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corrected "with fiber G" to "with total space G" and some minor spelling errors

Rene Thom put the same question to Sullivan during Sullivan's talk at IHES in 1980: "Why Lie groups are null-cobordant?" Sullivan answered: "It follows from Thom's theorem" (A manifold is null-cobordant precisely when each of its characteristic numbers vanish.)

Thom: "OK, but can you prove this without this heavy tool?"

After few minutes Sullivan said:

"Take a circle subgroup in the Lie group. Take all the cosets. This gives a fiber bundle with total space G (the Lie group) and fiber S^1. Take the associated disc bundle. Its boundary is G." Q.E.D.

An elementary proof that paralellizable manifold is null-cobordant can be found in a paper by Buoncristiano- Hacon. I am not sure it gives oriented null-cobordism.

András Szűcs
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