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The $℘$-function satisfies $℘'^2=4(℘-e_1)(℘-e_2)(℘-e_3)$. Supposing that none of $e_1,e_2,e_3$ is zero, show that the equation $℘(z)=0$ has two distinct solutions $z= ±a$. For any two distinct points $b,c∈ℂ/Γ$ write down a meromorphic function whose only poles are simple poles at $b$ and $c$. Solution. Since $℘'(a)^2=-4e_1e_2e_3≠0$, $a$ is not a ramification point, so $℘(z)=0$ has two distinct solutions, let one of them be $a$, then the other is $-a$, since $℘$ is even. The linear map $\frac{2a}{b-c}(z-b)+a$ maps $b,c$ to $a,-a$. So $\frac1{℘(\frac{2a}{b-c}(z-b)+a)}$ is a meromorphic function whose only poles are simple poles at $b$ and $c$.

Show that a sequence in $\ell^1$ is weakly convergent if and only if it is strongly convergent. [Hint: For given $\varepsilon>0$, construct inductively increasing sequences $n_k$ and $m_k$ such that $\sum_{j \leqslant m_{k-1}}|x_j^{(n_k)}|<\varepsilon / 8$ and $\sum_{j>m_k}|x_j^{(n_k)}|<\varepsilon / 8$. Then test the weak convergence against $b \in \ell^{\infty}$ given by $b_j=\operatorname{sign}(x_j^{(n_k)})$ for $m_{k-1}Strong and weak convergence in $\ell^1$ The 'if' part is clear. Assume that $(x^{(n)}) \subset \ell^1$ converges weakly to some $x$. We need to show that $x^{(n)}$ converges strongly to $x$ and replacing $x^{(n)}$ by $x^{(n)}-x$ we may assume without loss of generality that $x=0$. Arguing by contradiction, suppose that $x^{(n)}\nrightarrow 0$. Upon extracting a subsequence, we may assume without loss of generality that $\|x^{(n)}\|>\varepsilon>0$ for some $\varepsilon$. The weak convergence of $x^{(n)}$ implies that $x_j^{(n)} \rightarrow 0$ for each fixed $j$. Select the smallest $n_1$ such that $|x_1^{(n_1)}|<\varepsilon / 8$. Then select the smallest $m_1$ such that $\sum_{j>m_1}|x_j^{(n_1)}|<\varepsilon / 8$. Next, select the smallest $n_2>n_1$ such that $\sum_{j \leqslant m_1}|x_j^{(n_2)}|<\varepsilon / 8$, and then select the smallest $m_2 \geqslant m_1$ such that $\sum_{j>m_2}|x_j^{(n_2)}|<\varepsilon / 8$. Proceeding in this way, we obtain increasing sequence $(n_k)$ and $(m_k)$ such that $\sum_{j \leqslant m_{k-1}}|x_j^{(n_k)}|<\varepsilon / 8$ and $\sum_{j \geqslant m_k}|x_j^{(n_k)}|<\varepsilon / 8$. Now, define an element $b$ of $\ell^{\infty}=(\ell^1)^*$ by $b_j=1$ for $j \leqslant m_1$ and \[ b_j=\operatorname{sign}(x_j^{(n_k)}) \text { if } m_{k-1}m_k} b_j x_j^{(n_k)}, \end{align*} and so \[ \sum_{m_{k-1}m_k}|x_j^{(n_k)}| \\ & \leqslant \varepsilon / 8+b(x^{(n_k)})+\varepsilon / 4+\varepsilon / 8=b(x^{(n_k)})+\varepsilon / 2 . \end{align*} As $\varepsilon>0$, it follows that $b(x^{(n_k)}) \not \rightarrow 0$, contradicting the assumption that $x^{(n)} \rightharpoonup 0$.

Let $℘(z)$ be the Weierstrass ℘-function. Consider the meromorphic function $℘'(z)$ on $X=ℂ/Λ$. By considering $℘'$ as a map to $ℂℙ^1$, determine its degree and the number and indices of its ramification points. Is there a meromorphic function $f$ on $X$ such that $f'(z)=℘(z)$? Solution. By definition \[℘(z) = \frac{1}{z^2} + \sum_{ω∈Λ∖0} \left(\frac{1}{(z-ω)^2}-\frac{1}{ω^2}\right)\] Differentiate, \[℘'(z)=-2\sum_{ω∈Λ}\frac1{(z-ω)^3}\] There is a triple pole at $z=0$, so $0$ is a ramification point of index 3, and this is the only pole in $X$, so $\deg(℘')=3$ by looking at inverse image of $∞$. Over $∞$, $℘'(z)=-2z^3+O(1)$, so $∞$ is a ramification point of $℘'$ of index 3. Finite ramification points of $℘'$ satisfy $℘'' = 0$. $℘$ satisfies the following differential equation: $$ (℘')^2 = 4℘^3 - g_2℘ - g_3 $$ Differentiate, $$ 2℘'℘'' = 12℘^2℘' - g_2℘' $$ Since the zeros of $\wp' and $\wp'' are disjoint for a non-degenerate torus, we can divide by $\wp' where it is non-zero. $$ 2℘'' = 12℘^2 - g_2 $$ so $℘''=0⇔℘(z)=±\sqrt{\frac{g_2}{12}}$ has 4 solutions for $z$, since $℘$ has degree 2 and $℘'(z)≠0$. By Riemann-Hurwitz, \begin{align*}\sum_{\text{ramification point }r}(v_{℘'}(r)-1)&=\deg(℘')χ(ℂℙ^1)-χ(X)\\&=3×2-0=6,\end{align*} If $g_2≠0$ we have $\sqrt{\frac{g_2}{12}}≠-\sqrt{\frac{g_2}{12}}$, so $℘'$ have 4 ramification points of index 2. Riemann-Hurwitz, $(3-1)+4(2-1)=6$ If $g_2=0$ there are 2 zeros of $℘$ by Q4, also $℘'''=0$ at those points, so $℘'$ has 2 ramification points of index 3. Riemann-Hurwitz, $(3-1)+2(3-1)=6$ Suppose there is a meromorphic function $f$ on $X$ such that $f'(z)=℘(z)=z^{-2}+O(1)$, then $f=-z^{-1}+O(z)$ has a simple pole at 0 and no other poles, hence $\deg f=1$, $f:X→ℂℙ^1$ has degree 1. A degree 1 map is a biholomorphism, contradicts $X$ is not isomorphic to $ℂℙ^1$.

A sequence $(ℓ_n)$ in the dual space $X^*$ of a Banach space $X$ is said to be weak* convergent to $ℓ ∈ X^*$ if \[ ℓ_n(x) → ℓ(x) \text { for all } x ∈ X . \] (a) Show that weak* convergent sequences are bounded. Proof. $ℓ_n(x)$ bounded for all $x ∈ X$. By Uniform Boundedness Principle $(ℓ_n)$ is bounded. (b) Show that if $X$ is separable, then the closed unit ball of $X^*$ is weak* sequentially compact, i.e. every sequence $(ℓ_n)$ in $X^*$ with $\|ℓ_n\|_* ⩽ 1$ has a weak* convergent subsequence. %[Hint: Consider a countable dense subset $S=\{x_n, n ∈ ℕ\}$ and use a diagonal sequence argument to construct a subsequence $(ℓ_{n_k})$ such that $ℓ_{n_k}(x_m)$ is convergent for every $m$.] % Alaoglu https://dept.math.lsa.umich.edu/~adellape/PDFS/weakconvergence.pdf Proof. Let $\{x_k\}$ be the dense set in the separable space $X$. Let $\{ℓ_n\}$ be a sequence in the closed unit ball of $X^*$, so $\|ℓ_n\|_*≤ 1$, we need to find a weak* convergent subsequence. The sequence $\{ℓ_n(x_1)\}$ of real numbers is bounded ($∵|ℓ_n(x_1)|≤\|ℓ_n\|_*\|x_1\|≤\|x_1\|$) and thus contains a convergent subsequence which we denote as $\{ℓ_{n 1}(x_1)\}$ (Bolzano-Weierstrass). Similarly, $\{ℓ_{n 1}(x_2)\}$ is also bounded and contains a convergent subsequence $\{ℓ_{n 2}(x_2)\}$. Continuing in this fashion to extract convergent subsequences $\{ℓ_{n k}(x_k)\}$, we then form the diagonal sequence $\{ℓ_{n n}\}$ in $X^*$ which is a subsequence of $\{ℓ_n\}$. Note that as the way we build $\{ℓ_{n n}\}$, we have $\{ℓ_{n n}(x_k)\}$ converges. Thus the sequence $\{ℓ_{n n}\}$ converges on the dense subset $\{x_k\}$ of $X$. Next we need to prove that $\{ℓ_{n n}\}$ converges weak* to an element of $X^*$. Fix $x ∈ X$ and $ϵ>0$. Then for any $n, m, k$, we have \[ |ℓ_{n n}(x)-ℓ_{m m}(x)| ≤|ℓ_{n n}(x)-ℓ_{n n}(x_k)|+|ℓ_{n n}(x_k)-ℓ_{m m}(x_k)|+|ℓ_{m m}(x_k)-ℓ_{m m}(x)| \] First $|ℓ_{n n}(x)-ℓ_{n n}(x_k)|=|ℓ_{n n}(x-x_k)|≤\|ℓ_{n n}\|_*\|x-x_k\| ≤\|x-x_k\|$. Similarly, $|ℓ_{m m}(x_k)-ℓ_{m m}(x)| ≤\|x-x_k\|$. As $\{x_k\}$ is dense in $X$, choose $k$ so that $\|x_k-x\|<\frac{ϵ}{3}$. As $\{ℓ_{n n}(x_k)\}$ converges, $\{ℓ_{n n}(x_k)\}$ is Cauchy. Thus there exists $N$ such that for all $n, m>N$, we have $|ℓ_{n n}(x_k)-ℓ_{m m}(x_k)|<\frac{ϵ}{3}$. Thus we have $|ℓ_{n n}(x)-ℓ_{m m}(x)|<ϵ$. Then we know that $\{ℓ_{n n}(x)\}$ is Cauchy and converges. Define a functional $f$, $f(x)=\lim_n ℓ_{n n}(x)$. To prove $f∈X^*$ we prove that it is linear and bounded. Clearly $f$ is linear. To prove that $f$ is bounded, we have \begin{align*} |f(x)|&=|\lim ℓ_{n n}(x)|\\ & =\lim|ℓ_{n n}(x)|\\ & ≤\lim\|ℓ_{n n}\|\|x\|\\ & ≤ \|x\| \end{align*} Thus $f$ is linear and bounded. We have $ℓ_{n n}(x)$ converges to $f(x)$ for all $x$. Thus $\{ℓ_{n n}\}$ converges weak* and the closed unit ball in $X^*$ is weak* compact.

Let $1Proof. (⇒) By Proposition 4.3 $(x^{(n)})$ converges weakly implies it is bounded. The standard basis vectors $e_k$ belong to every $l^p$, and hence $e_k ∈ l^q=(l^p)^*$ where $q=\frac{p}{p-1}$. If $x^{(n)}=(x^{(n)}_k)_{k ∈ ℕ}$ and $x=(x_k)_{k ∈ ℕ}$ and $x^{(n)} ⇀ x$, then for each $j$ \[ x^{(n)}_j=⟨x^{(n)},e_j⟩→⟨x,e_j⟩=x_j. \] (⇐) For all $y∈l^q$ and $ϵ>0$, let $y^{(k)}$ be the sequence $y$ truncated at position $k$. Since $y^{(k)}→y$, ∃$k∈ ℕ$ such that $\|y-y^{(k)}\|_q\sup_{n∈ℕ}\|x-x^{(n)}\|_p<\frac{ϵ}{2}.$ By Hölder's inequality $$ |⟨y-y^{(k)},x-x^{(n)}⟩|≤ \|y-y^{(k)}\|_q⋅\|x-x^{(n)}\|_p<\frac{ϵ}2.$$ Since $y^{(k)}_j=0$ for $j>k$ and $x^{(n)}_j→x_j\ ∀j≤k$, there is $N∈ ℕ$ such that $|⟨y^{(k)},x-x^{(n)}⟩|<\frac{ϵ}{2}$ for $n≥N$. \[ |⟨y,x-x^{(n)}⟩|≤|⟨y-y^{(k)},x-x^{(n)}⟩|+|⟨y^{(k)},x-x^{(n)}⟩|<ϵ \] Since ϵ is arbitrary, $⟨y,x-x^{(n)}⟩→0$. Since $y$ is arbitrary, $x^{(n)}⇀x$.

Theorem 3.17. Let $C$ be a nonsigular cubic curve in $ℂ ℙ^2$. Then $C$ is equivalent under a projective transformation to the curve $y^2 z=x(x-z)(x-λ z)$ for some $λ ∈ ℂ \backslash\{0,1\}$. Proof. By Prop 3.16(c), $C$ has at least one point of inflection. Apply a projective transformation to make $(0,1,0)$ a point of inflection, with tangent line $z=0$. Let $C$ be defined by $P(x, y)$. Then\[P(0,1,0)=P_x(0,1,0)=0,   P_z(0,1,0) ≠ 0,   H_P(0,1,0)=0\] From Lemma 3.14 with $y,z$ reversed, get \[ y^2 H_P(x, y, z)=4\left|\begin{array}{lll} P_{x x} & P_x & P_{x z} \\ P_x & \frac{3}{2} P & P_z \\ P_{z x} & P_z & P_{z z} \end{array}\right| \] So $0=H_p(0,1,0)=4\left|\begin{array}{ccc}P_{x x}(0,1,0) & 0 & P_{x z}(0,1,0) \\ 0 & 0 & P_z(0,1,0) \\ P_{z x}(0,1,0) & P_z(0,1,0) & P_{z z}(0,1,0)\end{array}\right|$ $=-4 P_z(0,1,0)^2 P_{x x}(0,1,0)$, so $P_{x x}(0,1,0)=0 \text { as } P_z(0,1,0) ≠ 0$. Hence $P$ has no $y^3$ or $x y^2$ or $x^2 y$ term as \[ P(0,1,0)=P_x(0,1,0)=P_{x x}(0,1,0)=0\text {. } \] Thus $P(x,y, z)=yz(α x+β y+γ z)+Q(x, z)$, \[ β=P_z(0,1,0) ≠ 0 \text {. } \] Apply projective transformation $[x, y, z] ↦\left[x, y+\frac{α x+γ z}{2 β}, z\right]$. Takes $P$ to $β y^2z+\tilde{Q}(x, z)=: \tilde{P}(x, y, z)$ $C$ irreducible, so $z ∤ \tilde{P}$, so coefficient of $x^3$ in $\tilde{Q}$ is nonzero. Can rescale $x,y$ to get \[ \tilde{\tilde{P}}=y^2 z-(x-a z)(x-b z)(x-c z) \text {. } \] If $a=b$, say, the $[a,0,1]$ is a singular point❌ So $a, b, c$ are distinct. Finally apply projective transformation \[ [x, y, z] ⟼\left[\frac{x-a z}{b-a},(b-a)^{-3 / 2} y, z\right] \] to get $\tilde{\tilde{P}}(x, y, z)=y^2 z-x(x-z)(x-λ z)$, \[ λ=\frac{c-a}{b-a}   λ ≠ 0,1,   λ ∈ ℂ . \] The equation $y^2 z=x(x-z)(x-λ z)$ is called a normal form, ie. a standard way to write the cubic. ∎ Parameter count: cubics in $x,y,z=ℂ^{10}$ acted on by $GL(3, ℂ)$, dim 9. $⇒$ Expected dimension of cubic / projective tranformation, is $10-9=1$, parametrized by $λ ∈ \underbrace{ℂ∖\{0,1\}}_{\dim1}$

(i) Show that given any five points in $ℂℙ^2$ there is at least one conic passing through them. Solution. A conic is a curve of the form $a_1 x^2+a_2 x y+a_3 x z+a_4 y^2+a_5 y z+a_6 z^2$, and has 6 coefficients. Given 5 points in $ℙ^2$, each point specifies a linear constraint on the coefficients, and since there are 5 dimensions, by rank-nullity, dimension of image ≤5, so dimension of kernel≥1. (ii) Let $C$ be a quartic curve (i.e. of degree 4), with four singular points. By choosing an appropriate conic $D$ and using the strong form of Bézout's theorem (involving intersection multiplicities $I_p(C, D)$) prove that $C$ must be reducible. Solution. Let $C$ be a projective curve of degree 4 in $ℙ^2$ with four singular points. By (i) exist a conic $D$ containing the 4 singular points and another point $p$ of $C$. By Bézout's theorem, if the two curves have no common component, then they have 8 points of intersection counting multiplicities. Since the 4 singular points have multiplicity greater than 1, and $C$ and $D$ have an additional point $p$ of intersection, hence the sum of the intersection multiplicities is at least 9. Thus $C$ and $D$ must have a common component, which implies $C$ is reducible. (iii) Show that $y^4-4xzy^2-xz(z-x)^2=0$ is a quartic with three singular points. Solution. $P_x=-(3x^2+4y^2-4xz+z^2)z$ $P_y=4(y^2-2xz)y$ $P_z=-(x^2+4y^2-4xz+3z^2)x$ $P_x-P_z=(x - z) (x^2 - 6 x z + 4 y^2 + z^2)=0\implies z=x$ or $x^2 - 6 x z + 4 y^2 + z^2=0$ If $x=z$ then $P_x=0\implies y^2z=0\implies y=0$ [If $z=0$, by $P_y=0$, also $y=0$], solution is $[1,0,1]$. If $x^2 - 6 x z + 4 y^2 + z^2=0$…(*), substitute $4y^2+z^2=-x^2+6xz$ into $P_x=0$ $$-(x+z)2xz=0$$ so $x=-z$ [If $xz=0$ then $P_y=0\implies y=0$, but $[1,0,0]$ or $[0,0,1]$ don't satisfy (*).] $P_y=0\implies y=0$ or $±\sqrt{-2}x$, but $[1,0,-1]$ don't satisfy (*), so solution is $[1,±\sqrt{-2},-1]$. So three singular points $[1,0,1],[1,±\sqrt{-2},-1]$.

%https://math.stackexchange.com/questions/844053/in-a-uniformly-convex-banach-space-x-n-stackrelw-to-x-and-x-n-to-x Let $X$ be a Banach space. (a) Suppose that $x_n ⇀ x$ in $X$ and $ℓ_n → ℓ$ in $X^*$. Show that $ℓ_n(x_n) → ℓ(x)$. (b) Suppose in addition that $X$ is a Hilbert space. Show that if $x_n ⇀ x$ in $X$ and if $\|x_n\| →\|x\|$, then $x_n → x$. (c) Prove (b) when the assumption that $X$ is a Hilbert space is replaced by the assumption that $X$ is uniformly convex: for every $ε>0$, there exists $δ=δ(ε)>0$ such that if $\|x\|=\|y\|=1$ and if $\|x-y\| ⩾ ε$, then $\left\|\frac{1}{2}(x+y)\right\| ⩽(1-δ)$. Proof. (a) By Proposition 4.3 $x_n ⇀ x$ implies $(x_n)$ is bounded. By triangle inequality \begin{align*} |ℓ_n(x_n)-ℓ(x)|&≤ |ℓ_n(x_n)-ℓ(x_n)|+|ℓ(x_n)-ℓ(x)|\\ &≤ \|x_n\|⋅\|ℓ_n-ℓ\|+|ℓ(x_n)-ℓ(x)|\\ \end{align*} both terms$→0$, so $ℓ_n(x_n) → ℓ(x)$. (b) Since $⟨x_n,⋅⟩∈X^*$, $x_n ⇀ x$ implies $⟨x_n,x⟩→‖x‖^2$, \begin{align*}‖x_n-x‖^2&=‖x_n‖^2+‖x‖^2-2\operatorname{Re}⟨x_n,x⟩\\&→‖x‖^2+‖x‖^2-2‖x‖^2=0\end{align*} (c) The case $x = 0$ is immediate, since $‖x_n‖→ 0$ is just the norm convergence of $(x_n)$ to $0$. Suppose $x≠ 0$, wlog $‖x‖ = 1$. By the Hahn-Banach theorem, there is $λ∈X^*$ with $‖λ‖ = 1$ and $λ(x) = 1$. Since $x_n ⇀ x$, we have $λ(x_n) → λ(x) = 1$, and therefore $$\lim_{n→∞} λ\left(\frac{1}{2}(x_n+x)\right) = 1,$$ but $‖λ‖ = 1$, so $$\liminf_{n→∞} \left‖\frac{1}{2}(x_n+x)\right‖ ⩾ 1.$$ On the other hand, since $\left‖\frac{1}{2}(x_n+x)\right‖ ⩽ \frac{1}{2}(‖x_n‖ + ‖x‖)$, we have $$\limsup_{n→∞} \left‖\frac{1}{2}(x_n+x)\right‖ ⩽ 1,$$ so $$\lim_{n→∞} \left‖\frac{1}{2}(x_n+x)\right‖ = 1.$$ By the lemma, $x_n→x$. Lemma. In a uniformly convex space, for all sequences $(a_n)$ and $(b_n)$ with $\lim\limits_{n→∞} ‖a_n‖ = 1 = \lim\limits_{n→∞} ‖b_n‖$ and $\lim\limits_{n→∞} \left‖\frac{1}{2}(a_n+b_n)\right‖ = 1$ it follows that $\lim\limits_{n→∞} ‖a_n - b_n‖ = 0$. Proof. Suppose $‖a_n - b_n‖ \not→ 0$. Then there is an $ε > 0$ such that $‖a_{n_k} - b_{n_k}‖ > 2ε$ for subsequences $(a_{n_k})$ and $(b_{n_k})$. We may assume that this holds for the full sequences. For that $ε$, by uniform convexity, there is a $δ > 0$ such that $$\left‖\frac{1}{2}(u+v)\right‖ ⩽ 1 - δ\tag{1}\label{1}$$ for all $u,v$ with $‖u‖ = 1 = ‖v‖$ and $‖u-v‖ ⩾ ε$. Then, setting $α_n = \frac{1}{‖a_n‖}a_n$ and $β_n = \frac{1}{‖b_n‖}b_n$ ($a_n = 0$ or $b_n = 0$ can happen only finitely often, we drop these terms), we have \begin{align*} \left‖α_n-β_n\right‖ &⩾ ‖a_n-b_n‖ - ‖a_n-α_n‖ - ‖b_n-β_n‖\\ &= ‖a_n-b_n‖ - |1 - ‖a_n‖| - |1-‖b_n‖|\\ &> 2ε - |1 - ‖a_n‖| - |1-‖b_n‖|\\ &> ε \end{align*} for $n$ so large that $|1-‖a_n‖| < ε/2$ and $|1-‖b_n‖| < ε/2$. Then by (1), \begin{align*} \left‖\frac{1}{2}(a_n+b_n)\right‖ &⩽ \left‖\frac{1}{2}(α_n+β_n)\right‖ + \frac{1}{2}(‖α_n - a_n‖ + ‖β_n-b_n‖)\\ &⩽ 1-δ + \frac{1}{2}(‖α_n - a_n‖ + ‖β_n-b_n‖)\\ &⩽ 1 - \frac{δ}{2} \end{align*} for $n$ so large that $|1-‖a_n‖| < δ/2$ and $|1-‖b_n‖| < δ/2$, contradicting the assumption $\left‖\frac{1}{2}(a_n+b_n)\right‖ → 1$.

By a projective transformation, show how to take the cubic $x^3+y^3+z^3=0$ into the form $y^2 z=x(x-z)(x-λz)$, for $λ$ which you should determine. Solution.$$H=\det\begin{pmatrix} 6x&0&0\\0&6y&0\\0&0&6z \end{pmatrix}=6^3xyz$$ so $[0,-1,1]$ is an inflection point. with tangent line $3⋅0^2x+3(-1)^2y+3⋅1^2z=0⇔y+z=0$. The projective transformation $[x,y,z]↦[x,y,y+z]$ maps $[0,-1,1]$ to $[0,1,0]$ and maps the tangent line $y+z=0$ to $z=0$ and maps the curve to $x^3+y^3+(z-y)^3=0⇔3yz(y-z)+x^3+z^3=0$. Quadratic in $y$, complete the square: the projective transformation $[x, y, z] ↦\left[x, y-\frac{z}2, z\right]$ maps the curve to $3(y^2-(\frac{z}{2})^2)z+x^3+z^3=0⇔y^2z+\frac{x^3}3+\frac{z^3}{12}=0$. Rescaling $[x,y,z]↦[\root3\of{-4}x,\sqrt{12}y,z]$ the curve to $y^2z=x^3-z^3⇔y^2z=(x-z)(x-ωz)(x-ω^2z)$ where $ω=e^{i\frac{2π}3}$. so \[ λ=\frac{ω^2-1}{ω-1}=ω+1. \]