Over the reals the intersection of the orthogonal and symplectic groups in even dimension is isomorphic to the unitary group in the half dimension. See the answer here.

Over the reals the intersection of the orthogonal and symplectic groups in even dimension is isomorphic
  to the unitary group in the half dimension:$$ U(n) = O(2n, \mathbf{R}) \cap Sp(2n, \mathbf{R}).$$

To me, this is not so obvious. How do we see this without just saying "oh, that's trivial, cite the "2-out-of-3 property"?
 A: The result follows from carefully unwinding the identifications involved and the definitions of the various groups. Let $V = \mathbb{R}^{2n}$ and let us denote elements in $V$ by $(x_1, \ldots, x_n, y_1, \ldots, y_n)$. Define a map $\Phi \colon V \rightarrow \mathbb{C}^n$ by
$$ \Phi(x_1, \ldots, x_n, y_1, \ldots, y_n) = (x_1 + \sqrt{-1}y_1, \ldots, x_n + \sqrt{-1}y_n). $$
The map $\Phi$ is an $\mathbb{R}$-linear isomorphism that allows us to identify $V$ with $\mathbb{C}^n$, pullback all the structure of $\mathbb{C}^n$ to $V$ and think about $M_n(\mathbb{C})$ as sitting inside $M_{2n}(\mathbb{R})$. Let us make this explicit. Denote by $J$ the matrix
$$ J = \left( \begin{matrix} 0_{n \times n} & -I_{n \times n} \\ I_{n \times n} & 0_{n \times n} \end{matrix} \right) \in M_{2n}(\mathbb{R}). $$
Using the identification of $\Phi$, the linear map $T_J$ (the map represented in the standard basis by $J$) is the pullback of the natural complex structure on $\mathbb{C}^n$. A real $(2n)\times(2n)$ matrix $E \in M_{2n}(\mathbb{R})$ (whose elements correspond to a $\mathbb{R}$-linear map of $V$) belongs to $M_{n}(\mathbb{C})$ (whose elements correspond to a $\mathbb{C}$-linear map of $\mathbb{C}^n$) if and only if $EJ = JE$ (that is, $T_E$ commutes with $T_J$ and so the map $\Phi \circ T_E \circ \Phi^{-1}$ is complex linear).
If we write $E$ as a $2 \times 2$ block matrix with blocks of size $n$, we have that $E \in M_n(\mathbb{C})$ if and only if $E$ has the form
$$ E = \left( \begin{matrix} A & -B \\ B & A \end{matrix} \right) $$
with $A, B \in M_n(\mathbb{R})$ and the matrix $E$ is then identified with the matrix $A + \sqrt{-1}B \in M_n(\mathbb{C})$. The matrix $A + \sqrt{-1}B$ is unitary if and only if 
$$(A + \sqrt{-1}B)^{*} (A + \sqrt{-1}B) = (A^t - \sqrt{-1}B^t)(A + \sqrt{-1}B) = (A^tA + B^t B) + \sqrt{-1}(A^tB - B^tA) = I + \sqrt{-1} \cdot 0. $$
That is, if and only if $A^tA + B^t B = I$ and $A^t B - B^t A = 0$.

Now, let $E \in O(2n,\mathbb{R}) \cap \mathrm{Sp}(2n, \mathbb{R})$ and write
$$ E = \left( \begin{matrix} A & C \\ B & D \end{matrix} \right) $$
with $A,B,C,D \in M_n(\mathbb{R})$. Then $E^tE = I$ (because $E \in O(2n, \mathbb{R})$) and $E^tJE = J$ (because $E \in \mathrm{Sp}(2n, \mathbb{R})$). Multiplying the last equation by $E$ from the left, we get $EE^tJE = JE = EJ$ and so $E \in M_{n}(\mathbb{C})$. Thus, $C = -B$ and $D = A$. But then, writing the orthogonality condition we see that
$$ E^t E = \left( \begin{matrix} A^t & B^t \\ -B^t & A^t \end{matrix} \right) \left( \begin{matrix} A & -B \\ B & A \end{matrix} \right) = \left( \begin{matrix} A^tA + B^tB & -A^tB + B^tA \\ -B^tA + A^tB & B^tB + A^tA \end{matrix} \right) = \left( \begin{matrix} I & 0 \\ 0 & I \end{matrix} \right) $$
which shows that $A + iB$ is unitary and thus $E \in U(n)$. I'll leave you the details for the other direction.
