Various changes

Author: dalcde

IA_M/differential_equations.tex

@@ -344,7 +344,7 @@ \subsection{Chain rule}
   \delta f &= f(x+\delta x, y + \delta y) - f(x, y)\\
   &= f(x+\delta x, y + \delta y) - f(x + \delta x, y) + f(x+\delta x, y) - f(x, y)\\
   &= f_y(x + \delta x, y)\delta y + o(\delta y) + f_x(x, y)\delta x + o(\delta x)\\
-  &= (f_x(x, y) + o(1))\delta y + o(\delta y) + f_x(x, y)\delta x + o(\delta x)\\
+  &= (f_y(x, y) + o(1))\delta y + o(\delta y) + f_x(x, y)\delta x + o(\delta x)\\
   \delta f&= \frac{\partial f}{\partial x}\delta x + \frac{\partial f}{\partial y}\delta y + o(\delta x, \delta y)
 \end{align*}
 Take the limit as $\delta x, \delta y \to 0$, we have

IB_L/complex_analysis.tex

@@ -151,7 +151,7 @@ \subsection{Differentiation}
       \[
         u_x = \frac{x}{\sqrt{x^2 + y^2}},\quad u_y = \frac{y}{\sqrt{x^2 + y^2}}.
       \]
-      If we are not at the origin, then clearly both cannot vanish, but the partials of $v$ both vanish. Hence the Cauchy-Riemann equations do not hold and it is not differentiable outside of the origin.
+      If we are not at the origin, then clearly we cannot have both vanish vanishing, but the partials of $v$ both vanish. Hence the Cauchy-Riemann equations do not hold and it is not differentiable outside of the origin.
 
       At the origin, we can compute directly that
       \[

IB_M/markov_chains.tex

@@ -542,13 +542,13 @@ \subsection{Recurrence or transience}
   Consider $\Z^d = \{(x_1, x_2, \cdots, x_d): x_i \in \Z\}$. This generates a graph with $x$ adjacent to $y$ if $|x - y| = 1$, where $|\ph |$ is the Euclidean norm.
   \begin{center}
     \begin{tikzpicture}[scale=0.75]
-      \node at (-4, 0) {$d = 0$};
+      \node at (-4, 0) {$d = 1$};
       \draw (-2.5, 0) -- (2.5, 0);
       \foreach \x in {-2,-1,...,2} {
         \node [circ] at (\x, 0) {};
       }
       \begin{scope}[shift={(0, -3.5)}]
-        \node at (-4, 2) {$d = 1$};
+        \node at (-4, 2) {$d = 2$};
         \foreach \x in {-2, -1,...,2} {
           \foreach \y in {-2,-1,...,2} {
             \node [circ] at (\x, \y) {};

III_E/classical_and_quantum_solitons.tex

@@ -1576,7 +1576,7 @@ \subsubsection*{Behaviour of vortices as $r \to 0$}
 \subsection{Bogomolny/self-dual vortices and Taubes' theorem}
 As mentioned, we can think of the radial vortex solution as a collection of $N$ vortices all superposed at the origin. Is it possible to have separated vortices all over the plane? Naively, we would expect that the vortices exert forces on each other, and so we don't get a static solution. However, it turns out that in the $\lambda = 1$ case, there do exist static solutions corresponding to vortices at arbitrary locations on the plane.
 
-This is not obvious, and the proof requires some serious analysis. We will not do the analysis, which requires use of Sobolev spaces and PDE theory. However, we will do all the non-hard-analysis part. In particular, % we will obtain Bogomolny bounds as we did in the sine-Gordon case, and reduce the problem to finding solutions of a single scalar PDE, which can be understood with tools from calculus of variations and elliptic PDE.
+This is not obvious, and the proof requires some serious analysis. We will not do the analysis, which requires use of Sobolev spaces and PDE theory. However, we will do all the non-hard-analysis part. In particular, we will obtain Bogomolny bounds as we did in the sine-Gordon case, and reduce the problem to finding solutions of a single scalar PDE, which can be understood with tools from calculus of variations and elliptic PDEs.
 
 Recall that for the sine-Gordon kinks, we needed to solve
 \[

III_L/modular_forms_and_l_functions.tex

@@ -3100,7 +3100,7 @@ \section{\texorpdfstring{$L$}{L}-functions of eigenforms}
 \]
 However, we can't get rid of the $\varepsilon$.
 
-What we have established is a way to go from modular forms to $L$-functions, and we found that these $L$-functions satisfy certain functional equations. Now is it possible to go the other way round? Given any $L$-function, does it come from a modular form? This is known as the \term{converse problem}r One obvious necessary condition is that it should satisfy the functional equation, but is this sufficient?
+What we have established is a way to go from modular forms to $L$-functions, and we found that these $L$-functions satisfy certain functional equations. Now is it possible to go the other way round? Given any $L$-function, does it come from a modular form? This is known as the \term{converse problem}. One obvious necessary condition is that it should satisfy the functional equation, but is this sufficient?
 
 To further investigate this, we want to invert the Mellin transform.
 

III_M/algebraic_topology_iii.tex

@@ -3497,7 +3497,7 @@ \subsection{Compactly supported cohomology}
 \end{defi}
 Now if $f$ is proper, then it does induce a map $H_c^* (\ph)$ by the usual construction.
 
-From now on, we will assume all spaces are Hausdorff, so that all compact subsets are closed. This isn't too bad a restriction since we are ultimately interested in manifolds, which are by definition compact.
+From now on, we will assume all spaces are Hausdorff, so that all compact subsets are closed. This isn't too bad a restriction since we are ultimately interested in manifolds, which are by definition Hausdorff.
 
 Let $i: U \to X$ be the inclusion of an open subspace. We let $K \subseteq U$ be compact. Then by excision, we have an isomorphism
 \[
@@ -4070,7 +4070,7 @@ \subsection{\texorpdfstring{Poincar\'e}{Poincare} duality}
 Of course, we can get $H^*(\RP^n, \F_2)$ similarly.
 
 \subsection{Applications}
-We go through two rather straightforward applications, before we move on to bigger things like submanifolds.
+We go through two rather straightforward applications, before we move on to bigger things like the intersection product.
 
 \subsubsection*{Signature}
 We focus on the case where $d = 2n$ is even. Then we have, in particular, a non-degenerate bilinear form
@@ -4109,7 +4109,7 @@ \subsubsection*{Degree}
   \[
     f_*([M]) \in H_d(N, \Z) \cong \Z[N].
   \]
-  So $f_*([M]) = k \Z[N]$ for some $k$. This $k$ is called the \emph{degree} of $f$, written $\deg (f)$.
+  So $f_*([M]) = k [N]$ for some $k$. This $k$ is called the \emph{degree} of $f$, written $\deg (f)$.
 \end{defi}
 
 If $N = M = S^n$ and we pick the same orientation for them, then this recovers our previous definition.
@@ -4145,21 +4145,24 @@ \subsubsection*{Degree}
   This is a contradiction.
 \end{proof}
 
-\subsection{Submanifolds}
-\subsubsection*{Normal bundles}
-We restrict our attention to smooth manifolds, and then we can talk about the tangent bundle. Recall (from the example sheet) that an orientation of the manifold is the same as an orientation on the tangent bundle.
+\subsection{Intersection product}
+Recall that cohomology comes with a cup product. Thus, Poincar\'e duality gives us a product on homology. Our goal in this section is to understand this product.
 
-Let $M$ be a compact smooth $R$-oriented manifold, and $N \subseteq M$ be an $n$-dimensional $R$-oriented submanifold, $i: N \hookrightarrow M$ be the inclusion. Suppose $\dim M = d$ and $\dim N = n$. Then we obtain a canonical homology class
+We restrict our attention to smooth manifolds, so that we can talk about the tangent bundle. Recall (from the example sheet) that an orientation of the manifold is the same as an orientation on the tangent bundle.
+
+We will consider homology classes that come from submanifolds. For concreteness, let $M$ be a compact smooth $R$-oriented manifold, and $N \subseteq M$ be an $n$-dimensional $R$-oriented submanifold. Let $i: N \hookrightarrow M$ be the inclusion. Suppose $\dim M = d$ and $\dim N = n$. Then we obtain a canonical homology class
 \[
   i_*[N] \in H_n(M; R).
 \]
-We will abuse notation and write $[N]$ for $i_* [N]$. This may or may not be zero.
+We will abuse notation and write $[N]$ for $i_* [N]$. This may or may not be zero. Our objective is to show that under suitable conditions, the product of $[N_1]$ and $[N_2]$ is $[N_1 \cap N_2]$.
+
+To do so, we will have to understand what is the cohomology class Poinacr\'e dual to $[N]$. We claim that, suitably interpreted, it is the Thom class of the normal bundle.
 
-As before, we can arbitrarily pick a metric on $TM$, and then decompose
+Write $\nu_{N \subseteq M}$ for the normal bundle of $N$ in $M$. Picking a metric on $TM$, we can decompose
 \[
   i^*TM \cong TN \oplus \nu_{N \subseteq M},
 \]
-where $\nu_{N\subseteq M}$ is the normal bundle. Since $TM$ is oriented, we obtain an orientation on the pullback $i^*TM$. Similarly, $TN$ is also oriented by assumption. In general, we have the following result:
+Since $TM$ is oriented, we obtain an orientation on the pullback $i^*TM$. Similarly, $TN$ is also oriented by assumption. In general, we have the following result:
 \begin{lemma}
   Let $X$ be a space and $V$ a vector bundle over $X$. If $V = U \oplus W$, then orientations for any two of $U, W, V$ give an orientation for the third.
 \end{lemma}
@@ -4203,10 +4206,12 @@ \subsubsection*{Normal bundles}
 
 Under the identification $H_c^{d - n}(U; R) \cong H^{d - n}(\nu_{N \subseteq M}, \nu_{N \subseteq M}^\#; R)$, the above says that the image of $[N] \in H_n(N; R)$ in $H^{d - n}_c(U; R)$ is the Thom class of the normal bundle $\nu_{N \subseteq M}$.
 
-On the other hand, if we look at this composition via the bottom map, then $[N]$ gets sent to $D_M^{-1}([N])$. So we know that the Poincar\'e dual of a submanifold is the (extension by zero of the) normal Thom class.
+On the other hand, if we look at this composition via the bottom map, then $[N]$ gets sent to $D_M^{-1}([N])$. So we know that
+\begin{thm}
+  The Poincar\'e dual of a submanifold is (the extension by zero of) the normal Thom class.
+\end{thm}
 
-\subsubsection*{Intersection product}
-Now if we had two submanifolds $N, W \subseteq M$. Then the normal Thom classes give us two cohomology classes of $M$. We can then take the cup product, and see what we get. It turns out the answer is great. However, this happens only when the manifolds intersect nicely.
+Now if we had two submanifolds $N, W \subseteq M$. Then the normal Thom classes give us two cohomology classes of $M$. As promised, When the two intersect nicely, the cup product of their Thom classes is the Thom class of $[N \cap W]$. The niceness condition we need is the following:
 \begin{defi}[Transverse intersection]\index{transverse intersection}\index{intersect transversely}
   We say two submanifolds $N, W \subseteq M$ \emph{intersect transversely} if for all $x \in N \cap W$, we have
   \[

III_M/local_fields.tex

@@ -3897,7 +3897,7 @@ \subsection{Formal groups}
   Let $e \in \mathcal{E}_\pi$ be a Lubin--Tate series. Then there are unique power series $F_e(X, Y) \in \mathcal{O}_K[[X, Y]]$ such that
   \begin{align*}
     F_e(X, Y) &\equiv X + Y \mod (X + Y)^2\\
-    e(F_e(X, Y)) &= F_e(e(X), E(Y))
+    e(F_e(X, Y)) &= F_e(e(X), e(Y))
   \end{align*}
 \end{cor}
 

II_M/galois_theory.tex

@@ -2444,7 +2444,7 @@ \subsection{Insolubility of general equations of degree 5 or more}
 \end{proof}
 
 \begin{cor}
-  Let $K$ be a field, $h \in K[t]$. Let $L$ be the splitting of $h$ over $K$. Then $h$ can be solved by radicals if and only if $\Gal(L/K)$ is soluble.
+  Let $K$ be a field with $\Char K = 0$ and $h \in K[t]$. Let $L$ be the splitting of $h$ over $K$. Then $h$ can be solved by radicals if and only if $\Gal(L/K)$ is soluble.
 \end{cor}
 
 \begin{proof}

II_M/linear_analysis.tex

@@ -1320,7 +1320,7 @@ \subsection{Some applications}
   So $T$ is unbounded.
 
   We claim the graph of $T$ is closed. If so, then since $C([0, 1])$ is complete, the closed graph theorem implies $D(T)$ is not complete.
-  
+
   To check this, suppose we have $f_n \to f$ in the $C([0, 1])$ norm, and $f_n' \to g$, again in the $C([0, 1])$ norm. We want to show that $f' = g$.
 
   By the fundamental theorem of calculus, we have

IV_L/topics_in_number_theory.tex

@@ -817,7 +817,7 @@ \subsection{Hecke characters}
   \]
   By assumption, $\chi_v(x_v) = 1$ for all but finitely many $v$. So this is well-defined. These two operations are clearly inverses to each other.
 \end{proof}
-In general, we can write $\chi$ as 
+In general, we can write $\chi$ as
 \[
   \chi = \chi_\infty \chi^\infty,\quad \chi^\infty = \prod_{v \nmid \infty} \chi_v: K^{\times, \infty} \to \C^\times, \quad \chi_\infty = \prod_{v \mid \infty} \chi_v: K_\infty^\times \to \C^\times.
 \]
@@ -845,7 +845,7 @@ \subsection{Hecke characters}
 \begin{eg}
   The idele norm $|\ph|_{\A}: C_K \to \R^\times_{>0}$ is a character not of finite order. In the case $K = \Q$, we have $C_\Q = \R_{>0}^\times \times \hat{\Z}^\times$. The idele norm is then the projection onto $\R_{>0}^\times$.
 
-  Thus, if $\chi; C_\Q \to \C^\times$ is a Hecke character, then the restriction  to $\R_{>0}^\times$ is of the form $x \mapsto x^{s}$ for some $s$. If we write
+  Thus, if $\chi; C_\Q \to \C^\times$ is a Hecke character, then the restriction to $\R_{>0}^\times$ is of the form $x \mapsto x^{s}$ for some $s$. If we write
   \[
     \chi(x) = |x|_{\A}^s \cdot \chi'(x)
   \]
@@ -2521,7 +2521,7 @@ \subsection{Local Langlands correspondence}
     \item If a supercuspidal $\pi_0$ corresponds to the representation $\sigma_0$ of $W_F$, then the essentially square integrable representation $\pi = Q(\pi_0(-\frac{r-1}{2}), \ldots, \pi_0(\frac{r - 1}{2}))$ corresponds to $\sigma = \sigma_0 \otimes \Sym^{r - 1} \C^2$.
 
     \item If $\pi_i$ correspond to $\sigma_i$, where $\sigma_i$ are irreducible and unitary, then the tempered representation $\Ind_P^G(\pi_1 \otimes \cdots \otimes \pi_r)$ corresponds to $\sigma_1 \oplus \cdots \oplus \sigma_r$.
-%      
+%
 %      Tempered representations correspond to unitary representations of the form$\sigma = \sigma_1 \oplus \cdots \oplus \sigma_r$, where each $\sigma_i$ is irreducible and unitary, where if $\sigma_i$ corresponds to $\pi_i$, then the tempered representation is given by $\Ind_P^G(\pi_1 \otimes \cdots \otimes \pi_r)$.
     \item For general representations, if $\pi$ is the Langlands quotient of
       \[