# Coefficients of real k-theory with coefficients

Question: Calculate the group $\pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z)$.

Here $KO$ denotes the real k-theory spectrum and $M\mathbb Z/l\mathbb Z$ denotes a Moore Spectrum associated to the cyclic group of order $l$.

Assume for the following that $2$ divides $l$ (otherwise there are no problems).

Using the short exact sequence for Moore spectra one can easily calculated all the coefficients but $\pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z)$ . There one gets only a short exact sequence: $$\mathbb Z/2\mathbb Z \hookrightarrow \pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z) \twoheadrightarrow \mathbb Z/2\mathbb Z$$

Note that for $l=2$ we have: $\pi_{2}(KO \wedge M\mathbb Z/l\mathbb Z)\cong KO_{3}(\mathbb RP^2)$

But I'm neither able to calculate the latter group.

What techniques are used to solve problems like this?

Idee: Use $S$-Duality to calculate $\pi_{2}(KO \wedge M\mathbb Z/l\mathbb Z)\cong KO_{3}(C_l)$ where $C_l$ denotes a Moore space to the cyclic group of order $l$ (and $C_2\simeq\mathbb RP^2$).

The $S$-dual of $C_l$ is $\Sigma^{-3}C_l$ (use that $C_l$ can be realized as the cofibre of $S^1\to S^1$).

By $S$-duality we have: $KO_{3}(C_l) \cong KO^0(C_l)$.
The latter group can be calculated by comparing $C_l$ and $\mathbb RP^2$.

In the end we get that the coefficients are as follows:

Note that the extension splits for some $l$ which I find surprising.

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I guess you get the same extension problem if you just try to use the AHSS? – Drew Heard Apr 5 '13 at 14:18
This is the KO homology of the Moore spectrum. You can use the Adams spectral sequence to compute the $ko$ homology of the Moore spectrum where $ko$ is the connective cover of $KO$. Next you can invert the bott class to get periodic $K$-theory. – Sean Tilson Apr 7 '13 at 21:25

By $S$-duality the group $KO_3(\mathbb RP^2)$ is isomorphic to $KO^0(\mathbb RP^2)$. The latter group is not killed by $2$. Let $L$ be the nontrivial real line bundle on $\mathbb RP^2$. The bundle $L\oplus L$ on $\mathbb RP^2$ is nontrivial, even stably nontrivial, because its second Stiefel-Whiney class is not zero.
And in fact it's cyclic of order $4$. See the table at the bottom of the first page here: ir.library.osaka-u.ac.jp/dspace/bitstream/11094/6753/1/… – Mark Grant Apr 5 '13 at 14:46
@Tom: Is $\mathbb{R}P^2$ orientable over $KO$? I admit I am a bit out of my comfort zone. Nevertheless, it seems to me that in $KO^0(\mathbb{R}P^2)$ we have $-T(\mathbb{R}P^2)=\gamma_2 - 3\varepsilon$, where $\gamma_2$ is the tautological line bundle over $\mathbb{R}P^2$ (i.e. your $L$), and $\varepsilon$ is the trivial real line bundle. Thus, by Spanier duality, we conclude that the $S$-dual of $(\mathbb{R}P^2)_+$ is $\mathrm{Thom}(\gamma_2 - 3\varepsilon)=\Sigma^{-3}\mathbb{R}P^3$. It follows that $KO_3(\mathbb{R}P^2)=\widetilde{KO}^0(\mathbb{R}P^3)$. Did I make a mistake? – Ricardo Andrade Apr 6 '13 at 0:29
@Mark: That is a nice reference. It is actually quite simple to compute $\widetilde{KO}^0(\mathbb{R}P^2)=\mathbb{Z}/4$. The long exact sequence in reduced $KO$-cohomology associated with the cofibre sequence $S^1\overset{2}{\to}S^1\to\mathbb{R}P^2$ gives a short exact sequence $0\to\mathbb{Z}/2\to\widetilde{KO}^0(\mathbb{R}P^2)\to\mathbb{Z}/2\to 0$. By Tom's argument above, the tautological line bundle $L\in\widetilde{KO}^0(\mathbb{R}P^2)$ is not killed by 2. Hence the short exact sequence above cannot split, therefore $\widetilde{KO}^0(\mathbb{R}P^2)=\mathbb{Z}/4$ and is generated by $L$. – Ricardo Andrade Apr 6 '13 at 1:10
$\mathbb RP^2$ is the cofiber of $2:S^1\to S^1$. Its Spanier-Whitehead dual is the fiber of $2:S^{-1}\to S^{-1}$. Suspended twice, this becomes the fiber of $2:S^1\to S^1$, and when suspended once more this becomes the cofiber of $2:S^1\to S^1$, or $\mathbb RP^2$. – Tom Goodwillie Apr 6 '13 at 1:14
So I was thinking about the $S$-dual of $\mathbb RP^2$ in a different way from you, Ricardo. But there's no contradiction: stably $\mathbb RP^2_+$ is $\mathbb RP^2\vee S^0$ and stably $\Sigma^3\mathbb RP^3$ is $\Sigma^3\mathbb RP^2\vee S^3$. – Tom Goodwillie Apr 6 '13 at 1:27