$K(\mathbb R P^n)$ from $K(\mathbb C P^k)$

Question

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.

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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.

Answer 1

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.

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

Answer 2

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.)

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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 $u\oplus v=u+v\pm uv$.

Answer 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

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