Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
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
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
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
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:

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


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

share|improve this answer
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
@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 $u\oplus v=u+v\pm uv$.

share|improve this answer
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.

share|improve this answer
$\mathbb RP^n$ 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
@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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.