# Haar measure, the Peter-Weyl theorem and compact abelian groups

## Haar measure

We restrict our attention to groups that are locally compact Hausdorff and $\sigma$-compact.

Definition. (Radon measure) Let $X$ be a $\sigma$-compact locally compact Hausdorff space. The Borel $\sigma$-algebra $\mathcal{B}[X]$ on $X$ is the $\sigma$-algebra generated by the open subsets of $X$. A Borel measure is a countably additive nonnegative measure $\mu:\mathcal{B}[X]\to[0,+\infty]$ on the Borel $\sigma$-algebra. A Radon measure is a Borel measure  obeying three additional axioms:

(i) (Local finiteness) One has $\mu(K)<\infty$ for every compact set $K$.

(ii) (Inner regularity) One has $\mu(E)=\sup_{K\subset E,K\ \text{compact}}\mu(K)$ for every Borel measureable set $E$.

(iii) (Outer regularity)  One has $\mu(E)=\inf_{U\supset E,U\ \text{open}}\mu(U)$ for every Borel measureable set $E$.

Definition. (Haar measure) Let $G=(G,\cdot)$ be a $\sigma$-compact locally Hausdorff group. A Radon measure $\mu$ is left-invariant (resp. right-invariant) if one has $\mu(gE)=\mu(E)$ (resp. $\mu(Eg)=\mu(E) ) for all$latex g\in G$and Borel measureable sets $E$. A left-invariant Haar measure is a nonzero Radon measure which is left-invariant. Theorem. (Riesz representation theorem). Let $X$ be a $\sigma$-compact locally compact Hausdorff space. Then to every liner functional $I:C_c(X)\to \mathbf{R}$ whihc is nonnegative (thus $I(f)\geq 0$ whenever $f\geq 0$), one can associate a unique Radon measure $\mu$ such that $I(f)=\int_X f\,d\mu$ for all $f\in C_c(X)$. Conversely, for each Radon measure $\mu$, the functional $I_\mu:f\mapsto \int_X f\,d\mu$ is a nonnegative linera functional on $C_c(X)$. # Cayley-Hamilton theorem Theorem. (Cayley-Hamilton) Let $T$ be a $k$-linear endomorphism of a finite-dimensional vector space $V$ over a field $k$. Let $P_T(x)$ be the characteristic polynomial $\displaystyle P_T(x)=\det(x\cdot 1_V-T)$ Then $\displaystyle P_T(T)=0\in\mathrm{End}_k(V)$. # Uniform approximation by polynomials The famous Weierstrass approximation theorem states that any continuous real-valued function on a compact interval can be uniformly approximated by a polynomial with an error as small as one likes. This theorem is the first significant result in approximation theory of one real variable and plays a key role in the development of general approximation theory. There are several different ways to prove this important theorem, but in this blog, we present a proof which makes use of convolution. In particular, the convolution with an approximation to the identity is the important case on which this theorem is based. What is approximation theory? In this most general form, we say that approximation theory is dedicated to the description of elements in a topological space $X$, which can be approximated by elements in a subset of $X$. That is to say, given an elements $x\in X$, we want to approximate $x$ by an element $a$ of some subset $A$ of $X$. The elements of $A$ are nice'' ortractable” and we want to make $x$ and $a$ as closer as possible. In other words, approximation theory allows to characterize the closure of $A$ in $X$. A special case of approximation theory is the approximation of continuous function, which has significant importance in real analysis and partial differential equations. Continuous functions can be very badly behaved, for instance, they can be nowhere differentiable. One the other hand, functions such as polynomials are always very well behaved, in particular being always differentiable. Fortunately, while most continuous functions are not as well behaved as polynomials, they can always be uniformly approximated by polynomials; this important result is known as the Weierstrass approximation theorem, and is the subject of this paper. We wish to prove the following: Theorem. (Weierstrass approximation theorem) If $[a,b]$ is an interval, $f:[a,b]\to\mathbf{R}$ is a continuous function, and $\varepsilon>0$, then there exists a polynomial $P$ on $[a,b]$ such that $\|P-f\|_{\infty}\leq\varepsilon$ (i.e., $|P(x)-f(x)|\leq\varepsilon$ for all $x\in[a,b]$). Another way of stating the theorem is as follows. Let $C([a,b]\to\mathbf{R})$ be the space of continuous functions from $[a,b]$ to $\mathbf{R}$ with uniform metric $d_\infty$ and $P([a,b]\to\mathbf{R})$ be the space of all polynomials on $[a,b]$; this is a subspace of $C([a,b]\to\mathbf{R})$, since polynomials are continuous. The Weierstrass approximation theorem asserts that every continuous function is in the closure of $P([a,b]\to\mathbf{R})$: $\displaystyle \overline{P([a,b]\to\mathbf{R})}=C([a,b]\to\mathbf{R}).$ In particular, every continuous function on $[a,b]$ is uniform limit of polynomials. Another way of saying this is that the space of polynomials is dense in the space of continuous functions with uniform topology. The proof is fairly easy to remember in outline, once on has mastery over its ingredients. • We can composite a linear term and assume that $a=0$, $b=1$. Similarly, we can subtract a liner term and assume that $f(a)=f(b)=0$. • If one takes convolution of a continuous function on $[0,1]$ with an approximation to the identity, then the result is a uniformly good approximation. • If the approximation to the identity is a polynomial restricted to the domain $[-1,1]$, then the result convolution is a polynomial. # Any subspace of a separable metric space is separable Proposition. Let $X$ be a separable metric space, $A$ be a subpace of $X$, then $A$ is also separable., Proof. let $Q\subset X$ be a countable and dense subset. For each $q\in Q$ and $n\in\mathbf{N}$, choose $a_{q,n}\in A\cap B(q,1/n)$ if such intersection is nonempty. Then, $A^*=\bigcup_{q,n} \{a_{q,n}\}$ is a countable subset of $A$. Let $a\in A, 0<\varepsilon<1$. Then choose $q\in B(a,\varepsilon/4)\cap Q$. Now, choose $N$ such that $\varepsilon/4\leq N\leq\varepsilon/2$. Then, $B(q,1/N)\cap A\neq\emptyset$, so there exists $a^*\in A^*$, such that $d(a^*,q)<1/N$. Now $d(a^*,a)\leq d(a^*,q)+d(q,a)<\varepsilon/2+\varepsilon/4$. So $a^*\in B(a,\varepsilon)\cap A$. Hence, the closure of $A^*$ in $A$ is equal to $A$. # Primes in arithmetic progressions In analytic number theory, a fundamental problem is to understand the distribution of the primes number $\{2,3,5,\cdots\}$. One important result is Dirichelt’s theorem, which states that for any two positive coprime integers $a$ and $N$, the are infinitely many primes of the form $qN+a$, where $q$ is a non-negative integer. Theorem. Let $a$ and $N$ be fixed an integer with $a$ and $N$ coprime (thus $\gcd(a,N)=1$), then there are infinitely many primes $p$ in the arithmetic progression $\{qa+N:q\in\mathbf{N}\}$. Remark. If $\gcd(a,N)>1$, then there is at most one prime $p$ meeting the condition$latex $p=a$ mod $N$, since such $p$ would be divisible by the $\gcd$. Thus, the necessity of the gcd condition is obvious.

## Group characters

We first explain Dirichlet’s innovation, the use of group characters to isolate in a specified congruence class module $N$.

Definitoin. Let $G$ be a finite abelian group, the  dual group or group of character $\widehat{G}$ of a abelian group $G$ is
$\displaystyle\widehat{G}:=\{\text{group homomorphism}\ \chi:G\to\mathbf{C}^\times\}.$
This $\widehat{G}$ is itself an abelian group under the operation on character defined for $g\in G$ by
$\displaystyle(\chi_1\cdot\chi_2)(g):=\chi_1(g)\chi_2(g).$

Definition. Let $G$ be a cyclic group of order $n$ with specified generator $g$. Then $\widehat{G}$ is isomorphic to the group of $n$-th roots of unity $\mu_n$. That is, an $n$-th root of unity $\zeta$ gives the character $\chi$ such that
$\displaystyle\chi(g^\ell)=\zeta^\ell.$
In particular, $\widehat{G}$ is a cyclic group of order $n$.

Proposition.  Let $G=A\oplus B$ be a direct sum of finite abelian groups. Then there is a natural isomorphism of the dual groups
$\displaystyle \widehat{A}\oplus\widehat{B}\cong\widehat{G}$
by $(\chi_1,\chi_2)\mapsto \chi_1\cdot\chi_2$, where $\chi_1\cdot\chi_2\in \widehat{G}$ is the character defined by
$\displaystyle \chi_1\cdot\chi_2((a,b))=\chi_1(a)\chi_2(b).$

Remark.  Combining this proposition and the structure theorem for finite abelian groups, we conclude that $\widehat{G}\cong G$ for a finite abelian group $G$.

Theorem 1. (Dual version of cancellation) For $g$ in a finite ableian group,
$\displaystyle \sum_{\chi\in\widehat{G}}\chi(g)=\begin{cases} |G|& {\rm for}\ g=e,\\ 0& {\rm otherwise}. \end{cases}$

Proof. If $g=e$, then the sum counts the characters in $\widehat{G}$.
One the other hand, given $g\neq e$ in $G$, by previous proposition let $\chi_1$ be in $\widehat{G}$ such that $\chi_1(g)\neq 0$. The map on $\widehat{G}$
$\displaystyle\chi\mapsto \chi_1\cdot\chi$
is a bijection of $\widehat{G}$ to itself.

## Proof of  Dirichlet’s theorem

Definition.  A  Dirichlet character module $N$  is a group homomorphism
$\displaystyle \chi:(\mathbf{Z}/N\mathbf{Z})^\times\to \mathbf{C}^{\times}$
extended by $0$ to all of $\mathbf{Z}/N\mathbf{Z}$, that is by defining $\chi(a)=0$ if $a$ is not invertible modulo $N$. The trivial or principal character $\chi_0$ module $N$ is the character which takes only the value $1$ and $0$.

Remark. This extension by zero allows us to compose $\chi$ with the reduction-mod-$N$ map $\mathbf{Z}\to\mathbf{Z}/N\mathbf{Z}$ and also consider $\chi$ as a function on $\mathbf{Z}$. Even when extended by $0$ the function $\chi$ is still  multiplicative in the sense that
$\displaystyle \chi(mn)=\chi(m)\chi(n).$

Theorem. $\displaystyle\sum_{a\in\mathbf{Z}/N\mathbf{Z}}\chi(a)=\begin{cases} \varphi(N)& {\rm for}\ \chi=\chi_0,\\ 0& {\rm otherwise}, \end{cases}$
where $\varphi$ is the Euler’s totient function

Proof. The proof is easy, by changing variables.

Dirichlet’s  dual trick is to sum over character $\chi$ module $N$ evaluated at fixed $a$ in $(\mathbf{Z}/N\mathbf{Z})^\times$. From Theorem 1 we see that
$\displaystyle\sum_{\chi\ \text{mod}\ N} \chi(a)=\begin{cases} \varphi(N)& \text{for}\ a=1\ \text{mod}\ N,\\ 0& \text{otherwise}. \end{cases}$
We also have for $b$ invertible module $N$,
$\displaystyle \sum_{\chi\ \text{mod}\ N} \chi(a)^{-1}\chi(b)=\sum_{\chi\ \text{mod}\ N} \chi(a^{-1}b)=\begin{cases} \varphi(N)& \text{for}\ a=b\ \text{mod}\ N,\\ 0& \text{otherwise}. \end{cases}$

Definition. (Dirichlet $L$-function)  Given a character $\chi$ module $N$, the corresponding Dirichlet $L$-function is
$\displaystyle L(s,\chi):=\sum_{n=1}^\infty \frac{\chi(n)}{n^s},\quad \mathrm{Re}(s)>1.$
Since we have the multiplicative property $\chi(mn)=\chi(m)\chi(n)$, each $L$-function has an  Euler product expansion
$\displaystyle L(s,\chi)=\prod_{p\ \text{prime}}\frac{1}{1-\chi(p)p^{-s}}.$

The next step is to take logarithm of $L(s,\chi)$ and equate it to the logarithm of the Euler product. This must be done with some care since complex logarithms can be insidious.
As seen from the Euler product, the $L$-function $L(s,\chi)$ does not vanish in the half-plane $\mathrm{Re}(s)>1$, so the half plane being simple connected, the logarithm $\log L(s,\chi)$ is well defined (for instance, $\log L(s,\chi):=\int_{\gamma}\frac{L'(w,\chi)}{L(w,\chi)}\,dw+c$).

However, when taking the logarithm of the Euler product, one can not immediately infer that one arrive at the sum of the logarithm ($\log (zw)$ is not necessarily equal to $\log z+\log w$; they might different by $2\pi i$). Summing the principal logarithm of the factor in the Euler product, and using the branch it holds that
$\displaystyle -\log(1-z)=\sum_{m=1}^\infty \frac{z^m}{m}$
for $|z|<1$, we arrive at the following series
$\displaystyle G(s)=-\sum_{p}\log(1-\chi(p)p^{-s})=\sum_{p\ \text{prime}}\sum_{m\geq 1}\frac{\chi(p)^m}{mp^{ms}}$
which converges for $\mathrm{Re}(s)>1$, uniformly on compact set, as it is dominated by $\sum_{n\ge 1}\frac{1}{n^s}$. And since exponential function behaves better that the logarithm, i.e., it it continuous and $\exp(z+w)=\exp(z)\exp(w)$ always holds, this series is a logarithm of $L(s,\chi)$ (but may not be the principal one); that is, one has
$\displaystyle \exp(G(s))=L(s,\chi),$

We still use $\log (s,\chi)$ to denote this branch of logarithm, then we have
$\displaystyle \log L(s,\chi)=\sum_{p\ \text{prime},m\geq 1}\frac{\chi(p)^m}{mp^{ms}}=\sum_{p\ \text{prime}}\frac{\chi(p)}{p^s}+\sum_{p\ \text{prime},m\geq 2}\frac{\chi(p)^m}{mp^{ms}}.$
The second sum on the right will turn to be subordinate to the first, so we aim our attention at the first sum where $m=1$.

To pick out the primes $p$ with $p=a$ mod $N$, we use the sum over $\chi$ trick to obtain
$\displaystyle \sum_{\chi\ \text{mod}\ N}\chi(a)^{-1}\frac{\chi(p)}{p^s}=\begin{cases} \varphi(N)p^{-s}& \text{for}\ p=a\ \text{mod}\ N,\\ 0&\text{otherwise}. \end{cases}$
Thus
$\displaystyle \sum_{\chi\ \text{mod}\ N} \chi(a)^{-1}L(s,\chi)=\sum_{\chi\ \text{mod}\ N}\chi(a)^{-1}\sum_{p\ \text{prime},m\geq 1}\frac{\chi(p)^m}{mp^{ms}}\\ \quad\quad\quad\quad=\sum_{p=a\ \text{mod}\ N}\frac{\varphi(N)}{p^s}+\sum_{\chi\ \text{mod}\ N}\chi(a)^{-1}\sum_{p\ \text{prime},m\geq 2}\frac{\chi(p)^m}{mp^{ms}}$