From: H8401689@SUBMATHS.hku.hk
Date:          Tue, 3 Feb 1998 21:47:01 GMT+8
Subject:       degree of a mapping S^3->GL(n,H)

Dear Professor Davis,
I would like to express my sincere gratitude to helps and advice
from some topology experts in that I can settle
my problem: to calculate the degree of a mapping $S^3\to GL(n,H)$.
The following is the whole proof which is written in Amslatex 1.2.
I would be grateful if you post it for further comments and
modification.
Thank you.
Best regards, Lok-Shun.


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\begin{document}
\title{
Degree of a Mapping in $\pi_3(\gl_n(\mathbb{H}))$
}
\author{Siu Lok--Shun}
\maketitle
\begin{abstract}
In this note, first we show that
the third homotopy group of the general linear group over quaternions
is cyclic infinity and then show that the degree of the mapping
$f:S^3\to\gl_n(\mathbb{H})$, defined by $f(q)=q^m I_n$,
equals $mn$.
\end{abstract}
\begin{prop}[{\cite[\S 12.6]{fibrebundlebk}}]
The natural inclusion induces an isomorphism between
$\pi_3(Sp_1)$ and $\pi_3(Sp_k)$ for any $k\geq1$.
\end{prop}
\begin{prop}[{\cite[Theorem 2.6]{intro}}]
For any $k\geq1$,
$\pi_k(S^k)\approx\mathbb{Z}$ with the identity as a generator.
\end{prop}
Since $Sp_1$ is homeomorphic to $S^3$, in view of
the two propositions above, the following result follows readily.
\begin{cor}
\label{cor:spnz}
For any $k\geq1$,
$\pi_3(Sp_n)\approx\mathbb{Z}$ with
$q\mapsto\diag(q,1,\ldots)$ as a generator.
\end{cor}
\begin{prop}
Let $Y$, $Y'$ be two arcwisely connected Hausdorff spaces
with a single common point. Then
$$
\pi_k(Y\times Y')\approx\pi_k(Y)+\pi_k(Y').
$$
\end{prop}
\begin{proof}
The group isomorphism is induced from
the projections $\mu,\mu'$ of spaces $Y\times Y'$ onto $Y$ and $Y'$,
that is, $\eta(\alpha)=\mu(\alpha)+\mu'(\alpha)$
for any $\alpha:S^k\to Y\times Y'$.
For the detail, see \cite[Theorem 6.1 p. 42]{intro}.
\end{proof}
\begin{prop}[Iwasawa 1945 \cite{iwasawa},
M. Mimura 1995 \cite{mimurapaper}]
Any connected Lie Group $G$ is homeomorphic to the Cartesian product
of
a maximal compact subgroup $K$ and a subset which is homeomorphic with
a Euclidean space $\mathbb{R}^m$:
$$
G\approx K\times\mathbb{R}^m.
$$
Moreover all maximal compact subgroups are conjugate.
\end{prop}
Notice that the symplectic group $Sp_n$ is
a maximal compact subgroup of $\gl_n(\mathbb{H})$
and a Euclidean space is contractible, that is, homotopic
to a constant mapping,
hence $\pi_k(\mathbb R^m)=\mathbf{0}$ \cite[\S 1.14]{top&geo}.
Accordingly, follows from the two propositions above
and Corollary \ref{cor:spnz}, we have:
\begin{cor}
\label{cor:glnqz}
$\pi_k(\gl_n(\mathbb{H}))\approx\mathbb{Z}$ with
$q\mapsto\diag(q,1,\ldots)$ as a generator.
\end{cor}
Let $\alpha_i$ be a mapping sending $q\in S^3$ to
the diagonal matrix $\diag(d_i)$, where $d_j=q$ if $j=i$; 1,
otherwise.
\begin{lem}
$\alpha_1$ are homotopic to $\alpha_2$ in $\gl_2(\mathbb{H})$.
\end{lem}
\begin{proof}
The homotopy $[0,1]\times S^3\to\gl_n(\mathbb{H})$ is defined by
$$
\begin{bmatrix}
\cos t\pi/2&-\sin t\pi/2\\
\sin t\pi/2&\cos t\pi/2
\end{bmatrix}
\begin{bmatrix}
q&0\\0&1
\end{bmatrix}
\begin{bmatrix}
\cos t\pi/2&\sin t\pi/2\\
-\sin t\pi/2&\cos t\pi/2
\end{bmatrix}.
$$
\end{proof}
This result can be easily generalized.
\begin{lem}
\label{lem:1isoi}
For any $i=2,\ldots,n$, $\alpha_i$ are homotopic to $\alpha_1$
in $\gl_n(\mathbb{H})$.
\end{lem}
\begin{lem}[{\cite[Chapter V Lemma 3.17]{intro}}]
\label{lem:hspace}
Let $\circ$ be the multiplication of the topological group $G$.
Given two continuous mappings $f_1$, $f_2:$ $S^k\to G$
with the homotopy classes denoted by $[f_1]$, $[f_2]$.
If $f_1\circ f_2:$ $S^k\to G$ is defined by
$(f_1\circ f_2)(t)=f_1(t)\circ f_2(t)$, then
$[f_1\circ f_2]=[f_1]+[f_2]$.
\end{lem}
\begin{proof}
Let $f_0:S^k\to G$  defined by $f_0(x)=1$ for any $x$.
Then
$$
f_1+f_2\sim(f_1\circ f_0)+(f_0\circ f_2)
=(f_1+f_0)\circ(f_0+f_2)\sim f_1\circ f_2.
$$
\end{proof}
\begin{cor}
\label{cor:qmInmn}
The mapping $f:S^3\to\gl_n(\mathbb H)$ defined by $f(q)=q^m I_n$
corresponds the integer $mn$ in the homotopy group.
\end{cor}
\begin{proof}
Since $f(q)=q^m I_n=(\alpha_1(q))^m\cdots(\alpha_n(q))^m$,
the result follows
from Corollary \ref{cor:glnqz}, Lemma \ref{lem:1isoi}
and Proposition \ref{lem:hspace}.
\end{proof}
\begin{thebibliography}{99}
\bibitem{top&geo}
G. E. Bredon, Topology and Geometry, 1993, Springer.
\bibitem{intro}
W. V. D. Hodge, An Introduction to Homotopy Theory, 1953,
Cambridge at the Univ. Press.
\bibitem{fibrebundlebk}
D. Husemoller, Fibre Bundles, 1993, Springer.
\bibitem{iwasawa}
K. Iwasawa, {\it On some types of Topological Groups},
Annals of Math. \textbf{50} (1949) 507--558.
\bibitem{mimurapaper}
M. Mimura, {\it Homotopy Theory of Lie Groups},
in handbook of Algebraic Topology, 1995, 951--991, Elsevier.
\end{thebibliography}
\end{document}