# Homotopy groups of non-contractible manifolds

Motivated by a proof in a differential geometry book and so far my lack of knowledge in algebraic topology I would like to know the following :

Is it possible to have a compact non-contractible manifold $$M$$ with finite fundamental group and trivial homotopy groups $$\pi_k(M)$$ for $$k\geq 2$$?

Any help or reference is appreciated, thanks in advance.

• I suppose you mean to require your manifold to be connected as well. Commented Feb 20, 2021 at 23:02
• If you call your group $G$, then you're looking for the Eilenberg-Maclane space $K(G,1)$. For example, $K(\mathbb{Z}/2\mathbb{Z},1)$ is $\mathbb{R}P^\infty$, which is unfortunately not a finite-dimensional manifold. In fact, what you're looking for can't be achieved by a finite-dimensional manifold, so ana answer would be necessarily a non-compact space. Commented Feb 20, 2021 at 23:12
• "Non-contractible" is superfluous since any contractible space has trivial homotopy groups of all orders. Commented Feb 20, 2021 at 23:14

This is impossible even without assuming compactness (assuming you want $$M$$ to be connected). Suppose you had such a manifold $$M$$. Then $$M$$ would be a $$K(G,1)$$ space for the finite nontrivial group $$G=\pi_1(M)$$. However, any nontrivial finite group has infinite cohomological dimension, so $$M$$ would have nontrivial cohomology (with appropriate coefficients) in infinitely many degrees. This is a contradiction, since if $$M$$ is a manifold of dimension $$n$$ it can't have nontrivial cohomology above degree $$n$$.
This can also be achieved through Miller's theorem which states that the based mapping space $$Map_*(BG,X)$$ is weakly contractible if $$G$$ is a Lie group with a finite number of path components and $$X$$ is a nilpotent $$CW$$ complex with a finite number of cells.
In particular all maps $$BG \rightarrow X$$ are nullhomotopic.
Your $$M$$ is a $$K(\pi_1(M),1)$$ so also a $$B\pi_1(M)$$. Obviously $$G = \pi_1(M)$$ viewed as a discrete Lie group has finite number of path components and $$M$$ also has to be a finite $$CW$$ complex since it's compact. Thus the identity map $$M =BG \rightarrow M$$ is nullhomotopic and $$M$$ is contractible.
• You've not quite quoted Miller's theorem correctly. Note the appearance of the profinite completion in Theorem $2$ of McGibbon's paper. For instance it's known that there are uncountably many homotopy classes of maps $BS^1\rightarrow S^3$. Commented Feb 22, 2021 at 9:09