# Third differential in the homology AHSS

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)

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!

• 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 – Mark Grant May 8 '19 at 16:19
• 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 :( ) – Riccardo May 8 '19 at 17:12
• Can you work over $\mathbb{F}_p$ for each $p$ and then reassemble the results to recover what happens with integer coefficients? – John Palmieri May 8 '19 at 20:53
• 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? – Riccardo May 8 '19 at 22:12
• 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. – John Palmieri May 9 '19 at 19:51

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.

• Wow, I'll need some time to digest this super detailed answer, but let me thank you a lot! – Riccardo Jul 1 '19 at 4:48