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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

Feb 17, 1993, 5:43:18 PM2/17/93

to

In article <1993Feb16.2...@nas.nasa.gov> asi...@nas.nasa.gov

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.

(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 > 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 ?

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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