What Is a Kahler Manifold?

[s00mb]

TL;DR

Kahler Manifold question

Hi, I have just finished studying Riemannian Geometry and was moving on to trying to figure out what a Kahler manifold is. Using wikipedia's definition(probably a bad idea to start with) it says "Equivalently, there is a complex structure J on the tangent space of X at each point (that is, a real linear map from TX to itself with J^2 = −1) such that J preserves the metric g (meaning that g(Ju, Jv) = g(u, v)) and J is preserved by parallel transport. " I am not sure what it means by complex structure. Is j a subset of the tangent space with that property or is J the map itself J : TX -> TX with J^2=-1 ? I guess it could kind of be both but I was looking if someone could clarify this for me. Thanks.

 

[fresh_42]

The word complex refers to $J^2=- id$. Let's list what we have:

• $M$ is a smooth manifold
• $V(M)=\{\,T_pM\,|\,p\in M\,\}$ be the space of all smooth vector fields on $M$

Then we have two functions:

• $g\, : \,V(M) \times V(M) \longrightarrow C^\infty(M;\mathbb{R})$ a Riemannian metric on $V(M)\; , \;g(X,Y)(p)=g(X_p,Y_p)$
• $J\, : \,TM \longrightarrow TM$ a complex structure (a conjugation if you like)$\, : \,J^2(X)=-X$

and a compatibility together with a symplexity condition:

• $g(JX,JY)=g(X,Y)$ (the complex structure doesn't affect the metric)
• $\omega(X,Y)\, := \,g(JX,Y)$ is a symplectic form on $TM$, the Kähler form, i.e. a global, smooth, and closed $2-$form which is pointwise non degenerated

So all we did is considering an ordinary Riemannian smooth manifold, which has a symplectic structure on its tangent spaces that doesn't change the original Riemannian structure. It is an additional structure. But symplectic means we need a form of conjugation, for otherwise the term would not make sense. It has the same meaning, will say role as the imaginary unit has for symplectic vector spaces. Here our conjugation is called a complex structure, since imaginary would be confusing. It is not a subset, it is a multiplication with a complex unit, a mapping.

 

[WWGD]

But a symplectic structure may exist only in odd dimensions on the manifold. Edit: And I don't understand why we need conjugation to make sense of symplectic structures.

 

[fresh_42]

But a symplectic structure may exist only in odd dimensions on the manifold. Edit: And I don't understand why we need conjugation to make sense of symplectic structures.

I'm not sure about Kähler manifolds, but symplectic vector spaces are of even dimension!

... because the scalar product is alternating, and the complex structure kind of splits the sign on the two factors. Maybe "need" was a bit too strong, but it helps calculations a lot. Otherwise we would always have to consider the square. Of course one can do analysis on $\mathbb{R}[x]/\langle x^2+1 \rangle$, but $i$ makes it easier.

 

[WWGD]

I'm not sure about Kähler manifolds, but symplectic vector spaces are of even dimension!

... because the scalar product is alternating, and the complex structure kind of splits the sign on the two factors. Maybe "need" was a bit too strong, but it helps calculations a lot. Otherwise we would always have to consider the square. Of course one can do analysis on $\mathbb{R}[x]/\langle x^2+1 \rangle$, but $i$ makes it easier.

I think you mean i make it easier;).

 

[WWGD]

Nein mind, I mixed contact with symplectic. Stmplectic is even dimensional while contact is odd dimensional. Edit: you still have the issue of dimensional restrictions; odd instead of even though.

 

[s00mb]

Thanks makes sense now :)

 

[Ssnow]

In a Kähler manifold $M$ the three structures: complex $J$, Riemannian $g$ and symplectic $\omega$ are compatible in the sense of previous posts ... in fact there is an hermitian metric $h$ on the complex tangent space $TM$ such that $\omega=\Im{(h)}$ and $g=\Re{(h)}$. A famous Kähler manifold is the projective space $\mathbb{C}\mathbb{P}^{n}$ with the Fubini-Study metric $g_{FS}$.

Ssnow