3
$\begingroup$

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 $SU(3)$?

$\endgroup$
2
$\begingroup$

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 $Sp(6) \cap SL(3,\mathbb{C})=SU(3)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.