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Maclaurin's conic generation method Consider all triangles $ABC$ whose sides $\{AB, BC, CA\}$ pass correspondingly through three fixed points $\{C^*, A^*, B^*\}$ and two of its vertices $\{B, C\}$ lie correspondingly on two fixed lines $A_1X, A_1Y$. Then its vertex $A$ describes a conic passing through points $A_1, B^*, C^*$. The conic also passes through the intersection points $B_1, C_1$ of line-pairs $(A_1X, A^*B^*)$ and $(A_1Y, A^*C^*)$. The proof is modeled as an exercise in projective coordinates with respect to the fixed points $\{A^*, B^*, C^*\}$ and with coordinator the fixed point $A_1$. In this system $A_1=A^*+B^*+C^*$. Line $BC$ passing through $A^*$ has the form: $y-kz=0$ (with variable $k$), and lines $A_1X, A_1Y$ correspondingly $x-ay-bz=0$, $x-a'y-b'z=0$ with fixed $(a, b)$, $(a', b')$ satisfying$$1=a+b=a'+b'\tag{*}$$ since $A_1$ is on these lines. The coordinates of $\{B, C\}$ are correspondingly: $(ka+b, k, 1)$ and $(ka'+b', k, 1)$. Then lines $\{BC^*, CB^*\}$ obtain the form: $-kx+(ka+b)y=0$ and $-x+(ka'+b')z=0$. Eliminating $k$ from them leads to equation:$$ba'yz = (b'z-x)(ay-x)\tag{**}$$ representing a conic passing through $B^*(0,1,0)$, $C^*(0,0,1)$, and $A_1(1,1,1)$ (use (*)) as stated. The two additional points are the intersection points of line-pairs $(x-ay-bz=0, z=0)$ and $(x-a'y-b'z=0, y=0)$ which are easily calculated to be $(a,1,0)$ and $(b',0,1)$ and satisfy equation (**).

Let $X$ be a complex Hilbert space and $T ∈ ℬ(X)$ be normal (i.e. $T^* T=T T^*$). (a) Show that \[ r_σ(T)=‖T‖ \] Deduce that if $P$ is a polynomial, then \[ ‖P(T)‖=\sup_{λ ∈ σ(T)}|P(λ)| \] Solution. (6.7.1)Since $T$ is normal, $‖Tx‖=‖T^*x‖∀x∈X$ Apply this to $Tx$: $‖T^2x‖=‖T^*Tx‖$. By induction, $‖T^{2n}x‖=‖(T^*T)^nx‖∀x∈X∀n∈ℕ$, so $‖T^{2n}‖=‖(T^*T)^n‖∀n∈ℕ$. Since $T^*T$ is self-adjoint, $‖(T^*T)^n‖=‖T^*T‖^n$. By Gelfand's formula,$$r_σ(T)=\lim_{n→∞}‖T^{2n}‖^{\frac1{2n}}=\lim_{n→∞}‖T^*T‖^{\frac12}=‖T^*T‖^{\frac12}=‖T‖$$ For the second part: By the Spectral Mapping theorem, $σ(P(T))=P(σ(T))$. Since $P(T)$ is normal, $‖P(T)‖=\sup_{λ∈σ(P(T))}|λ|=\sup_{λ∈P(σ(T))}|λ|=\sup_{μ∈σ(T)}|P(μ)|$. (b) Let $P$ be a Laurent polynomial, i.e. $P(z)=\sum_k a_k z^k$ where the summation range is finite but may contains positive as well as negative powers. Show that if $T$ is unitary, then \[ ‖P(T)‖=\sup_{λ ∈ σ(T)}|P(λ)| . \] Solution. Let $P=\sum_{k=-n}^{n}a_kz^k,Q=\sum_{k=0}^{2n}a_{k-n}z^k$ so that $Q$ is polynomial and $P(z)=z^{-n}Q(z)$ Since $T$ is unitary, $σ(T)⊆\{λ∈ℂ∣|λ|=1\}$. By (a), we get \begin{align*} ‖P(T)‖&=‖T^{-n}Q(T)‖\\ &=\sup_{x∈X:‖x‖=1}‖T^{-n}(Q(T)(x))‖\\ &\overset{\text{isometry}}=\sup_{x∈X:‖x‖=1}‖Q(T)(x)‖\\ &=‖Q(T)‖\\ &\overset{\text{(a)}}=\sup_{λ∈σ(T)}|Q(λ)|\\ &\overset{|λ|=1}=\sup_{λ∈σ(T)}|P(λ)| \end{align*}

MSE If $X$ is a compact Riemann surface, and $f:X\to \mathbb C\cup \{\infty\}$ is a meromorphic function, then indeed $f$ has the same number of zeros and poles, with multiplicity counted. In the more general case, let $X, Y$ be Riemann surfaces, $X$ compact and $f:X\to Y$ be a non-constant holomorphic map and $f(x)=y$. Then in properly chosen local coordinates around $x$ and $y$, we can write $f(z)=z^k$. Then we define the ramification index at the point $x$ to be $v_x=k$. It's easy to see that $f^{-1}(y)$ is a finite set for all $y\in Y$. The function $d(y)=\sum\limits_{f(x)=y}v_x$ is defined. Now I will show that $d$ is a locally constant function on $Y$ and hence it's a constant. Then your question will be answered. For any $y\in Y$, let $f^{-1}(y)=\{x_1,\dots,x_m\}$. Choose a coordinate chart $U$ around $x$ and coordinate charts $V_i$ around $x_i$ such that $f(V_i)\subset U$. Via replacing $U$ by $\cap f(V_i)$ and replacing $V_i$ by $V_i\cap f^{-1}\left(\cap f(V_i)\right)$, we may assume that $f(V_i)=U$ for each $i$. Note that here I used the fact the holomorphic maps are open. By replacing $U$ by $U-f(X-\cup V_i)$, and intersecting $V_i$ with the preimage of the new $U$, we may assume that $f^{-1}(U)=\cup V_i$. (Here I used the fact that $f$ is a closed map.) Since locally a holomorphic map is $z\mapsto z^k$, by shrinking $U$ and intersecting $V_i$ and the preimage of the new $U$, finally we may assume that for each $y\neq y'\in U$, $y'$ has exactly $v_{x_j}$ preimages in $V_j$. Now clearly $d$ is constant in $U$.

Let $C$ be a nonsingular quartic curve in $ℂℙ^2$. (i) Let $HD(C)$ be the vector space of holomorphic differentials on $C$. Show using question 5 that there is an isomorphism $HD(C) ≅ ℂ^3=⟨x, y, z⟩$, such that if $0 ≠ ω$ is a holomorphic differential corresponding to $a x+b y+c z$ then the canonical divisor $(ω)$ is the hyperplane divisor corresponding to the line $a x+b y+c z=0$. Solution. By Riemann-Roch $\dim HD(C)=ℓ(κ)=g=3$. From the second example of theorem 30 $κ ∼ H$ and so we can obtain a holomorphic differential by writing \[\frac{Q(x, y, 1) d x}{\partial P / \partial y(x, y, 1)}\] for a homogeneous polynomial $Q(x, y, z)$ of degree 1 (hyperplane). The dimension of the space of homogeneous polynomials $Q(x,y,z)$ of degree 1 is 3, which equals $\dim HD(C)$, therefore every holomorphic differential is obtained from a polynomial this way. (ii) Let $p,q∈C$ be distinct points. Show that the vector subspace of $ω∈HD(C)$ vanishing at $p,q$ has dimension 1. Deduce that $ℓ(κ-p-q)=1$ for a canonical divisor $κ$. Solution. The vector subspace of $HD(C)$ vanishing at $p,q$ correspond to the line through $p,q$, so it has dimension 1. $f∈\mathcal{L}(κ-p-q)$ if any only if $(fω)=(f)+κ$ vanishes at $p,q$, if any only if $fω$ is in the vector subspace of $HD(C)$ vanishing at $p,q$, so $ℓ(κ-p-q)=1$. (iii) Show that $ℓ(p+q)=1$ for all distinct $p, q ∈ C$. Solution. By Riemann-Roch, $ℓ(p+q)-ℓ(κ-p-q)=\deg(p+q)+1-g$, by (ii) and $\deg(p+q)=2,g=3$, we get $ℓ(p+q)=1$.

$σ_{ap}(T) ⊇ ∂σ(T)$, where $∂σ(T)$ is the boundary of $σ(T)$ in the topology of ℂ. part II linear analysis, page 67 theorem 2.2 Proof. Let $λ ∈ ∂ σ(T)$. Then there is a sequence $λ_n ∉ σ(T)$ converging to $λ$. It follows from Corollary 5.2(iii) of the course that \[ \left\|\left(λ_n I-T\right)^{-1}\right\| → ∞   \text { as }   n → ∞ . \] Thus, there is a sequence $\left(x_n\right)$ of unit vectors such that \[ \left\|\left(λ_n I-T\right)^{-1} x_n\right\| → ∞   \text { as }   n → ∞ . \] Set \[ y_n=\frac{\left(λ_n I-T\right)^{-1} x_n}{\left\|\left(λ_n I-T\right)^{-1} x_n\right\|} . \] It is easy to check that $\left(y_n\right)$ is an approximate eigenvector for $λ$.

$f_n\overset{w*}⇀f$ if $f_n(x)→f(x)∀x∈X$ i.e. if $ι_x(f_n)→ι_x(f)$ i.e. $T(f_n)→T(f)∀T∈ι(X)⊆X^{**}$ $f_n\overset{w}⇀f$ if $T(f_n)→T(f)∀T∈X^{**}$ Example that weak convergence-∗ doesn't imply weak convergence Try to find a sequence $l_n∈X^*,l_n\overset{w*}⇀l,l_n\overset{w}↛l$. as $∃T∈X^{**}:T(l_n)↛T(l)$ while $∀T∈ι(X),T(l_n)→T(l)$. Try $X=L^1([0,1]),X^*=L^∞([0,1])$ so want sequence $f_n⊆L^∞$ s.t. $∫gf_n→∫gf ∀g∈L^1$ but $T∈(L^∞)^*,Tf_n↛0$ since $Tf_n=1∀n$. Take $f_n=1_{A_n}$ e.g. $A_n=[0,\frac{1}{n}]$ or anything $f(A_n)→0$ and $A_n⊇A_{n+1}…$ Then $|∫f_ng|≤∫|g|1_{A_n}→0$ since $1_{A_n}→0$ a.e. Try to find $T:Tf_n=1$. would like something like $f↦f(0)$ only ok on $C^0([0,1])$ $Y=C^0([0,1])⊆X$ $T_0:Y→ℝ$ $T_0(f)=f(0)$ Since $C^0⊆L^∞$ can extend $T_0$ by Hahn-Banach gives $T∈(L^∞)^*$. so $T(f)=\lim_{n→0}n\int_0^{\frac1n}f$

set of isomorphism classes of geometric objects made into a space of some kind(topological space, complex manifold) Example. The moduli space of complex tori $ℂ/Λ$ (≃moduli space of non-singular cubics up to projective transform.) $Λ=⟨w_1,w_2⟩_ℤ=$pairs $(w_1,w_2)/GL(2,ℤ)$ for $w_1/w_2∈ℂ∖ℝ$ moduli space of tori $ℂ/Λ$ $≅ℋ=\{z∈ℂ:\Im(z)>0\}$ $SL(2,ℤ)$ acts by Mobius transform $SL(2,ℤ)=⟨S,T⟩,S=\pmatrix{1&1\\0&1},T=\pmatrix{0&1\\-1&0}$ fundamental domain for 2 special tori: $ℂ/ℤ+iℤ$: $ℤ^4$ symmetry $ℂ/ℤ+e^{πi/3}ℤ$: $ℤ^6$ symmetry $C:x^3+y^3+z^3=0$ What $Λ$ is this $ℂ/Λ$ for? $ℂ/ℤ+e^{πi/3}ℤ$: $ℤ^6$ symmetry $(x,y,z)↦(e^{2πi/3}x,y,z)$ order 3 rotation fixes $(0,1,-1)$

Let $C$ be a nonsingular quartic curve in $ℂℙ^2$. (a) Let $H$ be a hyperplane divisor on $C$. Show that $ℓ(H)≥3$. Solution. By definition $H=\sum_{p∈C∩L}I_p(C,L)p$ for some line $L:a_0x+b_0y+c_0z=0$. For any $a,b,c∈ℂ$, $\frac{ax+by+cz}{a_0x+b_0y+c_0z}∈\mathcal{L}(H)$ and these functions are distinct, so $ℓ(H)≥3$. (b) Let $D$ be a divisor on $C$ with $\deg D=4$. Prove using the Riemann-Roch Theorem that $ℓ(D)=3$ if $D$ is a canonical divisor, and $ℓ(D)=2$ otherwise. Solution. $\deg(κ)=2g-2=4,\deg(κ-D)=4-4=0$. If $\mathcal{L}(κ-D)≠∅$, there is a meromorphic function $f∈\mathcal{L}(κ-D)$, so $(f)+κ-D$ is effective and degree 0, so $(f)+κ-D=0$, so $D$ is a canonical divisor. By Riemann-Roch, $ℓ(D)=g=3$. If $\mathcal{L}(κ-D)=∅$, then $ℓ(κ-D)=0$. By Riemann-Roch, $ℓ(D)=\deg D+1-g=2$, so $D$ is not canonical. Remark. If $D$ is a divisor such that $\deg D=0$, $ℓ(D)=0$ or $1$: • $ℓ(D)=1$ if $D$ is equivalent to $0$ • $ℓ(D)$ if $0$ is not equivalent to $0$. Proof: $D$ is equivalent to $0$, then ∃ meromorphic $f$ such that $(f^{-1})=D$ $g∈ℒ(D)$, then $(g)=-D=(f)⇒(\frac{g}{f})=0⇒g=λf$ for $λ∈ℂ^*$. (c) Deduce that every hyperplane divisor on $C$ is a canonical divisor. Solution. By Bézout's theorem $4=\sum_{p∈C∩L}I_p(C,L)$, so $\deg H=4$. By (a) $ℓ(H)≥3$, then by (b) $ℓ(H)=3$ and $H$ is a canonical divisor.

$\DeclareMathOperator{\im}{Im}$% Consider the right shift operator on sequences $R(x_1,x_2,…)=(0, x_1, x_2, …)$. Show that (a) As an operator on $ℓ^2$, $R$ satisfies $σ_p(R)=∅, σ_r(R)=\{λ:|λ|<1\}$ and $σ_{ap}(R)=σ_c(R)=\{λ:|λ|=1\}$. Solution 1. $x∈σ_p⇔Rx=λx⇔λx_1=0$ and $∀k≥1,λx_{k+1}=x_k$. If $λ=0$ then $x=0$, so $0∉σ_p$; if $λ≠0$ then $x_1=0⇒x_2=\frac{x_1}{λ}=0…⇒x=0$, so $λ∉σ_p$. So $σ_p(R)=∅$. Since $σ_p(R)=∅$, $R-λI$ is injective, $$λ∈σ_r(R)⇔\overline{\im(R-λI)}≠ℓ^2⇔\ker(R^*-\bar{λ}I)≠0⇔\bar{λ}∈σ_p(L)$$ in lecture computed $σ_p(L)=\{λ∈ℂ:|λ|<1\}$, so $σ_r(R)=\{λ∈ℂ:|λ|<1\}$. So $σ(R)⊇\{λ∈ℂ:|λ|<1\}$, but $σ(R)$ is closed, so $σ(R)⊇\{λ∈ℂ:|λ|≤1\}$. By Gelfand's formula $r_σ(R)=\|R\|=1$, so $σ(R)=\{λ∈ℂ:|λ|≤1\}$. So $σ_c(R)=σ(R)∖(σ_r(R)∪σ_p(R))=\{λ∈ℂ:|λ|=1\}$. $σ_{ap}⊇σ_c$ $∀|λ|<1,\|Rx-λx\|≥\|\|Rx\|-|λ|\|x\|\|=\|\|x\|-|λ|\|x\|\|=(1-|λ|)\|x\|$, so $λ∉σ_{ap}$. so $σ_{ap}=\{λ∈ℂ:|λ|=1\}$ Solution 2. $σ_p(R)=∅$ as in solution 1. $\im(R)⊆\{y∈ℓ^2:y_1=0\}=e_1^⟂$ is not dense, so $0∈σ_r$. If $|λ|<1$, $(y_1,y_2,…)≔(R-λI)(x_1,x_2,…)=(-λx_1, x_1-λx_2, x_2-λx_3, …)$, and \begin{align*}\sum_{i=1}^{n}λ^iy_i&=λ^1(-λx_1)+λ^2(x_1-λx_2)+λ^3(x_2-λx_3)+⋯+λ^n(x_n-λx_{n+1})\\&=-λ^{n+1}x_{n+1}→0\end{align*} so $\im(R-λI)⊆(λ,λ^2,…)^⟂$ is not dense, so $λ∈σ_r$. So $σ_r(R)⊇\{λ∈ℂ:|λ|<1\}$. So $σ(R)⊇\{λ∈ℂ:|λ|<1\}$, but $σ(R)$ is closed, so $σ(R)⊇\{λ∈ℂ:|λ|≤1\}$. By Gelfand's formula $r_σ(R)=\|R\|=1$, so $σ(R)=\{λ∈ℂ:|λ|≤1\}$. If $|λ|=1$, let $a_n=(λ^{-1},…,λ^{-n},0,…)$, then \begin{align*}e_1-\tfrac{λ}{n}a_n&=\left(1-\tfrac1n,-\tfrac{λ^{-1}}n,…,-\tfrac{λ^{1-n}}n,0,…\right) \\&=(λI-R)\left(λ^{-1}(1-\tfrac1{n}),λ^{-2}(1-\tfrac2n),…,λ^{-n}(1-\tfrac nn),0,…\right)\end{align*} so $e_1-\frac{λ}{n}a_n∈\im(λI-R)$. But $$‖e_1-(e_1-\tfrac{λ}{n}a_n)‖=\tfrac{‖a_n‖}{n}=\tfrac{1}{\sqrt{n}}→0$$ so $e_1∈\overline{\im(λI-R)}$. Similarly, for all $j≥1$, take $n≥j$, \begin{align*}e_j-\tfrac{λ^j}{n}a_n&=\left(-\tfrac{λ^{j-1}}n,-\tfrac{λ^{j-2}}n,…,-\tfrac{λ^1}n,1-\tfrac{λ^0}n,-\tfrac{λ^{-1}}n,…,-\tfrac{λ^{j-n}}{n},0,…\right) \\&=(λI-R)\left(-\tfrac{λ^{j-2}}n,-\tfrac{2λ^{j-3}}n,…,-\tfrac{(j-1)λ^0}n,λ^{-1}(1-\tfrac{j}{n}),λ^{-2}(1-\tfrac{2}n),…,λ^{j-n}(1-\tfrac{n}n),0,…\right)\end{align*} so $e_j-\frac{λ^j}{n}a_n∈\im(λI-R)$. But $$‖e_j-(e_j-\tfrac{λ^n}{n}a_n)‖=\tfrac{‖a_n‖}{n}=\tfrac{1}{\sqrt{n}}→0$$ so $e_j∈\overline{\im(λI-R)}$, but $\overline{\operatorname{span}}\{e_j\}=ℓ^2$, so $\overline{\im(λI-R)}=ℓ^2$, so $λ∈σ_c$. So $σ_c(R)=\{λ∈ℂ:|λ|=1\}$. [Another way: $λ∈σ_c(R)⇔λ∈σ(R)$ and $\overline{\im(R-λI)}=ℓ^2$ $⇔|λ|≤1$ and $\ker(R^*-\bar{λ}I)=0$ $⇔|λ|≤1$ and $\bar{λ}∉σ_p(L)=\{λ∈ℂ:|λ|<1\}$ so $σ_c(R)=\{λ∈ℂ:|λ|=1\}$.] (b) As an operator on $ℓ^{∞}$, $R$ satisfies $σ_p(R)=∅, σ_r(R)=\{λ:|λ| ⩽ 1\}$ and $σ_c(R)=∅$. Solution. By the same argument as (a), $σ_p(R)=∅$. Suppose $0<|λ|<1$, \[ (R-λ I) x=(1,0,0, …) \] for some $x ∈ ℓ^{∞}$, then \begin{align*} -λ x_1 & =1 \\ x_1-λ x_2 & =0 \\ x_2-λ x_3 & =0 \\ & … \end{align*} we obtain $x_k=-λ^{-k}$, but $|λ|<1$, so $|x_k|→∞$, contradicting $x∈ℓ^{∞}$, so $λ ∈ σ(R)$. Hence, $\{λ ∈ ℂ: 0<|λ|<1\} ⊂ σ(R)$, but $σ(R)$ is closed, so $σ(R)⊇\{λ∈ℂ:|λ|≤1\}$. By Gelfand's formula $r_σ(R)=\|R\|=1$, so $σ(R)=\{λ∈ℂ:|λ|≤1\}$. For $|λ|<1$, define a functional $f:ℓ^∞→ℂ,f(y)=\sum_{k=1}^∞λ^{k}y_k$. Can check $f$ is bounded and linear, $\ker(f)⊊ℓ^∞$ since $e_1∉\ker(f)$. \begin{align*}\sum_{k=1}^{n}λ^ky_k&=λ^1(-λx_1)+λ^2(x_1-λx_2)+λ^3(x_2-λx_3)+⋯+λ^n(x_n-λx_{n+1})\\&=-λ^{n+1}x_{n+1}→0\end{align*} so $\overline{\im(R-λ)}⊆\ker(f)$, so $\im(R-λ)$ not dense, so $λ∈σ_r(R)$. Let $a=(1,\frac{\bar{λ}}{|λ|},(\frac{\bar{λ}}{|λ|})^2,…)$ for $λ≠0$ and $a=(1,1,…)$ otherwise. We will show $B_{\frac12}(a)⊆ℓ^∞∖\im(R-λI)$. For $y∈\im(R-λI)$, $∃x∈ℓ^∞,(R-λI)(x)=y⇔y_1=λx_1,y_n=λx_n-x_{n-1}∀n≥2$ Summing up $λ^n$ times equation $n$ gives\[\sum_{k=1}^{n}λ^{k-1}y_k=λ^nx_n\tag{⋆}\] Assume that $y∈\im(R-λI),y=a+c∈B_{\frac{1}{2}}(a)$ so that $‖c‖<\frac{1}{2}$. $∃0≠x∈ℓ^∞$ s.t. $(R-λI)(x)=y=a+c$. (⋆)⇒for all $n∈ℕ$: \begin{align*} λ^nx_n&=\sum_{k=1}^{n}λ^{k-1}(a_k+c_k)\\ &=\sum_{k=1}^{n}λ^{k-1}(\tfrac{\bar{λ}}{|λ|})^{k-1}+\sum_{k=1}^{n}λ^{k-1}c_k\\ &=\sum_{k=1}^{n}|λ|^{k-1}+\sum_{k=1}^{n}λ^{k-1}c_k \end{align*} by triangle inequality \begin{align*} |λ^nx_n|&≥\sum_{k=1}^{n}|λ|^{k-1}-\sum_{k=1}^{n}|λ|^{k-1}|c_k|\\ &≥\frac12\sum_{k=1}^{n}|λ|^{k-1}\xrightarrow{n→∞}\frac12\sum_{k=1}^{∞}|λ|^{k-1} \end{align*} Case |λ|<1: RHS: $\frac12\sum_{k=1}^∞|λ|^{k-1}≥\frac12$ LHS: $|λ^nx_n|≤|λ|^n‖x‖_{\sup}\xrightarrow{n→∞}0$, contradiction! Case |λ|=1: $‖x‖_{\sup}≥|λ^nx_n|≥\frac12\sum_{k=1}^{∞}|λ|^{k-1}=∞$, contradiction! so $σ_r(R)=\{λ∈ℂ:|λ|≤1\}$. The question doesn't ask to compute $σ_{ap}(R)$.