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I am studying exceptional isomorphisms recently, which arise due to the coincidence in Dynkin diagram.

I saw two forms of expressing the exceptional isomorphisms, one is isomorphisms between the spin group and the corresponding group, established by spin or half-spin representations; the other is isomorphism from symmetric squares, exterior squares or tensor squares of the relevant group to the special orthogonal group. But in this case, I don't know how to construct the induced symmetric bilinear form on the vector space. And in the case ${\bigwedge}^2\operatorname{Sp}(4,\mathbb{C})\cong \operatorname{SO}(5,\mathbb{C})$, I think the dimension of exterior square of a 4-dimensional vector space is 6, and I am not clear how it gets mapped into $\operatorname{SO}(5)$.

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    $\begingroup$ As written, your equation doesn't make sense; one cannot take the exterior square of a group, and I think you don't mean to take the exterior square of the Lie algebra. I suspect what you mean is that $G = \operatorname{Sp}(4, \mathbb C)$ is isomorphic to $\operatorname{Spin}(5, \mathbb C)$—or you could say that it has a mapping onto $H = \operatorname{SO}(5, \mathbb C)$ that annihilates the centre. ($G$ and $H$ are not isomorphic.) … $\endgroup$
    – LSpice
    Commented Aug 22, 2022 at 16:28
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    $\begingroup$ … Then you try to show this using an isomorphism of ${\bigwedge}^2\mathbb C^4$ with $\mathbb C^5$ that intertwines the actions, but of course there isn't one, because ${\bigwedge}^2\mathbb C^4$ is 6D and $\mathbb C^5$ is 5D. Instead one way to proceed is to consider the 5D representation ${\bigwedge}^2(\mathbb C^4)^*/\mathbb C\omega$, where $\omega$ is the symplectic form—which is preserved by $G$ by definition! $\endgroup$
    – LSpice
    Commented Aug 22, 2022 at 16:29

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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\Ker{Ker}$Let's start with the double cover $\SL(4) \to \SO(6)$. Let $V$ be the standard $4$-dimensional representation of $\text{SL}(4)$, and let $\bigwedge^2 V$ be its tensor square. We have a non-degenerate symmetric bilinear form $\bigwedge^2 V \times \bigwedge^2 V \to \bigwedge^4 V \cong \mathbb{C}$ by $\langle \alpha, \beta \rangle = \alpha \wedge \beta$. (This is symmetric because even degree wedges commute with each other.) So the action of $\SL(4)$ preserves this symmetric bilinear form, and we get a map $\SL(4) \to \SO(6)$. This can be checked to be a double cover (the kernel is $\pm \mathrm{Id}$ and a computation with Lie algebras checks that the map is surjective).

Now, let $\omega : V \times V \to \mathbb{C}$ be a symplectic form and let $\Sp(4)$ be the subgroup of $\SL(4)$ preserving this form. Then $\omega$ induces a linear map $L_{\omega} : \bigwedge^2 V \to \mathbb{C}$, and $\Sp(4)$ preserves the $5$-dimensional space $\Ker(L_{\omega})$ inside $\bigwedge^2 V$. So $\Sp(4)$ maps to $\SO(5)$, sitting inside $\SO(6)$, and this is again a double cover.

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    $\begingroup$ Most of this was in LSpice's comments, but I wanted to spell out where the symmetric bilinear form comes from, and at that point it seemed like it deserved to be an answer. $\endgroup$
    – David E Speyer
    Commented Aug 22, 2022 at 18:21
  • $\begingroup$ We also chose dual realisations: I chose a quotient of ${\bigwedge}^2(\mathbb C^4)^*$, and you chose a subobject of ${\bigwedge}^2\mathbb C^4$. Anyway, as you say, it's certainly worth writing all these things down explicitly—it can be hard to find such "everyone knows" folklore expounded anywhere accessible. My comments: 1 2. $\endgroup$
    – LSpice
    Commented Aug 22, 2022 at 20:04

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