commit b868322b01444266972b7aec8a69a92c57a81399
parent fed8832ba997fdca91d156fd6d92bcedf2fd0066
Author: miksa <milutin@popovic.xyz>
Date: Wed, 3 Mar 2021 13:14:12 +0100
almost finished week3
Diffstat:
| M | week2.tex | | | 21 | ++++++++++----------- |
| A | week3.pdf | | | 0 | |
| A | week3.tex | | | 196 | +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ |
3 files changed, 206 insertions(+), 11 deletions(-)
diff --git a/week2.tex b/week2.tex
@@ -61,17 +61,17 @@
\end{definition}
\begin{question}
Does $/$ denote the complement, because one usually writes $\setminus$.
- \url{https://en.wikipedia.org/wiki/Quotient_space_(linear_algebra)}
+ $/$ denotes the Qutotient \url{https://en.wikipedia.org/wiki/Quotient_space_(linear_algebra)}
\end{question}
-READ THE WIKI In other words the balanced tensor product forms only elements of
-\begin{itemize}
- \item $E$ that preserver the \textit{left} representation of $A$ and
- \item $F$ that preserver the \textit{right} representation of $A$.
-\end{itemize}
-Which is the same saying:
-\begin{align*}
- E \otimes _A F = \left\{e a\otimes _A f = e \otimes _A a f: \;\;\; a \in A,\ e \in E,\ f \in F \right\}
-\end{align*}
+%In other words the balanced tensor product forms only elements of
+%\begin{itemize}
+% \item $E$ that preserver the \textit{left} representation of $A$ and
+% \item $F$ that preserver the \textit{right} representation of $A$.
+%\end{itemize}
+%Which is the same saying:
+%\begin{align*}
+% E \otimes _A F = \left\{e a\otimes _A f = e \otimes _A a f: \;\;\; a \in A,\ e \in E,\ f \in F \right\}
+%\end{align*}
\begin{definition}
Let $A$, $B$ be \textit{matrix algebras}. The \textit{Hilbert bimodule} for $(A, B)$ is given by
@@ -358,5 +358,4 @@ We conclude that.
a richer structure of morphism between matrix algebras.
\end{itemize}
-
\end{document}
diff --git a/week3.pdf b/week3.pdf
Binary files differ.
diff --git a/week3.tex b/week3.tex
@@ -0,0 +1,196 @@
+\documentclass[a4paper]{article}
+
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+
+\usepackage{mathptmx}
+
+\usepackage{subcaption}
+\usepackage[shortlabels]{enumitem}
+\usepackage{amsmath,amssymb}
+\usepackage{amsthm}
+\usepackage{bbm}
+\usepackage{graphicx}
+\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
+\usepackage[parfill]{parskip}
+
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}
+
+\theoremstyle{definition}
+\newtheorem{question}{Question}
+
+\theoremstyle{definition}
+\newtheorem{example}{Example}
+
+\theoremstyle{theorem}
+\newtheorem{theorem}{Theorem}
+
+\theoremstyle{theorem}
+\newtheorem{exercise}{Exercise}
+
+\theoremstyle{theorem}
+\newtheorem{lemma}{Lemma}
+
+\theoremstyle{definition}
+\newtheorem{solution}{Solution}
+
+\newtheorem*{idea}{Proof Idea}
+
+
+\title{Notes on \\ Noncommutative Geometry and Particle Physics}
+\author{Popovic Milutin}
+\date{Week 2: 26.02 - 4.03}
+
+\begin{document}
+
+\maketitle
+\tableofcontents
+\section{Excurse to Group Theory and Lie Groups}
+\subsection{Groups and Representations}
+ \begin{definition}
+ A Group $G$ is a set with a binary operation on $G$ satisfying.
+ \begin{enumerate}
+ \item $f, g \in G$ we have $fg = h \in G$.
+ \item $f(gh) = (fg) h$
+ \item $\exists\ e \in G\ \forall f\in G$ with $ef=fe=f$
+ \item $\forall f \in G\ \exists\ f^{-1}\in G$ with $ff^{-1}=f^{-1}f=e$
+ \end{enumerate}
+ \end{definition}
+
+ \begin{definition}
+ A Representation of a Group $G$ is a mapping, $D$ of elements of $G$ onto a set of \textit{linear
+ operators} such that:
+ \begin{enumerate}
+ \item $D(e) = 1$, $1$ is the identity operator in the space on which linear operators act
+ \item $D(g_1)D(g_2) = D(g_1g_2)$, the mapping is linear in group the group operation
+ \end{enumerate}
+ \end{definition}
+
+ Just by looking at symmetries of a Group we can find a nice representation, and if the group is finite we
+ can even find a matrix representation (Cheyley's Theorem). We all ready know a lot about linear algebra
+ which will then allow us to study these Groups very thoroughly and derive physical properties with
+ minimal information.
+
+
+\subsection{Lie Groups}
+ Group elements now depend \textit{smoothly} on a set \textit{continuous parameters} $g(\alpha) \in G$.
+ We are looking at continuous symmetries, e.g. a Sphere in $\mathbb{R}^3$ can be rotated in any direction
+ without changing. The collection of rotations forms a Lie group because the group elements are smoothly
+ differentiable.
+
+\subsubsection{Generators}
+ We parameterize $g(\alpha)|_{\alpha=0} = e$ and we assume that near the identity element, the group
+ elements can be described by a finite set of elements $\alpha_a$ for $a = 1,..,N$. For a representation
+ $D$ of this group, linear operators need to be parametrized the same way:
+ \begin{align}
+ D(\alpha)|_{\alpha=0} = 1
+ \end{align}
+
+ Because of the smoothness and continuity we can Taylor expand a representation near the identity:
+ \begin{align}
+ D(\alpha) &= 1 + id\alpha_a X_a + \cdots && \\
+ \text{with}&\;\; X_a = -i \frac{\partial D(\alpha)}{\partial \alpha_a}\bigg\arrowvert _{\alpha=0}
+ && \text{\footnote{Einstein Summation Convention, summation over repeated indices}}
+ \end{align}
+
+ We call $X_a$ the \textit{generators of the group}.
+ \begin{itemize}
+ \item If the parametrization is \textit{parsimonious}\footnote{parsimonious -
+ All parameters are needed to distinguish between group elements} then all
+ of $X_a$ will be independent.
+
+ \item If the representation is unitary then $X_a$ will be \textit{hermitian}, because of the
+ $i$ in the definition.
+
+ \item Sophus Lie showed how to derive generators without representations.
+ \end{itemize}
+
+ Now let us go in some fixed infinitesimal direction from the identity.
+ \begin{align}
+ D(d\alpha) = 1+ id\alpha _a X_a
+ \end{align}
+ Because of the group property of closure with respect to the group operation we can raise $D(d\alpha)$
+ to a large power and still get a group element.
+ \begin{align}
+ D(\alpha) = \lim_{k\rightarrow \infty}(1+i\frac{\alpha_a X_a}{k})^k = e^{i\alpha_a X_a}
+ \end{align}
+ This is called the \textit{exponential parameterization}. Looking at the expression we see that
+ group elements can be expressed in terms of generators, and generators form a vector space.
+ They are often referred to any element in the real linear space spanned by $X_a's$.
+
+\subsubsection{Lie Algebras}
+ Let us consider a parameter family of group elements created by one generator $X_a$:
+ \begin{align}
+ U(\lambda) = e^{i\lambda \alpha _a X_a}
+ \end{align}
+ We know for that for the same generator the group multiplication is linear meaning:
+ \begin{align}
+ U(\lambda _1)U(\lambda _2) = U(\lambda_1 + \lambda_2)
+ \end{align}
+ But if we multiply elements generated by two different generators the general case is
+ \begin{align}
+ e^{i\alpha_a X_a} e^{i\beta_b X_b} \neq e^{i (\alpha _a + \beta_b) X_a}
+ \end{align}
+ Yet because the exponentials are a representation of a group, and a group has closure under
+ group operation we know the above needs to be true for some $\delta _a$
+ \begin{align}
+ e^{i\alpha_a X_a} e^{i\beta_b X_b} = e^{i \delta _a X_a}
+ \end{align}
+ To further examine the exponent we rewrite the expression and Taylor expand $ln(1+K)$
+ to the second of $K = e^{i\alpha_a X_a} e^{i\beta_b X_b} -1$
+ \begin{align*}
+ i\delta _a X_a =& ln(1 + K) = K - \frac{K^2}{2} + \cdots \\
+ \text{and}\;\;\; K =&\ e^{i\alpha_a X_a} e^{i\beta_b X_b} -1 \\
+ =&\ (1 + i\alpha _a X_a - \frac{1}{2}(\alpha _a X_a)^2 + \cdots) \\
+ \cdot&\ (1 + i\beta _b X_b - \frac{1}{2}(\beta _b X_b)^2 + \cdots) -1 \\
+ =&\ i\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b \\
+ -&\ \frac{1}{2}(\alpha _a X_a)^2 - \frac{1}{2}(\beta _b X_b)^2 + \cdots
+ \end{align*}
+ So:
+ \begin{align*}
+ i\delta _a X_a =&\ i\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b \\
+ -&\ \frac{1}{2}(\alpha _a X_a)^2 - \frac{1}{2}(\beta _b X_b)^2 \\
+ +&\ \frac{1}{2}(\ai\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b)^2 \\
+ =&\ i\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b \\
+ -&\ \frac{1}{2}(\alpha _a X_a)^2 - \frac{1}{2}(\beta _b X_b)^2 \\
+ +&\ \frac{1}{2}(\alpha _a X_a)^2 + \frac{1}{2}(\beta _b X_b)^2 \\
+ +& \frac{1}{2}\alpha _a X_a \beta _b X_b + \frac{1}{2}\beta _b X_b \alpha _a X_a
+ \end{align*}
+ Because $X$'s are linear operators $\alpha _a X_a \beta _b X_b \neq \beta _b X_b \alpha _a X_a$.
+ These generators form an \textit{algebra under commutation} and we get
+ \begin{align*}
+ i\delta _a X_a =&\ i\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b \\
+ -&\ \frac{1}{2}[\alpha _a X_a, \beta _b X_b] + \cdots
+ \end{align*}
+ Thus rewriting the equation gives us
+ \begin{align*}
+ [\alpha _a X_a, \beta _b X_b] = -2i(\delta _c -\alpha _c -\beta _c) X_c \cdots \equiv i\gamma _c X_c
+ \end{align*}
+ Because this is true for all $\alpha$ and $\beta$, and considering the group closure, there exists some
+ \textit{real} $f_{abc}$ called the \textit{structure constant} satisfying.
+ \begin{equation}
+ \gamma _c = \alpha _a \beta _b f_{abc}
+ \end{equation}
+ Which is the same as.
+ \begin{equation}
+ [X_a, X_b] = i f_{abc} X_c
+ \end{equation}
+ This is called the \textit{Lie algebra of a group}
+ \newline
+ \newline
+ So $f$ is antisymmetric because $[A, B] = -[B, A]$, which means $f_{abc} = -f_{bac}$.
+ \newline
+ And $\delta$ can now be written as
+ \begin{equation}
+ \delta _a = \alpha _a + \beta _a - \frac{1}{2} \gamma _a \cdots
+ \end{equation}
+ Just by following the properties of Lie Groups (dependence on parameters and smoothness) in a fixed
+ direction near die identity to find physical statements. E.g.
+ $[\hat{r}_i, \hat{p}_j] = i \hslash \delta _{ij}$ tells us that we can't know the position
+ and the momentum of a particle exactly at a given time.
+
+
+
+
+\end{document}
