\documentclass[a4paper,12pt]{article}
%%%%%%%%%%
% Deutsch
%%%%%%%%%%
\usepackage{ngerman}
\usepackage{ucs}
\usepackage[utf8x]{inputenc}
%\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}

%%%
\usepackage{latexsym,amssymb,amsmath}
%\usepackage{mltex}
\usepackage{graphicx}
\usepackage{hyperref}
%\usepackage{amstext}
%\usepackage[english]{babel}
\usepackage[matrix,arrow]{xy}
\usepackage{enumerate}

\usepackage{bbold}


\parindent0mm

\frenchspacing
%\vfuzz=2pt \hfuzz=2pt
% \def \be{\begin{eqnarray*}}
% \def \ee{\end{eqnarray*}}
\def \ben{\begin{enumerate}[{\rm (a)}]}
\def \een{\end{enumerate}}
\def \beit{\begin{itemize}}
\def \eeit{\end{itemize}}
\def \bui#1#2{\mathrel{\mathop{\kern 0pt#1}\limits^{#2}}}
%\[\bui{\rightarrow}{b}\]
\def \buil#1#2{\mathrel{\mathop{\kern 0pt#1}\limits_{#2}}}
\def \bfll{\begin{flushleft}}
\def \efll{\end{flushleft}}
\def \bflr{\begin{flushright}}
\def \eflr{\end{flushright}}
\def \ligne{\noindent \rule{\textwidth}{0.5mm} \\}
\def \findemo{\hfill$\square$\\}
\def \fintdemo{\hfill$\surd$\\}
\def \grad{\mathop{\mathrm{grad}}}
\def \diam{\mathop{\mathrm{diam}}}
\def \ker{\mathop{\mathrm{ker}}}
\def \im{\mathop{\mathrm{im}}}
\def \injrad{\mathop{\mathrm{injrad}}}
\def \lra{\longrightarrow}
\def \lmt{\longmapsto}
\def \ovl{\overline}
\def \udl{\underline}
\def \wih{\widehat}
\def \wit{\widetilde}
\def \wnabla{\wit{\nabla}}
\def \cdotM{\buil{\cdot}{M}}
\def \diag{\mathrm{diag}}
\def \Id{\mathrm{Id}}
% \def \Q{\mathrm{Q}}
\def \ric{\mathrm{ric}}
\def \SO{\mathrm{SO}}
\def \so{\mathfrak{so}}
\def \Spin{\mathrm{Spin}}
\def \SU{\mathrm{SU}}
\def \su{\mathfrak{su}}
\def \Sym{\mathrm{Sym}}
\def \tr{\mathrm{tr}}
\def \T{\mathrm{T}}
\def \lala{\langle\!\langle}
\def \rara{\rangle\!\rangle}

\def\End{\mathop\mathrm{End}}

\def\cC{\mathcal{C}}
\def\cF{\mathcal{F}}
\def\cK{\mathcal{K}}
\def\cS{\mathcal{S}}

\def\mR{\mathbb{R}}
\def\mC{\mathbb{C}}
\def\mZ{\mathbb{Z}}
\def\mN{\mathbb{N}}

\let\mR\mR
\let\C\mC

\newcommand{\lquot}[2]{\raisebox{-0.5ex}{$#2$}\backslash\!\raisebox{0.5ex}{$#1$}}
\newcommand{\rquot}[2]{\raisebox{0.5ex}{$#1$}\!/\!\raisebox{-0.5ex}{$#2$}}
\newcommand{\pa}[1]{\left(#1\right)}
%\pa{argument}
\newcommand{\page}[1]{\ind \begin{minipage}[t]{15cm} #1 \end{minipage}}
% \newcommand{\la}{\langle}
% \newcommand{\ra}{\rangle}
\newcommand{\mnote}[1]{\marginpar{\tiny\em #1}}

\newtheorem{ethm}{Theorem}[section]
\newtheorem{edefi}[ethm]{Def\mbox{}inition}
\newtheorem{elemme}[ethm]{Lemma}
\newtheorem{erem}[ethm]{Note}
\newtheorem{erems}[ethm]{Notes}
\newtheorem{ecor}[ethm]{Corollary}
\newtheorem{prop}[ethm]{Proposition}
\newtheorem{eexemple}[ethm]{Example}
\newtheorem{eexemples}[ethm]{Examples}
\setlength{\textheight}{247mm}
%\setlength{\oddsidemargin}{-20mm}
%\setlength{\evensidemargin}{-20mm} \setlength{\textwidth}{200mm}
\setlength{\topmargin}{-1.5cm}

\let\al\alpha
\let\de\delta
\let\ep\epsilon
\let\ga\gamma
\let\la\lambda
\let\si\sigma

\def \Id{\mathrm{Id}}
\def \Q{\mathrm{Q}}
\def \SO{\mathrm{SO}}
\def \so{\mathfrak{so}}
\def \Spin{\mathrm{Spin}}
\def \SU{\mathrm{SU}}
\def \GL{\mathrm{GL}}
\def \su{\mathfrak{su}}
\def \Sym{\mathrm{Sym}}
\def \tr{\mathrm{tr}}
\def \T{\mathrm{T}}

\def\mRP{\mR\mathrm P}
\def\mCP{\mC\mathrm P}

\let\<\langle
\let\>\rangle
\def\lala{\langle\!\langle}
\def\rara{\rangle\!\rangle}

\let\witi\widetilde
\let\ti\tilde
\let\na\nabla
\let\pa\partial
\let\setminus\smallsetminus

\def \bui#1#2{\mathrel{\mathop{\kern 0pt#1}\limits^{#2}}}
%\[\bui{\rightarrow}{b}\]
\def \buil#1#2{\mathrel{\mathop{\kern 0pt#1}\limits_{#2}}}

\begin{document}
\pagestyle{empty}
Universit\"at Regensburg, Mathematik\hfill SoSe 2013\\
Prof. Dr. Bernd Ammann\hfill 21.05.2013\\
Dr. Mihaela Pilca\\

\begin{center}
{\bf Differential Geometry II \\
  Exercise Sheet no. 6} $ $\\
\end{center}



{\bf Exercise 1}\\
The goal of this exercise is to show that there is no matrix $A\in\mathfrak{gl}(2,\mathbb{R})\cong \mathbb{R}^{2\times 2}$ such that
$\mathrm{exp}(A)=\begin{pmatrix} -2 & 0 \\ 0 & -1 \end{pmatrix}$. Deduce a contradiction by considering the following cases:
\begin{itemize}
 \item[i)] $A$ is diagonalizable.
 \item[ii)] $A$ is triagonalizable, but not diagonalizable.
 \item[iii)] $A$ has no real eigenvalues. Hint: Consider the eigenvalues of $A$ and of $\mathrm{exp}(A)$.
\end{itemize}
Bonus points question: If $G$ is a connected compact Lie group, is the Lie group exponential map surjective?\\







{\bf Exercise 2}\\
We define
\[\mathcal{K}:=\{J\in \mathrm{End}(\mathbb{R}^{2n})\, |\, J^2=-\mathrm{Id}\}.\]

The elements of $\mathcal{K}$ are called complex structures on $\mathbb{R}^{2n}$. The group $\GL(2n, \mathbb{R})$ acts by conjugation on $\mathrm{End}(\mathbb{R}^{2n})$.
Show that $\mathcal{K}$ is an orbit of this action. Compute the isotropy group and write $\mathcal{K}$ as a homogeneous space.\\





{\bf Exercise 3}\\
Let $(V,[\cdot,\cdot])$ be a Lie algebra over a field $K$. An ideal is a vector subspace $W$, such that $[x,y] \in W$, for all $x\in W$ and $y\in V$.
Show the following:
\begin{itemize}
 \item[i)] The quotient space $V/W$ carries a unique Lie bracket, such that the projection $V \to V/W$ is a Lie algebra homomorphism. 
 \item[ii)] The kernel of a Lie algebra homomorphism is an ideal and conversely, each ideal is the kernel of a Lie algebra homomorphism.
 \item[iii)](Bonus points) Let now $K=\mathbb{R}$. Let $G$ be a Lie group and $H$ a normal subgroup of $G$ that is also a submanifold. Then the Lie algebra of $H$ is an ideal
 of the Lie algebra of $G$.
\end{itemize}



\vskip1cm
{\sl Hand in the solutions on
\textbf{Monday, May 27, 2013} before the lecture}.\\
\end{document}



