Skip to main content
MathOverflow

Return to Answer

corrected "with fiber G" to "with total space G" and some minor spelling errors
Source Link
edit approved Oct 15, 2014 at 4:27
Community Bot
edit approved Oct 15, 2014 at 4:27
Community Bot
  • 1
  • 2
  • 3

RheneRene Thom put the same question to Sullivan duaringduring 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 fibration with fiber bundle with total space G (the Lie group) and fiber S^1. Take the associtedassociated 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.

Rhene Thom put the same question to Sullivan duaring 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 fibration with fiber G (the Lie group) and fiber S^1. Take the associted 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.

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.

Sorry I wrote poset instead of coset. Is this clear now? I mean we consider the fibre bundle G \to G/S^1.
Source Link
András Szűcs
András Szűcs
  • 2.3k
  • 1
  • 22
  • 18

Rhene Thom put the same question to Sullivan duaring 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 posetscosets. This gives a fibration with fiber G (the Lie group) and fiber S^1. Take the associted 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.

Rhene Thom put the same question to Sullivan duaring 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 posets. This gives a fibration with fiber G (the Lie group) and fiber S^1. Take the associted 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.

Rhene Thom put the same question to Sullivan duaring 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 fibration with fiber G (the Lie group) and fiber S^1. Take the associted 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.

Source Link
András Szűcs
András Szűcs
  • 2.3k
  • 1
  • 22
  • 18

Rhene Thom put the same question to Sullivan duaring 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 posets. This gives a fibration with fiber G (the Lie group) and fiber S^1. Take the associted 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.