Georgi Lie Algebras Chapter 1 Solutions

==Problem 1==
Find the multiplication table for group with 3 elements and prove that it is unique.

==Solution==
Clear that if \(e,a\) are the elements of the group, then only other extra element allowed is - \(b=a^{-1}\). The multiplication table is as follows: \begin{equation} \begin{pmatrix} && e && a && b \\ e && e && a && b \\ a && a && b && e \\ b && b && e && a \end{pmatrix} \end{equation}

==Problem 2==
Find all essentially different possible multiplication tables for groups with four elements.

==Solution==
One is Klein Group 4. If \(a,b,c,e\) are four elements. \begin{equation} \begin{pmatrix} && e && a && b && c \\ e && e && a && b && c \\ a && a && e && c && b \\ b && b && c && e && a \\ c && c && b && a && e \end{pmatrix} \end{equation} Other is Symmetric Group of 4 elements with the following table. \begin{equation} \begin{pmatrix} && e && a && b & c \\ e && e && a && b && c \\ a && a && e && c && b \\ b && b && c && e && a \\ c && c && b && a && e \end{pmatrix} \end{equation}

==Problem 3==
Show that the representation \((1.135)\) is reducible.

==Solution==
Schur's lemma in special case states that if \(A\) is a complex matrix of order \(n\) that commutes with all matrices from $G(Here S_n)$, then A is scalar and the representation is reducible. Standard representation consists of single \(1\) in each row and column. Take all \(n\times n\) representation of the permutation group \(S_n\). Apply the first projection formula, \begin{equation} \phi = \frac{1}{|G|}\sum_{i=1}^n g \end{equation} where \(G=S_n\) and \(g \in End(V) = n \times n\) matrix of each group element. Clear that the Projection \(\phi\) consists of a an \(n\times n\) matrix with \(1\) in every row and column. Also it is trivial to show that this matrix \(\phi\) commutes with all group elements \(g\) (in matrix form). Thus \(\phi\) is a scalar matrix and the representation is reducible.

==Problem 4==
Suppose \(D_1\) and \(D_2\) are equivalent irreducible representations of finite group of \(G\) such that \(D_2(g) = S D_1(g) S^{-1}\) \(\forall g \in G\), what can you say about an Operator \(A\) that satisfies \(A D_1(g) = D_2(g) A\)?

==Solution==
We are given that both \(D_1,D_2\) are equivalent irreds. Consider a G-Module homomorphism \(A:D_1\rightarrow D_2\) given by \(AD_1(g)=D_2(g)A\) for all \(g \in G\). Then, \begin{align} A D_1(g) = D_2(g) A \\ = S D_1(g) S^{-1} A \\ \text { Or } S^{-1}A D_1(g) = D_1(g) S^{-1}A \end{align} Now, this can be considered as G-module homomorphism \(S^{-1}A:D_1 \rightarrow D_1\). As, per Schur's lemma, the operator \(S^{-1}A = \lambda I\). Therefore, \begin{align} S^{-1}A = \lambda I \\ A = \lambda S \end{align}

==Problem 1E==
Find the Group of all discrete rotations that leave a regular tetrahedron invariant by labelling \(4\) vertices and considering the rotations noas permutatof the four vertices. This defines four dimensional representation of a group. Find conjugacy classes and the character of the irreducible representations of the group.

==Solution==
Discrete rotations that leave the Tetrhedron invariant include,

• Identity
• Rotation around an axis thru vertex perpendicular to opposite plane by
\(\pm 120^o\). There are \(4\) such axis and \(2\) per axis giving a total of 8. If the Tetrahedron is labelled by \(1,2,3,4\), the permutations are \((1,3,2), (1,2,3), (1,4,3), (2,4,3), (1,2,4), (2,3,4),(1,3,4),(1,4,2) \).
• Three (3) \(180^o\) rotations that map edges to opposite edges.These permutations are \\\((1,3)(2,4),(1,4)(2,3),(1,2)(3,4)\).

Total number of elements that leave Tetrahedron invariant under discrete rotation are \(1+8+3=12\). This is isomorphic to Alternating Group of 4 elements. The conjugacy classes are \((), (1,2)(3,4), (1,2,3), (1,2,4)\). Using Sage Math or by using the character table computations, Irreducible characters: \begin{equation} \left(\begin{array}{rrrrr} Conj->& 1 & (123) & (132) & (12)(34)\\ U&1 & 1 & 1 & 1 \\ U'& 1 & \omega & \omega^2 & 1 \\ U^{''}&1 & \omega^2 & \omega & 1 \\ V & 3 & 0 & 0 & -1 \end{array}\right) \end{equation}

==Problem 1F==


==Solution==

• Each of the 4 blocks can move in \(x,y\) directions. Then there are \(8\) degrees of freedom of \(x,y\) coordinates. Call this \(D_2\).
• 4 Blocks arranged in a square has symmetries has Dihedral Group 4 symmetry
• \includegraphics[scale=5.0]{PermutationsColor.jpg}
• Dihedral group has \(8\) elements.