$S^n \to S^m \to B$ bundle: possible? Sphere bundles and bundles over spheres are everywhere and are excellent things to get one's hands dirty with.

(1a) But when can we have a bundle $S^n \to S^m \to B?$  It seems like requiring the total space of a sphere bundle to be a sphere is pretty restrictive.
(1b) Does the answer to (1a) depend on a choice of category (PL, TOP, etc)?

There's an $S^1-$ bundle over $CP^1$ with total space $S^3$, but that's the only example I can find.

(2) Do people know examples other than the one above?

 A: This question was studied in a paper by Browder who proves the following. 
Thereom. Consider any Serre fibration of $F\to S^n\to B$ where $F$, $B$ are connected polyhedra. Then $F$ is homotopy equivalent to $S^1$, $S^3$, or $S^7$. If $F=S^1$, then $B$ is homotopy equivalent to $CP^k$ with $2k+1=n$. If $F=S^7$, then $B$ is homotopy equivalent to $S^8$. 
One expects that if $F= S^3$, then $B$ is homotopic to a quaternionic projective space of a suitable dimension but Browder mentions (if I am reading it correctly) that this is not true, and there are other examples. 
 Caution: there might be something wrong with the above theorem because it rules out existence of $S^7$ Hopf bundles over the octonian projective plane. Is there such a bundle? Hatcher's comment above says there isn't, am I reading it right? Could someone clarify the situation? I care because Browder's theorem was used in some geometric problems involving boundary at infinity of nonnegatively curved manifolds. 
A: The Hopf fibrations give 4 such examples:
1) $S^3$ is an $S^1$ bundle over $\mathbb C \mathbb P^1 \cong S^2$, as you mentioned.
2) $S^7$ is an $S^3$ bundle over $\mathbb H \mathbb P^1 \cong S^4$, the quaternionic projective line.
3) $S^{15}$ is an $S^7$ bundle over $\mathbb O \mathbb P^1 \cong S^8$, the octonionic projective line.
4) $S^0$ is an $S^1$ bundle over $\mathbb R \mathbb P^1 \cong S^1$.
I don't know if there are any others.
A: Here's a sketch. Take a look at the Gysin sequence. This will tell you that $H^i(B)$ and $H^{i+n+1}(B)$ are isomorphic except in two dimensions, which gives you (if $n$ is odd) a restriction on the possible cohomology rings for $B$: truncated polynomial algebras or (polynomial algebra tensor an exterior algebra). The latter cannot happen because such a space would be an infinite CW-complex and cannot be the base space of a fiber bundle with total space $S^m$ (something just homotopy equivalent to $S^m$ would be possible, though.)
Hopf invariant 1 will then tell you that there's also a restriction on $n$. I'm guessing you'll end up with just the mentioned bundles over $RP^n$, $CP^n$, $HP^n$ and $OP^n$ (the latter for $n\leq 2$) and possibly some twisted versions of these.
A: More generally, there are bundles $S^0\to S^n\to \mathbb{R}P^n$ and $S^1\to S^{2n+1}\to\mathbb{C}P^n$ and $S^3\to S^{4n+3}\to\mathbb{H}P^n$.  There is also an "octonionic projective plane" $\mathbb{O}P^2$ but you have to construct it in a nonobvious way as the quotient $F_4/Spin(9)$ of exceptional Lie groups; more obvious constructions do not work because $\mathbb{O}$ is not associative.  This gives a bundle $S^7\to S^{23}\to\mathbb{O}P^2$.  I think it is known that there is no bundle $S^7\to S^{8k+7}\to B$ for $k>2$ but I do not know a proof of that.  In general, if we have $S^n\to S^m\to B$ with $n>0$ and $m>1$ then the homotopy long exact sequence of the fibration shows that $B$ is simply connected, so the Serre spectral sequence $H^{\ast}(B;H^\ast(S^n))\to H^\ast(S^m)$ has untwisted coefficients and we find that $H^\ast(B)=\mathbb{Z}[x]/x^k$ where $|x|=n+1$ and $m=(n+1)k-1$.  Adams's Hopf Invariant One theorem (applied to the attaching map for the $2(n+1)$-cell in $B$) now implies that $n\in\{1,3,7\}$.  If $n=1$ then the element $x$ gives a map $B\to K(\mathbb{Z},2)=\mathbb{C}P^\infty$ and we deduce that $B$ is homotopy equivalent to $\mathbb{C}P^{k-1}$.  I don't immediately see how to deal with the cases $n=3$ or $n=7$ but I suspect that all examples are at least homotopy equivalent to examples that have been mentioned already.  I don't know about homeomorphism or diffeomorphism.
