| Version 2 |
Version 1 |
| Let $G$ be a closed subgroup of $\operatorname{GL}(n,D)$, the general linear group over a division ring $D$. A |
Let $G$ be a closed subgroup of $\operatorname{GL}(n,D)$, the general linear group over a division ring $D$. A |
| \emph{one-parameter subgroup} of $G$ is a group homomorphism $$\phi\colon\mathbb{R}\to G$$ that is also a differentiable |
\emph{one-parameter subgroup} of $G$ is a group homomorphism $$\phi\colon\mathbb{R}\to G$$ that is also a differentiable |
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map at the same time. We view $\mathbb{R}$ additively and $G$ multiplicatively, so that $\phi(r+s)=\phi(r)\phi(s)$. This implies that $\operatorname{Im}(\phi)$ is a subgroup of $\operatorname{GL}(n,D)$ that is contained in the center of $\operatorname{GL}(n,D)$.
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map at the same time. We view $\mathbb{R}$ additively and $G$ multiplicatively, so that $\phi(r+s)=\phi(r)\phi(s)$.)$.
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| \textbf{Examples}. |
\textbf{Examples}. |
| \begin{enumerate} |
\begin{enumerate} |
| \item If $G=\operatorname{GL}(n,k)$, where $k=\mathbb{R}$ or $\mathbb{C}$, then any one-parameter subgroup has the form |
\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 |
$$\phi(t)=e^{tA},$$ 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 |
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 $\operatorname{GL}(n,k)$. |
one-to-one correspondence between one-parameter subgroups and tangent vectors of $\operatorname{GL}(n,k)$. |
| \item If $G=\operatorname{O}(n,\mathbb{R})\subseteq\operatorname{GL}(n,\mathbb{R})$, the orthogonal group over $R$, then |
\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: |
any one-parameter subgroup has the same form as in the example above, except that $A$ is skew-symmetric: |
| $A^{\operatorname{T}}=-A$. |
$A^{\operatorname{T}}=-A$. |
| \item If $G=\operatorname{SL}(n,\mathbb{R})\subseteq\operatorname{GL}(n,\mathbb{R})$, the special linear group over $R$, |
\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 |
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. |
$\operatorname{tr}$ is the trace operator. |
| \item If $G=\operatorname{U}(n)=\operatorname{O}(n,\mathbb{C})\subseteq\operatorname{GL}(n,\mathbb{C})$, the unitary |
\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 |
group over $C$, then any one-parameter subgroup has the same form as in the example above, except that $A$ is |
| skew-Hermitian: $A=-A^{*}=-\overline{A}^{\operatorname{T}}$ and $\operatorname{tr}(A)=0$. |
skew-Hermitian: $A=-A^{*}=-\overline{A}^{\operatorname{T}}$ and $\operatorname{tr}(A)=0$. |
| \end{enumerate} |
\end{enumerate} |
