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Let $V$ be a vector space of dim $n+1$. Let $p_0,\dots,p_{n+1}$ be $n+2$ points in $P(V)$, and let $v_0,v_1,\dots,v_{n+1}$ be the representative vectors in $V$ for $p_0,\dots,p_{n+1}$ respectively. Proposition. We say that $p_0,\dots,p_{n+1}$ are in general position if (i) no $n+1$ of $v_0,v_1,\dots,v_{n+1}$ are linearly dependent. [There are $n+2$ algebraic conditions.] or equivalently, (ii) no $n+1$ of $p_0,p_1,\dots,p_{n+1}$ lie in a hyperplane. [A hyperplane is a projective linear subspace $P(U)\subset P(V)$ with $\dim P(U)=\dim P(V)-1$.] Examples. $n=1$, $P(V)$ projective line. 3 points in $P(V)$ are in general position iff they are distinct. $n=2$, $P(V)$ projective plane. 4 points in $P(V)$ are in general position iff no three of them are collinear. Theorem 1.1. Let $V,W$ be vector spaces of dimension $n+1$, and let $p_0,p_1,\dots,p_{n+1},q_0,q_1,\dots,q_{n+1}$ be sets of points in general position in $P(V)$ and $P(W)$ respectively. Then there is a unique projective transformation $\tau:P(V)\to P(W)$ with $\tau(p_i)=q_i,i=0,1,\dots,n+1$. Remark. Points in general position are like a basis of a vector space. If $V,W$ are vector spaces of dim $n$, and $v_1,\dots,v_n,w_1,\dots,w_n$ are bases, there is a unique linear isomorphism $T:V\to W$ with $T(u_i)=w_i,i=1,\dots,n$. Dimension count: sets of points $(p_0,\dots,p_{n+1})\subset P(V)^{n+2}$ in general position has dimension $\dim V\cdot(n+2)=n(n+2)$. Group of projective linear transformations of $ℂP^n$ is $PGL(n,ℂ)=GL(n+1,ℂ)/ℂ^*$, so $\dim PGL(n+1,ℂ)=\dim GL(n+1,ℂ)-1=(n+1)^2-1=n(n+2)$.

Proposition 12.1. There are no integer solutions to $y^2+37=x^3$ Proof. Let $K=\mathbf{Q}(\sqrt{-37})$. It turns out that $h_K=2$; this is a question on Sheet 4. In particular, $\mathcal{O}_K$ does not have unique factorisation. The argument closely parallels the proof of Theorem 3.5, but we cannot use unique factorsation. The equation factors in $\mathcal{O}_K$ as $(y+\sqrt{-37})(y-\sqrt{-37})=x^3$. We do not have unique factorisation into elements of $\mathcal{O}_K$, only into ideals, so we think of this as an equation \[\tag{12.1} (y+\sqrt{-37})(y-\sqrt{-37})=(x)^3 \] of ideals. We are going to prove that the two ideals on the left are coprime. Suppose some prime ideal $\mathfrak{p}$ divides both terms on the LHS. Then $y+\sqrt{-37}, y-\sqrt{-37} \in \mathfrak{p}$, and so, taking the difference, $2 \sqrt{-37} \in \mathfrak{p}$. Therefore $\mathfrak{p} \mid(2 \sqrt{-37})$. (Here, of course, we are using the fact that containment and division of ideals are the same thing, Theorem 5.2.) Taking norms, we have \[\tag{12.2} N(\mathfrak{p}) \mid N(2 \sqrt{-37})=2^2 \cdot 37 . \] Also, since $\mathfrak{p} \mid(y+\sqrt{-37})$, we have $\mathfrak{p} \mid(x)^3$ and so \[\tag{12.3} N(\mathfrak{p}) \mid N\left((x)^3\right)=x^6 . \] We claim that neither 2 nor 37 divides $x$. If $2 \mid x$ then $8 \mid x^3$, so $y^2=x^3-37 \equiv 3\pmod 4$, a contradiction. If $37 \mid x$ then $37 \mid y$, and so $37^2 \mid y^2-x^3=37$. This is also a contradiction. From these facts and $(12.2),(12.3)$ we have $N(\mathfrak{p})=1$, which is impossible; therefore we are forced to conclude that $\mathfrak{p}$ does not exist, so the ideals $(y+\sqrt{-37}),(y-\sqrt{-37})$ are indeed coprime. Now we return to (12.1). By unique factorisation of ideals, both $(y+\sqrt{-37})$ and $(y-\sqrt{-37})$ are cubes of ideals. Suppose that $(y+\sqrt{-37})=\mathfrak{a}^3$. In particular, $[\mathfrak{a}]^3$ is trivial in the class group. However, we know that $h_K=2$, that is to say the class group has order 2. Therefore $[\mathfrak{a}]$ must itself be trivial, or in other words $\mathfrak{a}$ is a principal ideal. Thus we have an equation \[ (y+\sqrt{-37})=(a+b \sqrt{-37})^3 \] for some $a, b \in \mathbf{Z}$. This means that \[ y+\sqrt{-37}=u(a+b \sqrt{-37})^3 \] in $\mathcal{O}_K$, where $u$ is a unit. The only units are $\pm 1$; by replacing $a, b$ with $-a,-b$ if necessary, we may in fact assume that $u=1$. Expanding out and comparing coefficients of $\sqrt{-37}$ (which, of course, is irrational) we obtain \[ y=a\left(a^2-111 b^2\right), \quad b\left(3 a^2-37 b^2\right)=1 . \] The second of these implies that $b= \pm 1$ and hence that $3 a^2-37= \pm 1$, which is obviously impossible. This concludes the proof.

a bounded linear map T between Banach spaces is injective with closed range if and only if it is bounded below show that T is injective and has closed range ⟺∃c>0,∀x∈E,∥T(x)∥≥c∥x∥

MSE Proposition [Jacobson, Velez]: One has $G\cong \mathbb{Z}/n \rtimes (\mathbb{Z}/n)^{\times}$ if and only if $n$ is odd, or $n$ is even and $\sqrt{a}\not\in \mathbb{Q}(\zeta_n)$.

Laplacian (Example 4.22) Let $y \in \mathbb{R}^n$ and denote \[ G_y(x)=G_y^n(x):= \begin{cases}-\frac{1}{(n-2) \omega_{n-1}}|x-y|^{2-n} & \text { if } n \in \mathbb{N} \backslash\{2\} \\ \frac{1}{\omega_1} \log |x-y| & \text { if } n=2,\end{cases} \] $\Delta G_y=\delta_y \quad$ in $\mathscr{D}^{\prime}\left(\mathbb{R}^n\right)$ Cauchy-Riemann differential operators (Example 4.23) \[ \frac{\partial}{\partial z}:=\frac{1}{2}\left(\frac{\partial}{\partial x}-\mathrm{i} \frac{\partial}{\partial y}\right), \quad \frac{\partial}{\partial \bar{z}}:=\frac{1}{2}\left(\frac{\partial}{\partial x}+\mathrm{i} \frac{\partial}{\partial y}\right) \] \begin{aligned} \frac{\partial}{\partial \bar{z}}\left(\frac{1}{\pi z}\right) & =\delta_0 . \\ \frac{\partial}{\partial z}\left(\frac{1}{\pi \bar{z}}\right) & =\delta_0 . \end{aligned}

Example 8.15. Let $a ≥ b ≥ c$ be integers. We will compute $\Ext_{ℤ / 2^a}^*(ℤ / 2^b, ℤ / 2^c)$. Let $R=ℤ / 2^a ℤ$. Then $ℤ / 2^b / ℤ$ has projective resolution \[ ⋯ → R \xrightarrow{2^{a-b}} R \xrightarrow{2^b} R \xrightarrow{2^{a-b}} R \xrightarrow{2^b} R \xrightarrow{1} ℤ / 2^b ℤ . \] Therefore $\Ext_R^*(ℤ / 2^b, ℤ / 2^c)$ is the cohomology of \[ ⋯ ← ℤ / 2^c \stackrel{2^{a-b}}{←} ℤ / 2^c \stackrel{2^b}{←} ℤ / 2^c \stackrel{2^{a-b}}{←} ℤ / 2^c \stackrel{2^b}{←} ℤ / 2^c . \] Since $b>c$, the map $2^b: ℤ / 2^c → ℤ / 2^c$ is zero, so $\Ext_R^*(ℤ / 2^b, ℤ / 2^c)$ is the cohomology of \[ … ← ℤ / 2^c \stackrel{2^{a-b}}{←} ℤ / 2^c \stackrel{0}{←} ℤ / 2^c \stackrel{2^{a-b}}{←} ℤ / 2^c \stackrel{0}{←} ℤ / 2^c . \] Therefore, we have \[ \Ext_R^i(ℤ / 2^b, ℤ / 2^c)= \begin{cases}ℤ / 2^c & \text{ if } i=0 \\ \ker(f) & \text{ if } i ≥ 1 \text{ is odd } \\ \operatorname{coker}(f) & \text{ if } i ≥ 2 \text{ is even, }\end{cases} \]where $f$ is the map $ℤ / 2^c \xrightarrow{2^{a-b}} ℤ / 2^c$. Suppose that $a-b ≥ c$. Then $f=0$, so \[ \Ext_R^i(ℤ / 2^b, ℤ / 2^c)=ℤ / 2^c,   \text{ for all } i \] Suppose that $a-b

Given $R$-modules $A,B,C$ with projective resolutions $P→A,Q→B,T→C$, we see that \[\operatorname{Ext}^i_R(A,B)=H^i\operatorname{Tot}^Π\Hom_R(P,Q)=\text{chain homotopy classes of chain maps }P→Q[i]\] also \[\operatorname{Ext}^j_R(B,C)=H^j\operatorname{Tot}^Π_R\Hom_R(Q,T)=\text{chain homotopy classes of chain maps }Q→T[j]\] Note: shift by \(i\): $\Hom(Q,T[j])\overset\sim\to\Hom(Q[i],T[i+j])$ preserves chain homotopy \begin{tikzcd}P \ar[dr]\ar[rr] & & T[i+j]\\ &Q[i]\ar[ur]\end{tikzcd}gives a map $\operatorname{Ext}^i_R(A,B)×\operatorname{Ext}^j_R(B,C)→\operatorname{Ext}_R^{i+j}(A,C)$ is associative and biadditive. \(D^-(R\text{-mod}) \simeq K^-(\text{Projective }R\text{-mod})\) \begin{align*} & \Ext_R^•(A, A)=\oplus_i\Ext^i_R(A,A) \text{ is a graded ring. } \\ & \Ext^i(A, A) \times \Ext^j(A, A) \to \Ext^{i+j}(A, A) \\ \end{align*} For any $B$, $\Ext^•(A, B)=\oplus_i\Ext_R^i(A, B)$ is a graded module over this ring.

The Weingarten equations express the derivatives of the Normal using derivatives of the position vector. Let ${\bf x}:U\to\Bbb{R}^3$ be a Regular Patch, then the Shape Operator $S$ of $x$ is given in terms of the basis $\{{\bf x}_u, {\bf x}_v\}$ by \begin{align*} -S({\bf x}_u) &= {\bf N}_u = \frac{{fF-eG}}{{EG-F^2}}{\bf x}_u + \frac{{eF-fE}}{{EG-F^2}}{\bf x}_v\\ -S({\bf x}_v) &= {\bf N}_v = \frac{{gF-fG}}{{EG-F^2}}{\bf x}_u + \frac{{fF-gE}}{{EG-F^2}}{\bf x}_v \end{align*} where ${\bf N}$ is the Normal Vector, $E$, $F$, and $G$ the coefficients of the first Fundamental Form \[ ds^2 = E\,du^2 + 2F\,du\,dv + G\,dv^2 \] and $e$, $f$, and $g$ the coefficients of the second Fundamental Form given by \begin{align*} e &= -{\bf N}_u\cdot{\bf x}_u = {\bf N}\cdot{\bf x}_{uu}\\ f &= -{\bf N}_v\cdot{\bf x}_u = {\bf N}\cdot{\bf x}_{uv} = {\bf N}\cdot{\bf x}_{vu} = -{\bf N}_u\cdot{\bf x}_v\\ g &= -{\bf N}_v\cdot{\bf x}_v = {\bf N}\cdot{\bf x}_{vv} \end{align*}