, 2\pi]$ -- such a parametrization includes all and only $SU(2 , \mathbb C)$ group elements. That is actually a "closed box" in $\mathbb R ^3$ so to get a chart from it we need to exclude the box "boundary". This way we get a not global chart for $SU(2)$. Since we know it is homeomorphic to $\mathbb S^3$ we can cover it with at least two charts.

Apart the above chart, which is one of the other charts for it ? Thanks.

 

[martinbn]

Off topic: Can you quote a textbook that uses the notation $SU(2,\mathbb C)$?

 

[cianfa72]

Off topic: Can you quote a textbook that uses the notation $SU(2,\mathbb C)$?

I think it is not a standard notation, see for instance this lecture.

 

[cianfa72]

Btw the following map $\varphi: \mathbb C^2 \rightarrow \mathbb C^4$
$$(a,b)\mapsto \begin{bmatrix}
a & -\overline b \\
b & \overline a
\end{bmatrix}$$
is injective and continuous (as map from $\mathbb C^2$ to $\mathbb C^4$). The inverse map $\varphi^{-1}$ restricted from the image $A=\varphi(\mathbb C^2)$ to $\mathbb C^2$ should be the projection $\pi|_A$. Hence it is continuous in the subspace topology from $\mathbb C^4$ the set $A$ is endowed with. Therefore $\varphi$ is homeomorphis with its image $A$ (in the subspace topology).

Is the above correct ?

 

[cianfa72]

Any comment ? Thanks.

 

[pasmith]

TL;DR Summary: About the $SU(2, \mathbb C)$ parametrization using Euler angles.

Hi,
I found on some lectures the following parametrization of $SU(2, \mathbb C)$ group elements

\begin{pmatrix}
e^{i(\psi+\phi)/2}\cos{\frac{\theta}{2}}\ \ ie^{i(\psi-\phi)/2}\sin{\frac{\theta}{2}}\\
ie^{-i(\psi-\phi)/2}\sin{\frac{\theta}{2}}\ \ e^{-i(\psi+\phi)/2}\cos{\frac{\theta}{2}}
\end{pmatrix}
where $\theta, \psi$ and $\phi$ are the three Euler angles. In particular $\theta$ runs in the closed interval $[0, \pi]$ whereas $\psi$ and $\phi$ in the range $[0, 2\pi]$ -- such a parametrization includes all and only $SU(2 , \mathbb C)$ group elements. That is actually a "closed box" in $\mathbb R ^3$ so to get a chart from it we need to exclude the box "boundary". This way we get a not global chart for $SU(2)$. Since we know it is homeomorphic to $\mathbb S^3$ we can cover it with at least two charts.

Apart the above chart, which is one of the other charts for it ? Thanks.

Consider $\mathbb{R} \to \mathbb{S} : x \mapsto e^{ix}$. We can obtain a two-chart atlas for $\mathbb{S}$ from this by taking the domains to be open invervals of width $2\pi$ whose images between them cover $\mathbb{S}$; for example $(-\pi,\pi)$ which covers everything except -1 and $(0, 2\pi)$ which covers everything except 1.

Apply this logic to the above parametrization.

 

[cianfa72]

for example $(-\pi,\pi)$ which covers everything except -1 and $(0, 2\pi)$ which covers everything except 1.

You mean -1 and 1 as complex numbers (i.e. the pairs (-1,0) and (1,0) under the identification $\mathbb C \cong \mathbb R^2$).

As far as I can understand what you said, If we define $x =(\psi+\phi)/2$ and $y =(\psi - \phi)/2$ then we obtain a two-chart atlas for $SU(2)$ starting from the parametrization in post#1. However to get that atlas we need a similar approach for $\theta$ that runs in the closed interval $[0, \pi]$.

 

[fresh_42]

Off topic: Can you quote a textbook that uses the notation $SU(2,\mathbb C)$?

I try to use always the field. It is redundant in that case but a good habit in the general case, especially when groups with complex matrix entries are considered real manifolds. And it makes the mention of the characteristic unnecessary.

 

[martinbn]

I try to use always the field. It is redundant in that case but a good habit in the general case, especially when groups with complex matrix entries are considered real manifolds. And it makes the mention of the characteristic unnecessary.

But the point is the $\mathbb C$ is not the base field. If you view them as Lie groups then they are real groups (as in real manifolds) and the standard notation is $SU(2)$. If you view them as algebraic groups they are algebraic varieties over the real numbers and the standard notation is $SU(2,\mathbb C/\mathbb R)$. More generally $SU(2, E/ F)$ for a quadratic extension $E/ F$ are groups over the the field $F$. Over the quadratic extension $E$ the group splits and is isomorphic to $SL(2)$.

 

[fresh_42]

But the point is the $\mathbb C$ is not the base field.

Base field of what? $SU(2)=SU(2,\mathbb{C})$ has complex entries. If I consider it as a real manifold, I write $SU_\mathbb{R}(2,\mathbb{C})$ like I would indicate $V_\mathbb{F}$ as a vector space over $\mathbb{F}$ if there is a doubt about it. However, I try to avoid writing $SU_\mathbb{R}(2,\mathbb{C}),$ I prefer $\mathfrak{su}_\mathbb{R}(2,\mathbb{C})$ and let the group be what it is, complex. I generally do not like the fact that a) reals are most often automatically assumed to be the scalar field and b) that linear (in-)dependence almost never states the scalar field, which is sometimes a serious problem!

 

[martinbn]

Base field of what? $SU(2)=SU(2,\mathbb{C})$ has complex entries. If I consider it as a real manifold, I write $SU_\mathbb{R}(2,\mathbb{C})$ like I would indicate $V_\mathbb{F}$ as a vector space over $\mathbb{F}$ if there is a doubt about it. However, I try to avoid writing $SU_\mathbb{R}(2,\mathbb{C}),$ I prefer $\mathfrak{su}_\mathbb{R}(2,\mathbb{C})$ and let the group be what it is, complex. I generally do not like the fact that a) reals are most often automatically assumed to be the scalar field and b) that linear (in-)dependence almost never states the scalar field, which is sometimes a serious problem!

It has real dimension three. It cannot be complex since three is not even.

 

[fresh_42]

It has real dimension three. It cannot be complex since three is not even.

It is an algebraic group of matrices with complex entries like $GL(n,\mathbb{C})$ or $SL(n,\mathbb{C})$ are. I only use the same nomenclature that notes the field where the matrix entries are taken from.

Dimension kicks in if we consider them as Lie groups, i.e. consider the corresponding Lie algebra. And if we do so, it is even more important to distinguish the fields!

 

[martinbn]

It is an algebraic group of matrices with complex entries like $GL(n,\mathbb{C})$ or $SL(n,\mathbb{C})$ are. I only use the same nomenclature that notes the field where the matrix entries are taken from.

Dimension kicks in if we consider them as Lie groups, i.e. consider the corresponding Lie algebra. And if we do so, it is even more important to distinguish the fields!

As algebraic groups the unitary groups are not complex. They become isomorphic to the special linear groups over an algebraicly closed field (already over the quadratic extension).

 

[pasmith]

You mean -1 and 1 as complex numbers (i.e. the pairs (-1,0) and (1,0) under the identification $\mathbb C \cong \mathbb R^2$).

As far as I can understand what you said, If we define $x =(\psi+\phi)/2$ and $y =(\psi - \phi)/2$ then we obtain a two-chart atlas for $SU(2)$ starting from the parametrization in post#1. However to get that atlas we need a similar approach for $\theta$ that runs in the closed interval $[0, \pi]$.

You are only dealing with a quarter period in the $\theta$ dependence; the end points are distinct. You need only 4 charts whose domains are $(\theta,\psi,\phi) \in [0,\pi] \times A \times B$ where $A$ and $B$ are independently either $(-\pi,\pi)$ or $(0, 2\pi)$.

 

[cianfa72]

You are only dealing with a quarter period in the $\theta$ dependence; the end points are distinct. You need only 4 charts whose domains are $(\theta,\psi,\phi) \in [0,\pi] \times A \times B$ where $A$ and $B$ are independently either $(-\pi,\pi)$ or $(0, 2\pi)$.

Take for instance the chart $\varphi$ with domain $[0,\pi] \times (-\pi, \pi) \times (- \pi, \pi)$. Is this domain open in $\mathbb R^3$ ?

 

[jbergman]

Take for instance the chart $\varphi$ with domain $[0,\pi] \times (-\pi, \pi) \times (- \pi, \pi)$. Is this domain open in $\mathbb R^3$ ?

No. You have a closed interval for the first part.

 

[cianfa72]

No. You have a closed interval for the first part.

Exactly. Is there a way to "breakdown" that closed interval $[0,\pi]$ in open intervals to get overlapping charts for $SU(2)$ from open sets in $\mathbb R^3$ ?