Let us recall some background on the actions of Lie algebras of bilinear forms.
Let $$ be a $$-Lie algebra and let $V$ be a representation of $$, where $$ is some field with $\operatorname{char} ≠ 2$.
- An element $v$ of $V$ is called invariant if $xv = 0$ for all $x ∈ $.
Note that a non-zero invariant element always spans a one-dimensional trivial subrepresentation.
An irreducible representation must therefore not contain any non-zero invariant elements, unless it is the one-dimensional trivial representation.
- A bilinear form $β \colon V × V \to $ is called $$-invariant if
$$
β(xv, w) + β(v, xw) = 0
\qquad
\text{for all $x ∈ $ and all $v, w ∈ V$.}
$$
If $$ is the Lie algebra of a Lie group $G$, then this corresponds (roughly) to $β$ being $G$-invariant in the sense that
$$
β(gv, gw) = β(v, w)
\qquad
\text{for all $g ∈ G$, $v, w ∈ V$.}
$$
This notion of “$$-invariant” is a special case of the general notion of “invariant”.
- The vector space of bilinear maps $\newcommand{\Bil}{\operatorname{Bil}}\Bil^2(V, ℂ)$ becomes again a representation of $$ via
$$
(x β)(v, w) = β(xv, w) + β(v, xw)
\qquad
\text{for all $x ∈ $, $β ∈ \Bil^2(V, )$, $v, w ∈ V$.}
$$
A bilinear form $β$ is $$-invariant if and only if it invariant in the general sense.
We have some standard homomorphism/isomorphisms regarding $\Bil^2(V, )$.
- The usual isomorphism of vector spaces $\Bil^2(V, ) ≅ (V ⊗ V)^*$ is an isomorphism of representations.
The usual injective linear map $V^* ⊗ V^* \to (V ⊗ V)^*$ is a homomorphism of representations.
If $V$ is finite-dimensional, then it is thus an isomorphism of representations.
We can also restrict our attention to alternating bilinear forms, or symmetric bilinear forms.
Let $\newcommand{\Alt}{\operatorname{Alt}} \Alt^2(V, )$ be the linear subspace of $\Bil^2(V, )$ consisting of alternating bilinear forms.
This is $$-subrepresentation of $\Bil^2(V, )$.
The usual isomorphism of vector spaces $\Alt^2(V, ) ≅ (⋀^2 V)^*$ is an isomorphism of representations.
The usual injective linear map $⋀^2 (V^*) \to (⋀^2 V)^*$ is a homomorphism of representations.
If $V$ is finite-dimensional, then it is thus an isomorphism of representations.
Let $\newcommand{\Sym}{\operatorname{Sym}} \Sym^2(V, )$ be the linear subspace of $\Bil^2(V, )$ consisting of alternating bilinear forms.
This is $$-subrepresentation of $\Bil^2(V, )$.
The usual isomorphism of vector spaces $\newcommand{\Symp}{\operatorname{S}} \Sym^2(V, ) ≅ \Symp^2(V)^*$ is an isomorphism of representations.
The usual injective linear map $\Symp^2(V^*) \to \Symp^2(V)^*$ is a homomorphism of representations.
If $V$ is finite-dimensional, then it is thus an isomorphism of representations.
A bilinear form on a vector space $W$ is the same as a linear map $W \to W^*$.
If $W$ is finite-dimensional, then the linear map $W \to W^*$ is an isomorphism if and only if the corresponding bilinear form is non-degenerate.
We can do essentially the same for isomorphisms of representations $V ≅ V^*$.
- Every bilinear form $β$ on $V$ determines a linear map
$$
β' \colon V \longrightarrow V^* \,, \quad v \longmapsto β(v, -) \,.
$$
The bilinear form $β$ is $$-invariant if and only if this linear map $β'$ is a homomorphism of representations.
Consequently, if $V$ is finite-dimensional and $β$ is both $$-invariant and non-degenerate, then $β'$ is an isomorphism of representations $V ≅ V^*$.
As a consequence, we have the following:
Suppose that the representation $V$ is finite-dimensional, and that it admits a non-degenerate, $$-invariant bilinear form $β$.
There exist isomorphisms of representations
\begin{gather*}
\Bil^2(V, ) ≅ (V ⊗ V)^* ≅ V^* ⊗ V^* ≅ V ⊗ V \,, \\[0.5em]
\Alt^2(V) ≅ \Bigl( ⋀^2 V \Bigr)^* ≅ ⋀^2 V^* ≅ ⋀^2 V \,, \\[0.5em]
\Sym^2(V) ≅ \Symp^2(V)^* ≅ \Symp^2(V^*) ≅ \Symp^2(V) \,.
\end{gather*}
The bilinear form $β$ is an invariant element of $\Bil^2(V, )$, whence $V ⊗ V$ cannot be irreducible unless it the one-dimensional trivial representation.
This requires $V$ to be one-dimensional.
Suppose that $β$ is alternating.
It it then an invariant element of $\Alt^2(V, )$, whence $⋀^2 V$ cannot be irreducible unless it is one-dimensional and trivial.
This requires $V$ to be two-dimensional.
Suppose that $β$ is symmetric.
It it then an invariant element of $\Sym^2(V, )$, whence $\Symp^2(V)$ cannot be irreducible unless it is one-dimensional and trivial.
This requires $V$ to be one-dimensional.
We can pretty much directly apply this general argumentation to the given situation.
We only need to know that the form $Q$ is $_4(ℂ)$-invariant.
Depending on the given definition of $_4(ℂ)$, this is either true by definition, or a consequence of the fact that $Q$ is $\mathrm{Sp}_4(ℂ)$-invariant.
