# An integral cohomology operation related to Steenrod square?

Let $\beta: H^n(X, \mathbb{Z}_2)\to H^{n+1}(X, \mathbb{Z})$ be the Bockstein homomorphism. Is it possible to define a cohomology operation $f: H^{n+1}(X, \mathbb{Z})\to H^{n+k+1}(X, \mathbb{Z})$ such that \begin{eqnarray*}\beta \text{Sq}^k=f\beta\end{eqnarray*} where $\text{Sq}^k: H^n(X, \mathbb{Z}_2)\to H^{n+k}(X, \mathbb{Z}_2)$ is the $k$-th Steenrod square?

Added: It seems that when $k=2$ it is not possible. If we precompose both sides by $\rho_2:H^n(X, \mathbb{Z})\to H^n(X, \mathbb{Z}_2)$, the reduction mod 2 map, then the LHS becomes the third differential of the Atiyah-Hirzebruch spectral sequence which in general does not vanish, but the RHS vanishes because $\beta\rho_2=0$.

• Depends on what is the composite$$H^n(X,\mathbb Z)\to H^n(X,\mathbb Z/2)\xrightarrow{\mathrm{Sq}^k}H^{n+k}(X,\mathbb Z/2)\xrightarrow\beta H^{n+k+1}(X,\mathbb Z)$$ – მამუკა ჯიბლაძე Oct 28 '16 at 14:12

From the Adem relations we get $Sq^1Sq^{k-1}=(k-2)Sq^k$. It follows that for $k$ odd we have $Sq^k=Sq^1Sq^{k-1}$. Note that we have $Sq^1=\rho\beta$, where $\rho$ is mod 2 reduction, so if $k$ is odd $\beta Sq^k=\beta Sq^1Sq^{k-1}=\beta\rho\beta Sq^{k-1}=0$ since $\beta\rho\simeq \ast$. Therefore we are forced to take $f$ such that $f\beta=0$. Working stably we would then get a map $g:H\mathbb{Z}\rightarrow\Sigma^{k}H\mathbb{Z}$ such that $2g=f$. That is $f$ is divisible by 2. Alternatively, any map $H\mathbb{Z}\rightarrow\Sigma^{k}H\mathbb{Z}$ that is divisible by two defines an $f$ with the requried properties. There are many such examples odd-primary torsion classes that work, as well as higher 2-torsion, although it is clear that they do not exist for every odd $k$.

For $k$ even $\beta Sq^k$ is nontrivial since $Sq^1Sq^k=Sq^{k+1}$. Therefore $f\beta=\beta Sq^k$ must be non-trivial as well but this cannot happen. Assume to the contrary that such an $f$ exists. Then we have that $Sq^kSq^1=Sq^k\rho\beta\in H^{k+1}(K(\mathbb{Z}_2,n-1);\mathbb{Z}_2)$ and $Sq^1Sq^kSq^1=Sq^{k+1}Sq^1\in H^{k+2}(K(\mathbb{Z}_2,n-1);\mathbb{Z}_2)$ are both non-trivial by Cartan. It follows that

$0\neq Sq^{k+1}Sq^1=Sq^1Sq^{k}=\rho\beta Sq^kSq^1=\rho f\beta Sq^1=0$

Since $\beta Sq^1=0$. This is a contradiction and so $f$ cannot exist when $k$ is even.

Such an $f$ only exists in two cases: when $k=1$ (when $\beta Sq^1 = 0$) and when $k=0$ (when $\beta Sq^0 = \beta$). Here is a proof, which will take a little work with the Steenrod algebra.

Let's suppose that $X$ and $Y$ are either $H\Bbb Z$ or $H\Bbb Z/2$ (not necessarily the same). Then the map $$[X,Y]_t \to Hom_{A^*}(H^*Y, \Sigma^t H^*X),$$ sending a map to its (graded) effect on cohomology, is an isomorphism for $t \neq 0$. If one of $X$ or $Y$ is $H\Bbb Z/2$, it is an isomorphism for all $t$.

Let's prove this. This map is the edge morphism in the Adams spectral sequence $$Ext_{A^*}^s(H^*Y, \Sigma^t H^*X) \Rightarrow ([X,Y]^\wedge_2)_{t-s}$$ and so it suffices to show that the Adams spectral sequence degenerates.

If $Y = H\Bbb Z/2$, then $H^* Y$ is free over $A^*$ and there are no higher Ext-groups.

If $Y = H\Bbb Z$, then $H^* Y = A^* / Sq^1$, which has a free resolution $$\cdots \to A^* \xrightarrow{\cdot Sq^1}A^* \xrightarrow{\cdot Sq^1} A^* \to A^*/Sq^1 \to 0$$ (using the basis of $A^*$ in terms of admissible monomials). Applying $Hom_{A^*}(-, \Sigma^* H^*(X))$ to this resolution, we get the chain complex $$0 \to \Sigma^* H^*(X) \xrightarrow{Sq^1 \cdot} \Sigma^* H^*(X) \xrightarrow{Sq^1 \cdot} \Sigma^* H^*(X) \to \cdots$$ which gives us the Ext-groups (in particular, we could interpret this as the Bockstein spectral sequence from mod-2 cohomology to integral cohomology). If $X = H\Bbb Z/2$, then the basis of admissible monomials shows that this is exact except in degree zero, where it recovers $ker(Sq^1 \cdot -) = Im(Sq^1 \cdot -)$. If $X = H\Bbb Z$, then $H^*(X)$ has a basis consisting of admissible monomials we similarly find that this is exact except in degree zero: $A^* / Sq^1$ is a direct sum of $\Bbb Z/2$ in degree zero with a bunch of free modules over $\Bbb Z/2[Sq^1] / (Sq^1 \cdot Sq^1)$ on the admissible monomials $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$. Therefore, the higher Ext-groups vanish except for those contributing to $[X,Y]_0 = [H\Bbb Z, H\Bbb Z]_0$. The degree-zero part $Hom(A^*/Sq^1, A^*/Sq^1) = ker(Sq^1 \cdot -)$ has a basis consisting of $1$ and admissible monomials of the form $Sq^{2k+1} Sq^{n_1} \cdots Sq^{n_r}$.

In particular, we found in the course of the proof that the map $$[H\Bbb Z, H\Bbb Z]_n \to [H\Bbb Z, H\Bbb Z/2]_n,$$ sending $f$ to $\rho_2 \circ f$, is injective for $n \neq 0$. The image is parametrized by those elements of $A^* / Sq^1$ which are sums of admissible monomials $Sq^{2k+1} Sq^{n_1} \cdots Sq^{n_r}$; this corresponds to the cohomology operation $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2$.

Thus:

• The set $([H\Bbb Z, H\Bbb Z]_n)^\wedge_2$ is $2$-torsion for $n > 0$, and has a basis consisting of the cohomology operations $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2$ over $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$ admissible monomials in the Steenrod algebra with $n_r > 1$.

• The set $[H\Bbb Z/2, H\Bbb Z]_n$ has a basis of elements of the cohomology operations $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$ over $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$ admissible monomials in the Steenrod algebra.

• The subset of elements of the form $f \beta$ are thus sums of elements of the form $$\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2 \beta= \beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1.$$ The image of right multiplication by $\beta$, therefore, has a basis consisting of sums of elements of the form $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$, where $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$ is an admissible monomial in the Steenrod algebra. In particular, $\beta Sq^n$ is not of this form unless $n=1$ (when it is zero) or $n=0$ (when it is $\beta$).