6

Suppose $X$ is a topological manifold such that $\pi_1(X)\simeq G\times H$ for some nontrivial groups $G$ and $H$. Then can I always find topological manifolds $Y$ and $Z$ such that $X\cong Y\times Z$ and $\pi_1(Y)\simeq G$ and $\pi_1(Z)\simeq H$?

I thought the statement is strange enough to come up with a counterexample easily but I failed. The main reason is I don't know how to show a space cannot be decomposed as a product of two spaces or at least some "usual" argument (but I think one can use some fancy machinery in algebraic topology). I think many people believe this statement is false. But can you provide any counterexample?

Arctic Char
  • 16,007
one potato two potato
  • 4,981
  • 9
    Any finitely presented group is the fundamental group of a closed $4$-manifold. Consider $G=\mathbb Z^5$. If you could always decompose a group product as a product of manifolds, the corresponding $4$-manifold would have to decompose as a product of $5$ manifolds with fundamental group $\mathbb Z$, but this is impossible by dimension reasons. – Cheerful Parsnip Jan 11 '24 at 06:02
  • 1
    @CheerfulParsnip: That looks like an answer to me? – joriki Jan 11 '24 at 06:14
  • 5
    @CheerfulParsnip This should be an answer. It might be helpful to link to this post or some other as a reference. I am not enough of a topologist to have known this wonderful fact. – Ted Shifrin Jan 11 '24 at 06:15
  • Is "$\cong$" here supposed to mean homeomorphic or homotopy equivalent? – freakish Jan 11 '24 at 06:24
  • @freakish homeomorphic – one potato two potato Jan 11 '24 at 06:25
  • @onepotatotwopotato it will also be interesting to ask the same question up to homotopy equivalence – Kenta S Jan 11 '24 at 06:27
  • 1
    @CheerfulParsnip This is an answer to the question, not just a comment. Comments should only be used to clarify, not answer the question. See https://math.meta.stackexchange.com/a/36082/42781 for some reasons that it is best to post actual answers, not just comments. – John Palmieri Jan 11 '24 at 07:21
  • 4
    For closed aspherical manifolds, this is an open problem. (True in dimensions 2, 3.) – Moishe Kohan Jan 11 '24 at 16:48

3 Answers3

10

Any finitely presented group is the fundamental group of a closed 4-manifold. (See this question.) Consider $=\mathbb Z^5$. If you could always decompose a group product as a product of manifolds, the corresponding 4-manifold would have to decompose as a product of 5 manifolds with fundamental group $\mathbb Z$, but this is impossible by dimension reasons.

Cheerful Parsnip
  • 27,278
4

Another neat example worth mentioning is the $3$-dimensional Lens space $X=L(6;1)$ (or $L(n;m)$ for any coprime integers $n,m$ s.t. $n$ is not a prime power). Its fundamental group $\pi_1(X)=\mathbb{Z}/6\mathbb{Z}=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$ decomposes non-trivially, but this decomposition is not induced by any product decomposition of $X$ as a manifold. Indeed, one factor in such a decomposition would have to have dimension $1$, but the only compact $1$-manifold is $S^1$ whose fundamental group is not finite cyclic. (In fact, one can show a slightly stronger statement: $X$ does not non-trivially decompose as a product of CW-complexes.)

Let me also record that this example is minimal with respect to dimension. The torus has fundamental group $\mathbb{Z}\times\mathbb{Z}$ whose decomposition is induced by its own decomposition $T^2=S^1\times S^1$ and any other surface (not necessarily compact) has indecomposable fundamental group (this is not so easy), so there are no examples in dimension $2$.

Thorgott
  • 11,682
1

All manifolds will be connected. Here is a quite large class of non-compact counter-examples that is hopefully intuitive: Let $M,N$ be two compact, oriented manifolds s.t. $H_1(M;\mathbb{Z})\neq0$ (in particular, $\pi_1(M)\neq1$), $\pi_1(N)\neq1$ and $n=\dim(M)+\dim(N)\ge3$. Let $X=M\times N-\{\ast\}$ be the non-compact $n$-manifold obtained by removing any point from $M\times N$. The S-vK theorem and $n\ge3$ implies that $\pi_1(X)=\pi_1(M\times N)=\pi_1(M)\times\pi_1(N)$ decomposes non-trivially. However, I claim that $X$ does not decompose non-trivially as a product of manifolds.

Proof Sketch: Assume $X=Y\times Z$ with $Y,Z$ of dimensions $0<r,s<n$ resp. That $X$ is non-compact implies that at least one of $Y$ and $Z$ is non-compact.

Assume $Y$ and $Z$ are both non-compact. There is a field $F$ s.t. $H_1(M;F)\neq0$, hence $H_{\dim(M)-1}(M;F)\neq0$ by Poincaré duality and the UCT. The Künneth theorem then implies that $H_{n-1}(M\times N;F)\neq0$ since it contains this group as a summand. However, orientability implies that $H_{n-1}(X;F)=H_{n-1}(M\times N;F)$. Now, applying the Künneth theorem to $X=Y\times Z$ yields that $H_{n-1}(X;F)=\bigoplus_{i+j=n-1}H_i(Y;F)\otimes_FH_j(Z;F)=0$. Indeed, $i+j=n-1$ implies $i\ge r$ or $j\ge s$ and this implies the vanishing of $H_i(Y;F)$ resp. $H_j(Z;F)$ by non-compactness. A contradiction.

Assume $Y$ is non-compact and $Z$ is compact. The $k$-th "homotopy group at infinity" $\pi_k^{\infty}(X)$ is, roughly speaking, the inverse limit of the homotopy groups $\pi_k(X\setminus K)$, where $K$ ranges over the compact subsets of $X$. The compact subsets $L\times Z$ with $L\subseteq Y$ compact are cofinal among these, so we obtain $\pi_k^{\infty}(X)=\pi_k^{\infty}(Y)\times\pi_k(Z)$. However, $X$ is locally Euclidean at infinity since $M\times N$ is a manifold, so $\pi_k^{\infty}(X)=\pi_k(S^{n-1})$. Together, it follows that $Z$ is $(n-2)$-connected, but $Z$ is a compact manifold of dimension $<n$, so it altogether has dimension $n-1$. Thus, $Y$ is a non-compact $1$-manifold, i.e. $Y=\mathbb{R}$, but $\mathbb{R}\times Z$ clearly has two ends whereas $X$ only has one. A contradiction.

(I would like to weaken the assumption $H_1(M;\mathbb{Z})\neq0$ to $\pi_1(M)\neq1$, but I have not been able to adapt the argument.)

Thorgott
  • 11,682