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Jul 17, 2012 at 20:35 vote accept Akhil Mathew
Jul 17, 2012 at 20:35 answer added Akhil Mathew timeline score: 5
Jul 16, 2012 at 2:45 comment added Craig Westerland Away from $2 = | \mathbb{Z} / 2|$, the LHSSS for the fibration $E\mathbb{Z}/2 \times_{\mathbb{Z}/2} \mathbb{CP}^\infty \to B\mathbb{Z} / 2$, given by $H^*(\mathbb{Z} / 2, H^*(\mathbb{CP}^\infty))$ collapses at the $E_2$-term, and contains only 0-dimensional group cohomology (i.e., invariants). The generator of $\mathbb{Z} / 2$ acts on a a generator $x \in H^2(\mathbb{CP}^\infty)$ by multiplying by $-1$, and on the rest through ring homomorphisms. So the invariant subring is that which is generated by $x^2$, which happens to be the image of $H^*(\mathbb{HP}^\infty)$ under the natural map.
Jul 15, 2012 at 20:19 history edited Akhil Mathew CC BY-SA 3.0
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Jul 15, 2012 at 20:08 comment added Akhil Mathew @Craig: I don't really see how to get express the $E$-homology of $E \mathbb{Z}/2 \times_{\mathbb{Z}/2} \mathbb{CP}^\infty$ in terms of that of $\mathbb{CP}^\infty$. I'm a little tired now --- let me think about that more later. (As it happens, I'm mostly interested in the prime $2$.)
Jul 15, 2012 at 20:05 comment added Akhil Mathew @Sean: Thanks. Yes, that's what I meant to say.
Jul 15, 2012 at 20:05 history edited Akhil Mathew CC BY-SA 3.0
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Jul 15, 2012 at 16:53 comment added Sean Tilson The first sentence of the second paragraph contains an error, I think you mean two cell complex. The dimension of a cell complex is usually defined to be the dimension of the top cell, if it is finite.
Jul 15, 2012 at 16:23 comment added Craig Westerland Of course, the prime 2 is pretty important when talking about $\nu$...
Jul 15, 2012 at 16:21 comment added Craig Westerland Stably, and away from the prime 2, one can write $\mathbb{HP}^\infty$ as the Borel construction $E\mathbb{Z}/2 \times_{\mathbb{Z}/2} \mathbb{CP}^\infty$ for the action of complex conjugation on $\mathbb{CP}^\infty$. So perhaps you can do some sort of Galois-equivariant version of your analysis above for $\mathbb{CP}^2$?
Jul 15, 2012 at 14:28 history asked Akhil Mathew CC BY-SA 3.0