Assume V is the standard representation of $\mathfrak{sp}_4\mathbb{C}$ on $\mathbb{C}^4$.
By definition of symplectic group, the bilinear form $Q \in \wedge^2V^*$ is one-dimensional representation of $Sp_4\mathbb{C}$, hence $\mathfrak{sp}_4\mathbb{C}$. Also if we identify $v^* \in V^*$ with $Q(v,\cdot)$, we will find $V^* \cong V$:
We can define $\rho^*(g)(v^*):=Q(\rho(g)(v), \cdot), g \in Sp_4\mathbb{C}$, if $\rho(g)(v)=w$, then $\rho^*(g)(v^*)=Q(\rho(g)(v),\cdot)=w^*$.
Therefore $\wedge^2V^* \cong \wedge^2V$, and $\wedge^2V=W\oplus\mathbb{C}Q $, here $W$ is irreducible representation $\Gamma_{0,1}$.
Then we consider the tensor $\wedge^2V \otimes \wedge^2V$. (page 247 in Representation Theory: A First Course by Fulton and Harris). We have a natural map.
\begin{align*}
\wedge^2V \otimes \wedge^2V \to \wedge^4V\cong\mathbb{C}
\end{align*}
the map is symmetric(I think it means: $v_1\wedge v_2\wedge v_3 \wedge v_4=v_3 \wedge v_4\wedge v_1\wedge v_2)$, so:
\begin{align*}
\text{Sym}^2(\wedge^2V) \to \mathbb{C}
\end{align*}
$\underline{\text{Since the map doesn't depend on the choice of skew form Q, it can not contain}},$ $\underline{\text{ the subspace $\text{Sym}^2W \subset \text{Sym}^2(\wedge^2V)$ in its kernel}}$.
So it restricts to give a surjection
\begin{align*}
\varphi:\text{Sym}^2W \to \mathbb{C}
\end{align*}
$\underline{\text{From the principle that an irreducible representation cannot have two independent }}$ $\underline{\text{invariant bilinear forms, we see that $\text{Sym}^2W$ can contain at most one trivial}}$ $\underline{\text{summand}}$. We have
\begin{align*}
\text{Sym}^2W=\Gamma_{0,2}\oplus \mathbb{C}
\end{align*}
The question is I can not see the relation among bilinear form $Q$, $\text{Sym}^2(\wedge^2V)$ and $\text{Sym}^2W$.And I don't understand two underlined sentence
Best Answer
Like you write, the wedge product $\wedge : \bigwedge^2 V \otimes \bigwedge^2 V \to \bigwedge^4 V$ is symmetric, so we can regard it as a a symmetric $\bigwedge^4 V$-valued symmetric bilinear form $$\Phi : \textstyle{\operatorname{Sym}^2 \left(\bigwedge^2 V\right) \to \bigwedge^4 V} .$$ (Also you like you write, we identify $\bigwedge^4 V \leftrightarrow \Bbb C$ via the linear map characterized by $Q \wedge Q \mapsto 1$. This identification changes when we change the symplectic form $Q$ but the kernel of the resulting map $\operatorname{Sym}^2 \left(\bigwedge^2 V\right) \to \Bbb C$ does not.)
As you write, $\bigwedge^2 V$ decomposes into irreducible $\mathfrak{sp}_4$-representations as $$\textstyle{\bigwedge^2 V = W \oplus \Bbb C Q \cong W \oplus \Bbb C} ,$$ and we can realize $\Gamma_{0, 1} = W \subset \bigwedge^2 V$ concretely as the kernel of the map $$\textstyle{\bigwedge^2 V \to \bigwedge^4 V}, \qquad \alpha \mapsto \Phi(\alpha \odot Q) = \alpha \wedge Q .$$
With this description in hand, it's arguably easiest to see the first underlined claim using an explicit computation. For example, take $Q := v_1 \wedge v_2 + v_3 \wedge v_4$; then, $\beta := v_1 \wedge v_2 - v_3 \wedge v_4 \in W$ (as $\beta \wedge Q = 0$) and so $\beta \odot \beta \in \operatorname{Sym}^2 W$, but $\Phi(\beta \odot \beta) = \beta \wedge \beta \neq 0$, so $\operatorname{Sym}^2 W \not\subset \ker \Phi$.
It follows from the above decomposition of $\bigwedge^2 V$ into $\mathfrak{sp}_4$-representations that $$\textstyle{\operatorname{Sym}^2 \left(\bigwedge^2 V\right) \cong \operatorname{Sym}^2 (W \oplus \Bbb C) \cong \operatorname{Sym}^2 W \oplus W \oplus \Bbb C} .$$ In particular, the inclusion maps $W, \Bbb C \hookrightarrow \operatorname{Sym}^2 \left(\bigwedge^2 V\right)$ respectively are $$\alpha \mapsto \alpha \odot Q \qquad \textrm{and} \qquad z \mapsto z Q \odot Q .$$ Furthermore, since $\operatorname{Sym}^2 W \not\subset \ker \Phi$, $\operatorname{Sym}^2 W$ decomposes as a direct sum of $2$ (in fact, irreducible) $\mathfrak{sp}_4$-representations, namely, $\Gamma_{0, 2} = \ker \phi$, where $\varphi := \Phi\vert_{\operatorname{Sym^2} W} : \operatorname{Sym^2} W \to \bigwedge^4 V)$, and its distinguished $1$-dimensional complement. In summary, in weight notation $\Gamma_{a,b}$, the decomposition of $\operatorname{Sym}^2 \left(\bigwedge^2 \Gamma_{1, 0}\right)$ into irreducible $\mathfrak{sp}_4$-representations is: $$\textstyle{\operatorname{Sym}^2 \left(\bigwedge^2 \Gamma_{1, 0}\right) = \Gamma_{0,2} \oplus \Gamma_{0, 1} \oplus 2 \Gamma_{0, 0}} .$$
Remark Checking directly shows that the symmetric, bilinear form $\varphi$ on $\Gamma_{0,1} = W$ is nondegenerate, which defines an exceptional isomorphism $$\mathfrak{sp}_4 \cong \mathfrak{so}_5$$ (or, in terms of Dynkin diagrams, $B_2 \cong C_2$), and we can interpret $W$ as the standard (defining) representation of $\mathfrak{so}_5$. We can likewise reinterpret the objects in this discussion as $\mathfrak{so}_5$-invariant. For example, $V = \Gamma_{1, 0}$ is just the spinor representation of $\mathfrak{so}_5$, and $\operatorname{Sym}^2 W$ is the space of symmetric, bilinear forms on $W$. In particular, as an $\mathfrak{so}_5$-representation the latter decomposes into an invariant line (spanned by $\varphi$, interpreted as a $\Bbb C$-valued bilinear form) and an---in fact, irreducible---codimension-$1$ $\mathfrak{so}_5$-subrepresentation $\Gamma_{0, 2}$, namely the space of bilinear forms on $W$ that are tracefree with respect to $\varphi$. In particular, the presence of the invariant line in $\operatorname{Sym}^2 W$ is arguably easier to see from the orthogonal perspective.