OneParameterSubgroup 2004-12-15 01:33:07 2005-06-13 00:04:53 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $G$ be a Lie Group. A \emph{one-parameter subgroup} of $G$ is a group homomorphism $$\phi\colon\mathbb{R}\to G$$ that is also a differentiable map at the same time. We view $\mathbb{R}$ additively and $G$ multiplicatively, so that $\phi(r+s)=\phi(r)\phi(s)$. \textbf{Examples}. \begin{enumerate} \item If $G=\operatorname{GL}(n,k)$, where $k=\mathbb{R}$ or $\mathbb{C}$, then any one-parameter subgroup has the form $$\phi(t)=e^{tA},$$ where $A=\frac{d\phi}{dt}(0)$ is an $n\times n$ matrix over $k$. The matrix $A$ is just a tangent vector to the Lie group $\operatorname{GL}(n,k)$. This property establishes the fact that there is a one-to-one correspondence between one-parameter subgroups and tangent vectors of $\operatorname{GL}(n,k)$. The same relationship holds for a general Lie group. The one-to-one correspondence between tangent vectors at the identity (the Lie algebra) and one-parameter subgroups is established via the exponential map instead of the matrix exponential. \item If $G=\operatorname{O}(n,\mathbb{R})\subseteq\operatorname{GL}(n,\mathbb{R})$, the orthogonal group over $R$, then any one-parameter subgroup has the same form as in the example above, except that $A$ is skew-symmetric: $A^{\operatorname{T}}=-A$. \item If $G=\operatorname{SL}(n,\mathbb{R})\subseteq\operatorname{GL}(n,\mathbb{R})$, the special linear group over $R$, then any one-parameter subgroup has the same form as in the example above, except that $\operatorname{tr}(A)=0$, where $\operatorname{tr}$ is the trace operator. \item If $G=\operatorname{U}(n)=\operatorname{O}(n,\mathbb{C})\subseteq\operatorname{GL}(n,\mathbb{C})$, the unitary group over $C$, then any one-parameter subgroup has the same form as in the example above, except that $A$ is \PMlinkname{skew-Hermitian}{SkewHermitianMatrix}: $A=-A^{*}=-\overline{A}^{\operatorname{T}}$ and $\operatorname{tr}(A)=0$. \end{enumerate}