# A different approach towards spectral sequences.

I am somewhat acquainted with spectral sequences as in Weibel's book(the usual definition with many indices and pages), but I have found a different approach in the Stacks project.Link

Although I can sense that these are similar, I can't wrap my head around this new definition. Are they equivalent? Is one more general than the other?

## 1 Answer

I think these definitions are essentially equivalent. Weibel's definition is "obviously" a special case of SP's, the other direction is maybe a little less natural.

At first glance, Stacks Project's definition (basically) seems more general. Weibel's definition is probably the standard definition, and is equivalent to a special case in which SP's abelian category $$\mathcal A$$ is a category of the form $$\mathcal A = \bigoplus_{(i,j)\in\mathbb Z \times \mathbb Z} \mathcal A'$$ for some other abelian category $$\mathcal A'$$. Furthermore, the differentials in Weibel's definition must be of a very particular form - for instance on the $$E_2$$ page they are direct some of maps which go by a knight's move in $$\mathbb Z \times \mathbb Z$$.

Once you apply these two conditions, it looks like you arrive at something equivalent to Weibel's definition. However, watch out because Weibel allows the objects on the $$r$$-page to be isomorphic to the cohomology of the $$(r-1)$$-page, whereas SP says they are equal. This shouldn't make any difference up to isomorphism of spectral sequences though. And as long as we're pointing out insignificant details, Weibel also allows his spectral sequences to start at page $$a$$ rather than page $$1$$ like in SP.

Now let's try to come up with an SP-type spectral sequence that doesn't look like a Weibel-type spectral sequence, but then figure out how to make it into one. The example will show how to do it in general.

Let's take $$\mathcal A$$ to be abelian groups. Maybe $$E_0 = \mathbb Z/32$$ with $$d_0$$ given by multiplication by $$16$$. Then $$E_1 = 2\mathbb Z/16\mathbb Z$$. Now we may pick another differential, such as $$d_1$$ given by multiplication by $$4$$ for instance. Then $$E_2 = 4\mathbb Z/8\mathbb Z$$. At this point the spectral sequence would have to degenerate because there are no maps on $$\mathbb Z/2$$ which square to zero but the zero map itself. So, page by page it looks like this:

$$[\mathbb Z/32\mathbb Z \overset{\times 16}\longrightarrow \mathbb Z/32\mathbb Z], \ [2\mathbb Z/16\mathbb Z \overset{\times 4}\longrightarrow 2\mathbb Z/16\mathbb Z ],\ [4\mathbb Z/8\mathbb Z \overset{0}\longrightarrow 4\mathbb Z/8\mathbb Z], \ \ldots,\ E_{\infty} = [4\mathbb Z/8\mathbb Z \overset{0}\longrightarrow 4\mathbb Z/8\mathbb Z]$$

It actually looks a little like a Weibel spectral sequence if we imagine these diagrams are surrounded by many zeros, the only problem is that the differentials aren't pointing in the right directions. But there is an easy way to fix this. Instead of extending these diagrams by zeros, define a Weibel spectral sequence by $$E_r^{p,q} := E_r$$ - so we just put the same object at every node. Then define the differentials $$d_r^{p,q} = d_r$$ to be the same differential between every appropriate pair of nodes. What you get is a Weibel spectral sequence which converges to an $$\infty$$-page having $$E_\infty^{p,q} = E_\infty$$ for all $$p,q$$.

• Thanks for your trouble. The SP definiton looks neater, but I wonder if it makes a difference in calculations. – Jehu314 Jan 11 at 8:33
• I imagine this definition is most often used when you want to talk about things in a "coordinate free manner" or suppress notation. That is, you have a Weibel spectral sequence $E_r^{p,q}$ but instead of viewing it as a diagram of objects in $\mathcal A$ you view it as a single bi-graded object $E_r$. This change is just notational and won't make any difference in calculations. – Ben Jan 11 at 8:37