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I am trying to 'get my hands dirty', so to speak, with some of the calculations with the Adams spectral sequence in Ravenel's Complex Cobordism book, and I have a few questions (I hope it is OK to ask them in the one question).

In great generality for a Hopf algebroid $(A,\Gamma)$ we can define the cobar complex $C_\Gamma^*(M)$ by $C_\Gamma^*(M)=\overline{\Gamma}^{\otimes s} \otimes M$ where $\overline{\Gamma}=\text{ker} \epsilon: \Gamma \to A$ with coboundary map $$d_s(\gamma_1 \otimes \cdots \otimes \gamma_s \otimes m) = \cdots + \sum_{i=1}^s \gamma_1 \otimes \dots \otimes \gamma_{i-1} \otimes \psi(\gamma_i) \otimes \cdots \otimes \gamma_s \otimes m + \cdots$$

(where $\psi(\gamma_i)$ is the coproduct and I have omitted the first and last term for brevity) and then $\text{Ext}_{\Gamma}(A,M)$ is the cohomology of this cobar complex.

The first calculation (pp. 64-66) is the $E_2$ term for the calculation of $\pi_*(bo)$, which is equal to $ \text{Ext}_{\mathscr{A}(1)_*}(\mathbb{F}_2,\mathbb{F}_2) $. This is abutted to by a Cartan-Eilenberg Spectral sequence which has $E_2$ term equal to $\mathbb{F}_2[h_{10},h_{11},h_{20}]$, where $h_{i,j}$ corresponds to the class $[\overline{\xi}_i^{2^j}]$ in the cobar complex. The first claim is that $d_2(h_{20}) = h_{10}h_{11}$, and this follows from the fact that in the cobar complex $d(\xi_2) = \xi_1 \otimes \xi_1^2$, which in turn follows from the coproduct of the mod 2 Steenrod algebra. This gives $E_3$ term $\mathbb{F}_2(u,h_{10},h_{11})/(h_{10}h_{11})$ where $u$ corresponds to $h_{20}^2$. Again we can calculate $d(\overline{\xi}_2 \otimes \overline{\xi}_2) = \overline{\xi}_2 \otimes \xi_1 \otimes \xi_1^2 + \xi_1 \otimes \xi_1^2 \otimes \overline{\xi}_2 $ in the cobar complex. Ravenel then states

...the cobar complex is not commutative and when we add correcting terms to $\overline{\xi}_2 \otimes \overline{\xi}_2$ in the hope of getting a cycle we get instead $d(\overline{\xi}_2 \otimes \overline{\xi}_2 + \xi_1 \otimes \xi_1^2 \overline{\xi}_2 + \xi_1 \overline{\xi}_2\otimes \xi_1^2) = \xi_1^2 \otimes \xi_1^2 \otimes \xi_1^2$

which is used to conclude $d_3(u)=h_{11}^3$

Finally my questions:

1) Why are the correcting terms $\xi_1 \otimes \xi_1^2 \overline{\xi}_2$ and $\xi_1 \overline{\xi}_2\otimes \xi_1^2$?

2) (This may be answered by 1) Why does $\overline{\xi}_2 \otimes \overline{\xi}_2 + \xi_1 \otimes \xi_1^2 \overline{\xi}_2 + \xi_1 \overline{\xi}_2\otimes \xi_1^2$ represent $u$ in the cobar complex?

3) How can I calculate this differential? For example how do we calculate $d(\xi_1 \otimes \xi_1^2 \overline{\xi}_2$)?


That part 2 of this question concerns the May spectral sequence for calculating $\text{Ext}_\mathscr{A}(\mathbb{F}_2,\mathbb{F}_2)$. One can compute the $E_2$ term of the May spectral sequence to have generators (in the region $t-s \le 13$) $h_j = h_{1,j}$, $b_{i,j} = h_{i,j}^2$ and $x_7 = h_{20}h_{21} + h_{11}h_{30}$. There are some relations given without proof; $h_jh_{j+1} = 0, h_2b_{20} = h_0x_7$ and $h_2x_7 = h_0b_{21}$. I think that the relation $h_jh_{j+1}=0$ comes from the fact that $d_1(h_{2,j}) = h_jh_{j+1}$, but I am unsure where the other relations are coming from.

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a couple of criticisms, I think you mean $\pi_* bo$ instead of $bu$. You can also compute this $Ext$ group directly, for either case. $A(1)$ and $E(1)$ are simple enough algebras that the resolutions can be done by hand. (there is no room for differentials when you do it directly this way either.) –  Sean Tilson May 23 '12 at 18:46
    
@Sean - thank you, I have fixed the typo(s). I can do a resolution of $\mathbb{F}_2$ by hand for $\mathscr{A}(1)$, but I can't see how to do it for $\mathscr{A}(1)_*$ - I would be interested in seeing this! –  Drew Heard May 23 '12 at 23:21
    
@Drew: see math.wayne.edu/art for some $A(1)$-resolutions. –  John Palmieri May 24 '12 at 1:27
    
@John - thanks that is pretty neat! –  Drew Heard May 24 '12 at 1:45
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1 Answer 1

up vote 15 down vote accepted

One quick answer to 1 is trial and error; 2 is a typo: you don't mean $u = h_{11}^2$. For 3, unhelpfully, you just have to do the computation from the definitions. But I'm really answering the last question. I wish that people would use my original notations, and I wish that Doug had done so, since my notations are suggestive of the relevant powers and since this old man can't remember other people's notations. More importantly, the original notations make answers to questions like the one asked transparent. Write $R_j^i$ instead of $h_{ji}$ (note the order). Take the polynomial algebra over $\mathbf{F}_2$ on the $R_j^i$ and give it the differential $d(R_j^i) = \sum_{0 < k < j}R^{i+k}_{j-k}R^i_k$. I prove that the resulting homology is $E_2$ of the MSS. Then $h_i$ is represented by $R_1^i$ and $b^i_j$ is represented by $(R^i_j)^2$ (so $b^i_1 = (h_i)^2$). The differential on $R^i_2$ gives that $h_{i+1}h_i=0$. I don't know where the notation $x_7$ comes from; this is the guy I called $c_0$ that lives in the $8$-stem. I also called him $h_i(1)$ as part of a systematic naming of what I believe are ALL indecomposables in $E_2$. He is represented in $E_2$ by $R^0_2R^1_2 + R^1_1 R^0_3$. Compute $d(R^0_2R^0_3)$ and $d(R^1_2R^0_3)$. The relations $h_0c_0 = h_2b^0_2$ and $h_2c_0 = h_0 b^1_2$ in $E_2$ will appear before your eyes. Incidentally, I prove that $d_2(b^i_j) = h_{i+1}b^{i+1}_{j-1} + h_{i+j}b^i_{j-1}$ in the MSS. (See page II-6.3 of my thesis, which seems to be on Math Reviews now).

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Thank you very much! I've finally managed to locate your thesis, so I'll sit down this morning and try and work through the calculations. (Hopefully I've fixed in the typo's in the first question now. I guess my problem in part 3 is - how does one do the coproduct $\psi(\xi_1^2 \overline{\xi}_2)$?) –  Drew Heard May 23 '12 at 23:42
    
Ok, I've verified the relations in 2 now - thank you again. So it is really 'grunt work' - trying to find boundaries who have the right summand to produce a relation? –  Drew Heard May 24 '12 at 0:46
1  
@Drew: It's actually very standard to use: (a) $\psi(xy) = \psi(x) \psi(y)$; (b) $\psi(x) = x \otimes 1 + 1 \otimes x + d(x)$; hence (c) $d(xy) = x \otimes y + y \otimes x + \text{(correction)}$ to commute elements across each other in the cobar complex. –  Tyler Lawson May 24 '12 at 1:31
1  
@Tyler: Thanks for that. So are you saying, in the example I have in mind, $d[\xi_1 \vert \xi_1^2\xi_2 ]=[\xi_1 \vert \xi_1^2 \vert \xi_2] + [\xi_1 \vert \xi_2 \vert \xi_1^2]+\cdots$? –  Drew Heard May 24 '12 at 3:06
1  
@Drew: Right. In your initial formula for $d(\overline{\xi}_2 \otimes \overline{\xi}_2)$, the two elements would cancel with each other if you could commute tensor products of things, so you can decide to use differentials of these two listed elements to commute them; however, there's a correction factor in higher MSS filtration that's left over. –  Tyler Lawson May 24 '12 at 12:28
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