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?
Best Answer
Piotr, I think the question you really want to ask is:
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:
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:
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.