Here are various ways to define "Bockstein homomorphism:"

  1. Let $\beta_p:H^*(-,\mathbb{Z}_p) \to H^{*+1}(-,\mathbb{Z}_p)$ be the Bockstein homomorphism associated to the extension $$\mathbb{Z}_p\to\mathbb{Z}_{p^2}\to\mathbb{Z}_p,$$ it is an element of the mod $p$ Steenrod algebra $A_p$ where $p$ is a prime. If $p=2$, then $\beta_2=Sq^1$.

  2. Let $\beta_p'$:$H^*(-,\mathbb Z_p)\to H^{*+1}(-,\mathbb Z)$ be the Bockstein homomorphism associated to the extension $$\mathbb Z\stackrel{\cdot p}{\to}\mathbb Z\to\mathbb Z_p.$$

  3. Let $\beta_{2^n}: H^*(-,\mathbb Z_{2^n})\to H^{*+1}(-,\mathbb Z_2)$ be the Bockstein homomorphism associated to the extension $$\mathbb Z_2\to\mathbb Z_{2^{n+1}}\to\mathbb Z_{2^n}.$$ By the naturality of connecting homomorphism, $\beta_{2^{n+k}}\cdot2^k=\beta_{2^n}$ where $\cdot2^k: H^*(-,\mathbb Z_{2^n})\to H^*(-,\mathbb Z_{2^{n+k}})$ is induced from $\mathbb Z_{2^n}\stackrel{\cdot2^k}{\to}\mathbb Z_{2^{n+k}}$.

  • question (i) Since $\beta_2=Sq^1$ coincides with the Steenrod square, are there other additional coincidences of other "Generalized Square" (Pontryagin Square, Postnikov Square, etc) coincide with the above "Bockstein homomorphism" $\beta_p$, $\beta_p'$, $\beta_{2^n}$?

  • question (ii) Are there useful consistency formulas for these above "Bockstein homomorphism"?

For example, for Steenrod square, the total Stiefel-Whitney class $w=1+w_1+w_2+\cdots$ is related to the total Wu class $u=1+u_1+u_2+\cdots$ through the total Steenrod square $$ w=Sq(u),\ \ \ Sq=1+Sq^1+Sq^2+ \cdots . $$ Therefore, $w_n=\sum_{i=0}^n Sq^i (u_{n-i})$. The Steenrod squares have the following properties: $$ Sq^i(x_j) =0, \ i>j, \ \ Sq^j(x_j) =x_jx_j, \ \ Sq^0=1, $$

Do we have something similar for thse "Bockstein homomorphism?" $\beta_p$, $\beta_p'$, $\beta_{2^n}$?


I'm not sure if this is what you are asking, but you get useful relations among the Bockstein operators whenever you have a map between short exact coefficient sequences. For example, using the map of short exact sequences $\require{AMScd}$ \begin{CD} \mathbb{Z} @>\cdot p>> \mathbb{Z} @>\rho_p>> \mathbb{Z}/p\\ @V \rho_p V V @VV \rho_{p^2} V @VV = V\\ \mathbb{Z}/p @>>\cdot p> \mathbb{Z}/p^2 @>>\rho_p> \mathbb{Z}/p \end{CD} where $\rho_n$ denotes reduction mod $n$, and the associated map between long exact sequences of cohomology groups, one sees quickly that $\beta_p = \rho_p\circ \beta_p'$.

Similarly from the map of short exact sequences $\require{AMScd}$ \begin{CD} \mathbb{Z} @>\cdot 2^{k+1}>> \mathbb{Z} @>\rho_{2^{k+1}}>> \mathbb{Z}/2^{k+1}\\ @V \cdot 2 V V @VV = V @VV \rho_{2^k} V\\ \mathbb{Z} @>>\cdot 2^k> \mathbb{Z} @>>\rho_{2^k}> \mathbb{Z}/2^k \end{CD} one can deduce that $2\beta_{2^{k+1}}' = \beta_{2^k}'\circ \rho_{2^k}$. Thus for the Pontrjagin square $$ P_k:H^{2n}(X;\mathbb{Z}/2^k)\to H^{4n}(X;\mathbb{Z}/2^{k+1}), $$ which satisfies $\rho_{2^k}\circ P_k(x) = x^2$ for all $x\in H^{2n}(X;\mathbb{Z}/2^k)$, one sees that $2\beta_{2^{k+1}}'\circ P_k(x) = \beta_{2^k}'(x^2)$, giving a partial answer to your question (i).

  • $\begingroup$ Thank you very much +1, I appreciate it. $\endgroup$
    – wonderich
    Oct 14 '18 at 14:38

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