Contents

V. Balakrishnan | PDF | << prev

Connection between SO(3) and SU(2) :-

We know that in 3-Dimensional space, I would like to generate rotations (linear transformation of the coordinates, which leaves a point (origin) unchanged. Its linear and homogeneous. also \(|M_T| = 1\). (\(M_T\) = transformation matrix). These three conditions suffice to define rotations and they form a group, this group is called \(SO(3)\).

where `\(S\)' stands for \(|M_T| = +1\) (special)

`\(O\)' stands for orthogonal (because the transformations has to be orthogonal to ensure that distance b/w two points remain unchanged, \(|x_1 - x_2| = |x_1' - x_2'|\).

`\(3\)' : comes from 3-diamensions.

The set of \(3\times3\) matrices, which are orthogonal, with determinent \(+1\), they form a group. These matrices from a representation of the abstract group of rotations.

Rotations are operators in their abstract, but they are explicitly represented by set of orthogonal, \((3\times3)\) matrices with \(\det M_T = +1\).

Parameter space of the set of rotations

what is parameter space of set of rotations ?

In other words what are values of angles that specifies the possible rotations. Now these can be specified in many ways, as you know you can go to one coordinate system to a rotated one by specifieng three Euler angles, but you can specify those euler angles in many different ways, they all turn out to be equivalent with each other but there is nothing unique about it.

The most convenient way of specifying these rotations is to say that the rotations occur about some axis (\(\hat{n}\)) in space with respect to some fixed coordinate system through a certain amount of rotation. (angle).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
\draw[->] (0,0) -- (0,2.0);
\draw[->] (0,0) -- (2.4,0);
\draw[->] (0,0) -- (-1.2,-1.0);
\draw[->,thick] (0,0) -- (1.5,1.7) node[above right] {$\hat{n}$};
\draw[dashed] (0.62,0.55) ellipse (0.85 and 0.32);
\draw[->] (1.47,0.62) arc (0:300:0.85 and 0.32);
\end{tikzpicture}

\(R(\hat{n}, \psi)\) where \(\hat{n}\) is specified by \((\theta, \phi)\), so that

\[ \hat{n} = \left( \sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta \right) \] \[ 0 \le \theta \le \pi \ , \qquad 0 \le \phi < 2\pi \ , \qquad 0 \le \psi \le 2\pi \]
equation Locus
\(x = \) const plane
\(R = \) const w.r.t sphere
\(\theta = \) const. half cone
\(\phi = \) const half plane

Now we can specify Rotation using three variables \((\theta, \phi, \psi)\) and we can put these sets \((\theta, \phi, \psi)\) as point in certain space.

Lets model \((\theta, \phi, \psi)\) on sphere (solid sphere).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (0,0) circle (1.2);
\draw[dashed] (0,0) ellipse (1.2 and 0.4);
\draw[->] (0,0) -- (0,1.4);
\draw[->,thick] (0,0) -- (0.86,1.02);
\draw[dashed] (0,0) -- (0.86,-0.29);
\draw (0,0.62) arc (90:50:0.62);
\node at (0.34,0.78) {\scriptsize $\theta$};
\draw (0.62,0) arc (0:-19:0.62);
\node at (0.82,-0.1) {\scriptsize $\phi$};
\node[above right] at (0.86,1.02) {\scriptsize $\hat{n}$};
\end{tikzpicture}

If we cleverly choose \(R = 2\pi\) then length of arrow can represent amount of rotation \((\psi)\).

points in such solid sphere coorespondes to only one rotation. (But here point of origin & all points on sphere should be same, so this representation is not good).

But it is a fact of 3 Dimencion that rotating an object through \(\pi\) about \(\hat{n}\) is same as rotating about \((-\hat{n})\) through \(\pi\).

so \(R = \pi\) is enough.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.1,>=Stealth]
\draw (0,0) circle (1.1);
\draw[dashed] (0,0) ellipse (1.1 and 0.36);
\draw (-1.1,0) -- (1.1,0);
\draw[->] (0,0) -- (0.78,0.78);
\node at (0.3,0.42) {\scriptsize $\pi$};
\node[right] at (1.15,0.55) {$(\pi, \theta, \phi)$};
\node[right] at (2.6,0.55) {$\pi_1(V)$};
\node[right,align=left] at (3.5,0.55) {(Fundamental homotopy\\ group)};
\fill (0.78,0.78) circle (1.1pt);
\node[above] at (0.78,0.85) {\scriptsize $A$};
\fill (-0.78,-0.78) circle (1.1pt);
\node[below] at (-0.78,-0.85) {\scriptsize $B$};
\end{tikzpicture}

But there is further complication, \(A\)' and \(B\)' are mathematically identical points; in this parameter space because both coresponds to rotation by \(\pi\) about \(\hat{n}\) axis or its opposite and they physically cooresponds to the same rotation.

So the parameter space of \(SO(3)\) is complicated. It is a solid sphere, its radius is \(\pi\). Such a space can not be represented in 3-Ecludien dimensions. But we can look at all of its mathematical property given this property.

Connectivity of the parameter space

This space is connected. A connected space is one where you can go from any point in this space to any other point in space continuously, without leaving the space. so this space is certainty connected.

But it is not simply connected, a simply connected space is one where any path, any contineous close path in the space can be contineously deformed or shrunk to a point without leaving the space.

connected and simply connected :

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\begin{tikzpicture}[scale=0.9,>=Stealth]
\draw plot[smooth cycle,tension=0.8] coordinates {(-1.6,0.3) (-1.1,1.1) (0.3,1.3) (1.5,0.8) (1.7,-0.3) (0.8,-1.0) (-0.6,-0.9) (-1.5,-0.4)};
\draw[->] (-1.0,0.1) -- (-0.6,0.35);
\end{tikzpicture}

any closed path can be shrunk continuously to a point without leaving this paper. so this space is connected as well as simply connected.

Not connected, Not simply connected

so a space for example has pieces seperated.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.85,>=Stealth]
\draw plot[smooth cycle,tension=0.85] coordinates {(-0.5,0.5) (0.1,0.7) (0.5,0.2) (0.3,-0.5) (-0.35,-0.6) (-0.65,-0.05)};
\begin{scope}[xshift=2.6cm]
\draw plot[smooth cycle,tension=0.85] coordinates {(-0.5,0.5) (0.1,0.7) (0.5,0.2) (0.3,-0.5) (-0.35,-0.6) (-0.65,-0.05)};
\end{scope}
\draw[->] (1.35,-0.9) .. controls (2.1,-1.3) .. (3.0,-1.3);
\node[right,align=left] at (3.1,-1.3) {This seperation does not belong to the\\ space, so is not connected. (we can't join\\ the contineously)};
\end{tikzpicture}

connected but not simply connected :-

For example :-

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
\draw plot[smooth cycle,tension=0.8] coordinates {(-1.5,0.4) (-0.9,1.2) (0.4,1.3) (1.4,0.6) (1.4,-0.5) (0.4,-1.1) (-0.9,-0.9) (-1.5,-0.3)};
\draw plot[smooth cycle,tension=0.85] coordinates {(-0.55,0.75) (0.0,0.95) (0.45,0.5) (0.1,0.05) (-0.5,0.15)};
\draw[fill=black!12] plot[smooth cycle,tension=0.85] coordinates {(-0.35,0.65) (0.05,0.78) (0.3,0.45) (0.0,0.2) (-0.35,0.3)};
\draw plot[smooth cycle,tension=0.9] coordinates {(-0.75,-0.35) (-0.5,-0.2) (-0.3,-0.45) (-0.55,-0.65)};
\draw[->] (0.55,0.6) -- (1.9,0.6);
\node[right] at (1.9,0.6) {Hole};
\draw[->] (-0.7,-0.55) .. controls (-0.4,-1.3) .. (0.6,-1.55);
\node[right] at (0.6,-1.55) {can shrink to zero};
\node[right,align=left] at (2.0,-0.35) {space is connected (can go from\\ Any point to another point\\ without leaving the space)};
\end{tikzpicture}

path does not shrink to zero. so space is not simple connected.

(Q) If it is Not simply connected, what kind of connectivity it have ?

  1. Paths are equivalent to each other if you can deform one to the other, continueously. 
    \usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw plot[smooth cycle,tension=0.8] coordinates {(-1.5,0.4) (-0.9,1.2) (0.4,1.3) (1.4,0.6) (1.4,-0.6) (0.4,-1.2) (-0.9,-1.0) (-1.5,-0.3)};
\draw (-0.95,0.5) circle (0.28);
\node at (-0.95,0.5) {\scriptsize 1};
\draw (-0.1,0.7) circle (0.28);
\node at (-0.1,0.7) {\scriptsize 2};
\draw (-0.35,-0.05) circle (0.26);
\node at (-0.35,-0.05) {\scriptsize 3};
\draw (0.25,-0.5) circle (0.34);
\node at (0.25,-0.5) {\scriptsize 4};
\draw[fill=black!12] plot[smooth cycle,tension=0.9] coordinates {(0.1,-0.42) (0.35,-0.32) (0.45,-0.6) (0.2,-0.72)};
\draw[->] (0.6,-0.55) -- (1.75,-0.75);
\node[right] at (1.75,-0.75) {Hole};
\node[right,align=left] at (1.8,0.6) {all 3 paths are equivalent to each\\ other. $(1,2,3)$};
\end{tikzpicture}

Path (4) is not equivalent to path \((1,2,3)\) as path (4) can't be shrunk to a point.

(Q) what are classes of paths that can all be deform to each other.

All paths that can be deformed to each other forms a class known as equivalence class. (They are all equivalent to each other).

Fundamental homotopy group pi_1(V)

(Q) what are different equivalence classes of paths that you have in a space.

It turns out by obvious rule of composition of class, these equivalence class forms elements of a group. This group is called fundamental group of space.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw plot[smooth cycle,tension=0.8] coordinates {(-1.4,0.4) (-0.8,1.2) (0.5,1.3) (1.4,0.6) (1.3,-0.6) (0.3,-1.1) (-0.9,-0.9) (-1.4,-0.2)};
\draw (-0.1,0.35) circle (0.62);
\node at (-0.42,0.62) {\scriptsize 1};
\draw (0.35,-0.3) circle (0.35);
\node at (0.35,-0.02) {\scriptsize 2};
\draw[fill=black!12] plot[smooth cycle,tension=0.9] coordinates {(0.22,-0.35) (0.45,-0.25) (0.55,-0.52) (0.3,-0.62)};
\node[right,align=left] at (1.8,0.5) {Total path is composed\\ of path \textcircled{1} and path \textcircled{2}.};
\end{tikzpicture}

and together they form a element of a group (3rd path in a group), \(\pi(V')\).

example :-

Suppose the space is \(S^1\) (1-D circle). what are equivalence classes of paths ?

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw[thick] (0,0) circle (0.8);
\node[above right] at (0.6,0.6) {$S^1$};
\begin{scope}[xshift=3.4cm]
\draw (0,0) ellipse (0.45 and 0.32);
\end{scope}
\end{tikzpicture}

all these paths can shrunk to a point, while this complete circular path can no longer be shrunk to point.

Also winding of such path can also not be able to shrunk to a point, so

\[ \pi_1(S^1) = \mathbb{Z} \quad \text{(Integers)} \]

what is pi_1(S^2) :-

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (0,0) circle (1.0);
\draw[dashed] (0,0) ellipse (1.0 and 0.32);
\draw (0.05,0.55) ellipse (0.3 and 0.18);
\node[right,align=left] at (1.4,0.2) {all paths can shrunk to a\\ point without leaving the space\\ (2-D space).};
\end{tikzpicture}

``You can not lasso basket ball, rubber will slip off and shrunk to (\(\sim\)zero) again''.

\[ \pi_1(S^2) = \{1\} \ \text{ or } \ \{e\} = 0 \quad \text{(only one element)}. \]

Similarly, \(\pi_1(S^3) = 0\).

\(\pi_1(T^2)\) :-

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (0,0) ellipse (1.3 and 0.75);
\draw (0,0) ellipse (0.5 and 0.22);
\draw (-0.9,0.1) arc (180:360:0.9 and 0.45);
\draw (0.55,0.05) ellipse (0.24 and 0.5);
\end{tikzpicture}

so that

\[ T^2 \ \text{(torous)} \ = S^1 \otimes S^1 \] \[ \pi_1(T^2) = \mathbb{Z} \otimes \mathbb{Z} \]

Points on sphere :- (Rotation group points)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.1,>=Stealth]
\draw (0,0) circle (1.0);
\fill (0,0) circle (1.1pt);
\node[below right] at (0.03,-0.02) {\scriptsize $O$};
\draw[->] (0,0) -- (0.55,0.83);
\node[above right] at (0.55,0.83) {\scriptsize $A\ (R=\pi)$};
\draw[->] (0,0) -- (-0.55,-0.83);
\node[below] at (-0.55,-0.9) {\scriptsize $B$};
\node[left,align=right] at (-1.3,-0.5) {\scriptsize ($2\pi$ rotation)};
\end{tikzpicture}

since points \(A\) and \(B\) are identical this represents a closed path in that space (Hard to imagine, lol).

and it is completely distinct from other class of closed paths (which can be shrunk to a point). Because this path (\(B\)-\(O\)-\(A\) closed path) can not be shrunk to a point.

But I can rotate it again (\(2\pi\) angle \((\pi+\pi)\)) and shrink to zero by following trick.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.85,>=Stealth]
\draw (0,0) circle (1.0);
\draw (-0.62,-0.78) .. controls (-0.1,-0.1) and (0.1,0.1) .. (0.62,0.78);
\draw (-0.72,-0.68) .. controls (-0.2,-0.2) and (0.0,0.05) .. (0.5,0.86);
\node[above] at (0.6,0.86) {\scriptsize $B$};
\node[below] at (-0.7,-0.85) {\scriptsize $\hat{n}$};
\node[left,align=right] at (-1.5,0.6) {($4\pi$ rotation)\\ (shrinks to point)};
\draw[->] (1.3,0) -- (1.9,0);
\begin{scope}[xshift=3.2cm]
\draw (0,0) circle (1.0);
\draw (-0.5,-0.85) .. controls (0.15,-0.2) and (-0.15,0.3) .. (0.45,0.88);
\draw (-0.68,-0.72) .. controls (0.05,-0.05) and (-0.05,0.2) .. (0.3,0.94);
\end{scope}
\draw[->] (4.5,0) -- (5.1,0);
\begin{scope}[xshift=6.4cm]
\draw (0,0) circle (1.0);
\draw (-0.2,-0.97) .. controls (0.55,-0.3) and (0.35,0.35) .. (0.15,0.98);
\draw (-0.42,-0.9) .. controls (0.4,-0.15) and (0.2,0.4) .. (-0.05,0.99);
\end{scope}
\draw[->] (6.4,-1.6) -- (4.6,-1.6);
\begin{scope}[xshift=3.2cm,yshift=-1.6cm]
\draw (0,0) circle (0.72);
\end{scope}
\draw[->] (2.4,-1.6) -- (1.4,-1.6);
\begin{scope}[xshift=0.4cm,yshift=-1.6cm]
\draw (0,0) circle (0.6);
\end{scope}
\end{tikzpicture}

so there actually exists 2 class of objects in this parameter space, those that come to themselves after a rotation of \(2\pi\) and other objects which come to themselves after rotation of \(4\pi\).

Tensor and spinnor representations

This is the origin of \(\tfrac{1}{2}\) integer valued representations of rotation group.

\[ \begin{aligned} j &= 0, 1, 2, 3 \cdots && \text{(tensor representations)}\\ j &= \tfrac{1}{2}, \tfrac{3}{2}, \tfrac{5}{2} \cdots && \text{(spinnor representations)} \end{aligned} \]

so we know that in normal tensor representations

Tensor of rank `\(0\)' is a scalar.

Tensor of rank `\(1\)' is a vector. \(\cdots\) and so on.

The spinnors short of speak interpolate between these.

spinnors change sign on \(2\pi\) rotation, so they need one more \(2\pi\) rotation to get same orientation of object.

\(\pi_1(SO(3))\) is doubly connected \(= \mathbb{Z}_2\). (just two elements in group).

Lec 25 : Representation of SO(3) in SU(2)

let us complete the corespondence that was supposed to bring out between group of \(3\times3\) rotations and certain \(2\times2\) matrixes (\(SU(2)\)).

We recognize that rotations in \(3\times3\) space \(R\) is specified by unit vector \(\hat{n}(\theta,\phi)\) and \(\psi\) (amount of rotation) or \(R(\hat{n},\psi)\) led to parameter space (specified by \(\hat{n}, \psi\)) which was not simply connected and was doubly connected and there are two classes of closed paths in this space which could not be reduced to each other (two inequivalent classes of closed paths).

This was the main reason for the fact that we had single and double valued representations of the rotation group because we also saw that the second class of closed paths could be shrunk to a point (if we do such two rotations (\(4\pi\) total) instead of \(2\pi\)).

All those representations of the rotation group which transform such that when you went to a rotation of \(2\pi\), you return to the original state, are called tensor representations and others (rotation of \(4\pi\)) were called spinnor representations.

Representation of 3x3 Rotation group SO(3) in 2x2 matrix SU(2)

We normally represent position by three parameters in 3-D.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
\draw[->] (0,0) -- (0,1.6) node[above] {\scriptsize $z$};
\draw[->] (0,0) -- (1.9,0) node[right] {\scriptsize $y$};
\draw[->] (0,0) -- (-1.1,-1.0) node[below] {\scriptsize $x$};
\node[above right] at (0.05,1.35) {\scriptsize $(x,y,z)$};
\end{tikzpicture}

so that \(\vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\).

instead of writing \(\vec{r}\) in column matrix we can express another representation of position

\[ \vec{r}\cdot\vec{\sigma} = x\,\sigma_x + y\,\sigma_y + z\,\sigma_z = \begin{pmatrix} z & x-iy \\ x+iy & -z \end{pmatrix} \] \[ \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} , \qquad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} , \qquad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \]

So if we have \(\vec{r}\cdot\vec{\sigma}\) we can work back and find \(x, y, z\) as, (also, any \(2\times2\) matrix can be written in terms of unit matrix and pauli matrixes \(\sigma_i\))

So once \(\vec{r}\cdot\vec{\sigma}\) is given we can work back and find vector \(\vec{r}\) itself.

So there is a correspondence between vectors \(\vec{r}\) and matrices \(\vec{r}\cdot\vec{\sigma}\).

like wise we used Rotation matrix \((g) \in SO(3)\) (\(3\times3\) matrix)

\[ \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = g \begin{pmatrix} x \\ y \\ z \end{pmatrix} \ , \qquad \vec{r}\,' = g\,\vec{r} \]

in exactly same way we have \(2\times2\) matrix \((U)\) corresponding to element `\(g\)' such that \(U (\vec{r}\cdot\vec{\sigma}) U^{-1} = \vec{r}\,'\cdot\vec{\sigma}\).

Therefore this \(U\)' is representative of rotation matrix \(g\)'.

(Q) what kind of relation is there between \(g\)' and \(U\)' ?

Lec 25 (continued) : Stereographic projection and rotations

we can use stereographic projection to map surface of unit sphere to a plane (complex plane).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (-2.2,-0.45) -- (2.6,-0.45) -- (3.5,0.35) -- (-1.3,0.35) -- cycle;
\draw (0,0) circle (0.85);
\draw[dashed] (0,0) ellipse (0.85 and 0.28);
\fill (0,0.85) circle (1.1pt);
\node[above left] at (0.02,0.88) {\scriptsize $N\,(0,0,1)$};
\draw (0,0.85) -- (2.05,-0.45);
\fill (0.68,0.42) circle (1.0pt);
\node[above right] at (0.7,0.44) {\scriptsize $(\xi_1,\xi_2,\xi_3)$};
\fill (2.05,-0.45) circle (1.0pt);
\node[below] at (2.1,-0.5) {\scriptsize $z$};
\end{tikzpicture}

point on sphere \((\xi_1, \xi_2, \xi_3)\). For every point on sphere we have a point on plane and vice versa.

example :- The North pole is mapped on to `\(\infty\)'. The south pole is mapped on to origin. Equator \((\xi_1^2 + \xi_2^2 = 1)\) remains mapped to \((|z| = 1)\).

\[ \xi_1 = \frac{2x}{|z|^2+1} \ , \qquad \xi_2 = \frac{2y}{|z|^2+1} \ , \qquad \xi_3 = \frac{|z|^2-1}{|z|^2+1} \] \[ \left( \xi_1^2 + \xi_2^2 + \xi_3^2 = 1 \right) \quad \text{unit sphere} \]

So we map this Riemann sphere on this complex plane using stereographic projection, which are; \(x = \dfrac{\xi_1}{1-\xi_3}\) , \(y = \dfrac{\xi_2}{1-\xi_3}\).

This projection has many interesting property; example : It preserves/maps circles on spheres to either circles or straight lines.

Any latitude is mapped on the circle with cocentric origin, any longitude is mapped on the straight line passing through the origin. (does it pass through origin ?)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
\node[left] at (-2.4,0) {(1)};
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\node[above] at (0,0.85) {\scriptsize $N$};
\draw[dashed] (0,0.3) ellipse (0.74 and 0.22);
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\draw (0,0.8) -- (-1.05,-0.45);
\draw (0.3,-0.15) ellipse (0.95 and 0.28);
\draw[->] (2.1,0.1) -- (3.1,0.1);
\node[right] at (3.1,0.1) {circle on complex plane};
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
\node[left] at (-2.4,0) {(2)};
\draw (-2.1,-0.45) -- (2.0,-0.45) -- (2.9,0.35) -- (-1.2,0.35) -- cycle;
\draw (0,0) circle (0.8);
\fill (0,0.8) circle (1.0pt);
\node[above] at (0,0.85) {\scriptsize $N$};
\draw[dashed] (0,0) ellipse (0.28 and 0.8);
\draw (1.35,-0.35) -- (1.65,0.1);
\node[below right] at (1.5,-0.3) {\scriptsize line};
\end{tikzpicture}

Rotation in 3-D space and the Riemann sphere

Now Rotation in 3-D space would corresponds to rotating Riemann sphere and what does it do on the plane ? It induces of course a transformation on plane as well. It is easy to check (prove yourself) that if we rotate in physical 3-D space (rotate in \(x\)-\(y\) plane about \(\hat{z}\) axis) then the rotation matrix is (in simple form).

if, \(\hat{n} = \hat{e}_z\) , \(\psi = \gamma\)

\[ g \in SO(3) = \begin{pmatrix} \cos\gamma & \sin\gamma & 0 \\ -\sin\gamma & \cos\gamma & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

If we put that back to Riemann sphere and ask what that transformation corresponds to (the rotation in \(\xi_1\)-\(\xi_2\) plane about \(\xi_3\) axis) Then it is easy to see that

\[ U = \pm \begin{pmatrix} e^{i\gamma/2} & 0 \\ 0 & e^{-i\gamma/2} \end{pmatrix} \]

So the Transformation is represented as, \(U (\vec{r}\cdot\vec{\sigma}) U^{-1} = \vec{r}\,'\cdot\vec{\sigma}\), where

\[ \vec{r}\cdot\vec{\sigma} = x\sigma_1 + y\sigma_2 + z\sigma_3 = \begin{pmatrix} z & x-iy \\ x+iy & -z \end{pmatrix} \]

So the counterpart of `\(g\)' is to the pair of matrices

\[ U = \pm \begin{pmatrix} e^{i\gamma/2} & 0 \\ 0 & e^{-i\gamma/2} \end{pmatrix} \]

Similarly; The No transformation (i.e. No rotation at all) would corresponds to \((\gamma = 0)\)

\[ U = \pm \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \pm I \] \[ U (\vec{r}\cdot\vec{\sigma}) U^{-1} = I (\vec{r}\cdot\vec{\sigma}) I^{-1} = \vec{r}\cdot\vec{\sigma} = \vec{r}\,'\cdot\vec{\sigma} \]

Unitary and unimodular condition

The requirement that \(|r'| = |r|\) suffices that these matrices \((U)\) must be unitary and unimodular.

\[ \begin{aligned} \text{Unitary} &:- \quad U U^{\dagger} = I = U^{\dagger} U \quad \text{or} \quad \left( U^{\dagger} = U^{-1} \right)\\ \text{unimodular} &:- \quad |U| = +1 \end{aligned} \]

Important : All \(2\times2\) matrices, which satisfies these conditions would represent physical rotations in 3-D space.

A general such \(2\times2\) matrix has \(\begin{pmatrix} a+bi & c+di \\ e+fi & g+ih \end{pmatrix}\) 8 reel elements \((a,b,c,d,e,f,g,h)\).

\(U U^{\dagger} = I\) provides 4 conditions and further \(|U| = +1\) provides one more conditions. (So parameter left \(= 8-5 = 3\))

so we need \((8-5=3)\) 3 adependent parameters to specify rotation in physical 3-D space which is exactly what we need in \(SO(3)\), \(R(\hat{n},\psi)\), \(R(\theta,\phi,\psi)\).

upon solving one gets

\[ U = \begin{pmatrix} \alpha & \beta \\ -\beta^{*} & \alpha^{*} \end{pmatrix} , \qquad \left( |\alpha|^2 + |\beta|^2 = 1 \right) \ \text{ or } \ \alpha\alpha^{*} + \beta\beta^{*} = +1 \] \[ \left( \alpha_1^2 + \alpha_2^2 + \beta_1^2 + \beta_2^2 = 1 \right) , \qquad \alpha = \alpha_1 + i\alpha_2 \ , \quad \beta = \beta_1 + i\beta_2 \]
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