Show Sp($n$)$\subset$SU($2n$)

Question

On the Wikipedia page on symplectic groups, it is stated that Sp($n$)$\subset$SU($2n$). How can this be shown?

Answer 1

As $$\operatorname{Sp}(n) = \operatorname{Sp}(2n;\mathbb{C}) \cap \operatorname{U}(2n) $$you only have to show that $\det S =1$ for $S \in \operatorname{Sp}(2n;\mathbb{C})$.

This is surprisingly tricky. From the definition it follows that $$ \det(J)= \det (S J S^T) = \det(S)^2 \det(J)$$ so $$\det(S) = \pm 1.$$ So in the end, you either need to show that the symplectic group is path-connected (you have to find a path which is possible but cumbersome; I believe it is done in Stilwell's `Naive Lie Theory') or you introduce the notion of the Pfaffian, with which you can show that $$ \operatorname{pf}(J)= \operatorname{pf} (S J S^T) = \det(S) \operatorname{pf}(J)$$ so $\det S =1$.

Answer 2

There is another way to do it which is perhaps more concrete. Recall that you can write any quaternion $q$ in the form $q=a+bj$ where $a,b \in \mathbb{C}$. Similarly, one can write any matrix in $Sp(n)$ as $A+Bj$ for some matrices $n \times n$ comlex matrices $A$ and $B$. Then you can check that the embedding

Sp$(n) \rightarrow $ SU$(2n)$ which maps the element $A+Bj \in$ Sp$(n)$ to the element

\begin{pmatrix} A & B \\ -\overline{B} & \overline{A} \end{pmatrix}

is an injective Lie group homomorphism. Thus you can identify Sp$(n)$ as a subgroup of SU$(2n)$ via the image of this homomorphism.