Spheres fibred by spheres

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Daniel A. Asimov

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Feb 16, 1993, 3:55:07 PM2/16/93
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The following are well-known to researchers in this field:

1. A sphere S^n, where n is of the form n = 2*k - 1, is the total
space of a fibre bundle whose fibre is the circle S^1 (and with base
space a complex projective space).

2. A sphere S^n, where n is of the form n = 4*k - 1, is the total
space of a fibre bundle whose fibre is the sphere S^3 (and with base
space a quaternionic projective space).

3. S^15 is the total space of a bundle whose fibre is S^7
(and whose base space is S^8).

4. If n is of the form n = 8*k - 1, then S^n admits a continuous
field of tangent 7-planes.

(See N. Steenrod, Topology of Fibre Bundles, sections 20 and 27.)
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Question:
Do spheres S^n, where n is of the form n = 8*k - 1, for k > 2,
fibre with fibre = S^7 (over some manifold as base space) ?

E.g., is S^23 the total space of a fibre bundle whose fibre
is S^7 ?


Dan Asimov
Mail Stop T045-1
NASA Ames Research Center
Moffett Field, CA 94035-1000

asi...@nas.nasa.gov
(415) 604-4799

Geoffrey Mess

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Feb 17, 1993, 5:43:18 PM2/17/93
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In article <1993Feb16.2...@nas.nasa.gov> asi...@nas.nasa.gov
(Daniel A. Asimov) writes:

> Question:
> Do spheres S^n, where n is of the form n = 8*k - 1, for k > 2,
> fibre with fibre = S^7 (over some manifold as base space) ?
>
> E.g., is S^23 the total space of a fibre bundle whose fibre
> is S^7 ?
No. For then attaching a 24-ball by the bundle projection to the base of
the bundle would yield a space with integer cohomology ring Z[x]/(x^4)
having a single generator in dimension 8. Adams showed that Z[x]/(x^4)
with x of degree 8 cannot occur as an integer cohomology ring. The proof
is (reputedly-I haven't read it) very difficult, using secondary or
tertiary cohomology operations.


--
Geoffrey Mess
Department of Mathematics, UCLA
Los Angeles, CA.
ge...@math.ucla.edu
NeXTmail welcome.

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