Home

(a) Define $Φ\colon ℂℙ^1 × ℂℙ^1 → ℂℙ^3$ by \[Φ([x_0, x_1],[y_0, y_1])=[x_0 y_0, x_0 y_1, x_1 y_0, x_1 y_1] .\] Show that $Φ$ is well-defined and injective, and the image of $Φ$ is of the form $\{[z_0, z_1, z_2, z_3] ∈ ℂℙ^3: Q(z_0, …, z_3)=0\}$ for a homogeneous quadratic polynomial $Q$. Solution. Independent of choice of representative: $∀λ∈ℂ∖\{0\}$, \begin{gather*}Φ([λx_0,λx_1],[y_0, y_1])=[λx_0 y_0,λx_0 y_1,λx_1 y_0,λx_1 y_1] =Φ([x_0, x_1],[y_0, y_1])\\Φ([x_0,x_1],[λy_0,λy_1])=[λx_0 y_0,λx_0 y_1,λx_1 y_0,λx_1 y_1] =Φ([x_0, x_1],[y_0, y_1])\end{gather*} Ensure that image cannot be all zero. Two cases: i) $x_0≠0$ \[[\underbrace{y_0,y_1}_{\text{not both 0}},\frac{y_0x_1}{x_0},\frac{y_1x_1}{x_0}]\] ii) $x_0=0⇒x_1≠0$ \[[\frac{x_0y_0}{x_1},\frac{x_0y_1}{x_1},\underbrace{y_0,y_1}_{\text{not both 0}}]\] So $Φ$ is well-defined. If $[z_0,z_1,z_2,z_3]$ is in the image of $Φ$, two cases: i) $z_0,z_1$ not both zero \begin{gather*}[x_0, x_1]=[z_0, z_2]\cr[y_0, y_1]=[z_0,z_1]\end{gather*} ii) $z_0=z_1=0$ then $z_2,z_3$ not both zero, do similarly. So $[x_0, x_1],[y_0, y_1]$ are uniquely determined, so $Φ$ is injective. Let $Q(z_0, …, z_3)=z_0z_3-z_1z_2$. If $[z_0, z_1, z_2, z_3]$ is in image of $\Phi$, then $Q(z_0, …, z_3)=x_0y_0x_1y_1-x_0y_1x_1y_0=0$. Conversely if $[z_0, z_1, z_2, z_3] ∈ ℂℙ^3$ satisfy $Q(z_0, …, z_3)=0$, wlog $z_0≠0$. Since ${\color{red}z_2z_1}=z_0z_3$ \begin{align*}Φ([z_0, z_2], [z_0,z_1])&=[z_0z_0,z_0z_1,z_2z_0,{\color{red}z_2z_1}]\\&=[z_0z_0,z_0z_1,z_2z_0,z_0z_3]\\&=[z_0,z_1,z_2,z_3]\end{align*} so $[z_0, z_1, z_2, z_3]$ is in the image of $Φ$. (b) Write down a similar map $Ψ\colon ℂℙ^1 × ℂℙ^1 → ℂℙ^3$ with image \[\label1\tag1\{[z_0, z_1, z_2, z_3] ∈ ℂℙ^3: z_0^2+⋯+z_3^2=0\} .\] Solution. Rearrange to \[(z_0+iz_3)(z_0-iz_3)-(z_1+iz_2)(-z_1+iz_2)=0\] Define a map $f\colon ℂℙ^3 → ℂℙ^3$ \[f([z_0,z_1,z_2,z_3])=[z_0+iz_3,z_1+iz_2,-z_1+iz_2,z_0-iz_3]\] then [f^{-1}([z_0,z_1,z_2,z_3])=[i(z_0+z_3), i(z_1-z_2), z_1+z_2, z_0-z_3]\] Note that $z_0^2+⋯+z_3^2=Q(f(z_0,z_1,z_2,z_3))$, also \[Q∘f∘(f^{-1}∘Φ)=Q∘Φ=0\] so $Ψ$ can be defined $Ψ=f^{-1}∘Φ$. (c) What are lines contained in \eqref{1}? $Φ([x_0,x_1],ℂℙ^1)$ for fixed point $[x_0,x_1]$ $Φ(ℂℙ^1,[y_0,y_1])$ for fixed point $[y_0,y_1]$

(a) Define what it means for a subset $Y$ of $\mathbb{R}^n$ to be convex. Given a topological space $X$, show that any two maps $f, g: X \rightarrow Y$ are homotopic. Define when a space is contractible, and prove that every convex subset $Y$ of $\mathbb{R}^n$ is contractible. (b) Show that $S^2$ is not homeomorphic to $S^n$ for any $n \neq 2$. (c) Show that no open subset $U$ of $\mathbb{R}^2$ is homeomorphic to $\mathbb{R}^n$ for $n \neq 2$. [Hint: If $x \in U$, then prove that $U \backslash\{x\}$ retracts onto a circle.] [You may use without proof that $\pi_1\left(S^1\right) \cong \mathbb{Z}$ and $\pi_1\left(S^n\right)=0$ for $n>1$.]

Definition 6.25. A module $F$ is flat if $-⊗_R F$ is an exact functor. Lemma 6.26. Let $B$ be a left $R$-module. The following are equivalent:

1. $B$ is flat.
2. $\Tor_n^R(A, B)=0$ for all $n \geq 1$ and all left $R$-modules $A$.
3. $\Tor_1^R(A, B)=0$ for all left $R$-modules $A$.

Proof. Suppose that $B$ is flat. Let $F_• → A$ be a free resolution of $A$. Since $-⊗_R B$ is exact, the sequence \[ … → F_2 ⊗_R B → F_1 ⊗_R B → F_0 ⊗_R B → A ⊗_R B → 0 \] is exact, so the homology of \[ … → F_2 ⊗_R B → F_1 ⊗_R B → F_0 ⊗_R B → 0 \] vanishes in positive degree. Therefore, we have (1)⟹(2). The implication (2)⟹(3) is trivial. Finally, (3)⟹(1) follows from the long exact sequence of Tor, since for any short exact sequence $0 → X → Y → Z → 0$, we have that \[ 0=\operatorname{Tor}_1^R(Z, B) → X ⊗ B → Y ⊗ B → Z ⊗ B → 0 \] is exact.

Proposition. If Eisenstein works for the polynomial $f(x)=a_n x^n+\cdots+a_1 x+a_0 \in$ $\mathbb{Z}[x]$, with the prime $p$, then $p \mid \Delta(f)$. Proof. Since Eisenstein works, we have that $p \mid a_i$, for $0 \leq i \leq n-1$, and $p \nmid a_n$. Reducing modulo $p$ gives that $$ \bar{f}(x)=\overline{a_n} x^n \quad \text { in } \mathbb{Z} / p[x] . $$ Therefore $\bar{f}$ has a repeated root $x=\overline{0}$, modulo $p$. So we find, $\overline{\Delta(f)}=\Delta(\bar{f})=0$, in $\mathbb{Z} / p$. This means $\Delta(f)$ is divisible by $p$, as claimed.

Lemma: If $u \in \mathscr{D}^{\prime}(\mathbb{R})$ has order $k \in \mathbb{N}$, then $u^{\prime}$ has order $k+1$. (Note this is false if $k=0$.) Proof. It is easy to see that $u^{\prime}$ has order at most $k+1$. To see that the order is $k+1$ assume for a contradiction that the order is at most $k$. Then for any $R>0$ we find a constant $c=c_R \geq 0$ such that \[\tag4 \left|\left\langle u^{\prime}, \phi\right\rangle\right| \leq c \sum_{j=0}^k \max \left|\phi^{(j)}\right| \] holds for all $\phi \in \mathscr{D}(\mathbb{R})$ with support in $(-R, R)$. Take $\chi \in \mathscr{D}(-R, R)$ with $\int_{\mathbb{R}} \chi \mathrm{d} x=1$. For $\phi \in \mathscr{D}(-R, R)$ put $\varphi=\phi-c \chi$ with $c=\int_{\mathbb{R}} \phi \mathrm{d} x$. Then $\varphi \in \mathscr{D}(-R, R)$ and $\int_{\mathbb{R}} \varphi \mathrm{d} x=0$ and so $\psi(x)=\int_{-R}^x \varphi(t) \mathrm{d} t, x \in \mathbb{R}$, belongs to $\mathscr{D}(-R, R)$. Plugging it into (4) yields, after rearranging terms, the bound \[ |\langle u, \phi\rangle|=|\langle u', \psi\rangle|\leq c\sum_{j=0}^k \max \left|\psi^{(j)}\right|=c\max|\psi|+c\sum_{j=0}^{k-1} \max \left|\phi^{(j)}\right|\leq C \sum_{j=0}^{k-1} \max \left|\phi^{(j)}\right| \] for some constant $C=C_R$. The last $\le$ is because $\max|ψ|≤2R(\max|ϕ|+|c|\max|\chi|)≤2R(\max|ϕ|+2R|\phi|\max|\chi|)$ But this is contradicting the assumption that $u$ has order $k$.

Weyl's Lemma. Let $\Omega \subseteq \mathbb{R}^N$ be open. If $u \in \mathfrak{D}^{\prime}(\Omega ; \mathbb{R})$ (the space of Schwartz distributions on $\Omega$) satisfies $\Delta u=f \in C^{\infty}(\Omega ; \mathbb{R})$ in the sense that \[ \langle\Delta \psi, u\rangle=\langle\psi, f\rangle, \quad \psi \in C_{\mathrm{c}}^{\infty}(\Omega ; \mathbb{R}), \] then $u \in C^{\infty}(\Omega ; \mathbb{R})$.   Proof. Set \[ \gamma_t(x)=(4 \pi t)^{-\frac{N}{2}} \exp \left[-\frac{|x|^2}{4 t}\right] . \] Given $x_0 \in \Omega$, choose $r>0$ so that $\bar{B}\left(x_0, 3 r\right) \subset \subset \Omega$ and $\eta \in C_{\mathrm{c}}^{\infty}\left(B\left(x_0, 3 r\right) ;[0,1]\right)$ so that $\eta=1$ on $\bar{B}\left(x_0, 2 r\right)$. Set $v=\eta u$ and $w=\Delta v-\eta f$. Then $w$ is supported in $B\left(x_0, 3 r\right) \backslash \bar{B}\left(x_0, 2 r\right)$. Now take \[ v_t(x)=\gamma_t \star v(x) \equiv\left\langle\gamma_t(\cdot-x), v\right\rangle \\ w_t(x)=\gamma_t \star w(x)\equiv\left\langle\gamma_t(\cdot-x), w\right\rangle . \] For each $t>0, v_t$ is smooth. Moreover, \[ \dot{v}_t(x)=\left\langle\gamma_t(\cdot-x), \eta f\right\rangle+\left\langle\gamma_t(\cdot-x), w\right\rangle \] and so \[ v_t(x)=v_1(x)-\int_t^1 \gamma_\tau \star(\eta f)(x) d \tau-\int_t^1 w_\tau(x) d \tau . \]

Students in abstract algebra and number theory are usually interested to see that the arithmetic functions $\phi(n)$, Euler's function, and $d(n)$, the number of positive divisors of $n$, occur quite naturally in the solution of some group theory problems. It appears that the function $\sigma(n)$, the sum of the positive divisors of $n$, does also, in counting the number of subgroups of the dihedral group $D_n$. A formula is developed here for that number. Theorem. The number of subgroups of the dihedral group $D_n(n \geqq 3)$ is $d(n)+\sigma(n)$. Proof. When considered geometrically, $D_n$ consists of $n$ rotations and $n$ reflections of the regular $n$-gon. The subgroups of $D_n$ are of two types: (1) Those containing rotations only, and (2) those containing rotations and reflections. The subgroups of type 1 are simply the subgroups of $Z_n$, the cyclic group of order $n$, and the number of them is $d(n)$. The subgroups of type 2 contain an equal number of rotations and reflections, say $t$, of each. Now the $t$ rotations must comprise the unique subgroup of $Z_n$ of order $t$, whence $t \mid n$, but the $t$ reflections can be chosen in several ways. In fact, the axes of reflection form a star-shaped figure with equal central angles which can be positioned in the $n$-gon in $n / t$ ways. Thus for each divisor $t$ of $n$, there are $n / t$ subgroups of type 2 , and the total number of subgroups of type 2 must be $$ \sum_{t \mid n} n / t=\sum_{t \mid n} t=\sigma(n) . $$ This completes the proof of the theorem. It is of interest to note that when $p$ is a prime greater than 2, the number of subgroups of $D_p$ is simply $p+3$. Subgroups of dihedral groups (1)

Let $f$ be a tempered $\mathrm{L}^p$ function on $ℝ^n$ for some $p ∈[1, ∞]$. Define for each $j ∈ ℕ$, \[ f_j(ξ)=∫_{B_j(0)} f(x) \mathrm{e}^{-\mathrm{i} ξ ⋅ x} \mathrm{~d} x \] where $B_j(0)$ is the open ball in $ℝ^n$ with centre 0 and radius $j$. Explain why $f_j ∈ \mathrm{C}_0(ℝ^n)$ and so that it in particular can be considered as a tempered distribution on $ℝ^n$. Prove that $f_j → \hat{f}$ in $\mathscr{S}'(ℝ^n)$ as $j →+∞$. Hence find the limit of \[ ∫_{-j}^j x^k \mathrm{e}^{-\mathrm{i} ξ x} \mathrm{~d} x \] as $j →+∞$ for each $k ∈ ℕ_0$. Solution. Since a tempered $L^p$ function is in particular in $L^p_\text{loc}$ (since $L^p_\text{loc}⊆L^1_\text{loc}$) so in $L^1_\text{loc}$. By Riemann-Lebesgue lemma $f_j=\widehat{f1_{B_j(0)}} ∈ C_0⊆𝒮'$ Next, claim: $f1_{B_j(0)}→f$ in $𝒮'$ as $j→∞$. Proof. Note that for $ϕ∈𝒮$: $f1_{B_j(0)}ϕ-fϕ=f1_{ℝ^n∖B_j(0)}$ By result from lectures $∀k∈ℕ_0,∀q∈[1,∞],∃c=c(k,q,n)$ such that $‖(1+|⋅|^2)^{k/2}ψ‖_q≤c\bar{S}_{k+n+1,0}(ψ),∀ψ∈𝒮$. We use this bound with $k=m+1$ and $q=\frac{p}{p-1}$. \[\int_{ℝ^n}|f1_{ℝ^n∖B_j(0)}ϕ|dx=\int_{ℝ^n}\frac{|f|}{(1+|⋅|^2)^{m/2}}\frac{1_{ℝ^n∖B_j(0)}}{(1+|⋅|^2)^{1/2}}(1+|⋅|^2)^{(m+1)/2}|ϕ|dx\\ ≤⋯→0∎\] By $𝒮'$ continuity of $ℱ$, $f_j → \hat{f}$ in $\mathscr{S}'(ℝ^n)$ as $j →+∞$. Hence \[ \lim_{j→∞}∫_{-j}^j x^k \mathrm{e}^{-\mathrm{i} ξ x} \mathrm{~d} x=\widehat{x^k}=2π\mathrm{i}^kδ^{(k)}(ξ) \]

Let $ℙ(V)$ be a projective space of dimension 3, and let $L_1, L_2, L_3$ be non-intersecting projective lines in $ℙ(V)$. (a) Show that there are unique isomorphisms $α\colon L_1 → L_2, β\colon L_1 → L_3$ such that $x, α(x), β(x)$ are collinear for each $x ∈ L_1$. Solution. Let $L_j=P(U_j),j=1,2,3$. Then $U_1,U_2,U_3$ are vector subspaces of $ℝ^4$, $\dim U_j=2$. Since $L_2∩L_3=∅$, $U_2∩U_3=\{0\}$. Hence $ℝ^4=U_2⊕U_3$, counting dimensions. Let $v∈U_1∖\{0\}$ projects to $u_2∈U_2,u_3∈U_3$, $v=u_2+u_3$. If $u_2=0$ then $[v]∈ L_1∩ L_3$❌ If $u_3=0$ then $[v]∈ L_1∩ L_2$❌ The linear transformations $\tilde{α}(v)=u_2,\tilde{β}(v)=u_3$ induce projective transformations $α([v])=[u_2],β([v])=[u_3]$, then all of $v,u_2,u_3$ lie in $⟨u_2,u_3⟩$, so all of $[v],[u_2],[u_3]$ lie in the projective line $ℙ(⟨u_2,u_3⟩)$. For uniqueness, suppose $x, α'(x), β'(x)$ are collinear, choose representatives $v,u_2',u_3'$ of $x,α'(x),β'(x)$ then $v=C_1u_2'+C_2u_3'$, then $C_1u_2'+C_2u_3'=u_2+u_3$, then $C_1u_2'-u_2=u_3-C_2u_3'∈U_2∩U_3=\{0\}$, so $α'=α,β'=β$. (b) Suppose $ϕ\colon L_1 → L_1$ is a projective transformation of $L_1$. Show that there exists a unique projective transformation $Φ\colon ℙ(V) → ℙ(V)$ such that $Φ(L_j)=L_j$ and $Φ|_{L_1}=ϕ$. Solution. For $v∈V∖\{0\}$, write $v=u_2+u_3,u_2∈U_2,u_3∈U_3$. Since $\tilde{α}\colon U_1→U_2,\tilde{β}\colon U_1→U_3$ are isomorphisms, we define $Φ([v])≔[\tilde{α}ϕ\tilde{α}^{-1}u_2+\tilde{β}ϕ\tilde{β}^{-1}u_3]$. If $v∈U_1$ then $\tilde{α}^{-1}u_2=\tilde{β}^{-1}u_3=v$, so $Φ([v])=ϕ([v])$. If $v∈U_2$ then $u_2=v,u_3=0$, so $Φ([v])=[\tilde{α}ϕ\tilde{α}^{-1}v]∈L_2$. If $v∈U_3$ then $u_2=0,u_3=v$, so $Φ([v])=[\tilde{β}ϕ\tilde{β}^{-1}v]∈L_3$. Uniqueness: Take a basis $u,v$ for $U_2$ $e_1=\tilde{α}(u)$ $e_2=\tilde{α}(v)$ $e_3=\tilde{β}(u)$ $e_4=\tilde{β}(v)$ $\begin{bmatrix}a&b\\c&d\end{bmatrix}$matrix of transformation $ϕ$ on $u,v$ $\begin{bmatrix}a&b&0&0\\c&d&0&0\\0&0&a&b\\0&0&c&d\end{bmatrix}$matrix of transformation $Φ$ on $e_1,e_2,e_3,e_4$ $Φ(e_1+e_3)=a(e_1+e_3)+c(e_2+e_4)∈⟨e_1+e_3,e_2+e_4⟩$ $Φ(e_2+e_4)=b(e_1+e_3)+d(e_2+e_4)∈⟨e_1+e_3,e_2+e_4⟩$ But $u=e_1+e_3,v=e_2+e_4$, so $Φ$ fixes $⟨u,v⟩=U_1$ and $Φ|_{U_1}=ϕ$. $Φ(e_1)=ae_1+ce_2∈⟨e_1,e_2⟩$ $Φ(e_2)=be_1+de_2∈⟨e_1,e_2⟩$ so $Φ$ fixes $⟨e_1,e_2⟩=U_2$, similarly it fixes $⟨e_3,e_4⟩=U_3$.