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Galois Theory for Beginners The aim of this paper is to prove the unsolvability by radicals of the quintic (in fact of the general $n$th degree equation for $n\ge5$) using just the fundamentals of groups, rings and fields from a standard first course in algebra. The main fact it will be necessary to know is that if $\phi$ is a homomorphism of group $G$ onto group $G'$ then $G'\cong G/\ker\phi$, and conversely, if $G/H\cong G'$ then $H$ is the kernel of a homomorphism of $G$ onto $G'$. The concept of Galois group, which guides the whole proof, will be defined when it comes up. With this background, a proof of unsolvability by radicals can be constructed from just three basic ideas, which will be explained more fully below: 1. Fields containing $n$ indeterminates can be “symmetrized”. 2. The Galois group of a radical extension is solvable. 3. The symmetric group $S_n$ is not solvable The general equation of degree $n$ $$\tag{*} x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0=0 $$ The set of elements obtainable from $a_0, \ldots, a_{n-1}$ by $+,-,\times,\div$ is the field $\mathbb{Q}(a_0, \ldots, a_{n-1})$. If we denote the roots of (*) by $x_1, \ldots, x_n$, so that $$ (x-x_1) \cdots(x-x_n)=x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0 $$ then $a_0, \ldots, a_{n-1}$ are polynomial functions of $x_1, \ldots, x_n$ called the elementary symmetric functions: $$ a_0=(-1)^n x_1 x_2 \cdots x_n, \ldots, a_{n-1}=-(x_1+x_2+\cdots+x_n) . $$ The goal of solution by radicals is then to extend $\mathbb{Q}(a_0, \ldots, a_{n-1})$ by adjoining radicals until a field containing the roots $x_1, \ldots, x_n$ is obtained. For example, the roots $x_1, x_2$ of the quadratic equation lie in the extension of $\mathbb{Q}\left(a_0, a_1\right)=\mathbb{Q}\left(x_1 x_2, x_1+x_2\right)$ by the radical $$ \sqrt{a_1^2-4 a_0}=\sqrt{(x_1+x_2)^2-4 x_1 x_2}=\sqrt{(x_1-x_2)^2}= \pm(x_1-x_2) $$ In this case we get $\mathbb{Q}\left(x_1, x_2\right)$ itself as the radical extension $\mathbb{Q}(a_0, a_1, \sqrt{a_1^2-4 a_0})$, though in other cases a radical extension of $\mathbb{Q}(a_0, \ldots, a_{n-1})$ containing $x_1, \ldots, x_n$ is larger than $\mathbb{Q}(x_1, \ldots, x_n)$. In particular, the solution of the cubic equation gives a radical extension of $\mathbb{Q}(a_0, a_1, a_2)$ which includes imaginary cube roots of unity as well as $x_1, x_2, x_3$. In general, adjoining an element $\alpha$ to a field $F$ means forming the closure of $F \cup\{\alpha\}$ under $+,-,\times,\div$ (by a non-zero element), i.e., taking the intersection of all fields containing $F \cup\{\alpha\}$. The adjunction is called radical if some positive integer power $\alpha^m$ of $\alpha$ equals an element $f \in F$, in which case $\alpha$ may be represented by the radical expression $\sqrt[m]{f}$. The result $F(\alpha_1)(\alpha_2) \ldots(\alpha_k)$ of successive adjunctions is denoted by $F(\alpha_1, \ldots, \alpha_k)$ and if each adjunction is radical we say $F(\alpha_1, \ldots, \alpha_k)$ is a radical extension of $F$. It is clear from these definitions that a radical extension $E$ of $\mathbb{Q}(a_0,\dots,a_{n-1})$ containing $x_1,\dots,x_n$ is also a radical extension of $\mathbb{Q}(x_1,\dots,x_n)$, since $a_0,\dots,a_{n-1}\in\mathbb{Q}(x_1,\dots,x_n)$. Thus we also have to study radical extensions of $\mathbb{Q}(x_1,\dots,x_n)$. The most important property of $\mathbb{Q}(x_1,\dots,x_n)$ is that it is symmetric with respect to $x_1,\dots,x_n$, in the sense that any permutation $\sigma$ of $x_1,\dots,x_n$ extends to a bijection $\sigma$ of $\mathbb{Q}(x_1,\dots,x_n)$ defined by$$\sigma f(x_1,\dots,x_n)=f(\sigma x_1,\dots,\sigma x_n)$$for each rational function $f$ of $x_1,\dots,x_n$. Moreover, this bijection $\sigma$ obviously satisfies \begin{align} \sigma(f+g)&=\sigma f+\sigma g \\ \sigma(f g)&=\sigma f \cdot \sigma g \end{align}and hence is an automorphism of $\mathbb{Q}(x_1, \ldots, x_n)$. A radical extension $E$ of $\mathbb{Q}(x_1, \ldots, x_n)$ is not necessarily symmetric in this sense. For example, $\mathbb{Q}\left(x_1, \ldots, x_n, \sqrt{x_1}\right)$ contains a square root of $x_1$, but not of $x_2$, hence there is no automorphism exchanging $x_1$ and $x_2$. However, we can restore symmetry by adjoining $\sqrt{x_2}, \ldots, \sqrt{x_n}$ as well. The obvious generalization of this idea gives a way to “symmetrize” any radical extension $E$ of $\mathbb{Q}(x_1, \ldots, x_n)$: Theorem 1. For each radical extension $E$ of $\mathbb{Q}(x_1, \ldots, x_n)$ there is a radical extension $\bar{E} \supseteq E$ with automorphisms $\sigma$ extending all permutations of $x_1, \ldots, x_n$. Proof: For each adjoined element, represented by radical expression $e(x_1, \ldots, x_n)$, and each permutation $\sigma$ of $x_1, \ldots, x_n$, adjoin the element $e(\sigma x_1, \ldots, \sigma x_n)$. Since there are only finitely many permutations $\sigma$, the resulting field $\bar{E} \supseteq E$ is also a radical extension of $\mathbb{Q}(x_1, \ldots, x_n)$. This gives a bijection (also called $\sigma$) of $\bar{E}$ sending each $f\left(x_1, \ldots, x_n\right) \in \bar{E}$ (a rational function of $x_1, \ldots, x_n$ and the adjoined radicals) to $f\left(\sigma x_1, \ldots, \sigma x_n\right)$, and this bijection is obviously an automorphism of $\bar{E}$, extending the permutation $\sigma$.$\quad\blacksquare$ The reason for wanting an automorphism $\sigma$ extending each permutation of $x_1,\dots,x_n$ is that $a_0,\dots,a_{n-1}$ are fixed by such permutations, and hence so is every element of the field $\mathbb Q(a_0,\dots,a_{n-1})$. If $E\supseteq F$ are any fields, the automorphisms $\sigma$ of $E$ fixing all elements of $F$ form what is called the Galois group of $E$ over $F$, $\operatorname{Gal}(E/F)$. This concept alerts us to the following corollary of Theorem 1: Corollary. If $E$ is a radical extension of $\mathbb{Q}(a_0, \ldots, a_{n-1})$ containing $x_1, \ldots, x_n$ then there is a further radical extension $\bar{E} \supseteq E$ such that $\operatorname{Gal}\left(\bar{E} / \mathbb{Q}(a_0, \ldots, a_{n-1})\right)$ includes automorphisms $\sigma$ extending all permutations of $x_1, \ldots, x_n$. Proof: This is immediate from Theorem 1 and the fact that a radical extension of $\mathbb{Q}(a_0, \ldots, a_{n-1})$ containing $x_1, \ldots, x_n$ is also a radical extension of $\mathbb{Q}(x_1, \ldots, x_n)$.$\quad\blacksquare$ THE STRUCTURE OF RADICAL EXTENSIONS. So far we know that a solution by radicals of the general $n$th degree equation (*) entails a radical extension of $\mathbb{Q}\left(a_0, \ldots, a_{n-1}\right)$ containing $x_1, \ldots, x_n$, and hence a radical extension $\bar{E}$ with the symmetry described in the corollary above. This opens a route to prove nonexistence of such a solution by learning enough about $\operatorname{Gal}\left(\bar{E} / \mathbb{Q}(a_0, \ldots, a_{n-1})\right)$ to show that such symmetry is lacking, at least for $n \geq 5$. In the present section we shall show that the Galois group $\operatorname{Gal}\left(F(\alpha_1, \ldots, \alpha_k\right) / F)$ of any radical extension has a special structure, called solvability, inherited from the structure of $F(\alpha_1, \ldots, \alpha_k)$. Then in the next section we shall show that this structure is indeed incompatible with the symmetry described in the corollary. To simplify the derivation of this structure, we shall show that certain assumptions about the adjunction of radicals $\alpha_i$ can be made without loss of generality. First, we can assume that each radical $\alpha_i$ adjoined is a $p$th root for some prime $p$. E.g., instead of adjoining $\sqrt[6]{\alpha}$ we can adjoin first $\sqrt{\alpha}=\beta$, then $\sqrt[3]{\beta}$. Second, if $\alpha_i$ is a $p$th root we can assume that $F(\alpha_1, \ldots, \alpha_i)$ contains no $p$th roots of unity not in $F(\alpha_1, \ldots, \alpha_{i-1})$ unless $\alpha_i$ itself is a $p$th root of unity. If this is not the case initially we simply adjoin a $p$th root of unity $\zeta \neq 1$ to $F(\alpha_1, \ldots, \alpha_{i-1})$ before adjoining $\alpha_i$ (in which case $F(\alpha_1, \ldots, \alpha_{i-1}, \zeta)$ contains all the $p$th roots of unity: $1, \zeta, \zeta^2, \ldots, \zeta^{p-1}$). With both these modifications the final field $F(\alpha_1, \ldots, \alpha_k)$ is the same, and it remains the same if the newly adjoined roots $\zeta$ are included in the list $\alpha_1, \ldots, \alpha_k$. Hence we have: Any radical extension $F(\alpha_1, \ldots, \alpha_k)$ is the union of an ascending tower of fields $$ F=F_0 \subseteq F_1 \subseteq \cdots \subseteq F_k=F(\alpha_1, \ldots, \alpha_k) $$ where each $F_i=F_{i-1}(\alpha_i), \alpha_i$ is the $p_i$th root of an element in $F_{i-1}$, $p_i$ is prime, and $F_i$ contains no $p_i$th roots of unity not in $F_{i-1}$ unless $\alpha_i$ is itself a $p_i$th root of unity. Corresponding to this tower of fields we have a descending tower of groups $$ \operatorname{Gal}(F_k / F_0)=G_0 \supseteq G_1 \supseteq \cdots \supseteq G_k=\operatorname{Gal}(F_k / F_k)=\{\mathbf{1}\} $$ where $G_i=\operatorname{Gal}(F_k / F_i)=\operatorname{Gal}(F_k / F_{i-1}(\alpha_i))$ and $\bf1$ denotes the identity automorphism. The containments are immediate from the definition of $\operatorname{Gal}(E / B)$, for any fields $E \supseteq B$, as the group of automorphisms of $E$ fixing each element of $B$. As $B$ increases to $E$, $\operatorname{Gal}(E / B)$ must decrease to $\{\mathbf{1}\}$. The important point is that the step from $G_{i-1}$ to its subgroup $G_i$, reflecting the adjunction of the $p_i$th root $\alpha_{j i}$ to $F$, is “small” enough to be describable in group-theoretic terms: $G_i$ is a normal subgroup of $G_{i-1}$, and $G_{i-1} / G_i$ is abelian, as we shall now show. To simplify notation further we set $$ E=F_k, B=F_{i-1}, \alpha=\alpha_i, p=p_i, $$ so the theorem we want is: Theorem 2. If $E \supseteq B(\alpha) \supseteq B$ are fields with $\alpha^p \in B$ for some prime $p$, and if $B(\alpha)$ contains no $p$th roots of unity not in $B$ unless $\alpha$ itself is a $p$th root of unity, then $\mathrm{Gal}(E / B(\alpha))$ is a normal subgroup of $\mathrm{Gal}(E / B)$ and $\mathrm{Gal}(E / B) / \mathrm{Gal}(E / B(\alpha))$ is abelian. Proof: By the homomorphism theorem for groups, it suffices to find a homomorphism of $\operatorname{Gal}(E / B)$, with kernel $\operatorname{Gal}(E / B(\alpha))$, into an abelian group (i.e., onto a subgroup of an abelian group, which of course is also abelian). The obvious map with kernel $\operatorname{Gal}(E / B(\alpha))$ is restriction to $B(\alpha)$, $|_{B(\alpha)}$, since by definition $$\sigma \in \operatorname{Gal}(E / B(\alpha)) \Leftrightarrow \sigma|_{B(\alpha)} \text { is the identity map.} $$ The homomorphism property, $$ \sigma^{\prime} \sigma|_{B(\alpha)}=\sigma^{\prime}|_{B(\alpha)} \sigma|_{B(\alpha)} \text { for all } \sigma^{\prime}, \sigma \in \operatorname{Gal}(E / B) $$ is automatic provided $\left.\sigma\right|_{B(\alpha)}(b) \in B(\alpha)$ for each $b \in B(\alpha)$, i.e. provided $B(\alpha)$ is closed under each $\sigma \in \operatorname{Gal}(E / B)$. Since $\sigma$ fixes $B$, $\sigma|_{B(\alpha)}$ is completely determined by the value $\sigma(\alpha)$. If $\alpha$ is a $p$th root of unity $\zeta$ then $$ (\sigma(\alpha))^p=\sigma\left(\alpha^p\right)=\sigma\left(\zeta^p\right)=\sigma(1)=1, $$ hence $\sigma(\alpha)=\zeta^i=\alpha^i \in B(\alpha)$, since each $p$th root of unity is some $\zeta^i$. If $\alpha$ is not a root of unity then $$ (\sigma(\alpha))^p=\sigma\left(\alpha^p\right)=\alpha^p \quad \text { since } \quad \alpha^p \in B, $$ hence $\sigma(\alpha)=\zeta^j \alpha$ for some $p$th root of unity $\zeta$, and $\zeta \in B$ by hypothesis, so again $\sigma(\alpha) \in B(\alpha)$. Thus $B(\alpha)$ is closed as required. This also implies that $\left.\right|_{B(\alpha)}$ maps $\operatorname{Gal}(E / B)$ into $\operatorname{Gal}(B(\alpha) / B)$, so it now remains to check that $\operatorname{Gal}(B(\alpha) / B)$ is abelian. If $\alpha$ is a root of unity then, as we have just seen, each $\left.\sigma\right|_{B(\alpha)} \in \operatorname{Gal}(B(\alpha) / B)$ is of the form $\sigma_i$, where $\sigma_i(\alpha)=\alpha^i$, hence $$ \sigma_i \sigma_j(\alpha)=\sigma_i\left(\alpha^j\right)=\alpha^{i j}=\sigma_j \sigma_i(\alpha) . $$ Likewise, if $\alpha$ is not a root of unity then each $\left.\sigma\right|_{B(\alpha)} \in \operatorname{Gal}(B(\alpha) / B)$ is of the form $\sigma_i$ where $\sigma_i(\alpha)=\zeta^i \alpha$, hence $$ \sigma_i \sigma_j(\alpha)=\sigma_i\left(\zeta^j \alpha\right)=\zeta^{i+j} \alpha=\sigma_j \sigma_i(\alpha) $$ since $\zeta \in B$ and therefore $\zeta$ is fixed. Hence in either case $\operatorname{Gal}(B(\alpha) / B)$ is abelian.$\quad\blacksquare$ The property of $\operatorname{Gal}\left(F\left(\alpha_1, \ldots, \alpha_k\right) / F\right)$ implied by this theorem, that it has subgroups $\operatorname{Gal}\left(F\left(\alpha_1, \ldots, \alpha_k\right) / F\right)=G_0 \supseteq G_1 \supseteq \cdots \supseteq G_k=\{1\}$ with each $G_i$ normal in $G_{i-1}$ and $G_{i-1} / G_i$ abelian, is called solvability of $\operatorname{Gal}\left(F(\alpha_1, \ldots, \alpha_k) / F\right)$. NON-EXISTENCE OF SOLUTIONS BY RADICALS WHEN $\boldsymbol{n} \geq \mathbf{5}$. As we have said, this amounts to proving that a radical extension of $\mathbb{Q}(a_0, \ldots, a_{n-1})$ does not contain $x_1, \ldots, x_n$ or, equivalently, $\mathbb{Q}(x_1, \ldots, x_n)$. We have now reduced this problem to proving that the symmetry of the hypothetical extension $\bar{E}$ containing $x_1, \ldots, x_n$, given by the corollary to Theorem 1, is incompatible with the solvability of $\operatorname{Gal}\left(\bar{E} / \mathbb{Q}(a_0, \ldots, a_{n-1})\right)$, given by Theorem 2. Our proof looks only at the effect of the hypothetical automorphisms of $\bar{E}$ on $x_1, \ldots, x_n$, and hence it is really about the symmetric group $S_n$ of all permutations of $x_1, \ldots, x_n$. In fact, we are adapting a standard proof that $S_n$ is not a solvable group, given by Milgram in his appendix to Artin [1]. Theorem 3. A radical extension of $\mathbb{Q}(a_0, \ldots, a_{n-1})$ does not contain $\mathbb{Q}(x_1, \ldots, x_n)$ when $n \geqslant 5$. Proof: Suppose on the contrary that $E$ is a radical extension of $\mathbb{Q}\left(a_0, \ldots, a_{n-1}\right)$ which contains $\mathbb{Q}(x_1, \ldots, x_n)$. Then $E$ is also a radical extension of $\mathbb{Q}(x_1, \ldots, x_n)$ and by the corollary to Theorem 1 there is a radical extension $\bar{E} \supseteq E$ such that $G_0=\operatorname{Gal}\left(\bar{E} / \mathbb{Q}(a_0, \ldots, a_{n-1})\right)$ includes automorphisms $\sigma$ extending all permutations of $x_1, \ldots, x_n$. By Theorem 2, $G_0$ has a decomposition $$ G_0 \supseteq G_1 \supseteq \cdots \supseteq G_k=\{\mathbf{1}\} $$ where each $G_{i+1}$ is a normal subgroup of $G_i$ and $G_{i-1} / G_i$ is abelian. We now show that this is contrary to the existence of the automorphisms $\sigma$. Since $G_{i-1} / G_i$ is abelian, $G_i$ is the kernel of a homomorphism of $G_{i-1}$ onto an abelian group, and therefore $$ \sigma, \tau \in G_{i-1} \Rightarrow \sigma^{-1} \tau^{-1} \sigma \tau \in G_i $$ We use this fact to prove by induction on $i$ that, if $n \geq 5$, each $G_i$ contains automorphisms $\sigma$ extending all 3-cycles $(x_a, x_b, x_c)$. This is true for $G_0$ by hypothesis, and when $n \geq 5$ the property persists from $G_{i-1}$ to $G_i$ because $$ (x_a, x_b, x_c)=(x_d, x_a, x_c)^{-1}(x_c, x_e, x_b)^{-1}(x_d, x_a, x_c)(x_c, x_e, x_b) $$ where $a, b, c, d, e$ are distinct. Thus if there are at least five indeterminates $x_j$, there are $\sigma$ in each $G_i$ which extend arbitrary 3-cycles $(x_a, x_b, x_c)$, and this means in particular that $G_k \neq\{\mathbf1\}$. This contradiction shows that $\mathbb{Q}(x_1, \ldots, x_n)$ is not contained in any radical extension of $\mathbb{Q}(a_0, \ldots, a_{n-1})$ when $n \geq 5$.$\quad\blacksquare$

Proof of von Neumann’s Ergodic Theorem. $\def\ker{\operatorname{Ker}}\DeclareMathOperator{\im}{Im}$Let $X$ be a Hilbert space and let $U\colon X → X$ be a unitary operator. (a) Show that $\ker(I-U)=\ker(I-U^*)$; (b) Show that $X=\overline{\im(I-U)} ⊕ \ker(I-U)$; (c) Show that $\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x=x$ if $x ∈ \ker(I-U)$ and $\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x=0$ if $x ∈ \overline{\im(I-U)}$; (d) Deduce that, for each $x ∈ X$, \[\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x=P x,\] where $P$ is the orthogonal projection onto $\ker(I-U)$. (a) $\ker(I-U)=\ker(UU^*-U)=\ker(U(U^*-I))⊃\ker(I-U^*)$. $\ker(I-U^*)=\ker(U^*U-U^*)=\ker(U^*(U-I))⊃\ker(I-U)$. (b) By 1.5 for any $T∈B(X)$, $\overline{\im T}=(\ker T^*)^⟂$. Applying $X=\overline{\im T}⊕\ker T^*$ to $T=I-U$ \[X=\overline{\im(I-U)} ⊕ \ker(I-U^*)\overset{\text{(a)}}=\overline{\im(I-U)} ⊕ \ker(I-U)\] (c) If $x ∈ \ker(I-U)$, $x=U^nx$, so $\frac{1}{N} \sum_{n=1}^{N-1} U^n x=\frac{N-1}Nx→x$. If $x ∈ \im(I-U)$, let $x=y-Uy$. Since $U$ is isometric, $‖y‖=‖Uy‖=‖U^{n+1}y‖$, \[\frac{1}{N} \sum_{n=1}^{N-1} U^n x=\frac{1}{N} \sum_{n=1}^{N-1}U^ny-U^{n+1}y=\frac{Uy-U^{n+1}y}N→0\] Additionally we show this holds for $\overline{\im(I-U)}$, consider $x∈\overline{\im(I-U)}$ there exist $\tilde{x} ∈\im(I-U)$ s.t. $\|x-\tilde{x}\|<\frac{ϵ}{2}$ and large $N$ s.t. $\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n \tilde{x}\right\|<\frac{ϵ}{2}$ \begin{align*}\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n x\right\|&=\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n((x-\tilde{x})+\tilde{x})\right\|\\&≤\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n(x-\tilde{x})\right\|+\left\|\frac{1}{N} \sum_{n=1}^{N-1} U^n \tilde{x}\right\| \\&<\left(\frac{1}{N} \sum_{n=1}^{N-1}\left\|U^n\right\|\right)\frac{ϵ}{2}+\frac{ϵ}{2}\\&=\frac{N-1}N\frac{ϵ}{2}+\frac{ϵ}{2}<ϵ . \text { Using }\|U^n\|=1\end{align*} so $\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x=0$ for all $x ∈ \overline{\im(I-U)}$. Remark.(theorem 3.5) If $‖T_n‖≤M$ for all $n$, and $T_n→0$ for all $x$ in a dense set $D⊆X$, then $T_n→0$ for all $x$ in $X$. Not true if dropping the condition $‖T_n‖≤M$ for all $n$, example: $X=l^2,T_n((x_k))=nx_n,D=c_{00},T_n|_{c_{00}}→0$ but $T_n((\frac1k))=1$ (d) Using (b) and $Px∈ \ker(I-U),x-Px∈\ker(I-U)^⟂=\overline{\im(I-U)}$. Using (c) \begin{align*}\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n x&=\lim_{N → ∞} \frac{1}{N} \sum_{n=1}^{N-1} U^n(Px+(x-Px))\\&=P x+0=Px.\end{align*}

Let $t>0$ and put $G_t(x)=𝖾^{-t|x|^2}, x ∈ ℝ^n$, where $|x|=\sqrt{x_1^2+⋯+x_n^2}$ is the usual norm of $x$. Use the Fourier transform to find a formula for $G_s * G_t$ for all $s, t>0$. By Q2, for $x,ξ∈ ℝ$, \[\int_ℝ𝖾^{-tx^2}𝖾^{-𝗂ξx}𝖽x=\sqrt{π\over t}𝖾^{-{ξ^2\over4t}}\] By Fubini, for $x,ξ∈ ℝ^n$, \[\widehat{G_t}(ξ)=\prod_{i=1}^n\int_ℝ𝖾^{-tx_i^2}𝖾^{-𝗂ξ_ix_i}𝖽x_i=\left(π\over t\right)^{\frac n2}𝖾^{-{|ξ|^2\over4t}}\] so \[\widehat{G_s}(ξ)⋅\widehat{G_t}(ξ)=\left(π\over s\right)^{n\over2}𝖾^{-{|ξ|^2\over4s}}\left(π\over t\right)^{n\over2}𝖾^{-{|ξ|^2\over4t}}\] comparing with \[\widehat{G_{st\over s+t}}(ξ)=\left(π\over{st\over s+t}\right)^{\frac n2}𝖾^{-{|ξ|^2\over4s}}𝖾^{-{|ξ|^2\over4t}}\] we get \[\widehat{G_s}(ξ)⋅\widehat{G_t}(ξ)=\left(π\over s+t\right)^{n\over2}\widehat{G_{st\over s+t}}(ξ)\] by Convolution Rule \[G_s*G_t=\left(π\over s+t\right)^{n\over2}G_{st\over s+t}\]

Integral extensions of rings, when one of the rings is a field 5.2. going up Proposition. Suppose that $A, B$ are integral domains and $B$ is an integral extension of $A$. Then $B$ is a field iff $A$ is a field. Proof. Suppose $B$ is a field and $a \neq 0 \in A$. Then $B$ contains the inverse $a^{-1}$ of $a$, and since $B$ is an integral extension, $a^{-1}$ is a root of monic polynomial $f(x) \in A[x]$: \[a^{-n}+a_1 a^{1-n}+a_2 a^{2-n}+\cdots+a_n=0\] Multiplying by $a^{n−1}$, we get: \[a^{-1} + a_1 + a_2a+ \cdots+ a_na^{n−1} = 0.\] Solving for $a^{-1}$, we see that $a^{-1} \in A$. So, $A$ is a field. Conversely suppose that $A$ is field. Let $b \neq 0 \in B$. Then $b \in A[b]$ which is a finitely generated module. In other words, $A[b] \cong A^n$ is finite dimensional vector space over field. Multiplication by $b$ gives linear mapping: \[\mu_b :A[b] \cong A^n \to A^n\cong A[b].\] Since $B$ is an integral domain, the kernel of $\mu_b$ is $0$. This means that $\mu_b$ has rank $n$, and therefore an isomorphism of vector spaces. So, it is onto. So there is some $c \in B$ so that $\mu_b(c)=bc= 1$. So, $c=b^{-1} \in A[b] \subseteq B$ showing that $B$ is field.

[Generalized Functions 01] I. M. Gel'fand p34 2.5. Delta-Convergent Sequences There are many ways to construct a sequence of regular functions which converge to the $\delta$ function. All that is needed is that the corresponding ordinary functions $f_\nu(x)$ form what we shall call a deltaconvergent sequence, which means that they must possess the following two properties. (a) For any $M>0$ and for $|a| \leqslant M$ and $|b| \leqslant M$, the quantities \[ \left|\int_a^b f_p(\xi) d \xi\right| \] must be bounded by a constant independent of $a, b$, or $v$ (in other words, depending only on $M$ ). (b) For any fixed nonvanishing $a$ and $b$, we must have \[ \lim _{v \rightarrow \infty} \int_a^b f_v(\xi) d \xi= \begin{cases}0 & \text { for } aIntegral and Series Representations of the Dirac Delta Function

Part II Lecture Notes: Algebraic Topology, James Lingard Fundamental group of the circle To determine the fundamental group of the circle we first define the map $\pi: \mathbb{R} \rightarrow S^{1}$ by $\pi(x)=e^{2 \pi i x}$. $\pi$ is a covering map as defined below. We then apply the path-lifting and homotopy-lifting lemmas related to this covering map. We define a homomorphism $\sigma: \pi_{1}\left(S^{1}, 1\right) \rightarrow \mathbb{Z}$ as follows. Let $\alpha$ be a loop in $S^{1}$ based at 1 . This lifts uniquely to a path in $\mathbb{R}$ starting at 0 , whose endpoint must be an integer, and we define $\sigma(\alpha)$ to be this value. To show that $\sigma$ gives a well-defined map on $\pi_{1}\left(S^{1}, 1\right)$ we use the homotopy-lifting property. For if $\beta \sim \alpha$ is another loop in $S^{1}$ based at 1 then the homotopy from $\alpha$ to $\beta$ gives us a homotopy from the lift of $\alpha$ to the lift of $\beta$, and we then see that $\sigma(\alpha)=\sigma(\beta)$. It is easy to see that $\sigma$ is a homomorphism. Furthermore, $\sigma$ is injective, for if $\sigma(\alpha)=0$ then there is a straight line homotopy, relative to $\{0,1\}$, from the lift of $\alpha$ to the constant loop at 0 in $\mathbb{R}$, and this projects to show that $\alpha$ is homotopic to the constant loop in $\pi_{1}\left(S^{1}, 1\right)$. Finally, $\sigma$ is surjective, for the loop $s \mapsto e^{2 \pi i n s}$ in $S^{1}$ lifts to a path in $\mathbb{R}$ from 0 to $n$. Therefore we have shown that the fundamental group of the circle is isomorphic to $\mathbb{Z}$. The fundamental theorem of algebra Let $f$ be a polynomial of degree $n \geq 1$, and suppose that $f$ has no roots in $\mathbb{C}$. Then $f$ is a continuous map $\mathbb{C} \rightarrow \mathbb{C} \backslash\{0\}$. Since $\mathbb{C} \backslash\{0\}$ and $S^{1}$ are homotopy equivalent, they have isomorphic fundamental groups. In particular, loops in $\mathbb{C} \backslash\{0\}$ have a well-defined winding number, which is invariant under homotopy of loops (even if you change the base point). Now let $C_{r}$ be the circle of radius $r$ around 0 in $\mathbb{C}$, and consider the image of $C_{r}$ under $f$ as $r$ varies. These circles are obviously all homotopic, and a homotopy from $C_{r_{1}}$ to $C_{r_{2}}$ gives a homotopy from $f \circ C_{r_{1}}$ to $f \circ C_{r_{2}}$. Now the circle $C_{0}$ clearly projects to a single point, with winding number 0 , but for sufficiently large $R$ the image $f \circ C_{R}$ is homotopic to $\left(a_{n} x^{n}\right) \circ C_{R}$, which has winding number $n$. This contradicts the homotopy invariance of the winding number. Covering maps and covering spaces A covering map $\pi: X \rightarrow Y$ is a continuous map such that every $y \in Y$ has an open neighbourhood $U$ such that $\pi^{-1}(U)$ is a disjoint union of open subsets $U_{\alpha}$ such that the restriction of $\pi$ to each set $U_{\alpha}$ is a homeomorphism from $U_{\alpha}$ to $U$. We then say that $X$ is a covering space of $Y$. The path-lifting property For any covering space $X \rightarrow Y$ and any point $x \in X$, any path in $Y$ starting at $\pi(x)$ lifts to a unique path in $X$ starting at $x$. The path-lifting property is a special case of the homotopy-lifting property below, where the space $A$ is a point. The homotopy-lifting property For any covering space $X \rightarrow Y$, any other space $A$ and any continuous map $f^{\prime}: A \rightarrow X$, any homotopy from $f=\pi f^{\prime}: A \rightarrow Y$ to some other map $g: A \rightarrow Y$ lifts uniquely to a homotopy from $f^{\prime}$ to some other map $g^{\prime}: A \rightarrow X$. Computing fundamental groups using covering spaces For any covering space $\pi: X \rightarrow Y$ and any point $x \in X$, the induced group homomorphism $\pi_{*}: \pi_{1}(X, x) \rightarrow \pi_{1}(Y, \pi(x))$ is injective. This is proved using the homotopy-lifting property. If we have a covering space $\pi: X \rightarrow Y$, then $\pi_{1}(Y, y)$ acts naturally on the fibre $\pi^{-1}(y)$. Furthermore, if $X$ is path-connected then this action is transitive. In this case, the stabilizer of a point $x \in \pi^{-1}(y)$ is precisely the fundamental group $\pi_{1}(X, x)$ of $X$, and we have a bijection from the set of cosets of $\pi_{1}(X, x)$ in $\pi_{1}(Y, y)$ to the fibre $\pi^{-1}(y)$ given by $$ \pi_{1}(X, x) \sigma \longmapsto x \sigma $$ Therefore we can compute the order of the fundamental group of $Y$ if we know the order of the fundamental group of $X$. To determine the structure we need the following theorem: Let $X$ be a simply connected space. Let $G$ be a group of homeomorphisms which acts freely on $X$ in the sense that every point $x \in X$ has an open neighbourhood $U$ such that $U \cap g(U)=\emptyset$ for all $g \neq 1 \in G$. Let $Y=X / G$ be the space of orbits. Then the map $X \rightarrow Y$ is a covering map and the fundamental group of $Y$ at any base point is isomorphic to $G$. Fundamental group of the torus The $n$-torus $\left(S^{1}\right)^{n}$ has fundamental group $\mathbb{Z}^{n}$. We can prove this in two ways. Firstly, we can use the fact that $$ \pi_{1}(X \times Y,(x, y)) \cong \pi_{1}(X, x) \times \pi_{1}(Y, y) $$ Alternatively, we observe that $\left(S^{1}\right)^{n}$ is the quotient of $\mathbb{R}^{n}$ by the free action of $\mathbb{Z}^{n}$ acting by translation, and that $\mathbb{R}^{n}$ is simply connected, so the result follows by the theorem above. Fundamental group of real projective space The real projective space $\mathbb{R} \mathbb{P}^{n}$ can be thought of as the quotient of $S^{n}$ by the free action of the group $\mathbb{Z} / 2$, mapping each point in $S^{n}$ to its antipode. Therefore, if $n=0$ then $\mathbb{R}^{0}$ consists of just one point, and hence is simply connected. If $n=1$ then $\mathbb{R} \mathbb{P}^{1}$ is isomorphic to $S^{1}$ and so has fundamental group $\mathbb{Z}$. If $n \geq 2$ then $S^{n}$ is simply connected, and so the fundamental group of $\mathbb{R} \mathbb{P}^{n}$ is $\mathbb{Z} / 2$. The universal covering A space is locally contractible if every point has an open neighbourhood which is contractible. Also, we say that two covering spaces $X_{1}$ and $X_{2}$ of the same space $Y$ are isomorphic if there is a homeomorphism from $X_{1}$ to $X_{2}$ such that the composition $X_{1} \rightarrow X_{2} \rightarrow Y$ is the map $X_{1} \rightarrow Y$. Then we have the following theorem: Every connected, locally contractible space $Y$ has a unique simply connected covering space $X$, called the universal covering of $Y$. Moreover, the fundamental group of $Y$ acts freely on $X$, with $Y=X / \pi_{1}(Y, y)$. Therefore, the fundamental group of any reasonable space $Y$ may be computed by finding a simply connected covering space of $Y$ and applying this theorem. Finally, the following theorem classifies the connected covering spaces of a space $Y$. Let $Y$ be a connected, locally contractible space. Then there is a bijection between isomorphism classes of connected covering spaces of $Y$ and conjugacy classes of subgroups $H$ of the group $G=\pi_{1}(Y, y)$. The correspondence is defined by viewing $Y$ as $X / G$, where $X$ is the universal cover of $Y$, and, for each subgroup $H$ of $G$, forming the new covering space $X / H$ of $Y$.

page72 The annuli: consider $ℍ /⟨z↦λz⟩$ with $λ>1$. Then $\log z$ maps this onto \[ \{w ∈ ℂ: 0<\operatorname{Im} w<π, 0 ≤ \operatorname{Re} w ≤ \log λ\} \] with the sides $\operatorname{Re} w=0, \operatorname{Re} w=\log \lambda$ identified. Next, send this via the conformal map \[ w \mapsto \exp \left(2 \pi i \frac{w}{\log \lambda}\right) \] onto the annulus $\Delta_r:=\{r<|z|<1\}$ where $\log r=-\frac{2 \pi^2}{\log \lambda}$. We leave it to the reader to check that no two $\Delta_r$ are conformally equivalent (hence, the moduli space of tori $\left\{z \in \mathbb{C}^*: r_1<|z|Real and Complex Analysis (3rd Ed.) pg 291 For $014.22 Theorem. $A(r_1, R_1)$ is conformally equivalent to $A(r_2, R_2)$ if and only if $R_1/r_1 = R_2/r_2$. Proof: Suppose $R_1/r_1 = R_2/r_2$. Then there exists $k > 0$ such that $R_2 = k R_1$ and $r_2 = k r_1$. The map $f(z) = kz$ gives a conformal equivalence from $A(r_1, R_1)$ onto $A(r_2, R_2)$. Conversely, let $f$ be a biholomorphic map from $A(r_1, R_1)$ onto $A(r_2, R_2)$. By scaling if necessary, we may assume $r_1 = r_2 = 1$. Let $A_1 = A(1, R_1)$ and $A_2 = A(1, R_2)$. Fix some $1 0$ such that $A(1, 1 + δ) ∩ f^{-1}(C) = ∅$. Let $V = f(A(1, 1 + δ))$. Since $f$ is continuous, $V$ is connected, so either $V ⊆ A(1, r)$ or $V ⊆ A(r, R_2)$. By replacing $f$ with $R_2/f$, we may assume the first case holds. Claim: $|f(z_n)| → 1$ whenever $|z_n| → 1$. Proof of claim: Let $\{z_n\} ⊆ A(1, 1 + δ)$. Note that $\{f(z_n)\}$ does not have a limit point in $A_2$ since otherwise $\{z_n\}$ would have a limit point in $A_1$ by continuity of $f^{-1}$, contradicting $|z_n| → 1$. Hence $|f(z_n)| → 1$ or $|f(z_n)| → R_2$ (it must converge by continuity). The latter case is ruled out as $f(z_n) ∈ V ⊆ A(1, r)$. Similarly, we also have: Claim: $|f(z_n)| → R_2$ whenever $|z_n| → R_1$. Now set $α = \log R_2/ \log R_1$. Define $g: A_1 → ℝ$ by \[ g(z) = \log |f(z)|^2 - α \log |z|^2 = 2(\log |f(z)| - α \log |z|). \] We know that $\log |h|$ is harmonic whenever $h$ is holomorphic and nonzero, so $g$ is a harmonic function. By the two claims, $g$ extends to a continuous function on $\overline{A_1}$ vanishing on $∂ A_1$. This forces $g$ to vanish identically on $A_1$. In particular, \[ 0 = \frac{∂ g}{∂ z} = \frac{f'(z)}{f(z)} - \frac{α}{z}. \] Take some $1 0$ is an integer. Observe that \[ \frac{d}{dz}(z^{-α}f(z)) = z^{-α - 1}(-α f(z) + zf'(z)) = 0, \] thus $f(z) = Kz^{α}$ for some nonzero constant $K$. Since $f$ is injective, $α = 1$. Therefore $R_2 = R_1$.

$L_n(A⊗_R-)(B)≃L_n(-⊗_RB)(A)=\operatorname{Tor}^R_n(A,B)$ for all $n$. Proof: Choose a projective resolution $P_•\xrightarrow{ε}A$ and a projective resolution $Q_•\xrightarrow{η}B$. We can view $A,B$ as complexes concentrated in degree 0. Look at the double complexes $P⊗_RQ,A⊗_RQ,P⊗_RB$ we get morphisms of bicomplexes \begin{tikzcd} & P\otimes_R Q \arrow[dl,swap,"\varepsilon\otimes\mathrm{id}"] \arrow[dr,"\mathrm{id}\otimes\eta"] & \\ A\otimes_R Q & & P\otimes_R B \end{tikzcd} Since $\operatorname{Tot}$ is functorial, these maps in turn induce chain maps $$ f: \operatorname{Tot}\left(P_{\bullet} \otimes Q_{\bullet}\right) \rightarrow \operatorname{Tot}\left(M \otimes Q_{\bullet}\right) \cong M \otimes Q_{\bullet}, $$ and $$ g: \operatorname{Tot}\left(P_{\bullet} \otimes Q_{\bullet}\right) \rightarrow \operatorname{Tot}\left(P_{\bullet} \otimes N\right) \cong P_{\bullet} \otimes N . $$ We claim that $f$ and $g$ are quasi-isomorphisms, which would give the isomorphism on homology $L_n(A⊗_R-)(B)≃L_n(-⊗_RB)(A)=\operatorname{Tor}^R_n(A,B)$. Proof. To show that $\varepsilon\otimes\mathrm{id}$ is quasi-isomorphism, we need to show that $\operatorname{cone}(ε⊗Q)=\operatorname{Tot}C$ is acyclic. Since each $− ⊗ Q_q$ is an exact functor (because the $Q_q$ are projective, hence flat), the rows of $C$ are exact. Since $C$ is an upper half plane complex, the chain complex $\operatorname{Tot}(C_{••})$ is acyclic by the Acyclic Assembly Lemma. Thus, $\operatorname{cone}(f)$ is an acyclic chain complex, so by Corollary 8.4, we have that $f$ is a quasi-isomorphism.

VIGRE2010 REU Paper by Grant Rotskoff Definition 3.2. A simple region $R$ of a surface $S$ is a region such that $R$ is homeomorphic to the disk. Definition 3.3. Given a parametrization of a surface, $\mathbf{x}: U \rightarrow S$, we have the following quantities $$ E=\left\langle x_u, x_u\right\rangle, F=\left\langle x_u, x_v\right\rangle, G=\left\langle x_v, x_v\right\rangle . $$ These are the coefficients of the first fundamental form of a surface. Definition 3.4. Given a parametrization of a surface, $$ \mathrm{x}: U \to S $$ we call that parametrization orthogonal if $$ F=0 . $$ Definition 3.5. The geodesic curvature $k_g$ of a curve is a measure of the amount of deviance of the curve from the shortest are between two points on a surface. Definition 3.6. The Gaussian curvature $\kappa$ of a surface is an intrinsic measure of the curvature of a surface at a point. It is calculated by considering the maximal and minimal curvatures on the surface at a point. Formally, these values are multiplied to give $\kappa$. Theorem 3.7 (Gauss-Bonnet, Local). Let $R \subset \mathrm{x}(U)$ be a simple region of $S$ with orthogonal parametrization, and choose $\alpha: I \rightarrow S$ such that $\alpha(I)=\partial R$. Assume that $\alpha$ is positively oriented and parametrized piecewise by arc-length $s_i$. Let $\{\theta_i\}_{i=0}^k$ be the external angles of $\alpha$ at the vertices $\{\alpha(s_i)\}_{i=0}^k$, then $$ \sum_{i=0}^k \int_{s_i}^{s_{i+1}} k_{g(s)} d s+\iint_R \kappa d \sigma+\sum_{i=0}^k \theta_i=2 \pi . $$ Proof. We first let $u=u(s)$ and $v=v(s)$ be the expression of $\alpha$ in the parametrization $\mathrm{x}$. We recall that $$ k_{g(s)}=\frac{1}{2 \sqrt{E G}}\left(G_u \frac{d v}{d s}-E_v \frac{d u}{d s}\right)+\frac{d \varphi}{d s} $$ where we denote the differentiable function that measures the positive angle from $x$ to $\alpha^{\prime}(s)$ in $\left[s_i, s_{i+1}\right]$ as $\varphi\left(s_i\right)$. We now integrate the above expression, adding up the values for each $\left[s_i, s_{i+1}\right]$ : $$ \sum_{i=0}^k \int_{s_i}^{s_{i+1}} k_{g(s)}=\sum_{i=0}^k \int_{s_i}^{s_{i+1}} \frac{1}{2 \sqrt{E G}}\left(G_u \frac{d v}{d s}-E_v \frac{d u}{d s}\right) d s+\sum_{i=0}^k \int_{s_i}^{s_{i+1}} \frac{d \varphi_i}{d s} d s, $$ now, using the Gauss-Green theorem in the $u v$-plane on the right hand side of the above equation, we obtain the expression: $$ \iint_{\mathbf{x}^{-1}(R)} \frac{E_v}{2 \sqrt{E G}_v}+\frac{G_u}{2 \sqrt{E G}_u} d u d v+\sum_{i=0}^k \int_{s_i}^{s_{i+1}} \frac{d \varphi_i}{d s} d s $$ We note that by the Gauss Formula, $$ -\iint_{\mathbf{x}^{-1}(R)} \kappa \sqrt{E G} d u d v=-\iint_R \kappa d \sigma $$ also, recalling the Theorem of Turning Tangents, we know that, $$ \sum_{i=0}^k \int_{s_i}^{s_{i+1}} \frac{d \varphi_i}{d s} d s=\sum_{i=0}^k \varphi_i(s_i+1)-\varphi_i(s_i)= \pm 2 \pi-\sum_{i=0}^k \theta_i, $$ which we get because the theorem does not account for the discontinuities along the curve at the theta values. As we have assumed a positive orientation, we have, $$ \sum_{i=0}^k \int_{s_i}^{s_{i+1}} k_{g(s)} d s+\iint_R \kappa d \sigma+\sum_{i=0}^k \theta_i=2 \pi, $$ we note that we can obtain the opposite sign by assuming the opposite orientation, and we thus we have proven the local case of the Gauss-Bonnet Theorem. […Global…] Definition 5.7. The index $I$ of $v$ at the singular point $p$ is defined as follows. Let $\mathbf{x}$ be an orthogonal parametrization such that $\mathbf{x}(0,0)=p$ and the orientation is compatible with that of the surface $S$. Let $\alpha:[0, l]\to S$ be a closed simple regular parametrized curve such that it is the boundary of a simple region $R \subset S$, where the only singular point in $R$ is $p$. Now, we have a function $\varphi(t)$ with $t \in[0, l]$, such that it measures the angle from $\mathbf{x}_{\mathbf{u}}$ to the restriction of $v$ to $\alpha$, then, $$ 2 \pi I=\varphi(l)-\varphi(0)=\int_0^l \frac{d \varphi}{d t} d t . $$ Proposition 5.8. The index is independent of the choice of parametrization $\mathbf{x}$. Proposition 5.9. The index is independent of the choice of $\alpha$. Theorem 5.10 (Poincare-Hopf Index Theorem). The sum of the indices of a differentiable vector field $v$ with isolated singular points on a compact surface $S$ is equal to the Euler Characteristic of $S$. Proof. Let $S \subset \mathbb{R}^3$ be a compact surface and $v$ a differentiable vector field with exclusively isolated singular points. We notice that, due to compactness, these singular points must be finite in number otherwise there would exist a non-isolated singular point as a limit point for the others. We let $\{\mathbf x_a\}$ be a family of parametrizations such that each is compatible with the orientation of $S$. Now, we let $J$ be a triangulation of $S$ with the conditions that each $T$ in $J$ is contained in a coordinate neighborhood of $\{\mathbf{x_a}\}$, each triangle $T$ contains at most one singular point, and each triangle is positively oriented with no singular points on its boundary. Now, we apply the local Gauss-Bonnet theorem to each triangle and sum up the result. However, we recall that each triangle appears twice in this formulation, in opposite orientation, therefore, we have $$ \iint_S \kappa d \sigma-2 \pi \sum_{i=1}^k I_i=0, $$ to which we apply the most general form of Gauss-Bonnet to obtain, $$ \sum_{i=1}^k I_i=\frac{1}{2 \pi} \iint_S \kappa d \sigma=\chi(S) . $$ Remark 5.11. This result guarantees that a vector field on any surface homeomorphic to a sphere must have at least two isolated singular points, because the Euler characteristic of the 2-sphere is two. The solution is particularly remarkable because it shows that the sum of the indices of a vector field does not depend on $v$, but rather the topology of $S$, which is a non-intuitive idea. More tangibly, the Poincare-Hopf Index Theorem implies the previous theorem.