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I would like to compute the first few stable homotopy groups of $RP^2$.

I first thought to use the Atiyah-Hirzebruch Spectral Sequence, (see Davis & Kirk, pg. 242). Here is what I computed for the $E^2$ term of the spectral sequence: $$E^2_{p,q}=\begin{array}{|ccc} \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_2 \\ \mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_2 \\ \mathbb{Z} & \mathbb{Z}_2 & 0 \\\hline \end{array}$$

From this, I compute that the associated graded complex to $\pi_1^s(RP^2)$ is $\mathbb{Z}_2\oplus\mathbb{Z}_2$. (I think I made a mistake here with the local coefficients. I believe I showed the local coefficients act trivially, so it should just reduce to ordinary homology with coefficients in $\pi_q^s(S^0)$.) So either $\pi_1^s(RP^2)$ is $\mathbb{Z}_4$ or $\mathbb{Z}_2\oplus\mathbb{Z}_2$.

On the other hand, we know that $\pi_1^s(RP^2)=\pi_2(\Sigma RP^2)$ by the Freudenthal suspension theorem. Using the evident cell structure on $\Sigma RP^2$ consisting of a single 0-cell, a single 2-cell, and a single 3-cell, we see that $\pi_1(\Sigma RP^2)=0$ by cellular approximation. So by the Hurewicz theorem $\pi_2(\Sigma RP^2)\cong H_2(\Sigma RP^2) \cong H_1(RP^2) \cong \mathbb{Z_2}$.

Where am I going wrong using the AHSS? How does one compute $\pi_2^s(RP^2)$?

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2 Answers 2

up vote 13 down vote accepted

The version of the AHSS you wrote down converges to $\pi_*^s(\mathbb{RP}^2_+)$, i.e. with an extra basepoint. This splits canonically as $\pi_*^s(\mathbb{RP}^2) \oplus \pi_*^s(S^0)$, and the left hand column of your chart corresponds to the second summand. Thus, to get $\pi_*^s(\mathbb{RP}^2)$ you should throw that column away.

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Oh I see now. So the generalized homology theories for the spectral sequence are to be unreduced ones. Since $\pi_*^s$ is a reduced homology theory, you make it into an unreduced one by adding the basepoint. Everything works out quite nicely now. Thank you for correcting my error! I will see if I can get this to work for $\pi_2^s(RP^2)$. –  Glen M Wilson Nov 20 '12 at 19:31
3  
Alternatively, the SS should start with $\widetilde{H}_p(X; \pi_q(E))$. –  Oscar Randal-Williams Nov 20 '12 at 20:12
    
Even better! This seems to completely describe $\pi_k^s(RP^2)$ in terms of $\pi_*^s$. That is very nice indeed. –  Glen M Wilson Nov 20 '12 at 22:09
5  
Your AHSS collapses to the LES in stable homotopy associated to the cofiber sequence $S^1 \to S^1 \to RP^2$, where the first map is of degree two. That LES expresses $B = \pi^2_k(RP^2)$ as an extension, with $A = \pi^s_k(S^1)/2 = \pi^s_{k-1}/2$ as the subgroup and $C = \pi^s_{k-1}(S^1)[2] = \pi^s_{k-2}[2]$ as the quotient group. You may want to determine the group extension. For $b \in B$ a lift of $c \in C$, the multiple $2b \in B$ is the image of $a = \langle 2, c, 2 \rangle = \eta c$ in $A$; see Toda's orange book, Cor. II.3.7. –  John Rognes Nov 23 '12 at 11:05

Another approach will be to use Adams Spectral Sequence. This is a most suited to compute stable homotopy groups. An outline will be as follows.

As a module over steenrod algebra $\mathbb{RP}^{2}$ looks like two cells connected with $Sq^{1}$.

One can either compute the minimal resolution or do a May spectral sequence to compute the $E_{2}$ page of Adams Spectral Sequence. In fact, May Spectral Sequence is tailor made for such modules. All one has to do is run May SS after throwing away $h_{0}$ But after first few terms you want to probably want to see if you have any differentials in the Adams SS.

If you are going high enough to worry about Adams SS differentials, one can import the computations done for Sphere spectrum. Since $\mathbb{RP}^{2}$ is cofiber of $S \xrightarrow{2} S$ and multiplication by $2$ induces multiplication by $h_{0}$ in Adams SS. Also one has a long exact sequence $Ext_{A}^{s,t}(S) \xrightarrow{h_{0}} Ext_{A}^{s+1,t+1}(S) \to Ext_{A}^{s+1, t+1}(\mathbb{RP}^{2}) \to Ext_{A}^{s+1, t}(S)$

where $A$ is the mod 2 Steenrod Algebra. Using this method one can import the differentials Adams SS differentials for $S$, the sphere spectrum. One can easily recover homotopy groups $\mathbb{RP^{2}}$ from the knowledge of sphere. This obviously gives the answer easily up to $50^{th}$ stem, may be more, but I did not go beyond that. `

This approach will give all the extensions as well. Also for $\mathbb{RP}^{2}$ one has to worry about what happens at prime $2$ only. `

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