# How are these algebraic and geometric notions of homotopy of maps between manifolds related?

Let $M$ and $N$ be smooth manifolds, and $f,g: M \to N$ smooth maps. Denote by $(\Omega^\bullet M,\mathrm d_M)$ and $(\Omega^\bullet N, \mathrm d_N)$ the cdgas of de Rham forms in each manifold, and by $f^\ast, g^\ast : \Omega^\bullet N \to \Omega^\bullet M$ the pull-back of differential forms along each map. Note that $\Omega^\bullet$ is a fully-faithful functor from Manifolds to CDGAs.

I have been brought up with two not-obviously-the-same notions of "homotopy" between maps $f,g$:

A geometric homotopy between $f,g$ is a smooth map $H : M \times [0,1] \to N$ such that $H(-,0) = f$ and $H(-,1) = g$.

An algebraic homotopy between $f,g$ is a map $\eta: \Omega^\bullet N \to \Omega^{\bullet - 1} M$ of graded vector spaces such that $f^\ast - g^\ast = \eta \mathrm d_N + \mathrm d_M \eta$.

I believe that the following is true. Any geometric homotopy gives rise to an algebraic homotopy, and two geometric homotopies are homotopic iff the corresponding algebraic homotopies are homologous homotopic. Not every algebraic homotopy comes from a geometric homotopy; rather, it should be required to satisfy some (directly-checkable) condition that says roughly that it's an "antidifferential operator".

Unfortunately, I have been unable to really convince myself of either of the above beliefs. Probably this is textbook material, and so maybe my question is to be pointed to the correct textbook. But really my question is:

How, explicitly, are the above notions of homotopy between maps related? What extra conditions (if any?) should be put on an algebraic homotopy in order for it to be "geometric"?

It is somewhat embarrassing not to know the sharp relationship between the above concepts, but this is one of the many parts of mathematics that I have picked up largely from conversations and working on the examples that come from particular research questions, and not from ever formally learning such material.

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The claim "any geometric homotopy gives rise to an algebraic homotopy" is true. An explicit construction can be found in Lemma 20 here: thehcmr.org/issue1_2/poincare_lemma.pdf The obvious thing to try to do to get geometric homotopies from algebraic ones is to try to write down a vector field on $N$ and flow $M$ along it, but this must be hopeless if $M$ is not compact, right? So perhaps you want compactness as a hypothesis. –  Daniel Litt Jul 28 '11 at 19:00
If $M$ and $N$ are compact and simply connected, it seems like Hurewicz should imply that algebraic homotopy implies geometric homotopy, no? I'm not a topologist, but that may be the place to start. –  Stephen Jul 28 '11 at 19:21
I'm not sure I understand the meaning of "two geometric homotopies are homotopic iff the corresponding algebraic homotopies are homologous". To me, "homologous" for maps of chain complexes means that they induce the same map on homology. But these algebraic homotopies certainly aren't chain maps. –  Mark Grant Jul 28 '11 at 19:41
Oh, incidentally, @Daniel's linked article is quite good. So that answers one direction of the first question (in a way that I think I should have thought of). –  Theo Johnson-Freyd Jul 28 '11 at 23:23
er... that didn't typeset like I wanted. }:( –  some guy on the street Jul 29 '11 at 23:03

There is a simple way to understand the implication "geometric implies algebraic homotopy" if you remember that $\Omega^*(M \times I)$ is the (projective) tensor product of $\Omega^*(M)$ and $\Omega^*(I)$ of chain-complexes. Take the chain map $Int:\Omega^*(I) \to C^*(I)$ given by integration, where $C^*(I)$ is the simplicial cochain complex of the 1-simplex $I$ (of total dimension 3).
Then a geometric homotopy, composed with $Int$ gives a chain map $$\Omega^*(N) \to \Omega^*(M) \otimes C^*(I)$$ Unravelling this map into its 3 components, corresponding to the two 0-simplices and the 1-simplex of $I$, you get a triple $(f^*,g^*,\eta)$ which is precisely an algebraic homotopy.
It's very hard to go back since an algebraic homotopy is very weak information, it exists if and only if $f$ and $g$ induce the same map on de Rham cohomology (since these are chain complexes over a field).
One thing to do is to pull back the dga structure on $\Omega^*(I)$ to an $A_\infty$-structure on $C^*(I)$ via $Int$ (and a choice of a homotopy inverse). Instead of just an algebraic homotopy, you could then require an $A_\infty$ map $$\Omega^*(N) \to \Omega^*(M) \otimes C^*(I)$$ which extends your given pair $(f^*,g^*)$ on the boundary (note that by construction, an example comes from a geometric homotopy). If $M$ and $N$ are nilpotent, this should guarantee a homotopy on the "realifications" by rational (or better: real) homotopy theory.