# Can Bockstein Spectral Sequence detect multiple summands of the same power, in homology?

I understand that the differential $d^k$ of the Bockstein S.S. (mod p) is nonzero iff the homology $H_*(X;\mathbb{Z})$ has summand of the form $\mathbb{Z}/p^k$.

How about for multiple summands in the integral homology, say $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$. How does the Bockstein Spectral Sequence detect them? Merely knowledge that the differential is nonzero doesn't seem to be enough to detect the multiple summand.

Thanks for any help.

• If the differential is nonzero on two linearly independent classes, then each of those classes corresponds to a different summand. So I suppose you want to know the rank of the differential as a linear map; the rank should be the number of summands. – John Palmieri Nov 2 '17 at 14:29
• Thanks. Is there any reference that mentions this? I checked McCleary's book, it doesn't seem to have it. – yoyostein Nov 2 '17 at 14:32
• @JohnPalmieri Just to check regarding the dimension. Is the following true: if there is one summand $\mathbb{Z}/p$ in $d^r E^r_n$, then there is one summand $\mathbb{Z}/p^r$ in $H_{n-1}(X)$. Is that correct? Thanks! – yoyostein Nov 3 '17 at 2:11
• I'm not sure about the grading. Regarding references, look at the description of the spectral sequence in Mosher and Tangora, Cohomology Operations and Applications in Homotopy Theory. – John Palmieri Nov 3 '17 at 18:01

## 1 Answer

Each element in the image of $d_n$ corresponds to an element of order $p^n$ in the homology with integer coefficients. So it suffices to compute the image of $d_1=\beta$ on $H_*(X;\mathbb{Z}/p)$ to determine the number of summands isomorphic to $Z/p$ in $H_*(X;\mathbb{Z})$.