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From what I understand $\pi_n(S^2\vee S^2)\otimes\mathbb{Q}\neq 0$ for $n\geq 2$. My question is:

Is there a "hands-on" proof of this fact using differential forms?

I am sure I will receive answers like: that is Hilton's theorem or use Sullivan's minimal model or check the section in Bott and Tu about the rational homotopy theory.

However, all these answers are useless for me because I am an analyst and not topologist and in order to use this fact in my research I need a straightforward construction that I could use to get integral estimates of forms.

Rational homotopy theory of Sullivan is build on differential forms and you can represent elements in rational homotopy using the Hopf-Novokov integral as explained in:

Hardt, Robert; Rivière, Tristan Connecting rational homotopy type singularities. Acta Math. 200 (2008), no. 1, 15–83.

The points is that I do not really understand what they do and I do not even know if their machinery can be used to answer my question. Thus I am asking if there is a simple way to prove the result I am quoting using integration of differential forms.

To be more precise:

One can use differential forms to prove that $\pi_{4n-1}(S^{2n})\neq 0$ and that can be explained on a couple of pages with all details. Is there a similar proof of the fact I mentioned?

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    $\begingroup$ I am having trouble understanding what an answer could possibly look like. For instance, what form could a proof that $\pi_{1000} S^2 \vee S^2 \neq 0$ "via differential forms" take? What kind of things are you trying to estimate? I doubt I can answer your question but this might make it easier for me to appreciate. $\endgroup$
    – mme
    Jun 15, 2021 at 22:48
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    $\begingroup$ @mme Rational homotopy theory of Sullivan is build on differential forms and you can represent elements in rational homotopy using the Hopf-Novokov integral explained in Hardt, Robert; Rivière, Tristan Connecting rational homotopy type singularities. Acta Math. 200 (2008), no. 1, 15–83. The points is that I do not really understand what they do so I am asking if there is a simple way to explain it in that particular case. $\endgroup$ Jun 15, 2021 at 22:55
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    $\begingroup$ Differential forms on $S^2\vee S^2$? That's not a manifold, so what would that be? $\endgroup$ Jun 16, 2021 at 14:08
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    $\begingroup$ @FernandoMuro: This is an amazing idea of Sullivan. If $X$ is any simplicial complex, then you can define $\Omega^n(X)$ to be all $\omega$ where $\omega$ is an assignment of a polynomial differential $n$-form with rational coefficients on each simplex of $X$ (viewed as a subset of Euclidean space in the usual way, so "polynomial with rational coordinates" makes sense) that agree when you pass to faces. The usual $d$ operator works, making $\Omega^{\bullet}(X)$ into a cochain complex. A $\endgroup$
    – Sarah
    Jun 16, 2021 at 15:00
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    $\begingroup$ theorem of Sullivan shows that this actually compute the $\mathbb{Q}$-cohomology of $X$. In fact, $\Omega^{\bullet}(X)$ also has a ring structure (cup product), and is thus a differential graded algebra. Sullivan showed that all the information about the rational homotopy type of $X$ is encoded in it. $\endgroup$
    – Sarah
    Jun 16, 2021 at 15:00

3 Answers 3

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Piotr, I think the question you really want to ask is:

Given a map $f:S^n \to S^2 \vee S^2$, how do I prove that it is homotopically nontrivial?

This is the so-called homotopy period problem, and it has several solutions. I'll outline one way.

Every simply connected compact manifold (or cellwise smooth complex) $X$ has a Sullivan minimal model $\mathcal M_X$, which is a free, minimal differential graded algebra:

  • a differential graded algebra is a chain complex with a graded commutative multiplication satisfying the graded Leibniz rule (for example, the differential forms on a space)
  • it is free as a graded-commutative algebra (so for example, the powers of a generator are zero iff the generator is in odd degree)
  • it is minimal if indecomposable elements (linear combinations of generators) are never coboundaries.

The cohomology ring of $\mathcal M_X$ is the de Rham cohomology ring of $X$. Moreover there is an algebra homomorphism $m_X:\mathcal M_X \to \Omega^*X$ which induces an isomorphism on cohomology.

In the case of spheres and wedges of spheres, because the cohomology is concentrated in one dimension, this completely pins down $\mathcal M_X$—there are no choices involved in constructing it. So for example, for $S^n$, you have to start with one $n$-dimensional generator $x$ and $m_{S^n}(x)=\omega$ can be any form that integrates to 1 over $S^n$. Now if $n$ is odd, $x^2=0$ and we're done, $\mathcal M_{S^n}=\mathbb{R}\langle x \rangle$ and $m_{S^n}$ induces an isomorphism on cohomology. If $n$ is even, $x^2$ is nonzero, and so in order to induce an isomorphism on cohomology we have to kill it. That means there's also a $(2n-1)$-dimensional generator $y$ with $dy=x^2$. Now of course $m_{S^n}(x^2)=0$ so we can set $m_{S^n}(y)=0$ as well.

It turns out that the indecomposables in degree $k$ are naturally isomorphic to $\operatorname{Hom}(\pi_k(X),\mathbb{R})$. So if you're willing to assume this, we've proved that $\pi_{4n-1}(S^{2n})$ is nontrivial. Now the question is, how do you detect it, for a map $f:S^3 \to S^2$? Of course you already know the answer to this, but I want to give a slightly different answer.

This $f$ is homotopically trivial if and only if it extends to $D^4$. Now Sullivan tells us that asking if $f$ extends to $D^4$ is the same as asking if the map $f^*m_{S^2}:\mathcal M_{S^2} \to \Omega^*(S^3)$ lifts to $\Omega^*(D^4)$. One such lift would of course be $F^*m_{S^2}$, for a genuine filling $F:D^4 \to S^2$, but there could be many lifts that don't come from such a filling. For example, a lift which comes from a filling will always send the 3-dimensional generator $y$ to $0$, but a general lift need not do that.

How do you construct a lift $\phi$? Well, first you extend the form $f^*m_{S^2}(x)$ to a closed form $\phi(x)$ on $D^4$. There is no obstruction to doing this, by the Poincaré lemma. However, now $\phi(x)^2$ might not be zero, and we need a $\phi(y)$ such that $\phi(y)|_{S^3}=0$ and $d\phi(y)=\phi(x)^2$. You can build such a $\phi(y)$ if and only if $\int_{D^4} \phi(x)^2=0$. This $\int_{D^4} \phi(x)^2$ is the Hopf invariant. (This formula is related to the Whitehead formula via Stokes' theorem.)

Now $\mathcal M_{S^2 \vee S^2}$ is going to be much more complicated. You start by making sure that you have the right cohomology in degree 2, so you introduce two indecomposable generators $x_1$ and $x_2$. But now $\mathbb{R}\langle x_1,x_2 \rangle$ has cohomology in degree $4$ generated by $x_1^2, x_2^2, x_1x_2$, and you have to kill all of it by adding generators in degree $3$. Then that generates more cohomology in degree $5$ and you kill it by adding generators in degree $4$. And you just keep going, so it looks like this: $$\begin{align*}x_1, x_2 &\mid dx_1=dx_2=0 \\ y_{11},y_{12},y_{22} &\mid dy_{ij}=x_ix_j \\ z_1, z_2 &\mid dz_1=x_1y_{12}-x_2y_{11},dz_2=x_2y_{12}-x_1y_{22} \\ \vdots &\mid \vdots\end{align*}$$ (I might have gotten the signs wrong in the last row.)

But if someone gives you a map $f:S^n \to S^2 \vee S^2$, the resulting map $f^*m_{S^2 \vee S^2} \to \Omega^*S^n$ is still very simple: it sends $x_1$ and $x_2$ to the pullbacks of the volume forms on the two spheres, and everything else to zero. Now if you want to know whether your map is (rationally) trivial, you check if it can be extended "formally" to $D^{n+1}$, that is, whether there is a map $\phi:\mathcal M_{S^2 \vee S^2} \to \Omega^*D^{n+1}$ whose restriction to $\Omega^*S^n$ is $f^*m_{S^2 \vee S^2}$. So in the case of $S^4$, it goes like this:

  • You extend $f^*m_{S^2 \vee S^2}x_i$ to $\phi(x_i) \in \Omega^*D^5$. There's no obstruction.
  • You find primitives for $\phi(x_1)^2$, $\phi(x_1)\phi(x_2)$, and $\phi(x_2)^2$. There's no obstruction to doing this, and those are going to be your $\phi(y_{ij})$'s.
  • Now you have 5-forms $\phi(x_1y_{12}-x_2y_{11})$ and $\phi(x_2y_{12}-x_1y_{22})$ which are zero on the boundary $S^4$. The obstructions to finding primitives for these that are zero on the boundary are their integrals over $D^5$. Sullivan's theory says that these will be the same no matter how you chose the $\phi(y_{ij})$.
  • Now there are infinitely many additional generators of $\mathcal M_{S^2 \vee S^2}$, but luckily all their differentials are in degrees $\geq 6$, and therefore automatically go to zero under a map to $\Omega^*D^5$. So if there's no obstruction corresponding to 4-dimensional generators, we can send the generators in higher degrees to zero and be done.

The 5-dimensional obstructions are the equivalent of the Hopf invariant in this situation. Using Stokes' theorem repeatedly, you can also come up with an obstruction that lives entirely in $S^4$, just using primitives without extensions. But that's a bit harder to describe.

If you want to understand the proofs of these statements, I recommend Griffiths and Morgan's book. Section 3 of my paper "Plato's cave and differential forms" also contains a summary.

Addendum: The only things I really used about smooth forms in the above arguments are (1) the fact that they form a graded commutative DGA and (2) the Poincaré lemma. So if you want to think about other types of cochains, you can do that as long as they satisfy those two properties. In my work I have sometimes used flat forms instead of smooth ones.

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    $\begingroup$ Thank you Fedya. It will take me a while to digest your answer. I certainly want to learn more about Sullivan's theory. $\endgroup$ Jun 16, 2021 at 17:26
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    $\begingroup$ Thanks for the amazing answer. I almost didn't know anything about rational homotopy theory and I think I learnt a lot in this concise introduction :) $\endgroup$ Jul 21, 2021 at 9:54
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Here is an example of how to detect elements of $\pi_5 (S^2 \vee S^2)$.

Let $\omega_i : S^2 \to S^2 \vee S^2$ be the inclusion of the $i$-th wedge summand. Then how do we detect $[[[\omega_1,\omega_2],\omega_2],\omega_2] \in \pi_5 (S^2 \vee S^2)$? where here the bracket is the Whitehead bracket notation.

Given a map $f : S^5 \to S^2 \vee S^2$, take the pre-image of a point in either wedge summand (away from the wedge point). Generically this is a pair of disjoint framed $3$-manifolds in $S^5$. These framed manifolds are null-cobordant (for a generic map they would be torsion, but for our Whitehead bracket they are null). That said, even though they are null, they may not be disjointly null. Call the first manifold $M_1$ and the 2nd $M_2$.

So we have framed $4$-manifolds $W_1$ and $W_2$ with $\partial W_i = M_i$. If we use the notation $\delta M_i$ to denote a manifold whose boundary is $M_i$, i.e. the symbol means `a choice of cobounding manifold' then we could write $\delta M_i = W_i$. The generalized linking number we want to use to detect the non-triviality of the homotopy-class $[[[\omega_1, \omega_2], \omega_2], \omega_2]$ is:

$$\delta(\delta(\delta M_1 \cap M_2) \cap M_2) \cap M_2$$

i.e. we are computing this as a signed intersection number. Since $M_i$ is $3$ dimensional $\delta M_i$ is $4$-dimensional, so $\delta M_1 \cap M_2$ is $2$-dimensional, etc, one follows through the dimension argument and one sees the above intersection is a framed $0$-dimensional manifold, which up to cobordism is an integer.

I think if you were to put this into the language of differential forms, given a map $f : S^5 \to S^2 \vee S^2$, let $w_1, w_2$ be the area forms of the two $S^2$ wedge summands, then one computes

$$\int_{S^5} \delta \left( \delta \left( \left( \delta f^* w_1 \right) \wedge f^* w_2 \right) \wedge f^* w_2\right) \wedge f^* w_2$$

Where here $\delta$ denotes the dual to the `cobounding' operation, i.e. one is applying the Poincar'e Lemma to find an exterior anti-derivative.

So this is one integral that detects a very specific homotopy-class $S^5 \to S^2 \vee S^2$.

edit:

Regarding how to verify these statements, as Dev mentions, there are complete proofs in his paper. I think verifying the above statement about the higher linking numbers, expressed in terms of intersections of cobounding manifolds; the geometry behind these statements basically boil down to understanding the standard surgery decompositions of the sphere, i.e. $S^{n-1} \simeq S^{i-1} \times D^j \cup D^i \times S^{j-1}$ where $i+j=n$.

I think the differential form computations proceed similarly, using these standard surgery decompositions -- as the Whitehead products are basically defined in terms of these canonical surgery decompositions of spheres. The advantage to the intersection-theory approach is you can easily avoid technical issues in regard to the wedge point. Off the top of my head the forms approach seems a bit more technical to me, in that you need to use explicit collapse maps $D^n \to S^n$, in order to pull-back the volume forms. Other than that, the computation should be fairly mechanical.

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Ryan's answer generalizes. I prefer $\iota_1, \iota_2$ for the inclusions of wedge summands, and then $\omega_1, \omega_2$ for forms which generate cohomology supported on each of the wedge summands. Then I claim that a numerical homotopy invariant which evaluates non-trivially on $[[\cdots[\iota_1,\iota_2],\iota_2],\cdots],\iota_2]$ is

$f \mapsto \int_{S^n} d^{-1} \left( \cdots d^{-1} \left( d^{-1} \left(d^{-1} f^*(\omega_1) \wedge f^*(\omega_2)) \wedge f^*(\omega_2) \right) \wedge f^*(\omega_2) \right) \cdots \right) \wedge f^*(\omega_2),$ where each $d^{-1}$ denotes a choice of some form to cobound what is in each case a closed and thus exact form of degree less than $n$ on $S^n$. (As Sullivan and others point out, such choices can be made universally through a constructive proof of the Poincare Lemma.) Since there is such a Whitehead product in every degree greater than or equal to two, this claim answers the Question.

The proof that this integral is homotopy invariant is elementary, along the lines which Ryan gives in the example of his answer. There is also probably a more direct "hands-on" proof that this integral evaluates to one on the Whitehead product given, but I am going to cite my work with Ben Walter, namely https://arxiv.org/abs/0809.5084 (not the paper Ryan linked to, though I appreciate his seeing the relevance of our ideas). This integral is a case of a Hopf invariant, which is the main object of study of the paper. This integral is the Hopf invariant associated to the element of the bar construction $|\omega_1|\omega_2| \omega_2| \cdots |\omega_2|$. To see it evaluates as claimed I prefer to use the bracket-cobracket compatibility theorem, namely Theorem 1.12 of the paper, from which this is quick inductive combinatorics. While the present MO answer is thus not self-contained, note that Theorem 1.12 occurs just on page 7, and I would call its proof "hands-on".

The cited paper was written to solve exactly these types of questions. We give an algorithm called weight reduction which can be used to associate integral(s - non-unique) to any cycle in the (Lie coalgebraic) bar construction on the cochains (forms), which are homotopy periods. This association defines an isomorphism between the homology of this bar construction and the linear dual of homotopy. Though Sullivan in his seminal paper pointed to how one can produce complete rational homotopy periods (I wish we could call this "rational cohomotopy") through minimal models, what Ben and I find is that the Quillen functor approach is, perhaps surprisingly, more readily tied to geometry. The general principle which follows from our work is that one can detect all rational homotopy through numerical invariants which are "higher linking with correction," governed by the Lie coalgebraic bar construction. Thus, geometry, combinatorics, algebra and functorial formalism all work together as well as one could hope for in giving a sharp resolution to the homotopy periods question in the simply connected setting. (The non-simply connected case is ongoing research that I am having PhD students work on.) And as Ryan says, I have recorded lectures on this subject. The best place to see a summary of and links to the lectures is here https://pages.uoregon.edu/dps/GeometricAlgebraicTopology/ The first three lectures cover these Hopf invariants.

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