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I want to understand why Kähler manifolds are formal.
This was first proved by Deligne, Griffiths, Morgan, Sullivan


Let $\mathcal M$ be a minimal differential algebra and $H^*(\mathcal M)$ the cohomology of $\mathcal M$, viewed as a differential algebra with $d =0$.

Definition.

  1. $\mathcal M$ is formal if there is a map of differential algebras $\psi: \mathcal M \to H^*(\mathcal M)$ inducing the identity on cohomology.
  2. The homotopy type of a differential algebra $\mathcal A$ is a formal consequence of its cohomology if its minimal model is formal.
  3. The real (or complex) homotopy type of a manifold $M$ is a formal consequence of the cohomology $M$ if the de Rham homotopy type of the real (or complex) forms $\mathcal E$ is a formal consequence of its cohomology.

In section 6, the following (main) theorem is proved:

Let $M$ be a compact complex manifold for which the $dd^c$-lemma holds (e.g. $M$ Kähler, or $M$ a Moisezon space). Then the real homotopy type of $M$ is a formal consequence of the cohomology ring $H^*(M; \mathbb R)$

Let $\{\mathcal E^*_M,d\}$ be the de-Rham complex on $M$, $\{\mathcal E^c_M,d\}$ the subcomplex of $d^c$-closed forms and $\{H_{d^c},d\}$ the quotient complex $\mathcal E_M^c/d^c \mathcal E_M$.

Using the $dd^c$-lemma ($\partial \bar\partial$-lemma), it is an easy calculation, that the natural maps $$\{\mathcal E^*_M,d\} \stackrel i\leftarrow \{\mathcal E^c_M,d\} \stackrel p\to \{H_{d^c},d\} $$ are quasi-isomorphisms and that the differential on $H_{d^c}(M)$ vanishes.


In the proof, the theorem follows immediately from the above. I don't see how and I am a little confused that there were no reasons given why the theorem follows. Only "This proves the claim and consequently [part (1)] of the theorem".

The maps seem to be maps of differential algebras, so this is fine.
But as far as I see, the theorem only follows if we can replace $\{H_{d^c},d=0\}$ with $\{H_M,d=0\}$.


Update/ Solution:
I repeat: The theorem only follows if we can replace $\{H_{d^c},d=0\}$ with $\{H_M,d=0\}$.
But the above argument shows, that the cohomologies are isomorphic. More precisely, the isomorphism is induced by $i$ and $p$.

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A dg-algebra with vanishing differential is its own cohomology. The morphisms $i_*,\rho_*$ are isomorphisms between the cohomology of $(\mathscr{E}^*_M,d)$ -- which is $H^*(M)$ -- and the cohomology of $(H_{d^c}, d = 0)$ -- which is $H_{d^c}$, because the differential vanishes.

More generally if a dg-algebra $A$ is quasi-isomorphic to any dg-algebra $B$ with vanishing differential, then $B$ is isomorphic to the cohomology of $A$, pretty much by definition; hence $A$ is formal.

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  • $\begingroup$ Thank you. I already had everything in mind what you said, for some reason I just failed to make the last (surprisingly easy) step. It's clear now. $\endgroup$ – klirk Dec 7 '18 at 23:21

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