describe your research iii

So how much more complicated can this game get? What motivation do we have to study other homogeneous spaces of higher dimensions? How is combinatorics involved?

I think the first and second questions need to be answered together, as there is constant tension between the impulse to add additional complexity to mathematical structure and the need to explain to other people why they should care. Mathematicians have for the most part chosen a sort of middle path which is to build the tools that would enable the study of arbitrarily complicated spaces by classifying the components of which they can be built. Then, whenever motivation to study a particular case of these things comes along from physics, economics, computing, or other parts of mathematics (most often), the motivated researcher has some footing to make progress by putting together the pieces.

You could think of it sort of like being a plumber going to the hardware store looking for the right fittings for some pipes. Maybe you’re a physicist and you’re trying to study a system of particles that obey certain symmetries. So what do you do? You go to the group store of course, and see if they have the right group. Sometimes they don’t have the perfect one, but they have the right parts and documentation so that you can assemble the right one yourself without too much difficulty. Now that you have a sensible way to keep track of symmetries in your system, you can go back to worrying about predicting how the universe behaves.

It is on the one hand wonderfully remarkable that the business of classifying the simple pieces of which geometric spaces can be built is a tractable pursuit at all. On the other hand, it is a mathematicians job to make it tractable. If some definition of what constitutes a “simple piece” leads to an impossible classification, then it is not the right definition. While some people might have you believe that theorems like the classification of finite simple groups or semi-simple Lie algebras are miracles from gods, they have also come through people toying with and massaging definitions until the classification task began to seem sensible (though still possibly monumental).

I don’t mean to demean the examples above; I too am enamored of them, and most of my published research uses the classification of simple Lie algebras as sort of a starting point. But I think of integer factorization as the proto-problem for these deconstructive classification quests. The fact that natural numbers all break down into prime factors (they are “classified” by their prime divisors) really does seem like a miraculous piece of order in the universe, or at least an inescapable part of how human consciousness interprets it. That we don’t have a very efficient way to take a number and break it down into factors is astonishing given how fundamental this problem is.

There is a dilemma here that I’ve always sensed in mathematics but that no one really likes to speak about (if it registers in anyone else’s reality). Mathematics and academia at large are subject to economic and political realities. One of these realities is the need to have a career and to instantiate said career by publishing papers. This incentivizes the mathematician not to work on old, very difficult problems, but to invent new theories, ask questions possibly nobody else is asking, collect fruit from those fonts and write about them. This is a cynical view of what could also be romanticized as a tremendous freedom and creativity afforded to the profession by the fact that we’ve managed to convince the rest of the scientific and engineering fields that they need us to teach them calculus, statistics, and the like. (See: Mackey’s lecture on what it is to be a mathematician.) But there are also cases where solutions to old, difficult problems have eventually come through extended scenic detours.

Moreover, I don’t want to diminish the amount of work that has gone in to building up the vast amount of theory that exists today. It is not that it comes easily, but rather that it comes at all that appeals to the hungry mathematician. It can be a joyous experience to sit down with a problem, toy with it for hours, weeks or months, and then to finally resolve it. And it is absurd and incredible that human intellect is somehow at all suited to the process of assimilating information, experimenting, ordering it, and manipulating it in a way that reveals a “truer” structure. The pleasure that comes with this experience is, I guess, at least as strong a force behind the proliferation of mathematical theory. It is a good thing that we will never run out of problems.

But now I haven’t said anything about combinatorics.

describe your research ii

It might seem like were pretty far in the weeds by now but there’s a point to make here which is essential to my work. And we may as well introduce some of that noxious stuff — mathematical jargon. The collection of symmetries we’ve been talking about that can act in sequence or be done and then undone is usually called a group in mathematics. Maybe because it’s sort of like a group that gets together to play ball (though it might be with a line). By the way — not everyone in the group actually knows how to really play. There is always one character called the identity that just leaves the object as it is, moving nothing. This is still a symmetry though, in the same way 0 is a number, which is worth thinking about some.

An interesting question to ask is: where can points on the line go as the group bobbles it around? To simplify, how about where does a particular point, say zero, go? Where it ends up depends on the particular symmetry we apply, but considering all possibilities we see that zero can really go anywhere on the line. How? Well, if we want to take 0 to some value x we actually have two options: either apply the translation that adds x to every point, or apply the reflection through the value \frac{x}{2}. Since the point 0 can go anywhere, we say that its orbit under the group is the whole line.

We’ve seen that we can think of the members (elements) of our group as constituting two copies of the line. On the first line, each point is a translation. On the second line, each point is a reflection (which can actually be realized as a translation and a reflection through 0). The fact that we have two ways to take 0 to, say, \pi mirrors the fact that the group looks like two copies of the line.

So here’s a key idea: probably you would agree that the line is a natural enough geometric object. We need lines to get notions of distance and angle going in geometry! On the other hand, this group which is two copies of the line seems a bit funny and abstract. But it is still intimately attached to the line by keeping track of its symmetries. So what if there was a way to have both? That is, to have just the line in hand as an object but manifest in a way that also keeps track of the symmetry. This is what homogeneous spaces are for.

To explain how we get a homogeneous space here, I need to try to explain something in group theory called the orbit-stabilizer theorem. Now, I don’t really know how to do this without either bringing in a little notation or being extremely verbose (and likely confusing). But there is power in notation so let’s try that. We’ve observed that all reflections can be considered as reflections through zero together with a translation. Lets call the reflection through zero “flip,” and label translations by the value that is added to every point, x. Then every element of our group is a pair of the form (x, flip) or (x, no flip) according to whether it is a reflection or a translation.

Notice that the reflection through zero (0, flip) “fixes” zero, that is leaves it put, and other than the identity, which can be thought of as (0, no flip), it is the only element that does this. In group theory terms, these two elements form a subgroup which consists of the elements that fix 0. The orbit-stabilizer theorem says that if we clump these two elements together and think of them as a single object, and then do the same for all of the pairs (x, flip) and (x, no flip), then the resulting collection will be the same as the orbit of the point 0. But this is the just the line again! This should make some sense; in our group, we had two copies of the line and all we’ve done is glued those copies together at points that line up.

This “clumping” of objects according to a subgroup is called taking a quotient in group theory, and a homogeneous space is really just a quotient. It’s useful because now the coordinates on the line now come from the coordinates on the group, which facilitates studying how the symmetries act, especially in more complicated examples. A lot of my work is studying how these quotient spaces break up and fit together in pieces when you restrict your symmetry set. We’ll try to come back to this in the next post.

describe your research i

Response to a prompt from a non-mathematician friend.

What is symmetry? Usually we think of symmetry as how something looks — it’s the same on both sides of an axis, or you can fold an image along a crease but you only need one side to know how to complete the picture. You achieve this reflecting the half you see across the axis.

The word “symmetry” in the vernacular usually refers to this kind of reflective symmetry. In its origin, the word means something like “agreement in measure or form.” But mathematicians have deranged this word to turn it from a property that is into something you can do. This means instead of just observing the symmetry across an axis, we think of the symmetry as actually making the reflection happen. In the end, the resulting image is the same of course, but thinking of symmetries as actions is an important and useful step.

Now that symmetry is made into something kinetic, we can reconsider the forms symmetry can take. What is the fundamental result of applying a symmetry? Well, it’s really what we said already, ultimately the image or object to which the symmetry is applied appears the same. Taking this as our defining property allows us to include a few other kinds of symmetry we haven’t yet considered. In addition to reflective symmetry, there also rotational symmetry — like taking a wheel and spinning it on its axle — and translational symmetry.

Translational symmetry takes us immediately into the infinite. It is what it sounds like: you pick something up, slide it over, plunk it back down, and then it appears exactly the same as it did before. But I don’t mean it looks like the same object just in a different place. Rather, the whole montage looks exactly the same! Like if you had two photos of the scene where this happened side by side, a “before” and an “after,” they would look exactly the same. The thing is that no “finite” or “bounded” objects (as we are accustomed to) can have this symmetry. This is is because the bounded objects have extremities — a northernmost point, an southeasternmost point etc. If you take one of these and then drag further in that direction, you’ll always be cutting a new path, so the extremity can’t wind up back in a position occupied by the original object. Necessarily, the picture will look different.

But if you take something that extends infinitely — say an abstracted line, the continuum, our model of the real numbers — all of a sudden every point has somewhere back on the line to go when you slide. Here’s another funny thing that begins to emerge: the set of symmetries begins to look like the object it’s acting on. Considering the real line, every real number corresponds a symmetry of translation. That is, for a fixed real number a, we can send each number x in the line to x+a and have the whole line slide over (to the left or the right depending on whether a is negative or positive) and land back on top of itself. So the line has a line’s worth of symmetries! Does it have any others?

Well, again we have reflection. Pick a point in the line, and reflect the line about it. It’s not hard to see that everything goes back to somewhere else on the line. Moreover, points that were close together end up the same distance apart after applying the reflection. Symmetries that satisfy this property are called isometries, as they preserve the intrinsic geometry of the line.

So do we have another line’s worth of reflections to add to our list of symmetries? Well, you could say that if you want to think of these symmetries in isolation, as objects of a kind that are unable to interact with each other. But that would run counter to the notion we are developing of symmetries not as things that are but as things that do! A symmetry acts on the line, and once it has acted, another symmetry can act on the result and so on, kicking the points on the line this way and that.

To make the next point, let’s consider an example. Take the reflection through the number 1. Under this reflection, 1 stays put and 0 goes to 2. Maybe you can convince yourself that this is sufficient information to completely determine this symmetry. But here’s another way to do the same thing: reflect through 0, and then slide everything 2 to the right. With a bit more thought, you can probably convince yourself that the reflection through any point can be obtained similarly by a reflection through 0 followed by a slide, or a slide and and then a reflection through 0. What we say is that the reflections and translations altogether are generated by just the reflection through 0 together with the translations.

Mushrooms into corollaries

OK this is gonna be a quick oddball post, but I’m so excited about this that I can’t not tell the world.

Pandemic life has found me, like many, drinking much more coffee made at home, and also exploring new hobbies and ventures. One of these new ventures is experimenting with mushroom cultivation. There are myriad techniques at all levels of sophistication for the many varieties of edible and medicinal mushrooms, but I was instantly excited when I learned that one can grow oyster mushrooms on used coffee grounds with virtually no technical set-up.

I mean for real, cheap coffee that turns into free food?! This is like a math graduate student’s dream.

Colonizing mycelium

So I saved my spent coffee grounds in the freezer for a few weeks, got some oyster mushroom spawn, then inoculated the grounds in a sanitized plastic container and waited. Among cultivated mushrooms, oysters are known for being particularly vigorous and adaptable, growing on anything from logs to old clothes to coffee and tea. The rapidity with which they colonized the grounds was astounding; already the next day the mass was glowing with patches of white mycelium waking up and stretching its legs. Within a week, a fluffy cushion of the stuff coated the whole surface. A few more weeks go by (maintaining high moisture in a cool, dark place) and voila!

One day…

One day the fruiting bodies start to “pin” and then they double in size for a few more days until they are ready to harvest. The next night, mushroom fried rice for dinner. : D

… and the next! Look at this monster!!

As the myco-aware are eager to tell you, this is a large untapped resource that society in general and the math community in particular has been sleeping on. In our department, another graduate student used to collect used grounds from the coffee machine for compost, but with very little effort every department could put its spent theorem-precursor to good use by setting up its own mushroom growing operation (myceliated substrate makes great compost after the fact).

Growing food is a radical community-building and liberatory act. The origin of the word university is the combination of uni (one) and versus (turned), “turned into one”: basically, community. Subsidizing graduate student grocery budgets is one thing, but whenever we get back to being able to sharing pots of coffee, this would be a great project for pushing on universities to be a little more like communities and less like bureaucracies.

I basically followed the instructions here, and I recommend the source book (in the link) to anyone interested in learning more.

Renaming the math building at T. U. of Louisiana

**The following letter was delivered (Sep. 2020) to the board and senior administrators of the university on behalf of a group of graduate students and faculty in the math department, and other concerned community members.**

We, the undersigned members of the Mathematics Department and Tulane University community, call upon the senior university administration, President Mike Fitts, and his appointed Naming Review Task Force to undertake the renaming of

(1) Gibson Hall,

(2) the online university platform called Gibson, and

(3) any other monuments to Randall Lee Gibson within the university’s purview.

We make this petition with the facts in mind that Randall Lee Gibson

(1) was a slaver and sugar plantation owner in Terrebonne Parish,1 

(2) adhered to hardline beliefs in racial inequality, authored and published pro-slavery essays, ran for political office as a secessionist before the Civil War to preserve slavery in an independent South,1

(3) enlisted as a Confederate soldier upon outbreak of the Civil War, eventually rising to the rank of Brigadier General,3

(4) helped to restore former Confederates to political power in the backlash to Reconstruction and benefited from violent white terror campaigns meant to suppress black voters,4 and

(5) convinced Paul Tulane to “confine his bequest [to the university] to white persons.”4

We support the efforts of the Naming Review Task Force and President Fitts’ message that “racism has no place” on our campus, and so we insist upon the removal of all monuments to individuals aligned with slavery, racial segregation, and other forms of oppression. It is unacceptable that Tulane University continues to honor the name of a person that profited by and fought to protect chattel slavery. 

Our purpose is not to deny history, but rather to recognize it and connect its meaning to our present so that we may move beyond the moral deficiencies of our forebears.  Gibson Hall was named in honor of Randall Lee Gibson’s role as the first president of the Administrators of the Tulane Educational Fund, in which he oversaw the transformation of the public University of Louisiana into the private, exclusively white Tulane University of Louisiana. This conversion was made with explicit racialized intent through Paul Tulane’s act of donation.2

The university is much different now than it was in Gibson’s time, but its entanglement with white supremacy remains. The removal of monuments to oppressors is essential to our university’s project to become a more inclusive and equitable institution. With this aim in mind, we assert the necessity of renaming Gibson Hall.



[1] Allardice, Bruce S., and Lawrence Lee Hewitt, eds. Kentuckians in Gray: Confederate Generals and Field Officers of the Bluegrass State. University Press of Kentucky, 2015.

[2] Dyer, John Percy. Tulane: The biography of a university, 1834-1965. Harper & Row, 1966.

[3] United States Congress (1893-1894). Memorial address, and 2d session, 52d Cong. Memorial Addresses On the Life And Character of Randall Lee Gibson, (a Senator From Louisiana,): Delivered In the Senate And House of Representatives, March 1, 1893, And April 21, 1894. Washington: Govt. print off., 1894. 

[4] Sharfstein, Daniel J. The invisible line: A secret history of race in America. Penguin, 2011.

Rabinowitsch in Translation

One of our projects in number theory led us to thinking about the class number problem, which has a story too long and interesting to recount here. See the survey of Goldfeld from an old AMS Bulletin to get an overview. Briefly, the question is about the degree to which unique factorization holds in quadratic extensions of the rational numbers.  In any case, one of the first important results in the program is an old theorem of G. Rabinowitsch (Rabinowitz), which gives a testable criterion for whether the number field \Q(\sqrt{D}), with D a negative integer, possesses unique factorization. In modernish language (cf. Theorem 6 in the linked-to document), keeping this framework we have:

Theorem (Rabinowitsch, 1913): The field \Q(D) is a unique factorization domain (UFD) if and only if x^2-x+ m is prime for all x\in \{1,2, \dots, n\}, where D=1-4m.

The paper was published in German in Crelle’s journal over a century ago, and was somewhat hard to find on its own. We could not locate any other source where the content has been rewritten since then, so we translated and typeset the article (with the aid of various online translation tools). To preserve the spirit and style of the writing, some outdated and perhaps idiosyncratic (to author or translator, as the case may be) jargon has been allowed to survive. These designations are hopefully all made clear within the article so it is readable. Some notation has been modified for clarity.

Perhaps one day this material can be condensed and fully modernised. The article consists mostly of pleasant and clean elementary number theory, situated near the headwaters of one of the important achievements of 20th century number theory. It seems worthy of further propagation, so we post here our translation (no warranty included).


Spectral Sequences III

Bigraded Algebrae

McCleary introduces the concept of a differential graded algebra in section 1.3 (Definition 1.6, p. 11). These are algebras (over a field k), which tend to be \N-graded, and importantly carry with them a map d called a differential which is k-linear, shifts the degree of elements (in the grading) up by one, and satisfies a “Leibniz rule:”

    \[d(a\cdot a')=d(a)\cdot a'+(-1)^{\deg(a)} a\cdot d(a')\]

for a, a' in our algebra A^*. This is a twisted version of what is usually called Leibniz’ rule in calculus (which is basically just product rule), which coincides with how the differential works in the algebra of differential forms.

This idea is easily extended to the notion of a differential bigraded algebra (E^{*,*}, d), where now the elements are \N^2 graded (for the time being, later we’ll have \Z^2), but d remains a total-degree 1 mapping. That is,

    \[d: \bigoplus_{p+q=n} E^{p,q} \lra \bigoplus_{r+s=n+1}E^{r,s},\]

and d still satisfies the Leibniz rule

(1)   \begin{equation*} d(e\cdot e')=d(e)\cdot e'+(-1)^{p+q}e\cdot d(e') \end{equation*}

where e\in E^{p,q}.

A standard construction is to form a bigraded algebra by tensoring two graded algebras together. This would work with just component-wise multiplication, but to get a working differential that satisfies our version of the Leibniz rule 1 as well, we introduce an extra sign: we mean, supposing (A^*,d) and (B^*, d') are differential graded algebras, then we can assign E^{p,q}:=A^p\ox B^q, and furthermore

(2)   \begin{equation*}  (a_1\ox b_1 )\cdot (a_2\ox b_2):= (-1)^{(\deg a_2)(\deg b_1)}a_1a_2\ox b_1 b_2.\end{equation*}

Then if we define a differential d_\ox on E^{*,*} by

(3)   \begin{equation*}  d_\ox(a\ox b)=d(a)\ox b + (-1)^{\deg a} a \ox d'(b), \end{equation*}

then d_\ox satisfies the Leibniz rule 1. It is clarifying to check this, so we’ll record it here. Switching notation a bit, we will write (-1)^{\ab{a}} instead of (-1)^{\deg a}. To satisfy 1 we need

    \begin{align*}d_\ox((a_1\ox b_1)\cdot (a_2 \ox b_2))& =  d_\ox (a_1\ox b_1)\cdot (a_2 \ox b_2) + \\ &(-1)^{\ab{a_1}+\ab{b_1}} (a_1\ox b_1) \cdot d_\ox(a_2\ox b_2) \end{align*}

we then apply 3 to the individual terms on the right side above to get

    \begin{align*} d_\ox((a_1\ox b_1)\cdot (a_2 \ox b_2)) = [d(a_1)\ox b_1+((-1)^{\ab{a_1}} a_1 \ox d'(b_1)]\cdot (a_2\ox b_2)+\\ (-1)^{\ab{a_1}+\ab{b_1}} (a_1\ox b_1) \cdot [d(a_2)\ox b_2+(-1)^{\ab{a_2}} a_2\ox d'(b_2)]. \end{align*}

Now applying the multiplication rule 2 and distributing, we find

(4)   \begin{align*}  d_\ox((a_1\ox b_1)\cdot (a_2 \ox b_2)) = (-1)^{\ab{a_2}\ab{b_1}}d(a_1)a_2\ox b_1b_2 +(-1)^{\ab{a_1}+\\\ab{d'(b_1)}\ab{a_2}}a_1a_2\ox d'(b_1)b_2 + (-1)^{\ab{a_1}+\ab{b_1}+\ab{d(a_2)}\ab{b_1}} a_1 d(a_2)\ox b_1b_2 +\\ (-1)^{\ab{a_1}+\ab{a_2}+\ab{b_1} +\ab{b_1}\ab{a_2}} a_1a_2 \ox b_1d'(b_2). \end{align*}

To check the rule holds, we perform this computation by instead multiplying first and then applying the differential. That calculation looks like

    \begin{align*} d_\ox((a_1\ox b_1)\cdot (a_2 \ox b_2)) &= d_\ox((-1)^{\ab{a_2}\ab{b_1}} a_1 a_2 \ox b_1b_2) \\ &=(-1)^{\ab{a_2}\ab{b_1}}[d(a_1a_2)\ox b_1b_2+ (-1)^{\ab{a_1}+\ab{a_2}}a_1a_2\ox d'(b_1b_2)] \\ &=(-1)^{\ab{a_2}\ab{b_1}}[(d(a_1)a_2+(-1)^{\ab{a_1}}a_1d(a_2)\ox b_1b_2 +\\ &(-1)^{\ab{a_1}+\ab{a_2}}a_1a_2\ox(d'(b_1)b_2+(-1)^{\ab{b_1}}b_1d'(b_2))]  \qquad \\ &=(-1)^{\ab{a_2}\ab{b_1}}d(a_1)a_2\ox b_1b_2+(-1)^{\ab{a_1}+\ab{a_2}\ab{b_1}} a_1d(a_2)\ox b_1b_2+ \\ &(-1)^{\ab{a_1}+\ab{a_2}+\ab{a_2}\ab{b_1}} a_1 a_2\ox d'(b_1) b_2 +\qquad\\ & (-1)^{\ab{a_1}+\ab{a_2}+\ab{b_1} +\ab{a_2}\ab{b_1}}a_1a_2\ox b_1d'(b_2). \end{align*}

Finally, remarking that \ab{d'(b_1)}=\ab{b_1}+1 and \ab{d(a_2)}+1=\ab{a_2} shows that terms of the last line above match with those of 4, so everything checks out and (A^*\ox B^*, d_\ox) becomes a differential bigraded algebra.

A Chain Rule

Given the length and detail of section 1.3, surprisingly we find no glaring errors in this section, but the use of the differential becomes somewhat muddled in calculation in section 1.4. Again, perhaps as an undesirable side effect of the fact that we remain at the “informal stage,” it’s always difficult to keep track of what assumptions we’re working with in each example. Case in point, example 1.H, p. 20. The paragraph preceding definition 1.11 seems to indicate that all graded algebras are assumed to be graded commutative — at least for the rest of the section, one guesses, though the language is vague. Let’s try this here with a bit more force.

Assumption: All graded algebras are graded commutative for the rest of the post. This is to say, for all x,y in any A^*, we have x\cdot y =(-1)^{\ab{x}\ab{y}} y\cdot x. Now let’s have a look at the example. We suppose a spectral sequence of algebras (E_r^{*,*}, d_r) with E_2^{*,*}\cong V^*\ox W^*, converging to the graded algebra H^* which is \Q in degree 0 and \{0\} in all others.  The example asserts that if V^* is a graded commutative polynomial algebra in one generator/variable, then W^* is a graded commutative exterior algebra in one generator, and vice versa.

The first confusion appears in a restatement of the Leibniz rule near the bottom of page 20, except this time there are tensors involved. This appears to be a mixed use/abuse of notation, which was slightly different in the first edition of the book, but not more consistent. The idea is as follows. V* and W^* embed into V^*\ox W^* under the maps v \mapsto v \ox 1 and w \mapsto 1\ox w.  Then one can also write an element w\ox z \in V^*\ox W^* (mind the inexplicable inconsistent choice of letters) as

(5)   \begin{equation*}  w \ox z = (-1)^0 (w\ox 1)\cdot (1 \ox z)= (w\ox 1) \cdot (1 \ox z) \end{equation*}

since the degree of 1 is zero in each graded algebra. Note that this also allows us to regard V^*\ox W^* as graded commutative with the tensor product as multiplication between pure V^* and pure W^* elements, writing

    \[z\ox w:=(1\ox z) \cdot (w \ox 1)=(-1)^{\ab{z}\ab{w}} (w\ox z). \]

One can apply Leibniz rule to the product in 5 so that if V^*\ox W^* comes with a differential d, we get

    \[d(w\ox z) = d((w\ox 1) (1\ox z)) =d(w\ox 1)(1\ox z)+(-1)^{\ab{w}}(w\ox 1) d(1\ox z).\]

The thing is we really need not write the tensor product w\ox 1; it is just as correct to write w on it’s own, as we often do with polynomial algebras and so on. Then the above can be written instead as

    \[d(w\ox z) = d(w)\ox z +(-1)^{\ab{w}} w \ox d(z) \]

as McCleary does near the bottom of page 20. What makes this confusing is that up to this point we had only seen differentials acting on tensors by defining the bigraded differential from tensoring two differential graded algebras together, seen above. In this context, the differential of the bigraded algebra must act on an element of the algebra coming from V^* \ox W, it cannot act on just one side of the tensor. What’s different here is that the tensor product is actually the multiplication operation on each page of the spectral sequence. Thus, the restatement of the familiar rule with new notation.

Nevertheless, the next equality is also a bit confounding at first, partly because McCleary, goes back to writing the extra 1 in the tensor, suggesting that we need to pay attention to its effect. He says that if d_i(1\ox u) =\sum_j v_j\ox w_j, then

(6)   \begin{equation*}  d_i(1\ox u^k) = k \lt(\sum_j v_j \ox ( w_ju^{k-1})\rt) \end{equation*}

which looks sort of reasonable as it resembles something like a chain rule, d(u^k)=k u^{k-1} d(u). It is presented as if it should follow immediately from the Leibniz rule stated before. But this seems weird when the degree of u is odd. To be totally transparent about this, let’s illustrate the case where k=2, suppressing the subscript on the differential again, but maintaining the tensorial notation.

    \begin{align*} d(1\ox u^2) & =d((1\ox u)(1\ox u)) \\ & = d(1\ox u) (1\ox u) +(-1)^{\ab{u}} (1\ox u) d (1\ox u) \\ &= \sum_j (v_j \ox w_j)(1\ox u) + (-1)^{\ab{u}} \sum_j (1\ox u) (v_j \ox w_j) \\ &=\sum_j v_j\ox w_j u + (-1)^{\ab{u}} \sum_j (-1)^{\ab{u}\ab{v_j}} v_j \ox u w_j \\ &=\sum_j v_j\ox w_j u + (-1)^{\ab{u}} \sum_j (-1)^{\ab{u}\ab{v_j}+\ab{u} \ab{w_j}} v_j \ox  w_j u \\ &=\sum_j v_j\ox w_j u + (-1)^{\ab{u}} \sum_j  v_j \ox  w_j u \\ \end{align*}

where the last line follows since v_j\ox w_j has total degree \ab{u}+1, so the sign inside the sum there has exponent \ab{u}(\ab{u}+1) which is even. We see that if u has odd degree then, these terms cancel and we get 0. So you say “wait a minute, that’s not right, I wan’t my chain rule looking thing” until you eventually realize that if u has odd degree, since it’s sitting in a graded commutative algebra, u^2 is actually zero! And the same goes for all higher powers of u. Then, d(u^2)=d(0)=0 makes complete sense. Meanwhile, if u has even degree, the terms will pile up with positive sign and we get the chain rule looking thing that was claimed. So the statement 6 is in fact true, though it really breaks down into two distinct cases.

Going forward in the example, McCleary only really seems to use the chain rule (liberally mixing in the described sort of abuse of notation) on terms of even degree, so it’s tempting to think that it only applies there, but it is sort of “vacuously true” in odd degree as well. Oh well. Onwards.

Spectral Sequences II

Two Stripes

The next thing to address in McCleary is an apparent mistake on p. 9 of section 1.2. Here we again assume a first quadrant spectral sequence converging to a graded vector space H^*. This is mentioned at the beginning of the section, but it’s easy to forget that when a bold-faced and titled example (1.D) seems to be presenting a reset of assumptions, rather than building upon prior discussion. Furthermore, in this example, McCleary seems to be working again with the assumption from example 1.A that F^p+kH^p=0 for k>0. On the other hand, this can be seen as a consequence of the fact that our spectral sequence is limited to the first quadrant, provided the filtration is finite in the sense of Weibel’s Homological Algebra, p. 123 (F^mH^s=0 for some m). But then it would be unclear why McCleary took this as an additional assumption rather than as a consequence of prior assumptions in the first case. : /

The new part of this example is the assumption that E^{p,q}_2=0 unless q=0 or q=n, so all terms of the spectral sequence are to be found just in two horizontal stripes. In particular E_\infty^{p,q} is only possibly non-zero in these stripes, and since these correspond to filtration quotients, the filtration takes a special form.

First, we might look at the filtration on H^s where 0\leq s \leq n-1.  Note that the spectral sequence terms that give information about H^s are those along the diagonal line where p+q=s.  Since s\leq n-1, the only place where anything interesting might happen is when this line crosses the p-axis, i. e. when q=0. This forces p=s, so the only possible nonzero filtration quotient is

    \[E_\infty^{s,0}=F^sH^s/F^{s+1}H^s=F^sH^s=F^0H^s=H^s \]

working with the assumption that F^{s+1}H^s=0. So on the one hand, we get no interesting filtration of H^s for s<n, but on the other hand we can see exactly what it is from the spectral sequence limit.

Now we treat the case of H^{n+p}, where p\geq 0. I find this awkward notation again, preferring to reserve p for a pure arbitrary spectral sequence index, but since we are trying to address the mistake in this notation, we should keep it for now. The filtration of this vector space/cohomology is interesting when q=n and q=0, where the quotients are given by

    \[E^{p,n}_\infty=F^pH^{p+n}/F^{p+1}H^{p+n}\quad\text{and}\quad E^{p+n,0}_\infty=F^{p+n}H^{p+n}/0.\]

Every where else, successive quotients are 0, meaning the filtration looks like…

    \[0\sus F^{n+p}H^{n+p}= \dots =F^{p+1}H^{n+p} \sus F^pH^{n+p}=\dots=F^1H^{n+p}\sus H^0{n+p}\]

In the filtration on page 9, McCleary puts one of the (possibly) non-trivial quotients at F^nH^{n+p} instead of at F^pH^{n+p} where it should be.  That’s all I’m saying.

This situation is modeled on a spectral sequence for sphere bundles i.e. bundles where the fibers are spheres of a given dimension. The stripes coincide with the fact that a sphere \mathbb{S}^n has nontrivial cohomology only at H^n and H^0. This sort of computation is famous enough that it has a name: the Thom-Gysin sequence (or just Gysin sequence).

As a final remark on section 1.2, McCleary says that the sequence in example 1.C is the Gysin sequence. Example 1.C doesn’t exist, we mean example 1.D : )