Yesterday I watched my friend Jialin Wang defend her thesis, and as part of her background section she mentioned that the group $F_2 \times F_2$ is incoherent in the sense that it has a subgroup that’s finitely generated and not finitely presented. I was curious how one might prove something like this, and in the original paper (Stallings’s Coherence of 3-Manifolds Fundamental Groups) this fact is boiled down to an “exercise which can be performed with the help of [a] spectral sequence”. I’ve been slowly trying to make spectral sequences feel like friends, so this seemed like the perfect thing to work out quickly and turn into a blog post! I’ve been doing a lot of writing lately, with two papers that I want out by the end of the summer and a new result (which will be my thesis) that I want out by the end of the year and an NSF proposal1, and even more stuff that I’m not talking about yet… So everyone in my life has heard me do nothing but complain about writing for the last month, haha. Weirdly, though, I’ve been itching to write a blog post! Maybe because it’s so informal, or maybe because it’s something I know I can finish, or maybe it’s because I’m mainly sick of writing about the same thing all day. No matter what it is, I’m happy to be here, and happy to have the excuse to share something cool ^_^. There’s no way I can give an introduction to spectral sequences that’s better than Vakil’s notes, so I won’t even try. I highly recommend everyone give those a read at least once in your mathematical life, especially if you’re planning to do anything that might require you to actually use spectral sequences “in the wild”. Going forwards in this post, I’ll assume that you know the basics of what spectral sequences are, and how (roughly) to compute with them, but if you’re feeling brave and know a bit about homology you can probably already understand a fair amount of the post. First, though, a few words about our goal. We’re trying to show that a product of free groups, $G = F_2 \times F_2$, is not coherent. To do this, we need to find a subgroup of $G$ which is finitely generated but not finitely presented. Stalling’s original paper tells us that we should look at which is the kernel of the homomorphism This subgroup is obviously finitely generated (since we defined it in terms of $3$ generators!) so we need to show that it isn’t finitely presented! The key insight will be that Every finitely presented group has finitely generated $H_2$. $\ulcorner$ Recall that the group homology $H_\bullet(G;M)$ is isomorphic to the “usual” homology2 of its Eilenberg-MacLane Space $K(G,1)$ with coefficients in the local system associated to the $G$-module $M$. Now if $G = \langle x_1, \ldots, x_n \mid R_1, \ldots, R_m \rangle$ is finitely presented, we can explicitly build a $K(G,1)$ as follows: First add a loop for every generator $x_i$. Then each relation $R_i$ is a word in the generators, thus is a loop in our space, and we glue in a disk with boundary given by $R_i$. Note that this makes the loop vanish in $\pi_1$ so that we’ve forced the fundamental group of this space to be $G$. Finally we inductively add in higher cells to kill the higher homotopy groups, since we want our $K(G,1)$ to be aspherical. Now we compute $H_2(G;\mathbb{Z}) = H_2(K(G,1); \mathbb{Z})$ using this description of the cell structure of $K(G,1)$. Since $G = \langle x_1, \ldots, x_n \mid R_1, \ldots, R_m \rangle$ was finitely presented, we see that there’s $n$ many $1$-cells and $m$-many $2$-cells in $K(G,1)$. So the group of $2$-cycles is a subgroup of $\mathbb{Z}^m$, the free abelian group on our (finite) set of $2$-cells, and is itself finitely generated. Quotienting out the boundaries gives $H_2$, so we win since the quotient of a finitely generated group is still finitely generated.3 $\lrcorner$ So, to show that our \(N = \langle a, c, bd \rangle \trianglelefteq F\{a,b\} \times F\{c,d\}\) isn’t finitely presented, we just have to show its $H_2$ isn’t finitely generated. We can simplify the discussion by computing $H_2(N; \mathbb{Q})$ instead, since fields make homological algebra much easier and the dimension of $H_2(N; \mathbb{Q})$ (as a $\mathbb{Q}$-vector space) is a lower bound on the number of generators for $H_2(N;\mathbb{Z})$ (do you see why?4). So with this in mind It suffices to show that $H_2(N; \mathbb{Q})$ is not finite dimensional as a $\mathbb{Q}$-vector space! Unless otherwise stated, all homology groups have coefficients in $\mathbb{Q}$ (with the trivial action) for the rest of this post. If we could find some nice description of the isomorphism type of $N$ then we could maybe compute its $H_2$ directly… But why spend the effort looking? We already have a short exact sequence and the homologies of $F_2 \times F_2$ and $\mathbb{Z}$ should be easier to compute by hand. Experience shows there should be some way to relate the homologies of $N$, $F_2 \times F_2$, and $\mathbb{Z}$, and indeed we’re saved by the Hochschild-Serre Spectral Sequence! This says that whenever we have a short exact sequence we get a spectral sequence relating the homology of $G$ to the homologies of $Q$ and $N$. See Ch. VII.6 in Brown’s classic textbook for more details. Concretely this means that we can compute the homology of $G$ in terms of “nested” homology groups: $Q$ acts on $N$ by conjugation, and this induces an action of $Q$ on $H_q(N)$ – thus it makes sense to look at the homology of $Q$ with coefficients in $H_q(N)$! The spectral sequence gives a close relationship between $H_n(G)$ and the collection of “nested” homologies $H_p(Q; H_q(N))$ with $p+q = n$. Precisely, the $E^2$-page of the spectral sequence is In our case, we know that $Q = \mathbb{Z}$ has particularly simple homology. Recall that a $\mathbb{Q}\mathbb{Z}$-module is just a $\mathbb{Q}$-vector space $V$ with a $\mathbb{Z}$ action. That is, it’s just a vector space $V$ with a choice of automorphism $\varphi \in GL(V)$. For any $\mathbb{Q}\mathbb{Z}$-module $(V, \varphi)$, we compute \(H_\bullet(\mathbb{Z}; V) = \begin{cases} V_\mathbb{Z} = V \big / (1-\varphi) V & \bullet = 0 \\ V^\mathbb{Z} = \text{Ker}(1-\varphi) & \bullet = 1 \\ 0 & \text{otherwise} \end{cases}\) $\ulcorner$ Writing $\mathbb{Q}[t^\pm]$ for $\mathbb{Q}\mathbb{Z}$, we build a free resolution of $\mathbb{Q}$ This tells us that $H_\bullet(\mathbb{Z}; V)$ is the homology of where $t$ acts by the automorphism $\varphi$, giving the claim. $\lrcorner$ In case $V$ is finite dimensional (as a $\mathbb{Q}$ vector space), then it’s easy to see that $\dim V^\mathbb{Z} = \dim \text{Ker} (1 - \varphi)$ and $\dim V_\mathbb{Z} = \dim \left ( V \big / \text{Im}(1 - \varphi) \right ) = \dim V - \dim \text{Im}(1 - \varphi)$ are equal, so that these two vector spaces are isomorphic. In case $V$ is infinite dimensional, though, this can fail! Let $V = \mathbb{Q}[t^\pm]$, of countable dimension, and let $\varphi$ be the (invertible) “multiply by $t$” operator. The fixed points $V^\mathbb{Z}$ of this operator are the laurent polynomials $p$ so that $p = t \cdot p$ (read: so that $(1-t) \cdot p = 0$), and the only option is $p=0$. The co-fixed points $V_\mathbb{Z}$ are given by $V \big / (1-t)$ which is isomorphic to $\mathbb{Q}$. So $V^\mathbb{Z}$ is $0$-dimensional and $V_\mathbb{Z}$ is $1$-dimensional. When $V$ is finite dimensional as a $\mathbb{Q}$-vector space we compute as vector spaces. So if these are not isomorphic, then $V$ must be infinite dimensional! This lets us start evaluating the terms of our spectral sequence: becomes Moreover, we know that our $H_0(N; \mathbb{Q}) = \mathbb{Q}$ and $H_1(N; \mathbb{Q}) = N_\text{ab} \otimes \mathbb{Q} = \mathbb{Q}^3$, since the abelianization of $N = \langle a, c, bd \rangle$ is isomorphic to $\mathbb{Z}^3$. While we’re here we can compute that the conjugation action of $Q = \mathbb{Z}$ on $N$ induces the trivial action on $H_0(N)$ and $H_1(N)$. Since $Q = \mathbb{Z}$ is generated by the image of $b$, the conjugation action on $N$ is literally conjugation by $b$. On the generators we compute $a \mapsto b^{-1} a b = (bd)^{-1} a (bd)$ $c \mapsto b^{-1} c b = c$ $bd \mapsto b^{-1} (bd) b = bd$ In $H_0(N) = \mathbb{Q}$ the group $N$ doesn’t even make an appearance, so the induced action is trivial. On $H_1(N) = N_\text{ab} \otimes \mathbb{Q}$ we need to see what the conjugation action induces on the abelianization, but that becomes trivial since in $N_\text{ab}$ we have $(bd)^{-1} a (bd) = a$. Since the $\mathbb{Z}$-action is trivial on $H_0(N) = \mathbb{Q}$ and $H_1(N) = \mathbb{Q}^3$ we learn that $H_0(N)_\mathbb{Z} = H_0(N)^\mathbb{Z} = \mathbb{Q}$ $H_1(N)_\mathbb{Z} = H_1(N)^\mathbb{Z} = \mathbb{Q}^3$ So the $E^2$ page of our spectral sequence further reduces to The differential on the $E^2$ page points “up two, left one”, so we see that every differential is $0$. In fact it’s easy to see that all futher differentials vanish so that this is actually the $E^\infty$ page of our spectral sequence! General theory tells us that $H_n(G) = \bigoplus_{p+q = n} E^\infty_{pq}$, so we can compute $H_n(F_2 \times F_2)$ by summing over the $n$th diagonal in the above table. Of course, $H_n(F_2 \times F_2)$ is easy enough to compute by hand using the Künneth formula and the fact that $K(F_2, 1) = S^1 \vee S^1$ is a bouquet with two petals5. So we can compute it in two ways (directly via Künneth and “indirectly” via the spectral sequence) and compare to see what it tells us about $H_\bullet(N)$! In particular, we learn (the left isomorphism comes from Künneth and the right isomorphism comes from the spectral sequence): From the $H_2(F_2 \times F_2)$ computation, we learn that \(H_2(N)_\mathbb{Z}\) must be $1$ dimensional. But from the $H_3(F_2 \times F_2)$ computation we learn that $H_2(N)^\mathbb{Z}$ must be $0$ dimensional! Since the invariants and coinvariants have different diemnsions, our earlier discussion shows that $H_2(N)$ must be infinite dimensional! This means $N$ cannot have been finitely presented, as desired ^_^. Let’s take a second to reflect on what just happened, since there were a decent number of moving parts. We wanted to show that \(N = \langle a, c, bd \rangle \leq F\{a,b\} \times F\{c,d\}\) is not finitely presented. First, we showed that every finitely presented group $G$ has finitely generated $H_2(G;\mathbb{Z})$ (using a concrete model of $K(G,1)$) so that it suffices to show $H_2(N; \mathbb{Z})$ is infinitely generated. Since the dimension of $H_2(N;\mathbb{Q})$ is a lower bound for the number of $\mathbb{Z}$-generators, we can work over a field and show that $H_2(N; \mathbb{Q})$ is infinite dimensional. Next, we showed that $H_2(N)$ comes with a natural $\mathbb{Z}$ action, and argued that $H_2(N)$ must be infinite dimensional if the invariants and coinvariants $H_2(N)^\mathbb{Z}$ and \(H_2(N)_\mathbb{Z}\) have different dimensions. Finally, using the Hochschild-Serre spectral sequence, we were able to compute that $H_2(N)_\mathbb{Z}$ is one dimensional while $H_2(N)^\mathbb{Z}$ is zero dimensional. This shows that $H_2(N)$ must be infinite dimensional, and we win! This is a clever trick, and a fairly subtle one! It’s something I’ll have to try to remember, since the obvious approach is to try and compute the (co)invariants explicitly, but I’m not even sure6 how to compute the $\mathbb{Z}$-action on $H_2(N)$! This lets you get your hands on the infinite-dimensionality indirectly, which feels very useful. Thanks for hanging out, everyone! It’s wild to think that just a short week ago I was in Bozeman, Montana meeting a bunch of cool people and giving a talk about my thesis. Then all in a row over labor day weekend I had two little dinner parties and went to the beach to swim with leopard sharks! It wasn’t very productive, but it was extremely good for the soul, haha. Now I have a few short days to try and get more done before I fly to Chicago for the Fall School on Quantizations and Lagrangians. Take care all, and stay safe. We’ll talk soon 💖 On the off chance the NSF still exists next year ↩ Depending on which book you read, this is either a definition or a theorem. See, for instance, the Introduction or Chapter II.4 in Brown’s book on Group Cohomology. ↩ In fact, there’s a whole hierarchy of finiteness conditions on a group $G$. We say that a group $G$ is “of type $F_n$” if its $K(G,1)$ has a finite $n$-skeleton. That is, if there’s only finitely many $0$-cells, finitely many $1$-cells, …, and finitely many $n$-cells. In the body we really showed that being finitely presented means being type $F_2$… Well, we showed half of this. Showing the converse (that $F_2$-groups are finitely presented) isn’t so hard either, and uses essentially the same idea. Note that being type $F_2$ implies that $H_2$ is finitely generated, since (as we said in the main body) then $H_2$ is a quotient of a subgroup of a finitely generated abelian group. But even if $G$ isn’t of type $F_2$, then $H_2$ might “accidentally” be finitely generated, if we have a big generating set but then quotient out by a similarly big set of boundaries. In fact, this really does happen! Bestvina and Brady constructed a group whose $H_2$ is finitely generated (indeed, whose $H_n$ is finitely generated for all $n$) yet which is not finitely presented! See Morse Theory and Finiteness Properties of Groups ↩ The universal coefficient theorem promises $H_2(N;\mathbb{Q}) = H_2(N;\mathbb{Z}) \otimes \mathbb{Q}$, which kills any torsion subgroups (so we don’t see those generators) but keeps the free abelian part. ↩ If this isn’t obvious, it’s a fantastic exercise in algebraic topology! Can you compute the homology groups $H_\bullet \Big ( (S^1 \vee S^1) \times (S^1 \vee S^1) ; \mathbb{Q} \Big )$? Again, you’ll want the Künneth formula to handle the product, and then you’ll want something like Mayer-Vietoris to handle the wedge sums. To relate this to group homology, note that $K(G \times H, 1) \cong K(G, 1) \times K(H, 1)$, so that the Künneth formula also applies to group homology! ↩ Though I gave up almost immediately, since I want this post finished so I can go back to writing more important things ↩

I’ve spent the last week at CT2025, which has just come to a close. It was great getting to see so many old friends and meet so many new ones, and every time I go to a CT I’m reminded of just how much category theory there is in the world, as well as just how much I enjoy all of it! Right before this I was in Antwerp for some Noncommutative Geometry, where I learned a ton and met even more new friends! Then next week I go to Bonn for my third conference in a row. I’m trying to stay energetic, and thankfully I have a few days off between CT and QTMART to help me rest up! I want to write up a lot of things I learned over the last month, since I have a lot of new thoughts on noncommutative geometry, mirror symmetry, and deformation theory, all coming from just my time in Antwerp! I’ve also learned a lot at CT and talked to a lot of interesting people about interesting things, and I’m sure I’ll have even more to say after my time in Bonn. I think organizing all of those thoughts are going to take a while, though (if I end up writing them down at all), but today I have a quick observation inspired by a few lovely conversations I had with Clark Barwick at CT. One of the many questions I asked him was if there’s a conceptual reason the Sphere Spectrum (read: the homotopy groups of spheres) is so darn complicated. He gave me an answer that’s obvious in hindsight, but which totally rearranged the way I think about things: I think I internalized a while ago that “free” constructions are fairly concrete. After all, you look at the syntax of whatever object you’re interested in, quotient out by the relations you want to be true and you’re done! Plus, mapping out of a free thing is as simple as possible, since it’s a left adjoint! All you have to do is find a (usually simpler) map from your generating set to a structure of interest and let the magic of category theory build your (usually more complicated) map for you… Of course, this view is heavily influenced by the kind of free structures I have experience with, and the kinds of questions I was asking about them. I was thinking about free groups and monoids, which you can study with word combinatorics, free (dg-)algebras on (graded) vector spaces, which look like polynomials, free categories on graphs, free $k$-linear or dg-categories on categories, and free cauchy completions of these, all of which come from just looking at paths, linear combinations, concentrating things in degree $0$, or working with twisted closures to add shifts and cones and whatnot1. I was thinking about relatively free constructions like the universal enveloping algebra, with its PBW-basis, or the right angled artin group attached to a (reflexive, simple) graph… All of these constructions feel like friends to me, in part because I know how to compute with them. Why would the sphere spectrum – the free spectrum on a single point – be so different? The point is that I’ve internalized these constructions as being tractable because I’m usually mapping out of them, in the direction the category theory encourages. I’m also usually relying on serious “normal form” theorems that make computing with these things tractable, or I’m doing fairly simple combinatorics with my generating set before arguing that these extend in some obvious way to things defined on the whole free object. All of these constructions become much less friendly when you start mapping into them, or asking more difficult questions about their internal structure. In hindsight, I’ve even personally struggled with tons of questions about free structures in my research! Free groups are extremely interesting from basically any perspective, with deep questions about their first order theory (Tarski’s Problem), the combinatorics relating their generating sets (The Andrews-Curtis Conjecture), or the coarse geometry of their outer automorphisms. Free cauchy completions are obviously complicated when you want to understand them on their own terms! If you take an algebra $A$ and view it as a one-object dg-category, then its cauchy completion1 is its whole category of perfect complexes! An enormous chunk of representation theory is, in that lens, dedicated to nothing more than the study of a certain, complicated, free construction! I’ve personally given up2 on a problem about relatively free constructions in right angled artin groups! These interpolate between free and free-abelian groups, and geometric group theory is teeming with interesting open problems about raags. For instance, can you understand, at the level of the underlying graphs, when one raag will embed into another? I thought about this off and on for a year before I started working seriously with my current advisor, and I made almost no progress at all. I also spent some time working with the adjunctions It’s interesting to try and construct these explicitly, and to understand the essential images of the left adjoints. This amounts to understanding which essentially algebraic theories are actually algebraic, and which algebraic theories are actually props. One of the big difficulties here is that we have a relatively free construction which adds relations rather than just operations. Adding new operations tends to be a fairly mild thing to do – consider the free algebra on an abelian group, which sends $A$ to its tensor algebra $\bigoplus_n A^{\otimes n}$ where it’s easy to recover the $A$ you started with. If instead we want to add new relations or axioms, for instance by freely sending a group $G$ to its abelianization $G \big / [G,G]$, then we lose lots of information in this construction. After some conversations with John Baez and Todd Trimble I came quite close to characterizing the image of the left adjoint between finite product categories and symmetric monoidal categories by factoring it into a “lossy” construction adding new axioms forcing the monoidal unit to be terminal and a much simpler construction which freely adds new operations corresponding to the product projections. I’ve had to put that project on hold while I focus on my thesis work, but I really want to come back and finish it soon. Of course as soon as you’re interested in logic, you have to accept that free things are complicated! The freest version of any theory lives in its classifying category, where truth and provability coincide. Then proving anything at all about the free model gives immediate understanding about all other models of that theory! This is already true for groups, whose classifying finite-product category is just the category of finitely generated free groups and homomorphisms. We don’t usually think about it because the kinds of statements you prove in equational logic aren’t very deep. But if you look instead at the classifying topos for groups and ask geometric questions suddenly you’re able to do a lot more, and the game becomes much harder! Perhaps this is clearest in the semantics of programming languages, where the free model (often called the “term model” in this context3) is the programming language, and checking whether two terms are equal in this free model literally amounts to evaluating two programs and seeing if their respective values agree. Because of this, many important structural results about a programming language (such as canonicity) can be proven by building another model whose semantics you understand, and then producing a section of the unique map from the free model. Also coming from logic are various lattices, whose free models can be quite intricate. Famously the free modular lattice on $3$ generators has $28$ elements, while the free modular lattice on $4$ generators is infinite! Indeed, this lattice has an undecidable word problem4, so that no program can tell whether two descriptions of its elements are the same or not! Heyting algebras are extremely important for semantics of intuitionistic logic, yet already the free heyting algebra on one generator is infinite, and the free heyting algebra on two generators is famously complicated. The study of free heyting algebras is still ongoing and seems quite difficult (at least as an outsider). See, for instance, Almeida’s recent preprint Colimits and Free Constructions of Heyting Algebras through Esakia Duality. With all this in mind, it shouldn’t be surprising at all that the sphere spectrum is so complicated! It’s the free spectrum on a point, and as such the only “relations” it will have are those that hold in all spectra! But of course spectra should obviously be complicated – They control all possible (co)homology theories for all spaces! So in this sense one should expect the internal structure of the sphere spectrum to be quite complicated, since any simplification would persist to something true of all cohomology theories. Of course, it’s easy to be complicated without being interesting, and I still think it’s a bit of a miracle that the sphere spectrum should have all this intricate structure inside it. As I understand it, much of Chromatic Homotopy Theory came from trying to explain patterns in the homotopy groups of spheres, and this subject is now as famously intimidating to outsiders5 as it is famously fascinating once you put in the work to become an insider6. Thanks for reading all! This really was a quick one for once, since I already had a lot of these examples floating around in my head. I really had all of the tools to realize that free things are obviously complicated in general, especially their “internal structure” that doesn’t ride the coattails of the universal property, but for some reason I just didn’t put it together until my conversation with Clark. It’s always dangerous to say what I’m thinking about writing about, but I at least have one more short post planned from my time in Antwerp, and maybe a longer one too if I have the energy. I’m doing a ton of writing right now, since I have two half-finished papers that I want to submit by the end of the summer. I think I’ll be able to get it done, but between writing these and going to conferences it’s been a tiring month. It’s tiring in a fun way, though, and I really feel like I’ve been productive in a way that I haven’t felt in a little while. I’m excited to start crossing a lot of these projects off my long-term-todo-list, especially since I already have three more projects I want to start! Regardless, I hope you’re having a more restful summer than I am! Stay safe, all, and we’ll talk soon ^_^. Emily Roff, taken at the top of the main tower in Brno) Actually it’s not completely obvious to me that the cauchy completion of a dg-category should be its idempotent triangulated closure… The cauchy completion will certainly be idempotent complete and triangulated, but in the well-named paper Cauchy Completeness for DG-Categories, Nicolić, Street, and Tendas show that to be cauchy complete you also need to be closed under “cokernels of protosplit chain maps”… I think this is some kind of split idempotent condition? But I haven’t read the paper closely enough to know for sure. If you want to be guaranteed to be correct, instead of “cauchy completion” you can say “the idempotent closure of the twisted closure”. That’s still a free construction and it still gives you the derived category in the special case your dg-category is a ring. ↩ ↩2 At least for now ↩ Pun intended ↩ Which is made more interesting by the fact that if you look at the class of lattices coming from lattices of subgroups of abelian groups, the corresponding free lattice on $4$ generators does have solvable word problem! This is remarkable since (as I understand it) modular lattices are called that because they look like lattices of submodules (in particular, sublattices of a lattice of subgroups of an abelian group). This is apparently proven in Herrmann’s On the Equational Theory of Submodule Lattices, and I learned all this from Ralph Freese the comments of this n-Category Cafe post. ↩ Such as myself. ↩ Which it seems like I might start doing soon, for a project I’m not ready to talk about yet. ↩

Earlier this week my friend Shane and I took a day and just did a bunch of computations. In the morning we did some differential geometry, where he told me some things about what he’s doing with symplectic lie algebroids. We went to get lunch, and then in the afternoon we did some computations in derived algebraic geometry. I already wrote a blog post on the differential geometry, and now I want to write one on the derived stuff too! I’m faaaaar from an expert in this stuff, and I’m sure there’s lots of connections I could make to other subjects, or interesting general theorem statements which have these computations as special cases… Unfortunately, I don’t know enough to do that, so I’ll have to come back some day and write more blog posts once I know more! I’ve been interested in derived geometry for a long time now, and I’ve been sloooowly chipping away at the prerequisites – $\infty$-categories and model categories, especially via dg-things, “classical” algebraic geometry (via schemes), and of course commutative and homological algebra. I’m lucky that a lot of these topics have also been useful in my thesis work on fukaya categories and TQFTs, which has made the time spent on them easy to justify! I’ve just started reading a book which I hope will bring all these ideas together – Towards the Mathematics of Quantum Field Theory by Frédéric Paugam. It seems intense, but at least on paper I know a lot of the things he’s planning to talk about, and I’m hoping it makes good on its promise to apply its techniques to “numerous examples”. If it does, I’m sure it’ll help me understand things better so I can share them here ^_^. In this post we’ll do two simple computations. In both cases we have a family of curves where something weird happens at a point, and in the “clasical” case this weirdness manifests as a discontinuity in some invariant. But by working with a derived version of the invariant we’ll see that at most points the classical story and the derived story agree, but at the weird point the derived story contains ~bonus information~ that renders the invariant continuous after all! Ok, let’s actually see this in action! First let’s look at what happens when we intersect two lines through the origin. This is the example given in Paugam’s book that made me start thinking about this stuff again. Let’s intersect the $x$-axis (the line $y=0$) with the line $y=mx$ as we vary $m$. This amounts to looking at the schemes $\text{Spec}(k[x,y] \big / y)$ and $\text{Spec}(k[x,y] \big / y-mx)$. Their intersection is the pullback and so since everything in sight is affine, we can compute this pullback in $\text{Aff} = \text{CRing}^\text{op}$ as a pushout in $\text{CRing}$: Pushouts in $\text{CRing}$ are given by (relative) tensor products, and so we compute which is $k$ when $m \neq 0$ and is $k[x]$ when $m=0$, so we’ve learned that: When $m \neq 0$ the intersection of \(\{y=0\}\) and \(\{y=mx\}\) is $\text{Spec}(k)$ – a single point1. When $m = 0$ the intersection is $\text{Spec}(k[x])$ – the whole $x$-axis. This is, of course, not surprising at all! We didn’t really need any commutative algebra for this, since we can just look at it! The fact that the dimension of the intersection jumps suddenly is related to the lack of flatness in the family of intersections $k[x,y,m] \big /(y, y-mx) \to k[m]$. Indeed, this doesn’t look like a very flat family! We can also see it isn’t flat algebraically since tensoring with $k[x,y,m] \big / (y, y-mx)$ doesn’t preserve the exact sequence2 In the derived world, though, things are better. It’s my impression that here flatness is a condition guaranteeing the “naive” underived computation agrees with the “correct” derived computation. That is, flat modules $M$ are those for which $M \otimes^\mathbb{L} -$ and $M \otimes -$ agree for all modules $X$! I think that one of the benefits of the derived world is that we can pretend like “all families are flat”. I would love if someone who knows more about this could chime in, though, since I’m not confident enough to really stand by that. In our particular example, though, this is definitely true! To see this we need to compute the derived tensor product of $k[x,y] \big / (y)$ and $k[x,y] \big / (y-mx)$ as $k[x,y]$-algebras. To do this we need to know the right notion of “projective resolution” (it’s probably better to say cofibrant replacement), and we can build these from (retracts of) semifree commutative dg algebras in much the same way we build projective resolutions from free things! Here “semifree” means that our algebra is a free commutative graded algebra if we forget about the differential. Of course, “commutative” here is in the graded sense that $xy = (-1)^{\text{deg}(x) \text{deg}(y)}yx$. For example, if we work over the base field $k$, then the free commutative graded algebra on $x_0$ (by which I mean an element $x$ living in degree $0$) is just the polynomial algebra $k[x]$ all concentrated in degree $0$. Formally, we have elements $1, \ x_0, \ x_0 \otimes x_0, \ x_0 \otimes x_0 \otimes x_0, \ldots$, and the degree of a tensor is the sum of the degrees of the things we’re tensoring, so that for $x_0$ the whole algebra ends up concentrated in degree $0$. If we look at the free graded $k$-algebra on $x_1$, we again get an algebra generated by $x_1, \ x_1 \otimes x_1, \ x_1 \otimes x_1 \otimes x_1, \ldots$ except that now we have the anticommutativity relation $x_1 \otimes x_1 = (-1)^{1 \cdot 1} x_1 \otimes x_1$ so that $x_1 \otimes x_1 = 0$. This means the free graded $k$-algebra on $x_1$ is just the algebra with $k$ in degree $0$, the vector space generated by $x$ in degree $1$, and the stipulation that $x^2 = 0$. In general, elements in even degrees contribute symmetric algebras and elements in odd degrees contribute exterior algebras to the cga we’re freely generating. What does this mean for our example? We want to compute the derived tensor product of $k[x,y] \big / y$ and $k[x,y] \big / y-mx$. As is typical in homological algebra, all we need to do is “resolve” one of our algebras and then take the usual tensor product of chain complexes. Here a resolution means we want a semifree cdga which is quasi-equivalent to the algebra we started with, and it’s easy to find one! Consider the cdga $k[x,y,e]$ where $x,y$ live in degree $0$ and $e$ lives in degree $1$. The differential sends $de = y$, and must send everything else to $0$ by degree considerations (there’s nothing in degree $-1$). This cdga is semifree as a $k[x,y]$-algebra, since if you forget the differential it’s just the free graded $k[x,y]$ algebra on a degree 1 generator $e$! So this corresponds to the chain complex where $de = y$ is $k[x,y]$ linear so that more generally $d(pe) = p(de) = py$ for any polynomial $p \in k[x,y]$. If we tensor this (over $k[x,y]$) with $k[x,y] \big / y-mx$ (concentrated in degree $0$) we get a new complex where the interesting differential sends $pe \mapsto ey$ for any polynomial $p \in k[x,y] \big / y-mx$. Some simplification gives the complex whose homology is particularly easy to compute! $H_0 = k[x] \big / mx$ $H_1 = \text{Ker}(mx)$ We note that $H_0$ recovers our previous computation, where when $m \neq 0$ we have $H_0 = k$ is the coordinate ring of the origin3 and when $m=0$ we have $H_0 = k[x]$ is the coordinate ring of the $x$-axis. However, now there’s more information stored in $H_1$! In the generic case where $m \neq 0$, the differential $mx$ is injective so that $H_1$ vanishes, and our old “classical” computation saw everything there is to see. It’s not until we get to the singular case where $m=0$ that we see $H_1 = \text{Ker}(mx)$ becomes the kernel of the $0$-map, which is all of $k[x]$! The version of “dimension” for chain complexes which is invariant under quasi-isomorphism is the euler characteristic, and we see that now the euler characteristic is constantly $0$ for the whole family! Next let’s look at some kind of “hidden smoothness” by examining the singular curve $y^2 =x^3$. Just for fun, let’s look at another family of (affine) curves $y^2 = x^3 + tx$, which are smooth whenever $t \neq 0$. We’ll again show that in the derived world the singular fibre looks more like the smooth fibres. Smoothness away from the $t=0$ fibre is an easy computation, since we compute the jacobian of the defining equation $y^2 - x^3 - tx$ to be $\langle -3x^2 - t, 2y \rangle$, and for $t \neq 0$ this is never $\langle 0, 0 \rangle$ for any point on our curve4 (We’ll work in characteristic 0 for safety). Of course, when $t=0$ $\langle -3x^2, 2y \rangle$ vanishes at origin, so that it has a singular point there. To see the singularity, let’s compute the tangent space at $(0,0)$ for every curve in this family. We’ll do that by computing the space of maps from the “walking tangent vector” $\text{Spec}(k[\epsilon] \big / \epsilon^2)$ to our curve which deform the map from $\text{Spec}(k)$ to our curve representing our point of interest $(0,0)$. Since everything is affine, we turn the arrows around and see we want to compute the space of algebra homs so that the composition with the map $k[\epsilon] \big / \epsilon^2 \to k$ sending $\epsilon \mapsto 0$ becomes the map $k[x,y] \big / (y^2 - x^3 - tx) \to k$ sending $x$ and $y$ to $0$. Since $k[x,y] \big / (y^2 - x^3 - tx)$ is a quotient of a free algebra, this is easy to do! We just consult the universal property, and we find a hom $k[x,y] \big / (y^2 - x^3 - tx) \to k[\epsilon] \big / \epsilon^2$ is just a choice of image $a+b\epsilon$ for $x$ and $c+d\epsilon$ for $y$, so that the equation $y^2 - x^3 - tx$ is “preserved” in the sense that $(c+d\epsilon)^2 - (a+b\epsilon)^3 - t(a+b\epsilon)$ is $0$ in $k[\epsilon] \big / \epsilon^2$. Then the “deforming the origin” condition says that moreover when we set $\epsilon = 0$ our composite has to send $x$ and $y$ to $0$. Concretely that means we must choose $a=c=0$ in the above expression, so that finally: The tangent space at the origin of $k[x,y] \big / (y^2 - x^3 - tx)$ is the space of pairs $(b,d)$ so that $(d \epsilon)^2 - (b \epsilon)^3 - t(b \epsilon) = 0$ in $k[\epsilon] \big / \epsilon^2$. Of course, this condition holds if and only if $tb=0$, so that: When $t \neq 0$ the tangent space is the space of pairs $(b,d)$ with $b=0$, which is one dimensional. When $t = 0$ the tangent space is the space of pairs $(b,d)$ with no further conditions, which is two dimensional! Since we’re looking at curve, we expect the tangent space to be $1$-dimensional, and this is why we say there’s a singularity at the origin for the curve $y^2 = x^3$….. But what happens in the derived world? Now we want to compute the derived homspace. As before, a cofibrant replacement of our algebra is easy to find, it’s just $k[x,y,e]$ where $x$ and $y$ have degree $0$, $e$ has degree $1$ and and $de = y^2 - x^3 - tx$. Note that in our last example we were looking at quasifree $k[x,y]$-algebras, but now we just want $k$-algebras! So now this is the free graded $k$-algebra on 3 generators $x,y,e$, and our chain complex is: We want to compute the derived $\text{Hom}^\bullet(-,-)$ from this algebra to $k[\epsilon] \big / \epsilon^2$, concentrated in degree $0$. The degree $0$ maps are given by diagrams that don’t need to commute5! Of course, such maps are given by pairs $(a + b \epsilon, c + d \epsilon)$, which are the images of $x$ and $y$. As before, since we want the tangent space at $(0,0)$ we need to restrict to those pairs with $a=c=0$ so that $\text{Hom}^0(k[x,y] \big / y^2 - x^3 - tx, \ k[\epsilon] \big / \epsilon^2) = k^2$, generated by the pairs $(b,d)$. Next we look at degree $-1$ maps, which are given by diagrams which are given by a pair $r + s\epsilon$, the image of $e$. Again, these need to restrict to the $0$ map when we set $\epsilon=0$, so that $r=0$ and we compute $\text{Hom}^{-1}(k[x,y] \big / y^2 - x^3 - tx, \ k[\epsilon] \big / \epsilon^2) = k$, generated by $s$. So our hom complex is where the interesting differential sends degree $0$ to degree $-1$ and is given by $df = d_{k[\epsilon] \big / \epsilon^2} \circ f - f \circ d_{k[x,y] \big / y^2-x^3-tx}$. So if $f$ is the function sending $x \mapsto b \epsilon$ and $y \mapsto d \epsilon$ then we compute So phrased purely in terms of vector spaces we see our hom complex is (living in degrees $0$ and $-1$): So we compute $H^0 = \text{Ker} ((-t \ 0))$ $H^{-1} = \langle s \rangle \big / \text{Im}((-t \ 0))$ When $t \neq 0$, our map is full rank so that $H^0$ are the pairs $(b,d)$ with $b=0$ – agreeing with the classical computation. Then $H^{-1} = 0$, so again we learn nothing new in the smooth fibres. When $t=0$, however, our map is the $0$ map so that $H^0$ is the space of all pairs $(b,d)$ is two dimensional – again, agreeing with the classical computation! But now we see the $H^{-1}$ term, which is $1$ dimensional, spanned by $s$. Again, in the derived world, we see the euler characteristic is constantly $1$ along the whole family! There’s something a bit confusing here, since there seem to be two definitions of “homotopical smoothness”… On the one hand, in the noncommutative geometry literature, we say that a dga $A$ is “smooth” if it’s perfect as a bimodule over itself. On the other hand, though, I’ve also heard another notion of “homotopically smooth” where we say the cotangent complex is perfect. I guess it’s possible (likely?) that these should be closely related by some kind of HKR Theorem, but I don’t know the details. Anyways, I’m confused because we just computed that the curve $y^2 = x^3$ has a perfect tangent complex, which naively would make me think its cotangent complex is also perfect. But this shouldn’t be the case, since I also remember reading that a classical scheme is smooth in the sense of noncommutative geometry if and only if it’s smooth classically, which $y^2 = x^3$ obviously isn’t! Now that I’ve written these paragarphs and thought harder about things, I think I was too quick to move between perfectness of the tangent complex and perfectness of the cotangent complex, but I should probably compute the cotangent complex and the bimodule resolution to be sure… Unfortunately, that will have to wait for another day! I’ve spent probably too many hours over the last few days writing this and my other posts on lie algebroids. I have some kind of annoying hall algebra computations that are calling my name, and I have an idea about a new family of model categories which might be of interest… But checking that something is a model category is usually hard, so I’ve been dragging my feet a little bit. Plus, I need to start packing soon! I’m going to europe for a bunch of conferences in a row! First a noncommutative geometry summer school hosted by the institute formerly known as MSRI, then CT of course, and lastly a cute representation theory conference in Bonn. I’m sure I’ll learn a bunch of stuff I’ll want to talk about, so we’ll chat soon ^_^. Take care, all! In fact we know more! This $k$ is really $k[x,y] \big / (x=0,y=0)$, so we know the intersection point is $(0,0)$. ↩ Indeed after tensoring we get since here $k \cong k[m] \big / (m)$. But then we can simplify these to and indeed the leftmost map (multiplication by $m$) is not injective! The kernel is generated by $x$. ↩ Again, if you’re more careful with where this ring comes from, rather than just its isomorphism class, it’s $k[x,y] \big / (x,y)$, the quotient by the maximal ideal $(x,y)$ which represents the origin. ↩ The only place it could possibly be $\langle 0, 0 \rangle$ is when $y=0$, but the points on our curve with this property satisfy $x^3-tx=y^2=0$ so that when $t \neq 0$ the solutions are $(x,y) = (0,0), \ (\sqrt{t}, 0), \ (-\sqrt{t}, 0)$. But at all three of these points $\langle -3x^2 - t, 2y \rangle \neq \langle 0, 0 \rangle$. ↩ This is a misconception that I used to have, and which basically everyone I’ve talked to had at one point. Remember that dg-maps are all graded maps! Not just those which commute with the differential! The key point is that the differential on $\text{Hom}^\bullet(A,B)$ sends a degree $n$ map $f$ to so that $df = 0$ if and only if $d_B f = (-1)^n f d_A$ if and only if $f$ commutes with the differential (in the appropriate graded sense). This means that, for instance, $H^0$ of the hom complex recovers from all graded maps exactly $\text{Ker}(d) \big / \text{Im}(d)$, which are the maps commuting with $d$ modulo chain homotopy! ↩

While doing a computation with my friend Shane the other day, we realized we needed to explicitly compute a local chart near the identity of $SL_2(\mathbb{R})$. It took us longer than I’d like to admit to figure out how to do this (especially since it’s so geometrically obvious in hindsight), and so I want to write down the process for future grad students looking to just do a computation! If you want to see what Shane and I were actually interested in, you can check out the main post here. Ok, let’s hop right in! Say you have a $2$-manifold in $\mathbb{R}^3$ to start1: If we think of the purple manifold $M$ as being an open disk, representing a small neighborhood of some possibly larger 2-manifold, then we can see the projection onto the $xy$-plane is a diffeomorphism onto its image (an open disk in $\mathbb{R}^2$) while the projections onto the $yz$ and $xz$ planes are not open in $\mathbb{R}^2$! This is because the normal to $M$ is parallel to the $z$ axis inside $M$, (indeed, at the “top of the hill”) so the tangent plane at that point degenerates and projects to a line whenever we project onto a coordinate plane containing the $z$-direction. For a more computational example, let’s try the hyperboloid $x^2 + y^2 - z^2 = 1$, and let’s see what happens near a few points. The tangent plane at a point is controlled by the jacobian of the defining equation, which for us is $\langle 2x, 2y, -2z \rangle$. This gives us three (disconnected) charts: \(\{2x \neq 0\}\), \(\{2y \neq 0\}\), and \(\{-2z \neq 0\}\), which we can see visually here (and we also drop the unnecessary scalars): These turn into 6 honest-to-goodness charts where we turn the disconnected condition \(\{x \neq 0\}\) into the pair of connected conditions \(\{x \gt 0\}\) and \(\{x \lt 0\}\). Indeed it’s easy to see that the 6 connected components in the above pictures are all diffeomorphic to an open subset of $\mathbb{R}^2$, and we can see this algebraically by projecting onto the plane avoiding the nonzero coordinate. On \(\{x \gt 0 \}\), for example, we have an open set of the $yz$-plane, shown here in orange: Algebraically, we compute this chart by noting on \(\{x \gt 0 \}\), we can solve for $x$ and (using the positive square root, since $x \gt 0$) write our surface locally as which is diffeomorphic in the obvious way to its projection onto the $yz$-plane so this is one of our charts! Similarly, we can look at \(\{z \gt 0\}\), solve for $z$ and locally write our surface as which is diffeomorphic to \(\{ (x,y) \mid x^2 + y^2 \gt 1 \}\) – another chart. On the intersection of these charts, \(\{x, z \gt 0 \}\), it’s now easy to write down our transition maps (if one is so inclined): Here our charts are the diffeomorphisms so it’s easy to compose them to see our transition maps between these charts are As a (fun?) exercise, compute the \(\{y \gt 0 \}\) chart, and the other two transition maps. For another example, let’s take a look at $SL_2(\mathbb{R})$, which is defined to be \(\left \{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mid ad-bc = 1 \right \} \subseteq \mathbb{R}^4\). Then the jacobian of our defining map is $\langle d, -c, -b, a \rangle$, and we get charts corresponding to \(\{d \neq 0 \}\), \(\{-c \neq 0 \}\), \(\{ -b \neq 0 \}\), and \(\{ a \neq 0 \}\). In the \(\{d \neq 0\}\) chart, for instance, our defining equation looks like $a = \frac{1+bc}{d}$, so that $SL_2(\mathbb{R})$ looks locally like In the main post you can see how my friend Shane and I used this to compute the anchor map for a certain lie algebroid. Again, it makes a nice exercise to explicitly write out the various charts and transition maps What about a codimension 2 example? Let’s go back to our happy little hyperboloid, and intersect it with the surface $xyz = 1$. That is, we want to consider the manifold This is the levelset of the map $\mathbb{R}^3 \to \mathbb{R}^2$ sending $(x,y,z) \mapsto (x^2 + y^2 - z^2, \ xyz)$ taking value $(1,1)$. So we compute the jacobian and our charts are all the ways this matrix can have full rank. These conditions are: $2x \neq 0$ and $xz \neq 0$ $2x \neq 0$ and $xy \neq 0$ $2y \neq 0$ and $yz \neq 0$ $2y \neq 0$ and $xy \neq 0$ $-2z \neq 0$ and $yz \neq 0$ $-2z \neq 0$ and $xz \neq 0$ If we look at the \(\{2x \neq 0, \ xz \neq 0\}\) chart, we can ask sage to solve for $x$ and $z$ as functions of $y$: So as in the previous hyperboloid example, we need to break this into four charts, depending on whether $x$ and $z$ are positive or negative. Following the sage computation, in the \(\{x \gt 0, z \gt 0\}\) chart, we can write our curve as which, by projecting out the $y$ coordinate, is diffeomorphic to the open subset of $\mathbb{R}$ where all these square roots are defined. Ok, thanks for reading, all! This was extremely instructive for me, and hopefully it’s helpful to some of you as well! Sometimes it’s nice to just do some computations. Talk soon! That’s right. I finally bought an ipad. This has already dramatically improved my paper-reading experience (I love you Zotero) and my remote teaching experience, and I’m so glad that the example-drawing experience is shaping up to be everything I wanted it to be as well! ↩