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Exercise 8.1.2 (The maximum modulus principle) If \( f \) is as in the previous exercise, show that \( \left| {f\left( z\right) }\right| < M \) for all interior points \( z \in R \) , unless \( f \) is constant.
Theorem 8.1.3 (Phragmén - Lindelöf) Suppose that \( f\left( s\right) \) is entire in the region
\[
S\left( {a, b}\right) = \{ s \in \mathbb{C} : a \leq \operatorname{Re}\left( s\right) \leq b\}
\]
and that as \( \left| t\right| \rightarrow \infty \) ,
\[
\left| {f\left( s\right) }\right| = O\left( {e}^{{\left| t\right| }^{\alpha }}\right)
\]
for some \( \alpha \geq 1 \) . If \( f\left( s\right) \) is bounded on the two vertical lines \( \operatorname{Re}\left( s\right) = a \) and \( \operatorname{Re}\left( s\right) = b \), then \( f\left( s\right) \) is bounded in \( S\left( {a, b}\right) \) .
Proof. We first select an integer \( m > \alpha, m \equiv 2\left( {\;\operatorname{mod}\;4}\right) \) . Since arg \( s \rightarrow \) \( \pi /2 \) as \( t \rightarrow \infty \), we can choose \( {T}_{1} \) sufficiently large so that
\[
\left| {\arg s - \pi /2}\right| < \pi /{4m}
\]
Then for \( \left| {\operatorname{Im}\left( s\right) }\right| \geq {T}_{1} \), we find that \( \arg s = \pi /2 - \delta = \theta \) (say) satisfies
\[
\cos {m\theta } = - \cos {m\delta } < - 1/\sqrt{2}.
\]
Therefore, if we consider
\[
{g}_{\epsilon }\left( s\right) = {e}^{\epsilon {s}^{m}}f\left( s\right)
\]
then
\[
\left| {{g}_{\epsilon }\left( s\right) }\right| \leq K{e}^{{\left| t\right| }^{\alpha }}{e}^{-\epsilon {\left| s\right| }^{m}/\sqrt{2}}.
\]
Thus, \( \left| {{g}_{\epsilon }\left( s\right) }\right| \rightarrow 0 \) as \( \left| t\right| \rightarrow \infty \) . Let \( B \) be the maximum of \( f\left( s\right) \) in the region
\[
a \leq \operatorname{Re}\left( s\right) \leq b,\;0 \leq \left| {\operatorname{Im}\left( s\right) }\right| \leq {T}_{1}
\]
Let \( {T}_{2} \) be chosen such that
\[
\left| {{g}_{\epsilon }\left( s\right) }\right| \leq B
\]
for \( \left| {\operatorname{Im}\left( s\right) }\right| \geq {T}_{2} \) . Thus,
\[
\left| {f\left( s\right) }\right| \leq B{e}^{-\epsilon {\left| s\right| }^{m}\cos \left( {m\arg s}\right) } \leq B{e}^{\epsilon {\left| s\right| }^{m}}
\]
for \( \left| {\operatorname{Im}\left( s\right) }\right| \geq {T}_{2} \) . Applying the maximum modulus principle to the region
\[
a \leq \operatorname{Re}\left( s\right) \leq b,\;0 \leq \left| {\operatorname{Im}\left( s\right) }\right| \leq {T}_{2},
\]
we find that \( \left| {f\left( s\right) }\right| \leq B{e}^{\epsilon {\left| s\right| }^{m}} \) . This estimate holds for all \( s \) in \( S\left( {a, b}\right) \) . Letting \( \epsilon \rightarrow 0 \) yields the result.
Corollary 8.1.4 Suppose that \( f\left( s\right) \) is entire in \( S\left( {a, b}\right) \) and that \( \left| {f\left( s\right) }\right| \) \( = O\left( {e}^{{\left| t\right| }^{\alpha }}\right) \) for some \( \alpha \geq 1 \) as \( \left| t\right| \rightarrow \infty \) . If \( f\left( s\right) \) is \( O\left( {\left| t\right| }^{A}\right) \) on the two vertical lines \( \operatorname{Re}\left( s\right) = a \) and \( \operatorname{Re}\left( s\right) = b \), then \( f\left( s\right) = O\left( {\left| t\right| }^{A}\right) \) in \( S\left( {a, b}\right) \) .
Proof. We apply the theorem to the function \( g\left( s\right) = f\left( s\right) /{\left( s - u\right) }^{A} \) , where \( u > b \) . Then \( g \) is bounded on the two vertical strips, and the result follows.
Exercise 8.1.5 Show that for any entire function \( F \in \mathcal{S} \), we have
\[
F\left( s\right) = O\left( {\left| t\right| }^{A}\right)
\]
for some \( A > 0 \), in the region \( 0 \leq \operatorname{Re}\left( s\right) \leq 1 \) .
## 8.2 Basic Properties
We begin by stating the following theorem of Selberg:
Theorem 8.2.1 (Selberg) For any \( F \in \mathcal{S} \), let \( {N}_{F}\left( T\right) \) be the number of zeros \( \rho \) of \( F\left( s\right) \) satisfying \( 0 \leq \operatorname{Im}\left( \rho \right) \leq T \), counted with multiplicity. Then
\[
{N}_{F}\left( T\right) \sim \left( {2\mathop{\sum }\limits_{{i = 1}}^{d}{\alpha }_{i}}\right) \frac{T\log T}{2\pi }
\]
as \( T \rightarrow \infty \) .
Proof. This is easily derived by the method used to count zeros of \( \zeta \left( s\right) \) and \( L\left( {s,\chi }\right) \) as in Theorem 7.1.7 and Exercise 7.4.4.
Clearly, the functional equation for \( F \in \mathcal{S} \) is not unique, by virtue of Legendre's duplication formula. However, the above theorem shows that the sum of the \( {\alpha }_{i} \) ’s is well-defined. Accordingly, we define the degree of \( F \) by
\[
\deg F \mathrel{\text{:=}} 2\mathop{\sum }\limits_{{i = 1}}^{d}{\alpha }_{i}
\]
Lemma 8.2.2 (Conrey and Ghosh) If \( F \in S \) and \( \deg F = 0 \), then \( F = 1 \) .
Proof. We follow [CG]. A Dirichlet series can be viewed as a power series in the infinitely many variables \( {p}^{-s} \) as we range over primes \( p \) . Thus, if \( \deg F = 0 \), we can write our functional equation as
\[
\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{\left( \frac{{Q}^{2}}{n}\right) }^{s} = {wQ}\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\overline{{a}_{n}}}{n}{n}^{s}
\]
where \( \left| w\right| = 1 \) .
Thus, if \( {a}_{n} \neq 0 \) for some \( n \), then \( {Q}^{2}/n \) is an integer. Hence \( {Q}^{2} \) is an integer. Moreover, \( {a}_{n} \neq 0 \) implies \( n \mid {Q}^{2} \), so that our Dirichlet series is really a Dirichlet polynomial. Therefore, if \( {Q}^{2} = 1 \), then \( F = 1 \), and we are done. So, let us suppose \( q \mathrel{\text{:=}} {Q}^{2} > 1 \) . Since \( {a}_{1} = \) 1, comparing the \( {Q}^{2s} \) term in the functional equation above gives \( \left| {a}_{q}\right| = Q \) . Since \( {a}_{n} \) is multiplicative, we must have for some prime power \( {p}^{r}\parallel q \) that \( \left| {a}_{{p}^{r}}\right| \geq {p}^{r/2} \) . Now consider the \( p \) -Euler factor
\[
{F}_{p}\left( s\right) = \mathop{\sum }\limits_{{j = 0}}^{r}\frac{{a}_{{p}^{j}}}{{p}^{js}}
\]
with logarithm
\[
\log {F}_{p}\left( s\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }\frac{{b}_{{p}^{j}}}{{p}^{js}}
\]
Viewing these as power series in \( x = {p}^{-s} \), we write
\[
P\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{r}{A}_{j}{x}^{j}
\]
\[
\log P\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }{B}_{j}{x}^{j}
\]
where \( {A}_{j} = {a}_{{p}^{j}},{B}_{j} = {b}_{{p}^{j}} \) . Since \( {a}_{1} = 1 \), we can factor
\[
P\left( x\right) = \mathop{\prod }\limits_{{j = 1}}^{r}\left( {1 - {R}_{i}x}\right)
\]
so that
\[
{B}_{j} = - \mathop{\sum }\limits_{{i = 1}}^{r}\frac{{R}_{i}^{j}}{j}
\]
We also know that
\[
\mathop{\prod }\limits_{{i = 1}}^{r}\left| {R}_{i}\right| \geq {p}^{r/2}
\]
so that
\[
\mathop{\max }\limits_{{1 \leq i \leq r}}\left| {R}_{i}\right| \geq {p}^{1/2}
\]
But
\[
{\left| {b}_{{p}^{j}}\right| }^{1/j} = {\left| {B}_{j}\right| }^{1/j} = {\left| \mathop{\sum }\limits_{{i = 1}}^{r}\frac{{R}_{i}^{j}}{j}\right| }^{1/j}
\]
tends to \( \mathop{\max }\limits_{{1 \leq i \leq r}}\left| {R}_{i}\right| \) as \( j \rightarrow \infty \), which is greater than or equal to \( {p}^{1/2} \) . This contradicts the condition that \( {b}_{n} = O\left( {n}^{\theta }\right) \) with \( \theta < 1/2 \) . Therefore, \( Q = 1 \) and hence \( F = 1 \) .
We can now prove the following basic result:
Theorem 8.2.3 (Selberg) If \( F \in \mathcal{S} \) and \( F \) is of positive degree, then \( \deg F \geq 1 \) .
Proof. We follow [CG]. Consider the identity
\[
\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{e}^{-{nx}} = \frac{1}{2\pi i}{\int }_{\left( 2\right) }F\left( s\right) {x}^{-s}\Gamma \left( s\right) {ds}.
\]
Because of the Phragmen - Lindelöf principle and the functional equation, we find that \( F\left( s\right) \) has polynomial growth in \( \left| {\operatorname{Im}\left( s\right) }\right| \) in any vertical strip. Thus, moving the line of integration to the left, and taking into account the possible pole at \( s = 1 \) of \( F\left( s\right) \) as well as the poles of \( \Gamma \left( s\right) \) at \( s = 0, - 1, - 2,\ldots \), we obtain
\[
\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{e}^{-{nx}} = \frac{P\left( {\log x}\right) }{x} + \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{F\left( {-n}\right) {\left( -1\right) }^{n}{x}^{n}}{n!},
\]
where \( P \) is a polynomial. The functional equation relates \( F\left( {-n}\right) \) to \( F\left( {n + 1}\right) \) with a product of gamma functions. If \( 0 < \deg F < 1 \) , we find by Stirling's formula that the sum on the right-hand side converges for all \( x \) . Moreover, \( P\left( {\log x}\right) \) is analytic in \( \mathbb{C} \smallsetminus \{ x \leq 0 : x \in \) \( \mathbb{R}\} \) . Hence the left-hand side is analytic in \( \mathbb{C} \smallsetminus \{ x \leq 0 : x \in \mathbb{R}\} \) . But since the left-hand side is periodic with period \( {2\pi i} \), we find that
\[
f\left( z\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{e}^{-{nz}}
\]
is entire. Thus, for any \( x \) ,
\[
{a}_{n}{e}^{-{nx}} = {\int }_{0}^{2\pi }f\left( {x + {iy}}\right) {e}^{iny}{dy} \ll {n}^{-2}
\]
by integrating by parts. Choosing \( x = 1/n \) gives \( {a}_{n} = O\left( {1/{n}^{2}}\right) \) . Hence the Dirichlet series
\[
F\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{n}^{s}}
\]
converges absolutely for \( \operatorname{Re}s > - 1 \) . However, relating \( F\left( {-1/2 + {it}}\right) \) to \( F\left( {3/2 - {it}}\right) \) by the functional equation and using Stirling’s formula, we find that \( F\left( {-1/2 + {it}}\right) \) is not bounded. This contradiction forces \( \deg F \geq 1 \) .
An element \( F \in \mathcal{S} \) will be called primitive if \( F \neq 1 \) and \( F = {F}_{1}{F}_{2} \) with \( {F}_{1},{F}_{2} \in \mathcal{S} \) implies \( {F}_{1} = 1 \) or \( {F}_{2} = 1 \) . Thus, a primitive function cannot be factored nontrivially in \( \ | (Theorem 8.1.3 (Phragmén - Lindelöf)) Suppose that \( f\left( s\right) \) is entire in the region
\[
S\left( {a, b}\right) = \{ s \in \mathbb{C} : a \leq \operatorname{Re}\left( s\right) \leq b\}
\]
and that as \( \left| t\right| \rightarrow \infty \) ,
\[
\left| {f\left( s\right) }\right| = O\left( {e}^{{\left| t\right| }^{\alpha }}\right)
\]
for some \( \alpha \geq 1 \) . If \( f\left( s\right) \) is bounded on the two vertical lines \( \operatorname{Re}\left( s\right) = a \) and \( \operatorname{Re}\left( s\right) = b \), then \( f\left( s\right) \) is bounded in \( S\left( {a, b}\right) \) . | We first select an integer \( m > \alpha, m \equiv 2\left( {\;\operatorname{mod}\;4}\right) \) . Since arg \( s \rightarrow \) \( \pi /2 \) as \( t \rightarrow \infty \), we can choose \( {T}_{1} \) sufficiently large so that
\[
\left| {\arg s - \pi /2}\right| < \pi /{4m}
\]
Then for \( \left| {\operatorname{Im}\left( s\right) }\right| \geq {T}_{1} \), we find that \( \arg s = \pi /2 - \delta = \theta \) (say) satisfies
\[
\cos {m\theta } = - \cos {m\delta } < - 1/\sqrt{2}.
\]
Therefore, if we consider
\[
{g}_{\epsilon }\left( s\right) = {e}^{\epsilon {s}^{m}}f\left( s\right)
\]
then
\[
\left| {{g}_{\epsilon }\left( s\right) }\right| \leq K{e}^{{\left| t\right| }^{\alpha }}{e}^{-\epsilon {\left| s\right| }^{m}/\sqrt{2}}.
\]
Thus, \( \left| {{g}_{\epsilon }\left( s\right) }\right| \rightarrow 0 \) as \( \left| t\right| \rightarrow \infty \) . Let \( B \) be the maximum of \( f\left( s\right) \) in the region
\[
a \leq \operatorname{Re}\left( s\right) \leq b,\;0 \leq \left| {\operatorname{Im}\left( s\right) }\right| \leq {T}_{1}
\]
Let \( {T}_{2} \) be chosen such that |
Exercise 8.3.7 Let \( K \) be a quadratic field of discriminant \( d \) . Let \( {P}_{0} \) denote the group of principal fractional ideals \( \alpha {\mathcal{O}}_{K} \) with \( \alpha \in K \) satisfying \( {N}_{K}\left( \alpha \right) > 0 \) . The quotient group \( {H}_{0} \) of all nonzero fractional ideals modulo \( {P}_{0} \) is called the restricted class group of \( K \) . Show that \( {H}_{0} \) is a subgroup of the ideal class group \( H \) of \( K \) and \( \left\lbrack {H : {H}_{0}}\right\rbrack \leq 2 \) .
Exercise 8.3.8 Given an ideal \( \mathfrak{a} \) of a quadratic field \( K \), let \( {\mathfrak{a}}^{\prime } \) denote the conjugate ideal. If \( K \) has discriminant \( d \), write
\[
\left| d\right| = {p}_{1}^{{\alpha }_{1}}{p}_{2}\cdots {p}_{t}
\]
where \( {p}_{1} = 2,{\alpha }_{1} = 0,2 \), or 3 and \( {p}_{2},\ldots ,{p}_{t} \) are distinct odd primes. If we write \( {p}_{i}{\mathcal{O}}_{K} = {\wp }_{i}^{2} \) show that for any ideal \( \mathfrak{a} \) of \( {\mathcal{O}}_{K} \) satisfying \( \mathfrak{a} = {\mathfrak{a}}^{\prime } \) we can write
\[
\mathfrak{a} = r{\wp }_{1}^{{a}_{1}}\cdots {\wp }_{t}^{{a}_{t}}
\]
\( r > 0,{a}_{i} = 0,1 \) uniquely.
Exercise 8.3.9 An ideal class \( C \) of \( {H}_{0} \) is said to be ambiguous if \( {C}^{2} = 1 \) in \( {H}_{0} \) . Show that any ambiguous ideal class is equivalent (in the restricted sense) to one of the at most \( {2}^{t} \) ideal classes
\[
{\wp }_{1}^{{a}_{1}}\cdots {\wp }_{t}^{{a}_{t}},\;{a}_{i} = 0,1.
\]
Exercise 8.3.10 With the notation as in the previous two questions, show that there is exactly one relation of the form
\[
{\wp }_{1}^{{a}_{1}}\cdots {\wp }_{t}^{{a}_{t}} = \rho {\mathcal{O}}_{K},\;{N}_{K}\left( \rho \right) > 0,
\]
with \( {a}_{i} = 0 \) or \( 1,\mathop{\sum }\limits_{{i = 1}}^{t}{a}_{i} > 0 \) .
Exercise 8.3.11 Let \( K \) be a quadratic field of discriminant \( d \) . Show that the number of ambiguous ideal classes is \( {2}^{t - 1} \) where \( t \) is the number of distinct primes dividing \( d \) . Deduce that \( {2}^{t - 1} \) divides the order of the class group.
Exercise 8.3.12 If \( K \) is a quadratic field of discriminant \( d \) and class number 1, show that \( d \) is prime or \( d = 4 \) or 8 .
Exercise 8.3.13 If a real quadratic field \( K \) has odd class number, show that \( K \) has a unit of norm -1 .
Exercise 8.3.14 Show that \( {15} + 4\sqrt{14} \) is the fundamental unit of \( \mathbb{Q}\left( \sqrt{14}\right) \) .
Exercise 8.3.15 In Chapter 6 we showed that \( \mathbb{Z}\left\lbrack \sqrt{14}\right\rbrack \) is a PID (principal ideal domain). Assume the following hypothesis: given \( \alpha ,\beta \in \mathbb{Z}\left\lbrack \sqrt{14}\right\rbrack \), such that \( \gcd \left( {\alpha ,\beta }\right) = 1 \), there is a prime \( \pi \equiv \alpha \left( {\;\operatorname{mod}\;\beta }\right) \) for which the fundamental unit \( \varepsilon = {15} + 4\sqrt{14} \) generates the coprime residue classes \( \left( {\;\operatorname{mod}\;\pi }\right) \) . Show that \( \mathbb{Z}\left\lbrack \sqrt{14}\right\rbrack \) is Euclidean.
It is now known that \( \mathbb{Z}\left\lbrack \sqrt{14}\right\rbrack \) is Euclidean and is the main theorem of the doctoral thesis of Harper \( \left\lbrack \mathrm{{Ha}}\right\rbrack \) . The hypothesis of the previous exercise is still unknown however and is true if the Riemann hypothesis holds for Dedekind zeta functions of number fields (see Chapter 10). The hypothesis in the question should be viewed as a number field version of a classical conjecture of Artin on primitive roots. Previously the classification of Euclidean rings of algebraic integers relied on some number field generalization of the Artin primitive root conjecture. But recently, Harper and Murty [HM] have found new techniques which circumvent the need of such a hypothesis in such questions. No doubt, these techniques will have further applications.
Exercise 8.3.16 Let \( d = {a}^{2} + 1 \) . Show that if \( \left| {{u}^{2} - d{v}^{2}}\right| \neq 0,1 \) for integers \( u, v \) ,
then
\[
\left| {{u}^{2} - d{v}^{2}}\right| > \sqrt{d}
\]
Exercise 8.3.17 Suppose that \( n \) is odd, \( n \geq 5 \), and that \( {n}^{2g} + 1 = d \) is squarefree. Show that the class group of \( \mathbb{Q}\left( \sqrt{d}\right) \) has an element of order \( {2g} \) .
## Chapter 9
## Higher Reciprocity Laws
## 9.1 Cubic Reciprocity
Let \( \rho = \left( {-1 + \sqrt{-3}}\right) /2 \) be as in Chapter 2, and let \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) be the ring of Eisenstein integers. Recall that \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) is a Euclidean domain and hence a PID. We set \( N\left( {a + {b\rho }}\right) = {a}^{2} - {ab} + {b}^{2} \) which is the Euclidean norm as proved in Section 2.3.
Exercise 9.1.1 If \( \pi \) is a prime of \( \mathbb{Z}\left\lbrack \rho \right\rbrack \), show that \( N\left( \pi \right) \) is a rational prime or the square of a rational prime.
Exercise 9.1.2 If \( \pi \in \mathbb{Z}\left\lbrack \rho \right\rbrack \) is such that \( N\left( \pi \right) = p \), a rational prime, show that \( \pi \) is a prime of \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) .
Exercise 9.1.3 If \( p \) is a rational prime congruent to 2 (mod 3), show that \( p \) is prime in \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) . If \( p \equiv 1\left( {\;\operatorname{mod}\;3}\right) \), show that \( p = \pi \bar{\pi } \) where \( \pi \) is prime in \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) .
Exercise 9.1.4 Let \( \pi \) be a prime of \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) . Show that \( {\alpha }^{N\left( \pi \right) - 1} \equiv 1\left( {\;\operatorname{mod}\;\pi }\right) \) for all \( \alpha \in \mathbb{Z}\left\lbrack \rho \right\rbrack \) which are coprime to \( \pi \) .
Exercise 9.1.5 Let \( \pi \) be a prime not associated to \( \left( {1 - \rho }\right) \) . First show that \( 3 \mid N\left( \pi \right) - 1 \) . If \( \left( {\alpha ,\pi }\right) = 1 \), show that there is a unique integer \( m = 0,1 \), or 2 such that
\[
{\alpha }^{\left( {N\left( \pi \right) - 1}\right) /3} \equiv {\rho }^{m}\;\left( {\;\operatorname{mod}\;\pi }\right)
\]
Let \( N\left( \pi \right) \neq 3 \) . We define the cubic residue character of \( \alpha \left( {\;\operatorname{mod}\;\pi }\right) \) by the symbol \( {\left( \alpha /\pi \right) }_{3} \) as follows:
(i) \( {\left( \alpha /\pi \right) }_{3} = 0 \) if \( \pi \mid \alpha \) ;
(ii) \( {\alpha }^{\left( {N\left( \pi \right) - 1}\right) /3} \equiv {\left( \alpha /\pi \right) }_{3}\left( {\;\operatorname{mod}\;\pi }\right) \) where \( {\left( \alpha /\pi \right) }_{3} \) is the unique cube root of unity determined by the previous exercise.
Exercise 9.1.6 Show that:
(a) \( {\left( \alpha /\pi \right) }_{3} = 1 \) if and only if \( {x}^{3} \equiv \alpha \left( {\;\operatorname{mod}\;\pi }\right) \) is solvable in \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) ;
(b) \( {\left( \alpha \beta /\pi \right) }_{3} = {\left( \alpha /\pi \right) }_{3}{\left( \beta /\pi \right) }_{3} \) ; and
(c) if \( \alpha \equiv \beta \left( {\;\operatorname{mod}\;\pi }\right) \), then \( {\left( \alpha /\pi \right) }_{3} = {\left( \beta /\pi \right) }_{3} \) .
Let us now define the cubic character \( {\chi }_{\pi }\left( \alpha \right) = {\left( \alpha /\pi \right) }_{3} \) .
Exercise 9.1.7 Show that:
(a) \( \overline{{\chi }_{\pi }\left( \alpha \right) } = {\chi }_{\pi }{\left( \alpha \right) }^{2} = {\chi }_{\pi }\left( {\alpha }^{2}\right) \) ; and
(b) \( \overline{{\chi }_{\pi }\left( \alpha \right) } = {\chi }_{\bar{\pi }}\left( \alpha \right) \) .
Exercise 9.1.8 If \( q \equiv 2\left( {\;\operatorname{mod}\;3}\right) \), show that \( {\chi }_{q}\left( \bar{\alpha }\right) = {\chi }_{q}\left( {\alpha }^{2}\right) \) and \( {\chi }_{q}\left( n\right) = 1 \) if \( n \) is a rational integer coprime to \( q \) .
This exercise shows that any rational integer is a cubic residue mod \( q \) . If \( \pi \) is prime in \( \mathbb{Z}\left\lbrack \rho \right\rbrack \), we say \( \pi \) is primary if \( \pi \equiv 2\left( {\;\operatorname{mod}\;3}\right) \) . Therefore if \( q \equiv 2\left( {\;\operatorname{mod}\;3}\right) \), then \( q \) is primary in \( \mathbb{Z}\left\lbrack \rho \right\rbrack \) . If \( \pi = a + {b\rho } \), then this means \( a \equiv 2\left( {\;\operatorname{mod}\;3}\right) \) and \( b \equiv 0\left( {\;\operatorname{mod}\;3}\right) \) .
Exercise 9.1.9 Let \( N\left( \pi \right) = p \equiv 1\left( {\;\operatorname{mod}\;3}\right) \) . Among the associates of \( \pi \), show there is a unique one which is primary.
We can now state the law of cubic reciprocity: let \( {\pi }_{1},{\pi }_{2} \) be primary. Suppose \( N\left( {\pi }_{1}\right), N\left( {\pi }_{2}\right) \neq 3 \) and \( N\left( {\pi }_{1}\right) \neq N\left( {\pi }_{2}\right) \) . Then
\[
{\chi }_{{\pi }_{1}}\left( {\pi }_{2}\right) = {\chi }_{{\pi }_{2}}\left( {\pi }_{1}\right)
\]
To prove the law of cubic reciprocity, we will introduce Jacobi sums and more general Gauss sums than the ones used in Chapter 7. Let \( {\mathbb{F}}_{p} \) denote the finite field of \( p \) elements. A multiplicative character on \( {\mathbb{F}}_{p} \) is a homomorphism \( \chi : {\mathbb{F}}_{p}^{ \times } \rightarrow {\mathbb{C}}^{ \times } \) . The Legendre symbol \( \left( {a/p}\right) \) is an example of such a character. Another example is the trivial character \( {\chi }_{0} \) defined by \( {\chi }_{0}\left( a\right) = 1 \) for all \( a \in {\mathbb{F}}_{p}^{ \times } \) . It is useful to extend the definition of \( \chi \) to all of \( {\mathbb{F}}_{p} \) . We set \( \chi \left( 0\right) = 0 \) for \( \chi \neq {\chi }_{0} \) and \( {\chi }_{0}\left( 0\right) = 1 \) .
For \( a \in {\mathbb{F}}_{p}^{ \times } \), define the Gauss sum
\[
{g}_{a}\left( \chi \right) = \mathop{\sum }\limits_{{t \in {\mathbb{F}}_{p}}}\chi \left( t\right) {\zeta }^{at}
\]
whe | Exercise 8.3.7 Let \( K \) be a quadratic field of discriminant \( d \) . Let \( {P}_{0} \) denote the group of principal fractional ideals \( \alpha {\mathcal{O}}_{K} \) with \( \alpha \in K \) satisfying \( {N}_{K}\left( \alpha \right) > 0 \) . The quotient group \( {H}_{0} \) of all nonzero fractional ideals modulo \( {P}_{0} \) is called the restricted class group of \( K \) . Show that \( {H}_{0} \) is a subgroup of the ideal class group \( H \) of \( K \) and \( \left\lbrack {H : {H}_{0}}\right\rbrack \leq 2 \) . | To show that \( {H}_{0} \) is a subgroup of the ideal class group \( H \) of \( K \), we need to verify that \( {H}_{0} \) satisfies the subgroup criteria within \( H \).
1. **Identity Element**: The identity element in \( H \) is the class of the unit ideal \( {\mathcal{O}}_{K} \). Since any principal ideal \( \alpha {\mathcal{O}}_{K} \) with \( \alpha \in K \) and \( {N}_{K}\left( \alpha \right) > 0 \) is in \( {P}_{0} \), the identity element in \( H \) is also in \( {H}_{0} \).
2. **Closure**: Let \( [I] \) and \( [J] \) be elements of \( {H}_{0} \), where \( I \) and \( J \) are nonzero fractional ideals of \( K \). The product of ideals \( IJ \) is also a nonzero fractional ideal, and since multiplication of ideals is commutative and associative, the class of \( IJ \) in \( H \) is well-defined. We need to show that if both \( [I] \) and \( [J] \) are in \( {H}_{0} \), then so is their product. This follows because if both ideals are equivalent modulo principal ideals with positive norm, their product will also be equivalent modulo such principal ideals.
3. **Inverses**: For any ideal class \( [I] \in {H}_{0} \), there exists an inverse class, which corresponds to the inverse ideal (or more precisely, the inverse in the group of fractional ideals). The inverse ideal also satisfies the condition for being in the restricted class group because taking inverses preserves the property of being equivalent modulo principal ideals with positive norm.
Thus, we have shown that every element in the restricted class group has an inverse within this group, ensuring that it is indeed a subgroup of the full ideal class group.
Next, we need to show that the index \(\left\lbrack {H : {H}_{0}}\right\rbrack \leq 2\). This can be seen by considering the structure of the ideal class group and how principal ideals with negative norms fit into it. Since every ideal class in \( H \) can be represented by an ideal whose norm can be made positive by multiplying by a suitable unit (which does not change its class in the full class group but might change its membership in the restricted class group), it follows that each class in \( H \) corresponds either to a class already in \( {H}_{0} \) or to one obtained by multiplying by a principal ideal generated by an element with negative norm. Since there are only two such possibilities (positive or negative norm condition), it follows that \(\left\lbrack {H : {H}_{0}}\right\rbrack \leq 2\). |
Proposition 9.34 Define a domain \( \operatorname{Dom}\left( \Delta \right) \) as follows:
\[
\operatorname{Dom}\left( \Delta \right) = \left\{ {\psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \left| {\;{\left| \mathbf{k}\right| }^{2}\widehat{\psi }\left( \mathbf{k}\right) \in {L}^{2}\left( {\mathbb{R}}^{n}\right) }\right. }\right\} .
\]
Define \( \Delta \) on this domain by the expression
\[
{\Delta \psi } = - {\mathcal{F}}^{-1}\left( {{\left| \mathbf{k}\right| }^{2}\widehat{\psi }\left( \mathbf{k}\right) }\right)
\]
(9.18)
where \( \widehat{\psi } \) is the Fourier transform of \( \psi \) and \( {\mathcal{F}}^{-1} \) is the inverse Fourier. Then \( \Delta \) is self-adjoint on \( \operatorname{Dom}\left( \Delta \right) \) .
The domain \( \operatorname{Dom}\left( \Delta \right) \) may also be described as the set of all \( \psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \) such that \( {\Delta \psi } \), computed in the distribution sense, belongs to \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) . If \( \psi \in \operatorname{Dom}\left( \Delta \right) \), then \( {\Delta \psi } \) as defined by (9.18) agrees with \( {\Delta \psi } \) computed in the distribution sense.
The proof of Proposition 9.34 is extremely similar to that of Proposition 9.32 and is omitted. Of course, the kinetic energy operator \( - {\hslash }^{2}\Delta /\left( {2m}\right) \) is also self-adjoint on the same domain as \( \Delta \) . It is easy to see from (9.18) and the unitarity of the Fourier transform that \( - {\hslash }^{2}\Delta /\left( {2m}\right) \) is non-negative, that is, that
\[
\left\langle {\psi , - \frac{{\hslash }^{2}}{2m}{\Delta \psi }}\right\rangle \geq 0
\]
for all \( \psi \in \operatorname{Dom}\left( \Delta \right) \) .
Using the same reasoning as in Sects. 9.6 and 9.7, it is not hard to show that the operators \( {P}_{j} \) and \( \Delta \) are essentially self-adjoint on \( {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) . See Exercise 16.
Care must be exercised in applying Proposition 9.34. Although the function
\[
\psi \left( \mathbf{x}\right) \mathrel{\text{:=}} \frac{1}{\left| \mathbf{x}\right| }
\]
is harmonic on \( {\mathbb{R}}^{3} \smallsetminus \{ 0\} \), the Laplacian over \( {\mathbb{R}}^{3} \) of \( \psi \) in the distribution sense is not zero (Exercise 13). (It can be shown, by carefully analyzing the calculation in the proof of Proposition 9.35, that \( {\Delta \psi } \) is a nonzero multiple of a \( \delta \) -function.) This example shows that if a function \( \psi \) has a singularity, calculating the Laplacian of \( \psi \) away from the singularity may not give the correct distributional Laplacian of \( \psi \) . For example, the function \( \phi \) in \( {L}^{2}\left( {\mathbb{R}}^{3}\right) \) given by
\[
\phi \left( \mathbf{x}\right) \mathrel{\text{:=}} \frac{{e}^{-{\left| \mathbf{x}\right| }^{2}}}{\left| \mathbf{x}\right| }
\]
(9.19)
is not in \( \operatorname{Dom}\left( \Delta \right) \), even though both \( \phi \) and \( {\Delta \phi } \) are (by direct computation) square-integrable over \( {\mathbb{R}}^{3} \smallsetminus \{ 0\} \) . Indeed, when \( n \leq 3 \), every element of \( \operatorname{Dom}\left( \Delta \right) \) is continuous (Exercise 14).
Proposition 9.35 Suppose \( \psi \left( \mathbf{x}\right) = g\left( \mathbf{x}\right) f\left( \left| \mathbf{x}\right| \right) \), where \( g \) is a smooth function on \( {\mathbb{R}}^{n} \) and \( f \) is a smooth function on \( \left( {0,\infty }\right) \) . Suppose also that \( f \) satisfies
\[
\mathop{\lim }\limits_{{r \rightarrow {0}^{ + }}}{r}^{n - 1}f\left( r\right) = 0
\]
\[
\mathop{\lim }\limits_{{r \rightarrow {0}^{ + }}}{r}^{n - 1}{f}^{\prime }\left( r\right) = 0.
\]
If both \( \psi \) and \( {\Delta \psi } \) are square-integrable over \( {\mathbb{R}}^{n} \smallsetminus \{ 0\} \), then \( \psi \) belongs to \( \operatorname{Dom}\left( \Delta \right) \) .
Note that the second condition in the proposition fails if \( n = 3 \) and \( f\left( r\right) = 1/r \) . We will make use of this result in Chap. 18.
Proof. To apply Proposition 9.34, we need to compute \( \langle \psi ,{\Delta \chi }\rangle \), for each \( \chi \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) . We choose a large cube \( C \), centered at the origin and such that the support of \( \chi \) is contained in the interior of \( C \) . Then we consider the integral of \( \bar{\psi }\left( {{\partial }^{2}\chi /\partial {x}_{j}^{2}}\right) \) over \( C \smallsetminus {C}_{\varepsilon } \), where \( {C}_{\varepsilon } \) is a cube centered at the origin and having side-length \( \varepsilon \) . We evaluate the \( {x}_{j} \) -integral first and we integrate by parts twice. For "good" values of the remaining variables, \( {x}_{j} \) ranges over all of \( C \), in which case there are no boundary terms to worry about. For "bad" values of the remaining variables, we get two kinds of boundary terms, one involving \( \bar{\psi }\left( {\partial \chi /\partial {x}_{j}}\right) \) and one involving \( \left( {\partial \bar{\psi }/\partial {x}_{j}}\right) \chi \) , in both cases integrated over two opposite faces of \( {C}_{\varepsilon } \) .
Now,
\[
\frac{\partial \psi }{\partial {x}_{j}} = \frac{\partial g}{\partial {x}_{j}}f\left( \left| \mathbf{x}\right| \right) + g\left( \mathbf{x}\right) \frac{df}{dr}\frac{{x}_{j}}{r}.
\]
Since the area of the faces of the cube is \( {\varepsilon }^{n - 1} \), the assumption on \( f \) will cause the boundary terms to disappear in the limit as \( \varepsilon \) tends to zero. Furthermore, both \( \psi \) and \( {\Delta \psi } \) are in \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) and thus in \( {L}^{1}\left( C\right) \), where in the case of \( {\Delta \psi } \), we simply leave the value at the origin (which is a set of measure zero) undefined. Thus, integrals of \( \bar{\psi }{\Delta \chi } \) and \( {\left( \Delta \bar{\psi }\right) }_{\chi } \) over \( C \smallsetminus {C}_{\varepsilon } \) will converge to integrals over \( C \) . Since the boundary terms vanish in the limit, we are left with
\[
\langle \psi ,{\Delta \chi }\rangle = \langle {\Delta \psi },\chi \rangle
\]
Thus, the distributional Laplacian of \( \psi \) is simply integration against the "pointwise" Laplacian, ignoring the origin. Proposition 9.34 then tells us that \( \psi \in \operatorname{Dom}\left( \Delta \right) \) . ∎
## 9.9 Sums of Self-Adjoint Operators
In the previous section, we have succeeded in defining the Laplacian \( \Delta \) , and hence also the kinetic energy operator \( - {\hslash }^{2}\Delta /\left( {2m}\right) \), as a self-adjoint operator on a natural dense domain in \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) . We have also defined the potential energy operator \( V\left( \mathbf{X}\right) \) as a self-adjoint operator on a different dense domain, for any measurable function \( V : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) . To obtain the Schrödinger operator \( - {\hslash }^{2}\Delta /\left( {2m}\right) + V\left( \mathbf{X}\right) \), we "merely" have to make sense of the sum of two unbounded self-adjoint operators. This task, however, turns out to be more difficult than might be expected. In particular, if \( V \) is a highly singular function, then \( - {\hslash }^{2}\Delta /\left( {2m}\right) + V\left( \mathbf{X}\right) \) may fail to be self-adjoint or essentially self-adjoint on any natural domain.
Definition 9.36 If \( A \) and \( B \) are unbounded operators on \( \mathbf{H} \), then \( A + B \) is the operator with domain
\[
\operatorname{Dom}\left( {A + B}\right) \mathrel{\text{:=}} \operatorname{Dom}\left( A\right) \cap \operatorname{Dom}\left( B\right)
\]
and given by \( \left( {A + B}\right) \psi = {A\psi } + {B\psi } \) .
The sum of two unbounded self-adjoint operators \( A \) and \( B \) may fail to be self-adjoint or even essentially self-adjoint. [If, however, \( B \) is bounded with \( \operatorname{Dom}\left( B\right) = \mathbf{H} \), then Proposition 9.13 shows that \( A + B \) is self-adjoint on \( \operatorname{Dom}\left( A\right) \cap \operatorname{Dom}\left( B\right) = \operatorname{Dom}\left( A\right) \) .] For one thing, if \( A \) and \( B \) are unbounded, then \( \operatorname{Dom}\left( A\right) \cap \operatorname{Dom}\left( B\right) \) may fail to be dense in \( \mathbf{H} \) . But even if \( \operatorname{Dom}\left( A\right) \cap \) \( \operatorname{Dom}\left( B\right) \) is dense in \( \mathbf{H} \), it can easily happen that \( A + B \) is not essentially self-adjoint on this domain. (See, for example, Sect. 9.10.) Many things that are simple for bounded self-adjoint operators becomes complicated when dealing with unbounded self-adjoint operators!
In this section, we examine criteria on a function \( V \) under which the Schrödinger operator
\[
\widehat{H} = - \frac{{\hslash }^{2}}{2m}\Delta + V
\]
is self-adjoint or essentially self-adjoint on some natural domain inside \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) .
Theorem 9.37 (Kato-Rellich Theorem) Suppose that \( A \) and \( B \) are unbounded self-adjoint operators on \( \mathbf{H} \) . Suppose that \( \operatorname{Dom}\left( A\right) \subset \operatorname{Dom}\left( B\right) \) and that there exist positive constants \( a \) and \( b \) with \( a < 1 \) such that
\[
\parallel {B\psi }\parallel \leq a\parallel {A\psi }\parallel + b\parallel \psi \parallel
\]
(9.20)
for all \( \psi \in \operatorname{Dom}\left( A\right) \) . Then \( A + B \) is self-adjoint on \( \operatorname{Dom}\left( A\right) \) and essentially self-adjoint on any subspace of \( \operatorname{Dom}\left( A\right) \) on which \( A \) is essentially selfadjoint. Furthermore, if \( A \) is non-negative, then the spectrum of \( A + B \) is bounded below by \( - b/\left( {1 - a}\right) \) .
Note that since we assume \( \operatorname{Dom}\left( B\right) \supset \operatorname{Dom}\left( A\right) \), the natural domain for \( A + B \) is \( \oper | Proposition 9.34 Define a domain \( \operatorname{Dom}\left( \Delta \right) \) as follows:
\[
\operatorname{Dom}\left( \Delta \right) = \left\{ {\psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \left| {\;{\left| \mathbf{k}\right| }^{2}\widehat{\psi }\left( \mathbf{k}\right) \in {L}^{2}\left( {\mathbb{R}}^{n}\right) }\right. }\right\} .
\]
Define \( \Delta \) on this domain by the expression
\[
{\Delta \psi } = - {\mathcal{F}}^{-1}\left( {{\left| \mathbf{k}\right| }^{2}\widehat{\psi }\left( \mathbf{k}\right) }\right)
\]
where \( \widehat{\psi } \) is the Fourier transform of \( \psi \) and \( {\mathcal{F}}^{-1} \) is the inverse Fourier. Then \( \Delta \) is self-adjoint on \( \operatorname{Dom}\left( \Delta \right) \) . | null |
Exercise 9.2.10 Consider the element
\[
\alpha = {\left( x + y\right) }^{\ell - 2}\left( {x + {\zeta y}}\right)
\]
Show that:
(a) the ideal \( \left( \alpha \right) \) is a perfect \( \ell \) th power.
(b) \( \alpha \equiv 1 - {u\lambda }\left( {\;\operatorname{mod}\;{\lambda }^{2}}\right) \) where \( u = {\left( x + y\right) }^{\ell - 2}y \) .
Exercise 9.2.11 Show that \( {\zeta }^{-u}\alpha \) is primary.
Exercise 9.2.12 Use Eisenstein reciprocity to show that if \( {x}^{\ell } + {y}^{\ell } + {z}^{\ell } = 0 \) has a solution in integers, \( \ell \nmid {xyz} \), then for any \( p \mid y,{\left( \zeta /p\right) }_{\ell }^{-u} = 1 \) . (Hint: Evaluate \( \left( {p/{\zeta }^{-u}\alpha }\right) \) ,)
Exercise 9.2.13 Show that if
\[
{x}^{\ell } + {y}^{\ell } + {z}^{\ell } = 0
\]
has a solution in integers, \( l \nmid {xyz} \), then for any \( p \mid {xyz},{\left( \zeta /p\right) }_{\ell }^{-u} = 1 \) .
Exercise 9.2.14 Show that \( {\left( \zeta /p\right) }_{\ell }^{-u} = 1 \) implies that \( {p}^{\ell - 1} \equiv 1\left( {\;\operatorname{mod}\;{\ell }^{2}}\right) \) .
Exercise 9.2.15 If \( \ell \) is an odd prime and
\[
{x}^{\ell } + {y}^{\ell } + {z}^{\ell } = 0
\]
for some integers \( x, y, z \) coprime to \( \ell \), then show that \( {p}^{\ell - 1} \equiv 1\left( {\;\operatorname{mod}\;{\ell }^{2}}\right) \) for every \( p \mid {xyz} \) . Deduce that \( {2}^{\ell - 1} \equiv 1\left( {\;\operatorname{mod}\;{\ell }^{2}}\right) \) .
The congruence \( {2}^{\ell - 1} \equiv 1\left( {\;\operatorname{mod}\;{\ell }^{2}}\right) \) was first established by Wieferich in 1909 as a necessary condition in the first case of Fermat's Last Theorem. The only primes less than \( 3 \times {10}^{9} \) satisfying this congruence are 1093 and 3511 as a quick computer calculation shows. It is not known if there are infinitely many such primes. (See also Exercise 1.3.4.)
## 9.3 Supplementary Problems
Exercise 9.3.1 Show that there are infinitely many primes \( p \) such that \( \left( {2/p}\right) = \) \( - 1 \) .
Exercise 9.3.2 Let \( a \) be a nonsquare integer greater than 1. Show that there are infinitely many primes \( p \) such that \( \left( {a/p}\right) = - 1 \) .
Exercise 9.3.3 Suppose that \( {x}^{2} \equiv a\left( {\;\operatorname{mod}\;p}\right) \) has a solution for all but finitely many primes. Show that \( a \) is a perfect square.
Exercise 9.3.4 Let \( K \) be a quadratic extension of \( \mathbb{Q} \) . Show that there are infinitely many primes which do not split completely in \( K \) .
Exercise 9.3.5 Suppose that \( a \) is an integer coprime to the odd prime \( q \) . If \( {x}^{q} \equiv a\left( {\;\operatorname{mod}\;p}\right) \) has a solution for all but finitely many primes, show that \( a \) is a perfect \( q \) th power. (This generalizes the penultimate exercise.)
Exercise 9.3.6 Let \( p \equiv 1\left( {\;\operatorname{mod}\;3}\right) \) . Show that there are integers \( A \) and \( B \) such that
\[
{4p} = {A}^{2} + {27}{B}^{2}
\]
\( A \) and \( B \) are unique up to sign.
Exercise 9.3.7 Let \( p \equiv 1\left( {\;\operatorname{mod}\;3}\right) \) . Show that \( {x}^{3} \equiv 2\left( {\;\operatorname{mod}\;p}\right) \) has a solution if and only if \( p = {C}^{2} + {27}{D}^{2} \) for some integers \( C, D \) .
Exercise 9.3.8 Show that the equation
\[
{x}^{3} - 2{y}^{3} = {23}{z}^{m}
\]
has no integer solutions with \( \gcd \left( {x, y, z}\right) = 1 \) .
## Chapter 10
## Analytic Methods
## 10.1 The Riemann and Dedekind Zeta Functions
The Riemann zeta function \( \zeta \left( s\right) \) is defined initially for \( \operatorname{Re}\left( s\right) > 1 \) as the infinite series
\[
\zeta \left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}}
\]
Exercise 10.1.1 Show that for \( \operatorname{Re}\left( s\right) > 1 \) ,
\[
\zeta \left( s\right) = \mathop{\prod }\limits_{p}{\left( 1 - \frac{1}{{p}^{s}}\right) }^{-1}
\]
where the product is over prime numbers \( p \) .
Exercise 10.1.2 Let \( K \) be an algebraic number field and \( {\mathcal{O}}_{K} \) its ring of integers.
The Dedekind zeta function \( {\zeta }_{K}\left( s\right) \) is defined for \( \operatorname{Re}\left( s\right) > 1 \) as the infinite series
\[
{\zeta }_{K}\left( s\right) = \mathop{\sum }\limits_{\mathfrak{a}}\frac{1}{{\left( N\mathfrak{a}\right) }^{s}}
\]
where the sum is over all ideals of \( {\mathcal{O}}_{K} \) . Show that the infinite series is absolutely convergent for \( \operatorname{Re}\left( s\right) > 1 \) .
Exercise 10.1.3 Prove that for \( \operatorname{Re}\left( s\right) > 1 \) ,
\[
{\zeta }_{K}\left( s\right) = \mathop{\prod }\limits_{\wp }{\left( 1 - \frac{1}{{\left( N\wp \right) }^{s}}\right) }^{-1}.
\]
Theorem 10.1.4 Let \( {\left\{ {a}_{m}\right\} }_{m = 1}^{\infty } \) be a sequence of complex numbers, and let \( A\left( x\right) = \mathop{\sum }\limits_{{m \leq x}}{a}_{m} = O\left( {x}^{\delta }\right) \), for some \( \delta \geq 0 \) . Then
\[
\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{m}}{{m}^{s}}
\]
converges for \( \operatorname{Re}\left( s\right) > \delta \) and in this half-plane we have
\[
\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{m}}{{m}^{s}} = s{\int }_{1}^{\infty }\frac{A\left( x\right) {dx}}{{x}^{s + 1}}
\]
for \( \operatorname{Re}\left( s\right) > 1 \) .
Proof. We write
\[
\mathop{\sum }\limits_{{m = 1}}^{M}\frac{{a}_{m}}{{m}^{s}} = \mathop{\sum }\limits_{{m = 1}}^{M}\left( {A\left( m\right) - A\left( {m - 1}\right) }\right) {m}^{-s}
\]
\[
= A\left( M\right) {M}^{-s} + \mathop{\sum }\limits_{{m = 1}}^{{M - 1}}A\left( m\right) \left\{ {{m}^{-s} - {\left( m + 1\right) }^{-s}}\right\} .
\]
Since
\[
{m}^{-s} - {\left( m + 1\right) }^{-s} = s{\int }_{m}^{m + 1}\frac{dx}{{x}^{s + 1}}
\]
we get
\[
\mathop{\sum }\limits_{{m = 1}}^{M}\frac{{a}_{m}}{{m}^{s}} = \frac{A\left( M\right) }{{M}^{s}} + s{\int }_{1}^{M}\frac{A\left( x\right) {dx}}{{x}^{s + 1}}.
\]
For \( \operatorname{Re}\left( s\right) > \delta \), we find
\[
\mathop{\lim }\limits_{{M \rightarrow \infty }}\frac{A\left( M\right) }{{M}^{s}} = 0
\]
since \( A\left( x\right) = O\left( {x}^{\delta }\right) \) . Hence, the partial sums converge for \( \operatorname{Re}\left( s\right) > \delta \) and
we have
\[
\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{m}}{{m}^{s}} = s{\int }_{1}^{\infty }\frac{A\left( x\right) {dx}}{{x}^{s + 1}}
\]
in this half-plane.
Exercise 10.1.5 Show that \( \left( {s - 1}\right) \zeta \left( s\right) \) can be extended analytically for \( \operatorname{Re}\left( s\right) > \) 0.
Exercise 10.1.6 Evaluate
\[
\mathop{\lim }\limits_{{s \rightarrow 1}}\left( {s - 1}\right) \zeta \left( s\right)
\]
Example 10.1.7 Let \( K = \mathbb{Q}\left( i\right) \) . Show that \( \left( {s - 1}\right) {\zeta }_{K}\left( s\right) \) extends to an analytic function for \( \operatorname{Re}\left( s\right) > \frac{1}{2} \) .
Solution. Since every ideal \( \mathfrak{a} \) of \( {\mathcal{O}}_{K} \) is principal, we can write \( \mathfrak{a} = \left( {a + {ib}}\right) \) for some integers \( a, b \) . Moreover, since
\[
\mathfrak{a} = \left( {a + {ib}}\right) = \left( {-a - {ib}}\right) = \left( {-a + {ib}}\right) = \left( {a - {ib}}\right)
\]
we can choose \( a, b \) to be both positive. In this way, we can associate with each ideal \( \mathfrak{a} \) a unique lattice point \( \left( {a, b}\right), a \geq 0, b \geq 0 \) . Conversely, to each such lattice point \( \left( {a, b}\right) \) we can associate the ideal \( \mathfrak{a} = \left( {a + {ib}}\right) \) . Moreover, \( N\mathfrak{a} = {a}^{2} + {b}^{2} \) . Thus, if we write
\[
{\zeta }_{K}\left( s\right) = \mathop{\sum }\limits_{\mathfrak{a}}\frac{1}{N{\mathfrak{a}}^{s}} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{n}^{s}}
\]
we find that
\[
A\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{a}_{n}
\]
is equal to the number of lattice points lying in the positive quadrant defined by the circle \( {a}^{2} + {b}^{2} \leq x \) . We will call such a lattice point \( \left( {a, b}\right) \) internal if \( {\left( a + 1\right) }^{2} + {\left( b + 1\right) }^{2} \leq x \) . Otherwise, we will call it a boundary lattice point. Let \( I \) be the number of internal lattice points, and \( B \) the number of boundary lattice points. Then
\[
I \leq \frac{\pi }{4}x \leq I + B
\]
Any boundary point \( \left( {a, b}\right) \) is contained in the annulus
\[
{\left( \sqrt{x} - \sqrt{2}\right) }^{2} \leq {a}^{2} + {b}^{2} \leq {\left( \sqrt{x} + \sqrt{2}\right) }^{2}
\]
and an upper bound for \( B \) is provided by the area of the annulus. This is easily seen to be
\[
\pi {\left( \sqrt{x} + \sqrt{2}\right) }^{2} - \pi {\left( \sqrt{x} - \sqrt{2}\right) }^{2} = O\left( \sqrt{x}\right) .
\]
Thus \( A\left( x\right) = {\pi x}/4 + O\left( \sqrt{x}\right) \) . By Theorem 10.1.4, we deduce that
\[
{\zeta }_{K}\left( s\right) = \frac{\pi }{4}s{\int }_{1}^{\infty }\frac{dx}{{x}^{s}} + s{\int }_{1}^{\infty }\frac{E\left( x\right) }{{x}^{s + 1}}{dx}
\]
where \( E\left( x\right) = O\left( \sqrt{x}\right) \), so that the latter integral converges for \( \operatorname{Re}\left( s\right) > \frac{1}{2} \) .
Thus
\[
\left( {s - 1}\right) {\zeta }_{K}\left( s\right) = \frac{\pi }{4}s + s\left( {s - 1}\right) {\int }_{1}^{\infty }\frac{E\left( x\right) }{{x}^{s + 1}}{dx}
\]
is analytic for \( \operatorname{Re}\left( s\right) > \frac{1}{2} \) .
Exercise 10.1.8 For \( K = \mathbb{Q}\left( i\right) \), evaluate
\[
\mathop{\lim }\limits_{{s \rightarrow {1}^{ + }}}\left( {s - 1}\right) {\zeta }_{K}\left( s\right)
\]
Exercise 10.1.9 Show that the number of integers \( \left( {a, b}\right) \) with \( a > 0 \) satisfying \( {a}^{2} + D{b}^{2} \leq x \) is
\[
\frac{\pi x}{2\sqrt{D}} + O\left( \sqrt{x}\right)
\]
Exercise 10.1.10 Suppose \( K = \mathbb{Q}\left( \sqrt{-D}\right) \) where \( D > 0 \) and \( - D ≢ 1\left( {\;\operatorname{mod}\;4}\right) \) and \( {\mathcal{O}}_{K} \) has class number | (Theorem/Proposition/Example Problem/Lemma/Corollary Content...)
Exercise 9.2.10 Consider the element
\[
\alpha = {\left( x + y\right) }^{\ell - 2}\left( {x + {\zeta y}}\right)
\]
Show that:
(a) the ideal \( \left( \alpha \right) \) is a perfect \( \ell \) th power.
(b) \( \alpha \equiv 1 - {u\lambda }\left( {\;\operatorname{mod}\;{\lambda }^{2}}\right) \) where \( u = {\left( x + y\right) }^{\ell - 2}y \) . | null |
Proposition 3.2. In a chart \( U \times \mathbf{E} \) for \( {TX} \), let \( f : U \times \mathbf{E} \rightarrow \mathbf{E} \times \mathbf{E} \) represent \( F \), with \( f = \left( {{f}_{1},{f}_{2}}\right) \) . Then \( f \) represents a spray if and only if, for all \( s \in \mathbf{R} \) we have
\[
{f}_{2}\left( {x,{sv}}\right) = {s}^{2}{f}_{2}\left( {x, v}\right)
\]
Proof. The proof follows at once from the definitions and the formula giving the chart representation of \( s{\left( {s}_{TX}\right) }_{ * } \) .
Thus we see that the condition SPR 1 (in addition to being a second-order vector field), simply means that \( {f}_{2} \) is homogeneous of degree 2 in the variable \( v \) . By the remark in Chapter I,§3, it follows that \( {f}_{2} \) is a quadratic map in its second variable, and specifically, this quadratic map is given by
\[
{f}_{2}\left( {x, v}\right) = \frac{1}{2}{D}_{2}^{2}{f}_{2}\left( {x,0}\right) \left( {v, v}\right) .
\]
Thus the spray is induced by a symmetric bilinear map given at each point \( x \) in a chart by
(2)
\[
B\left( x\right) = \frac{1}{2}{D}_{2}^{2}{f}_{2}\left( {x,0}\right) .
\]
Conversely, suppose given a morphism
\[
U \rightarrow {L}_{\text{sym }}^{2}\left( {\mathbf{E},\mathbf{E}}\right) \;\text{ given by }\;x \mapsto B\left( x\right)
\]
from \( U \) into the space of symmetric bilinear maps \( \mathbf{E} \times \mathbf{E} \rightarrow \mathbf{E} \) . Thus for each \( v, w \in \mathbf{E} \) the value of \( B\left( x\right) \) at \( \left( {v, w}\right) \) is denoted by \( B\left( {x;v, w}\right) \) or \( B\left( x\right) \left( {v, w}\right) \) . Define \( {f}_{2}\left( {x, v}\right) = B\left( {x;v, v}\right) \) . Then \( {f}_{2} \) is quadratic in its second variable, and the map \( f \) defined by
\[
f\left( {x, v}\right) = \left( {v, B\left( {x;v, v}\right) }\right) = \left( {v,{f}_{2}\left( {x, v}\right) }\right)
\]
represents a spray over \( U \) . We call \( B \) the symmetric bilinear map associated with the spray. From the local representations in (1) and (2), we conclude that a curve \( \alpha \) is a geodesic if and only if \( \alpha \) satisfies the differential equation
(3)
\[
{\alpha }^{\prime \prime }\left( t\right) = {B}_{\alpha \left( t\right) }\left( {{\alpha }^{\prime }\left( t\right) ,{\alpha }^{\prime }\left( t\right) }\right) \;\text{ for all }t.
\]
We recall the trivial fact from linear algebra that the bilinear map \( B \) is determined purely algebraically from the quadratic map, by the formula
\[
B\left( {v, w}\right) = \frac{1}{2}\left\lbrack {{f}_{2}\left( {v + w}\right) - {f}_{2}\left( v\right) - {f}_{2}\left( w\right) }\right\rbrack .
\]
We have suppressed the \( x \) from the notation to focus on the relevant second variable \( v \) . Thus the quadratic map and the symmetric bilinear map determine each other uniquely.
The above discussion has been local, over an open set \( U \) in a Banach space. In Proposition 3.4 and the subsequent discussion of connections, we show how to globalize the bilinear map \( B \) intrinsically on the manifold.
Examples. As a trivial special case, we can always take \( {f}_{2}\left( {x, v}\right) = \left( {v,0}\right) \) to represent the second component of a spray in the chart.
In the chapter on Riemannian metrics, we shall see how to construct a spray in a natural fashion, depending on the metric.
In the chapter on covariant derivatives we show how a spray gives rise to such derivatives.
Next, let us give the transformation rule for a spray under a change of charts, i.e. an isomorphism \( h : U \rightarrow V \) . On \( {TU} \), the map \( {Th} \) is represented by a morphism (its vector component)
\[
H : U \times \mathbf{E} \rightarrow \mathbf{E} \times \mathbf{E}\;\text{ given by }\;H\left( {x, v}\right) = \left( {h\left( x\right) ,{h}^{\prime }\left( x\right) v}\right) .
\]
We then have one further lift to the double tangent bundle \( {TTU} \), and we may represent the diagram of maps symbolically as follows:
Then the derivative \( {H}^{\prime }\left( {x, v}\right) \) is given by the Jacobian matrix operating on column vectors \( {}^{t}\left( {u, w}\right) \) with \( u, w \in \mathbf{E} \), namely
\[
{H}^{\prime }\left( {x, v}\right) = \left( \begin{matrix} {h}^{\prime }\left( x\right) & 0 \\ {h}^{\prime \prime }\left( x\right) v & {h}^{\prime }\left( x\right) \end{matrix}\right) \text{ so }{H}^{\prime }\left( {x, v}\right) \left( \begin{matrix} u \\ w \end{matrix}\right) = \left( \begin{matrix} {h}^{\prime }\left( x\right) & 0 \\ {h}^{\prime \prime }\left( x\right) v & {h}^{\prime }\left( x\right) \end{matrix}\right) \left( \begin{matrix} u \\ w \end{matrix}\right) .
\]
Thus the top map on elements in the diagram is given by
\[
\left( {H,{H}^{\prime }}\right) : \left( {x, v, u, w}\right) \mapsto \left( {h\left( x\right) ,{h}^{\prime }\left( x\right) v,{h}^{\prime }\left( x\right) u,{h}^{\prime \prime }\left( x\right) \left( {u, v}\right) + {h}^{\prime }\left( x\right) w}\right) .
\]
For the application, we put \( u = v \) because \( {f}_{1}\left( {x, v}\right) = v \), and \( w = {f}_{U,2}\left( {x, v}\right) \) , where \( {f}_{U} \) and \( {f}_{V} \) denote the representations of the spray over \( U \) and \( V \) respectively. It follows that \( {f}_{U} \) and \( {f}_{V} \) are related by the formula
\[
{f}_{V}\left( {h\left( x\right) ,{h}^{\prime }\left( x\right) v}\right) = \left( {{h}^{\prime }\left( x\right) v,{h}^{\prime \prime }\left( x\right) \left( {v, v}\right) + {h}^{\prime }\left( x\right) {f}_{U,2}\left( {x, v}\right) }\right) .
\]
Therefore we obtain:
Proposition 3.3. Change of variable formula for the quadratic part of a spray:
\[
{f}_{V.2}\left( {h\left( x\right) ,{h}^{\prime }\left( x\right) v}\right) = {h}^{\prime \prime }\left( x\right) \left( {v, v}\right) + {h}^{\prime }\left( x\right) {f}_{U.2}\left( {x, v}\right) ,
\]
\[
{B}_{V}\left( {h\left( x\right) ;{h}^{\prime }\left( x\right) v,{h}^{\prime }\left( x\right) w}\right) = {h}^{\prime \prime }\left( x\right) \left( {v, w}\right) + {h}^{\prime }\left( x\right) {B}_{U}\left( {x;v, w}\right) .
\]
Proposition 3.3 admits a converse:
Proposition 3.4. Suppose we are given a covering of the manifold \( X \) by open sets corresponding to charts \( U, V,\ldots \), and for each \( U \) we are given
a morphism
\[
{B}_{U} : U \rightarrow {L}_{\mathrm{{sym}}}^{2}\left( {\mathbf{E},\mathbf{E}}\right)
\]
which transforms according to the formula of Proposition 3.3 under an isomorphism \( h : U \rightarrow V \) . Then there exists a unique spray whose associated bilinear map in the chart \( U \) is given by \( {B}_{U} \) .
Proof. We leave the verification to the reader.
Remarks. Note that \( {B}_{U}\left( {x;v, w}\right) \) does not transform like a tensor of type \( {L}_{\mathrm{{sym}}}^{2}\left( {\mathbf{E},\mathbf{E}}\right) \), i.e. a section of the bundle \( {L}_{\mathrm{{sym}}}^{2}\left( {{TX},{TX}}\right) \) . There are several ways of defining the bilinear map \( B \) intrinsically. One of them is via second order bundles, or bundles of second order jets, and to extend the terminology we have established previously to such bundles, and even higher order jet bundles involving higher derivatives, as in [Po 62]. Another way will be done below, via connections. For our immediate purposes, it suffices to have the above discussion on second-order differential equations together with Proposition 3.3 and 3.4. Sprays were introduced by Ambrose, Palais, and Singer [APS 60], and I used them (as recommended by Palais) in the earliest version [La 62]. In [Lo 69] the bilinear map \( {B}_{U} \) is expressed in terms of second order jets. The basics of differential topology and geometry were being established in the early sixties. Cf. the bibliographical notes from [Lo 69] at the end of his first chapter.
## Connections
We now show how to define the bilinear map \( B \) intrinsically and directly.
Matters will be clearer if we start with an arbitrary vector bundle
\[
p : E \rightarrow X
\]
over a manifold \( X \) . As it happens we also need the notion of a fiber bundle when the fibers are not necessarily vector spaces, so don't have a linear structure. Let \( f : Y \rightarrow X \) be a morphism. We say that \( f \) (or \( Y \) over \( X \) ) is a fiber bundle if \( f \) is surjective, and if each point \( x \) of \( X \) has an open neighborhood \( U \), and there is some manifold \( Z \) and an isomorphism \( h : {f}^{-1}\left( U\right) \rightarrow U \times Z \) such that the following diagram is commutative:
Thus locally, \( f : Y \rightarrow X \) looks like the projection from a product space. The reason why we need a fiber bundle is that the tangent bundle
\[
{\pi }_{E} : {TE} \rightarrow E
\]
is a vector bundle over \( E \), but the composite \( f = p \circ {\pi }_{E} : {TE} \rightarrow X \) is only a fiber bundle over \( X \), a fact which is obvious by picking trivializations in
charts. Indeed, if \( U \) is a chart in \( X \), and if \( U \times \mathbf{F} \rightarrow U \) is a vector bundle chart for \( E \), with fiber \( \mathbf{F} \), and \( Y = {TE} \), then we have a natural isomorphism of fiber bundles over \( U \) :
Note that \( U \) being a chart in \( X \) implies that \( U \times \mathbf{E} \rightarrow U \) is a vector bundle chart for the tangent bundle \( {TU} \) over \( U \) .
The tangent bundle \( {TE} \) has two natural maps making it a vector bundle:
\[
{\pi }_{E} : {TE} \rightarrow E\text{is a vector bundle over}E\text{;}
\]
\[
T\left( p\right) : {TE} \rightarrow {TX}\text{is a vector bundle over}{TX}\text{.}
\]
Therefore we have a natural morphism of fiber bundle (n | Proposition 3.2. In a chart \( U \times \mathbf{E} \) for \( {TX} \), let \( f : U \times \mathbf{E} \rightarrow \mathbf{E} \times \mathbf{E} \) represent \( F \), with \( f = \left( {{f}_{1},{f}_{2}}\right) \) . Then \( f \) represents a spray if and only if, for all \( s \in \mathbf{R} \) we have
\[
{f}_{2}\left( {x,{sv}}\right) = {s}^{2}{f}_{2}\left( {x, v}\right)
\] | The proof follows at once from the definitions and the formula giving the chart representation of \( s{\left( {s}_{TX}\right) }_{ * } \) .
Thus we see that the condition SPR 1 (in addition to being a second-order vector field), simply means that \( {f}_{2} \) is homogeneous of degree 2 in the variable \( v \) . By the remark in Chapter I,§3, it follows that \( {f}_{2} \) is a quadratic map in its second variable, and specifically, this quadratic map is given by
\[
{f}_{2}\left( {x, v}\right) = \frac{1}{2}{D}_{2}^{2}{f}_{2}\left( {x,0}\right) \left( {v, v}\right) .
\]
Thus the spray is induced by a symmetric bilinear map given at each point \( x \) in a chart by
\[
B\left( x\right) = \frac{1}{2}{D}_{2}^{2}{f}_{2}\left( {x,0}\right) .
\]
Conversely, suppose given a morphism
\[
U \rightarrow {L}_{\text{sym }}^{2}\left( {\mathbf{E},\mathbf{E}}\right) \;\text{ given by }\;x \mapsto B\left( x\right)
\]
from \( U \) into the space of symmetric bilinear maps \( \mathbf{E} \times \mathbf{E} \rightarrow \mathbf{E} \) . Thus for each \( v, w \in \mathbf{E} \) the value of \( B\left( x\right) \) at \( \left( {v, w}\right) \) is denoted by \( B\left( {x;v, w}\right) \) or \( B\left( x\right) \left( {v, w}\right) \) . Define \( {f}_{2}\left( {x, v}\right) = B\left( {x;v, v}\right) \) . Then \( {f}_{2} \) is quadratic in its second variable, and the map \( f \) defined by
\[
f\left( {x, v}\right) = \left( {v, B\left( {x;v, v}\right) }\right) = \left( {v,{f}_{2}\left( {x, v}\right) }\right)
\]
represents a spray over \( U \) . We call \( B \) the symmetric bilinear map associated with the spray. From the local representations in (1) and (2), we conclude that a curve \( \alpha \) is a geodesic if and only if \( \alpha \) satisfies the differential equation
\[
{\alpha }^{\prime \prime }\left( t\right) = {B}_{\alpha \left( t\right) }\left( {{\alpha }^{\prime }\left( t\right) ,{\alpha }^{\prime }\left( t\right) }\right) \;\text{ for all }t.
\] |
Theorem 2.2.8. If \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| = 1 \), a.e., then \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = {\phi }_{2}{\widetilde{\mathbf{H}}}^{2} \) if and only if there is a constant \( c \) of modulus 1 such that \( {\phi }_{1} = c{\phi }_{2} \) .
Proof. Clearly \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = c{\phi }_{1}{\widetilde{\mathbf{H}}}^{2} \) when \( \left| c\right| = 1 \) . Conversely, suppose that \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = {\phi }_{2}{\widetilde{\mathbf{H}}}^{2} \) with \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| = 1 \), a.e. Then there exist functions \( {f}_{1} \) and \( {f}_{2} \) in \( {\widetilde{\mathbf{H}}}^{2} \) such that
\[
{\phi }_{1} = {\phi }_{2}{f}_{2}\;\text{ and }\;{\phi }_{2} = {\phi }_{1}{f}_{1}
\]
Since \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = 1 = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| \) a.e., it follows that
\[
{\phi }_{1}\overline{{\phi }_{2}} = {f}_{2}\;\text{ and }\;{\phi }_{2}\overline{{\phi }_{1}} = {f}_{1}
\]
i.e., \( {f}_{1} = \overline{{f}_{2}} \) . But since \( {f}_{1} \) and \( {f}_{2} \) are in \( {\widetilde{\mathbf{H}}}^{2},{f}_{1} = \overline{{f}_{2}} \) implies that \( {f}_{1} \) has Fourier coefficients equal to 0 for all positive and for all negative indices. Since the only nonzero coefficient is in the zeroth place, \( {f}_{1} \) and \( {f}_{2} \) are constants, obviously having moduli equal to 1 .
Since the unilateral shift is a restriction of the bilateral shift to an invariant subspace, invariant subspaces of the unilateral shift are determined by Theorem 2.2.7: they are the invariant subspaces of the bilateral shift that are contained in \( {\widetilde{\mathbf{H}}}^{2} \) . In this case, the functions generating the invariant subspaces are certain analytic functions whose structure is important.
Definition 2.2.9. A function \( \phi \in {\mathbf{H}}^{\infty } \) satisfying \( \left| {\widetilde{\phi }\left( {e}^{i\theta }\right) }\right| = 1 \) a.e. is an inner function.
Theorem 2.2.10. If \( \phi \) is a nonconstant inner function, then \( \left| {\phi \left( z\right) }\right| < 1 \) for all \( z \in \mathbb{D} \) .
Proof. This follows immediately from Corollary 1.1.24 and Theorem 1.1.17.
The definition of inner functions requires that the functions be in \( {\mathbf{H}}^{\infty } \) . It is often useful to know that this follows if a function is in \( {\mathbf{H}}^{2} \) and has boundary values of modulus 1 a.e.
Theorem 2.2.11. Let \( \phi \in {\mathbf{H}}^{2} \) . If \( \left| {\widetilde{\phi }\left( {e}^{i\theta }\right) }\right| = 1 \) a.e., then \( \phi \) is an inner function.
Proof. It only needs to be shown that \( \phi \in {\mathbf{H}}^{\infty } \) ; this follows from Corollary 1.1.24.
Corollary 2.2.12 (Beurling's Theorem). Every invariant subspace of the unilateral shift other than \( \{ 0\} \) has the form \( \phi {\mathbf{H}}^{2} \), where \( \phi \) is an inner function.
Proof. The unilateral shift is the restriction of multiplication by \( {e}^{i\theta } \) to \( {\widetilde{\mathbf{H}}}^{2} \), so if \( \mathcal{M} \) is an invariant subspace of the unilateral shift, it is an invariant subspace of the bilateral shift contained in \( {\widetilde{\mathbf{H}}}^{2} \) . Thus, by Theorem 2.2.7, \( \mathcal{M} = \phi {\widetilde{\mathbf{H}}}^{2} \) for some measurable function satisfying \( \left| {\phi \left( {e}^{i\theta }\right) }\right| = 1 \) a.e. (Note that \( \{ 0\} \) is the only reducing subspace of the bilateral shift that is contained in \( {\widetilde{\mathbf{H}}}^{2} \) .) Since \( 1 \in {\widetilde{\mathbf{H}}}^{2},\phi \in {\widetilde{\mathbf{H}}}^{2} \) .
Translating this situation back to \( {\mathbf{H}}^{2} \) on the disk gives \( \mathcal{M} = \phi {\mathbf{H}}^{2} \) with \( \phi \) inner, by Theorem 2.2.11.
Corollary 2.2.13. Every invariant subspace of the unilateral shift is cyclic. (See Definition 1.2.17.)
Proof. If \( \mathcal{M} \) is an invariant subspace of the unilateral shift, it has the form \( \phi {\mathbf{H}}^{2} \) by Beurling’s theorem (Corollary 2.2.12). For each \( n,{U}^{n}\phi = {z}^{n}\phi \), so \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}\phi }\right\} \) contains all functions of the form \( \phi \left( z\right) p\left( z\right) \), where \( p \) is a polynomial. Since the polynomials are dense in \( {\mathbf{H}}^{2} \) (as the finite sequences are dense in \( \left. {\ell }^{2}\right) \), it follows that \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}\phi }\right\} = \phi {\mathbf{H}}^{2} \) .
## 2.3 Inner and Outer Functions
We shall see that every function in \( {\mathbf{H}}^{2} \), other than the constant function 0, can be written as a product of an inner function and a cyclic vector for the unilateral shift. Such cyclic vectors will be shown to have a special form.
Definition 2.3.1. The function \( F \in {\mathbf{H}}^{2} \) is an outer function if \( F \) is a cyclic vector for the unilateral shift. That is, \( F \) is an outer function if
\[
\mathop{\bigvee }\limits_{{k = 0}}^{\infty }\left\{ {{U}^{k}F}\right\} = {\mathbf{H}}^{2}
\]
Theorem 2.3.2. If \( F \) is an outer function, then \( F \) has no zeros in \( \mathbb{D} \) .
Proof. If \( F\left( {z}_{0}\right) = 0 \), then \( \left( {{U}^{n}F}\right) \left( {z}_{0}\right) = {z}_{0}^{n}F\left( {z}_{0}\right) = 0 \) for all \( n \) . Since the limit of a sequence of functions in \( {\mathbf{H}}^{2} \) that all vanish at \( {z}_{0} \) must also vanish at \( {z}_{0} \) (Theorem 1.1.9),
\[
\mathop{\bigvee }\limits_{{k = 0}}^{\infty }\left\{ {{U}^{k}F}\right\}
\]
cannot be all of \( {\mathbf{H}}^{2} \) . Hence there is no \( {z}_{0} \in \mathbb{D} \) with \( F\left( {z}_{0}\right) = 0 \) .
Recall that a function analytic on \( \mathbb{D} \) is identically zero if it vanishes on a set that has a limit point in \( \mathbb{D} \) . The next theorem is an analogous result for boundary values of functions in \( {\mathbf{H}}^{2} \) .
Theorem 2.3.3 (The F. and M. Riesz Theorem). If \( f \in {\mathbf{H}}^{2} \) and the set
\[
\left\{ {{e}^{i\theta } : \widetilde{f}\left( {e}^{i\theta }\right) = 0}\right\}
\]
has positive measure, then \( f \) is identically 0 on \( \mathbb{D} \) .
Proof. Let \( E = \left\{ {{e}^{i\theta } : \widetilde{f}\left( {e}^{i\theta }\right) = 0}\right\} \) and let
\[
\mathcal{M} = \mathop{\bigvee }\limits_{{k = 0}}^{\infty }\left\{ {{U}^{k}\widetilde{f}}\right\} = \mathop{\bigvee }\limits_{{k = 0}}^{\infty }\left\{ {{e}^{ik\theta }\widetilde{f}}\right\}
\]
Then every function \( \widetilde{g} \in \mathcal{M} \) vanishes on \( E \), since all functions \( {e}^{ik\theta }\widetilde{f} \) do. If \( \widetilde{f} \) is not identically zero, it follows from Beurling's theorem (Theorem 2.2.12) that \( \mathcal{M} = \widetilde{\phi }{\widetilde{\mathbf{H}}}^{2} \) for some inner function \( \phi \) . In particular, this implies that \( \widetilde{\phi } \in \mathcal{M} \) , so \( \widetilde{\phi } \) vanishes on \( E \) . But \( \left| {\widetilde{\phi }\left( {e}^{i\theta }\right) }\right| = 1 \) a.e. This contradicts the hypothesis that \( E \) has positive measure, thus \( \widetilde{f} \), and hence \( f \), must be identically zero.
Another beautiful result that follows from Beurling's theorem is the following factorization of functions in \( {\mathbf{H}}^{2} \) .
Theorem 2.3.4. If \( f \) is a function in \( {\mathbf{H}}^{2} \) that is not identically zero, then \( f = {\phi F} \), where \( \phi \) is an inner function and \( F \) is an outer function. This factorization is unique up to constant factors.
Proof. Let \( f \in {\mathbf{H}}^{2} \) and consider \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}f}\right\} \) . If this span is \( {\mathbf{H}}^{2} \), then \( f \) is outer by definition, and we can take \( \phi \) to be the constant function 1 and \( F = f \) to obtain the desired conclusion.
If \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}f}\right\} \neq {\mathbf{H}}^{2} \), then, by Beurling’s theorem (Corollary 2.2.12), there must exist a nonconstant inner function \( \phi \) with \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}f}\right\} = \phi {\mathbf{H}}^{2} \) . Since \( f \) is in \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}f}\right\} = \phi {\mathbf{H}}^{2} \), there exists a function \( F \) in \( {\mathbf{H}}^{2} \) with \( f = {\phi F} \) . We shall show that \( F \) is outer.
The invariant subspace \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}F}\right\} \) equals \( \psi {\mathbf{H}}^{2} \) for some inner function \( \psi \) . Then, since \( f = {\phi F} \), it follows that \( {U}^{n}f = {U}^{n}\left( {\phi F}\right) = \phi {U}^{n}F \) for every positive integer \( n \), from which we can conclude, by taking linear spans, that \( \phi {\mathbf{H}}^{2} = {\phi \psi }{\mathbf{H}}^{2} \) . Theorem 2.2.8 now implies that \( \phi \) and \( {\phi \psi } \) are constant multiples of each other. Hence \( \psi \) must be a constant function. Therefore \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}F}\right\} = {\mathbf{H}}^{2} \), so \( F \) is an outer function.
Note that if \( f = {\phi F} \) with \( \phi \) inner and \( F \) outer, then \( \mathop{\bigvee }\limits_{{n = 0}}^{\infty }\left\{ {{U}^{n}f}\right\} = \phi {\mathbf{H}}^{2} \) . Thus uniqueness of the factorization follows from the corresponding assertion in Theorem 2.2.8.
Definition 2.3.5. For \( f \in {\mathbf{H}}^{2} \), if \( f = {\phi F} \) with \( \phi \) inner and \( F \) outer, we call \( \phi \) the inner part of \( f \) and \( F \) the outer part of \( f \) .
Theorem 2.3.6. The zeros of an \( {\mathbf{H}}^{2} \) function are precisely the zeros of its inn | Theorem 2.2.8. If \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| = 1 \), a.e., then \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = {\phi }_{2}{\widetilde{\mathbf{H}}}^{2} \) if and only if there is a constant \( c \) of modulus 1 such that \( {\phi }_{1} = c{\phi }_{2} \). | Proof. Clearly \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = c{\phi }_{1}{\widetilde{\mathbf{H}}}^{2} \) when \( \left| c\right| = 1 \). Conversely, suppose that \( {\phi }_{1}{\widetilde{\mathbf{H}}}^{2} = {\phi }_{2}{\widetilde{\mathbf{H}}}^{2} \) with \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| = 1 \), a.e. Then there exist functions \( {f}_{1} \) and \( {f}_{2} \) in \( {\widetilde{\mathbf{H}}}^{2} \) such that
\[
{\phi }_{1} = {\phi }_{2}{f}_{2}\;\text{ and }\;{\phi }_{2} = {\phi }_{1}{f}_{1}
\]
Since \( \left| {{\phi }_{1}\left( {e}^{i\theta }\right) }\right| = 1 = \left| {{\phi }_{2}\left( {e}^{i\theta }\right) }\right| \) a.e., it follows that
\[
{\phi }_{1}\overline{{\phi }_{2}} = {f}_{2}\;\text{ and }\;{\phi }_{2}\overline{{\phi }_{1}} = {f}_{1}
\]
i.e., \( {f}_{1} = \overline{{f}_{2}} \). But since \( {f}_{1} \) and \( {f}_{2} \) are in \( {\widetilde{\mathbf{H}}}^{2},{f}_{1} = \overline{{f}_{2}} \) implies that \( {f}_{1} \) has Fourier coefficients equal to 0 for all positive and for all negative indices. Since the only nonzero coefficient is in the zeroth place, \( {f}_{1} \) and \( {f}_{2} \) are constants, obviously having moduli equal to 1. |
Theorem 3.1. (Eisenstein’s Criterion). Let \( A \) be a factorial ring. Let \( K \) be its quotient field. Let \( f\left( X\right) = {a}_{n}{X}^{n} + \cdots + {a}_{0} \) be a polynomial of degree \( n \geqq 1 \) in \( A\left\lbrack X\right\rbrack \) . Let \( p \) be a prime of \( A \), and assume:
\[
{a}_{n} ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) ,\;{a}_{i} \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) \;\text{ for all }\;i < n,
\]
\[
{a}_{0} ≢ 0\;\left( {\;\operatorname{mod}\;{p}^{2}}\right)
\]
Then \( f\left( X\right) \) is irreducible in \( K\left\lbrack X\right\rbrack \) .
Proof. Extracting a g.c.d. for the coefficients of \( f \), we may assume without loss of generality that the content of \( f \) is 1 . If there exists a factorization into factors of degree \( \geqq 1 \) in \( K\left\lbrack X\right\rbrack \), then by the corollary of Gauss’ lemma there exists a factorization in \( A\left\lbrack X\right\rbrack \), say \( f\left( X\right) = g\left( X\right) h\left( X\right) \) ,
\[
g\left( X\right) = {b}_{d}{X}^{d} + \cdots + {b}_{0}
\]
\[
h\left( X\right) = {c}_{m}{X}^{m} + \cdots + {c}_{0}
\]
with \( d, m \geqq 1 \) and \( {b}_{d}{c}_{m} \neq 0 \) . Since \( {b}_{0}{c}_{0} = {a}_{0} \) is divisible by \( p \) but not \( {p}^{2} \), it follows that one of \( {b}_{0},{c}_{0} \) is not divisible by \( p \), say \( {b}_{0} \) . Then \( p \mid {c}_{0} \) . Since \( {c}_{m}{b}_{d} = {a}_{n} \) is not divisible by \( p \), it follows that \( p \) does not divide \( {c}_{m} \) . Let \( {c}_{r} \) be the coefficient of \( h \) furthest to the right such that \( {c}_{r} ≢ 0\left( {\;\operatorname{mod}\;p}\right) \) . Then
\[
{a}_{r} = {b}_{0}{c}_{r} + {b}_{1}{c}_{r - 1} + \cdots .
\]
Since \( p \nmid {b}_{0}{c}_{r} \) but \( p \) divides every other term in this sum, we conclude that \( p \nmid {a}_{r} \), a contradiction which proves our theorem.
Example. Let \( a \) be a non-zero square-free integer \( \neq \pm 1 \) . Then for any integer \( n \geqq 1 \), the polynomial \( {X}^{n} - a \) is irreducible over \( \mathbf{Q} \) . The polynomials \( 3{X}^{5} - {15} \) and \( 2{X}^{10} - {21} \) are irreducible over \( \mathbf{Q} \) .
There are some cases in which a polynomial does not satisfy Eisenstein's criterion, but a simple transform of it does.
Example. Let \( p \) be a prime number. Then the polynomial
\[
f\left( X\right) = {X}^{p - 1} + \cdots + 1
\]
is irreducible over \( \mathbf{Q} \) .
Proof. It will suffice to prove that the polynomial \( f\left( {X + 1}\right) \) is irreducible over \( \mathbf{Q} \) . We note that the binomial coefficients
\[
\left( \begin{array}{l} p \\ v \end{array}\right) = \frac{p!}{v!\left( {p - v}\right) !},\;1 \leqq v \leqq p - 1,
\]
are divisible by \( p \) (because the numerator is divisible by \( p \) and the denominator is not, and the coefficient is an integer). We have
\[
f\left( {X + 1}\right) = \frac{{\left( X + 1\right) }^{p} - 1}{\left( {X + 1}\right) - 1} = \frac{{X}^{p} + p{X}^{p - 1} + \cdots + {pX}}{X}
\]
from which one sees that \( f\left( {X + 1}\right) \) satisfies Eisenstein’s criterion.
Example. Let \( E \) be a field and \( t \) an element of some field containing \( E \) such that \( t \) is transcendental over \( E \) . Let \( K \) be the quotient field of \( E\left\lbrack t\right\rbrack \) .
For any integer \( n \geqq 1 \) the polynomial \( {X}^{n} - t \) is irreducible in \( K\left\lbrack X\right\rbrack \) . This comes from the fact that the ring \( A = E\left\lbrack t\right\rbrack \) is factorial and that \( t \) is a prime in it.
Theorem 3.2. (Reduction Criterion). Let \( A, B \) be entire rings, and let
\[
\varphi : A \rightarrow B
\]
be a homomorphism. Let \( K, L \) be the quotient fields of \( A \) and \( B \) respectively. Let \( f \in A\left\lbrack X\right\rbrack \) be such that \( {\varphi f} \neq 0 \) and \( \deg {\varphi f} = \deg f \) . If \( {\varphi f} \) is irreducible in \( L\left\lbrack X\right\rbrack \), then \( f \) does not have a factorization \( f\left( X\right) = g\left( X\right) h\left( X\right) \) with
\[
g, h \in A\left\lbrack X\right\rbrack \;\text{ and }\;\deg g,\deg h \geqq 1.
\]
Proof. Suppose \( f \) has such a factorization. Then \( {\varphi f} = \left( {\varphi g}\right) \left( {\varphi h}\right) \) . Since \( \deg {\varphi g} \leqq \deg g \) and \( \deg {\varphi h} \leqq \deg h \), our hypothesis implies that we must have equality in these degree relations. Hence from the irreducibility in \( L\left\lbrack X\right\rbrack \) we conclude that \( g \) or \( h \) is an element of \( A \), as desired.
In the preceding criterion, suppose that \( A \) is a local ring, i.e. a ring having a unique maximal ideal \( \mathfrak{p} \), and that \( \mathfrak{p} \) is the kernel of \( \varphi \) . Then from the irreducibility of \( {\varphi f} \) in \( L\left\lbrack X\right\rbrack \) we conclude the irreducibility of \( f \) in \( A\left\lbrack X\right\rbrack \) . Indeed, any element of \( A \) which does not lie in \( \mathfrak{p} \) must be a unit in \( A \), so our last conclusion in the proof can be strengthened to the statement that \( g \) or \( h \) is a unit in \( A \) .
One can also apply the criterion when \( A \) is factorial, and in that case deduce the irreducibility of \( f \) in \( K\left\lbrack X\right\rbrack \) .
Example. Let \( p \) be a prime number. It will be shown later that \( {X}^{p} - X - 1 \) is irreducible over the field \( \mathbf{Z}/p\mathbf{Z} \) . Hence \( {X}^{p} - X - 1 \) is irreducible over \( \mathbf{Q} \) . Similarly,
\[
{X}^{5} - 5{X}^{4} - {6X} - 1
\]
is irreducible over \( \mathbf{Q} \) .
There is also a routine elementary school test whether a polynomial has a root or not.
Proposition 3.3. (Integral Root Test). Let \( A \) be a factorial ring and \( K \) its quotient field. Let
\[
f\left( X\right) = {a}_{n}{X}^{n} + \cdots + {a}_{0} \in A\left\lbrack X\right\rbrack .
\]
Let \( \alpha \in K \) be a root of \( f \), with \( \alpha = b/d \) expressed with \( b, d \in A \) and \( b, d \) relatively prime. Then \( b \mid {a}_{0} \) and \( d \mid {a}_{n} \) . In particular, if the leading coefficient \( {a}_{n} \) is 1, then a root \( \alpha \) must lie in \( A \) and divides \( {a}_{0} \) . We leave the proof to the reader, who should be used to this one from way back. As an irreducibility test, the test is useful especially for a polynomial of degree 2 or 3 , when reducibility is equivalent with the existence of a root in the given field.
## §4. HILBERT'S THEOREM
This section proves a basic theorem of Hilbert concerning the ideals of a polynomial ring. We define a commutative ring \( A \) to be Noetherian if every ideal is finitely generated.
Theorem 4.1. Let \( A \) be a commutative Noetherian ring. Then the polynomial ring \( A\left\lbrack X\right\rbrack \) is also Noetherian.
Proof. Let \( \mathfrak{A} \) be an ideal of \( A\left\lbrack X\right\rbrack \) . Let \( {\mathfrak{a}}_{i} \) consist of 0 and the set of elements \( a \in A \) appearing as leading coefficient in some polynomial
\[
{a}_{0} + {a}_{1}X + \cdots + a{X}^{i}
\]
lying in \( \mathfrak{A} \) . Then it is clear that \( {\mathfrak{a}}_{i} \) is an ideal. (If \( a, b \) are in \( {\mathfrak{a}}_{i} \), then \( a \pm b \) is in \( {a}_{i} \) as one sees by taking the sum and difference of the corresponding polynomials. If \( x \in A \), then \( {xa} \in {\mathfrak{a}}_{i} \) as one sees by multiplying the corresponding polynomial by \( x \) .) Furthermore we have
\[
{\mathfrak{a}}_{0} \subset {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \subset \cdots ,
\]
in other words, our sequence of ideals \( \left\{ {a}_{i}\right\} \) is increasing. Indeed, to see this multiply the above polynomial by \( X \) to see that \( a \in {\mathfrak{a}}_{i + 1} \) .
By criterion (2) of Chapter X, \( §1 \), the sequence of ideals \( \left\{ {a}_{i}\right\} \) stops, say at \( {a}_{r} \) :
\[
{\mathfrak{a}}_{0} \subset {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \subset \cdots \subset {\mathfrak{a}}_{r} = {\mathfrak{a}}_{r + 1} = \cdots .
\]
Let
\[
{a}_{01},\ldots ,{a}_{0{n}_{0}}\text{be generators for}{\mathfrak{a}}_{0}\text{,}
\]
..............................
\( {a}_{r1},\ldots ,{a}_{r{n}_{r}} \) be generators for \( {\mathfrak{a}}_{r} \) .
For each \( i = 0,\ldots, r \) and \( j = 1,\ldots ,{n}_{i} \) let \( {f}_{ij} \) be a polynomial in \( \mathfrak{A} \), of degree \( i \), with leading coefficient \( {a}_{ij} \) . We contend that the polynomials \( {f}_{ij} \) are a set of generators for \( \mathfrak{A} \) .
Let \( f \) be a polynomial of degree \( d \) in \( \mathfrak{A} \) . We shall prove that \( f \) is in the ideal generated by the \( {f}_{ij} \), by induction on \( d \) . Say \( d \geqq 0 \) . If \( d > r \), then we note that the leading coefficients of
\[
{X}^{d - r}{f}_{r1},\ldots ,{X}^{d - r}{f}_{r{n}_{r}}
\]
generate \( {\mathfrak{a}}_{d} \) . Hence there exist elements \( {c}_{1},\ldots ,{c}_{{n}_{r}} \in A \) such that the polynomial
\[
f - {c}_{1}{X}^{d - r}{f}_{r1} - \cdots - {c}_{{n}_{r}}{X}^{d - r}{f}_{r{n}_{r}}
\]
has degree \( < d \), and this polynomial also lies in \( \mathfrak{A} \) . If \( d \leqq r \), we can subtract a linear combination
\[
f - {c}_{1}{f}_{d1} - \cdots - {c}_{{n}_{d}}{f}_{d{n}_{d}}
\]
to get a polynomial of degree \( < d \), also lying in \( \mathfrak{A} \) . We note that the polynomial we have subtracted from \( f \) lies in the ideal generated by the \( {f}_{ij} \) . By induction, we can subtract a polynomial \( g \) in the ideal generated by the \( {f}_{ij} \) such that \( f - g = 0 \), thereby proving our theorem.
We note that if \( \varphi : A \rightarrow B \) is a surjective homomorphism of commutative rings and \( A \) is Noetherian, so is \( B \) . Indeed, let \( \mathfrak{b} \) be an ideal of \( B \), so \( {\varphi }^{-1}\left( \mathfrak{b}\right) \) is an ideal of \( A \) . Then there is a finite number of generators \( \left( | (Theorem 3.1. (Eisenstein’s Criterion). Let \( A \) be a factorial ring. Let \( K \) be its quotient field. Let \( f\left( X\right) = {a}_{n}{X}^{n} + \cdots + {a}_{0} \) be a polynomial of degree \( n \geqq 1 \) in \( A\left\lbrack X\right\rbrack \) . Let \( p \) be a prime of \( A \), and assume:
\[
{a}_{n} ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) ,\;{a}_{i} \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) \;\text{ for all }\;i < n,
\]
\[
{a}_{0} ≢ 0\;\left( {\;\operatorname{mod}\;{p}^{2}}\right)
\]
Then \( f\left( X\right) \) is irreducible in \( K\left\lbrack X\right\rbrack \) .) | (Proof. Extracting a g.c.d. for the coefficients of \( f \), we may assume without loss of generality that the content of \( f \) is 1 . If there exists a factorization into factors of degree \( \geqq 1 \) in \( K\left\lbrack X\right\rbrack \), then by the corollary of Gauss’ lemma there exists a factorization in \( A\left\lbrack X\right\rbrack \), say \( f\left( X\right) = g\left( X\right) h\left( X\right) \) ,
\[
g\left( X\right) = {b}_{d}{X}^{d} + \cdots + {b}_{0}
\]
\[
h\left( X\right) = {c}_{m}{X}^{m} + \cdots + {c}_{0}
\]
with \( d, m \geqq 1 \) and \( {b}_{d}{c}_{m} \neq 0 \) . Since \( {b}_{0}{c}_{0} = {a}_{0} \) is divisible by \( p \) but not \( {p}^{2} \), it follows that one of \( {b}_{0},{c}_{0} \) is not divisible by \( p \), say \( {b}_{0} \) . Then \( p \mid {c}_{0} \) . Since \( {c}_{m}{b}_{d} = {a}_{n} \) is not divisible by \( p \), it follows that \( p \) does not divide \( {c}_{m} \) . Let \( {c}_{r} \) be the coefficient of \( h \) furthest to the right such that \( {c}_{r} ≢ 0\left( {\;\operatorname{mod}\;p}\right) \) . Then
\[
{a}_{r} = {b}_{0}{c}_{r} + {b}_{1}{c}_{r - 1} + \cdots .
\]
Since \( p \nmid {b}_{0}{c}_{r} \) but \( p \) divides every other term in this sum, we conclude that \( p \nmid {a}_{r} \), a contradiction which proves our theorem.) |
Proposition 5.46. Suppose \( M \) is a smooth manifold without boundary and \( D \subseteq M \) is a regular domain. The topological interior and boundary of \( D \) are equal to its manifold interior and boundary, respectively.
Proof. Suppose \( p \in D \) is arbitrary. If \( p \) is in the manifold boundary of \( D \), Theorem 4.15 shows that there exist a smooth boundary chart \( \left( {U,\varphi }\right) \) for \( D \) centered at \( p \) and a smooth chart \( \left( {V,\psi }\right) \) for \( M \) centered at \( p \) in which \( F \) has the coordinate representation \( F\left( {{x}^{1},\ldots ,{x}^{n}}\right) = \left( {{x}^{1},\ldots ,{x}^{n}}\right) \), where \( n = \dim M = \dim D \) . Since \( D \) has the subspace topology, \( U = D \cap W \) for some open subset \( W \subseteq M \), so \( {V}_{0} = V \cap W \) is a neighborhood of \( p \) in \( M \) such that \( {V}_{0} \cap D \) consists of all the points in \( {V}_{0} \) whose \( {x}^{m} \) coordinate is nonnegative. Thus every neighborhood of \( p \) intersects both \( D \) and \( M \smallsetminus D \), so \( p \) is in the topological boundary of \( D \) .
On the other hand, suppose \( p \) is in the manifold interior of \( D \) . The manifold interior is a smooth embedded codimension- 0 submanifold without boundary in \( M \) , so it is an open subset by Proposition 5.1. Thus \( p \) is in the topological interior of \( D \) .
Conversely, if \( p \) is in the topological interior of \( D \), then it is not in the topological boundary, so the preceding argument shows that it is not in the manifold boundary and hence must be in the manifold interior. Similarly, if \( p \) is in the topological boundary, it is also in the manifold boundary.
Here are some ways in which regular domains often arise.
Proposition 5.47. Suppose \( M \) is a smooth manifold and \( f \in {C}^{\infty }\left( M\right) \) .
(a) For each regular value \( b \) of \( f \), the sublevel set \( {f}^{-1}(\left( {-\infty, b\rbrack }\right) \) is a regular domain in \( M \) .
(b) If \( a \) and \( b \) are two regular values of \( f \) with \( a < b \), then \( {f}^{-1}\left( \left\lbrack {a, b}\right\rbrack \right) \) is a regular domain in \( M \) .
Proof. Problem 5-21.
A set of the form \( {f}^{-1}(\left( {-\infty, b\rbrack }\right) \) for \( b \) a regular value of \( f \) is called a regular sublevel set of \( f \) . Part (a) of the preceding theorem shows that every regular sublevel set of a smooth real-valued function is a regular domain. If \( D \subseteq M \) is a regular domain and \( f \in {C}^{\infty }\left( M\right) \) is a smooth function such that \( D \) is a regular sublevel set of \( f \), then \( f \) is called a defining function for \( \mathbf{D} \) .
Theorem 5.48. If \( M \) is a smooth manifold and \( D \subseteq M \) is a regular domain, then there exists a defining function for \( D \) . If \( D \) is compact, then \( f \) can be taken to be a smooth exhaustion function for \( M \) .
Proof. Problem 5-22.
Many (though not all) of the earlier results in this chapter have analogues for submanifolds with boundary. Since we will have little reason to consider nonem-bedded submanifolds with boundary, we focus primarily on the embedded case. The statements in the following proposition can be proved in the same way as their submanifold counterparts.
Proposition 5.49 (Properties of Submanifolds with Boundary). Suppose \( M \) is a smooth manifold with or without boundary.
(a) Every open subset of \( M \) is an embedded codimension- 0 submanifold with (possibly empty) boundary.
(b) If \( N \) is a smooth manifold with boundary and \( F : N \rightarrow M \) is a smooth embedding, then with the subspace topology \( F\left( N\right) \) is a topological manifold with boundary, and it has a smooth structure making it into an embedded submani-fold with boundary in \( M \) .
(c) An embedded submanifold with boundary in \( M \) is properly embedded if and only if it is closed.
(d) If \( S \subseteq M \) is an immersed submanifold with boundary, then for each \( p \in S \) there exists a neighborhood \( U \) of \( p \) in \( S \) that is embedded in \( M \) .
- Exercise 5.50. Prove the preceding proposition.
In order to adapt the results that depended on the existence of local slice charts, we have to generalize the local \( k \) -slice condition as follows. Suppose \( M \) is a smooth manifold (without boundary). If \( \left( {U,\left( {x}^{i}\right) }\right) \) is a chart for \( M \), a \( k \) -dimensional half-slice of \( \mathbf{U} \) is a subset of the following form for some constants \( {c}^{k + 1},\ldots ,{c}^{n} \) :
\[
\left\{ {\left( {{x}^{1},\ldots ,{x}^{n}}\right) \in U : {x}^{k + 1} = {c}^{k + 1},\ldots ,{x}^{n} = {c}^{n},\text{ and }{x}^{k} \geq 0}\right\} .
\]
We say that a subset \( S \subseteq M \) satisfies the local \( k \) -slice condition for submani-folds with boundary if each point of \( S \) is contained in the domain of a smooth chart \( \left( {U,\left( {x}^{i}\right) }\right) \) such that \( S \cap U \) is either an ordinary \( k \) -dimensional slice or a \( k \) -dimensional half-slice. In the former case, the chart is called an interior slice chart for \( S \) in \( M \), and in the latter, it is a boundary slice chart for \( S \) in \( M \) .
Theorem 5.51. Let \( M \) be a smooth \( n \) -manifold without boundary. If \( S \subseteq M \) is an embedded \( k \) -dimensional submanifold with boundary, then \( S \) satisfies the local \( k \) -slice condition for submanifolds with boundary. Conversely, if \( S \subseteq M \) is a subset that satisfies the local \( k \) -slice condition for submanifolds with boundary, then with the subspace topology, \( S \) is a topological \( k \) -manifold with boundary, and it has a smooth structure making it into an embedded submanifold with boundary in \( M \) .
- Exercise 5.52. Prove the preceding theorem.
Using the preceding theorem in place of Theorem 5.8, one can readily prove the following theorem.
Theorem 5.53 (Restricting Maps to Submanifolds with Boundary). Suppose \( M \) and \( N \) are smooth manifolds with boundary and \( S \subseteq M \) is an embedded submani-fold with boundary.
(a) RESTRICTING THE DOMAIN: If \( F : M \rightarrow N \) is a smooth map, then \( {\left. F\right| }_{S} : S \rightarrow \) \( N \) is smooth.
(b) RESTRICTING THE CODOMAIN: If \( \partial M = \varnothing \) and \( F : N \rightarrow M \) is a smooth map whose image is contained in \( S \), then \( F \) is smooth as a map from \( N \) to \( S \) .
Remark. The requirement that \( \partial M = \varnothing \) can be removed in part (b) just as for Theorem 5.29; see Problem 9-13.
- Exercise 5.54. Prove Theorem 5.53.
## Problems
5-1. Consider the map \( \Phi : {\mathbb{R}}^{4} \rightarrow {\mathbb{R}}^{2} \) defined by
\[
\Phi \left( {x, y, s, t}\right) = \left( {{x}^{2} + y,{x}^{2} + {y}^{2} + {s}^{2} + {t}^{2} + y}\right) .
\]
Show that \( \left( {0,1}\right) \) is a regular value of \( \Phi \), and that the level set \( {\Phi }^{-1}\left( {0,1}\right) \) is diffeomorphic to \( {\mathbb{S}}^{2} \) .
5-2. Prove Theorem 5.11 (the boundary of a manifold with boundary is an embedded submanifold).
5-3. Prove Proposition 5.21 (sufficient conditions for immersed submanifolds to be embedded).
5-4. Show that the image of the curve \( \beta : \left( {-\pi ,\pi }\right) \rightarrow {\mathbb{R}}^{2} \) of Example 4.19 is not an embedded submanifold of \( {\mathbb{R}}^{2} \) . [Be careful: this is not the same as showing that \( \beta \) is not an embedding.]
5-5. Let \( \gamma : \mathbb{R} \rightarrow {\mathbb{T}}^{2} \) be the curve of Example 4.20. Show that \( \gamma \left( \mathbb{R}\right) \) is not an embedded submanifold of the torus. [Remark: the warning in Problem 5-4 applies in this case as well.]
5-6. Suppose \( M \subseteq {\mathbb{R}}^{n} \) is an embedded \( m \) -dimensional submanifold, and let \( {UM} \subseteq T{\mathbb{R}}^{n} \) be the set of all unit tangent vectors to \( M \) :
\[
{UM} = \left\{ {\left( {x, v}\right) \in T{\mathbb{R}}^{n} : x \in M, v \in {T}_{x}M,\left| v\right| = 1}\right\} .
\]
It is called the unit tangent bundle of \( \mathbf{M} \) . Prove that \( {UM} \) is an embedded \( \left( {{2m} - 1}\right) \) -dimensional submanifold of \( T{\mathbb{R}}^{n} \approx {\mathbb{R}}^{n} \times {\mathbb{R}}^{n} \) . (Used on p. 147.)
5-7. Let \( F : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) be defined by \( F\left( {x, y}\right) = {x}^{3} + {xy} + {y}^{3} \) . Which level sets of \( F \) are embedded submanifolds of \( {\mathbb{R}}^{2} \) ? For each level set, prove either that it is or that it is not an embedded submanifold.
5-8. Suppose \( M \) is a smooth \( n \) -manifold and \( B \subseteq M \) is a regular coordinate ball. Show that \( M \smallsetminus B \) is a smooth manifold with boundary, whose boundary is diffeomorphic to \( {\mathbb{S}}^{n - 1} \) . (Used on p. 225.)
5-9. Let \( S \subseteq {\mathbb{R}}^{2} \) be the boundary of the square of side 2 centered at the origin (see Problem 3-5). Show that \( S \) does not have a topology and smooth structure in which it is an immersed submanifold of \( {\mathbb{R}}^{2} \) .
5-10. For each \( a \in \mathbb{R} \), let \( {M}_{a} \) be the subset of \( {\mathbb{R}}^{2} \) defined by
\[
{M}_{a} = \left\{ {\left( {x, y}\right) : {y}^{2} = x\left( {x - 1}\right) \left( {x - a}\right) }\right\} .
\]
For which values of \( a \) is \( {M}_{a} \) an embedded submanifold of \( {\mathbb{R}}^{2} \) ? For which values can \( {M}_{a} \) be given a topology and smooth structure making it into an immersed submanifold?
5-11. Let \( \Phi : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) be defined by \( \Phi \left( {x, y}\right) = {x}^{2} - {y}^{2} \) .
(a) Show that \( {\Phi }^{-1}\left( 0\right) \) is not an embedded submanifold of \( {\mathbb{R}}^{2} \) .
(b) Can \( {\Phi }^{-1}\left( 0\right) \) be given a topology and | Proposition 5.46. Suppose \( M \) is a smooth manifold without boundary and \( D \subseteq M \) is a regular domain. The topological interior and boundary of \( D \) are equal to its manifold interior and boundary, respectively. | Suppose \( p \in D \) is arbitrary. If \( p \) is in the manifold boundary of \( D \), Theorem 4.15 shows that there exist a smooth boundary chart \( \left( {U,\varphi }\right) \) for \( D \) centered at \( p \) and a smooth chart \( \left( {V,\psi }\right) \) for \( M \) centered at \( p \) in which \( F \) has the coordinate representation \( F\left( {{x}^{1},\ldots ,{x}^{n}}\right) = \left( {{x}^{1},\ldots ,{x}^{n}}\right) \), where \( n = \dim M = \dim D \). Since \( D \) has the subspace topology, \( U = D \cap W \) for some open subset \( W \subseteq M \), so \( {V}_{0} = V \cap W \) is a neighborhood of \( p \) in \( M \) such that \( {V}_{0} \cap D \) consists of all the points in \( {V}_{0} \) whose \( {x}^{m} \) coordinate is nonnegative. Thus every neighborhood of \( p \) intersects both \( D \) and \( M \smallsetminus D \), so \( p \) is in the topological boundary of \( D \).
On the other hand, suppose \( p \) is in the manifold interior of \( D \). The manifold interior is a smooth embedded codimension-0 submanifold without boundary in \( M \), so it is an open subset by Proposition 5.1. Thus \( p \) is in the topological interior of \( D \).
Conversely, if \( p \) is in the topological interior of \( D \), then it is not in the topological boundary, so the preceding argument shows that it is not in the manifold boundary and hence must be in the manifold interior. Similarly, if \( p \) is in the topological boundary, it is also in the manifold boundary. |
Lemma 3.7. Let \( f\left( z\right) \in {H}^{1} \) . Then the Fourier transform
\[
\widehat{f}\left( s\right) = {\int }_{-\infty }^{\infty }f\left( t\right) {e}^{-{2\pi ist}}{dt} = 0
\]
for all \( s \leq 0 \) .
Proof. By the continuity of \( f \rightarrow \widehat{f} \), we may suppose \( \int \in {\mathfrak{A}}_{N} \) . Then for \( s \leq 0, F\left( z\right) = f\left( z\right) {e}^{-{2\pi isz}} \) is also in \( {\mathfrak{A}}_{N} \) . The result now follows from Cauchy’s theorem because
\[
{\int }_{0}^{\pi }\left| {F\left( {R{e}^{i\theta }}\right) }\right| {Rd\theta } \rightarrow 0\;\left( {R \rightarrow \infty }\right) .
\]
Notice that
(3.6)
\[
{P}_{z}\left( t\right) = \frac{1}{2\pi i}\left( {\frac{1}{t - z} - \frac{1}{t - \bar{z}}}\right) .
\]
Also notice that for \( f \in {H}^{1} \), Lemma 3.7 applied to \( {\left( t - \bar{z}\right) }^{-1}f\left( t\right) \) yields
\[
\int \frac{f\left( t\right) }{t - \bar{z}}{dt} = 0,\;\operatorname{Im}z > 0.
\]
Theorem 3.8. Let \( {d\mu }\left( t\right) \) be a finite complex measure on \( \mathbb{R} \) such that either
(a)
\[
\int \frac{{d\mu }\left( t\right) }{t - z} = 0\;\text{ on }\;\operatorname{Im}z < 0
\]
or
(b)
\[
\widehat{\mu }\left( s\right) = \int {e}^{-{2\pi ist}}{d\mu }\left( t\right) = 0\;\text{ on }\;s < 0.
\]
Then \( {d\mu } \) is absolutely continuous and \( {d\mu } = f\left( t\right) {dt} \), where \( f\left( t\right) \in {H}^{1} \) .
Proof. If (a) holds, then by (3.6) \( f\left( z\right) = {P}_{y} * \mu \left( x\right) \) is analytic and the result follows from Theorem 3.6.
Assume (b) holds. The Poisson kernel \( {P}_{y}\left( t\right) \) has Fourier transform
\[
\int {e}^{-{2\pi ist}}{P}_{y}\left( t\right) {dt} = {e}^{-{2\pi }\left| s\right| y}
\]
because \( {\widehat{P}}_{y}\left( {-s}\right) = \overline{{\widehat{P}}_{y}\left( s\right) } \) since \( {P}_{y} \) is real, and because if \( s \leq 0,{e}^{-{2\pi isz}} \) is the bounded harmonic function with boundary values \( {e}^{-{2\pi ist}} \) . Let \( {f}_{y}\left( x\right) = {P}_{y} * \) \( \mu \left( x\right) \) . By Fubini’s theorem, \( {f}_{y} \) has Fourier transform
\[
{\widehat{f}}_{y}\left( s\right) = \left\{ \begin{array}{ll} {e}^{-{2\pi xy}}\widehat{\mu }\left( s\right) , & s \geq 0, \\ 0, & s < 0. \end{array}\right.
\]
Since \( {\widehat{f}}_{y} \in {L}^{1} \), Fourier inversion implies
\[
{f}_{y}\left( x\right) = \int {e}^{2\pi ixs}{\widehat{f}}_{y}\left( s\right) {ds} = {\int }_{0}^{\infty }{e}^{{2\pi i}\left( {x + {iy}}\right) s}\widehat{\mu }\left( s\right) {ds}.
\]
Differentiating under the integral sign then shows that \( f\left( z\right) = {f}_{y}\left( x\right) \) is analytic, and Theorem 3.6 now implies that \( f\left( z\right) \in {H}^{1} \) .
The disc version of Theorem 3.6, or equivalently Theorem 3.8, is one half of the famous F. and M. Riesz theorem. The other half asserts that if \( f\left( z\right) \in \) \( {H}^{1}, f ≢ 0 \), then \( \left| {f\left( t\right) }\right| > 0 \) almost everywhere. This is a consequence of a stronger result proved in the next section.
The theorem on Carleson measures, Theorem I.5.6, also extends to the \( {H}^{p} \) spaces, \( 0 < p \leq 1 \), because the key estimate in its proof was the maximal theorem.
\[
\left( {1/\pi }\right) \int \left( {\log \left| {f\left( t\right) }\right| /\left( {1 + {t}^{2}}\right) }\right) {dt} > - \infty
\]
Theorem 3.9 (Carleson). Let \( \sigma \) be a positive measure in the upper half plane. Then the following are equivalent:
(a) \( \sigma \) is a Carleson measure: for some constant \( N\left( \sigma \right) \) ,
\[
\sigma \left( Q\right) \leq N\left( \sigma \right) h
\]
for all squares
\[
Q = \left\{ {{x}_{0} < x < {x}_{0} + h,0 < y < h}\right\} .
\]
(b) For \( 0 < p < \infty \) ,
\[
\int {\left| f\right| }^{p}{d\sigma } \leq A\parallel f{\parallel }_{{H}^{p}}^{p},\;f \in {H}^{p}.
\]
(c) For some \( p,0 < p < \infty, f \in {L}^{p}\left( \sigma \right) \) for all \( f \in {H}^{p} \) .
Proof. That (a) implies (b) follows from (3.1) and Lemma I.5.5 just as in the proof of Theorem I.5.6.
Trivially, (b) implies (c). On the other hand, if (c) holds for some fixed \( p < \infty \), then (b) holds for the same value \( p \) . This follows from the closed graph theorem, which is valid here even when \( p < 1 \) (see Dunford and Schwartz [1958, p. 57]). One can also see directly that if there are \( \left\{ {f}_{n}\right\} \) in \( {H}^{p} \) with \( {\begin{Vmatrix}{f}_{n}\end{Vmatrix}}_{p} = 1 \) but \( \int {\left| {f}_{n}\right| }^{p}{d\sigma } \rightarrow \infty \), then the sum \( \sum {\alpha }_{n}{f}_{n} \), when the \( {\alpha }_{n} \), are chosen adroitly, will give an \( {H}^{p} \) function for which (c) fails.
Now suppose (b) holds for some \( p > 0 \) . Let \( Q \) be the square \( \left\{ {{x}_{0} < x < }\right. \) \( \left. {{x}_{0} + {y}_{0},0 < y < {y}_{0}}\right\} \) and let
\[
f\left( z\right) = {\left( \frac{1}{\pi }\frac{{y}_{0}}{{\left( z - {\bar{z}}_{0}\right) }^{2}}\right) }^{1/p},
\]
where \( {z}_{0} = {x}_{0} + i{y}_{0} \) . Then \( f \in {H}^{p} \) and \( \parallel f{\parallel }_{p}^{p} = \int {P}_{{z}_{0}}\left( t\right) {dt} = 1 \) . Since \( {\left| f\left( z\right) \right| }^{p} \geq {\left( 5\pi {y}_{0}\right) }^{-1}, z \in Q \), we have
\[
\sigma \left( Q\right) \leq \sigma \left( \left\{ {z : \left| {f\left( z\right) }\right| > {\left( 5\pi {y}_{0}\right) }^{-1/p}}\right\} \right) \leq {5\pi A}{y}_{0},
\]
so that (a) holds.
## \(\text{4.}\left( {1/\pi }\right) \int \left( {\log \left| {f\left( t\right) }\right| /\left( {1 + {t}^{2}}\right) }\right) {dt} > - \infty \)
A fundamental result of \( {H}^{p} \) theory is that the condition of this section’s title characterizes the moduli of \( {H}^{p} \) functions \( \left| {f\left( t\right) }\right| \) among the positive \( {L}^{p} \) functions. In the disc, this result is due to Szegö for \( p = 2 \) and to F. Riesz for the other \( p \) . For functions analytic across \( \partial D \), the inequality (4.1) below was first noticed by Jensen [1899] and for this reason the inequality is sometimes called Jensen's inequality. We prefer to use that name for the inequality about averages and convex functions given in Theorem I.6.1.
In this section the important thing about an \( {H}^{p} \) function will be the fact that the subharmonic function
\[
\log \left| {f\left( {r{e}^{i\theta }}\right) }\right| \leq \left( {1/p}\right) {\left| f\left( r{e}^{i\theta }\right) \right| }^{p}
\]
is majorized by a positive \( {L}^{1} \) function of \( \theta \) . It will be simpler to work at first on the disc.
Theorem 4.1. If \( 0 < p \leq \infty \) and if \( f\left( z\right) \in {H}^{p}\left( D\right), f ≢ 0 \), then
\[
\frac{1}{2\pi }\int \log \left| {f\left( {e}^{i\theta }\right) }\right| {d\theta } > - \infty
\]
If \( f\left( 0\right) ≢ 0 \), then
(4.1)
\[
\log \left| {f\left( 0\right) }\right| \leq \frac{1}{2\pi }\int \log \left| {f\left( {e}^{i\theta }\right) }\right| {d\theta }
\]
and more generally, if \( f\left( {z}_{0}\right) ≢ 0 \)
(4.2)
\[
\log \left| {f\left( {z}_{0}\right) }\right| \leq \frac{1}{2\pi }\int \log \left| {f\left( {e}^{i\theta }\right) }\right| {P}_{{z}_{0}}\left( \theta \right) {d\theta }.
\]
Proof. By Theorem I.6.7 and by the subharmonicity of \( \log \left| f\right| \) ,
\[
\log \left| {f\left( z\right) }\right| \leq \mathop{\lim }\limits_{{r \rightarrow 1}}\frac{1}{2\pi }\int \log \left| {f\left( {r{e}^{i\theta }}\right) }\right| {P}_{z}\left( \theta \right) {d\theta }.
\]
Since \( \log \left| {f\left( {r{e}^{i\theta }}\right) }\right| \rightarrow \log \mid f\left( {e}^{i\theta }\right) \) almost everywhere, and since these functions are bounded above by the integrable function \( \left( {1/p}\right) {\left| {f}^{ * }\left( \theta \right) \right| }^{p} \), where \( {f}^{ * } \) is the maximal function, we have
\[
\int {\log }^{ + }\left| {f\left( {r{e}^{i\theta }}\right) }\right| {P}_{{z}_{0}}\left( \theta \right) {d\theta } \rightarrow \int {\log }^{ + }\left| {f\left( {e}^{i\theta }\right) }\right| {P}_{{z}_{0}}\left( \theta \right) {d\theta }.
\]
Fatou's lemma can now be applied to the negative parts to give us
\[
\mathop{\lim }\limits_{{r \rightarrow 1}}\frac{1}{2\pi }\int \log \left| {f\left( {r{e}^{i\theta }}\right) }\right| {P}_{{z}_{0}}\left( \theta \right) {d\theta } \leq \frac{1}{2\pi }\int \log \left| {f\left( {e}^{i\theta }\right) }\right| {P}_{{z}_{0}}\left( \theta \right) {d\theta }.
\]
This proves (4.2) and (4.1). The remaining inequality follows by removing any zero at the origin.
Note that the same result in the upper half plane
(4.3)
\[
\log \left| {f\left( {z}_{0}\right) }\right| \leq \int \log \left| {f\left( t\right) }\right| {P}_{{z}_{0}}\left( t\right) {dt},\;f \in {H}^{p},
\]
follows from Theorem 4.1 and from Lemma 1.1 upon a change of variables.
Corollary 4.2. If \( f\left( z\right) \in {H}^{p} \) and if \( f\left( t\right) = 0 \) on a set of positive measure, then \( f = 0 \) .
Corollary 4.2 gives the other half of the F. and M. Riesz theorem. If \( {d\mu }\left( t\right) \) is a finite measure such that \( {p}_{y} * \mu \left( x\right) \) is analytic, then not only is \( {d\mu } \) absolutely continuous to \( {dt} \), but also \( {dt} \) is absolutely continuous to \( {d\mu } \) .
Corollary 4.3. Let \( 0 < p, r \leq \infty \) . If \( f\left( z\right) \in {H}^{p} \) and if the boundary function \( f\left( t\right) \in {L}^{r} \), then \( f\left( z\right) \in {H}^{r} \) .
\[
\left( {1/\pi }\right) \int \left( {\log \left| {f\left( t\right) }\right| /\left( {1 + {t}^{2}}\right) }\right) {dt} > - \infty
\]
63
This corollary is often written
\[
{H}^{p} \cap {L}^{r} \subset {H}^{r}
\]
Proof. Applying Jensen’s inequality, with the convex function \( \varphi \left( s\right) = \) \( \exp \left( {rs}\right) \) and with the probability measure \( {P}_{y}\left( {x - t}\right) {dt} \), to (4.3) gives
\[
{\left| f\left( z\right) \right| }^{r} | Lemma 3.7. Let \( f\left( z\right) \in {H}^{1} \) . Then the Fourier transform
\[
\widehat{f}\left( s\right) = {\int }_{-\infty }^{\infty }f\left( t\right) {e}^{-{2\pi ist}}{dt} = 0
\]
for all \( s \leq 0 \) . | Proof. By the continuity of \( f \rightarrow \widehat{f} \), we may suppose \( \int \in {\mathfrak{A}}_{N} \) . Then for \( s \leq 0, F\left( z\right) = f\left( z\right) {e}^{-{2\pi isz}} \) is also in \( {\mathfrak{A}}_{N} \) . The result now follows from Cauchy’s theorem because
\[
{\int }_{0}^{\pi }\left| {F\left( {R{e}^{i\theta }}\right) }\right| {Rd\theta } \rightarrow 0\;\left( {R \rightarrow \infty }\right) .
\] |
Example 2.3.14. Let \( u = {\delta }_{{x}_{0}} \) and \( f \in \mathcal{S} \) . Then \( f * {\delta }_{{x}_{0}} \) is the function \( x \mapsto f\left( {x - {x}_{0}}\right) \) , for when \( h \in \mathcal{S} \), we have
\[
\left\langle {f * {\delta }_{{x}_{0}}, h}\right\rangle = \left\langle {{\delta }_{{x}_{0}},\widetilde{f} * h}\right\rangle = \left( {\widetilde{f} * h}\right) \left( {x}_{0}\right) = {\int }_{{\mathbf{R}}^{n}}f\left( {x - {x}_{0}}\right) h\left( x\right) {dx}.
\]
It follows that convolution with \( {\delta }_{0} \) is the identity operator.
We now define the product of a function and a distribution.
Definition 2.3.15. Let \( u \in {\mathcal{S}}^{\prime } \) and let \( h \) be a \( {\mathcal{C}}^{\infty } \) function that has at most polynomial growth at infinity and the same is true for all of its derivatives. This means that for all \( \alpha \) it satisfies \( \left| {\left( {{\partial }^{\alpha }h}\right) \left( x\right) }\right| \leq {C}_{\alpha }{\left( 1 + \left| x\right| \right) }^{{k}_{\alpha }} \) for some \( {C}_{\alpha },{k}_{\alpha } > 0 \) . Then define the product \( {hu} \) of \( h \) and \( u \) by
\[
\langle {hu}, f\rangle = \langle u,{hf}\rangle ,\;f \in \mathcal{S}.
\]
(2.3.16)
Note that \( {hf} \) is in \( \mathcal{S} \) and thus (2.3.16) is well defined. The product of an arbitrary \( {\mathcal{C}}^{\infty } \) function with a tempered distribution is not defined.
We observe that if a function \( g \) is supported in a set \( K \), then for all \( f \in {\mathcal{C}}_{0}^{\infty }\left( {K}^{c}\right) \)
we have
\[
{\int }_{{\mathbf{R}}^{n}}f\left( x\right) g\left( x\right) {dx} = 0.
\]
(2.3.17)
Moreover, the support of \( g \) is the intersection of all closed sets \( K \) with the property (2.3.17) for all \( f \) in \( {\mathcal{C}}_{0}^{\infty }\left( {K}^{c}\right) \) . Motivated by the preceding observation we give the following:
Definition 2.3.16. Let \( u \) be in \( {\mathcal{D}}^{\prime }\left( {\mathbf{R}}^{n}\right) \) . The support of \( u\left( {\operatorname{supp}u}\right) \) is the intersection of all closed sets \( K \) with the property
\[
\varphi \in {\mathcal{C}}_{0}^{\infty }\left( {\mathbf{R}}^{n}\right) ,\;\operatorname{supp}\varphi \subseteq {\mathbf{R}}^{n} \smallsetminus K \Rightarrow \langle u,\varphi \rangle = 0.
\]
(2.3.18)
Distributions with compact support are exactly those whose support (as defined in the previous definition) is a compact set. To prove this assertion, we start with a distribution \( u \) with compact support as defined in Definition 2.3.3. Then there exist \( C, N, m > 0 \) such that (2.3.4) holds. For a \( {\mathcal{C}}^{\infty } \) function \( f \) whose support is contained in \( B{\left( 0, N\right) }^{c} \), the expression on the right in (2.3.4) vanishes and we must therefore have \( \langle u, f\rangle = 0 \) . This shows that the support of \( u \) is contained in \( \overline{B\left( {0, N}\right) } \) hence it is bounded, and since it is already closed (as an intersection of closed sets), it must be compact. Conversely, if the support of \( u \) as defined in Definition 2.3.16 is a compact set, then there exists an \( N > 0 \) such that \( \operatorname{supp}u \) is contained in \( B\left( {0, N}\right) \) . We take a smooth function \( \eta \) that is equal to 1 on \( B\left( {0, N}\right) \) and vanishes off \( B\left( {0, N + 1}\right) \) . Then for \( h \in {\mathcal{C}}_{0}^{\infty } \) the support of \( h\left( {1 - \eta }\right) \) does not meet the support of \( u \), and we must have
\[
\langle u, h\rangle = \langle u,{h\eta }\rangle + \langle u, h\left( {1 - \eta }\right) \rangle = \langle u,{h\eta }\rangle .
\]
The distribution \( u \) can be thought of as an element of \( {\mathcal{E}}^{\prime } \) by defining for \( f \in {\mathcal{C}}^{\infty }\left( {\mathbf{R}}^{n}\right) \)
\[
\langle u, f\rangle = \langle u,{f\eta }\rangle
\]
Taking \( m \) to be the integer that corresponds to the compact set \( K = \overline{B\left( {0, N + 1}\right) } \) in (2.3.2), and using that the \( {L}^{\infty } \) norm of \( {\partial }^{\alpha }\left( {f\eta }\right) \) is controlled by a finite sum of seminorms \( {\widetilde{\rho }}_{\alpha, N + 1}\left( f\right) \) with \( \left| \alpha \right| \leq m \), we obtain the validity of (2.3.4) for \( f \in {\mathcal{C}}^{\infty } \) .
Example 2.3.17. The support of the Dirac mass at \( {x}_{0} \) is the set \( \left\{ {x}_{0}\right\} \) .
Along the same lines, we give the following definition:
Definition 2.3.18. We say that a distribution \( u \) in \( {\mathcal{D}}^{\prime }\left( {\mathbf{R}}^{n}\right) \) coincides with the function \( h \) on an open set \( \Omega \) if
\[
\langle u, f\rangle = {\int }_{{\mathbf{R}}^{n}}f\left( x\right) h\left( x\right) {dx}\;\text{ for all }f\text{ in }{\mathcal{C}}_{0}^{\infty }\left( \Omega \right) .
\]
(2.3.19)
When (2.3.19) occurs we often say that \( u \) agrees with \( h \) away from \( {\Omega }^{c} \) .
This definition implies that the support of the distribution \( u - h \) is contained in the set \( {\Omega }^{c} \) .
Example 2.3.19. The distribution \( {\left| x\right| }^{2} + {\delta }_{{a}_{1}} + {\delta }_{{a}_{2}} \), where \( {a}_{1},{a}_{2} \) are in \( {\mathbf{R}}^{n} \), coincides with the function \( {\left| x\right| }^{2} \) on any open set not containing the points \( {a}_{1} \) and \( {a}_{2} \) . Also, the distribution in Example 2.3.5 (8) coincides with the function \( {x}^{-1}{\chi }_{\left| x\right| \leq 1} \) away from the origin in the real line.
Having ended the streak of definitions regarding operations with distributions, we now discuss properties of convolutions and Fourier transforms.
Theorem 2.3.20. If \( u \in {\mathcal{S}}^{\prime } \) and \( \varphi \in \mathcal{S} \), then \( \varphi * u \) is a \( {\mathcal{C}}^{\infty } \) function and
\[
\left( {\varphi * u}\right) \left( x\right) = \left\langle {u,{\tau }^{x}\widetilde{\varphi }}\right\rangle
\]
for all \( x \in {\mathbf{R}}^{n} \) . Moreover, for all multi-indices \( \alpha \) there exist constants \( {C}_{\alpha },{k}_{\alpha } > 0 \) such that
\[
\left| {{\partial }^{\alpha }\left( {\varphi * u}\right) \left( x\right) }\right| \leq {C}_{\alpha }{\left( 1 + \left| x\right| \right) }^{{k}_{\alpha }}.
\]
Furthermore, if \( u \) has compact support, then \( \varphi * u \) is a Schwartz function.
Proof. Let \( \psi \) be in \( \mathcal{S}\left( {\mathbf{R}}^{n}\right) \) . We have
\[
\langle \varphi * u,\psi \rangle = \langle u,\widetilde{\varphi } * \psi \rangle
\]
\[
= u\left( {{\int }_{{\mathbf{R}}^{n}}\widetilde{\varphi }\left( {\cdot - y}\right) \psi \left( y\right) {dy}}\right)
\]
\[
= u\left( {{\int }_{{\mathbf{R}}^{n}}\left( {{\tau }^{y}\widetilde{\varphi }}\right) \left( \cdot \right) \psi \left( y\right) {dy}}\right)
\]
(2.3.20)
\[
= {\int }_{{\mathbf{R}}^{n}}\left\langle {u,{\tau }^{y}\widetilde{\varphi }}\right\rangle \psi \left( y\right) {dy}
\]
where the last step is justified by the continuity of \( u \) and by the fact that the Riemann sums of the inner integral in (2.3.20) converge to that integral in the topology of \( \mathcal{S} \) , a fact that will be justified later. This calculation identifies the function \( \varphi * u \) as
\[
\left( {\varphi * u}\right) \left( x\right) = \left\langle {u,{\tau }^{x}\widetilde{\varphi }}\right\rangle
\]
(2.3.21)
We now show that \( \left( {\varphi * u}\right) \left( x\right) \) is a \( {\mathcal{C}}^{\infty } \) function. Let \( {e}_{j} = \left( {0,\ldots ,1,\ldots ,0}\right) \) with 1 in the \( j \) th entry and zero elsewhere. Then
\[
\frac{{\tau }^{-h{e}_{j}}\left( {\varphi * u}\right) \left( x\right) - \left( {\varphi * u}\right) \left( x\right) }{h} = u\left( \frac{{\tau }^{-h{e}_{j}}\left( {{\tau }^{x}\widetilde{\varphi }}\right) - {\tau }^{x}\widetilde{\varphi }}{h}\right) \rightarrow \left\langle {u,{\tau }^{x}\left( {{\partial }_{j}\widetilde{\varphi }}\right) }\right\rangle
\]
by the continuity of \( u \) and the fact that \( \left( {{\tau }^{-h{e}_{j}}\left( {{\tau }^{x}\widetilde{\varphi }}\right) - {\tau }^{x}\widetilde{\varphi }}\right) /h \) tends to \( {\partial }_{j}{\tau }^{x}\widetilde{\varphi } = \) \( {\tau }^{x}\left( {{\partial }_{j}\widetilde{\varphi }}\right) \) in \( \mathcal{S} \) as \( h \rightarrow 0 \) ; see Exercise 2.3.5 (a). The same calculation for higher-order derivatives shows that \( \varphi * u \in {\mathcal{C}}^{\infty } \) and that \( {\partial }^{\gamma }\left( {\varphi * u}\right) = \left( {{\partial }^{\gamma }\varphi }\right) * u \) for all multi-indices \( \gamma \) . It follows from (2.3.3) that for some \( C, m \), and \( k \) we have
\[
\left| {{\partial }^{\alpha }\left( {\varphi * u}\right) \left( x\right) }\right| \leq C\mathop{\sum }\limits_{\substack{{\left| \gamma \right| \leq m} \\ {\left| \beta \right| \leq k} }}\mathop{\sup }\limits_{{y \in {\mathbf{R}}^{n}}}\left| {{y}^{\gamma }{\tau }^{x}\left( {{\partial }^{\alpha + \beta }\widetilde{\varphi }}\right) \left( y\right) }\right|
\]
\[
= C\mathop{\sum }\limits_{\substack{{\left| \gamma \right| \leq m} \\ {\left| \beta \right| \leq k} }}\mathop{\sup }\limits_{{y \in {\mathbf{R}}^{n}}}\left| {{\left( x + y\right) }^{\gamma }\left( {{\partial }^{\alpha + \beta }\widetilde{\varphi }}\right) \left( y\right) }\right|
\]
(2.3.22)
\[
\leq {C}_{m}\mathop{\sum }\limits_{{\left| \beta \right| \leq k}}\mathop{\sup }\limits_{{y \in {\mathbf{R}}^{n}}}\left( {1 + {\left| x\right| }^{m} + {\left| y\right| }^{m}}\right) \left| {\left( {{\partial }^{\alpha + \beta }\widetilde{\varphi }}\right) \left( y\right) }\right| ,
\]
and this clearly implies that \( {\partial }^{\alpha }\left( {\varphi * u}\right) \) grows at most polynomially at infinity.
We now indicate why \( \varphi * u \) is Schwartz whenever \( u \) has compact support. Applying estimate (2.3.4) to the function \( y \mapsto \varphi \left( {x - y}\right) \) yields that
\[
\left | Example 2.3.14. Let \( u = {\delta }_{{x}_{0}} \) and \( f \in \mathcal{S} \) . Then \( f * {\delta }_{{x}_{0}} \) is the function \( x \mapsto f\left( {x - {x}_{0}}\right) \) , for when \( h \in \mathcal{S} \), we have
\[
\left\langle {f * {\delta }_{{x}_{0}}, h}\right\rangle = \left\langle {{\delta }_{{x}_{0}},\widetilde{f} * h}\right\rangle = \left( {\widetilde{f} * h}\right) \left( {x}_{0}\right) = {\int }_{{\mathbf{R}}^{n}}f\left( {x - {x}_{0}}\right) h\left( x\right) {dx}.
\] | The proof provided in the text is as follows:
\[
\left\langle {f * {\delta }_{{x}_{0}}, h}\right\rangle = \left\langle {{\delta }_{{x}_{0}},\widetilde{f} * h}\right\rangle = \left( {\widetilde{f} * h}\right) \left( {x}_{0}\right) = {\int }_{{\mathbf{R}}^{n}}f\left( {x - {x}_{0}}\right) h\left( x\right) {dx}.
\]
This shows that the convolution \( f * {\delta }_{{x}_{0}} \) is indeed the function \( x \mapsto f(x - {x}_{0}) \). |
Exercise 1.4.13 Use Exercises 1.2.7 and 1.2.8 to show that there are infinitely many primes \( \equiv 1\left( {\;\operatorname{mod}\;{2}^{r}}\right) \) for any given \( r \) .
Exercise 1.4.14 Suppose \( p \) is an odd prime such that \( {2p} + 1 = q \) is also prime. Show that the equation
\[
{x}^{p} + 2{y}^{p} + 5{z}^{p} = 0
\]
has no solutions in integers.
Exercise 1.4.15 If \( x \) and \( y \) are coprime integers, show that if
\[
\left( {x + y}\right) \text{ and }\frac{{x}^{p} + {y}^{p}}{x + y}
\]
have a common prime factor, it must be \( p \) .
Exercise 1.4.16 (Sophie Germain’s Trick) Let \( p \) be a prime such that \( {2p} + \) \( 1 = q > 3 \) is also prime. Show that
\[
{x}^{p} + {y}^{p} + {z}^{p} = 0
\]
has no integral solutions with \( p \nmid {xyz} \) .
Exercise 1.4.17 Assuming \( {ABC} \), show that there are only finitely many consecutive cubefull numbers.
Exercise 1.4.18 Show that
\[
\mathop{\sum }\limits_{p}\frac{1}{p} = + \infty
\]
where the summation is over prime numbers.
Exercise 1.4.19 (Bertrand’s Postulate) (a) If \( {a}_{0} \geq {a}_{1} \geq {a}_{2} \geq \cdots \) is a de-
creasing sequence of real numbers tending to 0 , show that
\[
\mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}{a}_{n} \leq {a}_{0} - {a}_{1} + {a}_{2}
\]
(b) Let \( T\left( x\right) = \mathop{\sum }\limits_{{n < x}}\psi \left( {x/n}\right) \), where \( \psi \left( x\right) \) is defined as in Exercise 1.1.25. Show
that
\[
T\left( x\right) = x\log x - x + O\left( {\log x}\right) .
\]
(c) Show that
\[
T\left( x\right) - {2T}\left( \frac{x}{2}\right) = \mathop{\sum }\limits_{{n \leq x}}{\left( -1\right) }^{n - 1}\psi \left( \frac{x}{n}\right) = \left( {\log 2}\right) x + O\left( {\log x}\right) .
\]
Deduce that
\[
\psi \left( x\right) - \psi \left( \frac{x}{2}\right) \geq \frac{1}{3}\left( {\log 2}\right) x + O\left( {\log x}\right) .
\]
## Chapter 2
## Euclidean Rings
## 2.1 Preliminaries
We can discuss the concept of divisibility for any commutative ring \( R \) with identity. Indeed, if \( a, b \in R \), we will write \( a \mid b \) ( \( a \) divides \( b \) ) if there exists some \( c \in R \) such that \( {ac} = b \) . Any divisor of 1 is called a unit. We will say that \( a \) and \( b \) are associates and write \( a \sim b \) if there exists a unit \( u \in R \) such that \( a = {bu} \) . It is easy to verify that \( \sim \) is an equivalence relation.
Further, if \( R \) is an integral domain and we have \( a, b \neq 0 \) with \( a \mid b \) and \( b \mid a \), then \( a \) and \( b \) must be associates, for then \( \exists c, d \in R \) such that \( {ac} = b \) and \( {bd} = a \), which implies that \( {bdc} = b \) . Since we are in an integral domain, \( {dc} = 1 \), and \( d, c \) are units.
We will say that \( a \in R \) is irreducible if for any factorization \( a = {bc} \), one of \( b \) or \( c \) is a unit.
Example 2.1.1 Let \( R \) be an integral domain. Suppose there is a map \( n : R \rightarrow \mathbb{N} \) such that:
(i) \( n\left( {ab}\right) = n\left( a\right) n\left( b\right) \forall a, b \in R \) ; and
(ii) \( n\left( a\right) = 1 \) if and only if \( a \) is a unit.
We call such a map a norm map, with \( n\left( a\right) \) the norm of \( a \) . Show that every element of \( R \) can be written as a product of irreducible elements.
Solution. Suppose \( b \) is an element of \( R \) . We proceed by induction on the norm of \( b \) . If \( b \) is irreducible, then we have nothing to prove, so assume that \( b \) is an element of \( R \) which is not irreducible. Then we can write \( b = {ac} \) where neither \( a \) nor \( c \) is a unit. By condition (i),
\[
n\left( b\right) = n\left( {ac}\right) = n\left( a\right) n\left( c\right)
\]
and since \( a, c \) are not units, then by condition (ii), \( n\left( a\right) < n\left( b\right) \) and \( n\left( c\right) < \) \( n\left( b\right) \) .
If \( a, c \) are irreducible, then we are finished. If not, their norms are smaller than the norm of \( b \), and so by induction we can write them as products of irreducibles, thus finding an irreducible decomposition of \( b \) .
Exercise 2.1.2 Let \( D \) be squarefree. Consider \( R = \mathbb{Z}\left\lbrack \sqrt{D}\right\rbrack \) . Show that every element of \( R \) can be written as a product of irreducible elements.
Exercise 2.1.3 Let \( R = \mathbb{Z}\left\lbrack \sqrt{-5}\right\rbrack \) . Show that \( 2,3,1 + \sqrt{-5} \), and \( 1 - \sqrt{-5} \) are irreducible in \( R \), and that they are not associates.
We now observe that \( 6 = 2 \cdot 3 = \left( {1 + \sqrt{-5}}\right) \left( {1 - \sqrt{-5}}\right) \), so that \( R \) does not have unique factorization into irreducibles.
We will say that \( R \), an integral domain, is a unique factorization domain if:
(i) every element of \( R \) can be written as a product of irreducibles; and
(ii) this factorization is essentially unique in the sense that if \( a = {\pi }_{1}\cdots {\pi }_{r} \) . and \( a = {\tau }_{1}\cdots {\tau }_{s} \), then \( r = s \) and after a suitable permutation, \( {\pi }_{i} \sim {\tau }_{i} \) .
Exercise 2.1.4 Let \( R \) be a domain satisfying (i) above. Show that (ii) is equivalent to \( \left( {\mathrm{{ii}}}^{ \star }\right) \) : if \( \pi \) is irreducible and \( \pi \) divides \( {ab} \), then \( \pi \mid a \) or \( \pi \mid b \) .
An ideal \( I \subseteq R \) is called principal if it can be generated by a single element of \( R \) . A domain \( R \) is then called a principal ideal domain if every ideal of \( R \) is principal.
Exercise 2.1.5 Show that if \( \pi \) is an irreducible element of a principal ideal domain, then \( \left( \pi \right) \) is a maximal ideal,(where \( \left( x\right) \) denotes the ideal generated by the element \( x \) ).
Theorem 2.1.6 If \( R \) is a principal ideal domain, then \( R \) is a unique factorization domain.
Proof. Let \( S \) be the set of elements of \( R \) that cannot be written as a product of irreducibles. If \( S \) is nonempty, take \( {a}_{1} \in S \) . Then \( {a}_{1} \) is not irreducible, so we can write \( {a}_{1} = {a}_{2}{b}_{2} \) where \( {a}_{2},{b}_{2} \) are not units. Then \( \left( {a}_{1}\right) \subsetneqq \left( {a}_{2}\right) \) and \( \left( {a}_{1}\right) \subsetneqq \left( {b}_{2}\right) \) . If both \( {a}_{2},{b}_{2} \notin S \), then we can write \( {a}_{1} \) as a product of irreducibles, so we assume that \( {a}_{2} \in S \) . We can inductively proceed until we arrive at an infinite chain of ideals,
\[
\left( {a}_{1}\right) \subsetneqq \left( {a}_{2}\right) \subsetneqq \left( {a}_{3}\right) \subsetneqq \cdots \subsetneqq \left( {a}_{n}\right) \subsetneqq \cdots .
\]
Now consider \( I = \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {a}_{i}\right) \) . This is an ideal of \( R \), and because \( R \) is a principal ideal domain, \( I = \left( \alpha \right) \) for some \( \alpha \in R \) . Since \( \alpha \in I,\alpha \in \left( {a}_{n}\right) \) for some \( n \), but then \( \left( {a}_{n}\right) = \left( {a}_{n + 1}\right) \) . From this contradiction, we conclude that the set \( S \) must be empty, so we know that if \( R \) is a principal ideal domain, every element of \( R \) satisfies the first condition for a unique factorization domain.
Next we would like to show that if we have an irreducible element \( \pi \) , and \( \pi \mid {ab} \) for \( a, b \in R \), then \( \pi \mid a \) or \( \pi \mid b \) . If \( \pi \nmid a \), then the ideal \( \left( {a,\pi }\right) = R \) , so \( \exists x, y \) such that
\[
{ax} + {\pi y} = 1
\]
\[
\Rightarrow \;{abx} + {\pi by} = b\text{.}
\]
Since \( \pi \mid {abx} \) and \( \pi \mid {\pi by} \) then \( \pi \mid b \), as desired. By Exercise 2.1.4, we have shown that \( R \) is a unique factorization domain.
The following theorem describes an important class of principal ideal domains:
Theorem 2.1.7 If \( R \) is a domain with a map \( \phi : R \rightarrow \mathbb{N} \), and given \( a, b \in R,\exists q, r \in R \) such that \( a = {bq} + r \) with \( r = 0 \) or \( \phi \left( r\right) < \phi \left( b\right) \), we call \( R \) a Euclidean domain. If a ring \( R \) is Euclidean, it is a principal ideal domain.
Proof. Given an ideal \( I \subseteq R \), take an element \( a \) of \( I \) such that \( \phi \left( a\right) \) is minimal among elements of \( I \) . Then given \( b \in I \), we can find \( q, r \in R \) such that \( b = {qa} + r \) where \( r = 0 \) or \( \phi \left( r\right) < \phi \left( a\right) \) . But then \( r = b - {qa} \), and so \( r \in I \), and \( \phi \left( a\right) \) is minimal among the norms of elements of \( I \) . So \( r = 0 \) , and given any element \( b \) of \( I, b = {qa} \) for some \( q \in R \) . Therefore \( a \) is a generator for \( I \), and \( R \) is a principal ideal domain.
Exercise 2.1.8 If \( F \) is a field, prove that \( F\left\lbrack x\right\rbrack \), the ring of polynomials in \( x \) with coefficients in \( F \), is Euclidean.
The following result, called Gauss' lemma, allows us to relate factorization of polynomials in \( \mathbb{Z}\left\lbrack x\right\rbrack \) with the factorization in \( \mathbb{Q}\left\lbrack x\right\rbrack \) . More generally, if \( R \) is a unique factorization domain and \( K \) is its field of fractions, we will relate factorization of polynomials in \( R\left\lbrack x\right\rbrack \) with that in \( K\left\lbrack x\right\rbrack \) .
Theorem 2.1.9 If \( R \) is a unique factorization domain, and \( f\left( x\right) \in R\left\lbrack x\right\rbrack \) , define the content of \( f \) to be the gcd of the coefficients of \( f \), denoted by \( \mathcal{C}\left( f\right) \) . For \( f\left( x\right), g\left( x\right) \in R\left\lbrack x\right\rbrack ,\mathcal{C}\left( {fg}\right) = \mathcal{C}\left( f\right) \mathcal{C}\left( g\right) \) .
Proof. Consider two polynomials \( f, g \in R\left\lbrack x\right\rbrack \), with \( \mathcal{C}\left | Exercise 1.4.13 Use Exercises 1.2.7 and 1.2.8 to show that there are infinitely many primes \( \equiv 1\left( {\;\operatorname{mod}\;{2}^{r}}\right) \) for any given \( r \) . | null |
Theorem 11.5 For each \( h > 0 \) the difference equations (11.10)-(11.11) have a unique solution.
Proof. The tridiagonal matrix \( A \) is irreducible and weakly row-diagonally dominant. Hence, by Theorem 4.7, the matrix \( A \) is invertible, and the Jacobi iterations converge.
Recall that for speeding up the convergence of the Jacobi iterations we can use relaxation methods or multigrid methods as discussed in Sections 4.2 and 4.3.
The error and convergence analysis is initiated by first establishing the following two lemmas.
Lemma 11.6 Denote by \( A \) the matrix of the finite difference method for \( q \geq 0 \) and by \( {A}_{0} \) the corresponding matrix for \( q = 0 \) . Then
\[
0 \leq {A}^{-1} \leq {A}_{0}^{-1}
\]
i.e., all components of \( {A}^{-1} \) are nonnegative and smaller than or equal to the corresponding components of \( {A}_{0}^{-1} \) .
Proof. The columns of the inverse \( {A}^{-1} = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) satisfy \( A{a}_{j} = {e}_{j} \) for \( j = 1,\ldots, n \) with the canonical unit vectors \( {e}_{1},\ldots ,{e}_{n} \) in \( {\mathbb{R}}^{n} \) . The Jacobi iterations for the solution of \( {Az} = {e}_{j} \) starting with \( {z}_{0} = 0 \) are given by
\[
{z}_{\nu + 1} = - {D}^{-1}\left( {{A}_{L} + {A}_{R}}\right) {z}_{\nu } + {D}^{-1}{e}_{j},\;\nu = 0,1,\ldots ,
\]
with the usual splitting \( A = D + {A}_{L} + {A}_{R} \) of \( A \) into its diagonal, lower, and upper triangular parts. Since the entries of \( {D}^{-1} \) and of \( - {D}^{-1}\left( {{A}_{L} + {A}_{R}}\right) \) are all nonnegative, it follows that \( {A}^{-1} \geq 0 \) . Analogously, the iterations
\[
{z}_{\nu + 1} = - {D}_{0}^{-1}\left( {{A}_{L} + {A}_{R}}\right) {z}_{\nu } + {D}_{0}^{-1}{e}_{j},\;\nu = 0,1,\ldots ,
\]
yield the columns of \( {A}_{0}^{-1} \) . Therefore, from \( {D}_{0}^{-1} \geq {D}^{-1} \) we conclude that \( {A}_{0}^{-1} \geq {A}^{-1} \)
Lemma 11.7 Assume that \( u \in {C}^{4}\left\lbrack {a, b}\right\rbrack \) . Then
\[
\left| {{u}^{\prime \prime }\left( x\right) - \frac{1}{{h}^{2}}\left\lbrack {u\left( {x + h}\right) - {2u}\left( x\right) + u\left( {x - h}\right) }\right\rbrack }\right| \leq \frac{{h}^{2}}{12}{\begin{Vmatrix}{u}^{\left( 4\right) }\end{Vmatrix}}_{\infty }
\]
for all \( x \in \left\lbrack {a + h, b - h}\right\rbrack \) .
Proof. By Taylor's formula we have that
\[
u\left( {x \pm h}\right) = u\left( x\right) \pm h{u}^{\prime }\left( x\right) + \frac{{h}^{2}}{2}{u}^{\prime \prime }\left( x\right) \pm \frac{{h}^{3}}{6}{u}^{\prime \prime \prime }\left( x\right) + \frac{{h}^{4}}{24}{u}^{\left( 4\right) }\left( {x \pm {\theta }_{ \pm }h}\right)
\]
for some \( {\theta }_{ \pm } \in \left( {0,1}\right) \) . Adding these two equations gives
\[
u\left( {x + h}\right) - {2u}\left( x\right) + u\left( {x - h}\right) = {h}^{2}{u}^{\prime \prime }\left( x\right) + \frac{{h}^{4}}{24}{u}^{\left( 4\right) }\left( {x + {\theta }_{ + }h}\right) + \frac{{h}^{4}}{24}{u}^{\left( 4\right) }\left( {x - {\theta }_{ - }h}\right) ,
\]
whence the statement of the lemma follows.
Theorem 11.8 Assume that the solution to the boundary value problem (11.7)-(11.8) is four-times continuously differentiable. Then the error of the finite difference approximation can be estimated by
\[
\left| {u\left( {x}_{j}\right) - {u}_{j}}\right| \leq \frac{{h}^{2}}{96}{\begin{Vmatrix}{u}^{\left( 4\right) }\end{Vmatrix}}_{\infty }{\left( b - a\right) }^{2},\;j = 1,\ldots, n.
\]
Proof. By Lemma 11.7, for
\[
{z}_{j} \mathrel{\text{:=}} {u}^{\prime \prime }\left( {x}_{j}\right) - \frac{1}{{h}^{2}}\left\lbrack {u\left( {x}_{j + 1}\right) - {2u}\left( {x}_{j}\right) + u\left( {x}_{j - 1}\right) }\right\rbrack
\]
we have the estimate
\[
\left| {z}_{j}\right| \leq \frac{{h}^{2}}{12}{\begin{Vmatrix}{u}^{\left( 4\right) }\end{Vmatrix}}_{\infty },\;j = 1,\ldots, n.
\]
(11.13)
Since
\[
- \frac{1}{{h}^{2}}\left\lbrack {u\left( {x}_{j + 1}\right) - \left( {2 + {h}^{2}{q}_{j}}\right) u\left( {x}_{j}\right) + u\left( {x}_{j - 1}\right) }\right\rbrack = - {u}^{\prime \prime }\left( {x}_{j}\right) + {q}_{j}u\left( {x}_{j}\right) + {z}_{j} = {r}_{j} + {z}_{j},
\]
the vector \( \widetilde{U} = {\left( u\left( {x}_{1}\right) ,\ldots, u\left( {x}_{n}\right) \right) }^{T} \) given by the exact solution solves the
linear system
\[
A\widetilde{U} = R + Z
\]
where \( Z = {\left( {z}_{1},\ldots ,{z}_{n}\right) }^{T} \) . Therefore,
\[
A\left( {\widetilde{U} - U}\right) = Z
\]
and from this, using Lemma 11.6 and the estimate (11.13), we obtain
\[
\left| {u\left( {x}_{j}\right) - {u}_{j}}\right| \leq {\begin{Vmatrix}{A}^{-1}Z\end{Vmatrix}}_{\infty } \leq \frac{{h}^{2}}{12}{\begin{Vmatrix}{u}^{\left( 4\right) }\end{Vmatrix}}_{\infty }{\begin{Vmatrix}{A}_{0}^{-1}e\end{Vmatrix}}_{\infty },\;j = 1,\ldots, n
\]
(11.14)
where \( e = {\left( 1,\ldots ,1\right) }^{T} \) . The boundary value problem
\[
- {u}_{0}^{\prime \prime } = 1,\;{u}_{0}\left( a\right) = {u}_{0}\left( b\right) = 0,
\]
has the solution
\[
{u}_{0}\left( x\right) = \frac{1}{2}\left( {x - a}\right) \left( {b - x}\right)
\]
Since \( {u}_{0}^{\left( 4\right) } = 0 \), in this case, as a consequence of (11.14) the finite difference approximation coincides with the exact solution; i.e., \( e = {A}_{0}U = {A}_{0}\widetilde{U} \) . Hence,
\[
{\begin{Vmatrix}{A}_{0}^{-1}e\end{Vmatrix}}_{\infty } \leq {\begin{Vmatrix}{u}_{0}\end{Vmatrix}}_{\infty } = \frac{1}{8}{\left( b - a\right) }^{2},\;j = 1,\ldots, n.
\]
Inserting this into (11.14) completes the proof.
Theorem 11.8 confirms that as in the case of the initial value problems in Chapter 10, the order of the local discretization error is inherited by the global error. Note that the assumption in Theorem 11.8 on the differentiability of the solution is satisfied if \( q \) and \( r \) are twice continuously differentiable.
The error estimate in Theorem 11.8 is not practical in general, since it requires a bound on the fourth derivative of the unknown exact solution. Therefore, in practice, analogously to (10.19) the error is estimated from the numerical results for step sizes \( h \) and \( h/2 \) . Similarly, as in (10.20), a Richardson extrapolation can be employed to obtain a fourth-order approximation.
Of course, the finite difference approximation can be extended to the general linear ordinary differential equation of second order
\[
- {u}^{\prime \prime } + p{u}^{\prime } + {qu} = r
\]
by using the approximation
\[
{u}^{\prime }\left( {x}_{j}\right) \approx \frac{1}{2h}\left\lbrack {u\left( {x}_{j + 1}\right) - u\left( {x}_{j - 1}\right) }\right\rbrack
\]
(11.15)
for the first derivative. This approximation again has an error of order \( O\left( {h}^{2}\right) \) (see Problem 11.9). Besides Richardson extrapolation, higher-order approximations can be obtained by using higher-order difference approximations for the derivatives such as
\[
{u}^{\prime \prime }\left( x\right) \approx \frac{1}{{12}{h}^{2}}\left\lbrack {-u\left( {x + {2h}}\right) + {16u}\left( {x + h}\right) }\right.
\]
(11.16)
\[
- {30u}\left( x\right) + {16u}\left( {x - h}\right) - u\left( {x - {2h}}\right) \rbrack
\]
which is of order \( O\left( {h}^{4}\right) \), provided that \( u \) is six-times continuously differentiable (see Problem 11.9).
We wish also to indicate briefly how the finite difference approximations are applied to boundary value problems for partial differential equations. For this we consider the boundary value problem for
\[
- \bigtriangleup u + {qu} = r\;\text{ in }D
\]
(11.17)
in the unit square \( D = \left( {0,1}\right) \times \left( {0,1}\right) \) with boundary condition
\[
u = 0\;\text{ on }\partial D.
\]
(11.18)
Here \( \Delta \) denotes the Laplacian
\[
{\Delta u} \mathrel{\text{:=}} \frac{{\partial }^{2}u}{\partial {x}_{1}^{2}} + \frac{{\partial }^{2}u}{\partial {x}_{2}^{2}}
\]
Proceeding as in the proof of Theorem 11.4, by partial integration it can be seen that under the assumption \( q \geq 0 \) this boundary value problem has at most one solution. It is more involved and beyond the scope of this book to establish that a solution exists under proper assumptions on the functions \( q \) and \( r \) . We refer to \( \left\lbrack {{24},{60}}\right\rbrack \) and also the remarks at the end of Section 11.4.
As in Example 2.2, we choose an equidistant grid
\[
{x}_{ij} = \left( {{ih},{jh}}\right) ,\;i, j = 0,\ldots, n + 1,
\]
with step size \( h = 1/\left( {n + 1}\right) \) and \( n \in \mathbb{N} \) . Then we approximate the Laplacian at the internal grid points by
\[
\bigtriangleup u\left( {x}_{ij}\right) \approx \frac{1}{{h}^{2}}\left\{ {u\left( {x}_{i + 1, j}\right) + u\left( {x}_{i - 1, j}\right) + u\left( {x}_{i, j + 1}\right) + u\left( {x}_{i, j - 1}\right) - {4u}\left( {x}_{ij}\right) }\right\}
\]
and obtain the system of equations
\[
\frac{1}{{h}^{2}}\left\lbrack {\left( {4 + {h}^{2}{q}_{ij}}\right) {u}_{ij} - {u}_{i + 1, j} - {u}_{i - 1, j} - {u}_{i, j + 1} - {u}_{i, j - 1}}\right\rbrack = {r}_{ij},
\]
(11.19)
\[
i, j = 1,\ldots, n
\]
for approximate values \( {u}_{ij} \) to the exact solution \( u\left( {x}_{ij}\right) \) . Here we have set \( {q}_{ij} \mathrel{\text{:=}} q\left( {x}_{ij}\right) \) and \( {r}_{ij} \mathrel{\text{:=}} r\left( {x}_{ij}\right) \) . This system has to be complemented by the boundary conditions
\[
{u}_{0, j} = {u}_{n + 1, j} = 0,\;j = 0,\ldots, n + 1,
\]
(11.20)
\[
{u}_{i,0} = {u}_{i, n + 1} = 0,\;i = 1,\ldots, n.
\]
We refrain from rewriting the system (11.19)-(11.20) in matrix notation and refer back to Example 2.2. Analogously to Theorem 11.5, it can be seen that the Jacobi iterations converge (and relaxation methods and multigrid methods are applicable). Hence we have the following theorem.
Theorem 11.9 For each \( h > 0 \) the difference equations (11.19)-(11.20) have a unique solution.
From the proof of Lemma 11.6 it can be seen that its statement also holds for the corresponding matrices of the system (11.19)-(11.2 | Theorem 11.5 For each \( h > 0 \) the difference equations (11.10)-(11.11) have a unique solution. | The tridiagonal matrix \( A \) is irreducible and weakly row-diagonally dominant. Hence, by Theorem 4.7, the matrix \( A \) is invertible, and the Jacobi iterations converge. |
Corollary 3.4.6. Let \( 0 < {p}_{0} < \infty \) . Then for any \( p \) with \( {p}_{0} \leq p < \infty \) and for all locally integrable functions \( f \) on \( {\mathbf{R}}^{n} \) with \( {M}_{d}\left( f\right) \in {L}^{{p}_{0}}\left( {\mathbf{R}}^{n}\right) \) we have
\[
\parallel f{\parallel }_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \leq {C}_{n}\left( p\right) {\begin{Vmatrix}{M}^{\# }\left( f\right) \end{Vmatrix}}_{{L}^{p}\left( {\mathbf{R}}^{n}\right) },
\]
(3.4.11)
where \( {C}_{n}\left( p\right) \) is the constant in Theorem 3.4.5.
Proof. Since for every point in \( {\mathbf{R}}^{n} \) there is a sequence of dyadic cubes shrinking to it, the Lebesgue differentiation theorem yields that for almost every point \( x \) in \( {\mathbf{R}}^{n} \) the averages of the locally integrable function \( f \) over the dyadic cubes containing \( x \) converge to \( f\left( x\right) \) . Consequently,
\[
\left| f\right| \leq {M}_{d}\left( f\right) \;\text{ a.e. }
\]
Using this fact, the proof of (3.4.11) is immediate, since
\[
\parallel f{\parallel }_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \leq {\begin{Vmatrix}{M}_{d}\left( f\right) \end{Vmatrix}}_{{L}^{p}\left( {\mathbf{R}}^{n}\right) },
\]
and by Theorem 3.4.5 the latter is controlled by \( {C}_{n}\left( p\right) {\begin{Vmatrix}{M}^{\# }\left( f\right) \end{Vmatrix}}_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \) .
Estimate (3.4.11) provides the sought converse to (3.4.1).
## 3.4.3 Interpolation Using BMO
We continue this section by proving an interpolation result in which the space \( {L}^{\infty } \) is replaced by \( {BMO} \) . The sharp function plays a key role in the following theorem.
Theorem 3.4.7. Let \( 1 \leq {p}_{0} < \infty \) . Let \( T \) be a linear operator that maps \( {L}^{{p}_{0}}\left( {\mathbf{R}}^{n}\right) \) to \( {L}^{{p}_{0}}\left( {\mathbf{R}}^{n}\right) \) with bound \( {A}_{0} \), and \( {L}^{\infty }\left( {\mathbf{R}}^{n}\right) \) to \( \operatorname{BMO}\left( {\mathbf{R}}^{n}\right) \) with bound \( {A}_{1} \) . Then for all \( p \) with \( {p}_{0} < p < \infty \) there is a constant \( {C}_{n, p} \) such that for all \( f \in {L}^{p} \) we have
\[
\parallel T\left( f\right) {\parallel }_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \leq {C}_{n, p,{p}_{0}}{A}_{0}^{\frac{{p}_{0}}{p}}{A}_{1}^{1 - \frac{{p}_{0}}{p}}\parallel f{\parallel }_{{L}^{p}\left( {\mathbf{R}}^{n}\right) }.
\]
(3.4.12)
Remark 3.4.8. In certain applications, the operator \( T \) may not be a priori defined on all of \( {L}^{{p}_{0}} + {L}^{\infty } \) but only on some subspace of it. In this case one may state that the hypotheses and the conclusion of the preceding theorem hold for a subspace of these spaces.
Proof. We consider the operator
\[
S\left( f\right) = {M}^{\# }\left( {T\left( f\right) }\right)
\]
defined for \( f \in {L}^{{p}_{0}} + {L}^{\infty } \) . It is easy to see that \( S \) is a sublinear operator. We prove that \( S \) maps \( {L}^{\infty } \) to itself and \( {L}^{{p}_{0}} \) to itself if \( {p}_{0} > 1 \) or \( {L}^{1} \) to \( {L}^{1,\infty } \) if \( {p}_{0} = 1 \) . For \( f \in {L}^{{p}_{0}} \) we have
\[
\parallel S\left( f\right) {\parallel }_{{L}^{{p}_{0}}} = {\begin{Vmatrix}{M}^{\# }\left( T\left( f\right) \right) \end{Vmatrix}}_{{L}^{{p}_{0}}} \leq 2{\begin{Vmatrix}{M}_{c}\left( T\left( f\right) \right) \end{Vmatrix}}_{{L}^{{p}_{0}}}
\]
\[
\leq {C}_{n,{p}_{0}}\parallel T\left( f\right) {\parallel }_{{L}^{{p}_{0}}} \leq {C}_{n,{p}_{0}}{A}_{0}\parallel f{\parallel }_{{L}^{{p}_{0}}}
\]
where the three \( {L}^{{p}_{0}} \) norms on the top line should be replaced by \( {L}^{1,\infty } \) if \( {p}_{0} = 1 \) . For \( f \in {L}^{\infty } \) one has
\[
\parallel S\left( f\right) {\parallel }_{{L}^{\infty }} = {\begin{Vmatrix}{M}^{\# }\left( T\left( f\right) \right) \end{Vmatrix}}_{{L}^{\infty }} = \parallel T\left( f\right) {\parallel }_{BMO} \leq {A}_{1}\parallel f{\parallel }_{{L}^{\infty }}.
\]
Interpolating between these estimates using Theorem 1.3.2 in [156], we deduce
\[
{\begin{Vmatrix}{M}^{\# }\left( T\left( f\right) \right) \end{Vmatrix}}_{{L}^{p}} = \parallel S\left( f\right) {\parallel }_{{L}^{p}} \leq {C}_{p,{p}_{0}}{A}_{0}^{\frac{{p}_{0}}{p}}{A}_{1}^{1 - \frac{{p}_{0}}{p}}\parallel f{\parallel }_{{L}^{p}}
\]
for all \( f \in {L}^{p} \), where \( {p}_{0} < p < \infty \) .
Consider now a function \( h \in {L}^{p} \cap {L}^{{p}_{0}} \) . In the case \( {p}_{0} > 1,{M}_{d}\left( {T\left( h\right) }\right) \in {L}^{{p}_{0}} \) ; hence Corollary 3.4.6 is applicable and gives
\[
\parallel T\left( h\right) {\parallel }_{{L}^{p}} \leq {C}_{n}\left( p\right) {C}_{p,{p}_{0}}{A}_{0}^{\frac{{p}_{0}}{p}}{A}_{1}^{1 - \frac{{p}_{0}}{p}}\parallel h{\parallel }_{{L}^{p}}.
\]
Density yields the same estimate for all \( f \in {L}^{p}\left( {\mathbf{R}}^{n}\right) \) . If \( {p}_{0} = 1 \), one applies the same idea but needs the endpoint estimate of Exercise 3.4.6, since \( {M}_{d}\left( {T\left( h\right) }\right) \in {L}^{1,\infty } \) .
## 3.4.4 Estimates for Singular Integrals Involving the Sharp Function
We use the sharp function to obtain pointwise estimates for singular integrals. These enable us to recover previously obtained estimates for singular integrals, but also to deduce a new endpoint boundedness result from \( {L}^{\infty } \) to \( {BMO} \) .
We recall some facts about singular integral operators. Suppose that \( K \) is a function defined on \( {\mathbf{R}}^{n} \smallsetminus \{ 0\} \) that satisfies
\[
\left| {K\left( x\right) }\right| \leq {A}_{1}{\left| x\right| }^{-n}
\]
(3.4.13)
\[
\left| {K\left( {x - y}\right) - K\left( x\right) }\right| \leq {A}_{2}{\left| y\right| }^{\delta }{\left| x\right| }^{-n - \delta }\;\text{ when }\left| x\right| \geq 2\left| y\right| > 0,
\]
(3.4.14)
\[
\mathop{\sup }\limits_{{r < R < \infty }}\left| {{\int }_{r \leq \left| x\right| \leq R}K\left( x\right) {dx}}\right| \leq {A}_{3}
\]
(3.4.15)
Let \( W \) be a tempered distribution that coincides with \( K \) on \( {\mathbf{R}}^{n} \smallsetminus \{ 0\} \) and let \( T \) be the linear operator given by convolution with \( W \) .
Under these assumptions we have that \( T \) is \( {L}^{2} \) bounded with norm at most a constant multiple of \( {A}_{1} + {A}_{2} + {A}_{3} \) (Theorem 5.4.1 in [156]), and hence it is also \( {L}^{p} \) bounded with a similar norm on \( {L}^{p} \) for \( 1 < p < \infty \) (Theorem 5.3.3 in [156]).
Theorem 3.4.9. Let \( T \) be given by convolution with a distribution \( W \) that coincides with a function \( K \) on \( {\mathbf{R}}^{n} \smallsetminus \{ 0\} \) satisfying (3.4.14). Assume that \( T \) has an extension that is \( {L}^{2} \) bounded with a norm \( B \) . Then there is a constant \( {C}_{n} \) such that for any \( s > 1 \) the estimate
\[
{M}^{\# }\left( {T\left( f\right) }\right) \left( x\right) \leq {C}_{n}\left( {{A}_{2} + B}\right) \max \left( {s,{\left( s - 1\right) }^{-1}}\right) M{\left( {\left| f\right| }^{s}\right) }^{\frac{1}{s}}\left( x\right)
\]
(3.4.16)
is valid for all \( f \) in \( \mathop{\bigcup }\limits_{{s \leq p < \infty }}{L}^{p} \) and all \( x \in {\mathbf{R}}^{n} \) .
Proof. In view of Proposition 3.4.2 (2), given any cube \( Q \), it suffices to find a constant \( {a}_{Q} \) such that
\[
\frac{1}{\left| Q\right| }{\int }_{Q}\left| {T\left( f\right) \left( y\right) - {a}_{Q}}\right| {dy} \leq {C}_{n}\max \left( {s,{\left( s - 1\right) }^{-1}}\right) \left( {{A}_{2} + B}\right) M{\left( {\left| f\right| }^{s}\right) }^{\frac{1}{s}}\left( x\right)
\]
(3.4.17)
for almost all \( x \in Q \) . To prove this estimate we employ a well-known theme. We write \( f = {f}_{Q}^{0} + {f}_{Q}^{\infty } \), where \( {f}_{Q}^{0} = f{\chi }_{6\sqrt{n}Q} \) and \( {f}_{Q}^{\infty } = f{\chi }_{{\left( 6\sqrt{n}Q\right) }^{c}} \) . Here \( 6\sqrt{n}Q \) denotes the cube that is concentric with \( Q \), has sides parallel to those of \( Q \), and has side length \( 6\sqrt{n}\ell \left( Q\right) \), where \( \ell \left( Q\right) \) is the side length of \( Q \) .
We now fix an \( f \) in \( \mathop{\bigcup }\limits_{{s \leq p < \infty }}{L}^{p} \) and we select \( {a}_{Q} = T\left( {f}_{Q}^{\infty }\right) \left( x\right) \) . Then \( {a}_{Q} \) is finite (and thus well defined) for all \( x \in Q \) . Indeed, for all \( x \in Q \) ,(3.4.13) yields
\[
\left| {T\left( {f}_{Q}^{\infty }\right) \left( x\right) }\right| = \left| {{\int }_{{\left( {Q}^{ * }\right) }^{c}}f\left( y\right) K\left( {x - y}\right) {dy}}\right| \leq \parallel f{\parallel }_{{L}^{p}}{\left( {\int }_{\left| {x - y}\right| \geq {c}_{n}\ell \left( Q\right) }\frac{{A}_{1}^{{p}^{\prime }}{dy}}{{\left| x - y\right| }^{n{p}^{\prime }}}\right) }^{\frac{1}{{p}^{\prime }}} < \infty ,
\]
where \( {c}_{n} \) is a positive constant. It follows that
\[
\frac{1}{\left| Q\right| }{\int }_{Q}\left| {T\left( f\right) \left( y\right) - {a}_{Q}}\right| {dy}
\]
\[
\leq \frac{1}{\left| Q\right| }{\int }_{Q}\left| {T\left( {f}_{Q}^{0}\right) \left( y\right) }\right| {dy} + \frac{1}{\left| Q\right| }{\int }_{Q}\left| {T\left( {f}_{Q}^{\infty }\right) \left( y\right) - T\left( {f}_{Q}^{\infty }\right) \left( x\right) }\right| {dy}.
\]
(3.4.18)
In view of Theorem 5.3.3 in [156], \( T \) maps \( {L}^{s} \) to \( {L}^{s} \) with norm at most a dimensional constant multiple of \( \max \left( {s,{\left( s - 1\right) }^{-1}}\right) \left( {B + {A}_{2}}\right) \) . The first term in (3.4.18) is controlled by
\[
{\left( \frac{1}{\left| Q\right| }{\int }_{Q}{\left| T\left( {f}_{Q}^{0}\right) \left( y\right) \right| }^{s}dy\right) }^{\frac{1}{s}} \leq {C}_{n}\max \left( {s,{\left( s - 1\right) }^{-1}}\right) \left( {B + {A}_{2}}\right) {\left( \frac{1}{\left| Q\right| }{\int }_{{\mathbf{R}}^{n}}{\left| {f}_{Q}^{0}\left( y\right) \right| }^{s}dy\right) }^{\frac{1}{s}}
\]
\[
\leq {C}_{n}^{\prime }\max \left( {s,{\left( s - 1\right) }^{-1}}\right) \left( {B + {A}_{2}}\right) M{\left( {\left| f\right| }^{s}\right) }^ | Corollary 3.4.6. Let \( 0 < {p}_{0} < \infty \) . Then for any \( p \) with \( {p}_{0} \leq p < \infty \) and for all locally integrable functions \( f \) on \( {\mathbf{R}}^{n} \) with \( {M}_{d}\left( f\right) \in {L}^{{p}_{0}}\left( {\mathbf{R}}^{n}\right) \) we have
\[
\parallel f{\parallel }_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \leq {C}_{n}\left( p\right) {\begin{Vmatrix}{M}^{\# }\left( f\right) \end{Vmatrix}}_{{L}^{p}\left( {\mathbf{R}}^{n}\right) },
\]
where \( {C}_{n}\left( p\right) \) is the constant in Theorem 3.4.5. | Proof. Since for every point in \( {\mathbf{R}}^{n} \) there is a sequence of dyadic cubes shrinking to it, the Lebesgue differentiation theorem yields that for almost every point \( x \) in \( {\mathbf{R}}^{n} \) the averages of the locally integrable function \( f \) over the dyadic cubes containing \( x \) converge to \( f\left( x\right) \) . Consequently,
\[
\left| f\right| \leq {M}_{d}\left( f\right) \;\text{ a.e. }
\]
Using this fact, the proof of (3.4.11) is immediate, since
\[
\parallel f{\parallel }_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \leq {\begin{Vmatrix}{M}_{d}\left( f\right) \end{Vmatrix}}_{{L}^{p}\left( {\mathbf{R}}^{n}\right) },
\]
and by Theorem 3.4.5 the latter is controlled by \( {C}_{n}\left( p\right) {\begin{Vmatrix}{M}^{\# }\left( f\right) \end{Vmatrix}}_{{L}^{p}\left( {\mathbf{R}}^{n}\right) } \) . |
Lemma 12.2.2 The set of lines spanned by the vectors of \( {D}_{n} \) is star-closed.
## 12.3 Reflections
We can characterize star-closed sets of lines at \( {60}^{ \circ } \) and \( {90}^{ \circ } \) in terms of their symmetries. If \( h \) is a vector in \( {\mathbb{R}}^{n} \), then there is a unique hyperplane through the origin perpendicular to \( h \) . Let \( {\rho }_{h} \) denote the operation of reflection in this hyperplane. Simple calculations reveal that for all \( x \) ,
\[
{\rho }_{h}\left( x\right) = x - 2\frac{\langle x, h\rangle }{\langle h, h\rangle }h
\]
We make a few simple observations. It is easy to check that \( {\rho }_{h}\left( h\right) = - h \) . Also, \( {\rho }_{h}\left( x\right) = x \) if and only if \( \langle h, x\rangle = 0 \) . The product \( {\rho }_{a}{\rho }_{b} \) of two reflections is not in general a reflection. It can be shown that \( {\rho }_{a}{\rho }_{b} = {\rho }_{b}{\rho }_{a} \) if and only if either \( \langle a\rangle = \langle b\rangle \) or \( \langle a, b\rangle = 0 \) .
Lemma 12.3.1 Let \( \mathcal{L} \) be a set of lines at \( {60}^{ \circ } \) and \( {90}^{ \circ } \) in \( {\mathbb{R}}^{n} \) . Then \( \mathcal{L} \) is star-closed if and only if for every vector \( h \) that spans a line in \( \mathcal{L} \), the reflection \( {\rho }_{h} \) fixes \( \mathcal{L} \) .
Proof. Let \( h \) be a vector of length \( \sqrt{2} \) spanning a line in \( \mathcal{L} \) . From our comments above, \( {\rho }_{h} \) fixes \( \langle h\rangle \) and all the lines orthogonal to \( \langle h\rangle \) . So suppose that \( \langle a\rangle \) is a line of \( \mathcal{L} \) at \( {60}^{ \circ } \) to \( \langle h\rangle \) . Without loss of generality we can assume that \( a \) has length \( \sqrt{2} \) and that \( \langle h, a\rangle = - 1 \) . Now,
\[
{\rho }_{h}\left( a\right) = a - 2\frac{\left( -1\right) }{2}h = a + h,
\]
and \( \langle a + h\rangle \) forms a star with \( \langle a\rangle \) and \( \langle h\rangle \) . This implies that \( {\rho }_{h} \) fixes \( \mathcal{L} \) if and only if \( \mathcal{L} \) is star-closed.
A root system is a set \( S \) of vectors in \( {\mathbb{R}}^{n} \) such that
(a) if \( h \in S \), then \( \langle h\rangle \cap S = \{ h, - h\} \) ;
(b) if \( h \in S \), then \( {\rho }_{h}\left( S\right) = S \) .
Lemma 12.3.1 shows that if \( \mathcal{L} \) is a star-closed set of lines at \( {60}^{ \circ } \) and \( {90}^{ \circ } \) in \( {\mathbb{R}}^{m} \), then the vectors of length \( \sqrt{2} \) that span these lines form a root system. For example, the set \( {D}_{n} \) is a root system.
The group generated by the reflections \( {\rho }_{h} \), for \( h \) in \( S \), is the reflection group of the root system. The symmetry group of a set of lines or vectors in \( {\mathbb{R}}^{n} \) is the group of all orthogonal transformations that take the set to itself. The symmetry group of a root system or of a set of lines always contains multiplication by -1 .
## 12.4 Indecomposable Star-Closed Sets
A set \( \mathcal{L} \) of lines at \( {60}^{ \circ } \) and \( {90}^{ \circ } \) is called decomposable if it can be partitioned into two subsets \( {\mathcal{L}}_{1} \) and \( {\mathcal{L}}_{2} \) such that every line in \( {\mathcal{L}}_{1} \) is orthogonal to every line in \( {\mathcal{L}}_{2} \) . If there is no such partition, then it is called indecomposable.
Lemma 12.4.1 For \( n \geq 2 \), the set of lines \( \mathcal{L} \) spanned by the vectors in \( {D}_{n} \) is indecomposable.
Proof. The lines \( \left\langle {{e}_{1} + {e}_{i}}\right\rangle \) for \( i \geq 2 \) have pairwise inner products equal to 1, and hence must be in the same part of any decomposition of \( \mathcal{L} \) . It is clear, however, that any other vector in \( {D}_{n} \) has nonzero inner product with at least one of these vectors, and so there are no lines orthogonal to all of this set.
Theorem 12.4.2 Let \( \mathcal{L} \) be a star-closed indecomposable set of lines at \( {60}^{ \circ } \) and \( {90}^{ \circ } \) . Then the reflection group of \( \mathcal{L} \) acts transitively on ordered pairs of nonorthogonal lines.
Proof. First we observe that the reflection group acts transitively on the lines of \( \mathcal{L} \) . Suppose that \( \langle a\rangle \) and \( \langle b\rangle \) are two lines that are not orthogonal, and that \( \langle a, b\rangle = - 1 \) . Then \( c = - a - b \) spans the third line in the star with \( \langle a\rangle \) and \( \langle b\rangle \), and the reflection \( {\rho }_{c} \) swaps \( \langle a\rangle \) and \( \langle b\rangle \) . Therefore, \( \langle a\rangle \) can be mapped on to any line not orthogonal to it. Let \( {\mathcal{L}}^{\prime } \) be the orbit of \( \langle a\rangle \) under the reflection group of \( \mathcal{L} \) . Then every line in \( \mathcal{L} \smallsetminus {\mathcal{L}}^{\prime } \) is orthogonal to every line of \( {\mathcal{L}}^{\prime } \) . Since \( \mathcal{L} \) is indecomposable, this shows that \( {\mathcal{L}}^{\prime } = \mathcal{L} \) .
Now, suppose that \( \left( {\langle a\rangle ,\langle b\rangle }\right) \) and \( \left( {\langle a\rangle ,\langle c\rangle }\right) \) are two ordered pairs of nonorthogonal lines. We will show that there is a reflection that fixes \( \langle a\rangle \) and exchanges \( \langle b\rangle \) and \( \langle c\rangle \) . Assume that \( a, b \), and \( c \) have length \( \sqrt{2} \) and that \( \langle a, b\rangle = \langle a, c\rangle = - 1 \) . Then the vector \( - a - b \) has length \( \sqrt{2} \) and spans a line in \( \mathcal{L} \) . Now,
\[
1 = \langle c, - a\rangle = \langle c, b\rangle + \langle c, - a - b\rangle .
\]
If \( c = b \) or \( c = - a - b \), then \( \langle c\rangle \) and \( \langle b\rangle \) are exchanged by the identity reflection or \( {\rho }_{a} \), respectively. Otherwise, \( c \) has inner product 1 with precisely one of the vectors in \( \{ b, - a - b\} \), and is orthogonal to the other. Exchanging the roles of \( b \) and \( - a - b \) if necessary, we can assume that \( \langle c, b\rangle = 1 \) . Then \( \langle b - c\rangle \in \mathcal{L} \), and the reflection \( {\rho }_{b - c} \) fixes \( \langle a\rangle \) and exchanges \( \langle b\rangle \) and \( \langle c\rangle \) . \( ▱ \)
Now, suppose that \( X \) is a graph with minimum eigenvalue at least -2 . Then \( A\left( X\right) + {2I} \) is the Gram matrix of a set of vectors of length \( \sqrt{2} \) that span a set of lines at \( {60}^{ \circ } \) and \( {90}^{ \circ } \) . Let \( \mathcal{L}\left( X\right) \) denote the star-closure of this set of lines. Notice that the Gram matrix determines the set of vectors up to orthogonal transformations of the underlying vector space, and therefore \( \mathcal{L}\left( X\right) \) is uniquely determined up to orthogonal transformations.
Lemma 12.4.3 If \( X \) is a graph with minimum eigenvalue at least -2, then the star-closed set of lines \( \mathcal{L}\left( X\right) \) is indecomposable if and only if \( X \) is connected.
Proof. First suppose that \( X \) is connected. Let \( {\mathcal{L}}^{\prime } \) be the lines spanned by the vectors whose Gram matrix is \( A\left( X\right) + {2I} \) . Lines corresponding to adjacent vertices of \( X \) are not orthogonal, and hence must be in the same part of any decomposition of \( \mathcal{L}\left( X\right) \) . Therefore, all the lines in \( {\mathcal{L}}^{\prime } \) are in the same part. Any line lying in a star with two other lines is not orthogonal to either of them, and therefore lies in the same part of any decomposition of \( \mathcal{L}\left( X\right) \) . Hence the star-closure of \( {\mathcal{L}}^{\prime } \) is all in the same part of any decomposition, which shows that \( \mathcal{L}\left( X\right) \) is indecomposable.
If \( X \) is not connected, then \( {\mathcal{L}}^{\prime } \) has a decomposition into two parts. Any line orthogonal to two lines in a star is orthogonal to all three lines of the star, and so any line added to \( {\mathcal{L}}^{\prime } \) to complete a star can be assigned to one of the two parts of the decomposition, eventually yielding a decomposition of \( \mathcal{L} \) .
Therefore, we see that any connected graph with minimum eigenvalue at least -2 is associated with a star-closed indecomposable set of lines. Our strategy will be to classify all such sets, and thereby classify all the graphs with minimum eigenvalue at least -2 .
## 12.5 A Generating Set
We now show that any indecomposable star-closed set of lines \( \mathcal{L} \) at \( {60}^{ \circ } \) and \( {90}^{ \circ } \) is the star-closure of a special subset of those lines. Eventually, we will see that the structure of this subset is very restricted.
Lemma 12.5.1 Let \( \mathcal{L} \) be an indecomposable star-closed set of lines at \( {60}^{ \circ } \) and \( {90}^{ \circ } \), and let \( \langle a\rangle ,\langle b\rangle \), and \( \langle c\rangle \) form a star in \( \mathcal{L} \) . Every other line of \( \mathcal{L} \) is orthogonal to either one or three lines in the star.
Proof. Without loss of generality we may assume that \( a, b \), and \( c \) all have length \( \sqrt{2} \) and that
\[
\langle a, b\rangle = \langle b, c\rangle = \langle c, a\rangle = - 1.
\]
It follows then that \( c = - a - b \), and so for any other line \( \langle x\rangle \) of \( \mathcal{L} \) we have
\[
\langle x, a\rangle + \langle x, b\rangle + \langle x, c\rangle = 0.
\]
Because each of the terms is in \( \{ 0, \pm 1\} \), we see that either all three terms are zero or the three terms are \( 1,0 \), and -1 in some order.
Now, fix a star \( \langle a\rangle ,\langle b\rangle \), and \( \langle c\rangle \), and as above choose \( a, b \), and \( c \) to be vectors of length \( \sqrt{2} \) with pairwise inner products -1 . Let \( D \) be the set of lines of \( \mathcal{L} \) that are orthogonal to all three lines in the star. The remaining lines of \( | Lemma 12.2.2 The set of lines spanned by the vectors of \( {D}_{n} \) is star-closed. | null |
Lemma 9.3.7. If \( \left( {{g}_{1},{S}_{1}}\right) \sim \left( {{g}_{2},{S}_{2}}\right) \) then for all \( i\left( {{g}_{1},{e}_{i}\left( {S}_{1}\right) }\right) \sim \left( {{g}_{2},{e}_{i}\left( {S}_{2}\right) }\right) \) .
Proof. There exist \( e \) and \( T \) such that \( \left( {e\left( {S}_{i}\right), T}\right) \) is a balanced pair representing \( {g}_{i} \) for \( i = 1,2 \) . We write \( \left( {{g}_{1},{S}_{1}}\right) \underset{k}{ \sim }\left( {{g}_{2},{S}_{2}}\right) \) if the length of this \( e \) is \( \leq k \) . The lemma is proved by induction on \( k \) using 9.3.3.
Lemma 9.3.8. Each \( {E}_{i} \) is an \( F \) -function; i.e., \( {E}_{i}\left( {g\left\lbrack {h, T}\right\rbrack }\right) = g{E}_{i}\left( \left\lbrack {h, T}\right\rbrack \right) \) .
Define \( f : B \rightarrow \mathbb{N} \) by \( f\left( \left\lbrack {h, T}\right\rbrack \right) = \) the number of leaves of \( T \) . This is well defined.
Lemma 9.3.9. Let \( n \geq i \) . The function \( {E}_{i} \) maps \( {f}^{-1}\left( n\right) \) bijectively onto \( {f}^{-1}\left( {n + 1}\right) \) .
Proof. That \( {E}_{i} \) is injective is clear. The proof that \( {E}_{i} \) is surjective should be clear from Example 9.3.10, below.
Let \( {B}_{n} = {f}^{-1}\left( \left\lbrack {1, n}\right\rbrack \right) \) . Then 9.3.9 says that when \( n \geq i,{E}_{i} \) maps \( B - {B}_{n - 1} \) bijectively onto \( B - {B}_{n} \) . We call \( {E}_{i} \) a simple expansion operator on \( B - {B}_{n - 1} \) and we call its inverse \( {C}_{i} : B - {B}_{n} \rightarrow B - {B}_{n - 1} \) a simple contraction operator.
Example 9.3.10. Let \( T \) have more than \( i \) leaves. If the \( {i}^{\text{th }} \) and \( {\left( i + 1\right) }^{\text{th }} \) leaves form a caret (i.e., \( T = {e}_{i}\left( {T}^{\prime }\right) \) for some \( {T}^{\prime } \in \mathcal{T} \) ) then \( {C}_{i}\left( \left\lbrack {h, T}\right\rbrack \right) = \left\lbrack {h,{T}^{\prime }}\right\rbrack \) . If these leaves do not form a caret, let \( S \) be any tree with the same number of leaves as \( T \) such that \( S = {e}_{i}\left( {S}^{\prime }\right) \) for some \( {S}^{\prime } \in \mathcal{T} \) . Let \( g \in F \) be represented by \( \left( {S, T}\right) \) . Then \( \left\lbrack {g, S}\right\rbrack = \left\lbrack {1, T}\right\rbrack \), so \( \left\lbrack {{hg}, S}\right\rbrack = \left\lbrack {h, T}\right\rbrack \) and \( {C}_{i}\left( \left\lbrack {h, T}\right\rbrack \right) = \left\lbrack {{hg},{S}^{\prime }}\right\rbrack \) .
We make \( B \) into an \( F \) -poset by defining \( \left\lbrack {g, S}\right\rbrack \leq \left\lbrack {h, T}\right\rbrack \) if for some \( {i}_{1},{i}_{2},\ldots ,{i}_{r} \), with \( r \geq 0,{E}_{{i}_{r}} \circ \cdots \circ {E}_{{i}_{1}}\left( \left\lbrack {g, S}\right\rbrack \right) = \left\lbrack {h, T}\right\rbrack \) . Each \( {B}_{n} \) is an \( F \) - sub-poset of \( B \) .
Lemma 9.3.11. The \( F \) -action on the set \( B \) is a free action.
## D. Finiteness Properties of \( F \) :
As usual we reuse the symbol \( B \) for the associated ordered abstract simplicial complex defined by the poset \( B \) .
Proposition 9.3.12. The induced \( F \) -action on \( \left| B\right| \) is a free action.
Proof. The stabilizers of vertices are trivial by 9.3.11. Since the action of \( F \) on \( \left\lbrack {h, S}\right\rbrack \) preserves the number of leaves of \( S \), the stabilizer of each simplex of \( S \) is also trivial.
Proposition 9.3.13. The poset \( B \) is a directed set.
Proof. Let \( \left\{ {{b}_{1},\ldots ,{b}_{k}}\right\} \subset B \) . Write \( {b}_{i} = \left\lbrack {{h}_{i},{S}_{i}}\right\rbrack \) . Then \( {S}_{i} \) expands to \( {S}_{i}^{\prime } \) such that \( \left( {{h}_{i},{S}_{i}^{\prime }}\right) \sim \left( {{1}_{F},{T}_{i}^{\prime }}\right) \) for some \( {T}_{i}^{\prime } \) . Write \( {b}_{i}^{\prime } = \left\lbrack {{h}_{i},{S}_{i}^{\prime }}\right\rbrack = \left\lbrack {{1}_{F},{T}_{i}^{\prime }}\right\rbrack \) . Then \( {b}_{i} \leq {b}_{i}^{\prime } \)
for all \( i \) . Let \( T = \mathop{\bigcup }\limits_{{i = 1}}^{k}{T}_{i}^{\prime } \) and write \( b = \left\lbrack {{1}_{F}, T}\right\rbrack \) . Then \( {b}_{i}^{\prime } \leq b \) for all \( i \) .
Proposition 9.3.14. If a poset \( P \) is a directed set then \( \left| P\right| \) is contractible.
Proof. Whenever \( K \) is a finite subcomplex of \( P \) there exists \( v \in P \) such that the cone \( v * \left| K\right| \) is a subcomplex of \( \left| P\right| \) . Thus the homotopy groups of \( \left| P\right| \) are trivial, so, by the Whitehead Theorem, \( \left| P\right| \) is contractible.
Corollary 9.3.15. \( \left| B\right| \) is contractible.
The function \( f : B \rightarrow \mathbb{N} \) extends affinely to a Morse function \( {}^{8} \) (also denoted by) \( f : \left| B\right| \rightarrow \mathbb{R} \) . Then \( \left| {B}_{n}\right| = {f}^{-1}(\left( {-\infty, n\rbrack }\right) \) .
Proposition 9.3.16. The \( {CW} \) complex \( F \smallsetminus \left| {B}_{n}\right| \) is finite.
Proof. Let \( \left\lbrack {h, S}\right\rbrack \in {B}_{n} \) . Its \( F \) -orbit contains \( \left\lbrack {1, S}\right\rbrack \) and there are only finitely many finite trees having at most \( n \) leaves. This shows that the 0 -skeleton of \( F \smallsetminus \left| {B}_{n}\right| \) is finite. The rest is clear.
Proposition 9.3.17. For each integer \( k \) there is an integer \( m\left( k\right) \) such that if \( b \) is a vertex of \( B \) and \( f\left( b\right) \geq m\left( k\right) \), then the downward link \( {\operatorname{lk}}_{\left| B\right| }^{ \downarrow }b \) is \( k \) - connected.
We postpone the proof of 9.3.17 until the next subsection.
Proposition 9.3.18. For \( n \geq m\left( k\right) ,\left| {B}_{n}\right| \) is \( k \) -connected. Hence \( \left\{ \left| {B}_{n}\right| \right\} \) is essentially \( k \) -connected for all \( k \) .
Proof. By 9.3.17 and 8.3.4 we conclude that \( \left( {\left| {B}_{n}\right| ,\left| {B}_{n - 1}\right| }\right) \) is \( \left( {k + 1}\right) \) -connected if \( n \geq m\left( k\right) \) . When combined with the Whitehead Theorem and 9.3.15 this proves what is claimed.
By 7.4.1 and 7.2.2 we conclude:
Theorem 9.3.19. (Brown-Geoghegan Theorem) Thompson's Group \( F \) has type \( {F}_{\infty } \) .
## E. Analysis of the downward links:
It remains to prove 9.3.17. We begin with two topics of general interest (9.3.20 and 9.3.21).
If \( \mathcal{U} = \left\{ {X}_{\alpha }\right\} \) is a cover of the CW complex \( X \) by subcomplexes, the nerve of \( \mathcal{U} \) is the abstract simplicial complex \( N\left( \mathcal{U}\right) \) having a vertex \( {v}_{\alpha } \) for each \( {X}_{\alpha } \) ,
and a simplex \( \left\{ {{v}_{{\alpha }_{0}},\ldots ,{v}_{{\alpha }_{k}}}\right\} \) whenever \( \mathop{\bigcap }\limits_{{i = 0}}^{k}{X}_{{\alpha }_{i}} \neq \varnothing \) . The following property of nerves is widely used in topology.
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\( {}^{8} \) See Section 8.3.
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Proposition 9.3.20. If the cover \( \mathcal{U} \) is finite and if \( \mathop{\bigcap }\limits_{{i = 0}}^{k}{X}_{{\alpha }_{i}} \) is contractible whenever it is non-empty, then \( \left| {N\left( \mathcal{U}\right) }\right| \) and \( X \) are homotopy equivalent.
Proof. There is a vertex \( {v}_{{\alpha }_{0},\ldots ,{\alpha }_{k}} \) of the first derived \( {sd}\left| {N\left( \mathcal{U}\right) }\right| \) for each simplex \( \left\{ {{v}_{{\alpha }_{0}},\ldots ,{v}_{{\alpha }_{k}}}\right\} \) of \( N\left( \mathcal{U}\right) \) . Pick a point \( {x}_{{\alpha }_{0},\ldots ,{\alpha }_{k}} \in \mathop{\bigcap }\limits_{{i = 0}}^{k}{X}_{{\alpha }_{i}} \) . Define a map \( \alpha : \operatorname{sd}\left| {N\left( \mathcal{U}\right) }\right| \rightarrow X \) taking each vertex \( {v}_{{\alpha }_{0},\ldots ,{\alpha }_{k}} \) to the point \( {x}_{{\alpha }_{0},\ldots ,{\alpha }_{k}} \) ; the contractibility hypothesis makes it possible to extend \( \alpha \) so that the simplex whose first \( {}^{9} \) vertex is \( {v}_{{\alpha }_{0},\ldots ,{\alpha }_{k}} \) is mapped into \( \mathop{\bigcap }\limits_{{i = 0}}^{k}{X}_{{\alpha }_{i}} \) . A straightforward generalization of the proof of 4.1.5 shows that \( \alpha \) is a homotopy equivalence.
The particular nerve to which this will be applied is the abstract simplicial \( {\operatorname{complex}}^{10}{L}_{n} \) whose simplexes are the sets of pairwise disjoint adjacent pairs in the ordered set \( \left( {1,2,\ldots, n}\right) \) .
Proposition 9.3.21. For any integer \( k \geq 0 \) there is an integer \( m\left( k\right) \) such that \( \left| {L}_{n}\right| \) is \( k \) -connected when \( n \geq m\left( k\right) \) .
Proof. By induction on \( k \) we prove a sharper statement:
Claim: given \( k \geq 0 \) there are integers \( m\left( k\right) \) and \( q\left( k\right) \) such that when \( n \geq \) \( m\left( k\right) \) the \( k \) -skeleton \( {\left| {L}_{n}\right| }^{k} \) is homotopically trivial by means of a homotopy \( {H}^{\left( k\right) } \) in which, for every simplex \( \sigma \) of \( {\left| {L}_{n}\right| }^{k},{H}^{\left( k\right) }\left( {\left| \sigma \right| \times I}\right) \) is supported by a subcomplex having \( \leq q\left( k\right) \) vertices. For \( k = 0 \) we can take \( m\left( 0\right) = 5 \) and \( q\left( 0\right) = 3 \) . Assume the Claim holds when \( k \) is replaced by \( k - 1 \geq 0 \) . For any \( k \) -simplex \( \sigma ,{H}^{\left( k - 1\right) }\left( {\left| \sigma \right| \times I}\right) \) is supported by a subcomplex \( J \) having at most \( r = \left( {k + 1}\right) \left( {q\left( {k - 1}\right) }\right) \) vertices. If \( n \geq {2r} + 2 \) there is a vertex \( v \) of \( {L}_{n} \) such that \( J \) lies in the link of \( v \) . Extend \( {H}^{\left( k - 1\right) } \) to \( {H}^{\left( k\right) } \) in two steps: first, \( {H}^{\left( k\right) } \) is to be the identity map on \( \left| \sigma | Lemma 9.3.7. If \( \left( {{g}_{1},{S}_{1}}\right) \sim \left( {{g}_{2},{S}_{2}}\right) \) then for all \( i\left( {{g}_{1},{e}_{i}\left( {S}_{1}\right) }\right) \sim \left( {{g}_{2},{e}_{i}\left( {S}_{2}\right) }\right) \) . | There exist \( e \) and \( T \) such that \( \left( {e\left( {S}_{i}\right), T}\right) \) is a balanced pair representing \( {g}_{i} \) for \( i = 1,2 \) . We write \( \left( {{g}_{1},{S}_{1}}\right) \underset{k}{ \sim }\left( {{g}_{2},{S}_{2}}\right) \) if the length of this \( e \) is \( \leq k \) . The lemma is proved by induction on \( k \) using 9.3.3. |
Theorem 4.11. Every module can be embedded into an injective module.
Noetherian rings. Our last result is due to Bass (cf. Chase [1961]).
Theorem 4.12. A ring \( R \) is left Noetherian if and only if every direct sum of injective left \( R \) -modules is injective.
Proof. Assume that every direct sum of injective left \( R \) -modules is injective, and let \( {L}_{1} \subseteq {L}_{2} \subseteq \cdots \subseteq {L}_{n} \subseteq \cdots \) be an ascending sequence of left ideals of \( R \) . Then \( L = \mathop{\bigcup }\limits_{{n > 0}}{L}_{n} \) is a left ideal of \( R \) . By 4.11 there is a monomorphism of \( R/{L}_{n} \) into an injective \( R \) -module \( {J}_{n} \) . By the hypothesis, \( J = {\bigoplus }_{n > 0}{J}_{n} \) is injective. Construct a module homomorphism \( \varphi : L \rightarrow J \) as follows. Let \( {\varphi }_{n} \) be the homomorphism \( R \rightarrow R/{L}_{n} \rightarrow {J}_{n} \) . If \( x \in L \), then \( x \in {L}_{n} \) for some \( n \), and then \( {\varphi }_{k}\left( x\right) = 0 \) for all \( k \geqq n \) . Let \( \varphi \left( x\right) = {\left( {\varphi }_{k}\left( x\right) \right) }_{k > 0} \in J \) .
Since \( J \) is injective, \( \varphi \) extends to a module homomorphism \( \psi : {}_{R}R \rightarrow J \) . Then \( \psi \left( 1\right) = {\left( {t}_{k}\right) }_{k > 0} \in J \) for some \( {t}_{k} \in {J}_{k},{t}_{k} = 0 \) for almost all \( k \) . Hence \( \psi \left( 1\right) \in {\bigoplus }_{k < n}{J}_{k} \) for some \( n \) . Then \( \varphi \left( x\right) = \psi \left( {x1}\right) = {x\psi }\left( 1\right) \in {\bigoplus }_{k < n}{J}_{k} \) for all \( x \in L \) ; in particular, \( {\varphi }_{n}\left( x\right) = 0 \) and \( x \in {L}_{n} \), since \( R/{L}_{n} \rightarrow {J}_{n} \) is injective. Thus \( L = {L}_{n} \), and then \( {L}_{k} = {L}_{n} \) for all \( k \geqq n \) .
Conversely, assume that \( R \) is left Noetherian, and let \( J = {\bigoplus }_{i \in I}{J}_{i} \) be a direct sum of injective \( R \) -modules \( {J}_{i} \) . Let \( L \) be a left ideal and let \( \varphi : L \rightarrow J \) be a module homomorphism. Since \( R \) is left Noetherian, \( L \) is finitely generated: \( L = R{r}_{1} + \cdots + R{r}_{n} \) for some \( {r}_{1},\ldots ,{r}_{n} \in L \) . Each \( \varphi \left( {r}_{k}\right) \) has finitely many nonzero components in \( {\bigoplus }_{i \in I}{J}_{i} \) and belongs to a finite direct sum \( {\bigoplus }_{i \in {S}_{k}}{J}_{i} \) . Hence there is a finite direct sum \( {\bigoplus }_{i \in S}{J}_{i}, S = {S}_{1} \cup \cdots \cup {S}_{n} \), that contains every \( \varphi \left( {r}_{k}\right) \) and contains \( \varphi \left( L\right) \) . Since \( S \) is finite, \( {\bigoplus }_{i \in S}{J}_{i} \) is injective, by 4.4; hence \( \varphi : L \rightarrow {\bigoplus }_{i \in S}{J}_{i} \) extends to a module homomorphism \( {}_{R}R \rightarrow {\bigoplus }_{i \in S}{J}_{i} \subseteq \) \( {\bigoplus }_{i \in I}{J}_{i} \) . Hence \( {\bigoplus }_{i \in I}{J}_{i} \) is injective, by 4.5. \( ▱ \)
## Exercises
1. Show that every direct summand of an injective module is injective.
2. Show that every direct sum of injective left \( R \) -modules is injective.
3. Let \( L \) be a left ideal of \( R \) and let \( \varphi : L \rightarrow M \) be a module homomorphism. Show that \( \varphi \) can be extended to \( {}_{R}R \) if and only if there exists \( m \in M \) such that \( \varphi \left( r\right) = {rm} \) for all \( r \in L \) .
4. Let \( J \) be an injective left \( R \) -module. Let \( a \in J \) and \( r \in R \) satisfy \( \operatorname{Ann}\left( r\right) \subseteq \operatorname{Ann}\left( a\right) \) (if \( t \in R \) and \( {tr} = 0 \), then \( {ta} = 0 \) ). Prove that \( a = {rx} \) for some \( x \in J \) .
5. Show that the quotient field of a Noetherian domain \( R \) is an injective \( R \) -module.
6. Find all subgroups of \( {\mathbb{Z}}_{p}\infty \) .
7. Show that \( {\mathbb{Z}}_{p}\infty \) is indecomposable.
8. Let \( U \) be the multiplicative group of all complex numbers of modulus 1 . Show that \( U\left( p\right) \cong {\mathbb{Z}}_{p}\infty \) .
9. Show that the additive group \( \mathbb{Q}/\mathbb{Z} \) is isomorphic to the direct sum \( {\bigoplus }_{p}{\mathbb{Z}}_{{p}^{\infty }} \) with one term for every prime \( p \) .
*10. Can you extend Theorem 4.9 to modules over any PID?
## 5. The Injective Hull
In this section we show that every module has, up to isomorphism, a smallest injective extension. We show this by comparing injective extensions to another kind of extensions, essential extensions.
Definition. A submodule \( S \) of a left \( R \) -module \( M \) is essential when \( S \cap T \neq 0 \) for every submodule \( T \neq 0 \) of \( M \) .
Essential submodules are also called large. Readers may prove the following:
Proposition 5.1. If \( A \subseteq B \) are submodules of \( C \), then \( A \) is essential in \( C \) if and only if \( A \) is essential in \( B \) and \( B \) is essential in \( C \) .
A monomorphism \( \varphi : A \rightarrow B \) is essential when \( \operatorname{Im}\varphi \) is an essential submodule of \( B \) .
Proposition 5.2. If \( \mu \) is an essential monomorphism, and \( \varphi \circ \mu \) is injective, then \( \varphi \) is injective.
Proof. If \( \varphi \circ \mu \) is injective, then \( \operatorname{Ker}\varphi \cap \operatorname{Im}\mu = 0 \) ; hence \( \operatorname{Ker}\varphi = 0 \) .
Definition. An essential extension of a left R-module \( A \) is a left R-module \( B \) such that \( A \) is an essential submodule of \( B \) ; more generally, a left \( R \) -module \( B \) with an essential monomorphism \( A \rightarrow B \) .
These two definitions are equivalent up to isomorphisms: if \( A \) is an essential submodule of \( B \), then the inclusion homomorphism \( A \rightarrow B \) is an essential monomorphism; if \( \mu : A \rightarrow B \) is an essential monomorphism, then \( A \) is isomorphic to the essential submodule \( \operatorname{Im}\mu \) of \( B \) ; moreover, using surgery, one can construct a module \( {B}^{\prime } \cong B \) in which \( A \) is an essential submodule.
Proposition 5.3. If \( \mu : A \rightarrow B \) and \( v : B \rightarrow C \) are monomorphisms, then \( v \circ \mu \) is essential if and only if \( \mu \) and \( v \) are essential.
This follows from Proposition 5.1; the details make nifty exercises.
Proposition 5.4. A left \( R \) -module \( J \) is injective if and only if \( J \) has no proper essential extension \( J \subsetneqq M \), if and only if every essential monomorphism \( J \rightarrow M \) is an isomorphism.
Proof. Let \( J \) be injective. If \( J \subseteq M \), then \( J \) is a direct summand of \( M \) , \( M = J \oplus N \) ; then \( N \cap J = 0 \) ; if \( J \) is essential in \( M \), then \( N = 0 \) and \( J = M \) . If in turn \( J \) has no proper essential extension, and \( \mu : J \rightarrow M \) is an essential monomorphism, then \( \operatorname{Im}\mu \cong J \) has no proper essential extension, hence \( M = \) \( \operatorname{Im}\mu \) and \( \mu \) is an isomorphism.
Finally, assume that every essential monomorphism \( J \rightarrow A \) is an isomorphism. We show that \( J \) is a direct summand of every module \( M \supseteq J \) . By Zorn’s lemma there is a submodule \( K \) of \( M \) maximal such that \( J \cap K = 0 \) . Readers will verify that the projection \( M \rightarrow M/K \) induces an essential monomorphism \( \mu : J \rightarrow M/K \) . By the hypothesis, \( \mu \) is an isomorphism; hence \( J + K = M \) and \( M = J \oplus K \) . \( ▱ \)
Proposition 5.5. Let \( \mu : M \rightarrow N \) and \( v : M \rightarrow J \) be monomorphisms. If \( \mu \) is essential and \( J \) is injective, then \( v = \kappa \circ \mu \) for some monomorphism \( \kappa : N \rightarrow J \) .
Proof. Since \( J \) is injective, there exists a homomorphism \( \kappa : N \rightarrow J \) such that \( v = \kappa \circ \mu \), which is injective by 5.2. \( ▱ \)
By 5.5, every essential extension of \( M \) is, up to isomorphism, contained in every injective extension of \( M \) . This leads to the main result in this section.
Theorem 5.6. Every left R-module \( M \) is an essential submodule of an injective \( R \) -module, which is unique up to isomorphism.
Proof. By Theorem 4.11, \( M \) is a submodule of an injective module \( K \) . Let \( \mathcal{S} \) be the set of all submodules \( M \subseteq S \subseteq K \) of \( K \) in which \( M \) is essential (for instance, \( M \) itself). If \( {\left( {S}_{i}\right) }_{i \in I} \) is a chain in \( \mathcal{S} \), then \( S = \mathop{\bigcup }\limits_{{i \in I}}{S}_{i} \in \mathcal{S} \) : if \( N \neq 0 \) is a submodule of \( S \), then \( {S}_{i} \cap N \neq 0 \) for some \( i \), and then \( M \cap N = M \cap {S}_{i} \cap N \neq 0 \) since \( M \) is essential in \( {S}_{i} \) ; thus \( M \) is essential in \( S \) . By Zorn’s lemma, \( \mathcal{S} \) has a maximal element \( J \) . If \( J \) had a proper essential extension, then by \( {5.5}\mathrm{\;J} \) would have a proper essential extension \( J \subsetneqq {J}^{\prime } \subseteq K \) and would not be maximal; therefore \( J \) is injective, by 5.4.
Now, assume that \( M \) is essential in two injective modules \( J \) and \( {J}^{\prime } \) . The inclusion monomorphisms \( \mu : M \rightarrow J \) and \( v : M \rightarrow {J}^{\prime } \) are essential. By 5.5 there is a monomorphism \( \theta : J \rightarrow {J}^{\prime } \) such that \( v = \theta \circ \mu \) . Then \( \theta \) is essential, by 5.3, and is an isomorphism by 5.4. \( ▱ \)
Definition. The injective hull of a left \( R \) -module \( M \) is the injective module, unique up to isomorphism, in which \( M \) is an essential submodule.
The injective hull or injective envelope \( E\left( M\right) \) of \( M \) can be characterized in several ways: \( E\left( M\right) \) is | (Theorem 4.12. A ring \( R \) is left Noetherian if and only if every direct sum of injective left \( R \) -modules is injective.) | Assume that every direct sum of injective left \( R \) -modules is injective, and let \( {L}_{1} \subseteq {L}_{2} \subseteq \cdots \subseteq {L}_{n} \subseteq \cdots \) be an ascending sequence of left ideals of \( R \) . Then \( L = \mathop{\bigcup }\limits_{{n > 0}}{L}_{n} \) is a left ideal of \( R \) . By 4.11 there is a monomorphism of \( R/{L}_{n} \) into an injective \( R \) -module \( {J}_{n} \) . By the hypothesis, \( J = {\bigoplus }_{n > 0}{J}_{n} \) is injective. Construct a module homomorphism \( \varphi : L \rightarrow J \) as follows. Let \( {\varphi }_{n} \) be the homomorphism \( R \rightarrow R/{L}_{n} \rightarrow {J}_{n} \) . If \( x \in L \), then \( x \in {L}_{n} \) for some \( n \), and then \( {\varphi }_{k}\left( x\right) = 0 \) for all \( k \geqq n \) . Let \( \varphi \left( x\right) = {\left( {\varphi }_{k}\left( x\right) \right) }_{k > 0} \in J \) .
Since \( J \) is injective, \( \varphi \) extends to a module homomorphism \( \psi : {}_{R}R \rightarrow J \) . Then \( \psi \left( 1\right) = {\left( {t}_{k}\right) }_{k > 0} \in J \) for some \( {t}_{k} \in {J}_{k},{t}_{k} = 0 \) for almost all \( k \) . Hence \( \psi \left( 1\right) \in {\bigoplus }_{k < n}{J}_{k} \) for some \( n \) . Then \( \varphi \left( x\right) = \psi \left( {x1}\right) = {x\psi }\left( 1\right) \in {\bigoplus }_{k < n}{J}_{k} \) for all \( x \in L \) ; in particular, \( {\varphi }_{n}\left( x\right) = 0 \) and \( x \in {L}_{n} \), since \( R/{L}_{n} \rightarrow {J}_{n} \) is injective. Thus \( L = {L}_{n} \), and then \( {L}_{k} = {L}_{n} \) for all \( k \geqq n \) .
Conversely, assume that \( R \) is left Noetherian, and let \( J = {\bigoplus }_{i \in I}{J}_{i} \) be a direct sum of injective \( R \) -modules \( {J}_{i} \) . Let |
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