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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.