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

Factor Groups (Quotient Groups)

Let $G$ be a group and $N\leq G$. Recall the equivalence relation $\sim$ on $G$ defined by
$$a\sim b\ \mbox{if}\ ab^{-1}\in N$$ for any $a,b\in G$. Each equivalence class $[a]$ is identified with right coset $Na$. The quotient set $G/\sim$ is the set of all equivalence classes or equivalently all right cosets of the subgroup $N$ in $G$. A question we can ask is can we give a group structure to the quotient set $G/\sim$? The answer is affirmative if $N$ is a normal subgroup of $G$. This is what we are going to discuss in this lecture.

Let $N\vartriangleleft G$. Denote by $G/N$ the set of all right cosets of $N$ in $G$ i.e.
$$G/N=\{Na:a\in G\}.$$
For any $Na, Nb\in G/N$,
\begin{align*}
(Na)(Nb)&=N(aN)b\\
&=N(Na)b\\
&=N(Nab)\\
&=Nab.
\end{align*}
So, it appears that if $N\vartriangleleft G$, we may define an operation $\cdot$ on $G/N$ naturally from the binary operation on $G$ by the equation \begin{equation}
\label{eq:cosetoper}
Na\cdot Nb=Nab
\end{equation} for any $Na,Nb\in G/N$. But is this operation well-defined? To see that let $Na=Nc$ and $Nb=Nd$. Then $Nac^{-1}=N$ and $Nbd^{-1}=N$. So $ac^{-1}\in N$ and $bd^{-1}\in N$. Now,
\begin{align*}
N(ac)(bd)^{-1}&=N(ac)(d^{-1}b^{-1})\\
&=Na(cd^{-1})b^{-1}\\
&=a(Ncd^{-1})b^{-1}\\
&=aNb^{-1}\\
&=Nab^{-1}\\
&=N.
\end{align*}
That is, $Na\cdot Nb=Nab=Ncd=Nc\cdot Nd$. Hence, right coset operation given by the equation \eqref{eq:cosetoper} is well-defined. Conversely, if $N\leq G$ and right coset operation given by the equation \eqref{eq:cosetoper} is well-defined, then $N$ must be a normal subgroup of $G$. The verification of this is left to readers as an exercise. Furthermore, $Ne=N$ is an identity element in $G/N$ and for each $Na\in G/N$, $Na^{-1}\in G$ and $NaNa^{-1}=N$, hence $(Na)^{-1}=Na^{-1}$. Therefore, $(G/N,\cdot)$ is a group. This group is called a factor group or a quotient group of $G$ modulo $N$.

Theorem. If $N\vartriangleleft G$ then there exists an onto homomorphism (an epimorphism) $\psi:G\longrightarrow G/N$ such that $\ker\psi=N$, Such a homomorphism is called a natural homomorphism or a canonical homomorphism.

Proof. Define $\psi: G\longrightarrow G/N$ by
$$\psi(a)=Na$$
for any $a\in G$. If $a=b$ then $\psi(a)=Na=Nb=\psi(b)$ so $\psi$ is well-defined. Let $Na\in G/N$. Then $a\in G$ and $\psi(a)=Na$, so $\psi$ is onto. $\psi(ab)=Nab=NaNb=\psi(a)\psi(b)$, hence $\psi$ is a homomorphism. To show that $\ker\psi=N$,
\begin{align*}
a\in\ker\psi&\Longleftrightarrow \psi(a)=N\\
&\Longleftrightarrow Na=N\\
&\Longleftrightarrow a\in N.
\end{align*}

$G/N$ is the set of all right coset of $N$ in $G$, so $|G/N|=|G:N|=\frac{|G|}{|N|}$.

Example. Since $\mathbb{Z}$ is an abelian group, for any $n\in\mathbb{N}$, $n\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$. So, for any $n\in\mathbb{N}$, $\mathbb{Z}/n\mathbb{Z}$ is a factor group. Each $\mathbb{Z}/n\mathbb{Z}$ is indeed isomorphic to $\mathbb{Z}_n$.

The notion of factor groups can be used to prove an important theorem in group theory called Cauchy Theorem. First we study the following theorem as we will also need it (actually its corollary) to prove Cauchy Theorem.

Theorem. Let $G$ be a finite group and $a\in G$ with $|a|=n$. Then for any $k\in\mathbb{Z}$,

  1.  $|a^k|=\frac{n}{d}$ where $d=(k,n)$.
  2.  $|a^k|=n$ if and only if $(k,n)=1$.

Proof. Since 2 follows from 1, we prove 1 only. Since $d=(k,n)$, $n=n_1d$ and $k=k_1d$ for some $n_1,k_1\in\mathbb{Z}$ such that $(n_1,k_1)=1$.
$$(a^k)^{n_1}=a^{k_1n}=(a^n)^{k_1}=e.$$ So, if we let $|a^k|=s$, then $s|n_1$. On the other hand, $e=|a^k|^s=a^{ks}$, so $n|ks$ which implies that $n_1|k_1s$. Since $(n-1,k_1)=1$, $n_1|s$. Therefore, $s=n_1=\frac{n}{d}$.

Corollary. If a finite group $G$ has no nontrivial subgroup, then $G$ must be a cyclic group of a prime order.

Proof. Let $a\ne e\in G$. Then $\langle a\rangle\leq G$. Since $G$ has no nontrivial subgroup and $a\ne e$, $\langle a\rangle=G$ i.e. $G$ is a cyclic group. Let $|a|=n$. If $k\ne n$ and $k|n$. Then by the above theorem $|a^k|=\frac{n}{k}$ since $(n,k)=k$. Since $a^k\ne e$ and $G$ has no nontrivial subgroup, $\langle a^k\rangle=G$. This implies that $|a^k|=n$ and so $k=1$. Therefore, $n$ is a prime.

Theorem. [Cauchy] If $G$ a finite abelian group and and $p||G|$ where $p$ is a prime, then $G$ has an element of order $p$.

Proof. We prove the theorem by induction on $|G|$. If $|G|=1$, then there is no prime that divides $|G|$, so the theorem is vacuously true. Suppose that the statement is true for all abelian groups whose order is less than $|G|$. Let $\{e\}\not\leq N\not\leq G$. If $p||N|$, then by induction hypothesis there exists $a\in N$ such that $|a|=p$. Since $N\subset G$, we are done in this case. Now suppose that $p\not|N|$. Since $G$ is abelian, $N\vartriangleleft G$. Since $p|G$ and $p\not| N$, $p||G/N|<|G|$. So by induction hypothesis, $G/N$ has an element $Na$ of order $p$.
$$(Na)^p=Na^p=N\Longrightarrow a^p\in N$$
but $a\not\in N$ since $Na\ne N$. Let $|N|=m$. Then $e=(a^p)^m=(a^m)^p$. Let $b=a^m$. If $b\ne e$, then $b$ is an element of order $p$. What if $b=a^m=e$? If so, $(Na)^m=N$. Since $|Na|=p$, $p|m=|N|$. but by assumption $p\not||N|$. A contradiction! So, we are done if $G$ has a nontrivial subgroup. What if $G$ does not have a nontrivial subgroup? If so, by corollary above $G$ must be a cyclic group of a prime order. Since $p||G|$, $|G|=p$. In this case, any $a\ne e\in G$ is an element of order $p$.

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.