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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. –  Kerry Mar 11 '11 at 19:05
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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. –  Eric O. Korman Mar 11 '11 at 19:31
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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. –  Eric O. Korman Mar 11 '11 at 19:59
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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 $[x_0:\ldots:x_n]\mapsto [x_0:\ldots:x_n]$...? –  Aaron Mazel-Gee Mar 13 '11 at 8:52
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Whoa I just re-read your question and realized you already suggested this obvious map. Sorry about that. –  Aaron Mazel-Gee Mar 13 '11 at 14:49

3 Answers 3

Using the long exact sequence of the pair $(\mathbb{R}\mathbb{P}^{n},\mathbb{R}\mathbb{P}^{n-1})$. For n=odd, we should have the exact sequence $$..\mathbb{Z}\rightarrow K^{-1}(\mathbb{R}\mathbb{P}^{n})\rightarrow K^{-1}(\mathbb{R}\mathbb{P}^{n-1})\rightarrow 0\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{n})\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{n-1})\rightarrow \mathbb{Z}...$$ for n=even, we should have the exact sequence $$0\rightarrow K^{-1}(\mathbb{R}\mathbb{P}^{n})\rightarrow K^{-1}(\mathbb{R}\mathbb{P}^{n-1})\rightarrow \mathbb{Z}\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{n})\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{n-1})\rightarrow 0\rightarrow...$$

We wish to prove via induction that $K(\mathbb{R}\mathbb{P}^{n})=\mathbb{Z}/2^{n}\mathbb{Z}$. The base case $n=1,K(\mathbb{R}\mathbb{P}^{1})=K(S^{1})=0$ is established by the fact that $U(1)$ is connected. We proceed to the $n=2$ case.

We notice the middle coboundary map $\delta:K^{-1}(\mathbb{R}\mathbb{P}^{n-1})\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{n}/\mathbb{R}\mathbb{P}^{n-1})$. In this current case it is from $\mathbb{Z}$ to $\mathbb{Z}$. To ascertain $\delta$ we notice the following commuative diagram:

$$\begin{CD} S^{1} @> >> D^{2}\\ @VV V @VV V\\ \mathbb{R}\mathbb{P}^{1}@> >>\mathbb{R}\mathbb{P}^{2} \end{CD}$$

Hence $\delta$ is a $\times 2$ map. The exactness at $K^{-1}(\mathbb{R}\mathbb{P}^{2})$ implies $K^{-1}(\mathbb{R}\mathbb{P}^{2})=Ker(\delta)=0$, and $K^{0}(\mathbb{R}\mathbb{P}^{2})=\mathbb{Z}/2\mathbb{Z}$.

For $n=3$ we have the long exact sequence to be $$..\mathbb{Z}\rightarrow K^{-1}(\mathbb{R}\mathbb{P}^{3})\rightarrow 0 \rightarrow 0\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{3})\rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}...$$ The same commuative diagram

$$\begin{CD} S^{2} @> >> D^{3}\\ @VV V @VV V\\ \mathbb{R}\mathbb{P}^{2}@> >>\mathbb{R}\mathbb{P}^{3} \end{CD}$$

implies the coboundary map $K^{0}(\mathbb{R}\mathbb{P}^{2})\rightarrow K^{-1}(\mathbb{R}\mathbb{P}^{3}/\mathbb{R}\mathbb{P}^{2}\cong S^{3})$ is still $\times 2$.

Reorganize the sequence we have $$0\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{3})\rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}\rightarrow K^{-1}(\mathbb{R}\mathbb{P}^{3})\rightarrow 0$$

Hence $K^{0}(\mathbb{R}\mathbb{P}^{3})=\mathbb{Z}/2\mathbb{Z}$, $K^{-1}(\mathbb{R}\mathbb{P}^{3})=\mathbb{Z}$.

For $n=4$ we have the long exact sequence to be

$$...0\rightarrow K^{-1}(\mathbb{R}\mathbb{P}^{4})\rightarrow \mathbb{Z} \rightarrow \mathbb{Z}\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{4})\rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow 0...$$

Hence $K^{-1}(\mathbb{R}\mathbb{P}^{4})=0$, but we are not sure how to calculate $K^{0}(\mathbb{R}\mathbb{P}^{4})$. I think the map $\mathbb{Z}\rightarrow K^{0}(\mathbb{R}\mathbb{P}^{4})$ is $\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}$, both by the exact sequence and the geometrical picture. Hence $K^{0}(\mathbb{R}\mathbb{P}^{4})$ has two choices, either $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, or simply $\mathbb{Z}/4\mathbb{Z}$. So we need to show $K^{0}(\mathbb{R}\mathbb{P}^{4})$ have an element $x$ such that $2x\not=0$. But calculating the Stifel-Whitney class of $w(RP^{4}\oplus RP^{4})$ by Whitney summation formula implies it to be non-zero. Hence such an element do exist.

We conclude that $K^{0}(\mathbb{R}\mathbb{P}^{4})=\mathbb{Z}/4\mathbb{Z}$, $K^{-1}(\mathbb{R}\mathbb{P}^{4})=0$.

The induction scheme, totally analogous to the pervious arguments thus gives us:

$K^{0}(RP^{2k+1})=\mathbb{Z}/2^{k}\mathbb{Z}$, $K^{-1}(RP^{2k+1})=\mathbb{Z}$.

and

$K^{0}(RP^{2k})=\mathbb{Z}/2^{k}\mathbb{Z}$, $K^{-1}(RP^{2k})=0$.

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I don't understand the Stiefel-Whitney class argument. First of all $w(RP^n)$ vanishes if $n+1$ is a power of two. Secondly we are considering complex vector bundles and it is possible for the complexification of a non-trivial real vector bundle to be trivial (e.g. the complexification of the mobius bundle). –  Eric O. Korman Apr 6 '11 at 22:07
    
The second point is quite true; I did not thought about the complexification part. I will think about this in free time. –  Kerry Apr 6 '11 at 22:25
    
@Eric: BTW did you get any progress in my problem? I am still totally stuck in that one. Also Atiyah has a proof of $K(RP^{n})$ using tools I am not familiar with in his book as well. –  Kerry Apr 6 '11 at 22:26
    
@user7887: I spent some time on it but was not able to get anywhere. I was also unable to find what you referenced in my edition of Atiyah. –  Eric O. Korman Apr 7 '11 at 0:41
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@Eric: I think it(the calculation of $K(RP^{n})$ is around page 100-110 in my version. I could not really understand it(he used G-spaces and other things), but I hope you can understand it. –  Kerry Apr 7 '11 at 6:01

Let $c\colon\mathbb R P^{2n+1}\to \mathbb C P^n$ be the map from the post and let $v_2\colon\mathbb C P^n\to \mathbb C P^{N}$ be the degree 2 Veronese embedding. Their composition induces zero map on K-theory. So one gets a map $K_0(\mathbb R P^{2n+1})\gets K_0(\mathbb C P^n)/\operatorname{Im} v_2^*=\mathbb Z[t]/\langle t^{n+1},t^2-2t\rangle =\mathbb Z/2^n\mathbb Z$.

Finally, observe that by the spectral sequence from the post, this map is surjective$^*$ and both sides has the same size.

(The same computation is much more transparent in infinite-dimensional case: $\mathbb R P^{\infty}\to \mathbb C P^{\infty}\to \mathbb C P^{\infty}$ is a cofibration sequence which gives a short exact sequence $K_0(\mathbb C P^{\infty})\to K_0(\mathbb C P^{\infty})\to K_0(\mathbb R P^{\infty})\to 0$ where first map is $t\mapsto 2t-t^2$.)

$^*$) By functoriality of the spectral sequence, the map $K(B)\to K(E)$ is induced by the map $H(B;K(pt))\to H(B;K(F))$. So the coker consists of surviving elements from odd row. But by induction by column one can show, it seems, that any such element would give non-torsion element in K(E). (OK, I haven't quite checked details, but if you managed even to find size of K(E), it shouldn't be difficult.)


Perhaps, I should explain what is really going on here. There is a following observation (due to Quillen, probably). Let $h$ be a complex-oriented cohomology theory (it means, roughly speaking, that there exist Chern classes of complex vector bundles with values in $h$). Complex orientation implies $h(\mathbb{C}P^n)=h(pt)[u]/u^{n+1}$, where $u=c_1(\text{tautological bundle})$ and one can define a formal group law — a series $F\in h(pt)[[u,v]]$ s.t. $F(c_1(\xi),c_1(\eta))=c_1(\xi\otimes\eta)$ for any line bundles $\xi$, $\eta$. Explicitly, $F$ is just the inverse image of the generator under quadratic Veronese map $\mathbb{C}P^{\infty}\times \mathbb{C}P^{\infty}\to \mathbb{C}P^{\infty}$.

Let's write $a\oplus b$ instead of $F(a,b)$. Proposition. $h(\mathbb{R}P^{2n+1})=h(\mathbb{C}P^n)/u\oplus u$; more generally $h(S^{2n+1}/(\mathbb{Z}/n\mathbb{Z}))=h(\mathbb{C}P^n)/u\oplus u\oplus\dots\oplus u$.

Now, complex K-theory can be oriented by putting $c_1(\xi)=\pm 1\pm \xi$ and depending on the sings $u\oplus v=u+v\pm uv$.

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Thanks for the answer. I'm not familiar with the Veronese embedding. I found this en.wikipedia.org/wiki/Veronese_surface#Veronese_map but I don't think this is what you're talking about since this map doesn't always give a map $P^n \to P^{2n+1}$. Also, how do you see from the spectral sequence that the map is surjective? –  Eric O. Korman Apr 2 '11 at 2:13
    
Yes, this Veronese map (and you're right, $P^{2n+1}$ from the original answer should be $P^N$). The idea is that $v_2^*$(tautological bundle) is tensor square of tautological bundle. (I'll try to comment on surjectivity a little bit latter.) –  Grigory M Apr 2 '11 at 5:31

I could not comment on this now for unknown reason, so I will venture a proof, which means myself is not very sure. All the $K$ groups below are reduced $K$ groups as $\overline{K}$ is quite complicated to type.

For $n=2k+1$ we have $$K^{0}(S_{n},X)\rightarrow K^{0}(S_{n})\rightarrow K^{0}(X)\rightarrow K^{-1}(S^{n},X)\rightarrow K^{-1}(S_{n})\rightarrow K^{-1}(X)\rightarrow K^{0}(S_{n},X)..$$

From Bott periodicity we have $K^{0}(S^{n})=0$, $K^{1}(S^{n})=\mathbb{Z}$. Note $S_{n}/\mathbb{R}\mathbb{P}_{n}\cong S_{n}$. Hence we have:

$$0 \rightarrow 0\rightarrow K^{0}(X)\rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow K^{-1}(X)\rightarrow 0$$

The middle map from is $f:z\rightarrow 2z$. The exactness implies $K^{0}(X)=0$ and $K^{1}(X)=\mathbb{Z}/2\mathbb{Z}$. Now proceed from reduced K-group to normal K-group we get:

$K^{0}(X)=\mathbb{Z}$, $K^{1}(X)=\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$.

The $n=2k$ case should be smiliar. I am not sure if $K(\mathbb{S}\wedge \mathbb{R}\mathbb{P}^{n})=K(\mathbb{R}\mathbb{P}^{n+1})$ holds. If yes then it should be much easier to calculate.


The above computation may have unknown problems as $RP_{1}\cong S_{1}$ but the result $K$-groups are different.

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$\mathbb RP^n$ is not a subspace of $S^n$ so I don't see how you can consider the pair $(S^n, \mathbb RP^n)$. –  Eric O. Korman Mar 13 '11 at 3:05
    
Also, this probably doesn't affect your calculation but I think you need to be careful about non/reducedness when you're juggling things in the lexseq of a pair. –  Aaron Mazel-Gee Mar 13 '11 at 9:00
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@user7887 $\mathbb R P^n$ is a quotient of $S^n$, not a subspace. The long exact sequence is for a pair $(X,Y)$ where $Y \subset X$. –  Eric O. Korman Mar 14 '11 at 3:21
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No, if $P^n$ was a subspace of $S^n$ then that would mean that there was a map $j:P^n \rightarrow S^n$, such that $j$ is injective and continuous and has a ontinuous inverse on its image. That is $j$ would have to be an embedding. Then if we consider $S^2 \subset \mathbb R^3$, then $j$ is an embedding of $P^2$ into $\mathbb R^3$. Something which isn't possible. Also in my previous comment I meant $P^n$ to $S^n$. –  JSchlather Mar 14 '11 at 6:24
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An embedding of one closed connected $n$-dimensional manifold into another must be a homeomorphism. I think the easiest way to say it is that there's no continuous way to choose which point $p$ or $-p$ in $S^n$ you want to take for the image of a given point $\{p,-p\}$ in $\mathbb{RP}^n$. –  Aaron Mazel-Gee Mar 14 '11 at 7:00

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