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Let $X:=x^3$, $Y:=x^2y$, $Z:=xy^2$ and $W:=y^3$ be the 4 independent generators of $S^3\mathbb{C}^2$, and observe that the kernel of the natural epimorphism (total symmetrisation) $$ p:S^2S^3\mathbb{C}^2\longrightarrow S^6\mathbb{C}^2 $$ is the 3-dimensional subspace generated by $$ XZ-Y^2\, ,\quad XW-YZ\, ,\quad YW-Z^2\, . $$ These are precisely the generators of the ideal of the twisted cubic in $\mathbb{C}\mathbb{P}^3=\mathbb{P}(S^3\mathbb{C}^2)$.

QUESTION: is there a natural way to identify the subspace $\ker p$ of $S^2S^3\mathbb{C}^2$ with $S^2\mathbb{C}^2$?

Motivation & pseudo-answer

The 10-dimensional $\mathfrak{sl}_2(\mathbb{C})$-module $S^2S^3\mathbb{C}^2$ has two irreducible components, one of which is $S^6\mathbb{C}^2$, and $p$ is a $\mathfrak{sl}_2(\mathbb{C})$-module morphism, so I expect $\ker p$ to be a 3-dimensional submodule isomorphic to the other irreducible component, which is $S^2\mathbb{C}^2$.

The only way I could think of, in order to embed $S^2\mathbb{C}^2$ into $S^2S^3\mathbb{C}^2$, is to regard elements of the former as inner products $\langle\, \cdot\, ,\, \cdot\, \rangle$ on $\mathbb{C}^2$, and then extend them to $S^3\mathbb{C}^2$. For instance, $$ \langle\langle x^3,\, \cdot\, \rangle\rangle:=\langle x \, ,\, \cdot\, \rangle\langle x \, ,\, \cdot\, \rangle\langle x \, ,\, \cdot\, \rangle $$ (at the left-hand side there is a cubic polynomial, whereas at the right-hand side there is the associated trilinear form), and similarly for the other generators.

Unfortunately, the process $\langle\, \cdot\, ,\, \cdot\, \rangle\longmapsto \langle\langle\, \cdot\, ,\, \cdot\, \rangle\rangle$ does give an embedding, but not the desired one (though resembling much). And I cannot think of anything else!

Comment after Sasha's answer

$\ker p$ is isomorphic to $S^2\mathbb{C}$ via the correspondence $$ S^2\mathbb{C}\ni\left(\begin{array}{cc}a&b\\ b&c\end{array}\right)\mapsto a(XZ-Y^2)+b(XW-YZ)+c(YW-Z^2)\in\ker p\, . $$ The matrix of the bilinear form at the right-hand side is $$ \left(\begin{array}{cccc}0&0&a&b\\ 0&-2a&-b&c\\ a&-b&-2c&0\\ b&c&0&0\end{array}\right)\, , $$ whose determinant is exactly the square of the determinant of the matrix at the left-hand side. Sasha says that this guarantees that the map is also a $\mathfrak{sl}_2(\mathbb{C})$-module morphism, but I cannot see why.

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In the 3-dimensional space of quadrics through the twisted cubic curve consider the subset, parameterizing degenerate quadrics. An easy computation shows, these form a double smooth conic. So, the plane $P^2$ of quadrics comes with a distinguished conic, that identifies the corresponding 3-space with a symmetric square.

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  • $\begingroup$ Very concise and precise explanation - thanks! However, I had no doubts that the space which I called $\ker p$ identifies with a symmetric square. I wanted to see explicitly the identification, and understand to what extent is natural. As you can see from the comment above, I followed your reasoning, and I wrote down the identification. But I still can't see why it is an $\mathfrak{sl}_2(\mathbb{C})$-module morphism. $\endgroup$ – Giovanni Moreno Apr 14 '16 at 20:00
  • $\begingroup$ Assume you have a plane with a smooth conic in it. The conic then is isomorphic to $P^1$, and its embedding into the plane is the second Veronese. Therefore, the plane identifies with the projectivization of the symmetric square of the two-space, whose projectivization is $P^1$. $\endgroup$ – Sasha Apr 14 '16 at 20:10

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