every invariant subspace has an invariant complement, so one or both of the 6-dimensional subspaces $\Lambda_+^2 V$ and $\Lambda_-^2 V$ must have a 5-dimensional subspace invariant under the action of $\mathrm{Sp}(V)$

I get that the invariant subspace at the start of this quote is generated by the element $J$ preserved by $SU(4)$, but itâ€™s not clear what the subspace is. Is it the real span? Why the claim that **one or both** of $\Lambda_+^2 V$ and $\Lambda_-^2 V$ have an invariant subspace?

Added a proof that Spin(5) is isomorphic to Sp(2).

]]>added the following statement, but for the moment without good referencing:

The integral cohomology ring of the classifying space $B Spin(5)$ is spanned by two generators

the first fractional Pontryagin class $\tfrac{1}{2}p_1$

the linear combination $\tfrac{1}{2}p_2 - \tfrac{1}{2}(p_1)^2$ of the half the second Pontryagin class with half the cup product-square of the first Pontryagin class:

added the following statement; which appears as Lemma 2.1 in

- Raoul Bott, Alberto Cattaneo,
*Integral Invariants of 3-Manifolds*, J. Diff. Geom., 48 (1998) 91-133 (arXiv:dg-ga/9710001)

Let

$\array{ S^4 &\longrightarrow& B Spin(4) \\ && \big\downarrow^{\mathrlap{\pi}} \\ && B Spin(5) }$be the spherical fibration of classifying spaces induced from the canonical inclusion of Spin(4) into Spin(5) and using that the 4-sphere is equivalently the coset space $S^4 \simeq Spin(5)/Spin(4)$ (this Prop.).

Then the fiber integration of the triple cup power of the Euler class $\chi \in H^4\big( B Spin(4), \mathbb{Z}\big)$ (see this Prop) is twice the second Pontryagin class:

$\pi_\ast \left( \chi^3 \right) \;=\; 2 p_2 \;\;\in\;\; H^4\big( B Spin(5), \mathbb{Z} \big) \,.$ ]]>for ease of reference, and to go along with *SO(2), Spin(2), Pin(2), Spin(3), Spin(4)*