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Let $M$ be a smooth oriented manifold of dimension $n$. Let $(\mathcal{A} ^*(M),d)$ be the chain complex of differential forms on $M$. Endowing $M$ with an inner product gives us a hodge star operator which we could use to define the codiferential:

$$\delta : \mathcal{A} ^{k}(M) \to \mathcal{A} ^{k-1}(M), \delta: \omega \mapsto (-1)^k \star^{-1}d\star\omega $$

This gives a chain complex $(\mathcal{A} ^*(M),\delta)$.

We can now define the laplacian as $\triangle = d \delta + \delta d$. As a chain map, $\triangle : \mathcal{A}^*(M) \to \mathcal{A}^*(M)$ is null homotopic by constrution. By the exterior algebra functor I mean the functor that takes vector bundles over a manifold to their corresponding exterior algebra bundles.

1. Does there exist a map $\varphi: \Gamma(T^*M) \to \Gamma(T^*M)$ s.t. under the exterior algebra functor the induced map between graded algebras is $\triangle$?

To prove the Hodge theorem all we need to do is show that $i: Ker(\triangle) \to \mathcal{A}^*(M)$ is a quasi-isomorphism. From homological algebra we know that the chain map $i$ is a quasi-isomorphism iff its cone $Cone(i)$ is acyclic.

2. What properties should $\varphi: \Gamma (T^*M) \to \Gamma(T^*M)$ posses to induce a map whose kernel has an acyclic cone?

Could the hodge theorem be proved in this way? i.e. by exhibiting some function $\varphi$ satisfying (1) and proving that it posses (2)? (the latter part will no doubt involve some functional anaysis).

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  • $\begingroup$ Do you mean that $f$ is a map on sections of $T^\ast M$, or a vector bundle homomorphism, or something else? $\endgroup$ – Phillip Andreae Nov 23 '15 at 17:49
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    $\begingroup$ Re: your very last question. 95% of the proof of the Hodge theorem is analysis. I doubt some algebraic rephrasing of the theorem will avoid or reduce the amount of analysis you have to do or help with the proof. (I also don't understand what kind of object $f$ is supposed to be, as in Phillip's comment.) $\endgroup$ – user98602 Nov 23 '15 at 20:24
  • $\begingroup$ @MikeMiller I'm not trying to invent the wheel here. I know most of the content of hodge theorem is analysis anyways. I would just like to have a simple homological picture of what the theorem means. $\endgroup$ – Saal Hardali Nov 24 '15 at 4:22
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If I understand your first question correctly, you ask whether you can find a $\varphi \colon \Gamma(T^{*}M) \rightarrow \Gamma(T^{*}M)$ such that $\bigwedge^{k}(\varphi) \colon \bigwedge^k\Gamma(T^{*}M) \rightarrow \bigwedge^k\Gamma(T^{*}M)$ coincides with the Laplacian acting on $k$-differentiable forms under the identification of $\bigwedge^k\Gamma(T^{*}M) \cong \mathcal{A}^{k}(M)$ as $C^{\infty}(M)$-modules. The map $\varphi$ should be the Laplacian acting on one-forms but the Laplacian is a second-order differential operator and not a morphism of $C^{\infty}(M)$-modules (it is not $C^{\infty}(M)$ -linear) so it doesn't make sense to apply $\bigwedge$ to it as a functor from $C^{\infty}(M)$-modules.


On the most basic level, the Hodge decomposition is something that you can "expect" just from linear algebra. Consider the following observations:

  • Let $V$ be a finite dimensional vector space and $W \subseteq V$ a subspace. By choosing any subspace $W'$ that complements $W$ (that is, $W \oplus W' = V$) we obtain an isomorphism $$W' \hookrightarrow V \xrightarrow{\pi} (V /W).$$ More geometrically, we want to choose for each affine subspace $v + W$ a unique representative and so we choose some subspace $W'$ that is transversal to $W$ and send $v + W$ to the unique intersection point $(v + W) \cap W'$. To get a more canonical choice for $W'$, we can endow $V$ with an inner product and take $W' = W^{\perp}$. Geometrically, this amounts to sending each affine subspace $v + W$ to the unique point in $v + W$ that is the closest to the origin (with respect to the distance coming from the inner product).
  • Let $V$ be a finite dimensional vector space and $d \colon V \rightarrow V$ be an operator such that $d^2 = 0$. We can form the cohomology $\ker(d) / \mathrm{im}(d)$ and want to construct an operator $S \colon V \rightarrow V$ such that $$ \ker(S) \hookrightarrow \ker(d) \xrightarrow{\pi} \ker(d) / \mathrm{im}(d)$$ is an isomorphism. Endowing $V$ with an inner product, we can take the complement of $\mathrm{im}(d)$ inside $\ker d$ which is $\mathrm{im}(d)^{\perp} \cap \ker d = \mathrm{ker} (d^{*} ) \cap (\ker d)$ and so any operator $S$ with $\ker(S) = \ker(d) \cap \ker(d^{*})$ will do the trick.
    • One choice is to take $S = d + d^{*}$. This is a self-adjoint operator and clearly $\ker(d) \cap \ker(d^{*}) \subseteq \ker (S)$. But if $Sv = 0$ then $$ 0 = \left< Sv, Sv \right> = \left< dv, dv \right> + \left< dv, d^{*} v \right> + \left< d^{*} v, dv \right> + \left< d^{*}v, d^{*}v \right> = \left< dv, dv \right> + \left< v, (d^{*})^2v \right> + \left< v, d^2v \right> + \left< d^{*} v, d^{*} v \right> = \left< dv, dv \right> + \left< d^{*} v, d^{*} v \right>$$ so we see that $dv = 0$ and $d^{*}v = 0$.
    • Another choice is to take $L = S^2 = (d + d^{*})^2 = dd^{*} + d^{*}d$. The operator $L$ is also self-adjoint and since $S$ is self-adoint we have $$\ker(L) = \ker(S^2) = \ker(S) = \ker(d) \cap \ker(d^{*}).$$
  • If $$0 \rightarrow V^{0} \xrightarrow{d} V^{1} \xrightarrow{d} \ldots \xrightarrow{d} V^{n} \rightarrow 0$$ is a cochain complex of finite dimensional vector spaces, endow each $V^{i}$ with an inner-product and consider $V = \oplus V^i$ together with the direct sum inner-product. The discussion above proves that the inclusion $\ker(L) \hookrightarrow V$ is a quasi-isomorphism. Note that while $S$ is not a chain map (as $S(V^{i}) \subseteq V^{i-1} \oplus V^{i+1}$), the operator $L$ is a chain map of complexes.

Now, all of the discussion above can be repeated for the chain complex $\mathcal{A}^{*}(M)$. Unfortunately, $\mathcal{A}^{*}(M)$ is not finite dimensional. If $\mathcal{A}^{*}(M)$ together with the relevant inner-product would form a Hilbert space and $d, d^{*}$ were bounded operators, then the arguments above would still work. Unfourtunately, $\mathcal{A}^{*}(M)$ is not complete and the operators are not bounded. The role of the heavy analysis in the proof is to show that the arguments above essentially work even for the more problematic case. BTW, in this context, the first order operator $S$ (which serves as a square root of the Laplacian) is known as a Dirac operator.

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  • $\begingroup$ Thanks! That really cleared things up! $\endgroup$ – Saal Hardali Nov 29 '15 at 21:26

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