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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.

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.

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 $\mathrm{im}\,\psi=\ker N$. 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. $\mathrm{im}\,\phi\subset\ker\phi$ (the other direction $\ker\phi\subset\mathrm{im}\,\phi$ follows by computing the dimensions which turns out to be equal) and of course $\phi$ should be $\mathfrak{sl}_3(\mathbb{C})$-linear.

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.

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:

$$0\to\mathbb{C}\to V\otimes\wedge^2V\to S^2 V\otimes S^2(\wedge^2 V)\to S^3(S^2V)\to S^6V\to 0$$

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 $\mathrm{im}\,\psi=\ker N$. 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. $\mathrm{im}\,\phi\subset\ker\phi$ (the other direction $\ker\phi\subset\mathrm{im}\,\phi$ follows by computing the dimensions which turns out to be equal) and of course $\phi$ should be $\mathfrak{sl}_3(\mathbb{C})$-linear.

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 $\mathrm{im}\,\psi=\ker N$. 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. $\mathrm{im}\,\phi\subset\ker\phi$ (the other direction $\ker\phi\subset\mathrm{im}\,\phi$ follows by computing the dimensions which turns out to be equal) and of course $\phi$ should be $\mathfrak{sl}_3(\mathbb{C})$-linear.