This is part 5 of a series leading up to and exploring model categories. For the other parts see the series overview.
As promised in the previous part, we are going to calculate $\pi_4(S^3)$. I think we will have to use all of the machinery (plus some new) that we have been through during this series to do the calculation. What more could we possibly need you ask? Last time we developed the machinery to calculate the cohomology of the total space of a fibration, but we want to compute homotopy. Hence we need a method for translating cohmological information into homotopical information, which is what we are missing to be able to do the calculation. There may be other processes that I haven’t learned, but the process I know goes through two steps. First we must translate cohomology into homology. This is done through the so called cohomological universal coefficient theorem (cUCT). Then we need to translate from homology to homotopy. This is done through the Hurewicz theorem. I think of these two theorems together as sort of a Rosetta stone for algebraic topology. It makes us able (with some computation and restrictions of course) to move between the three fundamental theories of invariants we have in algebraic topology, which I find beautiful. There is one more thing we need, which is a starting point for our calculation. We need a good fibration to extract the information we want which we are able to translate into homotopy afterwards. Therefore we need a space in the fibration that does not complicate things when we translate into homotopy, i.e. we need a space in which we completely understand its homotopy groups. The “homotopy-easy” spaces I’m describing are called Eilenberg-MacLane spaces. In cohomology (and homology) theory we have the easy spaces being spheres because we completely understand their cohomological structure. They cam be thought of as the building blocks for (co)homology. The same type of space for homotopy is exactly theese Eilenberg-MacLane spaces, and they form the building blocks for homotopy groups in the same way as the spheres for (co)homology. Hence we can combine these spaces and spheres in a fibration and use that to compute cohomology and then relatively easily translate this to homotopy, which is exactly our plan for computing $\pi_4(S^3)$.
Expanding our tool belt
The first new tool, which we briefly mentioned above is called the Hurewicz theorem and allows us to move between homology and homotopy in certain nice cases. These nice cases happen when the topological space we are studying is very connected, i.e. at least both path connected and simply connected and often even the higher dimensional analog called n-connected. The level of connectedness can be formulated though the homotopy groups of the space, in particular we call a space $X$ for $n$–connected if $\pi_k(X) = 0$ for all $k\leq n$, here path connected corresponds to 0-connected, and simply connected corresponds to 1-connected.
Theorem (Hurewicz): Let $n\geq 2$. and $X$ be $(n-1)$-connected. Then $H_k(X) = 0$ for all $ 0 < k < n $ and the Hurewicz homomorphism $h_*: \pi_n(X) \rightarrow H_n(X)$ is an isomorphism.
We see that the absence of low degree homotopy groups also gives an absence of lower degree homology groups. The Hurewicz homomorphism is given by evaluating the induced map on homology from a homotopy class of a map $f:S^n\rightarrow X$ in the fundamental class of $S^n$, i.e. the choice of a generator for the group $H_n(S^n)\cong \mathbb{Z}$. If we denote the fundamental class by $[S^n]$, then it is given by $h_*([f])=f_([S^n])$. What exactly the maps are will not be important for us, but the fact that it is an isomorphism in nice cases is very useful.
Theorem (cUCT): Let $X$ be a topological space and $G$ an Abelian group. Then for any $i$ we have a short exact sequence
$$0\rightarrow Ext_{\mathbb{Z}}(H_{i-1}(X;\mathbb{Z}), G)\rightarrow H^i(X;G) \rightarrow Hom_{\mathbb{Z}}(H_i(X;\mathbb{Z}), G)\rightarrow 0.$$
This tells us that cohomology is almost just maps from homology into the coefficients, except we might be a little bit off, but in a manner we understand, i.e. up to some extension. We are going to rely quite heavily on this theorem in the calculation later, but there are a few extra bits we need. There is a theorem by Serre that says that the homology groups of a simply connected space are finitely generated if and only if the homotopy groups are finitely generated. We know that the homology groups of spheres are finitely generated, hence all of the homotopy groups are also finitely generated. In the short exact sequence in cUCT, we also need to know a little bit about when there are no homomorphisms between two groups, and how the Ext group works. If we have a finite group $A$, then the group group $Hom(A,\mathbb{Z})=0$. To have the converse statement we also need to know that A is finitely generated, i.e. if $A$ finitely generated then $Hom(A,\mathbb{Z})$ is a free group. Hence if $Hom(A,\mathbb{Z})=0$ then $A$ is a finite group. What we need to know about $Ext(A,\mathbb{Z})$ is that $Ext(\mathbb{Z}/2,\mathbb{Z})\cong \mathbb{Z}/2$, and that the converse holds if we have a finiteness condition, i.e. if $A$ is a finite group and $Ext(A,\mathbb{Z})=\mathbb{Z}/2$, then $A\cong \mathbb{Z}/2 .$
Definition: A topological space $X$ is called an Eilenberg-MacLane space of type $K(G,n)$, or just a $K(G,n)$-space for short, if we have $\pi_n(X) = G$ and all of its other homotopy groups are trivial.
The important cases for us in this computation is a $K(\mathbb{Z}, 3)$ space and a $K(\mathbb{Z}, 2)$ space. How do these spaces look? We know the simplest case, namely that the sircle $S^1$ is a $K(\mathbb{Z}, 1)$ space. This we know because it has the integers as a fundamental group, and as discussed in part 3, it can have no maps from higher dimensional spheres, hence all higher homotopy groups are trivial, and this is our definition of a $K(\mathbb{Z}, 1)$-space. In the computation later we need an explicit description of a $K(\mathbb{Z}, 2)$-space, so before we jump to computation, we remark that $\mathbb{C}P^{\infty}$ is in fact a $K(\mathbb{Z}, 2)$-space! This space is the space consisting of all complex lines passing through the origin in the infinite dimensional complex vector space $\mathbb{C}^{\infty}$. I’m not going to explain much about this space, but we will use it’s cohomology groups and it’s cohomology ring later.
Calculation of $\pi_4(S^3)$
I didn’t expect the preliminaries to be that long, but now the hard part is over (not really) and we are finally ready to compute! Lets start with the fibration $X\rightarrow S^3 \rightarrow K(\mathbb{Z},3)$. This fibration is not super easy to get, but the idea is to iteratively attach higher dimensional cells to $S^3$ in order to kill off higher homotopy groups in degrees higher than 3. When we kill off all the higher homotopy groups, we are only left with a homotopy group in degree three, which is the integers since it is “made” from the 3-sphere, and this is our definition of a $K(\mathbb{Z},3)$. This is sometimes referred to as “capping” of a space, giving a “capped” space. It is dual to the notion of a “killing space”, which is a cool name i thought i would mention. When we pass to the long exact sequence for this fibration we get
$$\cdots \rightarrow \pi_4(K(\mathbb{Z},3)) \rightarrow \pi_3(X) \rightarrow \pi_3(S^3) \rightarrow \pi_3(K(\mathbb{Z},3)) \rightarrow \pi_2(X) \rightarrow \pi_2(S^3) \rightarrow \cdots,$$
in which we know several of the groups. We know that $K(\mathbb{Z},3)$ only has homotopy in degree 3 and that $\pi_2(S^3) = 0$. We also know that the map $\pi_3(S^3) \rightarrow \pi_3(K(\mathbb{Z},3))$ is an isomorphism, since that is the way we constructed the space $K(\mathbb{Z},3)$. Further to the right we only have trivial groups. Hence we get
$$\pi_3(X) \cong \pi_2(X) \cong \pi_1(X) \cong \pi_0(X) \cong 0.$$
Further to the right we have that all of the homotopy groups of $K(\mathbb{Z},3)$ are trivial, and because we know that the long sequence is exact, we get that $0 \rightarrow \pi_n(X) \rightarrow \pi_n(S^3) \rightarrow 0$ is exact, hence the homotopy groups are isomorphic for all $n>3$. In particular we have $\pi_4(X) \cong \pi_4(S^3)$ which is what we are going to use in order to calculate it. As mentioned earlier, all homotopy groups of spheres are finitely generated, and hence all of the homotopy groups of $X$ are as well. We see that we have to compute the homotopy groups of $X$, and to do this we are going to use the Rosetta stone we described in the introduction, by first computing its cohomology, and then translating through homology and then finally into homotopy.
As we did in the Puppe sequence in part 3, we can extend the inclusion of the fibers into the total space to a fibration with fibers being the loop space of the base space in the original fibration, i.e. $\Omega K(\mathbb{Z},3)\rightarrow X \rightarrow S^3$. Have we made things even more complicated? How does the loop space of an Eilenberg-MacLane space look? We have actually made it simpler. In part 3 we discussed the suspension functor, and noticed that the suspension of a sphere is a new sphere in one dimension higher. Hence the suspension functor should shift the degrees of the homotopy groups up by one. We mentioned that the loop space functor is adjoint to the suspension, and hence is shifts the degrees of the homotopy groups down by one. We can also see this by the fibration $\Omega X\rightarrow PX \rightarrow X$, where $PX$ is the path space of $X$. The path space is contractible, and hence from it has only trivial homotopy groups, so when we pass to the long exact sequence of homotopy groups from the fibration we get that $\pi_{n+1}(X) \cong \pi_{n}(\Omega X)$. Because of this we get that $\pi_n(\Omega K(\mathbb{Z},3)) = 0$ for $n\geq 3$ and $\pi_2(\Omega K(\mathbb{Z},3)) = \mathbb{Z}$.
Hey, look at that, the loop space of the Eilenberg-MacLane space has only one non-trivial homotopy group, and thus by our definition it is an Eilenberg-MacLane space itself, namely a $K(\mathbb{Z}, 2)$-space. Luckily for us we already know a $K(\mathbb{Z}, 2)$-space, namely $\mathbb{C}P^{\infty}$. To summarize, our new fibration looks like $\mathbb{C}P^{\infty}\rightarrow X \rightarrow S^3$. Now we have something to put into a spectral sequence which we know computes the cohomology of $X$ from the cohomology of the base and the cohomology of the fibers. And, since $\mathbb{C}P^{\infty}$ is a nice CW-complex with one cell in each even degree, we know how its cohomology looks like, and even more important, we know how its cohomology ring looks like. We have $H^*(\mathbb{C}P^{\infty})\cong \mathbb{Z}[a_2]$, i.e. a polynomial ring with the generator in degree two. Lets throw all this information into the Serre spectral sequence. We get
Since our base space is a sphere, we naturally get two columns, and since $\mathbb{C}P^{\infty}$ has cohomology in every even degree, these two columns continue all the way up to infinity. Luckily for us, we only care about a small portion of the spectral sequence. Since there are 3 “steps” between the two columns, all of the differentials on the second page has a trivial group as either it’s domain or it’s codomain. Hence nothing happens at the second page. When we flip to the third page, the differentials are “long” enough to connect our two columns. We know that the spectral sequence computes the cohomology of the topological space $X$, and since there are only one group along each of the diagonals, i.e. either $\bigoplus_{p+q=n}E_r^{p,q} = E_r^{0,n}$ or $\bigoplus_{p+q=n}E_r^{p,q} = E_r^{3,n}$. By this we know that the n‘th cohomology group of $X$ is actually the group showing up in the spectral sequence along the n‘th diagonal. From the computation of the homotopy groups of $X$ earlier, we know that $\pi_3(X) = \pi_2(X)=0$ and by the Hurewicz theorem we know that this implies that $H_3(X) = H_2(X) = 0$, which by cUCT implies that $H^3(X) = 0$. We have a group on the third diagonal, namely $E_3^{3,0}$, and it is hit by only one single differential, which is the differential $d:E_3^{0,2}\rightarrow E_3^{3,0}$. Since all other differentials at the later pages are too long, we know that this differential is in fact the only differential hitting this group throughout the entire spectral sequence. Thus, we know that it has to be an isomorphism. Since the cohomology ring of $\mathbb{C}P^{\infty}$ has a generator a in the second degree, it lives in $E_3^{0,2}$ and it maps through $d_2$ to some generator of $E_3^{3,0}$, call this $x$, i.e. $d_2(a)=x$. The generator a lives in degree 2, so its square in the cohomology ring $a^2$ lives in degree 4, and hence lives in $E_3^{0,4}$. By the Leibniz rule that the differentials satisfy, we get that $d_4(a^2)= ax$ and that the map is multiplication by two. The cube of the generator $a^3$ lives in $E_3^{0,6}$ and maps to $a^2 x$ by multiplication by three and so on upwards. We can draw the information into the spectral sequence, just for a better visualization.
When we flip to the fourth page, all of the differentials are too long to pass between the columns, and hence we get that $E_4 = E_{\infty}$ which means that $H^4(X)=Ker(d_4)=Ker(\cdot 2) = 0$ and $H^5(X) = Cok(d_4)= Cok(\cdot 2) = \mathbb{Z}/2$. Now we’re getting somewhere! We see the first evidence of what is to become $\pi_4(S^3)$ after translating with the Rosetta stone. As described earlier, we need to translate into homology first, which we do by cUCT. We get the short exact sequence
$$0\rightarrow Ext_{\mathbb{Z}}(H_{3}(X), \mathbb{Z})\rightarrow H^4(X) \rightarrow Hom_{\mathbb{Z}}(H_4(X), \mathbb{Z})\rightarrow 0.$$
We figured out that $H^4(X) = 0$ and $H_3(X)=0$, and therefore we get $Ext_{\mathbb{Z}}(H_3(X),\mathbb{Z}) = Ext_{\mathbb{Z}}(0,\mathbb{Z}) = 0$. Since the sequence is exact we must have $0 = H^4(X;\mathbb{Z}) \cong Hom_{\mathbb{Z}}(H_4(X),\mathbb{Z})$. Then $Hom_{\mathbb{Z}}(H_4(X),\mathbb{Z}) = 0$ which implies that $H_4(X)$ is a finite group. This we know since the homotopy groups of $X$ are finitely generated (discussed previously), hence the homology groups of $X$ are as well which gives us that it is a finite group by the discussion in the paragraph about cUCT. In one degree higher we also get a short exact sequence from cUCT, namely the sequence
$$0\rightarrow Ext_{\mathbb{Z}}(H_{4}(X), \mathbb{Z})\rightarrow H^5(X) \rightarrow Hom_{\mathbb{Z}}(H_5(X), \mathbb{Z})\rightarrow 0.$$
Since $H_4(X)$ is finitely generated we know $Hom_{\mathbb{Z}}(H_5(X), \mathbb{Z}) = 0$, because it is either this or it is equal to $\mathbb{Z}/2$ and that option disappears since it is free. Now this means that $Ext_{\mathbb{Z}}(H_{4}(X), \mathbb{Z}) \cong H^5(X) \cong \mathbb{Z}/2$ and since we know that $H_4(X)$ is a finite group, we get that $H_4(X)\cong \mathbb{Z}/2$. Phew… Almost done now.
Thankfully, the last part of the translation from homology to homotopy is easier in this case. Recall that we figured out that $\pi_3(X) \cong \pi_2(X) \cong \pi_1(X) \cong \pi_0(X) \cong 0$ in the beginning when constructing our fibrations. Hence we know that $X$ is 3-connected, and by the Hurewicz theorem we have an isomorphism $H_4(X)\cong \pi_4(X)$ which means that we have $\pi_4(X) \cong \mathbb{Z}/2$. Our whole reason for doing this calculation with the space $X$ was that we figured out from the long exact sequence from the fibration that $\pi_4(X) \cong \pi_4(S^3)$, and by this we finally have our result, $\mathbb{Z}/2\cong H_4(X)\cong \pi_4(X) \cong \pi_4(S^3)$, or in short
$$\pi_4(S^3) \cong \mathbb{Z}/2 .$$
I really like this computation because it is rather difficult, and has a lot of moving parts, which makes it more fun! I don’t think the next posts will be this long and detailed, because this is maybe a bit much information in one post, even though it is just one computation in essence. Onward we will discuss cofibrations and weak equivalences, and then move into model category territory. I think my goal will be to show that the homotopy category of topological spaces is equivalent to the homotopy category of simplicial sets, which also enables me to finally start reading May’s book “Simplicial objects in algebraic topology”. Anyway, as usual I will leave off with an artwork by Anatoly Fomenko, this time his piece called “The method of killing spaces in homotopic topology”, which is dually relevant.