Talk:Hyperkähler manifold

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Initial discussions[edit]

A complex structure is, I think, an atlas of holomorphic maps to C^n. What I don't understand is how you can multiply two complex structures or form linear combinations. --MarSch 09:37, 7 September 2005 (UTC)

The atlas needs holomorphic transition maps as well. Locally, a complex structure is equivalent to choosing a linear isomorphism J on each fiber of the tangent bundle such that , where 1 is the identity map. So on each fiber the algebraic operations you're wondering about are those of matrix algebra. Orthografer 05:44, 29 March 2006 (UTC)

In mathematics, the name "hyperkahler" is more common, and "hyper-kahler" is used by un-initiates and rarely.

Google: Hyper-kahler 27100 Hyperkahler 38500 Hyper-Kahler 28 hyperkahler 98 --Tiphareth 18:45, 8 April 2006 (UTC)

The definition was somewhat confusing. It's wrong, for several reasons, to require that holonomy group is precisely Sp(n). Also, the word "holonomy group" is ambiguous - it's not clear whether we speak of local holonomy or of global. In the first case, we get Enriques surfaces, which are not hyperkaehler (neither Calabi-Yau either), but their local holonomy is Sp(n).

Outside of differential geometry, hyperkaehler manifolds are manifolds with quaternionic action on TM. In differential geometry, these are manifolds with holonomy contained in Sp(n). The latter definition is satisfying, though it's tough on people who don't know or like the holonomy.

--Tiphareth 19:31, 8 April 2006 (UTC)

I agree that the quaternionic action is what I naturally consider when I hear the word hyperkahler, but I've never gone through the equivalence in my head. For instance, I think it would be wonderful if you could explain in the page why it's wrong to say the holonomy group (which I have understood to be global, though I'm no expert) should be precisely Sp(n). I gave a rationale for excluding the 4-torus and I think your reason for including it would be a welcome addition, instead of simply changing to definition to taste. I've been harassing my hyperkahler friends to tell me things, but they always make me pin down a definition of hyperkahler first. Maybe this debate is worthy of inclusion in the article by someone who knows about it. Orthografer 04:27, 10 April 2006 (UTC)

The reason is too trivial to be included, I'm afraid.

We refer to the analogy between "Kaehler" and "hyperkaehler"; however, the Kaehler manifold, if defined in terms of the holonomy, is a manifold with holonomy contained in U(n), and not precisely U(n).

Otherwise we would have statements which are counter-intuitive: like a product or a submanifold of Kaehler/hyperkaehler manifolds will be no longer Kaehler/hyperkaehler.

Also, the other definitions of hyperkaehler (the local one in terms of I,J,K and the algebro-geometric one, in terms of the holomorphic symplectic form) are not valid, if we require holonomy =Sp(n).

Finally, consider the case when the global holonomy is Sp(n), and the local holonomy is a subgroup of Sp(n). You shall have a manifold which is hyperkaehler globally, but not hyperkaehler in a neighbourhood of any point. In mathematics, the definitions of "bla-bla-bla manifold" are almost always local, because the manifolds are defined in terms of a local atlas, and it's natural to define the structures on manifolds in terms of the same atlas. -- Tiphareth 13:13, 11 April 2006 (UTC)

Tiphareth: what do you think of my changes? Orthografer 20:39, 20 April 2006 (UTC)

I like your changes. Removed extra "|" in URL. --Tiphareth 16:42, 24 April 2006 (UTC)


Is there a reason that the spelling "hyperkähler" should not be dismissed as incorrect in favor of "hyper-Kähler"? — Preceding unsigned comment added by Seub (talkcontribs) 23:16, 12 March 2015 (UTC)


I'm not a fan of the definition given of a hyperkähler manifold in this article. By the way, it is only given in the preamble, is that normal?

The reason that I'm not a fan is the same reason that I wouldn't like the definition of a Kähler manifold as "a Riemannian manifold with holonomy contained in U(n)" (which is is not the definition given in the corresponding Wikipedia article).

Let me explain in case it's not clear: when you have a Kähler manifold, you know what the complex structure is. Saying that you have a Riemannian metric with holonomy in U(n) just tells you that there exists a complex structure compatible with the metric, turning your manifold into a Kähler manifold. However, several different complex structures can exist. And they can be "really different" in the sense that there might not exists self-automorphisms of your manifold that take a complex structure to the other. In practice, you really want to know what complex structure you are dealing with.

Similarly, when you have a hyperkähler manifold, you want to know hat I, J and K are (at least the triple, but a lot of the time each one of them, in my modest experience). So, in a nutshell, I would say that a Riemannian manifold with holonomy in Sp(k) is not a hyperKähler manifold per se, it is a Riemannian manifold which admits a compatible quaternionic (or hypercomplex) structure turning it into a hyperkähler manifold, however several different such quaternionic structures may exist.

I am not a seasoned contributor to Wikipedia, should I go ahead and edit the article, or would that be bad-mannered?--Seub (talk) 23:57, 13 March 2015 (UTC)

Assessment comment[edit]

The comment(s) below were originally left at Talk:Hyperkähler manifold/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Expand. There's hyperkahler quotients and (generalized) Gibbons-Hawking for a start. Geometry guy 21:06, 17 September 2008 (UTC)

Last edited at 21:06, 17 September 2008 (UTC). Substituted at 02:13, 5 May 2016 (UTC)