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(a) State the Seifert-van Kampen Theorem. (b) Give a cell decomposition of the Klein bottle $K$. Use this to give a presentation of $\pi_1(K)$. (c) Let $B$ be a small open ball in $K$. Give a presentation of $\pi_1(K \backslash B)$. What word does the boundary represent in $\pi_1(K \backslash B)$ ? (d) Let $S$ be a surface obtained by gluing together two copies of $K \backslash B$ along a homeomorphism of their boundaries. Compute $\pi_1(S)$ using the Seifert-van Kampen Theorem. (e) Prove that $\pi_1(S)$ is not abelian by constructing a surjective homomorphism $\phi: \pi_1(S) \rightarrow S_3$, where $S_3$ is the symmetric group on three letters. Using this, show that there is no covering map from $S$ to the torus.

Do there exist (non-Euclidean) equilateral n-gons whose angles are all right angles, for n≥6? For any $n>4$ we can get polygons with all right angles with $n$ sides in the hyperbolic plane: take a very small regular $n$-gon and gradually expand it until the angle defect gets bad/good enough. More precisely, for $r>0$ let $C_r$ be a circle in the hyperbolic plane with radius $r$ and let $P_r$ be an equally-spaced set of $n$ points on $C_r$. Let $\alpha_r$ be the interior angle at any vertex in (the polygon formed by) $P_r$. By the angle defect formula for the hyperbolic plane, we have $\lim_{r\rightarrow\infty}\alpha_r=0$ and $\lim_{r\rightarrow 0^+}\alpha_r={(n-2)\pi\over n}$, and moreover $\alpha_r$ is continuous as a function of $r$. By the intermediate value theorem, some $r_0$ has $\alpha_{r_0}={\pi\over 2}$ (note that we need ${(n-2)\pi\over n}>{\pi\over 2}$ here, which is where the assumption $n>4$ comes in).

Exercise 11.2: Show that $ℚ$ is a flat $ℤ$-module but is not projective. Solution: $ℚ=S^{-1} ℤ$ with $S=ℤ∖\{0\}$. Hence $ℚ$ is flat over $ℤ$. Localization is exact Next we prove that it is not projective. Consider the $ℤ$-module $ℚ / ℤ$, which by definition is the cokernel of the inclusion $ℤ ↪ ℚ$. For $n ∈ ℕ_{≥ 1}$ denote by $α_n: ℤ → ℚ / ℤ$ the unique $ℤ$-module homomorphism with $α_n(1) ≡ \frac{1}{n} \bmod ℤ$. Clearly, $\ker(α_n)=n ℤ$; hence $α_n$ factors via $\bar{α}_n$ : $ℤ / n ℤ → ℚ / ℤ$. By the UMP of the direct sum we get a $ℤ$-linear map $α:=⊕_n α_n: \bigoplus_{n ≥ 1} ℤ / n ℤ → ℚ / ℤ$. Clearly $α$ is surjective. Denote by $π: ℚ → ℚ / ℤ$ the quotient map. \begin{tikzcd}&\mathbb{Q} \arrow[d, "\pi"] \\ \oplus_{n \geq 1} \mathbb{Z}/n\mathbb{Z} \arrow[r, "\alpha"] & \mathbb{Q}/\mathbb{Z} \arrow[r] & 0\end{tikzcd} If $ℚ$ was a projective $ℤ$-module, then there would exist a map $π_1: ℚ →\bigoplus_{n ≥ 1} ℤ / n ℤ$, such that $α ∘ π_1=π$. But any $ℤ$-linear homomorphism $β: ℚ → ℤ / n ℤ$ is the zero map. (Since $β\left(\frac{a}{b}\right)=β\left(\frac{n a}{n b}\right)=n β\left(\frac{a}{n b}\right)=0$.) It follows that if $π_1: ℚ → \bigoplus_{n ≥ 1} ℤ / n ℤ$ is a $ℤ$-linear map, then its composition with any projection map $\bigoplus_{n ≥ 1} ℤ / n ℤ → ℤ / m ℤ$ is zero; hence $π_1$ is zero. Thus $α ∘ π_1=π$ would imply $π=0$, i.e. $ℚ / ℤ=0$, i.e. the inclusion $ℤ ↪ ℚ$ is surjective, which is absurd. Hence $ℚ$ is not projective as an $ℤ$-module. MSE

MSE We can view $F_7$ as the quotient ring $𝐙[ω]/(3+ω)𝐙[ω]$. $(3+ω)$ is a maximal ideal of $𝐙[ω]$, so $𝐙[ω]/(3+ω)𝐙[ω]$ is a field. $N(3+ω)=7$, so $𝐙[ω]/(3+ω)𝐙[ω]$ has order 7. When we do that the non-zero elements of $F_7$ are represented by the cosets of the sixth roots of unity, i.e. the cosets $(-\omega)^j+(3+\omega)\mathbf{Z}[\omega]$. This is similar in spirit to the discussion in the linked question. Things become more uniform, when instead of using the complex numbers we use the $p$-adic integers $\mathbf{Z}_p$ (not to be confused with the residue class ring $\mathbf{Z}/p\mathbf{Z}$. If $q=p^n$, and $\zeta$ is a root of unity of order $q-1$, then we have an isomorphism $$ F_q=\mathbf{Z}_p[\zeta]/p\mathbf{Z}_p[\zeta] $$ between the finite field $F_q$ and the quotient ring of an extension ring of the $p$-adic integers. That's a rather different animal, and this last paragraph may be meaningless to you unless you have the right background. The $p$-adic integeres themselves cannot be readily visualized geometrically, because the metric there is very weird. Every finite field is a residue field of a number field

Let $n \geq 1$. For any field $E$, define $$ \mu_n(E):=\{\rho \in E \mid \rho^n=1\} . $$ Note that the set $\mu_n(E)$ inherits a group structure from $E^*$. The group $\mu_n(E)$ is a finite cyclic group. Proof: We’ll use the fact that a finite group $G$ is cyclic iff for any divisor $d\mid\#G$, there is at most 1 subgroup in $G$ of order $d$. For any $d\mid\#\mu_n$ if there is a subgroup in $\mu_n(E)$ of order $d$, generated by $\zeta$, then $1,\zeta,\dots,\zeta^{d-1}$ are roots of $x^d-1$ in $\mu_n(E)$, but $x^d-1$ has at most $d$ roots in $E$, so these are all the roots, so there is only 1 subgroup in $\mu_n(E)$ of order $d$. $\Box$

Lemma 1. The equation $x^3+y^3+z^3=0$ has no non-trivial integer solutions. Proof. Assume that $(x, y, z)$ is a non-trivial solution in coprime integers. Let \[ \rho=\frac{-1+\sqrt{-3}}{2} \] be a third root of unity and $\mathscr{E}=\mathbb{Z}[\rho]$ be the ring of Eisenstein integers, which is a Euclidean ring. We assume first that $3 \nmid x y z$, so that both $x+y$, and $x^2-x y+y^2$ are cubes, say $x+y=s^3$, and \[ (x+\rho y)=(a+b \rho)^3 \] is the cube of a principal ideal. Exactly one of $x, y, z$ must be even – we shall thus assume that $y=2 v$ is even. Since the units of $\mathscr{E}$ are the sixth roots of unity, there is some $\delta \in⟨-\rho⟩$ such that \[ \begin{aligned} \delta(x+y \rho)=(a+b \rho)^3 & =a^3+b^3+3 a b \rho(a+b \rho) \\ & =a^3+b^3-3 a b^2+3 a b(a-b) \rho . \end{aligned} \] If $\delta= \pm 1$, then comparing coefficients implies that $y \equiv 0 \bmod 3$, contradicting our assumption. We thus have that \[ x+y \rho= \pm \rho^c(a+b \rho)^3, \text { with } c= \pm 1 . \] We have chosen $y \equiv 0 \bmod 2$, whereby $x \rho^{-c} \equiv w^3 \bmod 2 \mathscr{E}$ for some $w= \pm(a+$ $b \rho) \in \mathscr{E}$. But $x \equiv x+y=s^3 \bmod 2$, whence we may conclude that \[ \rho^{-c} \equiv(w / s)^3 \bmod 2 . \] The ideal $\mathfrak{p}:=(2) \subset \mathscr{E}$ is prime; let $\pi: \mathscr{E} \rightarrow \mathbb{F}_{2^2}$ be the natural projection. Since $c \not \equiv 0 \bmod 3$, it follows that $\pi(w / s) \in \mathbb{F}_{2^2}$ is a primitive 9-th root of unity, an impossibility. It remains, then, to consider the case where $3 \mid x y z$; we may suppose, without loss of generality, that $3 \mid z$. Appealing to (9), we find that there is a root of unity $\delta$ such that $(\alpha)=\delta\left(\frac{x+y \rho}{1-\rho}\right)=(a+b \rho)^3$ and $x+y=9 s^3$. Since $\frac{1-\rho}{1-\bar{\rho}}=-\rho$, we obtain after dividing the previous identity by its complex conjugate, that there is another root of unity, say $\delta^{\prime}$, such that \[\tag{8} \frac{x+y \rho}{x+y \bar{\rho}}=\frac{2 x-y+y \sqrt{-3}}{2 x-y-y \sqrt{-3}}=\delta^{\prime} \cdot\left(\frac{a+b \rho}{a+b \bar{\rho}}\right)^3 \equiv \delta^{\prime} \bmod 3 \sqrt{-3} \cdot \mathscr{E} . \] If $2 x \not \equiv y \bmod 3$, letting $y^{\prime} \equiv y /(2 x-y) \bmod 3$, the last identities imply that \[ \frac{1+y^{\prime} \sqrt{-3}}{1-y^{\prime} \sqrt{-3}} \equiv \delta^{\prime} \bmod 3 \sqrt{-3} \cdot \mathscr{E}, \] whence \[ y^{\prime} \equiv y \equiv 0 \bmod 3, \] contradicting the assumption that $3 \mid z$ and $\operatorname{gcd}(x, y, z)=1$. Finally, suppose that $2 x-y=-3 d$. Inserting this value into (8) yields \[ \frac{d \sqrt{-3}+y}{d \sqrt{-3}-y} \equiv \delta^{\prime} \bmod 3 \sqrt{-3} \cdot \mathscr{E}, \] whereby $\delta^{\prime}=-1$ and $d \equiv 0 \bmod 3$. Then $2 x-y \equiv x+y \equiv 0 \bmod 9$. Summing these two congruences, we find $3 x \equiv 0 \bmod 9$, and thus $3 \mid x$, a contradiction which completes the proof.

Definition 2.1. A triangle group, denoted by three parameters $(\alpha, \beta, \gamma)$ such that $2 \leq \alpha, \beta, \gamma$, is generated by the reflections over the edges of a triangle with angles $\pi / \alpha, \pi / \beta$, and $\pi / \gamma$. The group rule is composition of reflections. Note that if $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=1$, the triangle sits in Euclidean space. If the sum is greater than 1 , it sits in elliptic space and if it is less than 1 , it sits in hyperbolic space. We will focus on this third case because the first two are fairly limited (there are only three examples in Euclidean space). If we let the generators of our group be $a, b, c$ where $a$ is the reflection over the edge opposite angle $\pi / \alpha$ etc., we have the following relations: \[ a^2=b^2=c^2=1,(a b)^\gamma=(b c)^\alpha=(a c)^\beta=1 \] The first relation holds because applying any reflection twice is the identity, and the second relation holds because by reflecting over each edge incident to an angle repeatedly, we can rotate around that angle fully, again yielding the identity. Now, in the Klein Model of the hyperbolic plane, each reflection can be represented by a projective transformation, so we also have a corresponding group of matrices $A \in M_{3,3}$

Prove that $\mathrm{e}^x$ is not in $𝒮'(ℝ)$, but that $\mathrm{e}^{x+\mathrm{i} \mathrm{e}^x}$ is in $𝒮'(ℝ)$ provided we define \[ ⟨\mathrm{e}^{x+\mathrm{ie}^x}, ϕ⟩=\lim _{j →+∞} ∫_{-∞}^j \mathrm{e}^{x+\mathrm{ie}^x} ϕ(x) \mathrm{d} x \] for $ϕ ∈ 𝒮(ℝ)$. Solution. To prove $\mathrm{e}^x$ is not in $𝒮'(ℝ)$, there exists $φ∈𝒮(ℝ)$ such that \[∫_{-∞}^{+∞} φ(x)\mathrm{e}^x\,dx = +∞.\] {\color{gray}For example $φ(x) = \exp(-\sqrt{1+x^2})$.} Define $χ_j≝ρ*1_{(0,j+1)}$. Then $χ_j∈C^∞$, $1_{(1,j)}≤χ_j≤1_{(-1,j+2)}$ so $χ_j∈𝒟$. Take $ϕ_j=χ_je_{-1}∈𝒟$ and note $ϕ_j→ϕ=χe_{-1}$ in $𝒮$ where $χ=ρ*1_{(0,∞)}$. By $𝒮$-continuity of $E$, \[⟨E,ϕ⟩=\lim_{j→∞}⟨E,ϕ_j⟩=\lim_{j→∞}∫e^xχ_je^{-x}dx=∞\text{❌}\] Clearly $e^{x+ie^x}∈𝒟'∩C^∞$ and $(e^{ie^x})'=e^{ie^x}ie^x=ie^{x+ie^x}$ so $(\underbrace{-ie^{ie^x}}_{u(x)})'=e^{}$ Then $u∈C^∞$ and $|u(x)|=1$ therefore $u'∈𝒮$.

(a) Prove the localisation property of Fourier series: if two continuous $2 π$-periodic functions $f$ and $g$ are equal in an open interval containing $x_0$, their Fourier series either both converge at $x_0$ or both diverge at $x_0$. Solution. The function $f-g$ is 0 in $I$, so it is $1-$Hölder continuous, by theorem 2.7 $S_N(f)(x_0)-S_N(g)(x_0) → 0$ pointwise as $N→∞$, so $S_N(f)$ and $S_N(g)$ either both converge at $x_0$ or both diverge at $x_0$. (b) In the lecture, we will prove that there is a continuous function $f_0$ whose Fourier series diverges at 0. Use (a) to construct a continuous function $f_S$ whose Fourier series diverges at every point of a given finite subset $S=\{s_1Solution. Let $f_S$ be a 2π-periodic continuous function s.t. $f_S(x)=f_0(x-s_i)$ for all $x$ in a neighborhood $U_i$ of $s_i$ such that $U_i$ are disjoint. Since $f_0(x-s_i)$ diverge at $x=s_i$, by (a) $f_S$ diverge at $x=s_i$ for each $1≤i≤n$.