Tue Jun 5 22:16:21 BST 2018

III_L/positivity_in_algebraic_geometry.tex

@@ -696,7 +696,7 @@ \subsection{The intersection product}
   \[
     D_1 \cdot D_2 = \deg(D_1) \deg(D_2).
   \]
-  This is Bezout's theorem!
+  This is B\'ezout's theorem!
 \end{eg}
 \begin{ex}
   Let $C_1, C_2 \subseteq X$ be curves without common components. Then

III_M/algebraic_topology_iii.tex

@@ -2071,7 +2071,7 @@ \section{Cell complexes}
   Consider the diagram
   \[
     \begin{tikzcd}
-      H_n(D_\alpha^n, \partial D_\alpha^n) \ar[d, "(\Phi_\alpha)_*"] \ar[r, "\partial", "\sim"'] & H_{n - 1}(\partial D_\alpha^n)\ar[d, "(\varphi_\alpha)_*"] \ar[r, dashed] & \tilde{H}_{n - 1}(S^{n - 1}, \beta) \\
+      H_n(D_\alpha^n, \partial D_\alpha^n) \ar[d, "(\Phi_\alpha)_*"] \ar[r, "\partial", "\sim"'] & H_{n - 1}(\partial D_\alpha^n)\ar[d, "(\varphi_\alpha)_*"] \ar[r, dashed] & \tilde{H}_{n - 1}(S^{n - 1}_\beta) \\
       H_n(X^n, X^{n - 1}) \ar[r, "\partial"] \ar[rd, "d_n^\mathrm{cell}"] & H_{n - 1}(X^{n - 1}) \ar[d, "q"] & \tilde{H}_{n - 1}\left(\bigvee S_\gamma^{n - 1}\right)\ar[u, "\text{collapse}"]\\
       & H_{n - 1}(X^{n - 1}, X^{n - 2}) \ar[r, "\text{excision}", "\sim"'] & \tilde{H}_{n - 1}(X^{n - 1}/X^{n - 2}) \ar[u, equals]
     \end{tikzcd}

III_M/local_fields.tex

@@ -695,148 +695,107 @@ \subsection{Hensel's lemma}
 \end{eg}
 
 \begin{defi}[Primitive polynomial]\index{primitive polynomial}
-  If $K$ is a valued field and $F(x) = a_0 + a_1 x + \cdots + a_n x^n \in K[x]$ is a polynomial, we say that $F$ is \emph{primitive} if
+  If $K$ is a valued field and $f(x) = a_0 + a_1 x + \cdots + a_n x^n \in K[x]$ is a polynomial, we say that $f$ is \emph{primitive} if
   \[
     \max_i |a_i| = 1.
   \]
-  In particular, we have $F \in \mathcal{O}[x]$.
+  In particular, we have $f \in \mathcal{O}[x]$.
 \end{defi}
 The point of a primitive polynomial is that such a polynomial is naturally, and non-trivially, an element of $k[x]$. Moreover, focusing on such polynomials is not that much of a restriction, since any polynomial is a constant multiple of a primitive polynomial.
 
 \begin{thm}[Hensel's lemma]\index{Hensel's lemma}
-  Let $K$ be a complete valued field, and let $F \in K[x]$ be primitive. Put $f = F\bmod \mathfrak{m} \in k[x]$. If there is a factorization
+  Let $K$ be a complete valued field, and let $f \in K[x]$ be primitive. Put $\bar{f} = f\bmod \mathfrak{m} \in k[x]$. If there is a factorization
   \[
-    f(x) = g(x) h(x)
+    \bar{f}(x) = \bar{g}(x) \bar{h}(x)
   \]
-  with $(g, h) = 1$, then there is a factorization
+  with $(\bar{g}, \bar{h}) = 1$, then there is a factorization
   \[
-    F(x) = G(x) H(x)
+    f(x) = g(x) h(x)
   \]
   in $\mathcal{O}[x]$ with
   \[
-    g = G,\quad h = H\mod \mathfrak{m},
+    \bar{g} = g,\quad \bar{h} = h\mod \mathfrak{m},
   \]
-  with $\deg g = \deg G$.
+  with $\deg g = \deg \bar{g}$.
 \end{thm}
-Note that requiring $\deg g = \deg G$ is the best we can hope for --- we cannot guarantee $\deg h = \deg H$, since we need not have $\deg f = \deg F$.
-
-This is one of the most important results in the course. Unfortunately, the proof is not terribly enlightening.
+Note that requiring $\deg g = \deg \bar{g}$ is the best we can hope for --- we cannot guarantee $\deg h = \deg \bar{h}$, since we need not have $\deg f = \deg \bar{f}$.
 
+This is one of the most important results in the course.
 \begin{proof}
-  Put $d = \deg F$ and $m = \deg g$. Then we have $\deg h \leq d - m$.
-
-  We pick lifts $G_0, H_0 \in \mathcal{O}[x]$ of $g, h$ with $\deg G_0 = \deg g$, and $\deg H_0 \leq d - m$. Then we have
-  \[
-    F = G_0 H_0 \mod \mathfrak{m}.
-  \]
-  Our strategy is to add some correction terms to $G_0$ and $H_0$ so that the corresponding equation holds mod $\mathfrak{m}^2$, then lift to a result mod $\mathfrak{m}^3$ etc, \ldots and hopefully eventually get something in the limit.
-
-  However, this doesn't really work, since there is no guarantee that the limit converges! The trick is to pick and element $\pi \in \mathfrak{m}$, and then work modulo $\pi^k$. This is not hard, since we notice that
-  \[
-    \pi \mathcal{O} = \{x \in \mathcal{O}: |x| \leq |\pi|\}.
-  \]
-  Since our equations only have finitely many coefficients, we can always replace $\bmod \mathfrak{m}$ with $\bmod \pi$ for large enough $\pi$.
-
-  If we do this, then the correction terms we add in each iteration will be bounded by $|\pi|^k$, and thus tends to $0$. Now recall that a series converges iff the moduli of the terms tend to $0$! So we are done if this works.
-
-  Before we begin, we use the given condition $(g, h) = 1$ to obtain some $A, B \in \mathcal{O}[x]$ such that
+  Let $g_0, h_0$ be arbitrary lifts of $\bar{g}$ and $\bar{h}$ to $\mathcal{O}[x]$ with $\deg \bar{g} = g_0$ and $\deg \bar{h} = h_0$. Then we have
   \[
-    AG_0 + B H_0 \equiv 1 \mod \mathfrak{m}.
+    f = g_0 h_0 \mod \mathfrak{m}.
   \]
-  Then we know that
+  The idea is to construct a ``Taylor expansion'' of the desired $g$ and $h$ term by term, starting from $g_0$ and $h_0$, and using completeness to guarantee convergence. To proceed, we use our assumption that $\bar{g}, \bar{h}$ are coprime to find some $a, b \in \mathcal{O}[x]$ such that
   \[
-    F - G_0 H_0 \equiv AG_0 + BH_0 - 1 \equiv 0 \mod \mathfrak{m}.
+    ag_0 + bh_0 \equiv 1 \mod \mathfrak{m}.\tag{$\dagger$}
   \]
-  Thus, since the polynomials involved have finitely many coefficients, we can pick pick a $\pi \in \mathfrak{m}$ with large enough modulus such that
+  It is easier to work modulo some element $\pi$ instead of modulo the ideal $\mathfrak{m}$, since we are used to doing Taylor expansion that way. Fortunately, since the equations above involve only finitely many coefficients, we can pick an $\pi \in \mathfrak{m}$ with absolute value large enough (i.e.\ close enough to $1$) such that the above equations hold with $\mathfrak{m}$ replaced with $\pi$. Thus, we can write
   \[
-    F - G_0 H_0 \equiv AG_0 + BH_0 - 1 \equiv 0 \mod \pi.
+    f = g_0h_0 + \pi r_0,\quad r_0 \in \mathcal{O}[x].
   \]
-  The idea is to find successive approximations
-  \begin{align*}
-    G_{n - 1} &= G_0 + \pi P_1 + \cdots + \pi^{n - 1} P_{n - 1},\\
-    H_{n - 1} &= H_0 + \pi Q_1 + \cdots + \pi^{n - 1} Q_{n - 1}.
-  \end{align*}
-  that satisfy
+  Plugging in $(\dagger)$, we get
   \[
-    F \equiv G_{n - 1}H_{n - 1}\mod \pi^n.
+    f = g_0 h_0 + \pi r_0 (a g_0 + b h_0) + \pi^2 (\text{something}).
   \]
-  We then set
-  \begin{align*}
-    G &= G_0 + \pi P_1 + \pi^2 P_2 + \cdots,\\
-    H &= H_0 + \pi Q_1 + \pi^2 Q_2 + \cdots.
-  \end{align*}
-  We then have $G \equiv G_i \mod \pi^{i + 1}$, and similarly for $H$. So we have
+  If we are lucky enough that $\deg r_0 b < \deg g_0$, then we group as we learnt in secondary school to get
   \[
-    F \equiv GH \mod \pi^n
+    f = (g_0 + \pi r_0 b)(h_0 + \pi r_0 a) + \pi^2 (\text{something}).
   \]
-  for all $n$. As $|\pi^n| \to 0$ as $n \to \infty$, we must have $F = GH$.
-
-  We proceed by induction. Assume we already have $G_{n - 1}, H_{n - 1}$. We need to find $P_n, Q_n$ such that
+  We can then set
   \begin{align*}
-    G_n &= G_{n - 1} + \pi^n P_n\\
-    H_n &= H_{n - 1} - \pi^n Q_n
+    g_1 &= g_0 + \pi r_0 b\\
+    h_1 &= h_0 + \pi r_0 a,
   \end{align*}
-  satisfy
+  and then we can write
   \[
-    F - G_n H_n = 0\mod \pi^{n + 1}.
+    f = g_1 h_1 + \pi^2 r_1,\quad r_1 \in \mathcal{O}[x],\quad \deg g_1 = \deg \bar{g}.\tag{$*$}
   \]
-  Expanding $G_n$ and $H_n$ out, we get
+  If it is not true that $\deg r_0 b \leq \deg g_0$, we use the division algorithm to write
   \[
-    F - G_{n - 1}H_{n - 1} \equiv \pi^n (G_{m - 1} Q_n + H_{n - 1} P_n) \mod \pi^{n + 1}.
+    r_0 b = q g_0 + p.
   \]
-  We rearrange and divide by $\pi^n$ to obtain
-  \[
-    G_{n - 1} Q_n + H_{n - 1} P_n \equiv \frac{1}{\pi^n}(F - G_{n - 1}H_{n - 1}) \mod \pi.
-  \]
-  However, note that $G_{n - 1} = G_0 \mod \pi$ (and similarly for $H$), so we have
-  \[
-    G_0 Q_n + H_0 P_n \equiv \frac{1}{\pi^n}(F - G_{n - 1}H_{n - 1}) \mod \pi.
-  \]
-  Note that this division makes sense since by assumption, $F - G_{n - 1}H_{n - 1}$ is a multiple of $\pi^n$.
-
-  We write
-  \[
-    E_n = \frac{1}{\pi^n}(F - G_{n - 1}H_{n - 1}).
-  \]
-  Now recall that there are $A, B$ such that
-  \[
-    AG_0 + B H_0 \equiv 1\mod \pi.
-  \]
-  Multiplying by $E_n$ thus gives
+  Then we have
   \[
-    E_n \equiv AG_0 E_n + B H_0 E_n\mod \pi.
+    f = g_0 h_0 + \pi ((r_0a + q)g_0 + p h_0),
   \]
-  We might be tempted to set $Q_n$ and $P_n$ to be just $AE_n$ and $BE_n$ respectively, but we must control the degree of $P_n$ appropriately. Its degree must be less than $\deg g = \deg G_0$.
+  and then proceed as above.
 
-  Fortunately, the division algorithm comes to the rescue. Recall that $G_0$ is the lift of $g$ of the same degree, and thus its leading coefficient must be an element of $\mathcal{O} \setminus \mathfrak{m}$, i.e.\ a unit. So we can perform the division algorithm to write
+  Given the factorization $(*)$, we replace $r_1$ by $r_1(a g_0 + b h_0)$, and then repeat the procedure to get a factorization
   \[
-    B F_n = Q G_0 + P_n.
+    f \equiv g_2 h_2 \mod \pi^3,\quad \deg g_2 = \deg \bar{g}.
   \]
-  with $\deg P_n < \deg G_0$. This gives the desired $P_n$. Then we have
+  Inductively, we constrict $g_k, h_k$ such that
+  \begin{align*}
+    f &\equiv g_k h_k \mod \pi^{k + 1}\\
+    g_k &\equiv g_{k - 1} \mod \pi^k\\
+    h_k &\equiv h_{k - 1} \mod \pi^k\\
+    \deg g_k &= \deg \bar{g}
+  \end{align*}
+  Note that we may drop the terms of $h_k$ whose coefficient are in $\pi^{k + 1}\mathcal{O}$, and the above equations still hold. Moreover, we can then bound $\deg h_k \leq \deg f - \deg g_k$. It now remains to set
   \[
-    G_0 (A E_n + H_0 Q) + H_0 P_n \equiv E_n \mod \pi.
+    g = \lim_{k \to \infty} g_k,\quad h = \lim_{k \to \infty} h_k.\qedhere
   \]
-  Finally, let $Q_n = A E_n + H_0 Q$, where we drop all coefficients divisible by $\pi$. Then since $\deg E_n, \deg H_0 P_n \leq \deg F = m$, we must have $\deg Q_n \leq m - \deg G_0 = m - d$. So this has the desired degree as well. So done.
 \end{proof}
 
 \begin{cor}
-  Let $F(x) = a_0 + a_1 x + \cdots + a_n x^n \in K[x]$ where $K$ is complete and $a_0, a_n \not= 0$. If $F$ is irreducible, then
+  Let $f(x) = a_0 + a_1 x + \cdots + a_n x^n \in K[x]$ where $K$ is complete and $a_0, a_n \not= 0$. If $f$ is irreducible, then
   \[
     |a_\ell| \leq \max(|a_0|, |a_n|)
   \]
   for all $\ell$.
 \end{cor}
 
 \begin{proof}
-  By scaling, we can wlog $F$ is primitive. We then have to prove that $\max(|a_0|, |a_n|) = 1$. If not, let $r$ be minimal such that $|a_r| = 1$. Then $0 < r < n$. Moreover, we can write
+  By scaling, we can wlog $f$ is primitive. We then have to prove that $\max(|a_0|, |a_n|) = 1$. If not, let $r$ be minimal such that $|a_r| = 1$. Then $0 < r < n$. Moreover, we can write
   \[
-    F(x) \equiv x^r (a_r + a_{r + 1}x + \cdots + a_n x^{n - r})\mod \mathfrak{m}.
+    f(x) \equiv x^r (a_r + a_{r + 1}x + \cdots + a_n x^{n - r})\mod \mathfrak{m}.
   \]
-  But then Hensel's lemma says this lifts to a factorization of $F$, a contradiction.
+  But then Hensel's lemma says this lifts to a factorization of $f$, a contradiction.
 \end{proof}
 
 \begin{cor}[of Hensel's lemma]
-  Let $F \in \mathcal{O}[x]$ be monic, and $K$ complete. If $F \mod \mathfrak{m}$ has a simple root $\bar{\alpha} \in k$, then $F$ has a (unique) simple root $\alpha \in \mathcal{O}$ lifting $\bar{\alpha}$.
+  Let $f \in \mathcal{O}[x]$ be monic, and $K$ complete. If $f \mod \mathfrak{m}$ has a simple root $\bar{\alpha} \in k$, then $f$ has a (unique) simple root $\alpha \in \mathcal{O}$ lifting $\bar{\alpha}$.
 \end{cor}
 
 \begin{eg}
@@ -1023,7 +982,7 @@ \subsection{Extension of norms}
   \[
     N_{L/K}(\alpha) = \pm a_0^m,
   \]
-  and hence $|a_0| \leq 1$. 
+  and hence $|a_0| \leq 1$.
 
   By the corollary of Hensel's lemma, for each $i$, we have
   \[
@@ -2342,7 +2301,7 @@ \subsection{Unramified extensions}
 
   So we obtain a map
   \[
-    \Hom_{K\operatorname{-}\mathrm{Alg}}(L, M) \to \Hom_{k_K\operatorname{-}\mathrm{Alg}} (k_L, k_M)
+    \Hom_{K\text{-}\mathrm{Alg}}(L, M) \to \Hom_{k_K\text{-}\mathrm{Alg}} (k_L, k_M)
   \]
   To see this is bijective, we take a primitive element $\bar{\alpha} \in k_L$ over $k_K$, and take a minimal polynomial $\bar{f} \in k_K[x]$. We take a monic lift of $\bar{f}$ to $\mathcal{O}_k[x]$, and $\alpha \in \mathcal{O}_L$ the unique root of $f$ which lifts $\bar{\alpha}$, which exists by Hensel's lemma. Then by counting dimensions, the fact that the extension is unramified tells us that
   \[
@@ -2351,8 +2310,8 @@ \subsection{Unramified extensions}
   So we can construct the following diagram:
   \[
     \begin{tikzcd}
-      \varphi \ar[d, maps to] & \Hom_{K\operatorname{-}\mathrm{Alg}} \ar[d, "\cong"] \ar[r, "\text{reduction}"] & \Hom_{k_K\operatorname{-}\mathrm{Alg}} (k_L, k_M) \ar[d, "\cong"] & \bar{\varphi} \ar[d, maps to]\\
-      \varphi(\alpha) & \{x \in M: f(x) = 0\} \ar[r, "\text{reduction}"] & \{\bar{x} \in k_M: \bar{f}(\bar{x})\} & \bar\varphi(\bar{\alpha})
+      \varphi \ar[d, maps to] & \Hom_{K\text{-}\mathrm{Alg}}(L, M) \ar[d, "\cong"] \ar[r, "\text{reduction}"] & \Hom_{k_K\text{-}\mathrm{Alg}} (k_L, k_M) \ar[d, "\cong"] & \bar{\varphi} \ar[d, maps to]\\
+      \varphi(\alpha) & \{x \in M: f(x) = 0\} \ar[r, "\text{reduction}"] & \{\bar{x} \in k_M: \bar{f}(\bar{x}) = 0\} & \bar\varphi(\bar{\alpha})
     \end{tikzcd}
   \]
   But the bottom map is a bijection by Hensel's lemma. So done.
@@ -2495,7 +2454,11 @@ \subsection{Totally ramified extensions}
   \]
   So $[L:K] = e_{L/K} = n$. So $L/K$ is totally ramified and $\alpha$ is a uniformizer.
 \end{proof}
-In fact, more is true. We have $\mathcal{O}_L = \mathcal{O}_K[\pi_L]$.
+In fact, more is true. We have $\mathcal{O}_L = \mathcal{O}_K[\pi_L]$, since every element in $\mathcal{O}_L$ can be written as
+\[
+  \sum_{i \geq 0} a_i \pi_L^i,
+\]
+where $a_i$ is a lift of an element in $k_L = k_K$, which can be chosen to be in $\mathcal{O}_K$.
 
 \section{Further ramification theory}
 \subsection{Some filtrations}

III_M/symmetries_fields_and_particles.tex

@@ -1096,11 +1096,11 @@ \subsection{The exponential map}
 \[
   \exp(X) \exp(Y) = \exp(X + Y),
 \]
-since the usual proof assumes that $X$ and $Y$ commute. Instead, what we've got is the \emph{Baker-Campbell-Hausdorff formula}.
-\begin{thm}[Baker-Campbell-Hausdorff formula]
+since the usual proof assumes that $X$ and $Y$ commute. Instead, what we've got is the \term{Baker--Campbell--Hausdorff formula}.
+\begin{thm}[Baker--Campbell--Hausdorff formula]
   We have
   \[
-    \exp(X) \exp(Y) = \exp\left(X + Y + \frac{1}{2}[X, Y] + \frac{1}{12}([X, [X, Y]] - [Y, [X, Y]]) + \cdots\right).
+    \exp(X) \exp(Y) = \exp\!\left(\!X + Y + \frac{1}{2}[X, Y] + \frac{1}{12}([X, [X, Y]] - [Y, [X, Y]]) + \cdots\!\right).
   \]
 \end{thm}
 It is possible to find the general formula for all the terms, but it is messy. We will not prove this.
@@ -1257,11 +1257,10 @@ \subsection{Representations of Lie groups and algebras}
   \[
     [\ad_X, \ad_Y] = \ad_{[X, Y]}.
   \]
-  We compute
+  But the Jacobi identity says
   \[
-    [\ad_X, \ad_Y](Z) = [X, [Y, Z]] - [Y, [X, Z]] = -[Z, [X, Y]] = [[X, Y], Z] = \ad_{[X, Y]}(Z)
+    [\ad_X, \ad_Y](Z) = [X, [Y, Z]] - [Y, [X, Z]] = [[X, Y], Z] = \ad_{[X, Y]}(Z).\qedhere
   \]
-  by the Jacobi identity. So done.
 \end{proof}
 
 We will eventually want to find all representations of a Lie algebra. To do so, we need the notion of when two representations are ``the same''.
@@ -2777,7 +2776,7 @@ \subsection{The classification}
     0 & A^{(2)}
   \end{pmatrix},
 \]
-then in fact the Lie algebra is not simple, or even semi-simple.
+then in fact the Lie algebra is not simple.
 
 So we finally have
 \begin{prop}
@@ -2827,18 +2826,13 @@ \subsection{The classification}
 \begin{defi}[Dynkin diagram]\index{Dynkin diagram}
   Given a Cartan matrix, we draw a diagram as follows:
   \begin{enumerate}
-    \item We draw a node
+    \item For each simple root $\alpha_{(i)} \in \Phi_S$, we draw a node
       \begin{center}
         \begin{tikzpicture}
           \node [draw, fill=morange, circle, inner sep = 0, minimum size = 6] at (0, 0) {};
         \end{tikzpicture}
       \end{center}
-      for each simple root $\alpha_{(i)} \in \Phi_S$.
-    \item We join the nodes corresponding to $\alpha_{(i)}, \alpha_{(j)}$, with
-      \[
-        \max\{|A^{ij}|, |A^{ji}|\}
-      \]
-      many lines.
+    \item We join the nodes corresponding to $\alpha_{(i)}, \alpha_{(j)}$ with $A^{ij} A^{ji}$ many lines.
     \item If the roots have different lengths, we draw an arrow from the longer root to the shorter root. This happens when $A^{ij} \not= A^{ji}$.
   \end{enumerate}
 \end{defi}