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EDIT: I found a brief discussion of this in Husemoller's Fibre Bundles, chapter 16 section 12. Here to compute $\tilde K(\mathbb R P^{2n+1})$ he says to consider the map $$ \mathbb R P^{2n+1} = S^{2n+1}/\pm 1\to \mathbb C P^n = S^{2n+1}/U(1). $$ Under this map the canonical line bundle over $\mathbb C P^n$ pulls back to the complexification of the canonical line bundle over $\mathbb R P^{2n+1}$. Then he says from looking at the (Atiyah-Hirzebruch) spectral sequence we get that $\tilde K(\mathbb R P^{2n+1}) = \mathbb Z/2^n$. I don't see how looking at the spectral sequence helps (and what we learn from the map from $\tilde K(\mathbb C P^{n}) \to \tilde K(\mathbb R P^{2n+1})$). All I can see is that $\tilde K(\mathbb R P^{2n+1})$ is pure torsion (by the Chern character isomorphism) and that it has order $2^n$ (from the spectral sequence). I can't see why, for example, it isn't just a direct sum of $\mathbb Z/2$'s.


I found a homework assignment online from an old K-theory course and one of the problems says to compute $K(\mathbb R P^n)$ by using a suitable comparison map with $\mathbb C P^k$ and knowledge of $K(\mathbb C P^k)$.

I have attempted this but have not been able to get anywhere. The only map $\mathbb R P^n \to \mathbb C P^n$ I can think of is the one sending the equivalence class of $(x_0,\ldots,x_n) \in \mathbb R P^n$ to its equivalence class in $\mathbb C P^n$. Under this (I think) the tautological line bundle over $\mathbb C P^n$ (which generates $K(\mathbb C P^n)$) gets sent to the complexification of the tautological line bundle over $\mathbb R P^n$. But I really don't see where to go from here; if I had a map going the other way maybe I'd be able to say something but the map I have is neither injective nor surjective. I also can't see how torsion is going to come out of this: $K(\mathbb C P^n)$ is torsionfree but $K(\mathbb R P^n)$ isn't.

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    I am not very sure what you are computing about. For $K(\mathbb{C}\mathbb{P}^{n})$ we have the cell complex decomposition $\epsilon_{0},\epsilon_{2},\epsilon_{4}...$etc. This should give you $K_{1}( \mathbb{C}\mathbb{P}^{n})=0$ and $K_{0}(\mathbb{C}\mathbb{P}^{n})=\mathbb{Z}^{n}$. Now consider similar approach to $\mathbb{R}\mathbb{P}^{n}$ with cell complex $\epsilon_{0},\epsilon_{1}...$etc.2011-03-11
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    How do you compute K groups using cell structures? Is this the same as using the Atiyah-Hirzburch spectral sequence? If so, I think it is difficult to get $K(\mathbb RP^n)$ this way since you get a group extension problem.2011-03-11
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    Hi, this is standard. You use the sequence of the pair and Bott periodicity to reduce the long exact sequence to 6 item circular case (you may look it up in Hatcher). It is essentially a "double cone" argument. From this it should be quite simple. I do not think you need spectral sequence for this problem (in fact I do not know this).2011-03-11
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    I understand how to get the K group for $\mathbb C P^n$ in that way, but I do not think this will work for $\mathbb R P^n$. The complex case seems to involve crucially the fact that $CP^n$ has cells only in even dimensions so that the sequence breaks up into short exact sequences.2011-03-11
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    From the way you stated the problem it sounds like there really should be an actual map between $\mathbb{RP}^n$ and $\mathbb{CP}^n$, and it sounds like the discussion is diverging from this suggestion. I just tried it, and I agree that the exact hexagon won't do it alone. The fact that the $K^*(\mathbb{RP}^n)$ has torsion (supposedly?) and $K^*(\mathbb{CP}^n)$ doesn't should mean that the map on spaces is $\mathbb{RP}^n\rightarrow \mathbb{CP}^n$. And what could it be besides $[x_0:\ldots:x_n]\mapsto [x_0:\ldots:x_n]$...?2011-03-13
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    Whoa I just re-read your question and realized you already suggested this obvious map. Sorry about that.2011-03-13

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