Is there a fibration whose base, total and fibre is a Moore space?

Question

I want to find a example which base, total of fiber spaces is a Moore space.

I don't want trivial fibration.

For example the projection $p : M(G,n) \times M(H,n) \to M(G,n)$ and path fibration $\Omega M \to PM \to M$ where $M=M(G,n)$.

Answer 1

Spheres are Moore spaces, and there are the Hopf fibrations $S^0 \to S^1 \to S^1$, $S^1 \to S^3 \to S^2$, $S^3 \to S^7 \to S^4$, and $S^7 \to S^{15} \to S^8$.

I'm not so sure that the loop space of a Moore space is a Moore space, for example $\Omega S^3$ has non-trivial homology in all dimensions. Nor am I sure that the product of Moore spaces is a Moore space, for example $S^1 \times S^1$ has homology in degrees $1$ and $2$. It seems to me like this sort of thing should be quite rare.

Answer 2

The Serre spectral sequence seems like a useful approach for exploring this. Roughly speaking, given a fibration $F \to E \to B$, we make a table of $H^p(B; H^q(F))$ for all $p$ and $q$, and then there's an iterative process by which groups on any diagonal $p + q = l$ can cancel (partially or totally) with groups on the adjacent diagonal $p + q = l - 1$. At the end, each diagonal $p + q = l$ lists a set of composition factors for $H^l(E)$, under fairly mild conditions. (Notably we want $\pi_1(B)$ to act trivially on the higher homology groups, which is true in this setting.)

This is well suited to the problem: as we're talking about Moore spaces, the table is quite sparse. Since we ask for the total space to be a Moore space, this imposes the condition that only one diagonal (other than $l=0$) can contain a nonzero group at the end of the process.

Anyway, our starting table looks like

  q
  ^
r | H   ...   G⊗H  Tor(G,H)
  | .          .
  | .          .
0 | Z   ...    G
  -------------------------> p
    0          s     s+1

Groups on adjacent diagonals don't just get cancelled arbitrarily, though, but by certain maps ("differentials") which lead $2+n$ places left and $1+n$ places up ($n \geq 0$). So in fact the only groups that could possibly cancel in this picture are $G$ and $H$, and only if they're isomorphic and $r + 1 = s$. Recall that we're aiming to ensure that only one diagonal has anything nonzero in the end; so now one of $G \otimes H$ and $\mathrm{Tor}(G,H)$ had better be zero. But we've already found that $G \cong H$, and now this is only possible when $G$ is free. So the only such fibrations are of the form: $$ M(\mathbb{Z}^a, r) \to M(\mathbb{Z}^{a^2}, 2r+1) \to M(\mathbb{Z}^a, r+1). \qquad(*)$$ The sphere fibrations that Justin lists are all of this form.

So do all fibrations of this form exist? Well, take any map $M(\mathbb{Z}^{a^2}, 2r+1) \to M(\mathbb{Z}^a, r+1)$. Any map factors as a homotopy equivalence and a fibration, so we can assume that we've chosen a fibration (so long as we're not attached to a particular model for Moore spaces). We can now "reverse-engineer" the spectral sequence, going from our knowledge of $H_*(E)$ and $H_*(B)$ to deduce $H_*(F)$, and indeed we find that $F$ is $M(\mathbb{Z}^a, r)$. So a fibration of form $(*)$ exists, for all $a$ and $r$.

edit: It occurs to me that the coverings of the circle that Olivier mentioned are valid examples not of this form, and that's because the Serre SS assumes that the spaces involved are connected. Unless I've made a mistake, all fibrations involving three connected Moore spaces should take the form above.