Why is $\pi_{-*}F(H\mathbb{F}_p, H\mathbb{F}_p)$ the mod $p$ Steenrod algebra? (This is quite a common statement, seen, for instance, in EKMM.)

To be more precise, stable mod $p$ cohomology operations can be realized by maps of spectra (in particular, elements of $\pi_{-*}F(H\mathbb{F}_p, H\mathbb{F}_p)$), but why is this representation well-defined? (The existence of phantom maps makes me think that this is non-trivial.)

  • $\begingroup$ What's your definition of the Steenrod algebra? $\endgroup$ Jan 2, 2019 at 11:59
  • $\begingroup$ The algebra of stable cohomology operation in ordinary mod $p$ cohomology. $\endgroup$
    – user09127
    Jan 2, 2019 at 12:17
  • 1
    $\begingroup$ Sure, but what's your definition of stable cohomology operation? (Often the result you cite is taken as the definition of a stable cohomology operation...). In particular, are your cohomology theories defined on spaces or spectra? $\endgroup$ Jan 2, 2019 at 12:18
  • $\begingroup$ That's probably where the problem arise! I was wondering whether this was an issue. The cohomology operations are on spaces. $\endgroup$
    – user09127
    Jan 2, 2019 at 12:20

1 Answer 1


Recall that by representability of cohomology plus the Yoneda lemma, a cohomology operation $H^i→H^j$ is the same thing as a map $$ K(\mathbb{F}_p,i)→K(\mathbb{F}_p,j)\,.$$ Moreover, the suspension isomorphism $\sigma:H^i(X)\cong H^{i+1}(\Sigma X)$ is implemented by the counit of the suspension-loopspace adjunction $$\sigma: ΣK(\mathbb{F}_p,i)\cong ΣΩK(\mathbb{F}_p,i+1)→K(\mathbb{F}_p,i+1)$$ by sending a map $X\to K(\mathbb{F}_p,i)$ to $\Sigma X\to \Sigma K(\mathbb{F}_p,i)\to K(\mathbb{F}_p,i+1)$.

Putting all together, a stable cohomology operation $f:H^\ast→H^{\ast+d}$ is a collection of maps $$\{f_i:K(\mathbb{F}_p,i)→K(\mathbb{F}_p,i+d)\}_{i\ge 0}$$ together with a family of homotopy commutative diagrams

$$\require{AMScd} \begin{CD} \Sigma K(\mathbb{F}_p,i) @>{\Sigma f_i}>> \Sigma K(\mathbb{F}_p,i+d)\\ @V{\sigma}VV @V{\sigma}VV \\ K(\mathbb{F}_p,i+1) @>{f_{i+1}}>> K(\mathbb{F}_p,i+d+1) \end{CD}\,.$$

So, the group of stable cohomology operations of degree $d$ is precisely $$\mathscr{A}^d=\lim [K(\mathbb{F}_p,i),K(\mathbb{F}_p,i+d)]\cong \lim H^{i+d}(K(\mathbb{F}_p,i)$$

On the other hand, we have (basically by definition) $$\mathrm{Map}(H\mathbb{F}_p,\Sigma^dH\mathbb{F}_p)\cong \mathrm{ho}\lim \mathrm{Map}(K(\mathbb{F}_p,i),K(\mathbb{F}_p,i+d))\,.$$ where with $\mathrm{Map}$ I'm denoting the space of maps of pointed spaces and of spectra respectively. So we have a Milnor exact sequence $$0\to \lim{}^1 H^{i+d-1}(K(\mathbb{F}_p,i))\to [H\mathbb{F}_p,\Sigma^d H\mathbb{F}_p] \to \mathscr{A}^d\to 0$$ So the result you want follows if we can show that the $\lim{}^1$-term is trivial. This is easy enough to do directly (we know all cohomology groups involved), although I feel there should be some proof that works without this computational input.

EDIT: Dylan Wilson in the comments notes that we can say that the $\lim{}^1$ vanishes simply because all $\mathbb{F}_p$-vector spaces in the direct system are finite dimensional, without need for a detailed computation of the cohomology of EM spaces.

  • $\begingroup$ Isn't the square only commutative up to homotopy? $\endgroup$
    – user09127
    Jan 2, 2019 at 12:15
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    $\begingroup$ @user09127 Yes, all my commutative squares come with a choice of homotopy in the middle :). The precise statement is that $\mathrm{Map}(H\mathbb{F}_p,\Sigma^dH\mathbb{F}_p)\cong \mathrm{holim_i}\, \mathrm{Map}(K(\mathbb{F}_p,i),K(\mathbb{F}_p,i+d))$. $\endgroup$ Jan 2, 2019 at 12:17
  • $\begingroup$ Is that ture? (And what do you mean by Map?) $\endgroup$
    – user09127
    Jan 2, 2019 at 12:19
  • $\begingroup$ @user09127 By $\mathrm{Map}$ I mean the "mapping space". I fear you're going to have to detail a little bit more your background if you want more details. $\endgroup$ Jan 2, 2019 at 12:20
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    $\begingroup$ Lim^1 vanishes for any sequence of finite dimensional vector spaces, so you’re good as long as you know that these EM spaces are locally of finite type, which is much weaker than computing their cohomology directly. $\endgroup$ Jan 2, 2019 at 15:01

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