Notes on group and representation theory

We know that the order of the symmetric group (aka all permutations) $S_n$, is $n!$. We also learned that any permutation $\pi \in S_n$ can be written as a product of transpositions, which, if there are even number of them, then $\pi$ is an even permutation.

Apart from the fact that we call $A_n$ the alternating group, we can see that it is one, since identity $e=(12)(21)$ and a product of two even permutations are even, since the number of transpositions will add. We know that there are odd permutations too, so we know that $A_n$ is a subgroup of $S_n$, i.e. $A_n \subset S_n$.

But, we have not shown how many elements in $A_n$ there are. How to show this?

Our strategy will be to do a coset decomposition of $S_n$, then use Lagrange’s theorem. We will find that there are only two coset representatives (these are elements that label the disjoint cosets), so that there are exactly half the elements of $S_n$ in $A_n$.

First, we know that $A_n e = A_n$ is a valid right coset, and will contain all the even permutations in $S_n$. Let’s choose some odd permutation $d \in S_n$ as the next coset representative. We want to check that $A_n d$ contains the rest of $S_n$. Consider some other odd element $d’ \in S_n$. Then $A_n d’ = A_n d’ d^{-1} d$. But $d’ d^{-1}$ is an even permutation, since the number of transpositions will add and an odd number plus another odd number is an even number. Therefore $A_n d’ = A_n d$. So we see that $A_n d$ contains all the odd elements in $S_n$. All elements in $A_n d$ must be odd. So $A_n e$ and $A_n d$ are disjoint since the former contains only even elements and $A_n d$ contain only the odd elements, and $S_n = A_n e \cup A_n d$. There are two coset representatives, so $\lvert S_n \rvert = 2 \lvert A_n \rvert \implies \lvert A_n \rvert = n!/2$.

Consider the first operation, $(al)$. This means that $a \rightarrow k$ and $k \rightarrow a$. Let us write this in the two-line notation. Even though this is operating on elements $a$ and $l$ only, we can still write in elements $b c \cdots k$, as long as they stay in place. So we have: $$ (al) \leftrightarrow \begin{pmatrix} a & b & c & \cdots & k & l \ l & b & c & \cdots & k & a \end{pmatrix}. $$ Then, let’s apply the second operation, $(abc \cdots k)$. Here we just have $a \rightarrow b$, $b \rightarrow c, \cdots, k \rightarrow a$. Applying this on the previous cycle/permutation, we have $$ (abc \cdots k) (al) \leftrightarrow \begin{pmatrix} a & b & c & \cdots & k & l \ l & c & d & \cdots & a & b \end{pmatrix}. $$ Written in this form, it is not clear that it is a product of disjoint cycles. Let’s chase how each element is mapped. We have $a \rightarrow l$, $l \rightarrow b$. From $b$ to $k$ in fact we have the mapping we’d expect, $b \rightarrow c$, $c \rightarrow d$, and so on, until $j \rightarrow k$. Then we have $k \rightarrow a$. But we’re now back to $a$, so we have completed a cycle! We’ve also gone through all the elements, so it must be that there is only one cycle in this permutation. Summarising the above, we can write it as a single (and therefore disjoint) cycle: $$ (abc \cdots k) (al) = (a l b c \cdots k). $$

For fixed element $i$, let’s denote $H = \mathrm{Stab}_G(i)$. Since $H$ is a subgroup of $G$ (you should check this), let us consider the left coset decomposition of $G$: $$G = t_1 H \cup t_2 H \cup \cdots \cup t_n H,$$ where $t_1 = e$, the identity.

So we can write the orbit as $$\mathrm{Orb}_G(i) = \bigcup_{a=1}^n \left\{ g(i) | g \in t_a H \right\}.$$ We can also iterate over each coset like $$\mathrm{Orb}_G(i) = \bigcup_{a=1}^n \left\{ t_a(\phi(i)) | h \in H \right\}.$$ Since all $\phi(i) = i$, and sets only count unique elements, we only have one element contributing from each coset at most, so $$\mathrm{Orb}_G(i) = \bigcup_{a=1}^n \left\{ t_a(i) \right\} = \left\{ t_a(i) \vert a=1, \dots, n\right\}.$$ We’re almost there, since if $t_a(i)$ are unique with $a$, then we see that $\mathrm{Orb}_G(i)$ counts the number of cosets, so we can use Lagrange’s theorem to prove the result.

Let’s check this. Assume there exists $t_a(i) = t_b(i)$ where $a \neq b$. Then $t_b^{-1} (t_a(i)) = i$ so $t_b^{-1} t_a = h$. This means that $t_a = t_b h$ so $t_a H = t_b H$, so are the same cosets. But this is a contradiction because $t_a H$ and $t_b H$ were supposed to be disjoint. So $t_a(i)$ are unique.

Let us prove the second statement first. Define $$ f(t) := e^{t A} B e^{-t A}, $$ where $t$ is just a number. If we take the derivative, we have $$ \begin{aligned} \frac{df}{d t} & = A e^{t A} B e^{-t A} - e^{t A} B e^{-t A} A \\ & = [A, \, \cdot \,] f. \end{aligned} $$ This is the same operator at every `time’ $t$. Since we have $f(0) = B$, we can use the usual knowledge of differential equations and find that $f(t) = e^{t[A, \, \cdot \,]} B$.

For our specific case where $[A,B]$ commutes with both $A$ and $B$, let’s write out $e^{t[A, \, \cdot \,]} B$: $$ \begin{aligned} e^{t[A, \, \cdot \,]} B & = \sum_{n=0}^{\infty} \frac{t^n}{n!} [A, \, \cdot \,]^n B \\ & = B + t [A, B] + \sum_{n=2}^{\infty} \frac{t^n}{n!} [A, \, \cdot \,]^{n-1} [A, B]. \end{aligned} $$ But from our assumptions we know that $[A,B]$ commutes with $A$, so all $n \geq 2$ terms vanish. Therefore we have that $$ e^{tA} B e^{-tA} = e^{t[A, \, \cdot \,]} B = B + t[A, B].$$ Now let’s tackle the main result. Define $$ g(t) := e^{tA} e^{tB}, $$ where $t$ is just a number. Again differentiating, you should be able to show that $$ \begin{aligned} \frac{dg}{dt} & = (A + e^{tA} B e^{-tA}) g = g (B + e^{-tB} A e^{tB}) \\ & = (A + e^{t[A, \, \cdot \,]} B) g = g (B + e^{-t[B, \, \cdot \,]} A). \end{aligned} $$ Therefore we have $$ \frac{dg}{dt} = (A + B + t[A, B]) g. $$ Let us denote $T(t) := A + B + t[A,B]$. It’s easy to show that $[T(t), T(t’)]=0$ for all $t, t’$. Therefore there exists some $U$ independently diagonalises $T(t)$; defining $\overline{M} := S M S^{-1}$, $\overline{T}(x)$ is diagonal. We also have that $[g(0)]_{ij} = \delta_{ij}$, which stays diagonal upon similarity transformation. Therefore in the rotated frame everything stays in the diagonal, and we have $$ \frac{d \overline{g}_{ii}}{dt} = \overline{T}_{ii}(t) \overline{g}_{ii}. $$ Using the initial condition $\overline{g}_{ii}(0) = 1$, we have $$ \begin{aligned} \overline{g}_{ii}(t) & = \exp\left( \int_{0}^t dt’ \overline{T}_{ii} (t’) \right) = \exp\left( \sum_{jk} S_{ij} S^{-1}_{ki}\int_{0}^t dt’ \left(A_{jk} + B_{jk} + x [A, B]_{jk} \right) \right) \\ & = \exp\left( \sum_{jk} S_{ij} S^{-1}_{ki}\left(tA_{jk} + tB_{jk} + \frac{t^2}{2} [A, B]_{jk} \right) \right) \\ & = \exp\left( t \overline{A} + t\overline{B} + \frac{t^2}{2} \overline{[A, B]} \right)_{ii}. \end{aligned} $$ Transforming back to the normal coordinates by applying $S^{-1} \left( \, \cdot \, \right) S$, then setting $t=1$, we have $$ g(1) = \exp\left( {A} + {B} + \frac{1}{2} {[A, B]} \right) = \exp(A) \exp(B). $$