Manifold of odd dimension with infinite cyclic homology group in each dimension Yesterday in an exam I was asked if there exists a manifold of odd dimension $2n+1$ such that $H_{i}(X)=\mathbb{Z}$ for each $0\leq i\leq 2n+1$.
At first instance I thought that it was not the case such manifold could exist. But taking $n=0$ we would take $S^{1}$ as an odd dimension manifold with $H_{0}(S^{1})$ and $H_{1}(S^{1})$ isomorphic to $\mathbb{Z}$. I don't know if my teacher wanted the case for $n>0$ which I think is a more interesting case.
Does anyone know such manifold?
 A: Yes, such a manifold exists for every $n$.
As Eduardo Longa's comment and Thomas Preu's answer suggest, using products of spheres is a good idea. The missing ingredient is taking the connected sum of such manifolds.
First note that $S^{2k}\times S^{2n+1-2k}$ has homology $\mathbb{Z}$ in degrees $0$, $2k$, $2n+1-2k$, and $2n+1$ - these degrees are distinct if $k \neq 0$. For closed orientable manifolds $M$ and $N$ of the same dimension $d$, we have $\widetilde{H}_i(M\#N) \cong \widetilde{H}_i(M)\oplus\widetilde{H}_i(N)$ for $i < d$ and $H_d(M\# N) \cong \mathbb{Z}$. It follows that $X = {\large\#}_{k=1}^{n}(S^{2k}\times S^{2n + 1 - 2k})$ is a manifold with the desired homology.

One can ask the same question for even dimensions: is there a manifold $X$ of dimension $2n$ with $H_i(X) \cong \mathbb{Z}$ for $0 \leq i \leq 2n$? Since $H_{2n}(X) \cong \mathbb{Z}$, we see that if such a manifold exists, it must be connected, closed, and orientable.
The classification of surfaces shows that the answer is no for $n = 1$. More generally, no such manifold exists for $n$ odd. To see this, note that $X$ is closed and orientable, so we obtain an intersection pairing on $H^n(X)$ (after choosing an orientation). Since $n$ is odd, the pairing is skew-symmetric, so $H^n(X) \cong H_n(X)$ has even rank by the classification of skew-symmetric unimodular forms. In particular, we cannot have $H_n(X) \cong \mathbb{Z}$.
For $n$ even, such manifolds exist, e.g. $X = \mathbb{CP}^n\#{\large\#}_{i=1}^{n/2}(S^{2i-1}\times S^{2n-(2i-1)})$.
A: For the $n$-sphere $S^n$ we have $H_i(S^n)=\begin{cases}\mathbb{Z},&\text{ if }i\in\{0,n\}\\ \{0\},&\text{ otherwise.}\end{cases}$.
By the Künneth formula you see, e.g.:
$$
X^3=S^1\times S^2,\quad 
X^7=S^1\times S^2\times S^4,\quad 
X^{15}=S^1\times S^2\times S^4\times S^8, \quad\ldots 
$$
have the desired property in odd dimension $2^k-1$.
This is not quite what you want, as you want any odd dimension, but it provides at least a partial answer with an infinite family.
