Some calculations with the Adams spectral sequence and the cobar complex - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T17:15:54Z http://mathoverflow.net/feeds/question/97741 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97741/some-calculations-with-the-adams-spectral-sequence-and-the-cobar-complex Some calculations with the Adams spectral sequence and the cobar complex Drew Heard 2012-05-23T10:59:36Z 2012-05-24T01:15:15Z <p>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). </p> <p>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 $$d_s(\gamma_1 \otimes \cdots \otimes \gamma_s \otimes m) = \cdots + \sum_{i=1}^s \gamma_1 \otimes \dots \otimes \gamma_{i-1} \otimes \psi(\gamma_i) \otimes \cdots \otimes \gamma_s \otimes m + \cdots$$</p> <p>(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. </p> <p>The first calculation (pp. 64-66) is the $E_2$ term for the calculation of $\pi_*(bo)$, which is equal to <code>$\text{Ext}_{\mathscr{A}(1)_*}(\mathbb{F}_2,\mathbb{F}_2)$</code>. This is abutted to by a Cartan-Eilenberg Spectral sequence which has $E_2$ term equal to <code>$\mathbb{F}_2[h_{10},h_{11},h_{20}]$</code>, 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 <code>$\mathbb{F}_2(u,h_{10},h_{11})/(h_{10}h_{11})$</code> 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</p> <blockquote> <p>...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 $d(\overline{\xi}_2 \otimes \overline{\xi}_2 + \xi_1 \otimes \xi_1^2 \overline{\xi}_2 + \xi_1 \overline{\xi}_2\otimes \xi_1^2) = \xi_1^2 \otimes \xi_1^2 \otimes \xi_1^2$</p> </blockquote> <p>which is used to conclude $d_3(u)=h_{11}^3$</p> <p>Finally my questions:</p> <blockquote> <p>1) Why are the correcting terms $\xi_1 \otimes \xi_1^2 \overline{\xi}_2$ and $\xi_1 \overline{\xi}_2\otimes \xi_1^2$?</p> <p>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?</p> <p>3) How can I calculate this differential? For example how do we calculate $d(\xi_1 \otimes \xi_1^2 \overline{\xi}_2$)?</p> </blockquote> <hr> <p>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. </p> http://mathoverflow.net/questions/97741/some-calculations-with-the-adams-spectral-sequence-and-the-cobar-complex/97767#97767 Answer by Peter May for Some calculations with the Adams spectral sequence and the cobar complex Peter May 2012-05-23T15:14:12Z 2012-05-23T15:14:12Z <p>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 <code>$d(R_j^i) = \sum_{0 &lt; k &lt; j}R^{i+k}_{j-k}R^i_k$</code>. I prove that the resulting homology is $E_2$ of the MSS. Then $h_i$ is represented by <code>$R_1^i$</code> and <code>$b^i_j$</code> is represented by $(R^i_j)^2$ (so <code>$b^i_1 = (h_i)^2$</code>). The differential on $R^i_2$ gives that $h_{i+1}h_i=0$. I don't know where the notation <code>$x_7$</code> comes from; this is the guy I called <code>$c_0$</code> that lives in the $8$-stem. I also called him <code>$h_i(1)$</code> as part of a systematic naming of what I believe are ALL indecomposables in <code>$E_2$</code>. He is represented in <code>$E_2$</code> by <code>$R^0_2R^1_2 + R^1_1 R^0_3$</code>. Compute <code>$d(R^0_2R^0_3)$</code> and <code>$d(R^1_2R^0_3)$</code>. The relations <code>$h_0c_0 = h_2b^0_2$</code> and <code>$h_2c_0 = h_0 b^1_2$</code> in <code>$E_2$</code> will appear before your eyes. Incidentally, I prove that <code>$d_2(b^i_j) = h_{i+1}b^{i+1}_{j-1} + h_{i+j}b^i_{j-1}$</code> in the MSS. (See page II-6.3 of my thesis, which seems to be on Math Reviews now). </p>