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Let $g_j(x)=g(j x), x ∈(0,1)$, where $g$ is $T$-periodic and on $(0, T]$ is given by \[ g=-117\mathit1_{\left(0, \frac{T}{2}\right]}+117\mathit1_{\left(\frac{T}{2}, T\right]}= \begin{cases}-117 & \text { on }\left(0, \frac{T}{2}\right] \\ +117 & \text { on }\left(\frac{T}{2}, T\right]\end{cases} \] Clearly $\left\|g_j\right\|_1=117 \not → 0$. On Problem Sheet 3 you will be asked to prove that $g_j → 0$ in $𝒟'(0,1)$. Problem 3. Let $g ∈ \mathrm{L}_{\text{loc}}^1(ℝ)$ and assume that $g$ is $T$ periodic for some $T>0: g(x+T)=g(x)$ holds for almost all $x ∈ ℝ$. Define for each $j ∈ ℕ$ the function \[ g_j(x)=g(j x),   x ∈(0,1) . \] Prove that \[ g_j → \frac{1}{T} ∫_0^T g \mathrm{~d} x   \text { in }   𝒟'(0,1)   \text { as }   j → ∞ . \] Proof. (page 11) By density assume $g$ is piecewise continuous. If $\phi \in 𝒟(0,1)$, define \[ G(x)=\int_0^x g(t) d t,  x \in(0, T], \] Replacing $g$ with $g-\frac1T∫_0^T g \mathrm{~d} x$, we can assume that $G(T)=0$, so the $T$-periodic extension of $G$ is piecewise $C^1$ and $G'(x)=g(x)$ a.e. Integrate by parts, using $\phi(0)=\phi(1)=0$, \begin{aligned} \left\langle g_j, \phi\right\rangle&=\int_0^1 g(j x) \phi(x) d x\\ &={\left[\frac{G(j x)}{j} \phi(x)\right]_0^1-\frac1j\int_0^1 G(j x)\phi'(x) d x} \\ & \rightarrow 0 . \end{aligned}

Corollary 4.14. There exists a projective plane of order $p^n$. Proof. (b) Let $\mathbb{F}_{p^n}$ be a finite field of order $p^n$ and $V$ be a three-dimensional vector space over $\mathbb{F}_{p^n}$. Consider the projective plane $\mathbb{P}\left(\mathbb{F}_{p^n}\right)$. Fix a two dimensional subspace $U$ in $V$. Then $U=$ $\left\{a u+b v: a, b \in \mathbb{F}_{p^n}\right\}$ for some linearly independent vectors $u, v \in U$. Notice that when $a \neq 0$, $a u+b v$ and $u+\frac{b}{a} v$ define the same point in the projective plane, hence it suffices to count the number of choices of $u+c v$ for $c \in \mathbb{F}_{p^n}$, which is $p^n$. Adding the case when $a=0$, corresponding to the point $\langle v\rangle$, gives $p^n+1$ points passing through the line $U$. Hence $\mathbb{P}\left(\mathbb{F}_{p^n}\right)$ has order $p^n$.

苏家驹之代数的五次方程式解法不能成立之理由 華羅庚 1930年12月出版的《科學》15卷2期307-309頁 五次方程式經 Abel, Galois 之證明後，一般算學者均認為不可以代數解矣，而《學藝》七卷十號載有蘇君之《代數的五次方程式之解法》一文，羅欣讀之而研究之，於去年冬亦仿得「代數的六次方程式之解法」矣。羅對此欣喜異常，意為果能成立，則於算學史中亦可佔一席之地，惟自思若不將Abel言論駁倒，終不能完全此種理論，故羅沉思於Abel之論中。凡一閱月，見其條例精嚴，無懈可擊，後經本社編輯暗示遂從事於蘇君解法確否之工作，與六月中遂得其不能成立之理由，羅安敢自秘，特公之於世，尚祈示正焉。 解 法 簡 述 用 Sylvester 之分离消去法 (Dialytic method of elimination) 将普通形 $$ x^5+p_1 x^4+p_2 x^3+p_3 x^2+p_4 x+p_5=0 $$ 化为可解形 $$X^5+P_1 X^4+P_2 X^3+P_3 X^2+P_4 X+P_5=0$$(中有 $P_1=0, P_3=0, P_2^2=5 P_4$), 而 $x, X$ 有 $X=n_0+n_1 x+n_2 x^2+n_3 x^3+n_4 x^4$ 之關係. $P_1, P_2, P_3, P_4, P_5$ 为 $n_0, n_1, n_2, n_3, n_4$ 之一次、二次、三次、四次、五次齊次函數. $P_1=0$, 即 $n_0$ 可以 $n_1, n_2, n_3, n_4$ 之一齊次函數表之, 以之代入 $P_2, P_3, P_4, P_5$, 則得 $n_1, n_2, n_3, n_4$ 之二、三、四、五次齊次函數, 而 $P_3$ 之一般形可寫為 $$\tag I \begin{gathered} A_1 n_1^3+A_2 n_2^3+A_3 n_3^3+A_4 n_4^3+A_5 n_1^2 n_2+A_6 n_1^2 n_3+A_7 n_1^2 n_4+A_8 n_2^2 n_1+A_9 n_2^2 n_3 \\ + A_{10} n_2^2 n_4+A_{11} n_3^2 n_5+A_{12} n_3^2 n_2+A_{13} n_3^2 n_4+A_{14} n_4^2 n_1+A_{15} n_4^2 n_2+A_{16} n_4^2 n_3 \\ + A_{17} n_1 n_2 n_3+A_{18} n_1 n_2 n_4+A_{19} n_1 n_2 n_4+A_{20} n_2 n_3 n_4, \end{gathered} $$ 式中 $A_1, \cdots, A_{20}$ 为 $p_1, \cdots, p_5$ 之函數為已知者. 若令等於下式 $$\tag{II} \begin{gathered} (a_1 n_1+a_2 n_2)(a_3 n_1^2+a_4 n_2^2+a_5 n_3^2+a_6 n_4^2+a_7 n_1 n_2+a_8 n_1 n_3+a_9 n_1 n_4+a_{10} n_2 n_3 \\ + a_{11} n_2 n_4+a_{12} n_3 n_4)+(a_{13} n_3+a_{14} n_4)(a_{15} n_1^2+a_{16} n_2^2+a_{17} n_3^2+a_{18} n_4^2+a_{19} n_1 n_2\\ +a_{20} n_1 n_3+a_{21} n_1 n_4+a_{22} n_2 n_3+a_{23} n_2 n_4+a_{24} n_3 n_4) \end{gathered} $$ 式中 $a_1, \cdots, a_{24}$ 為未定係數. 再設 $a_1 n_1+a_2 n_2=0, a_{13} n_3+a_{14} n_4=0$, 代入 $P_2^2=5 P_4$ 式中, 則此式為 $n_2, n_4$之四次齊次函數, 解之, 則得 $n_2, n_4$ 之比值, 由此可作得 $n_0: n_1: n_2: n_3: n_4$ 之值, 故普通形可化为上之可解形, 換言之, 即五次方程式可得而解矣. 謬 誤 點 羅研究上意知其謬誤在 $P_3$ 中, 即 (I) 不能等於 (II) 也. 夫求未定係數 $a_1, \cdots, a_{24}$, 原文亦有求之之二十方程式, 羅為便利討探計, 特分之為四類, 轉錄於下: (一) $a_1 a_3=A_1$,$a_2 a_4=A_2$, $a_3 a_2+a_1 a_7=A_5$,$a_4 a_1+a_2 a_7=A_8$; (二) $a_{13} a_{17}=A_3$,$a_{14} a_{18}=A_4$, $a_{17} a_{14}+a_{13} a_{24}=A_{13}$,$a_{18} a_{13}+a_{14} a_{24}=A_{16}$; (三) $a_{13} a_{15}+a_1 a_3=A_6$,$a_{14} a_{15}+a_1 a_9=A_7$, $a_1 a_{11}+a_2 a_9=a_{18}-a_{14} a_{19}$,$a_2 a_{11}+a_{14} a_{16}=A_{10}$, $a_2 a_{10}+a_{13} a_{16}=A_9$,$a_1 a_{10}+a_2 a_8=a_{17}-a_{13} a_{19}$; (四) $a_1 a_5+a_{13} a_{20}=A_{11}$,$a_2 a_5+a_{13} a_{22}=A_{12}$, $a_2 a_{12}+a_{14} a_{22}=A_{19}-a_{13} a_{23}$,$a_1 a_{12}+a_{13} a_{21}=A_{20}-a_{14} a_{20}$, $a_1 a_6+a_{14} a_{21}=A_{14}$,$a_2 a_6+a_{14} a_{23}=A_{15}$. 依原所謂假 $a_7, a_{24}$ 則由 (一), (二) 得 $a_1, a_2, a_3, a_4, a_{13}, a_{14}, a_{17}, a_{18}$ 之值, 則第二類乃為 $a_8, a_{15}, a_9, a_{11}, a_{16}, a_{10}$ 之聯立一次方程式 (設 $a_{19}$ 為已知), 以行列式解之, 知其各分母悉為 $$ \Delta=\left|\begin{array}{cccccc} a_1 & a_{13} & 0 & 0 & 0 & 0 \\ 0 & a_{14} & a_1 & 0 & 0 & 0 \\ 0 & 0 & a_2 & a_1 & 0 & 0 \\ 0 & 0 & 0 & a_2 & a_{14} & 0 \\ 0 & 0 & 0 & 0 & a_{13} & a_2 \\ a_2 & 0 & 0 & 0 & 0 & a_1 \end{array}\right| . $$ 然 $$ \left|\begin{array}{cccccc} a_1 & a_{13} & 0 & 0 & 0 & 0 \\ 0 & a_{14} & a_1 & 0 & 0 & 0 \\ 0 & 0 & a_2 & a_1 & 0 & 0 \\ 0 & 0 & 0 & a_2 & a_{14} & 0 \\ 0 & 0 & 0 & 0 & a_{13} & a_2 \\ a_2 & 0 & 0 & 0 & 0 & a_1 \end{array}\right|=a_1\left|\begin{array}{ccccc} a_{14} & a_1 & 0 & 0 & 0 \\ 0 & a_2 & a_1 & 0 & 0 \\ 0 & 0 & a_2 & a_{14} & 0 \\ 0 & 0 & 0 & a_{13} & a_2 \\ 0 & 0 & 0 & 0 & a_1 \end{array}\right| -a_{13}\left|\begin{array}{ccccc} 0 & a_1 & 0 & 0 & 0 \\ 0 & a_2 & a_1 & 0 & 0 \\ 0 & 0 & a_2 & a_{14} & 0 \\ 0 & 0 & 0 & a_{13} & a_2 \\ a_2 & 0 & 0 & 0 & a_1 \end{array}\right|=0 $$ 而 $a_8, a_{15}, a_9, a_{11}, a_{16}, a_{10}=\delta / \Delta$. 因 $\Delta=0$, 故 $a_8, a_{15}, a_9, a_{11}, a_{16}, a_{10}$ 非不定即無限大, 故 (I) 等 (II) 之謬論不攻自破矣. 換言之, 即 $P_3$ 為零不能解得二一次式, 故此法亦不能解五次方程式也.

Part II Lecture Notes: Algebraic Topology, James Lingard Orientations Let $\Delta$ be an $n$-simplex with vertices $x_{1}, \ldots, x_{n+1}$. If $n \geq 0$ then an orientation of $\Delta$ is an equivalence class of orderings of the $x_{i}$, where the orderings $x_{1}, \ldots, x_{n+1}$ and $x_{\sigma(1)}, \ldots, x_{\sigma(n+1)}$ are equivalent if and only if $\sigma \in A_{n+1}$. If $n=0$ then an orientation of $\Delta$ is either 1 or -1 . An oriented simplex $\sigma$ is a simplex together with a choice of orientation. We write $-\sigma$ to mean the simplex with the opposite choice of orientation. We shall write $\left[x_{0}, \ldots, x_{n}\right]$ to mean the oriented simplex with the orientation corresponding to the given ordering of the vertices. An oriented simplex induces an orientation on each of its codimension-one faces in the following manner: - if $i$ is even, the induced orientation of $\left[x_{0}, \ldots, \hat{x}_{i}, \ldots, x_{n}\right]$ is the one corresponding to this ordering of the vertices, and - if $i$ is odd, the induced orientation is the opposite to the one corresponding to this ordering. We can check that this is well-defined. Chains and boundaries Let $X$ be a simplicial complex. We define the group $C_{k}(X)$ of $k$-chains on $X$ to be the free abelian group generated by the oriented $k$-simplices in $X$, modulo the relation that $(-1) \sigma=-\sigma$ for any oriented $k$-simplex $\sigma$. For any oriented $k$-simplex $\sigma$ in $X$, we define the boundary $\partial \sigma$ of $\sigma$ to be the sum of the codimension-one faces of $\sigma$, with the orientations induced from $\sigma$. So $\partial \sigma \in C_{k-1}(X)$. We can extend this definition to get a group homomorphism $\partial: C_{k}(X) \rightarrow C_{k-1}$ in the obvious manner. So we get the sequence $$ \cdots \stackrel{\partial}{\longrightarrow} C_{k+1}(X) \stackrel{\partial}{\longrightarrow} C_{k}(X) \stackrel{\partial}{\longrightarrow} C_{k-1}(X) \stackrel{\partial}{\longrightarrow} \cdots $$ In fact, the composition of any two of the consecutive homomorphisms is the zero homomorphism, or in other words, $\partial^{2}=0$. It suffices to prove that $\partial^{2} \sigma=0$ for any oriented $(k+1)$-simplex $\sigma$. We have $$ \begin{aligned} \partial^{2}\left[x_{0}, \ldots, x_{k+1}\right]= & \partial \sum_{i=0}^{k+1}(-1)^{i}\left[x_{0}, \ldots, \hat{x}_{i}, \ldots, x_{k+1}\right] \\ = & \sum_{i=0}^{k+1}(-1)^{i} \sum_{j=0}^{i-1}(-1)^{j}\left[x_{0}, \ldots, \hat{x}_{j}, \ldots, \hat{x}_{i}, \ldots, x_{k+1}\right] \\ & +\sum_{i=0}^{k+1}(-1)^{i} \sum_{j=i+1}^{k+1}(-1)^{j-1}\left[x_{0}, \ldots, \hat{x}_{i}, \ldots, \hat{x}_{j}, \ldots, x_{k+1}\right], \end{aligned} $$ where the terms in the final expression cancel in pairs. Homology groups Define the subgroup $Z_{k}(X) \leq C_{k}(X)$ of $k$-chains in $X$ to be the kernel of $\partial$, and define the subgroup $B_{k}(X) \leq C_{k}(X)$ of $k$-boundaries to be the image of $\partial$. Then since $\partial^{2}=0, B_{k}(X)$ is contained within $Z_{k}(X)$ and so we may define the $k$ th homology group of $X$ as the quotient $$ H_{k}(X)=Z_{k}(X) / B_{k}(X) $$ It is easy to see that $H_{0}(X)$ is a free abelian group whose rank is the number of connected components of $X$. Homology groups of an $n$-simplex If $X$ is a point, then by considering the groups of $k$-chains on $X$ we see immediately that $H_{0}(X)=\mathbb{Z}$ and $H_{i}(X)=0$ for $i \geq 1$. To compute the homology groups of an $n$-simplex for $n \geq 2$ we may use the following device. If $X \subset \mathbb{R}^{N}$ is a simplicial complex, define the cone on $X$ to be the simplicial complex $C X$ formed by embedding $\mathbb{R}^{N}$ in $\mathbb{R}^{N+1}$, choosing a point $v \in \mathbb{R}^{N+1} \backslash \mathbb{R}^{N}$ and then taking the union of all the line segments from $v$ to each of the points in $X$. Then we see that $C X$ is a simplicial complex, whose simplices are those in $X$, those formed as the convex hull of $v$ and a simplex in $X$, and the point $v$ itself. The homology groups of any cone are the same as those of a point. To prove this we define the homomorphism $d: C_{q}(C X) \rightarrow C_{q+1}(C X)$ which sends an oriented $q$-simplex $\sigma=\left[v_{0}, \ldots, v_{q}\right]$ in $C X$ to $\left[v, v_{0}, \ldots, v_{q}\right]$ if $\sigma$ is contained in $X$, or 0 otherwise. We can check that this is a well-defined homomorphism. Then we show that $$ (\partial \circ d)(\sigma)=\sigma-(d \circ \partial)(\sigma) $$ for any oriented $q$-simplex $\sigma$ with $q \geq 1$. But then if $z$ is any $q$-cycle in $C X$ with $q \geq 1$, $$ \partial(d(z))=z-d(\partial(z))=z-d(0)=z $$ and so $z$ is a boundary. Hence $H_{q}(C X)=0$. Now since an $(n+1)$-simplex is just the cone on an $n$-simplex, we have that for any simplex $X$, $H_{0}(X)=\mathbb{Z}$ and $H_{i}(X)=0$ for $i \geq 1$. Homology groups of $S^{n}$ If $n=0$ then the homology groups of $S^{n}$ are clearly $H_{0}\left(S^{0}\right)=Z^{2}, H_{i}\left(S^{0}\right)=0$ for $i \geq 1$. If $n \geq 1$ then we may triangulate $S^{n}$ is a simple way. Let $\Delta$ denote the simplicial complex which is the union of all the faces of an $(n+1)$-simplex, and let $\Sigma$ denote the same simplicial complex but without the single $(n+1)$-dimensional face. Then $\Sigma$ is a triangulation of $S^{n}$. Now the chain complex of $\Sigma$ is exactly the same as that of $\Delta$, except that we have $C_{n+1}(\Sigma)=0$, whereas $C_{n+1}(\Delta) \cong \mathbb{Z}$, generated by the single $(n+1)$-dimensional face. Therefore, for $i \leq(n-1)$ the homology groups $H_{i}(\Delta)$ and $H_{i}(\Sigma)$ are the same. But then we see that $H_{n}(\Sigma)$ is the kernel of the map $\partial: C_{n}(\Sigma) \rightarrow C_{n-1}(\Sigma)$, which is also the kernel of $\partial: C_{n}(\Delta) \rightarrow C_{n-1}(\Delta)$. But by the exactness of the sequence for $\Delta$, this is just the image of $\partial: C_{n+1}(\Delta) \rightarrow C_{n}(\Delta)$, which is isomorphic to $C_{n+1}(\Delta) \cong \mathbb{Z}$ as $\partial$ is injective. We will show that the homology groups of a space are independent of its triangulation, and hence we may conclude that $$ H_{i}\left(S^{n}\right)= \begin{cases}\mathbb{Z}^{2} & \text { if } n=0 \text { and } i=0 \\ \mathbb{Z} & \text { if } n \geq 1 \text { and } i=0 \text { or } n \\ 0 & \text { otherwise. }\end{cases} $$ Induced homomorphisms Let $f: X \rightarrow Y$ be a simplicial map. Then $f$ determines a homomorphism $$ f_{*}: C_{i}(X) \longrightarrow C_{i}(Y) $$ by $$ f_*([x_{0}, \ldots, x_{i}])= \begin{cases}{[f(x_0), \dots, f(x_i)]} & \text { if } f(x_0), \ldots, f(x_i) \text { are all distinct } \\ 0 & \text { otherwise }\end{cases} $$ We can check that $f_{*}$ is a chain map, that is, that $$ \partial \circ f_{*}=f_{*} \circ \partial $$ and so the following diagram commutes: \begin{CD} \cdots@>\partial>>C_{i+1}(X)@>\partial>>C_i(X)@>\partial>>C_{i-1}(X)@>\partial>>\cdots\\ &@VVf_*V@VVf_*V@VVf_*V\\ \cdots@>\partial>>C_{i+1}(Y)@>\partial>>C_i(Y)@>\partial>>C_{i-1}(Y)@>\partial>>\cdots \end{CD} It follows that $f_{*}$ maps $Z_{i}(X)$ into $Z_{i}(Y)$, and $B_{i}(X)$ into $B_{i}(Y)$, and so $f_{*}$ gives rise to a homomorphism, also denoted $f_{*}$, from $H_{i}(X)$ to $H_{i}(Y)$. Now it can be shown that barycentric subdivision of a complex does not change the homology groups (proof omitted - see Armstrong pp. 185-188). Thus if $f:|X| \rightarrow|Y|$ is any continuous map then we may define a homomorphism $$ f_{*}: H_{i}(X) \rightarrow H_{i}(Y) $$ as the composition $$ H_{i}(X) \cong H_{i}\left(X^{m}\right) \stackrel{s_{*}}{\longrightarrow} H_{i}(Y) $$ where $s: X \rightarrow Y$ is some simplicial approximation to $f$. It can be shown that this is welldefined, independent of the choice of $s$, by showing that any two "close" simplicial maps give rise to the same homomorphisms of homology groups. Functorial properties. Thus homology groups are invariant under homotopy equivalence. Applications $S^{m}$ and $S^{n}$ are not homotopy equivalent and thus $\mathbb{R}^{m}$ and $\mathbb{R}^{n}$ are not homeomorphic if $m \neq n$. The Brouwer fixed-point theorem.

Let $X=C([a, b])$, let $k:[a, b]×[a, b]→ℝ$ be continuous. Show that $T ∈ ℬ(X)$ defined by $T f(x)=∫_a^b k(x, t) f(t) d t$ is a compact operator. Method 1. Since $[a, b]×[a, b]$ is compact, $k$ is uniformly continuous, $∀ε>0∃δ>0∀x_1,x_2∈[a,b],|x_1-x_2|<δ$ then $|k(x_1,t)-k(x_2,t)|<ε$. $T(\bar B^X)$ is equicontinuous: for all $f∈\bar B^X$, \begin{align*}\left|Tf(x_1)-Tf(x_2)\right| & =\left|∫_a^b\left(k(x_1, t) f(t)-k(x_2, t) f(t)\right) d t\right| \\& ≤ ∫_a^b \overbrace{\left|k(x_1, t)-k(x_2, t)\right|}^{<ε} ⋅ \overbrace{|f(t)|}^{≤1} d t<(b-a)ε\end{align*} $T(\bar B^X)$ is uniformly bounded: \begin{align*}\sup_{‖f‖_∞≤1}‖T f‖_∞&=\sup_{‖f‖_∞≤1}\sup_{x∈[a,b]}\left|∫_a^b k(x, t) f(t) d t\right|\\& ≤ \sup_{‖f‖_∞≤1}\sup_{x∈[a,b]} ∫_a^b|k(x, t)|\underbrace{|f(t)|}_{≤‖f‖_∞}d t\\& ≤ \sup_{x∈[a,b]} ∫_a^b|k(x, t)| d t\\&≤(b-a)‖k‖_∞\end{align*} $⇒$ By Arzelà-Ascoli: $\overline{T(\bar B^X)}$ is compact $⇒ T$ is compact operator. Method 2. $X$ has an orthonormal basis $\{e_n\}$ by 1.13. For each $s ∈[a, b]$ and $n ∈ ℕ$, let $k_s∈X,k_s(t)≔k(s, t)$ \[T e_n(s)=∫_a^b k(s, t) e_n(t) d t=⟨k_s, e_n⟩,\] By 1.11 and continuity of $k_s$ \begin{align*}\sum_{n=1}^{\infty}‖T e_n‖_X^2 & =\sum_{n=1}^{\infty} \int_a^b\left|⟨k_s, e_n⟩\right|^2 d s\\&=\int_a^b \sum_{n=1}^{\infty}\left|⟨k_s, e_n⟩\right|^2 d s \\& =\int_a^b‖k_s‖_X^2 d s<\infty\end{align*} Since $\left\{e_n\right\}$ is an orthonormal basis, any $x∈X$ can be written as $x=\sum_{n=1}^{∞}⟨x, e_n⟩e_n$. For each $k ∈ ℕ$ define an operator $T_k ∈ B(X)$ by \[T_k x=T\left(\sum_{n=1}^k⟨x, e_n⟩e_n\right)=\sum_{n=1}^k⟨x, e_n⟩T e_n\] Clearly, $\dim\operatorname{im}(T_k)≤k$. Also, for any $x∈X$, using Cauchy-Schwarz inequality for $ℓ^2$ and 1.11 \begin{align*}‖(T_k-T) x‖ & =\left‖\sum_{n=1}^k⟨x, e_n⟩T e_n-\sum_{n=1}^{∞}⟨x, e_n⟩T e_n\right‖ \\& ≤ \sum_{n=k+1}^{∞}\left|⟨x, e_n⟩\right|‖T e_n‖ \\& ≤\left(\sum_{n=k+1}^{∞}\left|⟨x, e_n⟩\right|^2\right)^{1 / 2}\left(\sum_{n=k+1}^{∞}‖T e_n‖^2\right)^{1 / 2} \\& ≤‖x‖\left(\sum_{n=k+1}^{∞}\left‖T e_n\right‖^2\right)^{1 / 2}\end{align*} Hence \[‖T_k-T‖ ≤\left(\sum_{n=k+1}^{∞}\left‖T e_n\right‖^2\right)^{1 / 2}→0\] we have $T_k→T$, 1.15 shows that $T$ is compact.

Let $f(x)=𝖾^{-|x|}, x ∈ ℝ^n$, where $|x|=\sqrt{x_1^2+⋯+x_n^2}$ is the usual norm of $x$. (a) Compute the Fourier transform of $f$ when the dimension $n=1$. Deduce for $λ ≥ 0$ the identity \[𝖾^{-λ}=\frac{1}{π} ∫_{-∞}^{∞} \frac{1}{1+ξ^2} 𝖾^{𝗂 λ ξ} 𝖽 ξ\] (b) Using $\frac{1}{ξ^2+1}=∫_0^{∞} 𝖾^{-(1+ξ^2) t} 𝖽 t$ and (a) show that \[𝖾^{-λ}=∫_0^{∞} \frac{1}{\sqrt{π t}} 𝖾^{-t-\frac{λ^2}{4 t}} 𝖽 t\] holds for all $λ ≥ 0$. (c) Calculate the Fourier transform $\hat{f}$ in the general $n$-dimension, using the identity from (b) with $λ=|x|$. Solution. (a) When the dimension $n=1$, \[ \hat{f}(ξ)=∫_{-∞}^{∞}𝖾^{-|x|}𝖾^{-𝗂xξ}𝖽x =∫_0^{∞}𝖾^{-(1+𝗂ξ)x}𝖽x+∫_{-∞}^0𝖾^{-(-1+𝗂ξ)x}𝖽x =\frac1{1+𝗂ξ}+\frac1{-(-1+𝗂ξ)} =\frac2{1+ξ^2} \] By Fourier inversion formula \[𝖾^{-|x|}=\frac{1}{2π} ∫_{-∞}^{∞} \frac{2}{1+ξ^2} 𝖾^{𝗂xξ} 𝖽 ξ\] Take $x=λ ≥ 0$ \[𝖾^{-λ}=\frac{1}{π} ∫_{-∞}^{∞} \frac{1}{1+ξ^2} 𝖾^{𝗂 λ ξ} 𝖽 ξ\] (b) By Fubini and the Fourier transform of a Gaussian \begin{align*}𝖾^{-λ}&=\frac{1}{π} ∫_{-∞}^{∞} \frac{1}{1+ξ^2} 𝖾^{𝗂 λ ξ} \,𝖽ξ\\ &=\frac{1}{π} ∫_{-∞}^{∞} ∫_0^{∞} 𝖾^{-(1+ξ^2) t}𝖾^{𝗂 λ ξ} \,𝖽t𝖽ξ\\ &=\frac{1}{π}∫_0^{∞} 𝖾^{-t}∫_{-∞}^{∞}𝖾^{-tξ^2+𝗂 λ ξ}𝖽ξ𝖽t\\ &=\frac{1}{π}∫_0^{∞}𝖾^{-t}\sqrt{π\over t}𝖾^{-\frac{λ^2}{4t}}𝖽 t \\&=∫_0^{∞} \frac{1}{\sqrt{π t}} 𝖾^{-t-\frac{λ^2}{4 t}} 𝖽 t\end{align*} (c) Substitute (b) in $\hat{f}(ξ)$, by Fubini \begin{align*}\hat{f}(ξ)=\int_{x∈ℝ^n}𝖾^{-|x|}𝖾^{-𝗂ξ⋅x}𝖽x&=\int_{x∈ℝ^n}∫_0^{∞} \frac{1}{\sqrt{π t}} 𝖾^{-t-\frac{|x|^2}{4 t}}𝖾^{-𝗂ξ⋅x}𝖽t𝖽x\\ &=∫_0^{∞}\frac{1}{\sqrt{π t}} 𝖾^{-t}\int_{x∈ℝ^n}𝖾^{-\frac{|x|^2}{4 t}}𝖾^{-𝗂ξ⋅x}𝖽x𝖽t\end{align*} By 1.38 the Fourier transform of $𝖾^{-\frac{|x|^2}{2}}$ is $(2π)^{\frac{n}{2}}𝖾^{-\frac{|ξ|^2}{2}}$. By 1.12 [Dilation Rule] the Fourier transform of $𝖾^{-\frac{|x|^2}{4t}}$ is $(4πt)^{\frac{n}{2}}𝖾^{-t|ξ|^2}$, substitute in $\hat{f}(ξ)$: \[\hat{f}(ξ)=∫_0^{∞}\frac{(4πt)^{\frac{n}{2}}}{\sqrt{πt}}𝖾^{-t(1+|ξ|^2)}𝖽t=2^n π^{n-1\over2}(1 + |ξ|^2)^{-{n+1\over2}} Γ\left(\frac{n+1}2\right)\]

$p$ ramifies$⇔p∣Δ_K$ DISCRIMINANTS AND RAMIFIED PRIMES

By Hahn-Banach, for each $\varphi\in C([0,1])$ we have $\Phi\in C([0,1])''$ with $\|\Phi\|=1$ and $\Phi(\varphi)=\|\varphi\|$. If $C([0,1])$ is reflexive, then $\varphi=\operatorname{Ev}(f)$ for some $f\in C([0,1])$ with $\|f\|=1,\varphi(f) = \|\varphi\|$. Now look at $\displaystyle\varphi(f) := \int_0^{\frac 12} f(x)dx-\int_{\frac 12}^1f(x)dx$. We must have $f=\chi_{[0,\frac12]}-\chi_{[\frac12,1]}$ but $f\notin C([0,1])$.

Part II Lecture Notes: Algebraic Topology, James Lingard Simplices and simplicial complexes We say that $n+1$ points in $\mathbb{R}^{N}$ are in general position if the smallest affine-linear subspace containing them has dimension $n$. An $n$-simplex $\Delta$ in $\mathbb{R}^{N}$ is the convex hull of any $n+1$ points $x_{0}, \ldots, x_{n} \in \mathbb{R}^{N}$ in general position, that is $$ \Delta=\left\{a_{0} x_{0}+\cdots+a_{n} x_{n} \mid a_{i} \in \mathbb{R}, a_{i} \geq 0, \sum a_{i}=1\right\} $$ We call $x_{0}, \ldots, x_{n}$ the vertices of the simplex. A face of the simplex is the convex hull of a non-empty subset of the vertices. A simplicial complex in $\mathbb{R}^{N}$ is a finite collection of simplices in $\mathbb{R}^{N}$, such that whenever a simplex belongs to the collection then so do all of its faces, and whenever any two simplices in the collection have a non-empty intersection, their intersection is a face of both simplices. If $X$ is a simplicial complex we write $|X|$ to mean the geometric realisation of $X$, that is the topological space which is the union of all the simplices in $X .|X|$ is always a compact, metrizable space. Triangulations For a topological space $A$, a triangulation of $A$ is a simplicial complex $X$ together with a homeomorphism from $A$ to $|X|$. We say that $A$ is triangulable if it has some triangulation. Barycentric subdivision We define the barycentre of the simplex $\left[x_{0}, \ldots, x_{n}\right]$ to be $$ \frac{1}{n+1}\left(x_{0}+\cdots+x_{n}\right) $$ Starting with a simplex $\Delta$ in $\mathbb{R}^{N}$ we obtain a simplicial complex $\Delta^{1}$ by the following process, called barycentric subdivision. First we add a vertex $\hat{A}$ at the barycentre of each face $A$ of $\Delta$. Then we have a simplex through each set of vertices $\hat{A}_{0}, \ldots, \hat{A}_{k}$ if and only if after some reordering of these vertices we have $$ A_{0} \subset A_{1} \subset \cdots \subset A_{k} $$ The barycentric subdivision of a simplicial complex $X$ is defined to be the complex obtained by the barycentric subdivision of all the simplices in $X$. Clearly the geometric realisation of the barycentric subdivision of $X$ is the same as that of $X$. The mesh of a simplicial complex in $\mathbb{R}^{N}$ is the maximum of the diameters of its simplices. For any simplicial complex $X$, the mesh of the barycentric subdivision $X^{1}$ is at most $n /(n+1)$ times the mesh of $X$. Thus by repeatedly subdividing any simplicial complex we may make its mesh arbitrarily small. Simplicial maps and simplicial approximations For simplicial complexes $X$ and $Y$, A simplicial map $s: X \rightarrow Y$ is a function from $|X|$ to $|Y|$ which takes simplices of $X$ linearly onto simplices of $Y$. Any simplicial map is continuous (by the gluing lemma) and is completely determined by the images of the vertices in $X$, which are vertices in $Y$. Each point $x \in|X|$ lies in the interior of a unique simplex in $X$, called the carrier of $x$. A simplicial map $s: X \rightarrow Y$ is a simplicial approximation of a continuous map $f:|X| \rightarrow|Y|$ if $s(x)$ lies in the carrier of $f(x)$ for each point $x \in|X|$. Note that if $s$ is a simplicial approximation for $f$ then $s$ and $f$ are homotopic, by a straight line homotopy. The simplicial approximation theorem For any continuous map $f:|X| \rightarrow|Y|$ there is an $m \geq 0$ such that there exists a simplicial approximation $s: X^{m} \rightarrow Y$ of $f$. Proof Let $x$ be a vertex in $X$. Define the open star of $x$ to be the union of the interiors of all the simplices in $X$ which contain $x$. It is easy to prove that vertices $x_{0}, \ldots, x_{k}$ of $X$ span a simplex if and only if the intersection of their open stars is non-empty. First we prove the theorem in a special case. Suppose that for every vertex $x$ in $X$ there exists a vertex $y$ in $Y$ such that the open star of $x$ is mapped by $f$ into the open star of $y$. Then for every vertex $x$ in $X$, choose such a $y$ and define $s(x)=y$. By the above observation it follows that for any simplex $\left[x_{0}, \ldots, x_{k}\right]$ in $X$, the images $s\left(x_{0}\right), \ldots, s\left(x_{k}\right)$ span a simplex in $Y$, and hence we can extend our definition of $s$ linearly over $X$ to get a simplicial map $X \rightarrow Y$. Now suppose that $x \in|X|$. We need to show that $s(x)$ lies in the carrier of $f(x)$. Let $\left[x_{0}, \ldots, x_{k}\right]$ be the carrier of $x$. Then $x$ is in the intersection of the open stars of the $x_{i}$ and so, by our assumption, $f(x)$ is in the intersection of the open stars of the $s\left(x_{i}\right)$. Therefore $s\left(x_{0}\right), \ldots, s\left(x_{k}\right)$ span a face of the carrier of $f(x)$, so the carrier of $f(x)$ contains each of the $s\left(x_{i}\right)$, and so it contains $s(x)$. Thus $s$ is a simplicial approximation to $f$. Now let $f$ be an arbitrary map from $X$ to $Y$. We shall show that for some $m \geq 0, X^{m}$ satisfies the above condition. Now $|Y|$ is the union of the open stars $U_{y}$ of its vertices, so the sets $f^{-1}\left(U_{y}\right)$ form an open cover of $|X|$. Since $X$ is a compact metric space, by Lebesgue's lemma there exists $\epsilon>0$ such that any open subset of $|X|$ of diameter less than $\epsilon$ is contained in one of the sets $f^{-1}\left(U_{y}\right)$. But for $m$ sufficiently large, $X^{m}$ has mesh less that $\epsilon-$ and so $X^{m}$ satisfies the required condition.