The Homotopy Fiber of \(\mathrm{Sq}^1\)

We use the Serre spectral sequence to compute the homotopy fiber of the Steenrod square \(\mathrm{Sq}^1\colon K(\mathbb F_2,1)\to K(\mathbb F_2,2)\).

From our previous blog post, we recall the definitions of the Eilenberg–MacLane spaces and their basic properties.

Definition 1. Fix an abelian group \(A\) and a nonnegative integer \(n\). Then the Eilenberg–MacLane space \(K(A,n)\) is the unique simplicial abelian group (up to equivalence) whose normalized chain complex is only supported in position \(n\), where it is \(A\).

Remark 2. The Eilenberg–MacLane space \(K(A,n)\) for \(n\ge1\) is the unique space \(X\) up to homotopy for which \[\pi_i(X)=\begin{cases}A & \text{if }i=n, \\0 & \text{otherwise}.\end{cases}\]

Example 3. By Remark 2, we can see that \(K(A,0)\) is the discrete topological group \(A\).

Example 4. There is a universal covering \(S^\infty\to\mathrm{RP}^\infty\) with fiber \(\mathbb F_2\). Thus, \(\mathbb F_2\to S^\infty\to\mathrm{RP}^\infty\) is a homotopy fiber sequence, so we can use the homotopy long exact sequence to compute the homotopy groups of \(\mathrm{RP}^\infty\). (Note \(S^\infty\) is contractible!) It follows that \(K(\mathbb F_2,1)=\mathbb{RP}^\infty\).

Remark 6. In general, taking loops gives us a homotopy fiber sequence \[\Omega K(A,n+1)\to\Delta^0\to K(A,n+1).\] One can use the homotopy long exact sequence to identify \(\Omega K(A,n+1)=K(A,n)\). (This can also be shown directly using properties of the Dold–Kan correspondence.)

Remark 7. Here is what makes \(K(A,n)\)s important: they “represent” cohomology in the sense that \[[-,K(A,n)]=\mathrm H^n(-;A).\] Plugging in \(\mathrm{id}\colon K(A,n)\to K(A,n)\) gives us a canonical class \(\iota_n\in\mathrm H^n(K(A,n);A)\). Thus, by naturality, a homotopy class \(f\colon X\to K(A,n)\) of maps corresponds to \(f^*\iota_n\in\mathrm H^n(X;A)\).

In algebraic topology courses, one can spend quite a bit of time computing the cohomology of Eilenberg–MacLane spaces (with or without admitting). Remark 7 and the Yoneda lemma explains why one may want to do this: a class \(\alpha\in\mathrm H^n(K(A,m);B)\) corresponds to a homotopy class of maps \[\varphi\colon K(A,m)\to K(B,n)\] for which \(\varphi^*\iota_n=\alpha\). Such a map \(\varphi\) produces a natural transformation \(\mathrm H^m(-;A)\Rightarrow\mathrm H^n(-;B)\) defined by sending a map \(X\to K(A,m)\) to the composite \(X\to K(A,m)\to K(B,n)\). Conversely, we can of course recover a map \(K(A,m)\to K(B,n)\) from such a natural transformation by plugging in the identity.

Example 8. The cohomology ring \(\mathrm H^*(K(\mathbb F_2,1);\mathbb F_2)\) is \(\mathrm H^*(\mathbb{RP}^\infty;\mathbb F_2)=\mathbb F_2[x_1]\), where \(x_1\) is the unique nontrivial class in degree \(1\). In particular, because \(\iota_1\) is also a nontrivial class in degree \(1\), we conclude that \(\mathrm H^*(K(\mathbb F_2,1);\mathbb F_2)\cong\mathbb F_2[\iota_1]\). The cohomology class \(\iota_1^n\in\mathrm H^n(K(\mathbb F_2,1);\mathbb F_2)\) corresponds to the natural transformation \[\mathrm H^1(-;\mathbb F_2)\Rightarrow\mathrm H^n(-;\mathbb F_2)\] given by \(x\mapsto x^n\). Indeed, one simply needs to plug in \(\iota_1\).

Definition 9. Taking the cup product of a class \(x\) with itself produces a natural transformation \[\mathrm H^1(-;\mathbb F_2)\Rightarrow\mathrm H^2(-;\mathbb F_2).\] This natural transformation corresponds (by definition) to the first Steenrod square \(\operatorname{Sq}^1\colon K(\mathbb F_2,1)\to K(\mathbb F_2,2)\). By unravelling definitions, \(\operatorname{Sq}^{1*}(\iota_2)=\iota_1\cup\iota_1\). Indeed, this is a matter of tracking \(\operatorname{id}_{K(\mathbb F_2,1)}\) around the following commuting square. \[\begin{array}{ccc} {[K(\mathbb F_2,1),K(\mathbb F_2,1)]}&{\xrightarrow{\mathrm{Sq}^1\circ-}}&{[K(\mathbb F_2,1),K(\mathbb F_2,2)]}\\{} \downarrow && \downarrow\\{} \mathrm H^1(K(\mathbb F_2,1);\mathbb F_2)&\xrightarrow{\mathrm{Sq}^{1*}}&\mathrm H^2(K(\mathbb F_2,1);\mathbb F_2)\end{array}\]

We are interested in computing the homotopy fiber of \(\operatorname{Sq}^1\colon K(\mathbb F_2,1)\to K(\mathbb F_2,2)\).

Proposition 10. The homotopy fiber of \(\operatorname{Sq}^1\colon K(\mathbb F_2,1)\to K(\mathbb F_2,2)\) is \(K(\mathbb Z/4\mathbb Z,1)\).

Proof. We proceed in steps. Let the homotopy fiber be \(F\).

1. We claim that \(F\) takes the form \(K(A,1)\), where \(A\) is some group of order \(4\).

   To show this, we use Remark 2. By the homotopy long exact sequence, we have an exact sequence \[\pi_{i+1}K(\mathbb F_2,1)\to\pi_{i+1}K(\mathbb F_2,2)\to\pi_iF\to\pi_iK(\mathbb F_2,1)\to\pi_iK(\mathbb F_2,2).\] In particular, both terms neighboring \(\pi_iF\) immediately vanish if \(i\ne1\), so we see that \(F=K(\pi_1F,1)\). As for \(\pi_1F\), we see we have a short exact sequence \[0\to\mathbb F_2\to\pi_1F\to\mathbb F_2\to0,\] so the result follows.

2. It remains to determine \(\pi_1F\), which we know is an abelian group of order \(4\). Thus, by the Hurewisz theorem, we know \(\pi_1F=\mathrm H_1(F;\mathbb Z)\). Because this group is either \((\mathbb Z/2\mathbb Z)^2\) or \(\mathbb Z/4\mathbb Z\), it is enough to show \(\dim_{\mathbb F_2}\mathrm{Hom}(\mathrm H_1(F;\mathbb Z),\mathbb F_2)=1\), which is equivalent to checking that \(\dim_{\mathbb F_2}\mathrm H^1(F;\mathbb F_2)=1\) by the Universal coefficients theorem.

   We now use the cohomological Serre spectral sequence on our homotopy fiber sequence. Because \(K(\mathbb F_2,2)\) is simply connected, we get a spectral sequence \(E_\bullet\) for which \[E_2^{pq}=\mathrm H^p(K(\mathbb F_2,2);\mathbb F_2)\otimes\mathrm H^q(F;\mathbb F_2)\Rightarrow\mathrm H^{p+q}(K(\mathbb F_1,2);\mathbb F_2).\] Here are the terms of \(E_2\) we will be interested in. \[\begin{array}{ccc}\mathrm H^1(F;\mathbb F_2) & 0 & \mathrm H^1(F;\mathbb F_2)\iota_2 \\{} \mathbb F_2 & 0 & \mathbb F_2\iota_2\end{array}\] In particular, we have used the fact that \(\mathrm H^1(K(\mathbb F_2,2);\mathbb F_2)=0\), which follows from the Hurewisz theorem; for the same reason, we know that \(\mathrm H^2(K(\mathbb F_2,2);\mathbb F_2)=\mathbb F_2\iota_2\).

   To continue, note \(E_2^{10}\) and \(E_2^{20}\) only interact with \(d_2\colon E^{10}_2\to E_2^{02}\), so they converge as soon as we pass to \(E_3\). For example, this implies that \(\ker d_2=\mathbb F_2\). However, by construction of the Serre spectral sequence, \(E_\infty^{20}\) is the image of the map \(K(\mathbb F_2,1)\to K(\mathbb F_2,2)\) on \(\mathrm H^2(-;\mathbb F_2)\). But as in Definition 9, the map \[\operatorname{Sq}^1\colon\mathrm H^2(K(\mathbb F_2,2);\mathbb F_2)\to\mathrm H^2(K(\mathbb F_2,1);\mathbb F_2)\] is an isomorphism sending the generating class \(\iota_2\) to the generating class \(\iota_1\cup\iota_1\).

   Thus, \(\operatorname{coker}d_2=\mathbb F_2\), so \(d_2\) is the zero map. We conclude that \(\mathrm H^1(F;\mathbb F_2)=\mathbb F_2\). \(\blacksquare\).

Remark 11. One can alternatively consider the constant map \(K(\mathbb F_2,1)\to K(\mathbb F_2,2)\), but the homotopy fiber will now be \(K\left((\mathbb Z/2\mathbb Z)^2,1\right)\). Indeed, we again find that the fiber \(F\) takes the form \(K(\pi_1F,1)\), where \(\pi_1F\) is an abelian group of order \(4\). One can again use the Serre spectral sequence to calculate \(\mathrm H^1(F;\mathbb F_2)\). The difference is that this time the map \[\mathrm H^2(K(\mathbb F_2,2);\mathbb F_2)\to\mathrm H^2(K(\mathbb F_2,1);\mathbb F_2)\] is trivial, so \(d_2\colon E^{10}_2\to E_2^{02}\) must have trivial cokernel and therefore be surjective. It follows that \(\dim_{\mathbb F_2}\mathrm H^1(F;\mathbb F_2)-1=\dim_{\mathbb F_2}\mathrm H^1(K(\mathbb F_2,1);\mathbb F_2)\), so \(\dim_{\mathbb F_2}\mathrm H^1(F;\mathbb F_2)=2\) follows.