# Kähler manifolds are formal

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 $$p$$.

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.