I need some guidance in identifying the third differential in the homology AHSS for $\Omega_{\ast}^{\text{Spin}^c}(X)$ in degrees $\leq 4$.

Remember that $\pi_0(M\text{Spin}^c)=\Bbb Z$, $\pi_2(M\text{Spin}^c)=\Bbb Z$, $\pi_4(M\text{Spin}^c)=\Bbb Z\oplus \Bbb Z$ and zero in degree $1,3$. So if we consider the (homologically graded) AHSS $$ E_{p,q}^2 = H_p(X;\Omega_{q}^{\text{Spin}^c} ) \Rightarrow \Omega_{p+q}^{\text{Spin}^c}(X)$$ we immediately see that there are no second differentials and the third page looks like this (in the portion I'm interests in)

enter image description here

Can I conclude something akin to the cohomology AHSS (See here) where the first non-trivial differential is a stable cohomology operation? what bothers me is that having to deal with integer coefficients homology I don't have (in general) a perfect pairing with cohomology where the linked question would apply.

Is there an isomorphism in low degree homotopy group for $M\text{Spin}^c$ (I'm thinking of something like $M\text{Spin}$ is weakly homotopy to $ku$ up to degree $7$.) We could leverage this to compute the AHSS, but I can't find much online.

Any comments/observations are really appreciated! Thanks a lot!

  • 1
    $\begingroup$ If your $X$ is an orientable manifold, then maybe you can use that the homology SS is a module over the cohomology SS, as described in Ben Antieau's answer here: mathoverflow.net/a/186421/8103 $\endgroup$
    – Mark Grant
    May 8, 2019 at 16:19
  • $\begingroup$ Wow, sadly the space I had in mind is the classifying space of some finite group, far from being closed/oriented (and even a manifold :( ) $\endgroup$
    – Riccardo
    May 8, 2019 at 17:12
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    $\begingroup$ Can you work over $\mathbb{F}_p$ for each $p$ and then reassemble the results to recover what happens with integer coefficients? $\endgroup$ May 8, 2019 at 20:53
  • $\begingroup$ I don't know how does it work. It would be really interesting though! Do you have any example/references where I can check out how does this reasoning work? $\endgroup$
    – Riccardo
    May 8, 2019 at 22:12
  • $\begingroup$ A standard tool for relating mod $p$ homology to integral homology is the Bockstein spectral sequence. See pages.vassar.edu/mccleary/files/2011/04/MC10.fin_.pdf for example. $\endgroup$ May 9, 2019 at 19:51

1 Answer 1


The realification map $BU \to BSO$ lifts through $BSpin^c$, because the first Chern class provides an integral lift of the second Stiefel-Whitney class. Calculating with the Serre spectral sequence for the homotopy fiber sequence $K(\mathbb{Z}, 2) \to BSpin^c \to BSO$ I get that $H^*(BSpin^c; \mathbb{F}_2)$ agrees with $\mathbb{F}_2[w_2, w_4, w_6, w_7]$ in degrees $* \le 7$, so that $BU \to BSpin^c$ is $6$-connected after $2$-completion. (I have not checked this against other sources, so caveat emptor.) It follows that $MU \to MSpin^c$ is $6$-connected, also after $2$-completion. Hence $Spin^c$-bordism agrees with complex bordism in the range of degrees you say you are interested in, and the Atiyah-Hirzebruch spectral sequence for $MSpin^c_*(X)$ agrees with that for $MU_*(X)$, after $2$-completion, in vertical degrees $* \le 5$. (Do you care about odd primes?)

Since $MU$ splits $2$-locally as $BP \vee \Sigma^4 BP \vee \dots$, you may as well work out the spectral sequence for $BP_*(X)$. The differential $d^3_{n,0} : H_n(X; BP_0) \to H_{n-3}(X; BP_2)$ is the integral homology operation induced by the $k$-invariant $\tilde Q_1 : H\mathbb{Z} \to \Sigma^3 H\mathbb{Z}$ with homotopy fiber the second Postnikov section of $BP$. (This is the same as the second Postnikov section of $MU$, and of $ku$.) The differential $d^3_{n,2} : H_n(X; BP_2) \to H_{n-3}(X; BP_4)$ is induced by the $k$-invariant connecting $BP_2$ and $BP_4$, which is also $\tilde Q_1$. Here $\tilde Q_1$ is an integral lift of the Milnor primitive operation $Q_1 = [Sq^2, Sq^1]$.

Hence, apart from one of the two $\mathbb{Z}$-summands in bidegree $(0,4)$ in your picture, $MSpin^c_*(X)$ will agree after $2$-completion with $ku_*(X)$ for $* \le 4$. If you are interested in $X = BG$, the book by R.R. Bruner and J.P.C. Greenlees on $ku_*(BG)$ should be helpful.

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    $\begingroup$ Wow, I'll need some time to digest this super detailed answer, but let me thank you a lot! $\endgroup$
    – Riccardo
    Jul 1, 2019 at 4:48

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