# Minimum requirements for a Kähler manifold to be hyperkähler

In 'panoramic view of Riemmannian geometry' when introducing hyperkähler manifolds, Berger states, informally, that a hyperkähler manifold is a Riemmannian manifold which is Kähler for more than one different almost complex structures.

I was wondering whereas this was a theorem or just a 'catchphrase'. In other words, my question is : if a Riemannian manifold is Kähler for two different (linearly independent) almost complex structures, is it hyperkähler ?

Of course on has to ask the metric to be irreducible, since things such as $\mathbb{S}^2\times M$ where $M$ is hyperkähler has at least 6 (!) independent almost complex structures. And being $4n+2$ dimensional, they can't be hyperkähler !

I think this is true in (real) dimension 4 for the following reason : the holonomy group $G$ of a Kähler $(M^4,g)$ is contained in $U(2)$. It leaves the Kähler form invariant. If there are two almost complex structures, there are two Kähler forms, so $G$ must act trivially on a 2-dimensional subspace of $\Lambda^2T_p^*M$. Since the holonomy representation of $U(2)$ only preserves the Kähler form, $G$ is a strict subgroup of $U(2)$ and is therefore contained in $SU(2)=Sp(1)$, hence $M$ is hyperkähler.

My knowledge of representation theory is too scarce to see if this carry on in higher dimensions.

Suppose that the metric on $M^n$ has irreducible holonomy, is simply connected (or, slightly more generally, that the restricted holonomy $H^0$ acts irreducibly), and that there exist two independent parallel complex structures. Since there is at least one parallel complex structure $I$, the holonomy group $H^0$ is a subgroup of $\mathrm{U}(n/2)$.

If $H^0=\mathrm{U}(n/2)$ then $I$ generates the ring of parallel endomorphisms of the tangent bundle, and the only other parallel complex structure is $-I$, so the restricted holonomy must be a proper subgroup of $\mathrm{U}(n/2)$.

By Berger's classification (since it acts irreducibly), $H^0$ must be either $\mathrm{SU}(n/2)$ or $\mathrm{Sp}(n/4)$ or else the metric must be locally symmetric.

When $n=4$, one has $\mathrm{SU}(n/2)=\mathrm{Sp}(n/4)$, so if $H^0=\mathrm{SU}(2)$, the metric is hyperKähler. If $n>4$ and $H^0=\mathrm{SU}(n/2)$, then, again, $I$ generates the ring of parallel endomorphisms of the tangent bundle, and the only other parallel complex structure is $-I$, contrary to hypothesis, so the restricted holonomy can't be $\mathrm{SU}(n/2)$.

Finally, if the metric were locally symmetric, it would have to be locally isometric to one of the known irreducible Hermitian symmetric spaces, but looking at the list of these, one sees again that the ring of parallel endomorphisms of the tangent bundle is generated by $I$, so none of these admit a linearly independent parallel complex structure.

Thus, the only possibility is that the restricted holonomy is $\mathrm{Sp}(n/4)$, i.e., the metric is (locally) hyperKähler.

• Ok, so I guess my next move is to understand better the action of $U(n)$ on $\Lambda^2\mathbb{R}^{2n}$. I understood the $n=2$ case by bare hands computations with quaternions, but I'm not sure how to tackle the higher dimensions. Any reference ? – Thomas Richard Jul 27 '13 at 15:27
• You could always look in Besse's Einstein Manifolds, but there are many references for this. The irreducible decomposition of $\Lambda^2(\mathbb{R}^{2n}$ is well-understood: It splits into the (real-valued) $(1,1)$-forms (which then further decompose into the multiples of the Kähler form plus a compliment, the primitive $(1,1)$-forms, that is isomorphic to ${\frak{su}}(n)$ as a $\mathrm{U}(n)$-module) plus a $\mathrm{U}(n)$-irreducible piece that consists of the real parts of the $(2,0)$-forms. (The irreducibility of these summands is relatively easy but not obvious without some technology.) – Robert Bryant Jul 27 '13 at 17:08