abstract-algebragroup-theoryinfinite-groups
I'm trying to think of an extension to this question, which asks to show whether $\mathbb { C } ^ { * } / \mathbb { R } ^ { + } \simeq \mathbb { T } $. They do it using the First Isomorphism Theorem.
I think my postulate should be true because $ \mathbb { R } ^ { + } \simeq { R } ^ { * } $, (the positive reals under multiplication and the reals under multiplication ), but I'm having difficulty coming up with a homomorphism
$$ \phi : \mathbb { C } ^ { * } \mapsto \mathbb{T} $$
Any suggestions?
I'm writing the answer to my question but I'm not completely sure though
$$U(n) \simeq \frac{SU(n) \times U(1)}{\mathbb{Z}_{n}}$$
We use the first isomorphism theorem:
Let G and H be groups and let $f: G \longrightarrow G'$ be a group homomorphism. Then:
$$Im f \simeq \frac{G}{Ker f} $$
We have to do the following things:
\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*}
$$Ker f = \lbrace (S,e^{i\varphi}) \in (SU(n) \times U(1)) \: \vert \: f(S,e^{i\varphi}) = Se^{i\varphi} = e_{U(n)} = I \rbrace$$
\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:
$$S = e^{-i \frac{2 k \pi}{n}} I \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace$$
\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*}
$$Ker f = (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace $$
We 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 $\mathbb{Z}_{n}$
$$\phi: Ker f = (S,e^{i\varphi}) \rightarrow \mathbb{Z}_{n} : (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \mapsto k$$
$$\phi((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}})) = \phi(S_{1},e^{i\varphi _{1}}) + \phi(S_{2},e^{i\varphi _{2}})$$
\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*}
$$\Rightarrow \varphi _{1} = \varphi _{2} \Rightarrow S_{1} = S_{2}$$
$$Ker \phi = \lbrace (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \in Ker f \: \vert \: \phi(e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) = e_{\mathbb{Z}_{n}} = 0 \rbrace$$
$$\phi(e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) = k = 0 \Rightarrow Ker \phi = (e^{0} I,e^{0}) = (I,1)$$
$\phi$ is 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 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:
$$Im f \subseteq U(n) \quad \text{and} \quad U(n) \subseteq Im f$$
$$Im f \subseteq U(n) \Rightarrow \mid det (Se^{i\varphi}) \mid ^{2} = 1 \quad \forall S \in SU(n), \forall \varphi \in \left[ 0, 2\pi \right] $$
\begin{eqnarray*} \mid det (Se^{i\varphi}) \mid ^{2} & = & \mid e^{i n \varphi} \mid ^{2} \\ & = & 1 ^{2} \\ & = & 1 \end{eqnarray*}
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 $det(Y) = 1$
$$YY^{\dagger} = (m^{-1/n} X)(m^{-1/n} X)^{\dagger} = (e^{\frac{i\theta}{n}} X)(e^{\frac{i\theta}{n}} X)^{\dagger} = e^{\frac{i\theta}{n}} e^{\frac{-i\theta}{n}} XX^{\dagger} = I $$
$$det(Y) = det(m^{-1/n} X) = m^{-1} det(X) = m^{-1} m = 1 \Rightarrow Y \in SU(n)$$
The homomorphism is indeed a rather trivial one.
Consider the following map $\psi : S_n \to \mathbb P_{n \times n}$, where the latter is the set of permutation matrices of order $n$. This map is defined as follows : given $\phi \in S_n$, the $i$th column of $\psi(\phi)$ is the column vector with a $1$ in the $\phi(i)$th position, and $0$ elsewhere.
It is easy to see that $\psi(\phi)$ is indeed a permutation matrix, since a $1$ occurs in any position if and only if that position is described by $(\phi(i),i)$, for any $1 \leq i \leq n$.
This map is clearly multiplicative. The proof is awkward to write, but easy to understand. We must prove that $\psi(\alpha \circ \beta)=\psi(\alpha)\psi(\beta)$ for any permutations $\alpha,\beta$.
Note that since these are matrices, it's enough to show that each corresponding entry is equal. So let us take the entry $(i,j)$ of each matrix.
Then : $\psi(\alpha \circ \beta)_{ij} = 1$ if and only if $i = \alpha \circ \beta(j)$, as I said earlier. Otherwise, it is zero.
Note that by ordinary matrix multiplication, $(\psi(\alpha)\psi(\beta))_{ij} = \sum_{k=1}^n \psi(\alpha)_{ik}\psi(\beta)_{kj}$.
Now, we know that $\psi(\alpha)_{ik} = 1$ only when $i = \alpha(k)$. Similarly, $\psi(\beta)_{kj} = 1$ only when $k = \beta(j)$. Hence, their product is one precisely when both of these happen : $i = \alpha(k),k = \beta(j)$. If both these don't happen simultaneously, then whenever one of $\psi(\alpha)_{ik},\psi(\beta)_{kj}$ is one the other will be zero, so the whole sum will be zero. However, this is the same as saying that the sum is one exactly when $i = \alpha \circ\beta(j)$. This description matches with the decription given for $\psi(\alpha \circ \beta)_{ij}$ given earlier.
Hence, entry by entry these matrices are the same. Therefore the matrices are the same, and hence $\psi$ is a homomorphism between the two spaces, an isomorphism as it has trivial kernel and the sets are of the same cardinality.
Best Answer
Just use the same homomorphism, and then double the angle. Dividing by negatives means that we identify numbers on opposite sides of the origin. Double the angle, and that's not a problem anymore. After all, we can map the circle 2-1 to the circle.
(Or, as Tsemo Aristide's answer put it, going from the projective line to the circle is an angle-doubling)