# Bockstein homomorphism from $H^d(BG,Z_2)$ to $H^{d+1}(BG,Z)$, and Steenrod Square $Sq^1$

The Theorem 1.5 and 1.6 of

Brown, Edgar H., Jr. The cohomology of BSOn and BOn with integer coefficients. Proc. Amer. Math. Soc. 85 (1982), no. 2, 283–288.

give a general answer for $H^d(BSO_n,Z)$ and $H^d(BO_n,Z)$, in term of $\delta(w_i)$ and $p_i$, where $w_i$ are the Stiefel-Whitney classes.

Here $\delta$ is the natural map from $H^d(BG,Z_2)$ to $H^{d+1}(BG,Z)$. So $\delta$ looks like the Bockstein homomorphism (of some sort).

My question is that can we regard $\delta$ as the Steenrod Square $Sq^1$ in the above Theoroms?

• $\operatorname{Sq}^1=\operatorname{Bockstein}$: this is one of the definition (or theorem: there is but one stable operation of degree $1$). – Alex Degtyarev Nov 2 '14 at 12:42
• Doesn't $Sq^1$ map to $H^{d+1}(BG,Z_2)$ rather than $H^{d+1}(BG,Z)$? – ThiKu Nov 2 '14 at 12:44
• @AD: yes, but it's the Bockstein for the sequence $0\to Z_2\to Z_4\to Z_2\to 0$. – ThiKu Nov 2 '14 at 12:45
• There is another Bockstein $\beta$ (I think) that maps $H^d(BG,Z_2)$ to $H^{d+1}(BG,Z_2)$. I think $\beta =Sq^1$. But in the theorems, we only care about $\delta$ mod 2. So I wonder, can one say $\delta=\beta=Sq^1$ mod 2. – Xiao-Gang Wen Nov 2 '14 at 14:11
• Yes, that's correct. – Alex Suciu Nov 2 '14 at 16:13

## 1 Answer

The natural maps from $$0\to Z\to Z\to Z_2\to 0$$ to $$0\to Z_2\to Z_4\to Z_2\to 0$$ commute with the homomorphisms in the s.e.s. and hence also with the connecting homomorphisms of the long exact sequence, i.e. with the Bockstein homomorphisms.

In particular, the Bockstein of the first sequence corresponds to $Sq^1$ after mod 2 reduction.