The Cohomology of \(K(\mathbb Z,3)\)

We use the Serre spectral sequence to compute the first ten cohomology groups of \(K(\mathbb Z,3)\).

We will be very brief with our definitions, and we will freely use facts which we have not previously stated. I apologize in advance!

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. Here is what makes \(K(A,n)\)s important: they “represent” cohomology in the sense that \[[-,K(A,n)]=\mathrm H^n(-;A).\] Explicitly, for a space \(X\), maps \(X\to K(A,n)\) (considered up to homotopy) is in natural bijection with \(\mathrm H^n(X;A)\). Thus, by the Yoneda lemma, the identity map \(K(A,n)\to K(A,n)\) induces a canonical class \(\iota_n\in\mathrm H^n(K(A,n);A)\), and then a map \(f\colon X\to K(A,n)\) corresponds to the class \(f^*\iota_n\in\mathrm H^n(X;n)\). In general, identifying classes in \(\mathrm H^n(K(A,m);B)\) corresponds to finding natural transformations in cohomology.

Remark 3. It follows from Whitehead’s theorem that the Eilenberg–MacLane spaces are determined by their homotopy groups. Namely, \(K(A,n)\) is the unique space \(X\) up to homotopy for which \[\pi_i(X)=\begin{cases}A & \text{if }i=n, \\0 & \text{otherwise}.\end{cases}\] To summarize the proof, one uses the Hurewisz theorem to conclude that \(\mathrm H^n(X;A)\cong A^\lor\), so one can find a map \(X\to\mathrm K(A,n)\) inducing an isomorphism on \(\mathrm H^n(-;A)\). Another application of the Hurewisz theorem shows that \(f\) is an isomorphism on homotopy groups, so we are done by Whitehead’s theorem.

The preceding remark lets us identify some small Eilenberg–MacLane spaces and then compute their cohomology groups.

Example 4. Observe that the discrete topological group \(\mathbb Z\) has \(\pi_0(\mathbb Z)=\mathbb Z\) and no higher homotopy groups, so we see that \(K(\mathbb Z,0)=\mathbb Z\). Note that \(\mathbb Z\) is just a disjoint union of points, so \[\mathrm H^i(K(\mathbb Z,0))=\begin{cases}\mathbb Z & \text{if }i=0, \\0 & \text{if }i\ne0.\end{cases}\]

Example 5. The space \(S^1\) admits a universal cover \(\mathbb R\) with fiber \(\mathbb Z\). Thus, \(\mathbb R\to S^1\) is a Serre fibration, so \(\mathbb Z\to\mathbb R\to S^1\) is a homotopy fiber sequence, so we have a homotopy long exact sequence \[\pi_{i+1}(\mathbb R)\to\pi_{i+1}(\mathbb Z)\to\pi_i\left(S^1\right)\to\pi_i(\mathbb R).\] Note that \(\mathbb R\) is contractible, so the leftmost and rightmost terms vanish, so we see that \(\pi_i\left(S^1\right)=\pi_{i+1}(\mathbb Z)\) for all \(i\). We conclude that \(\pi_i\left(S^1\right)\) is supported in degree \(1\), where it is \(\mathbb Z\). We conclude that \(K(\mathbb Z,1)=S^1\), so \[\mathrm H^i(K(\mathbb Z,1))=\begin{cases}\mathbb Z & \text{if }i\in\{0,1\}, \\0 & \text{otherwise}.\end{cases}\]

Example 6. The space \(\mathbb{CP}^\infty\) admits a universal cover \(S^\infty\) with fiber \(S^1\). (To be explicit, one presents \(S^\infty\) as an infinite sequence of complex numbers supported in finitely many terms and whose norms sum to \(1\).) As before, we find that \(S^1\to S^\infty\to\mathbb{CP}^\infty\) is a homotopy fiber sequence, so we have a homotopy long exact sequence \[\pi_{i+1}(S^\infty)\to\pi_{i+1}\left(S^1\right)\to\pi_i\left(\mathbb{CP}^\infty\right)\to\pi_i(S^\infty).\] Note that \(S^\infty\) is contractible, so the leftmost and rightmost terms vanish, so we conclude as before that \(\pi_i\left(\mathbb{CP}^\infty\right)\) is only supported in degree \(2\), where it is \(\mathbb Z\). We conclude that \(K(\mathbb Z,2)=\mathbb{CP}^\infty\).

Remark 7. In general, the homotopy fiber sequence \[\Omega K(\mathbb Z,n+1)\to\Delta^0\to K(\mathbb Z,n+1)\] shows that \(\Omega K(\mathbb Z,n+1)=K(\mathbb Z,n)\). This more or less explains where the previous two examples come from: we are trying to find some space \(X\) fitting into a homotopy fiber sequence \(K(\mathbb Z,n)\to\Delta^0\to X\).

There are direct ways to compute the cohomology of \(\mathbb{CP}^\infty\), but to set ourselves up for later, we use the Serre spectral sequence.

Theorem 8 (Serre spectral sequence). Let \(F\to E\to B\) be a homotopy fiber sequence where \(B\) is simply connected. Then there is a spectral sequence \(E_\bullet\) for which \[E_2^{pq}=\mathrm H^p(B;\mathrm H^q(F;\mathbb Z))\Rightarrow\mathrm H^{p+q}(E;\mathbb Z).\]

There are versions for homology and for different coefficients and even weakening the simply connectedness hypothesis, but we will not need any of these.

Example 9. The homotopy fiber sequence \(S^1\to\Delta^0\to\mathbb{CP}^\infty\) gives rise to a Serre spectral sequence \(E_\bullet\) because \(K(\mathbb Z,2)\) is simply connected. For brevity, we relabel these spaces by \(F\to E\to B\). Noting that \(\mathrm H^*(S^1;\mathbb Z)=\mathbb Z\oplus\mathbb Z\) (where there is one \(\mathbb Z\) in only degrees \(0\) and \(1\)), we see that \(E_2\) looks like the following. \[\begin{array}{ccccc} \vdots & \vdots & \vdots & \vdots & \\{} 0 & 0 & 0 & 0 & \cdots \\{} \mathbb Z & \mathrm H^1(B) & \mathrm H^2(B) & \mathrm H^3(B) & \cdots \\{} \mathbb Z & \mathrm H^1(B) & \mathrm H^2(B) & \mathrm H^3(B) & \cdots\end{array}\] Thus, we see that our spectral sequence is supported in rows \(q\in\{0,1\}\), so only \(d_2\) may be nonzero, so \(E_3=E_4=\cdots=E_\infty\). However, \(E_\infty\) is the cohomology of a point and therefore needs to only be supported at \(\Delta^0\). We conclude that \(\mathrm H^1(B)=0\) and that all differentials are isomorphisms, so we find \[\mathrm H^i(\mathbb{CP}^\infty;\mathbb Z)=\begin{cases}\mathbb Z & \text{if }i\text{ is even}, \\0 & \text{if }i\text{ is odd}.\end{cases}\]

It turns out that the cohomological Serre spectral sequence is also a graded commutative ring at all steps, which we can use to recover ring structure.

Example 10. Continuing from Example 9, we let \(y_1\in\mathrm H^1(F)\) be a generator so that \(\mathrm H^*(F)=\mathbb Z[y_1]/\left(y_1^2\right)\). Then \(d_2\) is an isomorphism everywhere except at the origin, so we may find a generator \(x_2\in\mathrm H^2(B)\) with \(d_2(y_1)=x_2)\). Then we claim that \[\mathrm H^*(\mathbb{CP}^\infty;\mathbb Z)\stackrel?=\mathbb Z[x_2].\] Indeed, it is enough to check that \(x_2^k\) generators \(\mathrm H^{2k}(B)\) for each \(k\), which we do by induction on \(k\). There is nothing to do for \(k\in\{0,1\}\). The once the statement is true for \(k\), we note that \(x_2^ky_1\) is a generator for \(E_2^{2k,1}\mathrm H^{2k}(B)y_1\), so \[d_2\left(x_2^ky_1\right)=kx_2^{k-1}d_2(x_2)y_1+x_2^kd_2(y_1)=x_2^{k+1}\] generates \(\mathrm H^{2k+2}(B)\).

The previous examples might convince us that the cohomology of these Eilenberg–MacLane spaces is not very interesting. And indeed, if one only focuses on the rank of the cohomology, then it is not too hard to compute, and the method of Example 10 works in general.

However, there starts to be torsion in \(K(\mathbb Z,3)\).

Proposition 11. We compute the first ten cohomology groups of \(K(\mathbb Z,3)\) as follows. \[\mathrm H^i(K(\mathbb Z,3);\mathbb Z)=\begin{cases} \mathbb Z & \text{if }i\in\{0,3\}, \\\mathbb Z/2\mathbb Z & \text{if }i\in\{6,9\}, \\\mathbb Z/3\mathbb Z & \text{if }i=8, \\0 & \text{if }i\in\{1,2,4,5,7\}.\end{cases}\]

Proof. Remark 7 gives us the fiber sequence \(K(\mathbb Z,2)\to\Delta^0\to K(\mathbb Z,3)\), which we will again abbreviate to \(F\to E\to B\). Because \(B\) is simply connected, we receive a Serre spectral sequence \(E_\bullet\). Note that \(\mathrm H^*(F)=\mathbb Z[y_2]\) for some generating class \(y_2\) in degree \(2\) by Example 10, so \(E_2\) looks like the following. \[\begin{array}{ccccccc} \vdots & \vdots & \vdots & \vdots & \vdots \\{} \mathbb Zy_2^2 & \mathrm H^1(B)y_2^2 & \mathrm H^2(B)y_2^2 & \mathrm H^3(B)y_2^2 & \mathrm H^4(B)y_2^2 & \cdots \\{} 0 & 0 & 0 & 0 & 0 & \cdots \\{} \mathbb Zy_2 & \mathrm H^1(B)y_2 & \mathrm H^2(B)y_2 & \mathrm H^3(B)y_2 & \mathrm H^4(B)y_2 & \cdots \\{} 0 & 0 & 0 & 0 & 0 & \cdots \\{} \mathbb Z & \mathrm H^1(B) & \mathrm H^2(B) & \mathrm H^3(B) & \mathrm H^4(B) & \cdots \end{array}\] Let’s start with \(\mathrm H^1(B)\) and \(\mathrm H^2(B)\). Note that \(d_2=0\) for parity reasons, so \(E_2=E_3\). Now, \(E_\bullet\) converges to the cohomology of a point, so \(E_\infty\) is only supported at the origin, where it is \(\mathbb Z\). Thus, \(E_2^{10}=E_\infty^{10}\) and \(E_2^{20}=\) both must vanish, so \(\mathrm H^1(B)=\mathrm H^2(B)=0\). We now see that \(E_3\) currently looks like the following. \[\begin{array}{ccccccc} \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\{} \mathbb Zy_2^2 & 0 & 0 & \mathrm H^3(B)y_2^2 & \mathrm H^4(B)y_2^2 & \mathrm H^5(B)y_2^2 & \mathrm H^6(B)y_2^2 & \cdots \\{} 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\{} \mathbb Zy_2 & 0 & 0 & \mathrm H^3(B)y_2 & \mathrm H^4(B)y_2 & \mathrm H^5(B)y_2 & \mathrm H^6(B)y_2 & \cdots \\{} 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\{} \mathbb Z & 0 & 0 & \mathrm H^3(B) & \mathrm H^4(B) & \mathrm H^5(B) & \mathrm H^6(B) & \cdots \end{array}\] We may now compute \(\mathrm H^3(B)\). We see that \(E_4^{02}=E_4^{30}=0\), so \(d_3\colon\mathbb Zy_2\to\mathrm H^3(B)\) must be an isomorphism. Thus, we may choose a generator \(x_3\in\mathrm H^3(B)\) for which \(d_3(y_2)=x_3\). Note that this implies \[d_3\left(y_2^k\right)=ky_2^{k-1}d_2(y_2)=kx_3y_2^{k-1}.\] In particular, \(d_3\) is injective on \(E_3^{p0}\), so \(E_4^{p0}=0\).

Before continuing, we note that \(E_3^{40}\) and \(E_3^{50}\) can now be seen to interact with no nonzero differentials, so these entries have already converged, so we see that \(\mathrm H^4(B)=\mathrm H^5(B)=0\). While we’re here, we note that the same reasoning now shows that \(\mathrm H^7(B)=0\). Thus, \(E_3\) currently looks like the following. \[\begin{array}{ccccccccccc} \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\{} \mathbb Zy_2^2 & 0 & 0 & \mathbb Zx_3y_2^2 & 0 & 0 & \mathrm H^6(B)y_2^2 & 0 & \mathrm H^8(B)y_2^2 & \mathrm H^9(B)y_2^2 & \cdots \\{} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\{} \mathbb Zy_2 & 0 & 0 & \mathbb Zx_3y_2 & 0 & 0 & \mathrm H^6(B)y_2 & 0 & \mathrm H^8(B)y_2 & \mathrm H^9(B)y_2 & \cdots \\{} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\{} \mathbb Z & 0 & 0 & \mathbb Zx_3 & 0 & 0 & \mathrm H^6(B) & 0 & \mathrm H^8(B) & \mathrm H^9(B) & \cdots \end{array}\] The remaining calculations will be done in sequence.

• Now, \(E_3^{60}\) only interacts with the \(d_3\) differential, so \(E_3^{60}\) must be exactly the cokernel of the map \(d_3\colon\mathbb Zy_2^2\to\mathbb Zx_3y^2\). We computed this map to be \(y_2^2\mapsto2x_3y_2\) earlier, so we conclude that \(\mathrm H^6(B)\cong\mathbb Z/2\mathbb Z\). Note that we have shown that \(\mathrm H^6(B)\) is generated by \(d_3(x_3y_2)=x_3^2\), so we will write \(\mathrm H^6(B)=\mathbb Z/2\mathbb Zx_3^2\).
• Continuing, \(\mathrm H^8(B)\) can only interact with \(d_5\colon E_5^{34}\to E_5^{80}\), which we see must now be an isomorphism. To compute \(E_5^{34}\), we start with \(E_3^{34}=\mathbb Zx_3y_2^2\). Then the kernel of \(d_3\colon\mathbb Zx_3y_2^2\to\mathbb Z/2\mathbb Zx_3^2y_1\) is everything by a direct calculation, so we conclude that \(E_4^{34}\) is the cokernel of the map \(\mathbb Zy_2^3\to\mathbb Zx_3y_2^2\), which is exactly \(\mathbb Z/3\mathbb Z\).
• We now move on to \(\mathrm H^9(B)\), which can only interact with \(d_3\colon\mathbb Z/2\mathbb Zx_3^2y_2\to\mathrm H^9(B)\). Thus, \(\mathrm H^9(B)\) must be the cokernel of the map \(d_3\colon\mathbb Zx_3y_2^2\to\mathbb Z/2\mathbb Zx_3^2y_2\), which we see is exactly \(\mathbb Z/2\mathbb Z\).

The above calculations compete the proof. \(\blacksquare\)

One can continue the calculations a little further, but it will later require more input than just the Serre spectral sequence to continue.