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We just computed in class a few days ago that $$H^*(\mathbb{R}P^n,\mathbb{F}_2)\cong\mathbb{F}_2[x]/(x^{n+1}),$$ and it was mentioned that $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[x]$, but I (optimistically?) assumed that the cohomology ring functor would turn limits into colimits, and so $$\mathbb{R}P^\infty=\lim\limits_{\longrightarrow}\;\mathbb{R}P^n$$ would mean that we'd get the formal power series ring $$\lim\limits_{\longleftarrow}\;\mathbb{F}_2[x]/(x^{n+1})=\mathbb{F}_2[[x]].$$

My professor said that the reason we get $\mathbb{F}_2[x]$ is just that the cohomology ring, being the direct sum of the cohomology groups, can't have non-zero elements in every degree, which certainly makes sense.

Even though that should settle the matter, for some reason, I'm still having a bit of trouble making this make click for me. Is the explanation just that the cohomology ring functor doesn't act as nicely as I'd hoped? Is there an intuitive explanation of what's going on?

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I'll wait for a real expert to leave a precise answer, but...it seems that your intuition about the cohomology ring functor needs a little calibrating. How about this: you have more than just a ring structure on $H^*$ -- you have a graded algebra structure. Really you might as well view this as a sequence of functors into $R$-Mod together with a graded-commutative bilinear product on them. In particular though the property that each graded piece is finitely generated is fundamental and should be viewed as nice, I think.... –  Pete L. Clark Feb 20 '12 at 5:02
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...But I don't doubt that the completion of the cohomology ring functor appears elsewhere in algebraic topology. That's really what I'm waiting for an expert to weigh in on. –  Pete L. Clark Feb 20 '12 at 5:03
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(Above, when I say that each graded piece is finite-dimensional, I don't mean that this holds for $H^*(X,R)$ for all spaces $X$, but rather that it holds for sufficiently nice spaces, namely those homotopy equivalent to a CW-complex with finite $n$-skeleta for each fixed $n$.) –  Pete L. Clark Feb 20 '12 at 5:04
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Since $RP^\infty$ has a model with finitely many simplices in each dimension, its cohomology is of at most of countable dimension over the base field. –  Mariano Suárez-Alvarez Feb 20 '12 at 5:57
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@Pete: Ah, I see - the inverse limit of $\mathbb{F}_2[x]/(x^n)$ as graded $\mathbb{F}_2$-algebras is just $\mathbb{F}_2[x]$; whereas $\mathbb{F}_2[[x]]$ does not have a grading compatible with the maps to each $\mathbb{F}_2[x]/(x^n)$. –  Zev Chonoles Feb 20 '12 at 5:57
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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

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Dear Juan, independently of any technology, you probably know that $\mathbb RP^\infty$ can be constructed as a CW-complex with one cell in each dimension. It follows that the cellular complex has each of its homogeneous components a vector space of dimension $1$, and therefore its total dimension cannot be more than countable (in fact, the differential in the cellular complex with $\mod 2$-coefficients is zero, so the cohomology is of countable dimension).As the power series $\mathbb F_2[[x]]$ has uncountable dimension over $\mathbb F_2$, Lurie is either completing or wrong :) –  Mariano Suárez-Alvarez Feb 21 '12 at 0:47
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@Mariano - perhaps if we construct the polynomial ring not via a direct sum of cohomology groups but the direct product? As in Remark 1.2 in this? –  Juan S Feb 21 '12 at 1:18
    
That is one variant of "completing", actually. Since the Milnor exact sequence is clearly true in each degree, and inverse limits commute with products, with that definition of cohomology then the isomorphism with $\mathbb F_1[[x]]$ does hold. We can conclude that in the context of infinite dimensional complexes one needs to be precise about what exactly one is talking about:) –  Mariano Suárez-Alvarez Feb 21 '12 at 1:21
    
Cool. I've always secretly wondered about this very question myself, so it is interesting to me as well –  Juan S Feb 21 '12 at 1:25
    
I think we should be careful about declaring one or the other thing "wrong". It is a matter of how you define the cohomology ring... See Juan's comment about direct products vs direct sums. Also, we should note that when working with homogeneous elements (which we mostly do) there's no difference between the choices. –  Dylan Wilson Feb 21 '12 at 2:21
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