Spectral Sequences: Episode III – Revenge of the Serre

We'll slowly bridge the material about Steenrod squares from blog post #2 into spectral sequences. To build some motivation, we'll first look at $Sq^1 \circ Sq^1$ and see how this connects to differentials $d_r$ in the Serre spectral sequence (SSS).

How do Steenrod squares interact with each other? Without going too deep in the weeds, there are specific rules for how Steenrod squares compose with each other, and these are called the Adem Relations [1].

Theorem (Adem Relations): For integers $a, b$ such that $0 < a <2b$, the following relation holds:

where the binomial coefficients are taken mod 2. For large integers $a, b$ we can use Lucas's Theorem [2], where we basically look for any misalignment between the 1s in the base 2 version of $k$ and the 1s in the base 2 of $n$ for a binomial ${n\choose k}$. For example, consider ${10 \choose 6}$:

where we see that the first 1 from the right aligns with the 1 above it, but the second 1 does not, hence we can immediately conclude that

and one can confirm that, indeed, ${10 \choose 6} = 210$ which is even. For another example, consider ${13 \choose 5}$:

where in this case we get alignment with all the 1s at the bottom with 1s at the top, and can then conclude that

Note that since we have this initial alignment we can add any combination of 1s and 0s to the top and still get an odd number, so

where one could calculate the odd number ${173 \choose 5} = 1,\!218,\!218,\!079$. TL;DR: misalignment = even, alignment = odd.

Returning to Steenrod squares, let's look at $Sq^1 \circ Sq^1$. Recall that

Unfortunately we won't be able to flex Lucas's theorem much since 1 is not a big integer, but alas, let's use the Adem Relations to find $Sq^1 \circ Sq^1$. Note that we can use the Adem Relations since we have $1 < 2 \cdot 1$ where $a =1, b=1$ and $0<a<2b$:

where right off the bat we just have $k=0$ and no other iteration since $\left\lfloor 1/2 \right\rfloor = 0$, so the binomial coefficient is just ${0 \choose 1} = 0$, hence

i.e. $Sq^1 \circ Sq^1 = (Sq^1)^2 = 0$. Note that this $Sq^1 \circ Sq^1 = 0$ is a fundamental relation in the Steenrod algebra. Furthermore, note that $Sq^1 = \beta$ where $\beta$ is the Bockstein homomorphism [3]. The Bockstein homomorphism is the same mapping as $Sq^1$, where

So the Adem Relations confirm that $\beta^2 = (Sq^1)^2 = 0$. This ties into differentials where, for example in chain complexes, we have the classic result

We'll see as we proceed that this idea of a differential ($\beta^2=0$) is core to spectral sequences!

A spectral sequence is basically a vast generalization of this idea of differentials. Instead of a single differential, we get a sequence of them $(d_2, d_3, d_4, \ldots)$ that act on a grid of algebraic objects. In this post, we'll be exploring the Serre spectral sequence (SSS) [4, 5]. Before we introduce the SSS, let's briefly define fibrations.

Fibration: A map $p: E \rightarrow B$ is a fibration with fiber $F$ if it has the homotopy lifting property (HLP). We often denote these as $F \rightarrow E \rightarrow B$. If we have a fibration $p: E \rightarrow B$ then this means that for any space $Y$ and any map $g: Y \rightarrow E$, as well as any homotopy $H: Y \times I \rightarrow B$ (where this homotopy starts at $p \circ g$), then there exists a lifted homotopy $\widetilde{H}: Y \times I \rightarrow E$ that covers $H$. If the above is a bit jumbled, take a look at the following commutative diagram which is more clear (and organized):

Intuitively, this means that paths and homotopies in the base space $B$ can be "lifted" up to the total space $E$. One may wonder why the "total space" is denoted by $E$. This convention comes from Serre who, as a French mathematician, would've referred to the total space as "espace total" in his native language, hence the $E$. Finally, for any point $b \in B$, the preimage $p^{-1}(b)$ is homotopy equivalent to a single space $F$, the fiber. We denote this entire object by $F \rightarrow E \rightarrow B$. One of the most famous examples of fibrations comes from the Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$, where we can find a non-trivial homotopy group $\pi_3(S^2) \cong \mathbb{Z}$. With this, we're ready to tackle the SSS!

The Serre Spectral Sequence: For a fibration $F \rightarrow E \rightarrow B$, assuming $B$ is path-connected and simply-connected, there exists a (co)homological spectral sequence, which is a sequence of bigraded abelian groups (which we'll call pages) $\{ E_r^{p,q}, d_r \}$ (or $\{ E^r_{p,q}, d^r \}$ in the homological case) for $r \geq 2$ with the following properties:

$E_2$ Page: This is our starting point. The groups on this page are given by the base space $B$ with coefficients in the $q$-th (co)homology of the fiber $F$, i.e.

where $G$ can by any other coefficient group for $H^q(F;-)$ like $\mathbb{Z}, \mathbb{Z}_2,$ etc. Note that we use the notation $E_r^{p, q}$ for cohomology and the flipped notation $E_{p, q}^r$ for homology. Similarly, for homology we get the following for the $E^2$ page:

The index $p$ is the base degree and is the horizontal (left/right) axis, while the index $q$ is the fiber degree and is the vertical (up/down) axis. Note that we're assuming that $B$ is simply-connected since we want $\pi_1(B)$ to act trivially on the (co)homology of the fiber $H^*(F;G)$. This makes sure that things stay relatively, literally, and figuratively simple. Additionally, note that since we're usually dealing with path-connected spaces, at the left-most and bottom-most $(p, q) = (0, 0)$ we usually start off with:

The Differentials $d_r$: For each page $r \geq 2$, there's a map called the differential $d_r$ which is a collection of homomorphisms

The total degree is $p+q$ which is often denoted as $n = p+q$. Note that $d_r$ increases the total degree by 1 since we go from $p+q$ to $(p+r) + (q-r+1) = p+q+1$, so $n \rightarrow n+1$. Similar to $Sq^1 = \beta$, applying the differential twice gives zero, i.e.

Note that for the cohomology case (like above), the indices change the following way: $p \mapsto p+r$ and $q \mapsto q-r+1$, while in homology it's flipped: $p \mapsto p-r$ and $q \mapsto q+r-1$. For cohomological spectral sequences we use the $d_r$ notation, but for homological spectral sequences we use $d^r$, where

This is, unsurprisingly, all due to the duality between homology and cohomology. For differentials, specifically $d_2$, it may help to visualize how a knight moves on a chessboard like an L. For cohomological SSS we have $d_2$ which moves 2 units to the right and 1 unit down; and for homological SSS we have $d^2$ which moves 2 to the left and 1 up.

Higher Pages $E_{r+1}$: To build some intuition for the higher pages, let's consider $E_2^{p, q}$ and the incoming $d_2$ maps going into $E_2^{p, q}$, as well as the outgoing $d_2$ maps coming out of it:

The $E_3$ page is defined in a similar fashion as (co)homology:

with the following $d_2$ maps:

We can easily generalize this to the $E_{r+1}$ page:

Each subsequent page is the (co)homology of the previous page with respect to the previous page's differential.

Convergence: For any fixed $(p, q)$, the $d_r$ maps eventually either start at a zero group or land on a zero group as $r$ gets large. This means that the groups in $E_r$ stabilize. We call this stable page the $E_\infty$ page. The $E_\infty$ page is related to the (co)homology of the total space $H^*(E;G)$, which we denote like so:

There's a few details here that we'll skim over (like filtrations), but for well-behaved and "nice" total spaces $E$ we can usually get away with the following:

Furthermore, oftentimes we'll get to the $E_\infty$ page and just be left with a single non-zero group at some $(p_0, q_0)$ where $p_0+q_0 =n$, so we can often be left with a single term:

There's a lot of pieces here for the Serre spectral sequence, so let's see it in action! For the previous blog posts we've used $\mathbb{RP}^\infty$ as an example, but for the SSS we'll be looking at $\mathbb{CP}^\infty$ instead (since it's an easier case). For $\mathbb{RP}^\infty$ we had the CW complex $\mathbb{RP}^\infty = e^0 \cup e^1 \cup \cdots$, but for $\mathbb{CP}^\infty$ we have:

where we attach all the even-dimensional cells.

For the $\mathbb{RP}^\infty$ case we have the following fibration $S^0 \rightarrow S^\infty \rightarrow \mathbb{RP}^\infty,$ to compare/contrast to the fibration of $\mathbb{CP}^\infty$ we'll be working with

and note that we'll be proceeding with $\mathbb{Z}$ coefficients for the rest of this post for the (co)homology of $\mathbb{CP}^\infty.$

We'll begin by first exploring the homology $H_*(\mathbb{CP}^\infty)$ via the Serre spectral sequence and then transitioning to the cohomology.

Homological Serre Spectral Sequence: For this, we have the following three main ingredients:

• $E_{p,q}^2 \cong H_p(\mathbb{CP}^\infty ; H_q( S^1;\mathbb{Z} ))$
• $d^r: E_{p,q}^r \rightarrow E_{p-r, q+r-1}^r$
• $E_{p,q}^{\infty} \implies H_{p+q}(S^\infty;\mathbb{Z})$

The game plan is the following: we'll start off by looking at the $E^2$ page, this is where we usually always begin. Then we'll analyze the $E^\infty$ page. Then, we'll see that most $d^r$ differentials are zero maps, so we'll look for the few non-zero maps that are out there. These will help us find where the $E^r$ pages stabilize, i.e.

Then when we connect the $E^2$ page with the $E^\infty$ page via the differentials, we'll be able to solve the puzzle! Let's get started.

First, let's take a look at $E_{p,q}^2 \cong H_p(\mathbb{CP}^\infty ; H_q( S^1;\mathbb{Z} ))$. $S^1$ has pretty simple homology groups, where $H_0(S^1;\mathbb{Z}) \cong \mathbb{Z}$, $H_1(S^1;\mathbb{Z}) \cong \mathbb{Z}$, and in general $H_q(S^1;\mathbb{Z}) \cong 0$ for $q \geq 2$. Since the $q$-index moves vertically (up/down), these denote rows in the grid for the $E^2$ page. Since $H_q(S^1;\mathbb{Z}) \cong 0$ for $q \geq 2$, this means that the rows $q=2, q=3, \ldots$ are all zero. Let's look at the $q=0$ and $q=1$ rows:

Note that these groups are very subtly different since we usually let $1 \in \mathbb{Z} \cong H_0(S^1;\mathbb{Z}))$ be a (identity) generator different from the non-zero generator $\gamma \in \mathbb{Z} \cong H_1(S^1;\mathbb{Z})$. We get the following grid for $E^2$:

Here we see some differential maps $d^2: H_p(\mathbb{CP}^\infty) \rightarrow H_{p-2}(\mathbb{CP}^\infty)$, for $p \geq 2$, where we're mapping $d^2: E_{p, 0}^2 \rightarrow E_{p-2, 1}^2$. This is a good spot to be for $E^2$, we'll come back after looking at $E^\infty.$

Next, we're looking at $E^\infty$. Let's look at the homology of $S^\infty$ which is closely tied to $E^\infty$. Importantly, $S^\infty$ is a contractible and path-connected space, so we get the following homology groups:

We won't rigorously prove this, but one way to see this is to consider the following direct limit for the $0$th homology:

where we get $\mathbb{Z} \rightarrow \mathbb{Z}$ isomorphisms that eventually stabilize. Similarly for a fixed $k > 0$, we can in a hand-wavey fashion consider:

where we have a sequence like

such that we hit a non-zero group $\mathbb{Z}$ at $H_k(S^k) \cong \mathbb{Z}$, but then get zero everywhere else in $H_k(S^n)$ for any fixed $k$ such that $n > k > 0$. Note that in both of the arguments for $H_0$ and $H_{k\geq0}$ we're using the fact that homology commutes with the direct limit since homology is a covariant functor. Putting things together, we get:

for $n=0$, and $\bigoplus_{p+q =n} E_\infty^{p, q} \cong 0$ for all $n > 0$. This means that only at $(p,q) = (0,0)$ do we get a non-zero group in the $E^\infty$ page, namely $\mathbb{Z}$. So we get:

Now that we have both the $E^2$ and $E^\infty$ pages, let's connect them! The glue that connects these $E^r$ pages are the differentials $d^r$. Since the $d^r$ maps are responsible for annihilating almost all the groups from the $E^2$ page, let's look at which differentials are non-zero (everything else will just be zero maps).

• $d^2: E_{p,q}^2 \rightarrow E_{p-2, q+1}^2$ $\,|\,$ Let's see how this map works on our non-zero $q=0$ and $q=1$ rows. Spoiler alert: we've already seen this earlier in the grid for $E^2$. This map goes from groups in row $q=0$ to groups in row $q=1$ (and then shifts to the left by 2). For row $q=1$ we map to the zero row $q=2$, so note that these maps $H_p(\mathbb{CP}^\infty) \rightarrow 0$ are such that $\ker( H_p(\mathbb{CP}^\infty) \rightarrow 0) \cong H_p(\mathbb{CP}^\infty)$ since everything gets mapped to zero.
• $d^3: E_{p,q}^3 \rightarrow E_{p-3,q+2}^3$ $\,|\,$ Similarly, let's see how this $d^3$ map works. Recall that for the $E^2$ page we have that the rows $q=2, q=3, \ldots$ are all zero. A key property of spectral sequences is that any rows or columns that are entirely zero for some page $E^r$ must remain zero on all subsequent pages $E^{r+1}$ and so on. Thus, for $E^3$, $E^4$, etc, the all rows $q \geq 2$ are also zero. Note that this $d^3$ map goes from the row $q=0$ to the zero row $q=2$, and also from $q=1$ to the zero row $q=3$ in $E^3$. Since these are all maps to zero, then this means $d^3$ is a collection of zero maps. Furthermore, all $d^r$ for $r \geq 3$ are also forced to be zero since we'd be mapping $q=0$ to some zero row $q=r-1 \geq 2$ and $q=1$ to the zero row $q=r\geq3.$

From $d^3$ onward, we see that $d^r$ for $r \geq 3$ must all be zero. Since all the differentials from $E^3$ onwards are zero, then this implies that

where the pages all collapse such that $E^3 = E^\infty$. With this, we'll be able to connect $E^2$ to $E^\infty$ automatically with:

where $E_{p,q}^3 \cong E_{p,q}^\infty$. Recall that

or to visualize this in the grid for $E^\infty:$

Looking at $p \geq 2$ and $q=1$, we get the following incoming/outgoing maps for $E^2_{p,q}$:

where $\ker( d^2: E_{p, 1}^2 \rightarrow E_{p-2, 2}^2 ) \cong H_p(\mathbb{CP}^\infty)$ since this outgoing map maps everything in $E_{p, 1}^2 \cong H_p(\mathbb{CP}^\infty)$ to zero. So far we have

Importantly, since $E^3_{p,1} \cong E^\infty_{p,1}$ for all $p \geq 0$, then we see that the $q=1$ row of $E^3 = E^\infty$ is zero, thus $E^3_{p,1} \cong 0$ for all $p$. This implies that $\text{im}( d^2: E_{p+2, 0}^2 \rightarrow E_{p, 1}^2 ) \cong H_p(\mathbb{CP}^\infty)$, where $E_{p, q}^2 \cong H_p(\mathbb{CP}^\infty$ which means that this incoming $d^2$ map must be surjective.

We see a similar story for $q=0$. The formula for the $E^3$ page is:

The relevant sequence of maps for the term $E_{p,0}^2$ is:

Since any term with a negative $q$-index is zero, the incoming map from $E_{p+2, -1}^2$ is the zero map. Thus, its image is $0$, and the formula simplifies to:

We know $E_{p,0}^3 \cong E_{p,0}^\infty$, which is $0$ for all $p > 0$ (since $E_{0,0}^\infty \cong \mathbb{Z}$ is the only non-zero term on the bottom row). $E_{p,0}^3 = 0$ then implies that:

for $p > 0$. This means that the outgoing map $d^2: E_{p, 0}^2 \rightarrow E_{p-2, 1}^2$ must be injective for all $p>0.$

Putting this together with our previous result:

• The map $d^2: E_{p+2, 0}^2 \rightarrow E_{p, 1}^2$ is surjective for all $p \geq 0$. Importantly, note that (after a bit of re-indexing) this is equivalent to $d^2: E_{k,0}^2 \rightarrow E_{k-2,1}^2$ being surjective for $k \geq 2.$
• The map $d^2: E_{p, 0}^2 \rightarrow E_{p-2, 1}^2$ is injective for all $p > 0.$

The map $d^2: E_{p, 0}^2 \rightarrow E_{p-2, 1}^2$ is an isomorphism when the two conditions (i.) $p \geq 2$ (for surjectivity) and (ii.) $p > 0$ (for injectivity) are both satisfied at $p \geq 2$. Therefore, $d^2$ is an isomorphism for these indices.

Recalling that $E_{p, q}^2 \cong H_p(\mathbb{CP}^\infty; \mathbb{Z})$, we've found an isomorphism for $p \geq 2$:

Since $\mathbb{CP}^\infty$ is path-connected then we have $H_0(\mathbb{CP}^\infty;\mathbb{Z}) \cong \mathbb{Z}$, and since it's also simply-connected then we have $H_1(\mathbb{CP}^\infty;\mathbb{Z}) \cong 0$. Now we can put the $d^2$ isomorphism to use. Since $H_p(\mathbb{CP}^\infty;\mathbb{Z}) \cong H_{p-2}(\mathbb{CP}^\infty;\mathbb{Z})$, we get:

where we just get

With this, we've successfully used homological SSS to find $H_*(\mathbb{CP}^\infty)$! Next, we'll be using cohomological SSS to find the cohomology ring $H^*(\mathbb{CP}^\infty)$. We'll see that the cohomology side of this story is very similar to the homology one, but with an additional structure that comes from our good friend the cup product.

Cohomological Serre Spectral Sequence: We begin in pretty much the same way as homological SSS with the following three main ingredients:

• $E^{p,q}_2 \cong H^p(\mathbb{CP}^\infty ; H^q( S^1;\mathbb{Z} ))$
• $d_r: E^{p,q}_r \rightarrow E^{p+r, q-r+1}_r$
• $E^{p,q}_{\infty} \implies H^{p+q}(S^\infty;\mathbb{Z})$

Besides the subscript and superscript flipping, note the different ways the differentials in cohomological SSS move compared to the homological case, where we move $(p,q) \mapsto (p+r, q-r+1)$ in the cohomological case versus the homological one $(p,q) \mapsto (p-r, q+r-1).$

As we mentioned, the cohomological SSS has an additional product structure that is induced by the cup product. For $x \in E_r^{p,q}$ and $y \in E_r^{s, t}$ we have:

where $\deg(x) = p+q$. This ensures that the product is well-defined on $E_{r+1}$. Note that we'll use either $xy$ or $x \cdot y$ notation to denote this product which is induced by the cup product. On the $E_2$ page, however, for $x\in E_2^{p,q}$ and $y \in E_2^{s, t}$, the product operation $xy$ (or $x\cdot y$) is exactly the cup product. Furthermore, even for the higher pages $E_{r+1}$ for $r \geq 2$ the induced product inherits all the algebraic properties of the cup product, including the graded commutativity one:

where $\deg(x) = p+q$ for $x\in E_2^{p,q}$, and $\deg(y) = s+t$ for $y\in E_2^{s,t}.$

Same as before, let's start with the $E_2$ page. Note that, very similar to the homology case, we have the following for $H^q(S^1;\mathbb{Z})$

So, exactly as before, we immediately have that the rows $q\geq 2$ are all zero since $H^p(\mathbb{CP}^\infty ; 0) \cong 0$. Let's look at the $q=0$ and $q=1$ rows:

On $E_2$, we have the $d_2$ differential such that $d_2: E_2^{p,q} \rightarrow E_2^{p+2, q-1}$. On the $E_2$ grid, we'll be focusing on the non-zero rows $q=0$ and $q=1$. We get the following grid for the $E_2$ page:

Note that this is very similar to the homological $E^2$ grid with the arrows reversed. Additionally, same as the homological case, having the rows $q\geq 2$ be all zero implies that they are also all zero on the higher pages $E_{3}$ and up. Furthermore, having the rows $q \geq 2$ be all zero also implies that the differentials $d_r$ for $r \geq 3$ are also zero since they either "jump over" the non-zero $q=0,1$ rows or get a map like $0 \rightarrow H^{q}(-) \rightarrow 0$ which is also just zero.

On $E_\infty$, we have $H^*(S^\infty;\mathbb{Z})$. Same as the homology case, since $S^\infty$ is contractible, we get that $H^0(S^\infty;\mathbb{Z}) \cong \mathbb{Z}$ and $H^k(S^\infty;\mathbb{Z}) \cong 0$ for $k >0$. Similarly as before, we get

Since $d_r$ for $r \geq 3$ are all zero, we get that $E_3 = E_4 = \cdots = E_\infty$. We can go through the motions and parallel a very similar argument as the homological SSS to get that

is an isomorphism. This is similar to the previous $d^2$ isomorphism with some superscript/subscript flipping and re-indexing. OK, great so I guess we're done? Not so fast! We may have an isomorphism, but we still need to work out the product structure of $H^*(\mathbb{CP}^\infty).$

Let $1 \in E_2^{0,0} \cong H^0(\mathbb{CP}^\infty;\mathbb{Z})$ be the multiplicative identity, and let $\alpha \in E_2^{0,1} \cong H^0(\mathbb{CP}^\infty; \mathbb{Z}\langle\gamma\rangle)$ where $\gamma$ is the generator for $H^1(S^1;\mathbb{Z}) \cong \mathbb{Z}$. Since $\alpha \in E_2^{p,1}$ for $p=0$, let's apply $d_2$ to get $d_2(\alpha) \in E_2^{2,0}$. Note that, with the isomorphism, we can denote

where $x$ is the generator of $E_2^{2,0} \cong H^2(\mathbb{CP}^{\infty}; \mathbb{Z})$. Next, let's look at the $k$-th power of $x$, i.e. $x^k \in H^{2k}(\mathbb{CP}^\infty;\mathbb{Z})$. What happens when we do $\alpha \cdot x^k$?

Let's look at $d_2(\alpha \cdot x^k):$

Note that $d_2(x^k) = 0$ since $x^k \in E_2^{2k,0}$ then $d_2$ maps this to $E_2^{2k+2,-1}$ which is a zero row (since $q <0$). Since $d_2(x^k) = 0$ then this whole $(-1)^{\deg(\alpha)}\alpha \cdot d_2(x^k)$ term goes to zero. We then get

since $d_2(\alpha) = x$ as we had defined. Note that since $x^k \in E_2^{2k,0}$ and $\alpha \in E_2^{0,1}$ then $\alpha \cdot x^k \in E_2^{2k, 1}$ which follows from the bidegree conventions of the SSS. This means that $d_2$ maps to $E_2^{2k+2,0}$, hence

where $E_2^{2k+2,0} \cong H^{2k+2}(\mathbb{CP}^\infty;\mathbb{Z})$. Note that since $2k+2 = 2(k+1)$, we have the generator $x^{k+1} \in H^{2(k+1)}(\mathbb{CP}^\infty;\mathbb{Z})$ which is just the $(k+1)$-th power of the generator $x \in H^2(\mathbb{CP}^\infty;\mathbb{Z})$. Also note that $\deg(x) = 2$ since $x \in E_2^{2,0}.$

This shows (by induction) that the $d_2$ differential maps the generator of each group on the $q=1$ row isomorphically to a generator on the $q=0$ row. This complete cancellation is exactly what has to happen for the spectral sequence to collapse and correctly converge to the trivial cohomology of the total space $S^\infty$ on the $E_\infty$ page. Thus, the only way for the sequence to be consistent is if the cohomology of the base space $\mathbb{CP}^\infty$ has a structure that enables this cancellation, which is a polynomial ring generated by the single element $x$ of degree 2, i.e.

where $x\in H^2(\mathbb{CP}^\infty; \mathbb{Z})$ and $\deg(x)=2.$

This is very similar to the cohomology ring of $\mathbb{RP}^\infty$ from blog post #2, where we have

for $\alpha \in H^1(\mathbb{RP}^\infty;\mathbb{Z}_2)$. There's a deep geometric connection between the similarities of these two cohomology rings, but that's a story for a later episode. ■

[1] J. Adem, The iteration of the Steenrod squares in algebraic topology, Proc. Nat. Acad. Sci. U.S.A. 38 (1952), 720–726.

[2] R. P. Stanley, Enumerative Combinatorics, 2nd ed., Cambridge University Press, Cambridge, 2011.

[3] A. Hatcher, Algebraic Topology, Cambridge University Press, 2002.

[4] J. McCleary, A User's Guide to Spectral Sequences, 2nd ed., Cambridge University Press, Cambridge, 2000.

[5] A. Hatcher, Spectral Sequences in Algebraic Topology (course notes, freely available).