# Manifolds with non-vanishing vector field and vast homology

Let $n \ge 3$. Is there n-fold $M^n$ with both $\chi(M)=0$ and $\dim H_*(M,\mathbb{R}) \ge$ given number?

• How about $S^1 \times \dotsb \times S^1$? – user45861 Jun 21 '14 at 16:56
• Yes, take for example a disjoint union of infinitely many $S^1\times\mathbb{R}^{n-1}$. I believe it is true even if you wanted the manifold to be connected. – Peter Franek Jun 21 '14 at 17:07
• Construct an $n-1$-manifold with large homology in any way you like, then take the product with $S^1$. – Qiaochu Yuan Jun 21 '14 at 17:07

## 1 Answer

The product of two manifolds is a manifold. The Euler characteristic is multiplicative, so $\chi(M \times N) = \chi(M) \cdot \chi(N)$. Since $\chi(S^1) = 0$, it's sufficient to take the product of $S^1$ with an $(n-1)$-manifold that satisfies the homology dimension condition. For the latter, take the connected sum of enough copies of $\underbrace{S^1 \times \cdots \times S^1}_{(n-1) \text{ times}}$.

• This, however, doesn't answer the question if he has a fixed $n$ and a given large number. I'm not sure what you mean by "connected sum". – Peter Franek Jun 21 '14 at 17:23
• – Qiaochu Yuan Jun 21 '14 at 17:30
• @Peter The construction results in an $n$-manifold. Furthermore, taking connected sums allows you to add as many dimensions to homology as you want without affecting the dimension of the manifold. Look it up. – Ayman Hourieh Jun 21 '14 at 17:31
• @Peter $n\ge 3$ so we never take the connected sum of circles. – Ayman Hourieh Jun 21 '14 at 17:34
• Yes, I understand, sorry for the confusion. I misread it and thought you talk about "connected sum of circles". Nice answer, +1. – Peter Franek Jun 21 '14 at 17:35