The question came up while trying to shorten a paper I'm writing into submission-ready length.

Let $M$ be a compact simply connected manifold. By defininition, the rational homotopy groups of $M$ are the $\mathbb{Q}$-vector spaces $\pi_i(M)\otimes \mathbb{Q}$. (Since $\pi_i(M)$ is abelian for $i\geq 2$ and $\pi_1(M) = 0$, this definition makes sense). Such things are studied in rational homotopy theory, but I'm far from an expert. (Hence, I can't really gauge the difficulty of it.)

Here is my question:

Suppose $M^n$ is a smooth compact simply connected manifold and that the rational homotopy groups of $M$ are abstractly isomorphic to either the rational homotopy groups of $S^{n}$ or of $\mathbb{C}P^{n/2}$. Is the cohomology ring $H^\ast (M;\mathbb{Q})$ also abstractly isomorphic to the rational cohomology ring of $S^n$ or $\mathbb{C}P^{n/2}$?

A proof would be fine, but I'd also really like a reference.

I think it's true because if instead of working rationally, we just work with regular homotopy groups and $\mathbb{Z}$ coefficients in cohomology, then much more is true - $M$ must be homotopy equivalent to either $S^n$ or $\mathbb{C}P^{n/2}$. (I can fill in these details if anyone wishes.)

One the other hand, it is entirely possible for the homotopy groups of two simply connected manifolds to match and yet for the rational cohomology rings to be non-isomorphic. (For example, $\mathbb{C}P^2 \#\mathbb{C}P^2$ and $\mathbb{C}P^2 \# -\mathbb{C}P^2$). Also, there are plenty of spaces (e.g., the homogeneous space $SU(3)/SO(3)$) which has the same rational homotopy groups as $S^n$, yet is not homotopy equivalent to it (but the rational cohomology ring agrees with that of $S^n$ in all examples I know of.)


  • $\begingroup$ How do you actually compute the homotopy groups of $\Bbb{CP}^2 \# \pm\Bbb{CP}^2$ in degree greater than 2? Does this calculation extend to $\#_m \Bbb{CP}^2 \#_n -\Bbb{CP}^2$? $\endgroup$ – user98602 Jun 3 '15 at 18:26
  • $\begingroup$ @MikeMiller: Both spaces $B_{\pm} = \mathbb{C}P^2\sharp \pm \mathbb{C}P^2$ are the base space of a torus bundle $T^2\rightarrow S^3\times S^3\rightarrow B_{\pm}$. It follows from the LES homotopy sequence that, other than the fact that $\pi_2(B_{\pm}) = \mathbb{Z}^2$, $\pi_k(B_{\pm}) \cong \pi_k(S^3\times S^3)$. More generally, $\mathbb{C}P^n\sharp \pm \mathbb{C}P^n$ is the base space of a $T^2$ bundle with total space $S^3\times S^{2n-1}$, so the homotopy groups don't depend on the choice of $\pm$. $\endgroup$ – Jason DeVito Jun 4 '15 at 1:58
  • $\begingroup$ On the other hand, I have no idea what happens when you connect sum more than 2 pieces. I do know that such manifolds $M$ are rationally hyperbolic, which implies that the function $f(n) = \dim \bigoplus_{k= 0}^n \pi_k(M)\otimes\mathbb{Q}$ grows exponentially. A reference for the fact that two summands fit into such a bundle is Totaro's paper "Cheeger manifolds and the classification of biquotients". A reference for the case of more than two summands is combining Felix-Halperin-Thomas's Rational homotopy theory book (the section on rational ellipticity).... $\endgroup$ – Jason DeVito Jun 4 '15 at 2:01
  • $\begingroup$ together with a paper of mine in Diff. Geo. and Appl. about classification of 4 and 5-d biquotients which shows that the only rationally elliptic 4 manifolds are those with second betti number at most $2$. (I'm sure this was well known before my paper, but I don't remember a different reference off hand.) $\endgroup$ – Jason DeVito Jun 4 '15 at 2:04
  • $\begingroup$ Great, thanks! The FHT book seems like a bit of a tome, but probably one worth diving into. $\endgroup$ – user98602 Jun 4 '15 at 2:16

The answer is yes.

Let's start with $S^n$, with $n$ even (the $n$ odd case is much simpler). The idea is to use rational homotopy theory. We have $\pi_n\otimes\mathbb Q=\mathbb Q$, $\pi_{2n-1}\otimes\mathbb Q=\mathbb Q$, and the rest is $0$. The Sullivan minimal model of your manifold is thus $A=\mathbb Q[x,y]$ with $\deg x=n$ and $\deg y=2n-1$ ($y$ has odd degree, so it's an anti-commutative variable; a better notation might be $\bigwedge(x,y)$). As for the differential on $A$: $dx=0$, since there is nothing non-linear of degree $n+1$ in $A$, and $dy=x^2$ (why? we can't have $dy=0$: that would mean $d=0$, so the cohomology would be $A$, i.e. non-zero in arbitrarily large dimension. So $dy=cx^2$ for some $c\neq0$, and we just redefine $y$ to $y/c$). $A$ is thus fully determined, and so is thus the rational cohomology of $M$, since it is the cohomology of $A$.

Let's try $\mathbb CP^n$. Its minimal Sullivan model is $\bigwedge(x,y)$ with $\deg x=2$, $\deg y=2n+1$, and the differential is $dx=0$, $dy=x^{n+1}$. The question is now whether the same graded algebra, but with a different differential, can be the minimal Sullivan model of a manifold. But $dx=0$ is necessary, as there is nothing of degree $3$, and again $dy=cx^{n+1}$ for some $c$ and we can check $c\neq 0$ and redefine $y$ to $y/c$. So the differential is the only possible, and thus again yes, the rational cohomology (even the rational homotopy type) must be the same.

For completeness: if $n$ is odd then the minimal model of $S^n$ is $\bigwedge(x)$ with $\deg x=n$ and $d=0$. There is no way to change $d$ to something non-zero.

So in the end: the reason is that we can't change the differential on the minimal models of $S^n$ and $\mathbb CP^n$ is such a way that they could still be minimal models of manifolds.

As for references, I can recommend the book "Algebraic models in geometry" as an intro to rational homotopy theory. I don't know a reference for your question (it's simple enough so I could solve it).

  • $\begingroup$ Thanks for the answer. This does seem simple enough that instead of black-boxing it, I should really understand it. I need to run through a text book to see how to recognize a DGA is the Sullivan model for a space $X$, but then the rest, as you showed, is really easy to verify. $\endgroup$ – Jason DeVito Oct 2 '13 at 17:14

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