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Suppose $x \in Ext^{s,t}_A(k,k)$ and let $\mathcal{C} = \cdots \to C_s \to \cdots \to C_0 \to \to k \to 0$ be a resolution so that $x$ is represented by a cocycle $C_s \to \Sigma^t k$.

To compute $Sq^i(x)$, find a 'small' extension $0 \to \Sigma^t k \to M_{s-1} \to \cdots \to M_0 \to k \to 0$ realizing x and call it $\mathcal{X}$. Then $x$ is represented by a chain map $\mathcal{C} \to \mathcal{X}$ with the cocycle $C_s \to \Sigma^t k$ at one end and the identity of $k$ at the other. The cocommutative Hopf algebra structure of $A$ makes $\mathcal{X} \otimes_k \mathcal{X}$ a complex of $A$-modules, and there is a chain map $\chi : \mathcal{C} \to \mathcal{X} \otimes_k \mathcal{X}$ covering $1_k$ whose other end $C_{2s} \to \Sigma^{2t} k$ is a cocycle representing $x^2$. The composite $\tau \chi$ of this chain map with the twist map $\tau : \mathcal{X} \otimes \mathcal{X} \to \mathcal{X} \otimes \mathcal{X}$ is another lift of $1_k$, so there is a chain homotopy $\chi_1 : \mathcal{C} \to \mathcal{X} \otimes \mathcal{X}$ between them. This gives a cocycle $C_{2s-1} \to \Sigma^{2t} k$ which represents $x \cup_1 x = Sq^{s-1}(x)$. Repeat to get $\chi_2$, giving a cocycle $C_{2s-2} \to \Sigma^{2t} k$ representing $x \cup_2 x = Sq^{s-2}(x)$, etc. If the extension $\mathcal{X}$ is small, this is a remarkable remarkably effective way to compute these operations, ${\it{\text{ if you have a resolution to work with}}}$.

Two minute exercise: show that $Sq^0(h_0) = h_1$ in $Ext_{A(1)}(F_2,F_2)$ using this method and the resolution

$C_1 = \Sigma^1 A(1) \oplus \Sigma^2 A(1) \to C_0 = A(1)$ by $\iota_1 \mapsto Sq^1$, $\iota_2 \mapsto Sq^2$, and

$C_2 = \Sigma^2 A(1) \oplus \Sigma^4 A(1) \to C_1$ by $\iota_2 \mapsto Sq^1 \iota_1$ and $\iota_4 \mapsto Sq^1 Sq^2 \iota_1 + Sq^2 \iota_2$.

What have we done? The universal case is a $C_2$-equivariant map $\mathcal{W} \otimes \mathcal{C} \to \mathcal{C} \otimes \mathcal{C}$ extending the diagonal $\mathcal{C} \to \mathcal{C} \otimes \mathcal{C}$ coming from the cocommutative coproduct of $A$. This is what May's Springer LNM V. 168 article needs to construct Steenrod operations. Unfortunately, if $\mathcal{C}$ is a production strength resolution, not just some little toy, then $\mathcal{C} \otimes \mathcal{C}$ is a monster, and we do not want to be trying to lift maps over its differential. (Maybe you, dear reader, see a way out of this. If so, this is great news!)

Christian's clever observation is that we can do this one cocycle at a time, and compute the map $\mathcal{W} \otimes \mathcal{C} \to \mathcal{X} \otimes \mathcal{X}$ if we can find small extensions $\mathcal{X}$ representing the cocycles of interest. This step is not algorithmic, because the easy answers are too large, essentially as large as $\mathcal{C}$ itself. On the other hand it is quite feasible by hand in low degrees. Sean Tilson and I have carried this out for $c_0$, $d_0$, $e_0$, $f_0$ and $n$ (in the BMT notation for $Ext_A(F_2,F_2)$), perhaps another one or two. I suppose we should convert the results from computer code to TeX one day. I was interested in this in order to unambiguously determine which of the two possible elements in $Ext^{4,22}$ found by my computer program was $f_0 = Sq^1(c_0)$ and which was $f_0 + h_1^3 h_4$. Similarly for $f_1 \in Ext^{4,44}$; this follows by $Sq^0$ from identifying $f_0$, and as Christian notes, $Sq^0$ is easy since that is induced by the Frobenius in the dual of $A$, hence by restriction along the dual of the Frobenius, $A \to D(A)$, where $D(A)$ is the 'double' of $A$, $D(A)_{2n} = A_n$ and $D(A)_{2n+1} = 0$. On Milnor basis elements this is $Sq(2r_1, \ldots, 2r_k) \mapsto Sq(r_1, \ldots, r_k)$ while basis elements with any odd entries go to 0, so it is easy to automate.

Finally, since what I thought was going to be a paragraph or two has already grown absurdly long, I might as well go whole hog and mention a favorite problem. The complex $\mathcal{X} \otimes \mathcal{X}$ is one way to construct a complex representing $x^2$. The Yoneda composite of $\mathcal{X}$ with itself is a much 'slimmer' one. Unfortunately, we cannot recognize $\tau$ in it. If we could, we could then automate the calculation of the $Sq^i$. This is the problem of a description of the squaring operations in terms of Yoneda product rather than tensor product and a solution would be great to have.

1

Suppose $x \in Ext^{s,t}_A(k,k)$ and let $\mathcal{C} = \cdots \to C_s \to \cdots \to C_0 \to \to k \to 0$ be a resolution so that $x$ is represented by a cocycle $C_s \to \Sigma^t k$.

To compute $Sq^i(x)$, find a 'small' extension $0 \to \Sigma^t k \to M_{s-1} \to \cdots \to M_0 \to k \to 0$ realizing x and call it $\mathcal{X}$. Then $x$ is represented by a chain map $\mathcal{C} \to \mathcal{X}$ with the cocycle $C_s \to \Sigma^t k$ at one end and the identity of $k$ at the other. The cocommutative Hopf algebra structure of $A$ makes $\mathcal{X} \otimes_k \mathcal{X}$ a complex of $A$-modules, and there is a chain map $\chi : \mathcal{C} \to \mathcal{X} \otimes_k \mathcal{X}$ covering $1_k$ whose other end $C_{2s} \to \Sigma^{2t} k$ is a cocycle representing $x^2$. The composite $\tau \chi$ of this chain map with the twist map $\tau : \mathcal{X} \otimes \mathcal{X} \to \mathcal{X} \otimes \mathcal{X}$ is another lift of $1_k$, so there is a chain homotopy $\chi_1 : \mathcal{C} \to \mathcal{X} \otimes \mathcal{X}$ between them. This gives a cocycle $C_{2s-1} \to \Sigma^{2t} k$ which represents $x \cup_1 x = Sq^{s-1}(x)$. Repeat to get $\chi_2$, giving a cocycle $C_{2s-2} \to \Sigma^{2t} k$ representing $x \cup_2 x = Sq^{s-2}(x)$, etc. If the extension $\mathcal{X}$ is small, this is a remarkable effective way to compute these operations, ${\it{\text{ if you have a resolution to work with}}}$.

Two minute exercise: show that $Sq^0(h_0) = h_1$ in $Ext_{A(1)}(F_2,F_2)$ using this method and the resolution

$C_1 = \Sigma^1 A(1) \oplus \Sigma^2 A(1) \to C_0 = A(1)$ by $\iota_1 \mapsto Sq^1$, $\iota_2 \mapsto Sq^2$, and

$C_2 = \Sigma^2 A(1) \oplus \Sigma^4 A(1) \to C_1$ by $\iota_2 \mapsto Sq^1 \iota_1$ and $\iota_4 \mapsto Sq^1 Sq^2 \iota_1 + Sq^2 \iota_2$.

What have we done? The universal case is a $C_2$-equivariant map $\mathcal{W} \otimes \mathcal{C} \to \mathcal{C} \otimes \mathcal{C}$ extending the diagonal $\mathcal{C} \to \mathcal{C} \otimes \mathcal{C}$ coming from the cocommutative coproduct of $A$. This is what May's Springer LNM V. 168 article needs to construct Steenrod operations. Unfortunately, if $\mathcal{C}$ is a production strength resolution, not just some little toy, then $\mathcal{C} \otimes \mathcal{C}$ is a monster, and we do not want to be trying to lift maps over its differential. (Maybe you, dear reader, see a way out of this. If so, this is great news!)

Christian's clever observation is that we can do this one cocycle at a time, and compute the map $\mathcal{W} \otimes \mathcal{C} \to \mathcal{X} \otimes \mathcal{X}$ if we can find small extensions $\mathcal{X}$ representing the cocycles of interest. This step is not algorithmic, because the easy answers are too large, essentially as large as $\mathcal{C}$ itself. On the other hand it is quite feasible by hand in low degrees. Sean Tilson and I have carried this out for $c_0$, $d_0$, $e_0$, $f_0$ and $n$ (in the BMT notation for $Ext_A(F_2,F_2)$), perhaps another one or two. I suppose we should convert the results from computer code to TeX one day. I was interested in this in order to unambiguously determine which of the two possible elements in $Ext^{4,22}$ found by my computer program was $f_0 = Sq^1(c_0)$ and which was $f_0 + h_1^3 h_4$. Similarly for $f_1 \in Ext^{4,44}$; this follows by $Sq^0$ from identifying $f_0$, and as Christian notes, $Sq^0$ is easy since that is induced by the Frobenius in the dual of $A$, hence by restriction along the dual of the Frobenius, $A \to D(A)$, where $D(A)$ is the 'double' of $A$, $D(A)_{2n} = A_n$ and $D(A)_{2n+1} = 0$. On Milnor basis elements this is $Sq(2r_1, \ldots, 2r_k) \mapsto Sq(r_1, \ldots, r_k)$ while basis elements with any odd entries go to 0, so it is easy to automate.

Finally, since what I thought was going to be a paragraph or two has already grown absurdly long, I might as well go whole hog and mention a favorite problem. The complex $\mathcal{X} \otimes \mathcal{X}$ is one way to construct a complex representing $x^2$. The Yoneda composite of $\mathcal{X}$ with itself is a much 'slimmer' one. Unfortunately, we cannot recognize $\tau$ in it. If we could, we could then automate the calculation of the $Sq^i$. This is the problem of a description of the squaring operations in terms of Yoneda product rather than tensor product and a solution would be great to have.