$U(n) \simeq \frac{SU(n) \times U(1)}{\mathbb{Z}_{n}}$ isomorphism

I'm trying to proof the following isomorphism

$$U(n) \simeq \frac{SU(n) \times U(1)}{\mathbb{Z}_{n}}$$

So I'm using the first Isomorphism theorem: http://en.wikipedia.org/wiki/Isomorphism_theorem

It's easy to show that the following map is an homomorphism:

$$f: SU(n) \times U(1) \rightarrow U(n): (S,e^{i\varphi}) \mapsto e^{i \varphi} S$$

But I'm having troube to show that:

$$Ker f = \mathbb{Z}_{n}$$

How am I supposed to do that ?

I'm writing the answer to my question but I'm not completely sure though

Intro

What do we want to show ?

$$U(n) \simeq \frac{SU(n) \times U(1)}{\mathbb{Z}_{n}}$$

Groups definition

• $U(n)$ = the group of $n\times n$ unitary matrices $\Rightarrow$ $U \in U(n): UU^{\dagger} = U^{\dagger}U = I \Rightarrow \mid det (U) \mid ^{2} = 1$
• $U(1) =$ the group of $1\times 1$ unitary matrices $\Rightarrow U(1) = \lbrace e^{i\varphi} \mid \varphi \in \left[ 0, 2\pi \right] \rbrace$
• $SU(n) =$ the group of $n\times n$ unitary matrices with determinant 1
• $\mathbb{Z}_{n} =$ the cyclic group of n integers modulo n $\Rightarrow$ $\mathbb{Z}_{n} =$ the set $\lbrace 0,1,2,...,n-1 \rbrace$ with the operation of addition modulo n. \end{itemize}

Isomorphism theorem

We use the first isomorphism theorem:

Let G and H be groups and let $f: G \longrightarrow G'$ be a group homomorphism. Then:

$$Im f \simeq \frac{G}{Ker f} $$

What do we have to do ?

We have to do the following things:

1. Find a map $f: SU(n) \times U(1) \rightarrow U(n)$ and show that it's a homomorphism
2. Show that $Ker f = \mathbb{Z}_{n}$
3. Show that $Im f = U(n)$ which is equivalent to show that f is surjective

We find and homomorphism between $SU(n) \times U(1)$ and $U(n)$

\begin{eqnarray*} f: && SU(n) \times U(1) \rightarrow U(n) \\ && (S,e^{i\varphi}) \mapsto Se^{i\varphi} \end{eqnarray*}

$f$ is a homomorphism if :

\begin{eqnarray*} f((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}} )) = f(S_{1},e^{i\varphi _{1}})f(S_{2},e^{i\varphi _{2}} )&& \forall \varphi _{1}, \varphi _{2} \in U(1) \\ && \forall S_{1},S_{2} \in SU(n) \end{eqnarray*}

It's easy to show:

\begin{eqnarray*} f((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}})) &=& f(S_{1}S_{2},e^{i\varphi _{1}}e^{i\varphi _{2}} ) \\ &=& S_{1}S_{2} e^{i\varphi _{1}} e^{i\varphi _{2}} \\ &=& e^{i\varphi _{1}} S_{1} e^{i\varphi _{2}} S_{2} \\ &=& f(S_{1},e^{i\varphi _{1}})f(S_{2},e^{i\varphi _{2}}) \end{eqnarray*}

We show that $Ker f = \mathbb{Z}_{n}$

$$Ker f = \lbrace (S,e^{i\varphi}) \in (SU(n) \times U(1)) \: \vert \: f(S,e^{i\varphi}) = Se^{i\varphi} = e_{U(n)} = I \rbrace$$

Let's find S

\begin{eqnarray*} e^{i\varphi}S&=& I \\ S &=& e^{-i\varphi} I\\ det(S) &=& det(e^{-i\varphi} I) \\ 1 &=& e^{-in \varphi} \\ 1 &=& \cos (n\varphi ) - i \sin (n \varphi) \end{eqnarray*}

$$ \left\{ \begin{array}{r c l} \cos (n \varphi ) &=& 1\\ \sin (n \varphi ) &=& 0\\ \end{array} \right. \Rightarrow n \varphi = 2k \pi \Rightarrow \varphi = \frac{2k \pi}{n} \quad \text{with} \quad k \in \mathbb{Z}$$

So the matrices S are:

$$S = e^{-i \frac{2 k \pi}{n}} I \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace$$

Let's find $e^{i\varphi}$

\begin{eqnarray*} e^{i\varphi}S&=& I \\ e^{i\varphi}e^{-i \frac{2 k \pi}{n}} I&=& I \\ e^{i\varphi}&=& e^{i \frac{2 k \pi}{n}} \\ \end{eqnarray*}

Ker f

$$Ker f = (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace $$

We show that Ker $f$ is isomorph to $\mathbb{Z}_{n}$

We denote by $\mathbb{Z}_{n}$ the cyclic group of integers modulo n. Let's not forget that:

• The group law $\cdot$ of Ker f is given by: $(S_{1},e^{i\varphi _{1}})\cdot(S_{2},e^{i\varphi _{2}}) = (S_{1}S_{2},e^{i(\varphi _{1} + \varphi _{2})})$

• The group law $\cdot$ of $\mathbb{Z}_{n}$ is given by: $k_{1} \cdot k_{2} = k_{1} + k_{2}$ \end{itemize}

Let's show that Ker f is isomorph to $\mathbb{Z}_{n}$

$$\phi: Ker f = (S,e^{i\varphi}) \rightarrow \mathbb{Z}_{n} : (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \mapsto k$$

$\phi$ is an homomorphism

$$\phi((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}})) = \phi(S_{1},e^{i\varphi _{1}}) + \phi(S_{2},e^{i\varphi _{2}})$$

\begin{eqnarray*} \phi((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}}))&=& \phi(S_{1}S_{2},e^{i(\varphi _{1} + \varphi _{2})}) \\ &=& \phi (e^{-i \frac{2 (k_{1} + k_{2}) \pi}{n}} I,e^{i \frac{2 (k_{1} + k_{2}) \pi}{n}}) \\ &=& k_{1} + k_{2} \\ \phi(S_{1},e^{i\varphi _{1}}) + \phi(S_{2},e^{i\varphi _{2}})&=& (e^{-i \frac{2 k_{1} \pi}{n}} I,e^{i \frac{2 k_{1} \pi}{n}}) + (e^{-i \frac{2 k_{2} \pi}{n}} I,e^{i \frac{2 k_{2} \pi}{n}}) \\ &=& k_{1} + k_{2} \end{eqnarray*}

$\phi$ is injective

• $\phi$ is injective $\Leftrightarrow \forall \varphi_{1},\varphi_{2} \in \left[ 0, 2\pi \right] \: \text{and} \: \forall S_{1},S_{2} \in SU(n): \phi (S_{1},e^{i\varphi _{1}}) = \phi (S_{2},e^{i\varphi _{2}}) \Rightarrow \varphi_{1} = \varphi_{2} \: \text{and} \: S_{1} = S_{2}$
  \begin{eqnarray*} \phi (S_{1},e^{i\varphi _{1}}) & = & \phi (S_{2},e^{i\varphi _{2}})\\ S_{1}e^{i\varphi _{1}} & = & S_{2}e^{i\varphi _{2}} \\ e^{i(\varphi _{1} - \varphi _{2})} S_{1} & = & S_{2} \\ det( e^{i(\varphi _{1} - \varphi _{2})} S_{1}) & = & det( S_{2} ) \\ e^{in(\varphi _{1} - \varphi _{2})} & = & 1 \end{eqnarray*}

$$\Rightarrow \varphi _{1} = \varphi _{2} \Rightarrow S_{1} = S_{2}$$

• We can also use the fact that $\phi$ is injective if and only if $Ker \phi = \lbrace e_{Ker f} \rbrace = \lbrace (I,1)\rbrace$

$$Ker \phi = \lbrace (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \in Ker f \: \vert \: \phi(e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) = e_{\mathbb{Z}_{n}} = 0 \rbrace$$

$$\phi(e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) = k = 0 \Rightarrow Ker \phi = (e^{0} I,e^{0}) = (I,1)$$

$\phi$ is surjective

• $\phi$ is injective $\Leftrightarrow \forall k \in \mathbb{Z}_{n}, \exists (e^{i \frac{2 k \pi}{n}} I,e^{-i \frac{2 k \pi}{n}}) \in Ker f : \phi(e^{i \frac{2 k \pi}{n}} I,e^{-i \frac{2 k \pi}{n}}) = k$

• Another way to show $\phi$ is surjective is by noting that fact that $\phi$ is an injective map between 2 finites sets with the same number of elements so $\phi$ is surjective

We show that f is surjective

How do we show that $Im f = U(n)$ ?

We want to show that $Im f = U(n)$ where: $ Im f = \lbrace Se^{i\varphi} \vert S \in SU(n), \varphi \in \left[ 0, 2\pi \right] \rbrace$

To do that, we can show that:

$$Im f \subseteq U(n) \quad \text{and} \quad U(n) \subseteq Im f$$

$Im f \subseteq U(n)$

$$Im f \subseteq U(n) \Rightarrow \mid det (Se^{i\varphi}) \mid ^{2} = 1 \quad \forall S \in SU(n), \forall \varphi \in \left[ 0, 2\pi \right] $$

\begin{eqnarray*} \mid det (Se^{i\varphi}) \mid ^{2} & = & \mid e^{i n \varphi} \mid ^{2} \\ & = & 1 ^{2} \\ & = & 1 \end{eqnarray*}

$U(n) \subseteq Im f$

It's equivalent to show that: $\forall X \in U(n), \exists (Y,z) \in SU(n) \times U(1) \quad \text{such as} \quad f(Y,z) = Yz = X$

\begin{eqnarray*} X \in U(n)&\Rightarrow& \vert det(X) \vert ^{2} = 1 \Rightarrow det(X) = e^{i\theta} \equiv m \\ &\Rightarrow& X^{\dagger}X = XX^{\dagger} = I \\ \end{eqnarray*}

Let's write X as: $X = zY = m^{1/n} Y$ where $Y = m^{-1/n}X$ and let's show that $Y \in SU(n)$ and $m^{1/n} \in U(1)$:

• $Y \in SU(n) \Leftrightarrow YY^{\dagger} = Y^{\dagger} Y = I$ and $det(Y) = 1$

  $$YY^{\dagger} = (m^{-1/n} X)(m^{-1/n} X)^{\dagger} = (e^{\frac{i\theta}{n}} X)(e^{\frac{i\theta}{n}} X)^{\dagger} = e^{\frac{i\theta}{n}} e^{\frac{-i\theta}{n}} XX^{\dagger} = I $$

$$det(Y) = det(m^{-1/n} X) = m^{-1} det(X) = m^{-1} m = 1 \Rightarrow Y \in SU(n)$$

• $m^{1/n} = e^{\frac{i\theta}{n}} \in U(1) $

If I assume that x is just a scalar when I'm trying to find Ker f like I did with Tobias Kildetoft using the nth roots of unity group. Isn't easier to assume directly that x can be written as $e^{i \varphi}$ ?

We show that $Ker f = \mathbb{Z}_{n}$

$$Ker f = \lbrace (S,x) \in (SU(n) \times U(1)) \: \vert \: f(S,x) = Sx = e_{U(n)} = I \rbrace$$

Let's find S

\begin{eqnarray*} xS&=& I \\ S &=& x^{-1}I \end{eqnarray*}

Let $x^{-1} = y$. S is a diagonal matrix with $y$ as diagonal entries \

As $S \in SU(2)$ then

\begin{eqnarray*} det(S)&=& det(yI) \\ &=& y^{n} \\ &=& 1 \end{eqnarray*}

$n^{th}$ roots of 1

How can we solve this equation ? We want to solve $y^{n} = 1$ with $\lambda \in \mathbb{C}$. Let's write both side of equation in polar form:

$$y^{n} = (r e^{i \theta} )^{n} = r e^{i \theta n}$$ $$1 = (r e^{i \theta} ) = 1 e^{i 0}$$

$y^{n}$ and $1$ are equal if and only if:

• $r^{n} = 1$. As r is positif, we have r = 1
• $n\theta = 0 + 2k\pi$ where $k \in \mathbb{Z}$ so $\theta = \frac{2 k \pi}{n} $

We can visualize this geometrically. The points will lie on the unit circle and they will be equally spaced on the unit circle every $\frac{2 \pi}{n}$ radians.

So our equation have $n-1$ different solutions as we go back to where we started once we reach $k = n$

So the $n^{th}$ roots of unity are given by:

$y = e^{i \frac{2 k \pi}{n}}$ with $k = \lbrace 0,1,...,n-1 \rbrace$

So the matrices S are:

$$S = e^{-i \frac{2 k \pi}{n}} I \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace$$

It's easy to show that this form a group

Let's find x

\begin{eqnarray*} xS&=& I \\ xe^{i \frac{2 k \pi}{n}} I&=& I \\ x&=& e^{-i \frac{2 k \pi}{n}} \\ \end{eqnarray*}

Ker f

$$Ker f = (e^{i \frac{2 k \pi}{n}} I,e^{-i \frac{2 k \pi}{n}}) \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace $$