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$SU(2)$ can be seen as a subgroup of $SO(5)$ through the following chain of subgroups

$$ SU(2) \subset SO(4) \subset SO(5). $$

If we identify $SU(2)\cong Sp(1)$, does the inclusion $Sp(1) \to SO(5)$ factor through $Spin(5) \cong Sp(2) \to SO(5)$ as the standard embedding of $Sp(1)$ in $Sp(2)$. I understand, that there is a lift of $Sp(1) \to SO(5)$ to $Sp(1) \to Sp(2)$, but I can not see that this is the standard embedding.

I already asked this also at mathstackex.

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  • $\begingroup$ Over the reals, the irreducibles of $SU(2)$ are $V_n$ of dimension $n$ for $n\ge 1$ odd (which factor through $SO(3)$) and $W_n$ for $n\ge 4$ mutliple of $4$, of dimension $4n$ and faithful. Therefore in dimension 5 there's a single faithful representation namely $W_4\oplus V_1$. I guess it's also unique (at least its image) up to conjugation in $SO(5)$. So probably the answer is yes iff the composite map $Sp(1)\to Sp(2)\to SO(5)$ (where the left-hand is the standard embedding) is injective. $\endgroup$
    – YCor
    Commented Apr 9, 2018 at 19:15

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Yes. The rep of $Sp(2)$ on $\mathbb{C}^5$ that induces the covering map is the complement to the line spanned by the symplectic form in $\bigwedge{}^{2}\mathbb{C}^4$. What you are asking is whether there is another line in this space which is invariant under $Sp(1)$, that is, if the action of $Sp(1)$ on the wedge square $\bigwedge{}^{2}\mathbb{C}^4$ fixes a 2-d subspace. Since the representation on $ \mathbb{C}^4$ of $Sp(1)\cong SU(2)$ is the defining rep on $\mathbb{C}^2$ and two trivials, you get two invariant 2-forms from the symplectic form on $\mathbb{C}^2$ and wedging the two trivials.

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