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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?

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How about the Hopf bundles? Those are the only bundles where the fibre, total space and base are all spheres -- a result, I believe, due to Frank Adams? Experts here will correct me if I'm wrong. –  José Figueroa-O'Farrill Nov 3 '10 at 2:21
    
José's comment came at the same time as my answer. I believe him that they're the only examples. –  Spiro Karigiannis Nov 3 '10 at 2:29
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Here we're allowing the base to be arbitrary. –  Romeo Nov 3 '10 at 2:46
    
The answer to (1b) is that the category does matter. Milnor found seven smooth 7-manifolds, all different in DIFF but all PL homeomorphic to $S^7$, which are $S^3$-bundles over $S^4$. [Jose: is your quotation of Adams precisely correct? Isn't there an infinite sequence of $S^3$-bundles over $S^4$ with Euler class $\pm 1$ but varying $p_1$, and a subsequence for which the total space is $S^7$ even in DIFF?] –  Tim Perutz Nov 3 '10 at 4:28
    
Oops. I didn't read closely enough. Of course, if the base need not be a sphere either, there are the more general Hopf fibrations that Neil describes very well in his answer. –  Spiro Karigiannis Nov 3 '10 at 12:28
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4 Answers

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.

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One can construct $OP^2$ by attaching a 16-cell to $S^8$ via the Hopf bundle map $S^{15}\to S^8$, and this Hopf bundle can be constructed just using the octonions. (Details can be found in Example 4.47 of my algebraic topology book.) So Lie groups aren't needed to construct $OP^2$. If one had a bundle $S^7 \to S^{23} \to OP^2$, couldn't one attach a 24-cell to $OP^2$ via the map $S^{23} \to OP^2$ to construct an $OP^3$ (which doesn't exist)? –  Allen Hatcher Nov 3 '10 at 9:21
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There are exotic $\mathbb{HP}^n$s X such that X is 1-connected and has the same integral cohomology ring as $\mathbb{HP}^n$, but not the same homotopy type. There is also up to homotopy a fibration $S^3\to S^{4n-3}\to X$. This can be found in an article by McGibbon: jstor.org/stable/1998715?seq=3 I don't know if these X can be chosen as manifolds and the fibrations as actual bundles. Probably at least the first part should be accesible to the methods of surgery since we have here a simply connected Poincare duality space. –  Lennart Meier Nov 3 '10 at 9:27
    
Allen is right. I remember reading somewhere (probably the article on Octonions by Baez, that although $\mathbb O \mathbb P^n$ exists for $n=1$ and $n=2$ only, the associated Hopf fibration $S^7 \to S^{8n+7} \to \mathbb O \mathbb P^n$ exists only for $n=1$. –  Spiro Karigiannis Nov 3 '10 at 12:31
    
@Niel: very nice! –  Romeo Nov 4 '10 at 0:43
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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.

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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.

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These are certainly the only examples where all three spaces are spheres. But certainly there are more: all complex projective spaces have circle bundles whose total spaces are spheres. –  José Figueroa-O'Farrill Nov 3 '10 at 2:33
    
What's the map $S^9 \to CP^4$ giving a circle bundle? –  Romeo Nov 3 '10 at 2:36
    
@Spiro: thanks for adding the other 2 classics! –  Romeo Nov 3 '10 at 2:37
    
$CP^4 = (C^5-0)/C^*$, and since $C^5 \simeq R^10$, we can get an isomorphism $(C^5-0)/C^* \simeq S^9/U(1)$. So take $S^9 \subset C^5$ and then quotient by the natural diagonal action of the complex numbers of unit length. –  David Roberts Nov 3 '10 at 2:54
    
that should be $R^{10}$ not $R^10$. $C^*$ is the nonzero complex numbers. –  David Roberts Nov 3 '10 at 2:54
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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.

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