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edit approved Dec 23, 2016 at 8:12

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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 $\alpha(\phi)\not=0$.

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):

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 $\alpha(\phi)\not=0$.

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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created Dec 21, 2016 at 14:01

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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):

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 $\alpha(\phi)\not=0$.