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'one-parameter subgroup'
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| Title of object: |
one-parameter subgroup |
| Canonical Name: |
OneParameterSubgroup |
| Type: |
Definition |
| Created on: |
2004-12-15 01:33:07 |
| Modified on: |
2004-12-15 02:12:55 |
| Classification: |
msc:22E10, msc:22E15 |
| Synonyms: |
one-parameter subgroup=1-parameter subgroup |
Revision comment (for changes between this and next version):
Generalized from a GL(n) over a division ring to any Lie group.
Also, a one parameter subgroup need not lie in the center of the group. |
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Content:
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
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 $\operatorname{GL}(n,D)$.
\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 $\operatorname{GL}(n,k)$.
\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} |
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