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  1. $\mathcal{P}_2$ is Pontryagin square $$H^{2i}(M,\mathbb Z_{2^k})\to H^{4i}(M,\mathbb{Z}_{2^{k+1}}).$$

  2. $\mathfrak{P}$ is the Postnikov square $$H^2(M,\mathbb Z_3)\to H^5(M,\mathbb Z_9).$$


Comments about question (i)

We know for $\mathbb Z_2$-valued cocycles $z_n$, $$ Sq^{n-k}(z_n) \equiv z_n\cup_{k} z_n $$ is always a cocycle. Here $Sq$ is called the Steenrod square.
More generally $h_n \cup_{k} h_n$ is a cocycle if $n+k =$ odd and $h_n$ is a cocycle.

If we define a generalized Steenrod square for cochains $c_n$: $$ \tilde Sq^{n-k} c_n \equiv c_n\cup_{k} c_n + c_n\cup_{k+1} d c_n . $$ We can check $$ d \tilde Sq^{k} c_n = d( c_n\cup_{n-k} c_n + c_n\cup_{n-k+1} d c_n ) $$ $$ = \tilde Sq^k d c_n, \;\;\;\; k=\text{odd} $$ $$ =\tilde Sq^k d c_n +(-)^{n} 2 \tilde Sq^{k+1} c_n , \;\;\;\; k=\text{odd}. $$ This $$\tilde Sq^{2} c_2 \equiv c_2\cup_{0} c_2 + c_2\cup_{1} d c_2$$ almost is the same as the Pontryagin square $\mathcal{P}_2$ above, for ($i=1,k=2$ above) $$H^{2}(M,\mathbb Z_{2})\to H^{4}(M,\mathbb{Z}_{4}),$$ for $$\mathcal{P}_2 (x_2) \equiv x_2\cup_{0} x_2 + x_2\cup_{1} d x_2.$$

Are these generalized Steenrod squares known? ($\tilde Sq^{n-k} c_n$) Where can I find more discussions along this?


Comments about question (ii) 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?" $\beta_p$, $\beta_p'$, $\beta_{2^n}$?

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You may be interested in the following paper:

Massey, W. S., Pontryagin squares in the Thom space of a bundle, Pac. J. Math. 31, 133-142 (1969). ZBL0188.28504.

Massey proves an analogue for the Pontryagin square of Thom's formula $w_k=\Phi^{-1}Sq^k(u)$ for the Stiefel-Whitney classes, which seems relevant to your question (ii).

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  • $\begingroup$ thanks so much for the precious ref +1! $\endgroup$ – wonderich Oct 16 '18 at 13:11

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