We recall the definition of the semidirect product of two groups. Suppose $N$ and $H$ are two groups and $H$ acts on $N$ by $\vphi$, i.e. we have a homomorphism $\func{\vphi}{H}{\mathrm{Aut}(N)}$, $h \mapsto \vphi_h$. The semidirect product $N \rtimes_\vphi H$ is defined by endowing the set $N \times H$ the multiplication \[ (n_1, h_1) \cdot (n_2, h_2) = (n_1 \vphi_{h_1}(n_2), h_1 h_2). \] Identifying $N$ with $N \times \{1\}$ and $H$ with $\{1\} \times H$, $N$ and $H$ are subgroups of $N \rtimes_\vphi H$ with $N$ normal.
Theorem (Splitting Lemma). Let $G$ be a group, $H$ be a subgroup of $G$ and $N$ be a normal subgroup of $G$. Then $G$ is isomorphic to a semidirect product $N \rtimes_\vphi H$ if and only if
(i) there exist a short exact sequence $1 \rightarrow N \overset{\iota}\rightarrow G \overset{\pi}{\rightarrow} H \rightarrow 1$ and
(ii) a homomophism $\func{\alpha}{H}{G}$ such that $\pi \circ \alpha = \mathrm{id}_H$.
Proof: $(\Leftarrow)$ Suppose $G = N \rtimes_\vphi H$. To prove (i) let $\iota$ be the inclusion map of $N$ into $G$, which is a injective homomorphism. Define $\func{\pi}{G}{H}$ by $\pi((n, h)) = h$. Clearly $\pi$ is a surjective homomorphism and its kernel is $N \times \{1\} \cong N$. For (ii), define $\func{\alpha}{H}{G}$ by $\alpha(h) = (1, h)$. Then $\alpha$ is a homomorphism with $\pi(\alpha(h)) = \pi((1, h)) = h$ for all $h \in H$.
$(\Rightarrow)$ Define a homomorphism $\func{\vphi}{H}{\mathrm{Aut}(N)}$ by $\vphi_h(n) = \iota^{-1}(\alpha(h)\iota(n)\alpha(h)^{-1})$, note that since $N \lhd G$ we have $gng^{-1} \in N$ whenever $g \in G$ and $n \in N$, so $\vphi$ is well-defined. We claim that $G \cong N \rtimes_\vphi H$. Define $\func{\Psi}{N \rtimes_\vphi H}{G}$ by $\Psi((n, h)) = \iota(n)\alpha(h)$. It is a homomorphism since \[ \begin{align*}\Psi((n_1, h_1)\cdot(n_2, h_2)) &= \iota(n_1)\alpha(h_1)\iota(n_2)\alpha(h_1)^{-1}\alpha(h_1 h_2) \\
&= \iota(n_1)\alpha(h_1)\iota(n_2)\alpha(h_2) \\
&= \Psi((n_1, h_1))\Psi((n_2, h_2)). \end{align*} \] Now define $\func{\Phi}{G}{N \rtimes_\vphi H}$ by $\Phi(g) = (\iota^{-1}(g (\alpha \circ \pi)(g^{-1})), \pi(g))$. It is well-defined since \[ \pi(g(\alpha \circ \pi)(g^{-1})) = \pi(g)(\pi \circ \alpha)(\pi(g)^{-1}) = \pi(g)\pi(g)^{-1} = 1 \] and thus $g(\alpha \circ \pi)(g^{-1}) \in \ker(\pi) = \mathrm{im}(\iota)$. It is a homomorphism as
\[ \begin{align*} \Phi(g_1)\Phi(g_2) &= (\iota^{-1}(g_1 (\alpha \circ \pi)(g_1^{-1})), \pi(g_1)) \cdot (\iota^{-1}(g_2 (\alpha \circ \pi)(g_2^{-1})), \pi(g_2)) \\
&= (\iota^{-1}(g_1 (\alpha \circ \pi)(g_1^{-1})) \vphi_{\pi(g_1)}(\iota^{-1}(g_2 (\alpha \circ \pi)(g_2^{-1}))), \pi(g_1)\pi(g_2)) \\
&= (\iota^{-1}(g_1 (\alpha \circ \pi)(g_1^{-1})) \iota^{-1}(\alpha(\pi(g_1))g_2(\alpha \circ \pi)(g_2^{-1})\alpha(\pi(g_1))^{-1}), \pi(g_1 g_2)) \\
&= (\iota^{-1}(g_1 g_2 (\alpha \circ \pi)(g_2^{-1} g_1^{-1})), \pi(g_1 g_2)) \\
&= \Phi(g_1 g_2).
\end{align*}\] Now \[ \begin{align*}\Phi \circ \Psi(n, h) &= \Phi(\iota(n)\alpha(h)) = \Phi(\iota(n))\Phi(\alpha(h)) \\
&= (\iota^{-1}(\iota(n)(\alpha \circ \pi \circ \iota)(n^{-1})), (\pi \circ \iota)(n)) \cdot (\iota^{-1}(\alpha(h)(\alpha \circ \pi \circ \alpha)(h^{-1})), (\pi \circ \alpha)(h)) \\
&= (n, 1) \cdot (1, h) = (n, h)
\end{align*} \] and \[ \begin{align*} \Psi \circ \Phi(g) &= \Psi((\iota^{-1}(g (\alpha \circ \pi)(g^{-1})), \pi(g))) \\
&= \iota(\iota^{-1}(g (\alpha \circ \pi)(g^{-1})))\alpha(\pi(g)) = g.
\end{align*} \] Hence $\Phi$ and $\Psi$ are inverse to each other and thus are isomorphisms.
Showing posts with label Algebra. Show all posts
Showing posts with label Algebra. Show all posts
Wednesday, December 19, 2012
Tuesday, December 18, 2012
A proof of Cayley-Hamilton Theorem using Zariski topology
The notation $\mathbb{A}^n$ is used to denote the space $\mathbb{C}^n$ with the Zariski topology.
We note the following basic properties of the Zariski topology.
Proposition 1. Nonempty open subsets of $\mathbb{A}^n$ is dense.
Lemma 2. Polynomial maps from $\mathbb{A}^n$ to $\mathbb{A}^m$ is continuous.
We proceed to prove the Cayley-Hamilton Theorem. Identify $M_n(\mathbb{C})$ with $\mathbb{A}^{n^2}$. For any matrix $A$, let $p_A(\lambda)$ denotes its characteristic polynomial. The Cayley-Hamilton Theorem is the statement that the map $\Psi: \mathbb{A}^{n^2} \rightarrow \mathbb{A}^{n^2}$, $\Psi(A) = p_A(A)$ vanishes identically. Let $D$ be the subset of $\mathbb{A}^{n^2}$ of matrices with distinct eigenvalues.
Lemma 3. $D$ is nonempty and open.
Proof: Clearly $D$ is not empty. The complement of $D$ is the collection of n x n matrices with repeated eigenvalues. But a matrix has repeated eigenvalues if and only if the discriminant of its characteristic polynomial vanishes. Hence $D$ is closed.
It follows from Proposition 1 and Lemma 3 that $D$ is dense in $\mathbb{A}^{n^2}$. Note that $\Psi$ is a polynomial map, so it is continuous by Lemma 2. Hence it suffices to show that $\Psi$ vanishes on $D$. Let $A \in D$. As $A$ is diagonalizable, there exists an invertible matrix $P$ such that $P^{-1}AP$ is diagonal. Note that $p_A(A) = p_{P^{-1}AP}(P^{-1}AP)$. So we may assume $A$ itself is diagonal. If $A = \mathrm{diag}(\lambda_1, \ldots, \lambda_n)$, then $p_A(\lambda) = \prod_{i = 1}^n (\lambda - \lambda_i)$, thus it is clear that $p_A(A) = 0$.
We note the following basic properties of the Zariski topology.
Proposition 1. Nonempty open subsets of $\mathbb{A}^n$ is dense.
Lemma 2. Polynomial maps from $\mathbb{A}^n$ to $\mathbb{A}^m$ is continuous.
We proceed to prove the Cayley-Hamilton Theorem. Identify $M_n(\mathbb{C})$ with $\mathbb{A}^{n^2}$. For any matrix $A$, let $p_A(\lambda)$ denotes its characteristic polynomial. The Cayley-Hamilton Theorem is the statement that the map $\Psi: \mathbb{A}^{n^2} \rightarrow \mathbb{A}^{n^2}$, $\Psi(A) = p_A(A)$ vanishes identically. Let $D$ be the subset of $\mathbb{A}^{n^2}$ of matrices with distinct eigenvalues.
Lemma 3. $D$ is nonempty and open.
Proof: Clearly $D$ is not empty. The complement of $D$ is the collection of n x n matrices with repeated eigenvalues. But a matrix has repeated eigenvalues if and only if the discriminant of its characteristic polynomial vanishes. Hence $D$ is closed.
It follows from Proposition 1 and Lemma 3 that $D$ is dense in $\mathbb{A}^{n^2}$. Note that $\Psi$ is a polynomial map, so it is continuous by Lemma 2. Hence it suffices to show that $\Psi$ vanishes on $D$. Let $A \in D$. As $A$ is diagonalizable, there exists an invertible matrix $P$ such that $P^{-1}AP$ is diagonal. Note that $p_A(A) = p_{P^{-1}AP}(P^{-1}AP)$. So we may assume $A$ itself is diagonal. If $A = \mathrm{diag}(\lambda_1, \ldots, \lambda_n)$, then $p_A(\lambda) = \prod_{i = 1}^n (\lambda - \lambda_i)$, thus it is clear that $p_A(A) = 0$.
Friday, May 4, 2012
Additive subgroups of the reals and irrational rotation
We present a useful characterization of the additive subgroups of $\mathbb{R}$.
Theorem. Let $G$ be a subgroup of $(\mathbb{R}, +)$. Then precisely one of the following holds:
(i) $G = d\mathbb{Z}$ for some $d \in \mathbb{R}$;
(ii) $G$ is dense in $\mathbb{R}$.
Proof: If $G = \{0\}$, then $G = 0\mathbb{Z}$. Assume $G$ is nontrivial. Let \[ d = \inf\{g: 0 < g \in G\}.\] Note that $d \geq 0$ since for every nonzero $y \in G$, $ky \in G$ and $ky > 0$ for some $k \in \mathbb{Z}$.
Case 1: $d > 0$. We claim that $d \in G$. If not, then for every $\epsilon > 0$ there exists $g \in G$ with $g \in (d, d + \epsilon)$. It follows that for every $\epsilon > 0$ there are $g, h \in G$ such that $0 < g - h < \epsilon$, thus there exists $x \in G$ with $0 < x < \epsilon$ (let $x = g - h$). This contradicts to $d > 0$. We also have $d = \min\{|g|: 0 \neq g \in G\}$. Now fix any $0< g \in G$ and consider $S = \{g - nd: n = 0, 1, 2, \ldots\}$. The set $S$ is discrete, we can find $n \geq 0$ such that $g - nd \geq 0$ and $g - (n + 1)d < 0$. Since $g - nd, g - (n +1)d \in G$, $g - (n + 1)d \leq -d$, so $g - nd \leq 0$ and $g = nd$. Therefore $G = d\mathbb{Z}$.
Case 2: $d = 0$. Arguing as in the beginning of case 1, we find that 0 is a cluster point of $G \cap \mathbb{R}_+$. Let $(a, b)$ be any nonempty bounded open interval of $\mathbb{R}$. Take $\epsilon = b - a$ and pick some $g \in G \cap (0, \epsilon)$. For sufficiently large $n$, $ng \geq a$, and we take the smallest such $n$. Then $ng - g < a$, i.e. $ng \in (a - g, a]$, thus $ng + g \in (a, a + g] \subseteq (a, b)$. Therefore $G \cap (a, b) \neq \emptyset$. It follows that $G$ is dense in $\mathbb{R}$.
The above result can be used to prove the density of the orbit of an irrational rotation in the unit circle.
Corollary. (i) For an irrational $t$, $\mathbb{Z} + t\mathbb{Z}$ is dense in $\mathbb{R}$.
(ii) For an irrational $t$, $\{e^{2\pi int}: n \in \mathbb{Z}\}$ is dense in the unit circle.
Proof: (i) Note that $\mathbb{Z} + t\mathbb{Z}$ is a subgroup of $(\mathbb{R}, +)$. If $\mathbb{Z} + t\mathbb{Z} = a\mathbb{Z}$ for some $a \in \mathbb{R}$, then $1 = an$ for some $n \in \mathbb{Z}$, so $a \in \mathbb{Q}$. But $t = am$ for some $m \in \mathbb{Z}$, so $t \in \mathbb{Q}$, contradiction. By Theorem, $\mathbb{Z} + t\mathbb{Z}$ must be dense in $\mathbb{R}$.
(ii) Let $e^{2\pi i x}$ be a point on the unit circle. By (i), there exist sequences $m_k, n_k$ of integers such that $|m_k + n_k t - x| \rightarrow 0$. Then $|e^{2 \pi i n_k t} - e^{2 \pi i x}| = |e^{2\pi i x}||e^{2 \pi i (n_k t - x)} - 1| = |e^{2\pi i (m_k + n_k t - x)} - 1| = O(|m_k + n_k t - x|) \rightarrow 0$. Hence $\{e^{2\pi int}\}_{n \in \mathbb{Z}}$ is dense in the unit circle.
Proof: If $G = \{0\}$, then $G = 0\mathbb{Z}$. Assume $G$ is nontrivial. Let \[ d = \inf\{g: 0 < g \in G\}.\] Note that $d \geq 0$ since for every nonzero $y \in G$, $ky \in G$ and $ky > 0$ for some $k \in \mathbb{Z}$.
Case 1: $d > 0$. We claim that $d \in G$. If not, then for every $\epsilon > 0$ there exists $g \in G$ with $g \in (d, d + \epsilon)$. It follows that for every $\epsilon > 0$ there are $g, h \in G$ such that $0 < g - h < \epsilon$, thus there exists $x \in G$ with $0 < x < \epsilon$ (let $x = g - h$). This contradicts to $d > 0$. We also have $d = \min\{|g|: 0 \neq g \in G\}$. Now fix any $0< g \in G$ and consider $S = \{g - nd: n = 0, 1, 2, \ldots\}$. The set $S$ is discrete, we can find $n \geq 0$ such that $g - nd \geq 0$ and $g - (n + 1)d < 0$. Since $g - nd, g - (n +1)d \in G$, $g - (n + 1)d \leq -d$, so $g - nd \leq 0$ and $g = nd$. Therefore $G = d\mathbb{Z}$.
Case 2: $d = 0$. Arguing as in the beginning of case 1, we find that 0 is a cluster point of $G \cap \mathbb{R}_+$. Let $(a, b)$ be any nonempty bounded open interval of $\mathbb{R}$. Take $\epsilon = b - a$ and pick some $g \in G \cap (0, \epsilon)$. For sufficiently large $n$, $ng \geq a$, and we take the smallest such $n$. Then $ng - g < a$, i.e. $ng \in (a - g, a]$, thus $ng + g \in (a, a + g] \subseteq (a, b)$. Therefore $G \cap (a, b) \neq \emptyset$. It follows that $G$ is dense in $\mathbb{R}$.
The above result can be used to prove the density of the orbit of an irrational rotation in the unit circle.
Corollary. (i) For an irrational $t$, $\mathbb{Z} + t\mathbb{Z}$ is dense in $\mathbb{R}$.
(ii) For an irrational $t$, $\{e^{2\pi int}: n \in \mathbb{Z}\}$ is dense in the unit circle.
Proof: (i) Note that $\mathbb{Z} + t\mathbb{Z}$ is a subgroup of $(\mathbb{R}, +)$. If $\mathbb{Z} + t\mathbb{Z} = a\mathbb{Z}$ for some $a \in \mathbb{R}$, then $1 = an$ for some $n \in \mathbb{Z}$, so $a \in \mathbb{Q}$. But $t = am$ for some $m \in \mathbb{Z}$, so $t \in \mathbb{Q}$, contradiction. By Theorem, $\mathbb{Z} + t\mathbb{Z}$ must be dense in $\mathbb{R}$.
(ii) Let $e^{2\pi i x}$ be a point on the unit circle. By (i), there exist sequences $m_k, n_k$ of integers such that $|m_k + n_k t - x| \rightarrow 0$. Then $|e^{2 \pi i n_k t} - e^{2 \pi i x}| = |e^{2\pi i x}||e^{2 \pi i (n_k t - x)} - 1| = |e^{2\pi i (m_k + n_k t - x)} - 1| = O(|m_k + n_k t - x|) \rightarrow 0$. Hence $\{e^{2\pi int}\}_{n \in \mathbb{Z}}$ is dense in the unit circle.
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