Sp(2n, R) = SL(2n, R)

Question

I have proved that Sp(2n,R) is a subgroup of SL(2n,R). But is there an equality? If no, what counter example can do we have? Thanks

Answer 1

Even the dimensions are different in general. We have $$ \dim Sp(2n)=n(2n+1),\; \dim SL(2n)=(2n)^2-1. $$ For $n=3$ we have $\dim Sp(6)=21$, but $\dim SL(4)=35$. The dimensions coincide with the vector space dimensions of the associated Lie algebras.

References: Finding the dimension of the symplectic group

Dimension of $SL(n,\mathbb{R})$ and some other Lie groups

Answer 2

For the case of $n=1$, we have Sp$(2,\mathbb{R}) = $SL$(2,\mathbb{R})$.