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Let $G=\operatorname{SL}_6$ act on $V=\Lambda^3 \mathbb C^6$. I would like to find the ring of invariants $\mathbb C[V]^G$. There is an obvious invariant $$Sq: V \to \mathbb C, \quad \omega \mapsto \omega^2 \in \Lambda^6 \mathbb C^6 \simeq \mathbb C,$$ with the last isomorphism $\operatorname{SL}_6$-invariant. Update: $Sq=0$, but there is a unique invariant $\alpha$ of degree $4$, not $2$ (see the comment of Robert Bryant below).

My idea how to prove $\mathbb C[V]^G = \mathbb C[\alpha]$ is to consider $$W=\mathbb Cx_1 \wedge x_2 \wedge x_3 \oplus \mathbb Cx_4 \wedge x_5 \wedge x_6 \subset V.$$ If a general $G$-orbit intersects $W$, then a map $f: \mathbb C[V]^G \to \mathbb C[W]$ is injective. But the image lies in $\mathbb C[W]^{\mathbb C^* \rtimes \mathbb Z/2\mathbb Z}$, where the action of $\mathbb C^*$ on $W$ is given by $$t(w_1 \cdot x_1 \wedge x_2 \wedge x_3 + w_2 \cdot x_4 \wedge x_5 \wedge x_6)=tw_1 \cdot x_1 \wedge x_2 \wedge x_3 + t^{-1}w_2 \cdot x_4 \wedge x_5 \wedge x_6,$$ and the action of $\mathbb Z/2\mathbb Z$ is given by $$1(w_1 \cdot x_1 \wedge x_2 \wedge x_3 + w_2 \cdot x_4 \wedge x_5 \wedge x_6)=w_2 \cdot x_1 \wedge x_2 \wedge x_3 - w_1 \cdot x_4 \wedge x_5 \wedge x_6.$$ Now it is easy to check that $\mathbb C[W]^{\mathbb C^* \rtimes \mathbb Z/2\mathbb Z} \simeq \mathbb C[\beta]$, where $\beta=w_1^2 w_2^2$ is an invariant of degree $4$. Therefore the question is whether a general $G$-orbit intersects $W$ (and also how to construct $\alpha$ with $f(\alpha)=\beta$), and here I am stuck. Could you help me?

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    $\begingroup$ It does. This is a well-known result in invariant theory. $\endgroup$ – Sasha Dec 20 '16 at 20:45
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    $\begingroup$ @Sasha, could you give me a reference or a name for this result? $\endgroup$ – evgeny Dec 20 '16 at 20:57
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    $\begingroup$ Vinberg-Popov, Invariant theory (VINITI, Algebraic geometry - IV), and references therein. $\endgroup$ – Sasha Dec 20 '16 at 21:11
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    $\begingroup$ @evgeny I hate to have to point this out, but if $\omega$ is a $3$-form then $\omega^2 \equiv 0$ (after all, $\omega\wedge\eta = -\eta\wedge\omega$ when $\omega,\eta\in\Lambda^3$), so your attempt to construct an invariant this way fails. In fact, there is a nonzero invariant polynomial under $\mathrm{SL}(6,\mathbb{C})$ and it does generate the ring of invariants, but it is irreducible of degree $4$, not degree $2$. $\endgroup$ – Robert Bryant Dec 20 '16 at 21:49
  • $\begingroup$ @evgeny, you should edit further, since all expectations about $C[sq]$ make little sense now. $\endgroup$ – YCor Dec 21 '16 at 7:29
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The principal isotropy group is $H=SL(3)\times SL(3)$: it has the right dimension (namely 16) and occurs as an isotropy group (namely of a general element of $W$). Now it is a general result of Luna-Richardson that the restriction map $\mathbb C[V]^G\to\mathbb C[W]^N$ is an isomorphism where $W=V^H$ and $N=N_G(H)/H$.

For a concrete construction of $\alpha$ it might be easiest to consider $V$ as a symplectic vector space with symplectic form $$\langle\omega_1,\omega_2\rangle:=\frac{\omega_1\wedge\omega_2}{x_1\wedge\ldots\wedge x_6}.$$ Then there is a moment map given by $$m:V\to\mathfrak g^*:\omega\mapsto[\xi\mapsto\langle\omega,\xi\omega\rangle].$$ Then $\alpha$ is the pullback of an invariant quadratic form on $\mathfrak g^*$. Concretely: Let $\xi_i$ be a basis of $\mathfrak g$ and let $\xi^i$ be its dual basis (i.e. with ${\rm tr}(\xi_i\xi^j)=\delta_{ij}$). Then $$ \alpha(\omega)=\sum_i\langle\omega,\xi_i\omega\rangle\langle\omega,\xi^i\omega\rangle.$$

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Here's an alternate construction of the invariant $\alpha$ that seems a little simpler (and, besides, gets used in proving the normal forms for $3$-forms in $6$-variables). The details may be found in Nigel Hitchin The geometry of three-forms in six and seven dimensions, section 2.1.

Let $S$ be a vector space over $\mathbb{C}$ of dimension $6$, and fix an isomorphism $\Lambda^6(S^*)= \mathbb{C}$ (i.e., choose a volume form). Then there is an induced natural isomorphism $$ S = S\otimes \Lambda^6(S^*) = \Lambda^5(S^*) .\tag1 $$ Given $\phi\in\Lambda^3(S^*)$, define a mapping $J_\phi:S\to S$ by the rule $$ J_\phi(s) = (\iota_s\phi)\wedge\phi $$ where $\iota_s\phi\in\Lambda^2(S^*)$ is the interior product of $s\in S$ with $\phi$. It is easy to see that the trace of $J_\phi\in \mathrm{End}(S,S) = S\otimes S^*$ vanishes identically. However, if one sets $$ \alpha(\phi) = \tfrac16\,\mathrm{tr}\bigl((J_\phi)^2\bigr),\tag2 $$ then one finds that $\alpha(\phi)$ does not vanish identically, and it is obviously a quartic polynomial in the coefficients of $\phi$. In fact, one has the identity $$ (J_\phi)^2 = \alpha(\phi)\,\mathrm{Id}_S\,,\tag3 $$ and this identity can be used to put $\phi$ in normal form with respect to the eigenspaces of $J_\phi$ when $\alpha(\phi)\not=0$.

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  • $\begingroup$ This actually is the same construction as in Friedrich's answer. The map $\phi \mapsto J_\phi$ is just the moment map, and the trace of the square is just the Killing form. $\endgroup$ – Sasha Dec 21 '16 at 14:28
  • $\begingroup$ @Sasha I agree that it's similar (though the normalization of $\alpha$ is different), but thinking of it this way yields a bit more information: The fact that $(J_\phi)^2$ is a multiple of the identity map on $S$ turns out to be very useful in deriving the normal forms of the orbits, even the degenerate ones. $\endgroup$ – Robert Bryant Dec 21 '16 at 15:58
  • $\begingroup$ Honestly, I don't see any difference (except for the normalization). But definitely, you wrote the map in a more explicit way and explained more of its properties. $\endgroup$ – Sasha Dec 21 '16 at 16:47
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    $\begingroup$ @Sasha I guess the difference in point of view for me is that Friedrich's construction of $\alpha$ is a construction that produces a quartic invariant for any symplectic representation $V$ of any semisimple Lie group $G$ (via its Killing form) (of course, it might vanish identically in some situations), while, from my point of view, the space $S$ for which $V=\Lambda^3(S^*)$ is the fundamental object, and interpreting $J_\phi$ as an element of $\mathrm{End}(S)$ (so that it can be squared) is something that you wouldn't see if all you had was $V$. $\endgroup$ – Robert Bryant Dec 22 '16 at 21:05
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Another elementary description of the degree four invariant is as $$ \epsilon_{i_1i_2i_3i_4i_5i_6}\ \epsilon_{i_7i_8i_9i_{10}i_{11}i_{12}} \ \omega_{i_1i_2i_3\ }\ \omega_{i_4i_5i_7}\ \omega_{i_6i_8i_9}\ \omega_{i_{10}i_{11}i_{12}} $$ where indices are summed from 1 to 6. The epsilon notation is as in this MO answer.

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