Home

\href{https://ajwilson.co.uk/files/maths/luroth-problem.pdf{Theorem 1.}} (Lüroth) Suppose we have an inclusion of fields $k \subset L \subseteq k(t)$, where $k \neq L$ and $k(t)$ is the field of rational functions in one variable $t$. Then $L$ is isomorphic to $k(t)$. We now sketch an elementary algebraic proof of Lüroth's theorem. For details see [Waerden] Section 63. Considering any element $\lambda \in L \backslash k$, we observe that $t$ is an algebraic element of $k(\lambda)$ and so an algebraic member of $L$. Next, examine the polynomial \[ f(z)=z^n+a_1 z^{n-1}+\cdots+a_n \in L[z] \] where the $a_i$ are rational functions in $x$. Multiplying through by the lowest common denominator yields polynomials and we may write \[ f(x, z)=b_0(x) z^n+b_1(x) z^{n-1}+\cdots+b_n(x) \] Let the degree of $f$ with respect to $x$ be $m$. Note that not all the coefficients $a_i=\frac{b_i}{b_0}$ in $f(x)$ can be independent of $x$, since that would imply that $x$ is algebraic with respect to $k$. Thus at least one of the terms $a_i=\theta$ must be dependent on $x$. To complete the proof, one uses elementary field extension properties to show that (i) $m=n$, and (ii) $\theta$, as a function of $x$, is of degree $m$. It follows that \[ [k(x): k(\theta)]=m=[k(x): L] \] and as $L \supseteq k(\theta)$, we have $[L: k(\theta)]=1$. Therefore \[ L=k(\theta) \cong k(t) \] by a change of variables.

\begin{align*} r&=(a\cos u\sin v,a\sin u\sin v,a\left(\cos v+\ln \tan \frac {v}{2}\right)+bu)\\ r_u&=(-a\sin u\sin v,a\cos u\sin v,b)\\ r_v&=(a\cos u\cos v,a\sin u\cos v,a\cos v\cot v)\\ E=r_u⋅r_u&=a^2\sin^2 v+b^2\\ F=r_u⋅r_v&=ab\cos v\cot v\\ G=r_v⋅r_v&=a^2\cot^2 v\\ r_u×r_v&=(a\cos v(a\cos u\cos v - b\sin u), a\cos v(a\sin u\cos v + b\cos u), -a^2\sin v\cos v)\\ |r_u×r_v|&=a\cos v\sqrt{a^2 + b^2}\\ n=\frac{r_u×r_v}{|r_u×r_v|}&=\frac{(a\cos u\cos v - b\sin u,a\sin u\cos v + b\cos u, -a\sin v)}{\sqrt{a^2 + b^2}}\\ r_{uu}&=(-a\cos u\sin v,-a\sin u\sin v,0)\\ r_{uv}&=(-a\sin u\cos v,a\cos u\cos v,0)\\ r_{vv}&=(-a\cos u\sin v,-a\sin u\sin v,-a\cos v(1+\csc^2 v))\\ e=r_{uu}⋅n&=-\frac{a^2 \cos v \sin v}{\sqrt{a^2 + b^2}}\\ f=r_{uv}⋅n&=\frac{ab \cos v}{\sqrt{a^2 + b^2}}\\ g=r_{vv}⋅n&=\frac{a^2 \cot v}{\sqrt{a^2 + b^2}}\\ K=\frac{eg-f^2}{EG-F^2}&=-\frac1{a^2+b^2} \end{align*}

thm 1.17 Prove codim(Im(I+K))<∞ for a compact operator By compactness of $K$, the closed unit ball in $\ker(I+K)$ is compact, so $\ker(I+K)$ has finite dimension. Therefore $X = \ker(I+K) \oplus S$ for some closed subspace $S$ of $X$. The operator $I + K$ restricted to $S$ has an inverse $T : \operatorname{Im}(I+K) \to S$. You can show (by way of contradiction, using compactness of $K$) that $T$ is continuous at $0$ (and hence continuous). Then $\operatorname{Im}(I + K)$ is closed. To demonstrate that $\operatorname{Im}(I + K)$ has finite codimension, suppose to the contrary that the codimension of $\operatorname{Im}(I + K)$ is infinite. There is a sequence $\operatorname{Im}(I + K) = S_0 \subsetneq S_1 \subsetneq S_2 \subsetneq \cdots$ of closed subspaces of $X$ with $\dim(S_{n+1}/S_n) = 1$ for every $n$. Riesz's lemma gives normalized vectors $x_n\in S_n$ such that $\operatorname{dist}(x_n,S_{n-1}) \ge 0.5$. If $n > m$, then $(I + K)x_n, (I + K)x_m$, and $x_m$ belong to $S_{n-1}$, forcing $$\|Kx_n - Kx_m\| = \|(I + K)x_n - (I + K)x_m + x_m - x_n\| \ge \operatorname{dist}(x_n,S_{n-1}) \ge 0.5$$ Consequently, $(Kx_n)$ has no convergent subsequence, contradicting compactness of $K$.

$\DeclareMathOperator{\pv}{pv}\DeclareMathOperator{\sgn}{sgn}$The Hilbert transform of $ϕ ∈ 𝒮(ℝ)$ is defined as \[ ℋ(ϕ)≝\pv\left(\frac{1}{π x}\right) * ϕ . \] In lectures we have seen that $ℋ(ϕ)$ is a moderate $\mathrm{C}^{∞}$ function and so in particular a tempered distribution and that hereby $ℋ: 𝒮(ℝ) → 𝒮'(ℝ)$ is a linear operator. (a) Show that $\|ℋ(ϕ)\|_2=\|ϕ\|_2$ for all $ϕ ∈ 𝒮(ℝ)$. Next, show that $ℋ$ admits a unique extension to a continuous linear operator $ℋ_0:𝖫^2(ℝ)→𝖫^2(ℝ)$. Finally, show that $ℋ_0$ is an isometric isomorphism and that $ℋ_0^2=-\mathrm{Id}_{𝖫^2}$. Solution. By Example 1.60, $\widehat{\pv(\frac1x)}(ξ)=−𝗂π\sgn(ξ)$, so $\widehat{\pv(\frac1{πx})}(ξ)=−𝗂\sgn(ξ)$. By convolution rule, $\widehat{ℋ(ϕ)}=-𝗂\sgn(ξ)\hat{ϕ}$, so $\|\widehat{ℋ(ϕ)}\|_2=\|\hat{ϕ}\|_2$. By Plancherel’s Theorem $\|ℋ(ϕ)\|_2=\frac1{\sqrt{2π}}\|\widehat{ℋ(ϕ)}\|_2,\|ϕ\|_2=\frac1{\sqrt{2π}}\|\hat{ϕ}\|_2$. So $\|ℋ(ϕ)\|_2=\|ϕ\|_2$. Since $𝒮(ℝ)$ is dense in $𝖫^2(ℝ)$, $ℋ$ has a unique extension to $𝖫^2(ℝ)$, it's continuous since $\|ℋ(ϕ)\|_2=\|ϕ\|_2$. \begin{align*} \widehat{ℋ_0^2(ϕ)}(ξ) & =-𝗂\sgn(ξ)\widehat{ℋ_0(ϕ)}(ξ) \\ &=(-𝗂\sgn(ξ))^2 \hat{ϕ}(ξ) \\ &=-\hat{ϕ}(ξ) \end{align*} Taking the inverse Fourier transform $ℋ_0^2=-I$. So $ℋ_0^{-1}=-ℋ_0$. So $ℋ_0$ is an isometric isomorphism. (b) Using that $𝖫^2(ℝ)$ is a Hilbert space show that also the $𝖫^2$ based Sobolev space of order $s ∈ ℝ$, $𝖧^s(ℝ)$, is a Hilbert space. Show also that $𝒮(ℝ)$ is dense in $𝖧^s(ℝ)$. Solution. To show $𝖧^s(ℝ)$ is complete, we want to show that if $\{u_i\}$ is a Cauchy sequence in $𝖧^s$ then it converges to some $u∈𝖧^s$ \[ \|u_i-u_j\|_{𝖧^s}=\|(1+|x|^2)^{s/2}(\widehat{u_i}-\widehat{u_j})\|_2 \] so $(1+|x|^2)^{s/2}\widehat{u_i}$ is a Cauchy sequence in $𝖫^2$, by $𝖫^2$ completeness, $(1+|x|^2)^{s/2}\widehat{u_i}$ converges. To show $𝒮(ℝ)$ is dense in $𝖧^s(ℝ)$, for $ϕ∈𝖧^s(ℝ)$, $(1+|x|^2)^{s/2}\hat{ϕ}∈𝖫^2(ℝ)$. Since $𝒮(ℝ)$ is dense in $𝖫^2(ℝ)$, $∃f∈𝒮(ℝ),‖f-(1+|x|^2)^{s/2}\hat{ϕ}‖_2<ε$. Since $f∈𝒮(ℝ),(1+|x|^2)^{-s/2}f∈𝒮(ℝ).$ Let $ψ∈𝒮,\hat{ψ}=(1+|x|^2)^{-s/2}f$. $‖ψ-ϕ‖_{𝖧^s}=‖(\hat{ψ}-\hat{ϕ})(1+|x|^2)^{s/2}‖_2<ε$. (c) Show that $\|ℋ(ϕ)\|_{𝖧^s}=\|ϕ\|_{𝖧^s}$ for all $ϕ ∈ 𝒮(ℝ)$ and all $s ∈ ℝ$. Prove that $ℋ$, for each $s ∈ ℝ$, admits a unique extension to a continuous linear operator $ℋ_s: 𝖧^s(ℝ) → 𝖧^s(ℝ)$. Show that $ℋ_s$ is an isometric isomorphism and that $ℋ_s^2=-\mathrm{Id}_{𝖧^s}$. Finally, show that for $sSolution. First part basically the same as (a). To show for $s

Let $C$ be the curve in $ℂ^2$ with equation $y^2=x^6-1$. Show how to extend this to a curve $C$ of degree 6 in $ℂℙ^2$. How many points lie on the line $z=0$? Is the curve nonsingular? Solution. $p(x,y)≔y^2-x^6+1=0$ extends to the curve $P(x,y,z)≔z^6 p\left(\frac{x}{z}, \frac{y}{z}\right)=y^2z^4-x^6+z^6=0$ in $ℂℙ^2$. Plugging in $z=0$ we get $x^6=0$, so $C$ intersects the line $z=0$ at a point $[0,1,0]$ of multiplicity 6. $\frac{∂P}{∂x}=-6x^5,\frac{∂P}{∂y}=2yz^4,\frac{∂P}{∂z}=4y^2z^3+6z^5$ all vanish at $[0,1,0]$, so the curve is singular.

category theory semidirect product Definition 1. Let $G$ be a group, $X ⊂ G$, and $g ∈ G$. Set \[gXg^{-1} =\{y ∈ G\mid y = gxg^{-1}\text{ for some }x ∈ X\}.\] We say that $g$ normalizes $X$ if $gXg^{-1} = X$. Let $A \subset G$. We say that $A$ normalizes $X$ if $a$ normalizes $X$ for every $a \in A$. Let $H \leq G$. The normalizer of $X$ in $H$ is \[ N_H(X)=\left\{h \in H \mid h X h^{-1}=X\right\} . \] Proposition 1. Let $G$ be a group, $H \leq G$, and $X \subset G$. Then $N_H(X) \leq H$. Proof. Since $1 \in H$, and $1 x 1^{-1}=x$ for all $x \in X$, we know $1 \in N_H(X)$. Let $h_1, h_2 \in N_H(X)$. Then \[ h_1 h_2 X\left(h_1 h_2\right)^{-1}=h_1\left(h_2 X h_2^{-1}\right) h_1^{-1}=h_1 X h_1^{-1}=X, \] so $h_1 h_2 \in N_H(X)$. Let $h \in H$, so that $X=h X h^{-1}$. Multiply both sides by $h^{-1}$ on the left and by $h$ on the right to get \[ h^{-1} X h=h^{-1} h X h^{-1} h=X \] thus $h^{-1} \in N_H(X)$. □ Definition 2. Let $G$ be a group and let $X, Y \subset G$. Set \[ X Y=\{x y \in G \mid x \in X \text { and } y \in Y\}\\\quad X^{-1}=\left\{x^{-1} \in G \mid x \in X\right\} . \] Proposition 2. Let $G$ be a group and let $H, K \leq G$. Then $H K \leq G$ if and only if $H K=K H$. Proof. If $M \leq G$, then $M^{-1}=M$. Thus if $H K \leq G$, then $H K=(H K)^{-1}=$ $K^{-1} H^{-1}=K H$. Suppose $H K=K H$. Let $h_1, h_2 \in H$ and $k_1, k_2 \in K$ so that $h_1 k_1$ and $h_2 k_2$ are arbitrary members of $H K$. Since $H K=K H$, there exists $k_3 \in K$ such that $k_1 h_2=h_2 k_3$. Then $h_1 k_1 h_2 k_2=h_1 h_2 k_3 k_2 \in H K$. Let $h \in H$ and $k \in K$ so that $h k$ is an arbitrary member of $H K$. Then $(h k)^{-1}=k^{-1} h^{-1} \in K H=H K$. Thus $H K \leq G$. □ [...]

(a) State the Seifert-van Kampen Theorem. Hence give and prove a formula for the fundamental group of $X \vee Y$ in terms of the fundamental groups of two path-connected cell complexes $X$ and $Y$. (i) Let $X$ be a path-connected cell complex and $a, b$ be two points in $X$. Let $Y$ be the quotient space of $X$ where $a$ is identified with $b$. Compute the fundamental group of $Y$. Justify your answer using a sketch. (ii) Let $X$ be a path-connected cell complex and $D_{a}, D_{b}$ be two disjoint disks embedded in $X$. Let $Z$ be the space obtained by attaching a cylinder $S^{1} \times[0,1]$ to $X$ such that $S^{1} \times\{0\}$ and $S^{1} \times\{1\}$ are identified with the boundaries of $D_{a}$ and $D_{b}$ respectively. Compute the fundamental group of $Z$. Justify your answer using a sketch. Answers: [bookwork] Formula: $\pi_{1}(X \vee Y, a)=\pi_{1}(X, a) * \pi_{1}(Y, a)$ with $a$ the common base point of $X$ and $Y$. Proof: As $X$ and $Y$ are cell complexes $a$ has a contractible neighbourhoods $U_{X}$ in $X$ and $U_{Y}$ in $Y$. Apply Seifert-van Kampen to the cover $\left\{\left\{X \vee U_{Y}\right\},\left\{U_{X} \vee Y\right\}\right\}$. (i) $Y$ is homotopy equivalent to $Y_{1}$ which is $X$ with an interval attached to $a, b$ via its endpoints. Moving $a$ to $b$ via a chosen path in $X$ between the two points, $Y_{1}$ is seen to be homotopy equivalent to $X \vee S^{1}$. Hence $$ \pi_{1}(Y)=\pi_{1}(X) * \mathbb{Z} $$ (ii) $Z$ is homotopy equivalent to the space $Z_{1}$ where the two discs are contracted to points in $X$. So $Z_{1}$ is $X$ with a sphere $S^{2}$ attached at two points, say $a$ and $b$. Pick a (direct) path in $S^{2}$ between those two and collapse it to get $Z_{2}$. The result is homotopy equivalent to $Y \vee S^{2}$ where $Y$ is as in part (i). Hence $$ \pi_{1}(Z)=\pi_{1}(Y) * \pi_{1}\left(S^{2}\right)=\pi_{1}(Y)=\pi_{1}(X) * \mathbb{Z} $$ (b) Let $\langle S \mid R\rangle$ be a finite presentation of a group $G$. Describe how to construct a path-connected, compact space $X$ with fundamental group $G$. (i) Let $K$ be the fundamental group of the Klein bottle. Give a finite presentation of $K$. Briefly justify your answer. (ii) Show that there is an injective homomorphism $\rho: \mathbb{Z} \rightarrow K$ and there are surjective homomorphisms $\gamma_{1}: K \rightarrow \mathbb{Z} / 2 \mathbb{Z}$ and $\gamma_{2}: K \rightarrow \mathbb{Z}$. Describe maps of spaces that induce $\rho, \gamma_{1}$ and $\gamma_{2}$ as maps on fundamental groups. Briefly justify your answer. Answers: [bookwork] (i) The Klein bottle is a square with sides $a, b, a, b^{-1}$. We can construct is as a wedge of two $S^{1}-a$ and $b$ - with a 2 -cell attached with boundary attaching map $a b a b^{-1}$. Hence $$ \pi_{1}(K)= $$ (ii) Define $\gamma_{1}: \pi_{1}(K) \rightarrow \mathbb{Z} / 2 \mathbb{Z}$ by sending $a \mapsto 1$ and $b \mapsto 0$; define $\gamma_{2}: \pi_{1}(K) \rightarrow \mathbb{Z}$ by sending $a \mapsto 0$ and $b \mapsto 1$. These define a homomorphisms as $a b a b^{-1} \mapsto 0$. Define $\rho: \mathbb{Z} \rightarrow \pi_{1}(K)$ via $1 \mapsto b$; this is injective as composition with $\gamma_{2}$ is the identity. Contracting the sides in the square labelled $b$ to a point gives a representation of the projective plane and hence a map $K \rightarrow \mathbb{R} P^{2}$ inducing $\gamma_{1}$ on fundamental groups. Projection of the square onto the $b$ coordinate defines a map $K \rightarrow S^{1}$ which induces $\gamma_{2}$. The inclusion of $S^{1}$ as the edge defined by $b$ induces $\rho$.

converting between the Poincaré disc model and the upper half plane model Example: Consider the unit disc $D=\left\{x+i y \in \mathbf{C} \mid x^2+y^2<1\right\}$ with first fundamental form \[ \frac{4\left(d x^2+d y^2\right)}{\left(1-x^2-y^2\right)^2} \] and the upper half plane $H=\{u+i v \in \mathbf{C} \mid v>0\}$ with the first fundamental form \[ \frac{d u^2+d v^2}{v^2} . \] The Möbius transformation \begin{align*} H&\to D\\ w&\mapsto z=\frac{w-i}{w+i} \end{align*} is a bijection. We shall show it is an isometry. \[ \frac{d z}{d w}=\frac{1}{w+i}-\frac{(w-i)}{(w+i)^2}=\frac{2 i}{(w+i)^2} \] so \[ \left|\frac{d z}{d w}\right|^2=\frac{4}{(w+i)^2 \left(w^*-i\right)^2} \] so \[ |dw|^2=|dz|^2/\left|\frac{d z}{d w}\right|^2=\frac{(w+i)^2 \left(w^*-i\right)^2}4|dz|^2 \] substituting into $v^{-2}|d w|^2$ gives \begin{align*}v^{-2}|d w|^2&=\left(\frac{w-w^*}{2 i}\right)^{-2}|d w|^2\\&=\left(\frac{w-w^*}{2 i}\right)^{-2}\frac{(w+i)^2 \left(w^*-i\right)^2}4|dz|^2\\&=-\left(w-w^*\right)^{-2}(w+i)^2 \left(w^*-i\right)^2|dz|^2\\&=4\left(1-\frac{(w-i) \left(w^*+i\right)}{(w+i) \left(w^*-i\right)}\right)^{-2}|dz|^2\\&=4\left(1-|z|^2\right)^{-2}|d z|^2\end{align*} so this Möbius transformation gives us an isometry from $H$ to $D$ as required.

Proposition 27. A set $K \subset \mathcal{H}$ in a separable Hilbert space is compact if and only if it is bounded, closed and has equi-small tails with respect to any (one) complete orthonormal basis. Proof. We already know that a compact set in a metric space is closed and bounded. Suppose the equi-smallness of tails condition fails with respect to some orthonormal basis $e_k$. This means that for some $\epsilon>0$ and all $p$ there is an element $u_p \in K$, such that \[ \sum_{k>p}\left|\left(u_p, e_k\right)\right|^2 \geq \epsilon^2 \] Consider the subsequence $\left\{u_p\right\}$ generated this way. No subsequence of it can have equi-small tails (recalling that the tail decreases with $p$ ). Thus, by Lemma 25, it cannot have a convergent subsequence, so $K$ cannot be compact if the equismallness condition fails. Thus we have proved the equi-smallness of tails condition to be necessary for the compactness of a closed, bounded set. It remains to show that it is sufficient. So, suppose $K$ is closed, bounded and satisfies the equi-small tails condition with respect to an orthonormal basis $e_k$ and $\left\{u_n\right\}$ is a sequence in $K$. We only need show that $\left\{u_n\right\}$ has a Cauchy subsequence, since this will converge $(\mathcal{H}$ being complete) and the limit will be in $K$ (since it is closed). Consider each of the sequences of coefficients $\left(u_n, e_k\right)$ in $\mathbb{C}$. Here $k$ is fixed. This sequence is bounded: \[ \left|\left(u_n, e_k\right)\right| \leq\left\|u_n\right\| \leq C \]