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Jan 12, 2013 at 2:00 history edited Chris Gerig CC BY-SA 3.0
edited title
S Jan 11, 2013 at 17:46 vote accept Mathboy
Jan 11, 2013 at 17:35 vote accept Mathboy
S Jan 11, 2013 at 17:46
Jan 11, 2013 at 11:05 vote accept Mathboy
Jan 11, 2013 at 17:35
Jan 11, 2013 at 1:02 answer added Mike Usher timeline score: 3
Jan 10, 2013 at 18:36 comment added Spiro Karigiannis I edited the question, so the above remarks are no longer relevant.
Jan 10, 2013 at 18:36 history edited Spiro Karigiannis CC BY-SA 3.0
edited grammar, changed "cohomology group" to "cohomology class"
Jan 10, 2013 at 12:21 answer added Peter Michor timeline score: 3
Jan 10, 2013 at 9:36 comment added Mathboy Dear Serge, yes, I mean the de Rham cohomology class of f.
Jan 10, 2013 at 9:31 comment added Serge Lvovski Did you mean de Rham cohomology clacc of $f$?
Jan 10, 2013 at 8:44 history edited Mathboy CC BY-SA 3.0
I Edited the question according to the comments of Serge and Tagging. Thanks them!
Jan 10, 2013 at 8:39 comment added Mathboy Yes, actually it should be called the de Rham cohomology. Thanks!
Jan 9, 2013 at 17:49 comment added Tim Perutz Should "Hodge homology group" read "de Rham cohomology class"?
Jan 9, 2013 at 15:43 history edited Mathboy CC BY-SA 3.0
added 39 characters in body; edited title
Jan 9, 2013 at 15:37 comment added Mathboy Tanks serge. I think I made a mistake here. I will edit my question.
Jan 9, 2013 at 14:24 comment added Serge Lvovski The "well known result" you refer to cannot be true if understood literally: for example, if $Y\subset X$ is a smooth submanifold, then the set of all covectors at points $y\in Y$ that annihilate the tangent subspace $T_yY\subset T_yX$, is a Lagrangian submanifold as well.
Jan 9, 2013 at 13:50 history edited Mathboy CC BY-SA 3.0
added 2 characters in body
Jan 9, 2013 at 13:44 history asked Mathboy CC BY-SA 3.0