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I posted this question some days ago at math.stackexchange, but didn't receive an answer.

I have two questions:

  • I wonder whether anyone has taken the time to compute $KO_*(\mathbb{R}P^\infty)$?

The standard tools to compute these Groups in the complex case rest on the requirement for the cohomology theories $E$ to be complex orientable. Naturally, I looked up whether something like real orientable cohomology theories exist in the literature but found out that $KO$ is not real-oriented. Anyway, there is a way to "circumvent" Snaiths theorem for the spectrum $K$ if one is only interestd in the algebra of cooperations, in the sense that one can show that $$K_*(\mathbb{C}P^\infty) \xrightarrow{i_*} K_*K$$ is an injection of rings, where $i$ is induced from the inclusion $\mathbb{C}P^\infty \simeq BU(1) \hookrightarrow BU$. In fact, one only needs to invert the Bott element $\beta$ to turn it into an isomorphism, so it is a localization. This can be concluded from

Robert M. Switzer. Algebraic topology—homotopy and homology. Classics in Mathematics. Springer-Verlag, Berlin, 2002. Reprint of the 1975 original [Springer, New York; MR0385836 (52 #6695)].

17.33, which states

$K_*K$ is generated over $\mathbb{Z}[u,u^{-1},v^{-1}]$ by the polynomials $\{p_1,p_2,\ldots\}$.

By a process reminiscent of

J. F. Adams. Stable homotopy and generalised homology. University of Chicago Press, Chicago, Ill.-London, 1974. Chicago Lectures in Mathematics.

p. 44 we can describe the relations of the generators of $\beta_i$ of $K_*(\mathbb{C}P^\infty) = K_* \{\beta_0 , \beta_1 , \ldots \}$ such that

$$\beta_1\beta_n = n \beta_n +(n+1)\beta_{n+1}$$

and by setting

$$\binom{x}{i} = \frac{x(x-1)\cdots (x-(i-1))}{i!} \in \mathbb{Q}[x]$$

with $x:=\beta_1$ one can see that

$K_*(\mathbb{C}P^\infty)\otimes \mathbb{Q}$ is the polynomial algebra $K_* \otimes \mathbb{Q}[x]$ over $K_*\otimes \mathbb{Q} = \mathbb{Q} [t,t^{-1}]$ and $K_*(\mathbb{C}P^\infty)$ can be identified with the subalgebra of $K_* \otimes \mathbb{Q}[x]$ generated by $\binom{x}{i}$ for $i=0,1,2, \ldots$,

where we set $\binom{x}{0}=1$.

  • While snaiths theorem works on the spectrum level and the aforementioned result follows, I wonder whether a similar result holds in the real case, i.e. $$ KO_*(\mathbb{R}P^\infty)[\alpha^{-1}] \cong KO_*KO$$ for some element $\alpha \in KO_*(\mathbb{R}P^\infty)$?
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    $\begingroup$ Regarding Snaith's theorem: the direct analogue of Snaith's construction, using the generator you might expect from $\Bbb{RP}^\infty$ by analogy, will result in a ring spectrum $R$ that has an orientation for real vector bundles. This means that it can't quite be $KO$. For example, it would mean that it would have a map of ring spectra $MO \to R$. Because of known structure about $MO$, this would force $R$ to be a sum of shifts of the Eilenberg-Mac Lane spectrum $H\Bbb{F}_2$. $\endgroup$ – Tyler Lawson Jan 28 at 19:21
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    $\begingroup$ The $p$–completion of $KO_* KO$ is known and easy to state: it is the algebra of continuous functions from the $p$–adic group $(\mathbb Z_p^\times / C_2)$ to $(KO^\wedge_p)_*$. Given the size of domain and codomain, I believe this to be too hopelessly large to be caught by inverting an element in (the $p$–completion of) Bob's answer. $\endgroup$ – Eric Peterson Jan 28 at 22:23
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There are many ways to do this. One elementary approach is to use the Adams spectral sequence

$Ext_A(H^*ko \wedge RP^\infty, F_2) \cong Ext_{A(1)}(H^*RP^\infty,F_2) \Rightarrow ko_*(RP^\infty) $

and invert the Bott map. An $A(1)$ resolution giving the answer (since $E_2 = E_\infty$) can be seen at the bottom of the page

http://www.rrb.wayne.edu/art/index.html

We see the groups, starting with $KO_0$, are $0, Z/2, Z/2, Z/2^\infty, 0, 0, 0, Z/2^\infty$ and repeat by Bott periodicity, with the evident action of the coefficients $KO_*$. (This action follows from the homological algebra of the $Ext$ calculation.)

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    $\begingroup$ A cute trick for computing $ko_*(\mathbb{R}P^{\infty})$ is explained in section 4 of Mahowald-Milgram's "Operations which detect..." (academic.oup.com/qjmath/article/27/4/415/1531880) $\endgroup$ – Dylan Wilson Jan 28 at 14:25
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    $\begingroup$ Right: this uses the fact that the cofiber of the reduced transfer $RP^\infty \to S$ has cohomology which is 'made up of' $A(1)//A(0)$'s, so that smashing it with $ko$ gives a wedge of $HZ$'s in degrees 0,4,8, .... As a $ko$-module, however, the Bott map and hence also the 4 dim class in $ko_*$ act nontrivially, so the $Z$ summands in $ko_*$ map by an Adams filtration preserving map to this cofiber. The $\eta$'s in the kernel, and the finite quotients in degrees 4n shifted down one, then assemble to give the answer. $\endgroup$ – Robert Bruner Jan 28 at 15:10
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    $\begingroup$ There is one extension to determine, from filtration 1 to filtration 2, easily shown to be non-zero, and then Bott periodicity says all the others are non-zero as well. $\endgroup$ – Robert Bruner Jan 28 at 15:11

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