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

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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 $\det S =1$.

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