Plethysm of $S^3(S^2V)$ as $\mathfrak{sl}_3(\mathbb{C})$-module I have asked this question in MSE before, but have not got any answer. So here I am asking it again with some more detail.
I believe that the following sequence of $\mathfrak{sl}_3(\mathbb{C})$-modules is exact:
$$0\to\mathbb{C}\to V\otimes\wedge^2V\stackrel\phi\to S^2 V\otimes S^2(\wedge^2 V)\stackrel\psi\to S^3(S^2V)\stackrel{N}\to S^6V\to 0$$
where $V=\mathbb{C}^3$ and $S^2$ denotes the symmetric square. The exactness of this sequence can be shown easily using eigen value diagram. But I want to show the exactness explicitly by constructing the maps and showing exactness of those maps.
Let's call the third and fourth map $\phi$ and $\psi$ respectively, the fifth map is the natural map (call it $N$). I also know what $\psi$ is, it is given by the followig $$\psi ((s\cdot t)\otimes (u\wedge v)\cdot(w\wedge z))=((u\cdot w)\cdot(v\cdot z)-(u\cdot z)\cdot(v\cdot w))\cdot (s\cdot t)$$
and I have checked that $\operatorname{im}\psi=\ker N$. But I am not sure what $\phi$ is. 
So my question is, how to define $\phi$ so that $\ker\phi$ is isomorphic to $\mathbb{C}$ and $\psi\circ\phi$ is zero i.e. $\operatorname{im}\phi\subset\ker\psi$ (the other direction $\ker\psi\subset\operatorname{im}\phi$ follows by computing the dimensions which turns out to be equal) and of course $\phi$ should be $\mathfrak{sl}_3(\mathbb{C})$-linear.
 A: Under the action of $\mathfrak{sl}_3(\mathbb{C})$, $\Lambda^2V\simeq V^\ast$. This isomorphism can be written explicitly in terms of the invariant volume form $\omega\in\Lambda^3V\simeq \mathbb{C} $, by $(v_1,v_2\wedge v_3)=v_1\wedge v_2\wedge v_3=\lambda \omega$ (in other words, for $\theta\in V^\ast$, $\theta\mapsto \iota_\theta\omega\in\Lambda^2V$).
Thus $V\otimes\Lambda^2V\simeq V\otimes V^\ast = \mathfrak{gl}_3(\mathbb{C}) = \mathfrak{sl}_3(\mathbb{C})\oplus \mathbb{C}$. Moreover, $S^2V\otimes S^2(\Lambda^2 V)\simeq S^2V\otimes S^2 V^\ast = \mathfrak{gl}(S^2V)$. The latter space contains a subalgebra isomorphic to $\mathfrak{sl}_3(\mathbb{C})$, which is spanned by the representation matrices of the induced action on $S^2V$. 
Your map $\phi$ is then given by composing the following operations: first use the isomorphism $\Lambda^2V\simeq V^\ast$, then remove trace of the resulting operator by $A\mapsto A-\text{Tr}(A)/3 I$, then send the resulting matrix to its induced matrix on $S^2V$, and finally use the isomorphism $\Lambda^2V\simeq V^\ast$ to obtain an element of $S^2V\otimes S^2(\Lambda^2 V)$.
The kernel property is clear, because $S^3S^2V$ contains no submodule isomorpic to $\mathfrak{sl}_3(\mathbb{C})$. (This follows from highest weight calculations).
