Let $(z_1,\overline{z}_1,\dots, z_n,\overline{z}_n)$ be the coordinates on $\mathbb{R}^{2n}$ given by the identification $\mathbb{R}^{2n} \simeq \mathbb{C}^n$. Define $$ g =\left| dz_1 \right|^2+ \dots + \left| dz_n \right|^2 \\ \omega = \frac{i}{2} (dz_1 \wedge d \overline{z}_1+ \dots + dz_n \wedge d \overline{z}_n) \\ \gamma=dz_1 \wedge \dots \wedge dz_m $$ The subgroup of $GL(2n)$ that stabilizes $g$ and $\omega$ is $U(n)$ ("2-out-of-3-property"). The subgroup of $GL(2n)$ that additionally stabilizes $\gamma$ is $SU(n)$.

In one source an $SU(3)$ structure on a manifold $M$ is given by specifying a $2$-form $\omega$ and a $3$-form $\gamma$ so that in any point $p \in M$ there exists an element in the frame bundle $L \in GL(M)_p$ that pulls back $\omega$ and $\gamma$ to the forms above.

Question: A priori, this should not be an $SU(3)$-structure, because the stabilizer of the forms $\omega$ and $\gamma$ from above is bigger than $SU(3)$. Is this correct?

Is it maybe assumed that the manifold is a Riemannian manifold, and that $L \in GL(M)_p$ also pulls back the Riemannian metric to the $g$ defined above? I can see that this would then be an $SU(3)$-structure. Or is this a specialty of (complex) dimension 3, that here the stabilizer of $\omega$ and $\gamma$ is already equal to $SU(3)$?


The answer is: A carefully chosen $3$-form and $2$-form in fact do have stabiliser $SU(3)$.

Let $V$ be a real vector space of dimension 6 over $\mathbb{R}$ and let $e^1$, $\dots$, $e^6$ be a basis of $V^*$. Define $$ \phi = e^1 \wedge e^3 \wedge e^5- e^1 \wedge e^4 \wedge e^6- e^2 \wedge e^3 \wedge e^6- e^2 \wedge e^4 \wedge e^5. $$ (This is $Re (dz^1 \wedge dz^2 \wedge dz^3)$ after choosing a suitable complex structure on $V$) The oriented stabiliser of $\phi$ is $SL(3,\mathbb{C})$, i.e. $$ GL_+(V) \cap Stab_{GL(V)} \phi = SL(3,\mathbb{C}). $$ (A proof for this can be found in the appendix of R. Bryant: On the geometry of almost complex 6-manifolds) If $\sigma \in \Lambda^2 V^*$ is non-degenerate, it has stabiliser $Sp(6)$ the endomorphism which stabilise both forms are $Sp(6) \cap SL(3,\mathbb{C})=SU(3)$.


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