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I'm looking at Kahler geometry at the moment and admiring how it manages to do so much with clean global algebraic arguments. One of the big exceptions to all this, however, is the proof of the Kahler identities $$ [\Lambda,\overline{\partial}]=-i \partial^\ast, ~~~~~~ [\Lambda,\partial]=-i \overline{\partial}^\ast. $$ In the two standard references, Voisin, and Griff + Harris, the identities are proved using arguments that are local and somewhat analaytic. Does there exists anywhere a nice global algebraic proof?

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Look at R O Wells Differential Analysis on Complex manifolds 3 edn.He has a global algebraic proof of the Kahler identities due to Henryk Hecht. –  Mohan Ramachandran May 10 '11 at 19:08
These identities are themselves local statements. A global proof would, in my opinion, ultimately hide the essential content. –  John Pardon Jun 25 at 0:01

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up vote 11 down vote accepted

I don't know of a proof that would really be characterized as global, and if I saw one I would immediately try to figure out how it's really local. There is, however a proof along different lines than that in GH or Voison, and seems more enlightening to me. Huybrechts in his Complex Geometry book gives one that is more representation theoretic/linear algebraic. First he proves a formula (due to Weil?): for a primitive k-form $\alpha \in P^k$ we have:

$\ast L^j (\alpha) = (-1)^\frac{k(k+1)}{2} \frac{j!}{(n-k-j)!} L^{n-k-j}I(\alpha)$

Here n is of course the complex dimension of the manifold, and $I$ is the operator induced on forms by the almost complex structure (page 37 of Huybrechts). This underappreciated formula seems to only be found in this book.

He then uses this together with the purely linear algebraic Lefschetz decomposition on $\alpha$ and $d\alpha$ to prove the commutation relation in the form $[\Lambda, d] = -d^c$. The proof is a bit calculational though. See page 121-122.

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Since the Kähler identities do not involve coordinates, it is natural to expect that there is a coordinate-free proof. However, it is completely unreasonable to expect that the coordinate-free proof is in any way more elegant.

Here is my rationale: if you read a proof using coordinates, and \it{this proof does not use any special property of the chosen coordinates}, then you can translate the argument into a coordinate-free one.

But the argument for the Kaehler identities is NOT: choose an arbitrary coordinate system and do a stupid calculation. If this were the case, Deligne, Sullivan, Griffiths and Morgan would surely have avoided coordinates. Instead, the argument is (in my opinion this argument is marvelous):

  1. check the identities on $C^n$ with the standard metric;
  2. at a given point $x \in M$, there is a coordinate system, mapping $x $ to $0$, such that the $1$-jet of the metric/complex structure is the ($1$-jet of the) standard metric/complex structure at $0$. Here Kählerness comes in, and if I remember correctly, exponential coordinates do the job.
  3. Since the Kähler identities involve only the $1$-jet of the metric/complex structure and completely invariant things, the formulae are valid on a general Kähler manifold.

The standard proof of point 1 is a bit messy, and my suggestion is to make it slicker by using the symmetries of $C^n$. Observe that all expressions on $C^n$ are translation-invariant and invariant under $U(n)$. Moreover, the effect of scaling $C^n$ and conjugation is easy to figure out. These properties should be enough to force the Kähler identities.

In a completely coordinate-free proof, you do no longer have these special coordinates at hand, and thus I expect the proof, though possible, to be much more complicated.

EDIT: as RdN points out, Huybrechts gives a coordinate-free proof in his book. Huybrecht's proof is about as elegant and complicated as the standard proof.

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It is often convenient to make calculations in nice local coordinates, but my understanding is that using such coordinates is rarely, if ever, necessary. –  John Pardon Jun 25 at 0:04

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