John Carlos Baez on Nostr: npub12n4k2…c5dmn - to compute the de Rham cohomology of SO(n) we should probably ...
npub12n4k27w8jyj29f0jrpmupmzx8rpvttg6c4z0jws7xfu9kky0cp6q8c5dmn (npub12n4…5dmn) - to compute the de Rham cohomology of SO(n) we should probably use the fact that these cohomology groups form a graded-commutative algebra (as de Rham cohomology groups of any manifold always do). Moreover, for SO(n) I believe we get the exterior algebra on generators of degree 4k-1 where 4k-1 goes 3, 7, ... all the way up to the largest number less than the dimension of SO(n), which I guess is n(n-1)/2.
But the proof of this fact, assuming I've got it right, is not trivial! It's part of a general theory of the de Rham cohomology of compact Lie groups. Looking around, I see people pointing toward this:
Mark Reeder, On the cohomology of compact Lie groups, L'Enseignement Math 41 (1995), 181-200.
though the basic ideas are a lot older.
(2/n)
Published at
2024-10-04 05:22:59Event JSON
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"content": "nostr:npub12n4k27w8jyj29f0jrpmupmzx8rpvttg6c4z0jws7xfu9kky0cp6q8c5dmn - to compute the de Rham cohomology of SO(n) we should probably use the fact that these cohomology groups form a graded-commutative algebra (as de Rham cohomology groups of any manifold always do). Moreover, for SO(n) I believe we get the exterior algebra on generators of degree 4k-1 where 4k-1 goes 3, 7, ... all the way up to the largest number less than the dimension of SO(n), which I guess is n(n-1)/2. \n\nBut the proof of this fact, assuming I've got it right, is not trivial! It's part of a general theory of the de Rham cohomology of compact Lie groups. Looking around, I see people pointing toward this:\n\nMark Reeder, On the cohomology of compact Lie groups, L'Enseignement Math 41 (1995), 181-200. \n\nthough the basic ideas are a lot older.\n\n(2/n)",
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