Group Theory 11: The Isomorphism Theorems

The following theorem is called the Fundamental Homomorphism Theorem or the First Isomorphism Theorem.

Theorem. Let $G$ and $G’$ be groups, and $\varphi:G\longrightarrow G’$ an epimorphism (onto homomorphism). Then $G/K\cong G’$ where $K=\ker\varphi$.

Proof.

Let $\gamma: G\longrightarrow G/K$ be the canonical homomorphism i.e. for any $a\in G$, $\gamma(a)=Ka$. Define $\psi: G/K\longrightarrow G’$ by
$$\psi(Ka)=\varphi(a)$$for each $a\in G$. Then

1. $\psi$ is well-defined:\begin{align*}Ka=Kb&\Rightarrow ab^{-1}\in K\\&\Rightarrow\varphi(ab^{-1})=e’\\&\Rightarrow\psi(Ka)=\varphi(a)=\varphi(b)=\psi(Kb).\end{align*}
2. $\psi$ is a homomorphism:$$\psi(KaKb)=\psi(Kab)=\varphi(ab)=\varphi(a)\varphi(b)=\psi(Ka)\psi(Kb).$$
3. $\psi$ is one-to-one:\begin{align*}\psi(Ka)=\psi(Kb)&\Rightarrow\varphi(a)=\varphi(b)\\&\Rightarrow\varphi(ab^{-1})=e’\\&\Rightarrow ab^{-1}\in K\\&\Rightarrow Ka=Kb.\end{align*}
4. $\psi$ is onto: Let $b\in G’$. Then there exists $a\in G$ such that $\varphi(a)=b$ since $\varphi$ is onto. Now, $Ka\in G/K$ and $\psi(Ka)=\varphi(a)=b$.

Example. Let $S^1$ be the unit circle centered at the origin. Then it can be represented in terms of complex numbers as
$$S^1=\{e^{2x\pi i}:x\in[0,1)\}.$$Define a map $\varphi: (\mathbb{R},+)\longrightarrow(S^1,\cdot)$ by $$\varphi(x)=e^{2\pi ix}$$for each $x\in\mathbb{R}$. Then $\varphi$ is an onto homomorphism. The kernel of $\varphi$ is$$\ker\varphi=\mathbb{Z}.$$Hence, by the Fundamental Homomorphism Theorem
$$\mathbb{R}/\mathbb{Z}\cong S^1.$$Note that $\mathbb{R}/\mathbb{Z}$ can be viewed as the quotient set $\mathbb{R}/\sim$ where $\sim$ is an equivalence relation on $\mathbb{R}$ defined as follows: For all $x,y\in\mathbb{R}$,
$$x\sim y\ \mbox{if}\ x-y=n$$for some $n\in\mathbb{Z}$.

Theorem [Correspondence Theorem]. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Let $K=\ker\varphi$, $H’\leq G’$ and $H=\varphi^{-1}(H’)=\{a\in G: \varphi(a)\in H\}$. Then $K\subset H\leq G$ and $H/K\cong H’$. If $H’\triangleleft G’$, then $H\triangleleft G’$.

Proof. Since $e\in K\subset H$, $H\ne\emptyset$. Let $a,b\in H$. Then $\varphi(a),\varphi(b)\in H’$. Since $H’\leq G’$, $\varphi(a)\varphi(b)^{-1}=\varphi(ab^{-1})\in H’$ and so, $ab^{-1}\in H$. Hence, $H\leq G$. Since $e’\in H’$, $K\subset H$. Let $\psi=\varphi|_H$. Then $\psi$ is a homomorphism from $H$ onto $H’$ and $\ker\psi=\ker\varphi=K$. Therefore, by the Fundamental Homomorphism Theorem $H/K\cong H’$.

Suppose that $H’\triangleleft G’$. Let $a\in G$ and $h\in H$. Then $\varphi(a)\in G’$ and $\varphi(h)\in H’$. Since $H’\triangleleft G’$, $\varphi(a)\varphi(h)\varphi(a)^{-1}=\varphi(aha^{-1})\in H’$ which implies that $aha^{-1}\in H$. Hence, $H\triangleleft G$.

Theorem [The Second Isomorphism Theorem]. Let $H\leq G$ and $N\triangleleft G$. Then $HN\leq G$, $H\cap N\triangleleft G$ and
$$H/H\cap N\cong HN/N.$$

Proof. Let $a,b\in HN$. Then $a=h_1n_1$ and $b=h_2n_2$ for some $h_1,h_2\in H$ and $n_1,n_2\in N$. Then
\begin{align*}
ab^{-1}&=h_1n_1(h_2n_2)^{-1}\\
&=h_1n_1(n_2^{-1}h_2^{-1})\\
&=h_1h_2^{-1}h_2(n_1n_2^{-1})h_2^{-1})\in HN.
\end{align*}
So, $HN\leq G$. Clearly $H\cap N\triangleleft G$, in particular $H\cap N\triangleleft H$. Also clearly $N\leq HN$. Let $n_1\in N$. Then $\forall hn\in HN$,
$$(hn)n_1(hn)^{-1}=h(nn_1n^{-1})h^{-1}\in N.$$This means that $N\triangleleft HN$. Define $\varphi:H\longrightarrow HN/N$ by
$$\varphi(h)=Nh$$for each $h\in H$. Then $\varphi$ is clearly well-defined, a homomorphism. Let $Nhn\in HN/N$. Then $Nhn=hnN=hN=Nh$ and $\varphi(h)=Nh=Nhn$. So, $\varphi$ is onto.
\begin{align*}
h\in \ker\varphi&\Leftrightarrow \varphi(h)=Nh=N\\
&\Leftrightarrow h\in N\\
&\Leftrightarrow h\in H\cap N.
\end{align*}
So, $\ker\varphi=H\cap N$. Therefore, by the Fundamental Homomorphism Theorem
$$H/H\cap N\cong HN/N.$$

Theorem [The Third Isomorphism Theorem]. Let $\varphi: G\longrightarrow G’$ be an epimorphism. Let $K=\ker\varphi$, $N’\triangleleft G’$, and $N=\varphi^{-1}(N’)=\{a\in G: \varphi(a)\in N’\}$. Then
$$G/N\cong G’/N,$$or equivalently
$$G/N\cong(G/K)/(N/K).$$

Proof. Define $\psi: G\longrightarrow G’/N’$ by
$$\psi(a)=N’\varphi(a)$$for each $a\in G$. Then

1. $\psi$ is well-defined:\begin{align*}a=b\in G &\Rightarrow\varphi(a)=\varphi(b)\\&\Rightarrow \psi(a)=N’\varphi(a)=N’\varphi(b)=\psi(b).\end{align*}
2. $\psi$ is a homomorphism:\begin{align*}\psi(ab)&=N’\varphi(ab)\\&=N’\varphi(a)\varphi(b)\\&=N’\varphi(a)N’\varphi(b)\\&=\psi(a)\psi(b).\end{align*}
3. $\psi$ is onto: Let $N’c\in G’/N’$. Then $c\in G’$. Since $\varphi$ is onto, there exists $a\in G$ such that $\varphi(a)=c$. $N’c=N’\varphi(a)=\psi(a)$.
4. $\ker\psi=N$:\begin{align*}a\in\ker\psi&\Leftrightarrow N’\varphi(a)=N’\\&\Leftrightarrow\varphi(a)\in N’\\&\Leftrightarrow a\in N.\end{align*}

Therefore, by the Fundamental Homomorphism Theorem we obtain
$$G/N\cong G’/N’.$$Since $\varphi:G\longrightarrow G’$ is an epimorphism, by the Fundamental Homomorphism Theorem, $G’\cong G/K$. Since $N=\varphi^{-1}(N’)$, by the Correspondence Theorem, $N’\cong N/K$. Hence, $G’/N’\cong(G/K)/(N/K)$ i.e. $G/N\cong(G/K)/(N/K)$.

Functonal Analysis 7: Further Properties of Normed Spaces

Normed spaces are not necessarily finite dimensional. So it is important to understand the notion of a basis for an infinite dimensional normed space. Suppose that there is a basis of a normed space $X$ as an infinite sequence $(e_n)$ in $X$. Then any $x\in X$ can be represented as the infinite superposition of the $e_n$’s

\label{eq:superposition}
x=\sum_{j=1}^\infty\alpha_je_j,

where $\alpha_1,\alpha_2,\cdots$ are scalars.
In order for this to make sense, we need to make sure that the infinite sum in \eqref{eq:superposition} converges. Thus we have the following definition of a basis for an infinite dimensional normed space.

Definition. Suppose that a normed space $X$ contains a sequence $(e_n)$ with property that $\forall x\in X$, there exists uniquely a sequence of scalars $(\alpha_n)$ such that
$$||x-\sum_{j=1}^n\alpha_je_j||\rightarrow 0\ \mbox{as}\ n\to\infty.$$
Then $(e_n)$ is called a Schauder basis for $X$. The infinite sum $\displaystyle\sum_{j=1}^\infty\alpha_je_j$ is called the expansion of $x$.

Example. $\ell^p$ has a Schauder basis $(e_n)$, where $e_n=(\delta_{nj})$.

Theorem. If a normed space has a Schauder basis, it is separable i.e. it has a countable dense subset.

Proof. Recall that $D$ is a dense subset of $X$ if $\bar D=X$. This equivalent to saying that $\forall \epsilon>0$, $\forall x\in X$, $B(x,\epsilon)\cap D\ne\emptyset$.

Let $D$ be the set of all possible finite linear combinations (superpositions) of the $e_n$’s. Then $D$ is countable. Let $x\in X$. Then $\exists$ a sequence of scalars $(\alpha_n)$ such that $x=\displaystyle\sum_{j=1}^\infty\alpha_je_j$. Given $\epsilon>0$, $\exists$ a positive integer $N$ such that
$$||x-(\alpha_1e_1+\cdots+\alpha_ne_n)||<\epsilon$$
for all $n\geq N$. This implies that $\alpha_1e_1+\cdots+\alpha_ne_n\in B(x,\epsilon)\cap D$ for all $n\geq N$.

One question mindful readers may have is does every separable Banach space have a Schauder basis? The answer is negative and a counterexample can be found in

Enflo, P. (1973), A counterexample to the approximation property. Acta Math. 130, 309–317.

We finish this lecture with the following theorem.

Theorem. [Completion] Let $X$ be a normed space. Then there exists a Banach space $\hat X$ and an isometry from $X$ onto $W\subset\hat X$ which is dense in $\hat X$. The space $\hat X$ is unique up to isometries.

Irreducible Representations of $\mathrm{U}(1)$

A representation $\rho$ of a group $G$ on a vector space $V$ always has the subspace $\{0\}$ and $V$ itself as invariant subspaces. Here a subspace $W\subset V$ is invariant means that $\rho(g)W\subset W$ for every $g\in G$. if $\rho$ has no other invariant subspaces, we say that it is irreducible.

Theorem. If $G$ is compact, every representation of $G$ is equivalent to a direct sum of irreducible representations.

This theorem is important for physicists since most Lie groups that are important in physics are compact. The theorem also says that if $G$ is compact, irreducible representations are the building blocks of other representations.

Example. For each $n\in\mathbb{Z}$, define $\rho_n:\mathrm{U}(1)\longrightarrow\mathrm{GL}(1,\mathbb{C})$ by
$$\rho_n(e^{i\theta})v=e^{i n\theta}v.$$
Then each $\rho_n$ is irreducible since $\mathbb{C}$ has no nontrivial vector subspaces. What is really important about this example is that any complex 1-dimensional representation is equivalent to $\rho_n$ for some $n\in\mathbb{Z}$. (Prove this!)

Schur’s Lemma. Let $\rho: G\longrightarrow\mathrm{GL}(V)$ be an irreducible complex representation and let $\phi: V\longrightarrow V$ an interwining map of $V$ with itself (i.e. $\phi(\rho(g)v)=\rho(g)(\phi(v))$ for all $g\in G$, $v\in V$). Then $\phi=\lambda I$ for some $\lambda\in\mathbb{C}$.

Suppose the group $G$ is abelian and $g\in G$. Then
\begin{align*}
\rho(g)(\rho(g’)v)&=\rho(gg’)v\\
&=\rho(g’g)v\\
&=\rho(g’)(\rho(g)v)
\end{align*}
for all $g’\in G$ , $v\in V$. Since $\rho(g)$ is an interwining map of $V$ with itself, $\rho(g)$ is a scalar multiple of $I$ by Schur’s Lemma. So every subspace of $V$ is invariant and hence $\rho$ is 1-dimensional. This means that any irreducible representation of $\mathrm{U}(1)$ is equivalent to one the $\rho_n$. Since $\mathrm{U}(1)$ is compact, any finite dim representation of $\mathrm{U}(1)$ is given as a direct sum of the $\rho_n$.

In quantum mechanics the electric charge of a particle is assumed to be a (integer) multiple of a certain unit charge $q$ i.e. charge is quantized. (This was indeed the case as confirmed by experiments.) In terms of representation, this means that a particle with charge $nq$ transforms according to $\rho_n$ of $\mathrm{U}(1)$. If we move a particle of charge $nq$ around a loop $\gamma$ in spacetime, its wave function is multiplied by a certain phase
$$e^{-\frac{i}{\hbar}nq\oint_\gamma A}\in\mathrm{U}(1)$$
where $A$ is the vector potential or more generally a connection as
$$\rho_n\left(e^{-\frac{i}{\hbar}q\oint_\gamma A}\right)v=e^{-\frac{i}{\hbar}nq\oint_\gamma A}v.$$

Proposition. The tensor product $\rho_n\otimes\rho_m$ is equivalent to $\rho_{n+m}$.

I will leave the proof of this proposition as an exercise. This proposition has an interesting physical implication. If we have two particles corresponding to two different representations of a group, a bound state corresponds to the tensor product of the two representations. The lectric charge of such a bound state is the sum of the charges of the constituents.

References:

[1] John Baez, Javier P. Muniain, Gauge Fields, Knots and Gravity, World Scientific 1994

[2] Brian C. Hall, Lie Groups, Lie Algebras, and Representations, An Elementary Introduction, Springer-Verlag 2003

SouthernMiss Math Forum

Recently I have put up a math forum site, called SouthernMiss Math Forum. This is an online meeting place where math faculty members, undergrad students, and grad students can discuss math outside of classrooms. This forum is also open to high school students. In fact, any math lovers are welcome to participate.

Have something to say about mathematics or have a math question? Go right ahead and also hear what other people have to say.