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Apr 13, 2017 at 12:19 history edited CommunityBot
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Nov 7, 2015 at 21:17 vote accept Ali Taghavi
Oct 23, 2015 at 10:30 answer added Liviu Nicolaescu timeline score: 4
Oct 21, 2015 at 19:30 answer added Sebastian Goette timeline score: 2
Oct 21, 2015 at 14:57 comment added Ali Taghavi @SebastianGoette thanks.for your elegance argument using Thom isomorphism.
Oct 21, 2015 at 8:46 comment added Sebastian Goette Regarding the cell structure obtained by gluing - it is not clear that you get the same as for $\mathbb C P^n$. But that does not matter if you are interested in homology only, because the resulting CW complex would only have cells in even dimensions. Then the cellular complex would have trivial differential, and hence be the same as for $\mathbb C P^n$ with the usual cell structure.
Oct 21, 2015 at 7:58 comment added Ali Taghavi could you please more explain about cell structure and gluing cells?
Oct 21, 2015 at 7:57 answer added Achim Krause timeline score: 5
Oct 21, 2015 at 7:49 history edited Ali Taghavi CC BY-SA 3.0
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Oct 21, 2015 at 7:32 comment added Ali Taghavi Moreover are we sure that these gluing maps induce the same maps as the $\mathbb{C}P^{n},s$( in relative homologies $H_{n}(X_{n+1}, X_{n})$)?
Oct 21, 2015 at 7:26 comment added Ali Taghavi @SebastianGoette thank you very much for your very interesting comment. For such cell-gluing, are you considering the Thom space of the bundle?
Oct 21, 2015 at 6:32 comment added Sebastian Goette You clearly cannot have a one-point remainder. Because if you had, then you would obtain $G(2,n+1)$ from $G(2,n)$ by gluing in a $2(n-1)$-cell. Inductively, $G(2,n)$ would have the same homology as $\mathbb C P^{n-2}$, but already $G(2,3)\cong\mathbb R P^2\not\cong\mathbb C P^1$.
Oct 21, 2015 at 6:32 history edited Ali Taghavi CC BY-SA 3.0
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Oct 21, 2015 at 6:23 history edited Ali Taghavi CC BY-SA 3.0
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Oct 21, 2015 at 6:09 history edited Ali Taghavi CC BY-SA 3.0
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Oct 21, 2015 at 5:54 history asked Ali Taghavi CC BY-SA 3.0