Why isn't $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[[x]]$?

Solution 1:

This is probably something that should be a comment, mainly because I'm not sure it is correct. Feel free to tell me it is rubbish!

1) I have definitely seen it written $H^*(\mathbb{R} P^\infty; \mathbb{F}_2) \simeq \mathbb{F}_2[\![ x ]\!]$. For example see Jacob Lurie's notes here.

2) Cohomlogy doesn't play well with inverse limits in general. There is the Milnor exact sequence, which is this case should give:

$$0 \to \text{lim}^1 H^{\ast-1}(\mathbb{R} P^\infty;\mathbb{F}_2) \to H^*(\mathbb{R} P^\infty;\mathbb{F}_2) \to \underset{n}{\text{lim}} H^*(\mathbb{R}P^\infty;\mathbb{F}_2) \to 0$$

where $\text{lim}^1$ is the first dervied functor of the inverse limit.

In this case the $\text{lim}^1$ terms should all be zero by the Mittag-Leffler criteria (which follows since the maps $H^*(\mathbb{R} P^{n+1};\mathbb{F}_2)\to H^*(\mathbb{R} P^{n};\mathbb{F}_2)$ are surjective). Thus we can conclude that $$H^*(\mathbb{R} P^\infty; \mathbb{F}_2) \simeq \mathbb{F}_2[\![ x ]\!]$$

As an aside this is the calculation usually given to show $E^*(\mathbb{C} P^\infty) \simeq E_*[\![ x ]\!]$ for any complex oriented cohomology theory $E$, which basically sets up the whole relationship between complex oriented cohomology theories and formal group laws.

Edit: Please see the comments by Mariano below.

Possibly also this Math Overflow link is relevant