# Non-vanishing $\mathrm{lim}^1$-term for the cohomology of a CW-complex

Let $h$ be an additive cohomology theory. If we want to compute $h^*(X)$ for an infinite CW-complex $X$, a standard method is to use the Milnor sequence

$$0 \to \mathrm{lim}^1_k h^{n-1}(X^{(k)}) \to h^n(X) \to \mathrm{lim}_k h^n(X^{(k)}) \to 0,$$

where $X^{(k)}$ is the $k$-skeleton of $X$. If $h$ is singular cohomology, then the $\mathrm{lim}^1$-term always vanishes.

What are examples of CW-complexes $X$ and cohomology theories $h$, where this $\mathrm{lim}^1$-term does not vanish?

• – André Henriques Feb 10 '14 at 5:13
• – Eric Wofsey Feb 10 '14 at 5:23
• Just for the record: Akhil Mathew discusses in the blog post linked by André an example by Gray with $h$ stable cohomotopy, $X = \mathbb{CP^\infty}$ and $n = 3$. This is a great example and the proof is actually quite simple. – Lennart Meier Feb 12 '14 at 16:14

For our CW-complex I'm going to take $X = \Bbb{CP}^\infty$ (as a based space), whose skeleta are $\Bbb{CP}^n$. The cohomology theory will be more difficult to construct.

For any $k \geq 1$, let $E_k$ be the spectrum which is the homotopy fiber of the map $$Sq^{2^k} \cdots Sq^8 Sq^4 Sq^2: \Sigma^2 H\Bbb{Z}/2 \to \Sigma^{2^{k+1}} H\Bbb{Z}/2$$ so that, for any space $Y$, we have a long exact sequence in cohomology with at least the terms $$\cdots \to H^1(Y) \to H^{2^{k+1}-1}(Y) \to (E_k)^0(Y) \to H^2(Y) \stackrel{Sq^{2^k} \cdots Sq^2}{\longrightarrow} H^{2^{k+1}}(Y) \to \cdots$$ (The cohomology operation is actually taking the $(2^k)$'th power of the elements in degree 2.) In particular, we find that: $$(E_k)^0(\Bbb{CP}^n) = \begin{cases} \Bbb{Z}/2 &\text{if }n < 2^{k}\\ 0 &\text{if }n \geq 2^{k}\\ \end{cases}$$

Now I'll put all these cohomology theories together by taking $E = \bigvee E_k$, so that $E^0(Y) = \bigoplus (E_k)^0(Y)$ for $Y$ a finite CW-complex.

We then find that $$E^0(\Bbb{CP}^n) = \bigoplus_{n < 2^k} \Bbb{Z}/2.$$ This is a decreasing family of abelian groups with a nonzero $lim^1$-term. To prove that, you can use the short exact sequence of inverse systems $$0 \to \bigoplus_{n < 2^k} \Bbb{Z}/2 \to \bigoplus_{k \geq 1} \Bbb{Z}/2 \to \bigoplus_{n \geq 2^k} \Bbb{Z}/2 \to 0,$$ which gives a nontrivial $lim^1$-sequence $$0 \to 0 \to \bigoplus_{k \geq 1} \Bbb{Z}/2 \to \prod_{k \geq 1} \Bbb{Z}/2 \to lim^1 \to 0.$$ This illustrates the basic issue: we can explicitly build our cohomology theory so that no cohomology classes survive to the whole of $\Bbb{CP}^\infty$, but it can take arbitrarily long to figure it out. You can imagine how to build more general examples than this, but this one is convenient to calculate.

• Great! Thanks! I looked more for examples of usual cohomology theories and crazy space, but it is nice to know that one has even to be careful with $\mathbb{CP}^\infty$ in general. – Lennart Meier Feb 10 '14 at 16:18

Take $X$ to be the Moore space for the group $\mathbb Z[\frac1p]$ in dimension $n$ (realized by a telescope of $n$-spheres mapping to each other vie the times-$p$ map), and take $h^*$ to be ordinary cohomology with coefficients in $\mathbb Z$.

The universal coefficient theorem gives you a sort exact sequence $$0\longrightarrow \mathrm{Ext}(H_n(X,\mathbb Z),\mathbb Z) \longrightarrow H^{n+1}(X,\mathbb Z) \longrightarrow \mathrm{Hom}(H_{n+1}(X,\mathbb Z),\mathbb Z)\longrightarrow 0.$$ The last term is zero, and so we get

$$H^{n+1}(X,\mathbb Z)=\mathrm{Ext}(H_n(X,\mathbb Z),\mathbb Z)=\mathrm{Ext}(\mathbb Z[\tfrac1p],\mathbb Z)=\mathbb Z_p/\mathbb Z,$$ the quotient of the $p$-adics by the integers.

That same computation can be done using the Milnor sequence. The limit of $$\mathbb Z\stackrel p \leftarrow\mathbb Z\stackrel p \leftarrow\mathbb Z\stackrel p \leftarrow\ldots$$ is zero, but the $lim^1$ of that inverse system is not zero. It's again $\mathbb Z_p/\mathbb Z$.

• I was aware of this (nice) example. But this is not the filtration by $p$-skeleta. – Lennart Meier Feb 10 '14 at 14:27