The symplectic group ${\mathrm{Sp}}_{2n}\left(𝔽\right)$

Last updates: 5 November 2011

The symplectic group ${\mathrm{Sp}}_{2n}\left(𝔽\right)$

Let $V$ be a vector space oover $𝔽$ and let $⟨,⟩:V\otimes V\to 𝔽$ be a skew symmetric bilinear form. The symplectic group is

 $\mathrm{Sp}\left(⟨,⟩\right)=\left\{g\in \mathrm{GL}\left(V\right)\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}⟨g{v}_{1},g{v}_{2}⟩=⟨{v}_{1},{v}_{2}⟩\right\}.$
The Lie algebra of $\mathrm{Sp}\left(⟨,⟩\right)$ is
 $\mathrm{𝔰𝔭}\left(⟨,⟩\right)=\left\{g\in \mathrm{𝔤𝔩}\left(V\right)\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}⟨g{v}_{1},{v}_{2}⟩+⟨{v}_{1},g{v}_{2}⟩=0\right\}.$

By Gram-Schmidt there exists a basis $\left\{{e}_{1},\dots ,{e}_{n},{e}_{1}^{*},\dots ,{e}_{n}^{*}\right\}$ of $V$ such that

 $⟨{e}_{i},{e}_{j}^{*}⟩={\delta }_{ij}\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}⟨{e}_{i},{e}_{j}⟩=0,\phantom{\rule{2em}{0ex}}\text{so that}\phantom{\rule{2em}{0ex}}J=\left(\begin{array}{cc}0& \begin{array}{ccc}1& & 0\\ & \ddots & \\ 0& & 1\end{array}\\ \begin{array}{ccc}-1& & 0\\ & \ddots & \\ 0& & -1\end{array}& 0\end{array}\right)$
is the matrix of $⟨,⟩$ with respect to the basis $\left\{{e}_{1},\dots ,{e}_{n},{e}_{1}^{*},\dots ,{e}_{n}^{*}\right\}$. Using the basis $\left\{{e}_{1},\dots ,{e}_{n},{e}_{1}^{*},\dots ,{e}_{n}^{*}\right\}$ to identify $\mathrm{GL}\left(V\right)$ with ${\mathrm{GL}}_{2n}\left(𝔽\right)$,
 $\text{identifies}\phantom{\rule{2em}{0ex}}\mathrm{Sp}\left(⟨,⟩\right)\phantom{\rule{2em}{0ex}}\text{with}\phantom{\rule{0.5em}{0ex}}{\mathrm{Sp}}_{2n}\left(𝔽\right)=\left\{g\in {\mathrm{GL}}_{2n}\left(𝔽\right)\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}gJ{g}^{t}=J\right\}$
and
 $\text{identifies}\phantom{\rule{2em}{0ex}}\mathrm{𝔰𝔭}\left(⟨,⟩\right)\phantom{\rule{2em}{0ex}}\text{with}\phantom{\rule{0.5em}{0ex}}{\mathrm{𝔰𝔭}}_{2n}\left(𝔽\right)=\left\{g\in {\mathrm{𝔤𝔩}}_{2n}\left(𝔽\right)\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}gJ+J{g}^{t}=0\right\}.$

A maximal compact subgroup of ${\mathrm{Sp}}_{2n}\left(ℂ\right)$ is

 $\mathrm{Sp}\left(n\right)={U}_{2n}\left(ℂ\right)\cap {\mathrm{Sp}}_{2n}\left(ℂ\right).$
The group $\mathrm{Sp}\left(n\right)is a compact, connected, and simply connected real Lie group.$

HW: Show that ${\mathrm{Sp}}_{2}\left(ℂ\right)={\mathrm{SL}}_{2}\left(ℂ\right)$.

Notes and References

These notes were influenced by the Wikipedia articles ????. They were prepared for lectures and working seminars in Representation Theory at University of Melbourne in 2008-2011.

References

[St] R. Steinberg, Lectures on Chevalley groups, Notes prepared by John Faulkner and Robert Wilson, Yale University, New Haven, Conn., 1968. iii+277 pp. MR0466335.