• #Module 13, Lecture 34: The rotation group and all that
  • #Outline :-
  • #The rotation group :-
• #Module 13, Lecture 35
  • #Outline :-
• #Module 13, Lecture 36
  • #outline :-
  • #parameter space of SO(3) :-
  • #Example :-
  • #For parameter space of SO(3) :-
• #See also

overview | playlist | << prev

1. Rotations of the coordinate axis
2. orthogonality of rotation matrices.
3. proper and im-proper rotations
4. Generators of infinitesimal rotations in 3-D
5. Lie algebra of generators
6. Rotation generators in 3-D transforms like a vector.
7. The general rotation matrix in 3-dimensions.

We could just as well consider rotations in N-Dimensions so frequently we will talk about N-Dimensional euclidean space or ocassionally we will talk about rotation in 3-D space (euclidean space).

It turns out that there are major differences between rotations in even-dimensions or rotation in odd dimensions.

The notation used here will be of cartesian tensors, the index notation with summation convension. so indices \(i = 1,2,3\ldots N\) in N-Dimensional case or \(i=1,2,3\) in 3-D case.

So let us look at simplest case (i.e. 3-dimencional rotation). coordinate of a point is represented as \(X_i\) \(i=1,2,3\) and we knows that under rotation of coordinate axis, when origen is fixed, the distances do not change.

and the rotation is a linear transformation, in the sense that

\[ X_i \longrightarrow X_i' = R_{ij} X_j \] \[ \downarrow \]

• a) homogeneous transformation as origin does not change.
• b) linear transformation as each new coordinate is linear combination of old coordinates.

If \(\vec{r} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\) then
\(R = \begin{pmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{pmatrix}\)

\[ \vec{r}\,' = R\,\vec{r} \]

ex, for rotation along \(z\) axis (in x-y plane) about angle `\(\varphi\)'.

\[ R = \begin{pmatrix} \cos\varphi & \sin\varphi & 0 \\ -\sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

But more general way of writing a rotation is to parametrise this rotation in some suitable form.

\[ R(\hat{n},\varphi) \;\longrightarrow\; \text{Rotation matrix for axis of rotation } \hat{n}(\theta,\phi) \]

by amount of rotation `\(\varphi\)'

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Note :-
"Rotation about an axis is artifact of 3-D (rotation about an axis is peculiar to 2-D) The reason is we could have rotated in 2-D (plane)/(euclidean plane) i.e the rotation occurs about origin (0,0) and there is no third dimension at all (axis)

If we rotate in 4-D for \(\vec{r} = (X_1\, X_2\, X_3\, X_4)\), The rotation can be done in \(X_1\)-\(X_2\) plane leaving \(X_3\)-\(X_4\) subspace unchanged. So we do not have any unique axis about which such rotation can be performed."

The more general way of rotation is to say rotation in a plane so for 2-D one plane exist and 2 axis

for 3-D -- 3 mutually orthogonal planes exist, and 3 axis

for 4-D -- 6 mutually " planes exist, and 4 axis.

So, for three dimensions only, no. of mutually orthogonal planes is equal to no. of axis.

\[ \left( \text{No of } \perp \text{ planes} \Rightarrow \frac{n(n-1)}{2} = n \quad \text{only for } n=3 \right) \]

technically

So If we say rotation about \(\hat{z}\) axis, we mean (rotation in x-y plane.

\[ \left\{ \begin{aligned} &\text{for 4-D we have 6 mutually orthogonal planes.}\\ &i =, \quad 1 \quad 2 \quad 3 \quad 4 \qquad (N_{C_2}) \text{ planes}\\ &\left. \begin{aligned} &12 \quad 13 \quad 14\\ &23 \quad 24\\ &34 \end{aligned} \right\} \text{ set of } \perp \text{ planes. in 4-D.} \end{aligned} \right\} \]

(Q) Can we associate an axis with rotation always ?

for even dimensional case \(n = 2, 4, 6 \cdots\)

we can not associate unique axis wito define rotation.

So we can not always associate an unique axis of rotation for N-D (even) euclidean (euclidean) space. | "But we can always associate an unique axis for odd dimensional euclidean space."

we could have rotation about an axis in 4-D space, but all rotations in 4-D need not have unique axis associated with them. (i.e. we can rotate about \(x_1\) axis leavng \(x_1\) unchanged but there are other rotations where we can not associate a unique axis).

Lets come back to rotation in 3-D. what more we can say about such rotations?

In general these rotations form a group. because of these

1. combination of two rotations \(R_1 R_2 = R_3\) is again a rotation.
2. Every rotation has an inverse \(R R^{-1} = I = R^{-1} R\) \[ \left( \begin{aligned} &R^{-1} \text{ is inverse rotation such that in succession we}\\ &\text{obtain } (R R^{-1}) = I \ (\text{i.e. No rotation at all}). \end{aligned} \right) \]

(?) (So its clear that there is correspondence between set of abstract rotations in N-dimensional space and set of matrices.) How?.

In N-Dimensional space matrices are \((n \times n)\) dimension. and are orthogonal with \(\det(M) = +1\).

\(R\) is an orthogonal matrix.

also set of \((n\times n)\) orthogonal matrix forms group.

hence \(R\) is an element of \(O(n)\)

\(\hookrightarrow\) group of \((n\times n)\) Orthogonal matrices.

orthogonality is easily established.

under a rotation :

\[ X_i X_i = X_i' X_i' \qquad -(1) \]

\(\begin{pmatrix} \text{distance} \\ \text{b/w two points} \end{pmatrix} \longrightarrow\)

\[ \begin{aligned} X_i' &= R_{ij} X_j , \qquad x_i' = R_{ik} X_k\\ X_i' X_i' &= R_{ij} R_{ik} X_j X_k\\ &= R^{T}_{ji} R_{ik} X_j X_k = (R^T R)_{jk} X_j X_k \qquad -(2) \end{aligned} \]

on the other hand

\[ X_i X_i = X_i' X_i' = X_j X_j = \delta_{jk} X_j X_k \qquad -(3) \]

equating equation (3) and (2)

\[ \begin{aligned} (R^T R)_{jk} X_j X_k &= \delta_{jk} X_j X_k\\ \Rightarrow \quad (R^T R)_{jk} &= \delta_{jk}\\ \Rightarrow \quad R^T R &= I \qquad\qquad (R^T = R^{-1}) \end{aligned} \]

So a matrix whose inverse is its transpose is called orthogonal matrix. So \(R\) is orthogonal matrix.

(we arrived at this result from our condition that distance between any two points should be unchanged under rotation)

Note :- for finite dimensional matrices left inverse = right inverse if they exists. \(R^T R = R R^T = I\).

\[ \begin{aligned} &\Rightarrow \quad |R^T R| = 1\\ &\text{as} \quad |R^T| = |R| , \text{ so } \ |R|^2 = 1\\ &\Rightarrow \quad |R| = \pm 1 \end{aligned} \]

If, *\(\det(R) = +1\)* , then `\(R\)' represents proper rotations. because \(|I|\) itself is \(+1\).

all Rotation matrix with \(\det(R) = +1\) can be constructed by moving contineously from the Identity. So we start with orignal coordinate system, and we want to go to proper rotation \(R\), we can do so by sucession of infinitesimal transformations, each of which has determinent \(+1\).

So this property of connected rotation is what defines proper rotation.

On the other hand there are matrices whose determinent is \((-1)\) \((|R| = -1)\) and they are discontineous transformations.

There is no way of reaching \(R\) such that \(|R| = -1\) by contineously moving from identity by making succession of infinitesimal transformations. There has to be discrete transformation somewhere. It would still be orthogonal but it can not have \(|R| = +1\)

ex

\[ \left. \begin{aligned} x_1' &= -x_1\\ x_2' &= x_2\\ x_3' &= x_3 \end{aligned} \right\} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \quad \left\{ \begin{aligned} &\text{Reflection}\\ &\text{transformation} \end{aligned} \right. \] \[ x' = R X \]

here `\(R\)' represents improper rotation.

Once we have performed improper rotation \((R_1)\) any proper rotation \((R_2)\) made. Can not make proper transformation.

\[ \begin{aligned} R_1 R_2 &= R_3\\ |R_1 R_2| &= |R_3|\\ |R_1||R_2| &= -1 \times 1 = -1 \Rightarrow |R_3| = -1 \Rightarrow R_3 \text{ is improper Rotation.} \end{aligned} \]

So this set \(O(n)\) has two disjoint pieces.

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If \(f, g \in SO(n)\)

1. \(f \cdot g = h \in SO(n)\)
2. \(f \cdot f^{-1} = I\) and \(I \in SO(n)\)

So \(|R| = +1\) forms group among themselves called \(SO(n)\).

where as set of elements with \((\det R = -1)\) does not form group as

\[ \text{if} \quad f, g \in \{|R| = -1\} \]

1. \(f \cdot g \notin A\)
2. \(f f^{-1} = I\) but \(I \notin A\) as *\(|I| = +1\)*. \(A = \{|R| = -1\}\) \(\downarrow\) set of all matix \(|R| = -1\) and are orthogonal

So `\(A\)' does not form group.

"Therefore the parameter space of \(O(n)\) is disconnected, there is a piece which is proper (that forms a subgroup) and there is a piece which has \(\{|R| = -1\}\). which are

\[ (\text{proper transformation}) \times (\text{improper transformation}) = (\text{improper transf.})'' \]

we have to be careful as, in even -- dimensional space parity transformation is not improper transformation. but is proper rotation.

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\[ \left. \begin{aligned} x_1' &= -x_1\\ x_2' &= -x_2 \end{aligned} \right\} \Rightarrow \quad x' = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \] \[ |R| = +1 \]

\(\hookrightarrow\) proper transformation

but reflection is improper transformation.

\[ R(\hat{e}_3, \varphi) = \begin{bmatrix} \cos\varphi & \sin\varphi & 0 \\ -\sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

\(\downarrow\) represents rotation about \(\hat{z}\) axis.

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let us see what infinitensimal rotation does to first order.

\[ R(\hat{e}_3, \delta\varphi) = \begin{pmatrix} \cos\delta\varphi & \sin\delta\varphi & 0 \\ -\sin\delta\varphi & \cos\delta\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

\(\downarrow\) first order \(\cos\delta\varphi = \underline{1}\)

\[ = \begin{pmatrix} 1 & \delta\varphi & 0 \\ -\delta\varphi & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] \[ R(\hat{e}_3, \delta\varphi) = I_{3\times 3} + \delta\varphi \underbrace{\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}}_{\displaystyle \downarrow} \]

plays role of generators of these rotations.

Repeated action by these generators (matrices) will lead to the rotation \(R\).

In Q.M it turn out that these generators of rotation correspond to physical quantities, namely the components of angular momentum. and in Q.M we like to represent these generators as hermitian operators. so let us write

\[ \begin{aligned} R(\hat{e}_3, \delta\varphi) &= I_{3\times 3} + i(\delta\varphi) \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\\ &= I + i J_3 (\delta\varphi) \end{aligned} \]

where \(J_3 = \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\) is hermitian.

i.e. \(\left( J_3^{\dagger} = J \right)\).

we need to make sure that repeated action of these generators indeed makes a finite rotation.

This means that for a finite rotation amount \(\varphi\)' we can reach to it in \(n\)' infiniteasimal rotations by `\(\delta\varphi\)' such that

\[ \underset{n \to \infty}{lt} \; n(\delta\varphi) = \varphi \]

This implies

\[ R(\hat{e}_3, \varphi) = (I + i J_3(\delta\varphi))\,(I + i J_3(\delta\varphi))\,(I + i J_3(\delta\varphi)) \cdots N \text{ times} \] \[ = \left( I + i J_3(\delta\varphi) \right)^n \] \[ \delta\varphi \to \frac{\varphi}{n} \] \[ R(\hat{e}_3, \varphi) = \lim_{n \to \infty} \left( I + \frac{i J_3 \varphi}{n} \right)^n \]

if \((i J_3 \varphi = X)\) \(I \ll 1\)

\[ \begin{aligned} &= \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\\ &= e^{x} \end{aligned} \] \[ R(\hat{e}_3, \varphi) = e^{i J_3 \varphi} \qquad \left( I + \frac{i J_3 \varphi}{1} + \frac{(i J_3 \varphi)^2}{2!} + \frac{(i J_3 \varphi)^3}{3!} + \cdots \right) \] \[ = \begin{pmatrix} \cos\varphi & \sin\varphi & 0 \\ -\sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

It turns out that the matrices like \(J_3\) obey lie algebra (obey certain algebric relation) and forms group too which is called as Lie group.

what is \(R(\hat{e}_1, \varphi) = ?\)

\[ R(\hat{e}_1, \delta\varphi) = I + i J_1 (\delta\varphi) \qquad (\text{Infinitesimal rotation}) \]

(what is \(J_1 = ?\))

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\[ \begin{aligned} x_1' &= x_1\\ x_2' &= x_2 \cos\varphi + x_3 \sin\varphi\\ x_3' &= -x_2 \sin\varphi + x_3 \cos\varphi \end{aligned} \] \[ R = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\varphi & \sin\varphi \\ 0 & -\sin\varphi & \cos\varphi \end{pmatrix} \] \[ R(\hat{e}_1, \delta\varphi) = I + i(\delta\varphi) \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{pmatrix} \]

\(\hookrightarrow\) \(J_1\)

Similarly,

\[ R(\hat{e}_2, \delta\varphi) = I + i(\delta\varphi) J_2 \]

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\[ \begin{aligned} x_1' &= x_1 \cos\varphi - x_3 \sin\varphi\\ x_2' &= x_2\\ x_3' &= x_1 \sin\varphi + x_3 \cos\varphi \end{aligned} \] \[ \begin{pmatrix} x_1' \\ x_2' \\ x_3' \end{pmatrix} = \begin{pmatrix} \cos\varphi & 0 & -\sin\varphi \\ 0 & 1 & 0 \\ \sin\varphi & 0 & \cos\varphi \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \] \[ R(\hat{e}_2, \delta\varphi) = \begin{pmatrix} 1 & 0 & -\delta\varphi \\ 0 & 1 & 0 \\ \delta\varphi & 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \] \[ = I + i\delta\varphi \begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{pmatrix} \] \[ J_2 = \begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{pmatrix} \]

If the no of mutually orthogonal planes is 3, then there are three generators, corresponding to infinitensimal rotations in each of three planes.

Interesting thing is, these generators \((J_1, J_2, J_3)\) they form lie algebra i.e.

1. They form LVS among themselves such that
2. commutator of any two of them belong to same LVS. \[ [J_j , J_k] = i \epsilon_{jk\ell} J_{\ell} \]

(Infinitesimal transformation) generators

where,

\[ \left\{ \begin{aligned} &\epsilon_{jk\ell} = \epsilon_{k\ell j} = \epsilon_{\ell jk} = 1\\ &j,k,\ell = 1,2,3\\ &\epsilon_{123} = \epsilon_{231} = \epsilon_{312} = 1\\ &\epsilon_{213} = \epsilon_{321} = \epsilon_{132} = -1\\ &\epsilon_{112} = \epsilon_{212} = \epsilon_{322} = \epsilon_{122} = 0\\ &(\text{if any two are same}) \end{aligned} \right. \] \[ \left\{ \begin{aligned} &\epsilon_{jk\ell} \text{ has 27 components, 21 are zero, and}\\ &\text{3 of them are } +1 \text{ and 3 of them are } -1. \end{aligned} \right\} \]

Lie algebra of \(SO(3)\) is represented as "\(so(3)\)".

General Rotation about \((\hat{n})\) axis :-

\[ R(\hat{n},\varphi) = e^{\,i (J_1 n_1 + J_2 n_2 + J_3 n_3)\varphi} \]

where,

\[ \begin{aligned} J_1 n_1 + J_2 n_2 + J_3 n_3 &= \text{Infinitesimal generator in } \hat{n} \text{ direction}\\ \varphi &= \text{amount of rotation} \end{aligned} \]

The difficulty is in fact that \(J_i J_j\) do not commute with each other. so we have difficulty in exponentiation of \((J_1 n_1 + J_2 n_2 + J_3 n_3)\).

\[ \text{for, } [A,B] \neq 0 \qquad e^{A+B} \neq e^{A} \cdot e^{B} \] \[ \downarrow \] \[ \left( I + \frac{(A+B)}{1!} + \frac{(A+B)^2}{2!} + \cdots \right) \] \[ \Rightarrow \left( I + \frac{(A+B)}{1} + \frac{A^2}{2!} + \frac{B^2}{2!} + \frac{AB}{2!} + \frac{BA}{2!} \right) + \cdots \] \[ \neq e^{A} \cdot e^{B} \]

"and this problem that \([J_j J_k] \neq 0\) is at root of all the phenomena we see in 3-D euclidean space."

Because it essentially says rotation about two different axis do not commute with each other.

i.e.

\[ R_1(\hat{e}_1, \varphi_1) R_2(\hat{e}_2, \varphi_1) \neq R_2(\hat{e}_2 \varphi) R_1(\hat{e}_1 \varphi) \]

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    \draw[->] (-0.95,-0.55) arc[start angle=200, end angle=-40, radius=0.32];
    \node[right] at (-0.35,-0.85) {$R_x$};
  \end{scope}
  \draw[->] (2.6,0.6) -- (3.4,0.6);
  % frame 2
  \begin{scope}[shift={(4.4,0)}]
    \draw[->] (0,0) -- (0,1.5) node[above] {$y$};
    \draw[->] (0,0) -- (1.8,0) node[right] {$-z$};
    \draw[->] (0,0) -- (-0.8,-1.4) node[below] {$x$};
    \draw[pattern=north east lines] (-0.2,-0.15) -- (-0.45,-0.6) -- (-0.6,-0.5) -- (-0.35,-0.05) -- cycle;
    \draw[->] (-0.45,0.75) arc[start angle=200, end angle=-30, radius=0.3];
  \end{scope}
  \draw[->] (6.9,0.6) -- (7.7,0.6);
  % frame 3
  \begin{scope}[shift={(9.0,0)}]
    \draw[->] (0,0) -- (0,1.5) node[above] {$y$};
    \draw[->] (0,0) -- (1.8,0) node[right] {$x$};
    \draw[->] (0,0) -- (-0.8,-1.4) node[below] {$z$};
    \draw (-0.5,0) -- (0,0);
    \draw[pattern=north east lines] (0.1,0) rectangle (0.8,0.18);
  \end{scope}
\end{tikzpicture}

we can see both rotations do not end up having different state. (rotation).

On the other hand if we are on a plane, If we rotate by angle \(\varphi\) and subsequently by angle \(\varphi'\). The result is rotation by angle \((\varphi + \varphi')\).

we could have added them in either order.

So Rotation group in 2-D which is group of \(2\times 2\) orthogonal matrices, \((SO(2))\) is called abelian group.

\[ R(\varphi) R(\varphi') = R(\varphi + \varphi') = R(\varphi') R(\varphi) \]

But that is not true in 3-D euclidean space because

\[ [J_j J_k] = i \epsilon_{jk\ell} J_{\ell} \neq 0. \]

Because of this \(e^{\,i(J_1 n_1 + J_2 n_2 + J_3 n_3)\varphi}\) is not trivial to find out.

In N-Dimensions, this problem is much worse, as we have \(n_{C_2}\) such generators and general transformation can be non-trivial to find out.

(These generators \((J_1 J_2 J_3)\) itself transform like vectors under rotation

"vector in 3-D is set of 3 elements which transform the exactly same way coordinates/axis transform under rotation".

\[ \vec{a} = (a_1\, a_2\, a_3) \]

under rotation \(\vec{a} \longrightarrow \vec{a}\,'\)

\[ \text{if} \quad x_i' = R_{ij} X_j \] \[ \text{and if} \quad a_i' = R_{ij} a_j \]

then we call \(\vec{a}\) "vector".

The physical law is invarient when it is expressed in terms of quantities whose transformation law is prescribed to us.

So the laws are form invarient provided they are expressed in terms of covarient objects (objects whose transformation law is already encoded in the object itself).

It turns out that \(J = (J_1, J_2, J_3)\) where \(\vec{J}\) transforms exactly as coordinates themselves under rotation.

1. The finite rotation formula for a vector
2. The general form of elements of \(U(2)\) and \(SU(2)\)
3. Relation between the groups \(SO(3)\) and \(SU(2)\). \[ R(\hat{n}, \psi) \;=\; e^{i(J_1 n_1 + J_2 n_2 + J_3 n_3)\psi} \;=\; e^{i(\vec{J}\cdot\hat{n})\psi} \qquad (n_1^2 + n_2^2 + n_3^2 = 1) \]

let \(M = i(\vec{J}\cdot\hat{n})\)

\[ J_1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{pmatrix} \qquad J_2 = \begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{pmatrix} \qquad J_3 = \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \] \[ M = i(J_1 n_1 + J_2 n_2 + J_3 n_3) = i \begin{pmatrix} 0 & -n_3 i & i n_2 \\ i n_3 & 0 & -i n_1 \\ -i n_2 & i n_1 & 0 \end{pmatrix} \] \[ M = \begin{pmatrix} 0 & n_3 & -n_2 \\ -n_3 & 0 & n_1 \\ n_2 & -n_1 & 0 \end{pmatrix} \]

or \(M_{ij} = \epsilon_{ijk} n_k\) (\(M_{ij}\) = matrix element)

\[ \begin{aligned} M^2_{ij} &= M_{il} M_{lj} \\ &= (\epsilon_{ilk} n_k)(\epsilon_{ljm} n_m) \\ &= -\epsilon_{lik}\epsilon_{ljm} n_k n_m \\ &= -(\delta_{ij}\delta_{km} - \delta_{im}\delta_{kj}) n_k n_m \\ &= n_i n_j - \delta_{ij} \end{aligned} \]

(\(\epsilon_{ilk} = -\epsilon_{lik}\))

\[ \begin{aligned} (M^3)_{ij} &= (M^2)_{ik} M_{kj} \\ &= -M_{ij} \quad \text{(or)} \quad (M^3 = -M) \end{aligned} \] \[ e^{iM\psi} \;=\; a I + b M + c M^2 \] \[ R_{ij}(\hat{n},\psi) = \boxed{(\cos\psi)\,\delta_{ij} + (1-\cos\psi)\, n_i n_j + (\sin\psi)\,\epsilon_{ijk} n_k} \] \[ x_i \to x_i' = R_{ij} x_j = (\cos\psi) x_i + (1-\cos\psi)(\vec{r}\cdot\hat{n}) n_i + (\sin\psi)\,\epsilon_{ijk} x_j n_k \] \[ \vec{r} \xrightarrow{\;R(\hat{n},\psi)\;} \vec{r}\,' = \underline{(\cos\psi)\,\vec{r} + (1-\cos\psi)(\vec{r}\cdot\hat{n})\,\hat{n} + (\sin\psi)(\vec{r}\times\hat{n})} \]

(Finite rotation formula)

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\begin{tikzpicture}[scale=0.8,>=Stealth]
\coordinate (O) at (0,0);
\draw[->] (O) -- (-0.9,0.9) node[above left] {$\hat{n}$};
\draw[->] (O) -- (-0.15,1.1);
\draw[->] (O) -- (1.1,0.8) node[right] {$\vec{r}$};
\draw[->] (O) -- (-1.2,0.25);
\draw[->] (O) -- (-1.0,-0.7);
\draw[->] (O) -- (1.2,-0.15);
\end{tikzpicture}

we can see that first term \((\cos\psi)\vec{r}\) is along the vector itself, second term has component along axis of rotation \((\hat{n})\) and third component is perpendicular to plane of \(\vec{r}\) and \(\hat{n}\).

If \(\vec{r}\cdot\hat{n} = |\vec{r}|\) (\(\vec{r}\) is along \(\hat{n}\)) then there is no rotation at all. for vector \(\vec{r}\).

\[ \begin{aligned} \vec{r}\,' &= (\cos\psi)\,\vec{r} + (1-\cos\psi)(\vec{r}\cdot\hat{n})\,\hat{n} \\ &= (\cos\psi)\,\vec{r} + (1-\cos\psi)\,|\vec{r}|\,\hat{n} \qquad {\small \left(\begin{aligned} \hat{n} &= \hat{r} \\ |\vec{r}|\,\hat{n} &= \vec{r} \end{aligned}\right)} \\ &= (\cos\psi + 1 - \cos\psi)\,\vec{r} \;=\; \vec{r} \end{aligned} \]

so any set of 3 no. which satisfy finite rotation formula is a vector.

but `J' which transforms like vector does not obey finite rotation formula; as this is not the rule of transformation for operators.

\[ \vec{r} \xrightarrow{\;R(\hat{n},\psi)\;} \vec{r}\,' \] \[ R(\hat{n},\psi) = (\cos\psi)\,\delta_{ij} + (1-\cos\psi)\, n_i n_j + (\sin\psi)\,\epsilon_{ijk} n_k \]

where,

\[ R(\hat{n},\psi) = e^{i(\vec{J}\cdot\hat{n})\psi} \qquad \text{and} \qquad [J_k, J_l] = i\,\epsilon_{klm} J_m \]

(\(\hat{J}\) need not be \(3\times3\) matrix. \(\hat{J}\) can have any dimensionality depends on object it acts upon.)

we know that \(J_1, J_2, J_3\) involves \(\sigma_i\) (pauli) matrix in them. and it turns out that if \(J_j = \tfrac{1}{2}\sigma_j\) where \(J_i\) = angular momentum operators. for spin(\(\tfrac{1}{2}\)).

\[ \begin{aligned} [J_i J_j] &= \tfrac{1}{4}\,[\sigma_i, \sigma_j] = \tfrac{1}{4}\, 2i\,\epsilon_{ijk}\sigma_k \\ &= i\,\epsilon_{ijk}\left(\tfrac{\sigma_k}{2}\right) \end{aligned} \] \[ \boxed{[J_i J_j] = i\,\epsilon_{ijk} J_K} \qquad \textcircled{1} \]

These operators \(J_j\) satisfies exactly same lie algebra as generators of rotation group \([J_k J_l] = i\epsilon_{klm} J_m\).

_(The fact is rotation in 3-D space is characterized by lie algebra of infinitesimal generators \((J_1, J_2, J_3)\).)_

any set of matrices which satisfies eq. ① represents rotation group in 3-D. (any set of 3 objects of any dimensionality which satisfies lie algebra, is represents rotation in 3-D). For instance we have ket vector \(|\psi\rangle\), describing state of particle/system and if we change coordinate axis by rotation about \(\hat{n}\) axis by angle \(\psi\). Then \(|\psi\rangle\) changes to \(|\psi'\rangle\) in changed frame.

\[ |\psi'\rangle = U(R)\,|\psi\rangle \]

\(\hookrightarrow\) unitary transformation which depends on \(R(\hat{n},\psi)\).

But \(U(R)\) turns out to be

\[ U(R) = e^{i(\vec{J}\cdot\hat{n})\psi} \qquad \left.\right\}\ \text{\small Rotation operator in hilbert space.} \]

as in general \(|\Psi\rangle\) is infinite dimensional vector, so \(\vec{J}\) has to be \(\infty\times\infty\) dimensional matrices.

Here rotation in coordinate system (in 3D) has induced a transformation in hilbert space given by the operator \(U\). (where `\(U\)' turns out to be infinite dimensional matrix).

ex.

Since \([J_i J_j] = i\,\epsilon_{ijk} J_K\) where \(\left(J_j = \tfrac{1}{2}\sigma_j\right)\) \(\left(\vec{J}_j \text{ is } 2\times2 \text{ matrix}\right)\)

Therefore \(\exists\) \((2\times2)\) matrix representation of the group \(SO(3)\).

It is called *\(SU(2)\)* \(\hookrightarrow\) unitary matrices.

\[ \boxed{U(\hat{n},\psi) = e^{i(\vec{\sigma}\cdot\hat{n})\frac{\psi}{2}}} \qquad \left.\right\}\ \text{\small only acts upon }2\times2\text{ matrix or }(2\times1)\text{ matrix} \] \[ U(\vec{r}\cdot\sigma)\,U^{-1} = (\vec{r}\,'\cdot\sigma) \]

(as the matrices \(\left(\tfrac{\sigma_i}{2} = J_i\right)\) satisfies lie algebra, \(U\) represents rotation in 3-D space.) But, we need 3 piece of information to specify vector in 3-D so it has to be act upon \(2\times2\) matrix rather than \((2\times1)\) matrix.

\[ U^\dagger = e^{-i(\sigma^\dagger\cdot\hat{n})\frac{\psi}{2}} \qquad (\sigma^\dagger = \sigma) \]

as, \(UU^\dagger = U^\dagger U = I\) ; \(\Rightarrow\) \(U^\dagger = U^{-1}\) \(\to\) `\(U\) is unitary matrix. \((2\times2)\).

But,

(Q) But how we implement \((2\times2)\) rotation matrix on 3-D space?

For any 3-D vector \(\vec{a} = (a_1, a_2, a_3)\)

we can construct \(\vec{a}\cdot\vec{\sigma} = a_1\sigma_1 + a_2\sigma_2 + a_3\sigma_3\)

\[ (\vec{a}\cdot\vec{\sigma}) = \begin{pmatrix} a_3 & a_1 - i a_2 \\ a_1 + i a_2 & -a_3 \end{pmatrix} \]

where coordinate \((x_1, x_2, x_3)\) are represented as:
\(\begin{pmatrix} x_3 & x_1 - i x_2 \\ x_1 + i x_2 & -x_3 \end{pmatrix}\)

just as

\[ \vec{r} \xrightarrow{\;R(\hat{n},\psi)\;} \vec{r}\,' \] \[ (\vec{r}\cdot\sigma) \xrightarrow{\;U\;} (\vec{r}\,'\cdot\vec{\sigma}) \] \[ U(\vec{r}\cdot\sigma)U^{-1} = (\vec{r}\,'\cdot\sigma) \]

which is,:-

\[ e^{i(\sigma\cdot\hat{n})\psi/2}\,(\vec{r}\cdot\vec{\sigma})\, e^{-i(\sigma\cdot\hat{n})\psi/2} = (\vec{r}\,'\cdot\vec{\sigma}) \qquad \text{---}\,\textcircled{2} \]

exponential of \((2\times2)\) matrix. \(e^{i(\sigma\cdot\hat{n})\frac{\psi}{2}}\) :-

\[ \boxed{\,e^{i(\sigma\cdot\hat{n})\psi/2} = I\left(\cos(\psi/2)\right) + i(\sigma\cdot\hat{n})\sin\frac{\psi}{2}\,} \qquad \text{---}\,\textcircled{3} \] \[ \left\{\text{In general.}\quad e^{i(\vec{\sigma}\cdot\vec{a})\theta} = I(\cos\theta) + \frac{i(\vec{\sigma}\cdot\vec{a})}{|a|}\sin\theta \right\} \qquad \text{---}\,\textcircled{4} \]

If we put back eq ③ in eq ② we get exactly the finite rotation formula.

So there is a connection between every \(3\times3\) orthogonal matrix with \((\det R = +1)\) of \(SO(3)\) grp has \(2\times2\) matrix of \(SU(2)\) grup.

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\begin{tikzpicture}[scale=0.8,>=Stealth]
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\node at (0,1.85) {$SO(3)$};
\node (R) at (-0.35,0.45) {$R$};
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\node (U) at (4.4,0.6) {$U$};
\node (mU) at (4.4,-0.9) {$-U$};
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\draw[->] (mU) to[out=180,in=-20] (R);
\node[align=center] at (-1.9,-1.5) {$R = \left(\underline{e^{i(\vec{J}\cdot\hat{n})\psi}}\right)$};
\node[align=center] at (7.6,-0.6) {$U = \left(\underline{e^{i(\sigma\cdot\hat{n})\frac{\psi}{2}}}\;,\;\underline{e^{-i(\sigma\cdot\hat{n})\frac{\psi}{2}}}\right)$};
\node at (3.3,-2.4) {2 to 1 homomorphism from $SU(2)$ to $SO(3)$};
\end{tikzpicture}

If we change \(U \to (-U)\)

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

So for every one element of \(SO(3)\) \(\exists\) 2 such unitary matrix of \(SU(2)\) group.

Every element of \(SO(3)\) group is parameterised by three parameters \((\theta, \phi, \psi)\) or \((\hat{n}(\theta,\phi), \psi)\). That should be same in \(SU(2)\).

What are no. of independent parameters in \(SU(2)\).

\[ \text{General } (2\times2) \text{ matrix} = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \]

\(M\) has to be unitary.

\[ M^\dagger = \begin{pmatrix} \alpha^* & \gamma^* \\ \beta^* & \delta^* \end{pmatrix} \] \[ M M^\dagger = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}\begin{pmatrix} \alpha^* & \gamma^* \\ \beta^* & \delta^* \end{pmatrix} = \begin{pmatrix} |\alpha|^2 + |\beta|^2 & \alpha\gamma^* + \beta\delta^* \\ \gamma\alpha^* + \delta\beta^* & |\gamma|^2 + |\delta|^2 \end{pmatrix} = I \]

(\(M\) to be unitary matrix)

\[ M M^\dagger = I \] \[ \Rightarrow \quad \begin{aligned} |\alpha|^2 + |\beta|^2 &= 1 \\ |\gamma|^2 + |\delta|^2 &= 1 \end{aligned} \quad , \quad \begin{aligned} \alpha\gamma^* + \beta\delta^* &= 0 \\ \gamma\alpha^* + \delta\beta^* &= 0 \end{aligned} \quad \begin{aligned} \gamma^* &= \frac{-\beta\delta^*}{\alpha} \\ \Rightarrow \ \delta &= \frac{-\gamma\alpha^*}{\beta^*} \end{aligned} \]

also \(|M| = +1\)

\[ \Rightarrow \quad \alpha\delta - \beta\gamma = e^{i\theta} \qquad {\small (\text{How?}).} \]

\(\longrightarrow\) \(|e^{i\theta}| = 1\) \(\left\{\text{}e^{i\theta}\text{ can be any complex \underline{no}.}\right.\)

\[ \alpha\left(\frac{-\gamma\alpha^*}{\beta^*}\right) - \beta\left(\frac{-\beta^*\delta}{\alpha^*}\right) \] \[ \frac{-|\alpha|^2\gamma}{\beta^*} - |\beta|^2\gamma \;=\; -(|\alpha|^2 + |\beta|^2)\frac{\gamma}{\beta^*} \;=\; -\frac{\gamma}{\beta^*} \] \[ \delta = \frac{-\alpha^*\gamma}{\beta^*} = \alpha^* e^{i\theta} \qquad \gamma = -\beta^* e^{i\theta} \]

The general \((2\times2)\) unitary matrix must be of the form,

\[ \begin{pmatrix} \alpha & \beta \\ -\beta^* e^{i\theta} & \alpha^* e^{i\theta} \end{pmatrix} , \qquad \text{with } |\alpha|^2 + |\beta|^2 = 1 \]

No of real parameters \((\alpha_1 + i\alpha_2)\), \((\beta_1 + i\beta_2)\) = 4 parameters but \(|\alpha|^2 + |\beta|^2 = 1\) makes it 3 parameters, we also have `\(\theta\)' as 4th parameter again.

If, \(|M| = +1\) (then \(\theta = 0\)) \(\Rightarrow\) we have 3 parameters in \(SU(2)\) where as original \(U(2)\) has four parameters.

Hence in \(SU(2)\) group, the most general matrix is

\[ \begin{pmatrix} \alpha & \beta \\ -\beta^* & \alpha^* \end{pmatrix} \qquad \begin{aligned} &\text{with } |\alpha|^2 + |\beta|^2 = 1 \\ &\text{or } \ \alpha_1^2 + \alpha_2^2 + \beta_1^2 + \beta_2^2 = 1 \end{aligned} \]

So the claim is now that we have two such unitary matrices for a element of \(SO(3)\).

1. 2 to 1 homomorphism b/w \(SU(2)\) and \(SO(3)\)
2. The parameter space of \(SU(2)\) and \(SO(3)\)
3. Double connectivity of \(SO(3)\)
4. The universal covering group of a lie group.
5. The groups \(SO(n)\) and \(spin(n)\)
6. Tensor and spinnor representations.
7. parameter space of \(U(n)\) and \(SU(n)\)
8. A bit about fundamental group (or first homotopy group) of a space

Note :- \(SU(2)\) is simply connected and connected group.

we have already seen that parameter space of \(SU(2)\) elements/group is four-dimensional.

\[ \alpha_1^2 + \alpha_2^2 + \beta_1^2 + \beta_2^2 = 1 \]

which represents hypersphere / 4-D sphere.

Just as we denote a 3-D sphere by \(S^2\) we denote 4-D sphere by \(S^3\) (\(S\to\) stands for surface).

\(S^2 \to\) surface of balloon

3-D surface --- \(S^3\).

So \(S^3\) is connected (i.e. every other point can be reached from any point without leaving the space). \(S^3\) is also simply connected as any closed loop can be shrunk to a point. (absence of holes in \(S^3\) surface/space).

\(*\) 2 to 1 homomorphism between \(SU(2)\) and \(SO(3)\).

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\begin{tikzpicture}[scale=0.8,>=Stealth]
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\draw[->] (0.4,-0.85) to[out=15,in=195] (5.7,0.45);
\node at (7.9,1.3) {$(\hat{n},\psi)$};
\node at (0.4,-2.3) {\underline{$SU(2)$}};
\node at (6.2,-2.3) {\underline{$SO(3)$}};
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(connected space. \(SO(3)\) has \(|R| = +1\) ; proper rotation)

\(\{I_2, -I_2\}\) is kernel of \(SU(2)\). It is set of elements which are mapped on to \(I_{3\times3}\) in \(SO(3)\).

\(\{I_2, -I_2\}\) also form a group by themselves. (group of order 2)

\[ I\,I = I, \quad I.(-I) = -I \;, \quad (-I)(+I) = -I \;, \quad (-I)(-I) = I. \]

So they form a cyclic group. (\(Z_2\)).

\(SO(3)\) is isomorphic (one to one connection) Not to \(SU(2)\) but is isomorphic upto a sign \((\pm)\).

\[ SO(3) \simeq SU(2)/Z_2 \qquad \to \text{order } \textcircled{2} \]

\(\hookrightarrow\) cyclic group of order (2) \(\{I, -I\}\) \(\swarrow\) quotiented with cyclic group of order 2

neighborhood of \(I_2\) is same as neighbourhood of \(I_3\) but the global topology is different.

elements of \(SO(3)\) are parameterized by \((\hat{n}, \psi)\) where \(\hat{n}(\theta,\phi)\) is axis of rotation about \(\hat{n}(\theta,\phi)\). This space can be realized by solid sphere of radius \(2\pi\). where \(\hat{n}\) of point in sphere gives \((\theta,\phi)\) and lenght of radius/vector provides amount of rotation.

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\draw (0,0) circle (1.0);
\draw[->] (0,0) -- (0.15,0.85);
\node[left] at (0.05,0.55) {$\hat{n}$};
\node at (2.0,1.5) {$\hat{n}(\theta,\phi)$};
\node at (4.6,0.4) {$0 < |\psi| \le 2\pi$};
\end{tikzpicture}

But It is fact of 3-Dimension that Rotation about \(\hat{n}\) by \(\pi\) is same as rotation about \((-\hat{n})\) by \(\pi\)

\[ R(\hat{n},\pi) = R(-\hat{n},\pi) \]

So radius of \(\pi\) is sufficient.

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\node[above] at (0.78,1.1) {$A$};
\node[right] at (0.85,1.15) {$R(\hat{n},\pi)$};
\draw[->] (0,0) -- (-0.75,-1.05);
\node at (-0.15,-0.75) {$\pi$};
\fill (-0.75,-1.05) circle (1.3pt);
\node[below] at (-0.72,-1.15) {$B$};
\node[right] at (-0.6,-0.8) {$-\hat{n}$};
\node[left] at (-1.1,-1.4) {$R(-\hat{n},\pi)$};
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points \(A\) and \(B\) must be same

So we can not represent this parameter space in 3-D. we need higher dimension to

represent/realize the object.

There is also a 2-D object that can not be embeded in 3-D space like klein bottle.

Similerly we can not model this in 3-D space.

The space/parameter space of \(SO(3)\) is connected but what kind of connectivity it have?

How to find connectivity of any such objects; weather they are simply connected, doubly connected \(\ldots\)?

what we do is to draw, & map loops on this space and if these loops can be shrunk to a point (contineously without leaving the space) then its simply connected. If these loops can not shrunk to point contineously then it means space is not simply connected and the connectivity depends on how many classes of such paths we can find.

"It will turn out that \(SO(3)\) is doubly connected There are exactly two classes of closed paths which are inequivalent to each other".

lets look at simple example, by considering group of rotations in a plane (one generator).

\[ SO(2) \] \[ R = \begin{pmatrix} \cos\psi & \sin\psi \\ -\sin\psi & \cos\psi \end{pmatrix} \qquad (0 \le \psi < 2\pi) \] \[ |R| = 1. \]

\(\to\) parameter space is one-dimensional \((S^1)\) \(\hookrightarrow\) parameter space of \(SO(2) = S^1\)

So all the rotations in \(SO(2)\) are parameterized by points on circle \((S^1)\).

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\draw (0,0) -- (0.9,0);
\draw (0,0) -- (0.55,0.71);
\node at (0.28,0.52) {$\psi$};
\end{tikzpicture}

So, parameter space of \(SO(2)\) is \(S^1\).

we can see that parameter space of \(SO(2)\) i.e. \(S^1\) is connected. (we can go from any point to another point in \(S^1\) without leaving \(S^1\) (contineously))

Is this parameter space \((S^1)\) simply connected ? (No)

let us map figure \(A\) to circle \((S^1)\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
\draw (0,0) ellipse (2.1 and 0.75);
\draw (1.9,0) circle (0.75);
\draw[->] (-1.5,0.45) -- (0.6,0.55);
\draw[->] (1.3,0.5) to[out=20,in=110] (2.5,0.0);
\draw[->] (0.9,-0.35) -- (-1.4,-0.35);
\node at (-1.7,-1.2) {$A$};
\draw[->] (1.9,-0.8) to[out=-80,in=160] (2.6,-1.4);
\node[right] at (2.6,-1.5) {This loop can be shrunk to a point};
\end{tikzpicture}

But there is another way of mapping figure/object \(A\) on surface \(S^1\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
\draw (0,0) ellipse (0.7 and 0.9);
\node at (0,-1.35) {$A$};
\draw[->] (0.9,0.6) -- (2.8,0.6);
\draw[line width=1.1pt] (3.9,0) circle (0.85);
\draw (3.9,0) -- (4.9,0);
\draw (3.9,0) -- (4.5,0.6);
\node at (4.25,0.18) {$\psi$};
\node[right,align=left] at (5.4,0.9) {So $SO(2)$ is not simply connected};
\node[right,align=left] at (5.4,0.1) {as,};
\node[right,align=left] at (5.4,-0.8) {This loop can not be shrunk to\\ a point; without cutting it.};
\end{tikzpicture}

we can map it twice (2 times)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.5,baseline=-2pt]
\draw[line width=1.0pt] (0,0) ellipse (0.35 and 0.5);
\end{tikzpicture}

on the \(S^1\), and There is no way to unwind too.

we can also map it in opposite direction (\(-1\) times) (\(-2\) times) .. so on.

It means all closed paths, i.e. maps of loop on \(S^1\), can be put into classes and each classes is specified by the integer (corresponding to no. of times it winds around).

\[ n = (\infty, -6, -5, -4, \ldots\, 0, 1, 2 \ldots, \infty). \]

These classes are in correspondence with one to one integers but the integers form group under addition.

"which means, the classes of maps (closed loops on \(S^1\)), they themselves form a group called fundamental group of space".

Note :- In general --

If we look at *any space \(V\)* and ask what are all distinct classes of maps of \(S^1\) on to \(V\) which is denoted by \(\pi_1(V)\) and it is called fundamental group of that space.

\[ \pi_1(V) = \text{fundamental \underline{group} of that space} \] \[ \text{or, (first homotopy group of space } (V) \]

So what is group composition law of \(\pi_1(V)\) --

1. If we wind around 6 times and then we wind loop further 7 times, total is something which winds around 13 times.

If this first homotopy group \(\pi_1(V)\) has only one element (Identity element) then space is said to be simply connected.

If it contains more than one element then it is non-trivial case. (doubly connected, triply connected -- infnitely connected).

So, \(V = S^1\), i.e. \((SO(2))\)

1. what is first homotopy group of \(S^1\) \(\Rightarrow\) \(\pi_1(S^1)\) = group of integers \[ \pi_1(SO(2)) \sim \pi_1(S^1) = \underline{Z} \]

(\(\hookrightarrow\) contains \(\infty\)-elements ; so it is infnitely connected.)

1. first homotopy group of \(S^2 = \pi_1(S^2)\)

as any loop on surface \(S^2\) can be shrunk to a point

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.5,baseline=-2pt]
\draw (0,0) circle (0.8);
\draw[dashed] (0,0) ellipse (0.8 and 0.25);
\draw[dashed] (0,0.35) ellipse (0.72 and 0.2);
\draw[dashed] (0,-0.35) ellipse (0.72 and 0.2);
\end{tikzpicture}

So, \(\pi_1(S^2) = 0\) (group with one element).

1. Similarly \(\pi_1(S^n) = 0\) for \((n \ge 2)\) or \((n > 1)\)

So \(\pi_1(S^3) = 0\) \(\Rightarrow\) \(\pi_1(SU(2)) = 0\)

\(S^3\) i.e. \(SU(2)\) parameter space of \(SU(2)\) is simply connected

\(\Rightarrow\) \(SU(2)\) is simply connected. as \(\pi_1(SU(2)) = \pi_1(S^3) = 0\) \(\downarrow\) has only one element

1. \(\pi_1(T^2)\) or \(\pi_1(S^1 \otimes S^1)\)

   \usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
   \begin{tikzpicture}[scale=0.8,>=Stealth]
   \draw (0,0) ellipse (2.0 and 1.05);
   \draw (0,0.05) ellipse (0.95 and 0.4);
   \fill (-2.0,-0.15) circle (1.2pt);
   \node[right] at (2.5,0.1) {$T^2 = S^1 \otimes S^1$};
   \end{tikzpicture}
   

   \usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
   \begin{tikzpicture}[scale=0.8,>=Stealth]
   \draw (0,0) circle (0.85);
   \draw[dashed] (0,0) circle (1.25);
   \fill (-0.1,0.05) circle (1.0pt);
   \draw (0.55,0.75) ellipse (0.28 and 0.42);
   \draw (0.95,0.55) ellipse (0.3 and 0.2);
   \node[right] at (1.25,0.45) {$S^1_b$};
   \node[below] at (-0.15,-1.35) {$S^1_a$};
   \node[right,align=left] at (3.2,0.2) {To wind path on torus we need to\\ know how many times we wind around $S^1_a$\\ and $S^1_b$ so we need two integers.};
   \end{tikzpicture}
   

So,

\[ \pi_1(T^2) = Z \otimes Z \]

Includes \(\infty\)-no of elements so Its connectivity is infinite.

(as there are infinite classes of closed paths that can not be reduced to each other).

Instead of a rotation matrix in 2-D.

\[ R = \begin{pmatrix} \cos\psi & \sin\psi \\ -\sin\psi & \cos\psi \end{pmatrix} \underline{\hspace{2cm}} \text{element of } SO(2) \]

let us associate with this rotation\(^{(R)}\) a complex no \(\underline{e^{i\psi}}\) \(\left(e^{2\pi i} + 1 = 0\right)\).

Then it is clear that there is one to one correspondance between \(2\times2\) orthogonal matrices \(SO(2)\) and points on unit circle. but points on unit circle also form a group, group of \(1\times1\) matrices, and unitary matrices.

\[ \begin{pmatrix} \cos\psi & \sin\psi \\ -\sin\psi & \cos\psi \end{pmatrix} \;\simeq\; e^{i\psi} \qquad \text{(one to one correspondance)} \] \[ \underline{SO(2)} \hspace{4.5cm} \underline{U(1)} \]

as we have seen \(\pi_1(S^1)\) is infinitely connected.

we can do the following :-

1. Take our circle \(S^1\) and roll it out on real number line.

   \usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
   \begin{tikzpicture}[scale=0.8,>=Stealth]
   \draw (-2.2,0) -- (6.0,0);
   \draw (0,0.7) circle (0.7);
   \draw[->] (0.35,0.35) arc (-45:200:0.45);
   \fill (0,0) circle (1.2pt);
   \node[below] at (0,-0.05) {$A$};
   \draw[->] (0.9,1.2) -- (2.1,1.2);
   \node at (3.0,0) {$\times$};
   \node at (5.2,0) {$\times$};
   \end{tikzpicture}
   

point `\(A\)' will mapped on Real no line infinite times.

So Real number line provide cover for \(S^1\).

\[ U(1) = SO(2) \simeq R^1/Z \qquad \text{just as} \qquad SO(3) \simeq SU(2)/Z_2 \]

\(\to\) \(R^1\) is called universal covering group of \(SO(2)\), because it is simply connected on infinite no. real line.

\(\to\) Every lie group has got a universal covering group, which is simply connected; (whose parameter space is simply connected).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
\draw (0,0) circle (1.5);
\fill (0.35,1.46) circle (1.3pt); \node[above] at (0.35,1.5) {$A$};
\fill (-0.45,-1.43) circle (1.3pt); \node[below] at (-0.45,-1.5) {$A'$};
\draw (-0.45,-1.43) .. controls (0.1,-0.9) and (-0.4,-0.4) .. (0.0,0.1)
      .. controls (0.4,0.6) and (0.1,1.0) .. (0.35,1.46);
\draw (0.75,-0.35) circle (0.4);
\fill (0.4,-0.4) circle (1.2pt); \node[left] at (0.35,-0.4) {$B$};
\fill (1.1,-0.3) circle (1.2pt); \node[right] at (1.15,-0.3) {$B'$};
\node[left] at (-2.0,0.5) {(RP$_3$)};
\node[left,align=right] at (-2.2,1.3) {projective space};
\draw[->] (-2.4,1.15) to[out=-70,in=150] (-1.95,0.7);
\end{tikzpicture}

1. \(A'A\) is closed path but can not shrung to a point on \(2\pi\) rotation \((\pi + \pi)\).
2. path \(BB'\) in sphere \(S^2\) is closed path and can be shrunk to a point.

Note

\(\to\) \(A'A\) can also be shrunk to a point by following trick

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.75,>=Stealth]
% first circle
\begin{scope}
\draw (0,0) circle (1.3);
\fill (-0.35,1.25) circle (1.2pt); \node[above] at (-0.4,1.3) {$Q$};
\fill (0.2,1.28) circle (1.2pt); \node[above] at (0.25,1.32) {$P$};
\fill (-0.35,-1.25) circle (1.2pt); \node[below] at (-0.4,-1.3) {$P'$};
\fill (0.2,-1.28) circle (1.2pt); \node[below] at (0.25,-1.32) {$Q'$};
\draw[->] (-0.35,-1.25) .. controls (-0.1,-0.5) and (-0.5,0.3) .. (-0.35,1.25);
\draw[->] (0.2,-1.28) .. controls (0.4,-0.4) and (0.0,0.4) .. (0.2,1.28);
\node at (0,-2.2) {($4\pi$ rotation)};
\end{scope}
\draw[->] (2.0,0) -- (2.9,0);
% second circle
\begin{scope}[xshift=4.6cm]
\draw (0,0) circle (1.3);
\fill (-0.45,1.22) circle (1.2pt); \node[above] at (-0.5,1.28) {$Q$};
\fill (0.3,1.26) circle (1.2pt); \node[above] at (0.35,1.32) {$P$};
\fill (-0.45,-1.22) circle (1.2pt); \node[below] at (-0.5,-1.3) {$P'$};
\fill (0.3,-1.26) circle (1.2pt); \node[below] at (0.35,-1.32) {$Q'$};
\draw[dashed,->] (-0.45,-1.22) .. controls (0.5,-0.2) and (0.1,0.6) .. (0.3,1.26);
\draw[dashed,->] (0.3,-1.26) .. controls (-0.4,-0.2) and (-0.1,0.5) .. (-0.45,1.22);
\end{scope}
\draw[->] (6.6,0) -- (7.5,0);
% third circle
\begin{scope}[xshift=9.2cm]
\draw (0,0) circle (1.3);
\fill (-0.85,0.98) circle (1.2pt); \node[above left] at (-0.85,0.98) {$Q$};
\fill (1.25,0.35) circle (1.2pt); \node[right] at (1.3,0.4) {$P$};
\fill (-1.28,-0.2) circle (1.2pt); \node[left] at (-1.33,-0.2) {$P'$};
\fill (0.9,-0.94) circle (1.2pt); \node[below right] at (0.9,-0.98) {$Q'$};
\draw[dashed,->] (-1.28,-0.2) -- (1.25,0.35);
\draw[dashed,->] (0.9,-0.94) -- (-0.85,0.98);
\draw[->] (0,-1.9) -- (0,-2.7);
\draw (0,-3.9) circle (0.9);
\fill (-0.9,-3.9) circle (1.2pt);
\fill (0.9,-3.9) circle (1.2pt);
\end{scope}
\end{tikzpicture}

Therefore there are only two classes of closed paths which can be shrunk to point. (1) \((\le 2\pi)\) rotation ② \(4\pi\) rotation

\(\uparrow\) (general loops) (like \(BB'\)

These are objects in this space (3-D) which will not come back to their orignal value from \(2\pi\) rotation but gauranteed to come back at \(4\pi\) rotation. (spinors). of coordinate system/axes

\[ \pi_1(SO(3)) \;=\; \pi_1(RP_3) \;=\; \pi_1\!\left(SU(2)/Z_2\right) \;=\; Z_2 \]

So, \(SO(3)\) is doubly connected. (parameter space of)

1. what is connectivity of space \(SO(n)\) for \((n > 3)\). \[ SO(n) \simeq spin(n)/Z_2 \qquad \to \text{doubly connected} \] \[ Spin(3) = SU(2) \qquad \text{(only for } n = 3) \] \[ Spin(6) = SU(4) \qquad spin(n) \text{ is simply connected.} \]

(parameter space of rotations in higher dimensions \(n \ge 3\) is also doubly connected)

No of parameters in \(SU(n) = n^2 - 1\)

No of * in \(U(n) = n^2\)

(\(SO(n)\) has \(^n C_2\) parameters. and analogue of \(SU(2)\) for \(SO(n)\) is \(spin(n)/Z_2\) & it so happens that \(spin(3) = SU(2)\) for \(n > 3\))

Note :-

It turns out that

\[ \pi_1(SU(n)) = 0 \qquad \text{(simply connected)} \]

i.e. parameter space of *\(SU(n)\)* is simply connected. for any \(n\).

universal covering group for \(SO(3)\) is \(SU(2)\).

Idea of universal covering group is extremely useful, because it ignores the complications in global topology of group space and focuses on simply-connected portions.

parameter space of \(SU(n)\): If it is a \((n\times n)\) matrix it should have \(n^2\) elements but with complex no entries it has \(2n^2\) elements \(U^\dagger U = I\) gives \(n^2\) conditions, which reduces parameters to \(n^2\) and \(|U| = +1\) also decreases no of parameters to \(\underline{n^2 - 1}\). (\(U(n)\) has \(n^2\) elements/parameter but \(SU(n) = n^2 - 1\) elements/parameter)

local property of group/parameter space is decided by generators. (of infinitesimal rotation)

let us write general generator for \(SO(n)\) :-

\[ J_{jk} \;(= -J_{kj}) \] \[ [J_{jk}, J_{lm}] = \delta_{jl} J_{km} + \delta_{km} J_{jl} - \delta_{jm} J_{kl} - \delta_{kl} J_{jm} \]

For (3-D), \(J_{12} \to J_3\), \(J_{23} \to J_1\), \(J_{31} \to J_2\)

\(\downarrow\) (rotation in 1-2 plane)

(Q) What is \(SU(1)\) ?

Ans. \(U(1)\) has 1 element/parameter

whereas \(SU(1)\), \(|U| = +1\), requires 0 parameters.

so its constant \(1\times1\) matrix with \(|M| = 1\)

\(SU(1)\) has one element only \(= [1]\)

Note : In general

\[ U(n) = U(1) \otimes SU(n) \] \[ \pi_1(U(n)) = \pi_1(U(n)) \otimes \underbrace{\pi_1(SU(n))}_{\textstyle = 0} \] \[ = Z \]

\(U(n)\) is infnitely connected.

The same circle of ideas -- the parameter space of \(SO(3)\), its double connectivity, the fundamental homotopy group \(\pi_1\), and the explicit \(2\times2\) representation \(U(\vec{r}\cdot\vec{\sigma})U^{-1} = \vec{r}\,'\cdot\vec{\sigma}\) of a rotation -- is worked out again from the quantum mechanics notebook (Lec 25) on a separate page :

• Group Theory -- SO(3) and SU(2)