# Artwork for a T-shirt

OK, we are still planning to have the Stacks project workshop this summer. I am slowly trying to work on having a new t-shirt or sweat shirt. I asked my youngest and middle son to make some artwork. Here is what they came up with for the front and back of the shirt.

I’ve ordered 3 shirts with these pictures on it for me and my sons; I will post a picture here when we get them. But a question I have is what text (if any) to put with these pictures. What I came up with was

• Front: Climbing mathematics…
• Back: You can keep going

This doesn’t really work very well. Instead what one of my sons came up with was

• On the stack
• No looking back

and I think it’s better. What do you think? Do you have a suggestion?

# Determinants of perfect complexes

There is an interesting question on mathoverflow by Damian Rössler.

His question reminds me of an approach to constructing determinants of perfect complexes that I’ve been meaning to think about. The goal is to do a minimum of computation. For the impatient: the outcome of the discussion below is that you can probably construct determinants in this manner, but that it is of very limited use — in particular it wouldn’t answer Damian’s question and related questions.

Aside: Currently, the Stacks project discusses determinants of perfect complexes only when they have tor amplitude in [-1, 0], see here and here.

Let S be a scheme. Denote Pic(S) the category whose objects are invertible O_S-modules and whose morphisms are isomorphisms of invertible modules. An object E of the derived category D(O_S) is called perfect if it is locally on S quasi-isomorphic to a bounded complex of finite free O_S-modules. Denote Perf-Iso(S) the category whose objects are perfect objects of D(O_S) and whose morphisms are isomorphisms E → E’ of perfect objects of D(O_S). If s ∈ S is a point we denote E ⊗ κ(s) the derived pullback of E to s viewed as the spectrum of its residue field. We often think of E ⊗ κ(s) as an object of D(κ(s)), i.e., as a complex of κ(s) vector spaces.

Let E be an object of D(O_S). We will say E is mighty fine if (a) E is perfect, (b) the cohomology sheaves of E are finite locally free, and (c) for any point s ∈ S the restriction E ⊗ κ(s) of E to s has no two consecutive nonzero cohomology spaces. In other words, if H^n(E ⊗ κ(s)) is nonzero, then both H^{n – 1}(E ⊗ κ(s)) and H^{n + 1}(E ⊗ κ(s)) are zero. Conditions (a) and (c) actually imply condition (b) as follows from the following lemma.

Lemma: Let E be a perfect object of D(O_S). Let s ∈ S be a point such that E ⊗ κ(s) of E to s has no two consecutive nonzero cohomology spaces. Then there is an open neighbourhood U of s in S such that E|_U is mighty fine.

To prove the lemma you can use Lemma 068U to split the complex into a direct sum of vector bundles in a neighbourhood of s in S. (The lemma applies because a perfect complex is pseudo-coherent, see Lemma 0658.)

If E is mighty fine, then we define det(E) as the alternating tensor product of the determinants of the cohomology sheaves of E. This makes sense because these sheaves are finite locally free. The construction is clearly a functor on the category of mighty fine complexes and isomorphisms in D(O_S) between these. Also, it is compatible with pullbacks.

Having made these definitions we impose the following requirements on our determinant: A determinant is a rule that to every scheme S assigns a functor det : Perf-Iso(S) → Pic(S) satisfying the following requirements

1. det is compatible with pullbacks,
2. if E is mighty fine, then det(E) is as above,
3. if a : E → E’ is an isomorphism of mighty fine objects, then det(a) is as above.

We claim that this already almost completely pins down the determinant. To explain this we need another definition.

Let E be an object of D(O_S). We will say E is workable if there exists a quasi-compact open immersion j : U → S which is dense and schematically dense such that E|_U is mighty fine. Observe that by the first axiom of determinants above there always is a canonical map

det(E) ⟶ j_*det(E|_U)

and by our assumptions on j we find that this map is injective! Since we have already constructed the right hand side, we conclude, for example, that if a : E → E’ is an isomorphism in D(O_S), then there is absolutely no choice for the map det(a) : det(E) → det(E’). It must be the unique map det(a) : det(E) → det(E’) compatible with the aready defined map det(a|_U) provided it exists. Moreover, still in the workable case, we’re going to define det(E) as a submodule of j_*det(E|_U) constructed affine locally (see below).

Thus it is useful to have many workable perfect complexes.

Lemma: Let E be a perfect object of D(O_S). For every s ∈ S there exists an open neighbourhood U of s in S and a morphism f : U → V and a workable perfect object E’ in D(O_V) and an isomorphism a : E|_U → f^*E’.

Proof. Locally on S the object E is given by a bounded complex of finite free O_S-modules. Thus locally on S the object E is the pullback of the universal complex on the variety of complexes V = V(r). Here r = (r_i) is a sequence of nonnegative integers almost all zero and V(r) is the scheme parametrizing sequences of size r_{i + 1} × r_i matrices M_i with M_i M_{i – 1} = 0 for all i. The variety of complexes V is a finite type scheme over Z which is reduced by Theorem 10.2 in a paper by Musili and Sheshadri of 1983. The irreducible components of the variety of complexes V are known to be Cohen-Macaulay and normal (although this doesn’t seem to be particularly helpful for us) and moreover these components are distinguished by the ranks of the maps in the complex at their generic points. Finally, it is easy to see that at each generic point one obtains a mighty fine complex — because it is easy to show that over a field any bounded complex of finite dimensional vector spaces which has two consecutive nonzero cohomology spaces can be deformed to a complex with different (smaller) betti numbers for its cohomology. EndProof

Lemma. Let U be a scheme. Consider morphisms f_i : U → V_i, i = 1, 2 where V_i is a variety of complexes with universal complex E_i and an isomorphism a : f_1^*E_1 → f_2^*E_2 in D(O_S). Then, Zariski locally on U, there exists a morphism of schemes g : U → W and smooth morphisms g_i : W → V_i, i = 1, 2 and a map of complexes b : g_1^*E_1 → g_2^*E_2 which is a quasi-isomorphism such that f_i = g_i o g and a = g^*b. (In particular the complexes g_i^*E_i are workable as the pullback of a workable complex by a flat morphism of schemes.)

Proof sketch. Say V_1 = V(r1) and V_2 = V(r2) for some rank vectors r1, r2. Zariski locally on U the isomorphism a is given by a map of complexes between the pullbacks of the universal complexes. Thus we can let W be the scheme of finite type over Z parametrizing maps of complexes with rank vectors r1 and r2 which are moreover quasi-isomorphisms. Then deformation theory of complexes shows that both forgetful maps W → V_1 and W → V_2 are smooth (this uses that we are versally deforming E_1 on V_1 and E_2 on V_2). EndProof

These lemmas show that it suffices to produce a sufficiently canonical determinant for the pullback of the universal complexes over the varieties of complexes by smooth morphisms. I think this reduces us to the algebra problem formulated below. The main point I am trying to make with this post is that now we’re looking for the existence of certain lattices in modules and we are not checking any diagrams commute!

Algebra problem: Given a ring A, a nonzerodivisor f in A, a bounded complex of finite free A-modules M such that M_f is mighty fine, construct a rank 1 locally free A-submodule det(M) ⊂ det(M_f) where det(M_f) is the alternating tensor product of the determinants of the cohomology modules of M_f over A_f which is compatible with pullbacks and with quasi-isomorphisms between such complexes.

For example, if A is a dvr and f is a power of the uniformizer, then we have the lattices H^i(M)/torsion ⊂ H^i(M_f) and we can take the corresponding lattice in the determinant. A similar construction works if A is a valuation ring. If the varieties of complexes are semi-normal (I don’t know if this is true, do you?), then we’d only have to look at Noetherian semi-normal rings A and our choice for dvrs would be enough.

Drawback of this method: One of the properties you want is that given an object (E, F) of the filtered derived category of S such that E and gr E are perfect, then we have a canonical isomorphism det(E) = ⊗ det(gr_i E). However, I do not think it is true that a “versally deformed” pair (E, F) has E and gr E workable. (I think I have a counter example.) Thus the method fails to produce the desired isomorphism in a sort of straightforward manner. Similarly, this method also wouldn’t answer Damian Rössler question… Sigh!

# Purity (part 2)

Let f : X → Y be a dominant, finite type morphism of integral Noetherian schemes. We assume X is normal and Y regular. Let Sing(f) be the closed set of points of X where f isn’t smooth.

Question: Is codim Sing(f) ≤ 1 + dimension of generic fibre of f?

The discussion in the previous post shows that the answer is yes when f is a morphism between smooth varieties over a field all of whose fibres have the same dimension and that the bound given is best possible.

Dolgachev proved the answer to the question is yes in case X is a local complete intersection over Y (which happens for example if both X and Y are smooth over a common base scheme). There is a paper of Rolf k\”allstr\”om which has this result and much more.

As we’ve seen in the previous post, the answer to the question is “yes” when f is generically finite, i.e., when the dimension of the generic fibre is 0.

In the rest of this post we discuss the case where the generic fibre has dimension 1.

Let f : X → Y be as above with generic fibre of dimension 1. Let x be a generic point of an irreducible component of Sing(f) with image y in Y. We want to show that dim O_{X, x} is at most 2. To get a contradiction, assume this is not the case.

1. dim O_{X, x} = dim O_{Y, y} + 1 – trdeg κ(x)/κ(y) by the dimension formula.
2. If dim O_{X, x} > 3, then dim O_{Y, y} > 2 and we can find a regular divisor Y_0 in Y passing through y = f(x) and thereby reduce the dimension (lots of details missing, but I think this probably can be made to work).
3. Assume dim O_{X, x} = 3 so dim O_{Y, y} = 2 + t where t is the transcendence degree of κ(x) over κ(y).

At the moment I have nothing intelligent to say in the case t > 0; suggestions are welcome. Assume t = 0.

In this case x is a closed point of the fibre. In particular, we see that, after shrinking X we may assume x is the only singular point of the fibre X_y. An argument similar to the one in miracle flatness shows that O_{Y, y} → O_{X, x} must be flat (here we use X is assumed to be normal).

Thus we see that we have to show something like this: given the formal germ C of an isolated reduced curve singularity over an algebraically closed field k there cannot be a flat deformation D/A of C over a 2-dimensional complete regular local ring A with residue field k which smooths out the singularity in all directions (no singular fibres except for the central one).

By an argument, which I think is due to Deligne, such a deformation can be globalized. In other words, given D/A we can find a proper flat family of geometrically connected curves X over Spec(A) and a closed point x of the closed fibre of X such that the completion of X at x is D and such that X is smooth over Spec(A) everywhere except at x. Essentially the way I think about this is that you first “attach” smooth projective curves to the germ C over k and then you use that deformations of curves always are smooth over the deformations of their singularities. In this construction we may and do assume the genera of the irreducible components of the closed fibre are > 1 and a fortiori that the genus of the generic fibre is > 1.

Thus we reduce to showing: there cannot be a flat proper family X of geometrically connected curves over a 2 dimensional complete local regular Noetherian ring A which is smooth except at finitely many points of the special fibre such that all irreducible components of the special fibre have genus > 1.

By a result of Moret-Bailly, given such an X there exists a different family X’ over A which agrees with X over the punctured spectrum and whose special fibre is proper smooth over k. A simple argument (using the genera of components being > 1) shows that X and X’ are isomorphic as schemes over A as desired.

Edit 4-28-2020 The answer is no, see Purity (Part 3)

# Purity (part 1)

Let f : X → Y be a dominant, finite type morphism of integral Noetherian schemes. We assume X is normal and Y regular. In this and the next blog post we define Sing(f) to be the closed set of points of X where f isn’t smooth and we define Disc(f) to be the image of Sing(f) in Y.

In this setting a purity result is one that gives an upper bound on the codimension of Sing(f) in X or a lower bound on the codimension of Disc(f) in Y.

When the relative dimension of f is 0, then the Stacks project has Zariski-Nagata purity of branch locus. If f is generically finite, then the Stacks project has purity of ramification locus and over fields we even have that the complement of Sing(f) is affine in X (Tag 0ECA).

But what if the relative dimension is > 0?

For the rest of this blog post, let’s say f has relative dimension m ≥ 0 in the sense that every irreducible component of every fibre has dimension m.

The simplest case is when Y and X are smooth over a field k with Y of dimension n and X of dimension n + m. In fact, let’s assume Y and X are both affine spaces over k. Picking coordinates x_1, …, x_{n + m} on X and coordinates y_1, …, y_n on Y and say f = (f_1, …, f_n) is given in coordinates by y_i = f_i(x_1, …, x_{n + m}) for some polynomials f_i. Then of course the locus Sing(f) is the set of points of X where the n × (n + m) matrix of partial derivatives (d f_i / d x_j) does not have maximal rank. We conclude immediately that codim Sing(f) ≤ m + 1. Taking the image in Y we find codim Disc(f) ≤ m + 1.

In fact, this is all you can say! Here are some examples to show this is true. (We assume the characteristic of k is not 2.)

Relative dimension m = 0.

(A) For n ≥ 1 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_n^2). Then Sing(f) = V(x_n) and Disc(f) = V(y_n).

Relative dimension m = 1.

(A) For n ≥ 1 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_nx_{n + 1}). Then Sing(f) = V(x_n, x_{n + 1}) and Disc(f) = V(y_n). Here Disc(f) is a divisor in Y and the map Sing(f) → Disc(f) is an isomorphism. This is the usual example of a family of nodal curves.

(B) For n ≥ 2 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_1x_n + x_{n + 1}^2). Then Sing(f) = V(x_1, x_{n + 1}) and Disc(f) = V(y_1, y_n). Here Disc(f) has codimension 2 in Y, the fibres over points of Disc(f) are nonreduced, and the map Sing(f) → Disc(f) has fibres of dimension 1. Thanks to Will Sawin for finding this example.

Relative dimension m = 2.

(A) For n ≥ 1 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_n^2 + x_{n + 1}^2 + x_{n + 2}^2). Here Sing(f) = V(x_n, x_{n + 1}, x_{n + 2}) and Disc(f) = V(y_n). Here Disc(f) is a divisor in Y and the map Sing(f) → Disc(f) is an isomorphism. This is the usual example of a family of nodal surfaces.

(B) For n ≥ 2 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_1 x_n + x_{n + 1}^2 + x_{n + 2}^2). Here Sing(f) = V(x_1, x_{n + 1}, x_{n + 2}) and Disc(f) = V(y_1, y_n). Here Disc(f) has codimension 2 in Y, the fibres of f over points of Disc(f) are planes meeting in lines, and the nap Sing(f) → Disc(f) has fibres of dimension 1.

(C) For n ≥ 3 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_1 x_n + x_2 x_{n + 1} + x_{n + 2}^2). Here Sing(f) = V(x_1, x_2, x_{n + 2}) and Disc(f) = V(y_1, y_2, y_n). Here Disc(f) has codimension 3 in Y, the fibres of f over points of Disc(f) are nonreduced, and the map Sing(f) → Disc(f) has fibres of dimension 2.

It seems pretty clear that we can keep going in this manner for arbitrary m. Enjoy!

Edit 3/8/2020 Will Sawin gives this general example: the map from A^{2n+m} with coordinates (x_1,…,x_n, y_1,…,y_n,z_1,….z_m) to A^{n+1} given by (x_1,…,x_n , x_1y_1+ …. + x_ny_n + z_1^2 + …. + z_m^2) is a map with (n+m-1)-dimensional fibers which has a codimension n+1 locus on the base where the codimension of the singularities of the fiber is exactly m-1 (the locus where the x_i and z_i all vanish), showing that all such pairs appear sharply.

# Applications to Stacks project workshop

The deadline for applications to the Stacks project workshop is January 31st. Don’t miss out. You can sign up by filling out a form on the website. Email me if you have any questions about how it works. You can also read more about what happened during the previous workshop here.

You can sign up by filling out a form on the website. Please read about the previous workshop to see if attending a workshop like this is the right thing for you. Feel free to contact me if you have questions, suggestions, or comments.

# Update

OK, I’ve worked through your comments on the Stacks project once more. This is good because my semester starts here tomorrow and I won’t have nearly as much time to do these things during the semester. However, I’ll try to be a bit more attentive to your comments as they come by, because I did find some to be quite a bit more interesting than I initially thought they would be. Also, maybe we can have an ongoing discussion this way too if you like. Anyway, please keep them coming!

# Cohomology and motives

This summer I am trying to write a little bit about Weil cohomology theories for the Stacks project. My motivation is that I want to add the theorem that over a field of characteristic zero de Rham cohomology is one. And in turn, I became motivated to do this, because of the discussions I had with our graduate students in my course Ask me anything. Because after all: what is a “good cohomology theory” for algebraic varieties?

While working on this, I found that I had never properly understood the reason for all the conditions imposed by Kleiman in his paper. I still haven’t understood, I think. In this post I will discuss a basic question I had as I worked through this material and the answers I came up with so far.

Let k be an algebraically closed field; this will be the base field for our smooth projective varieties X, Y, Z. Let F be a field of characteristic 0; this will be the coefficient field for our cohomologies. Feel free to assume F is algebraically closed too. A Weil cohomology theory H^* assigns a graded commutative F-algebra H^*(X) to X in a contravariant manner. It comes with additional structure, namely trace maps and cycle classes, and it has to satisfy a bunch of axioms:

(A) Poincare duality: if dim(X) = d the trace map H^{2d}(X)(d) —> F is an isomorphism, H^i nonzero only for 0 ≤ i ≤ 2d, H^i is finite dimensional and the pairing H^i(X) x H^{2d – i}(X)(d) —> F defined using cup product and the trace map is nondegenerate,

(B) Kunneth formula: H^*(X x Y) = ⨁ H^i(X) ⊗ H^j(Y) and this is compatible with trace maps.

(C) Cycle classes are compatible with pullbacks, pushforwards, intersection products, and trace of the cycle class of a point on Spec(k) is 1.

Remark 1: I’ve written the axioms slightly differently from Kleiman in his paper “Algebraic cycles and the Weil conjecture”. In particular I am keeping track of Tate twists and I am adding the axiom that trace is compatible with Kunneth (which you can deduce from the other axioms — but will become important later).

Remark 2: In some references compatibility of cycle classes and pushforwards is omitted. But without this compatibility I immediately get stuck, for example I don’t know how to compute the class of the diagonal without this axiom. If you know a proof of this compatibility from the other axioms, please drop me an email.

Let M_k be the symmetric monoidal category of Chow motives over k (using rational equivalence). Let h(X) be the motive of X and recall that h(-) is contravariant. Let us write 1(1) for the Tate motive, i.e., the inverse of the invertible object h^2(P^1) of M_k.

A key consequence of the axioms (A, (B), (C) is that H^*(X) = G(h(X)) for some symmetric monoidal functor G from M_k to the symmetric monoidal category of graded F-vector spaces such that G(1(1)) is nonzero only in degree -2.

In fact, let’s start with a G as above and consider the contravariant functor H^*(X) = G(h(X)) from smooth projective varieties to graded F-vector spaces. What can we say about H^*? It turns out that you get cup products, trace maps, cycles classes and all the axioms (A), (B), (C) as formulated above except for possibly “the trace map … isomorphism” and “H^i nonzero only for 0 ≤ i ≤ 2d”.

Question: Can we make an example of a G where the corresponding H^*(X) does not have the “correct” betti numbers?

For example, we can ask whether there could be a G such that H^*(X) is nonzero in negative degrees for some X? Or we could ask whether we can find a G such that H^0(X) has dimension > 1. I think the answer is “yes” when k is the complex numbers and conjecturally “no” when the ground field k is the algebraic closure of a finite field.

Remark 3: When k is the algebraic closure of a finite field, a paper of Katz and Messing proves the betti numbers are the same for any Weil cohomology theory satisfying more axioms (namely some Lefschetz type thing). As far as I can tell, this result doesn’t apply to answer my question above, since I’m even weakening the assumptions of a Weil cohomology theory.

Construction of a weird G when both k and F are the field of complex numbers. Namely, let (H’)^* be the functor which sends X to (H’)^*(X) = ⨁ H^{p, q}(X) viewed as a bigraded vector space over F. This is a Weil cohomology theory: by Hodge theory it is just the same as sending X to its usual cohomology with complex coefficients. By the discussion above we get G’ such that for every motive M we have a bigrading G'(M) = ⨁ (G’)^{p, q}(M). (One could say that the target category is the category of graded complex Hodge structures, suitably defined as a symmetric monoidal category.) Of course as a graded F-vector space we have that the degree of the summand (G’)^{p, q}(X) is p + q, in other words, we have

(G’)n(M) = ⨁ n = p + q (G’)p, q(M)

Now I am going to define a weird G by setting G(M) equal to the graded F-vector space with degree n part equal to

Gn(M) = ⨁ n = 3p – q (G’)p, q(M)

Observe that G(1(1)) is still in degree -2! (A technical point is that because I chose 3 and -1 odd the functor G is still compatible with the commutativity constraints which you have to check in order to see that G is a symmetric monoidal functor.) Now, if E is an elliptic curve then H^*(E) = G(h(E)) has betti numbers dim H^{-1}(E) = 1, dim H^0(E) = 1, dim H^1(E) = 0, dim H^2(E) = 1, dim H^3(E) = 1 and everything else zero. Cool!

On the other hand, let p be a prime number and suppose that k is the algebraic closure of the field with p elements. Then I think conjecturally all objects in M_k ⊗ F are semi-simple and all simple objects are “1 dimensional”. Moreover, I think that those simple guys are classified by Weil p-numbers α in F up to multiplication by roots of unity (these α are algebraic numbers, but aren’t necessarily algebraic integers, and have a weight which is an integer which can be negative). Say M(α) is the motive corresponding to α. Now the point is that given an α of weight zero, its nth root for any n > 1 is also a Weil p-number of weight zero. Thus the 1-dimensional F-vector space G(M(α)) had better be in degree 0. And this then fixes the degrees for all motives because the motive corresponding to α = p^{-1} is 1(1) which we are requiring to be in degree -2.

Enjoy!

# Workshop announcement

Next summer (2020) sometime in late July or early August Wei Ho, Pieter Belmans, and I are going to have another Stacks project workshop. If you google around you’ll find the website for the previous iteration of the workshop; this one will be structured similarly. But we’ve only just started working on the organization and everything is still in flux. Feel free to email me with suggestions for things we could do, or mentors we could invite, etc. Thanks!

This semester (Spring 2019) I taught our first year, second semester algebraic geometry course for graduate students as a series of ask me anything sessions. This blog post is a report on how it worked and whether I would advise you to do the same when you teach your course.

Some background info: The enrollment was 10 students (including 4 undergraduates) but the course was attended by a few additional graduate students. Here is a link to the course webpage.

Why did I do this? First of all, I have taught algebraic geometry for graduate students many, many times and I have never taught it the same way twice. When I teach these courses, I am trying to “level up” our graduate students: I am trying to show that there is a well defined language of algebraic geometry with very sound foundations, that you can enjoy building more complicated things on top of this, that you can read about fantastic new methods/results in papers and books, and that you can do all of these things yourself actually. Maybe I can put it simply this way: I want to get them to the point where it would be fun for me (and hopefully others) to have them as graduate students. So what is more fun than a student who comes up with good questions? My second motivation was a more personal one, namely I wanted to have a bit more direct contact than usual with our first year graduate students.

How did it work? Well, I simply went to the lecture room and asked if there were any questions. After an awkward silence, usually lasting not more than one or two minutes, somebody would ask a question. Most of the time I would answer. Then another question, and so on. Often the answers were longish because of the backtracking necessary to give sufficient background in order for the answer to make sense.

Initially, the students were reading along and asked questions which allowed me to more or less develop basics of scheme theory. I tried to stimulate questions from people who preferred to think about only classical varieties, as I think this is a perfectly sound way to do algebraic geometry. Roughly in the middle of the course, in one of the lectures I limited myself to 3 minutes per answer. This lecture I think worked rather well, because I ended up getting more different questions from more different people in the room. Later I introduced another ingredient, namely, 20 questions for schemes! The students surprisingly quickly guessed I was thinking about the degree 5 Fermat surface over the field with 7 elements. I repeated this the next time, where the student, again surprisingly quickly, found that I was thinking of the spectrum of the dual numbers over the complex numbers.

An ingredient of the course which I quite enjoyed was that I was able to talk about the de Rham complex and its cohomology, both in characteristic p and in characteristic 0. Namely, the natural question arose to what extend we can find the “usual cohomology” if you are an algebraic geometer who isn’t willing to use inequalities between real numbers (such as myself). I was then also able to tie this in with coherent duality and the relationship between dualizing sheaves and sheaves of differentials.

For each of the lectures, I wrote, from memory, an account of the mathematics discussed during each meeting. You can find these on the webpage of the course linked above. As you can see, we covered an acceptable range of topics during the lectures. Also, I gave problem sets based on the topics discussed previously. (Since you do not know what the students are going to ask, you cannot give problems foreshadowing the upcoming material.) Finally, I decided to give an oral exam to the graduate students and assigned final papers to the undergraduates.

What worked well?

1. The students were amazing: they asked good questions whose answers they really desired to know.
2. You don’t need to prepare the lectures!
4. You can cover more diverse topics.
5. Different answers can be at different levels (some of the students may already know more theory and you can occasionally discuss material which is more advanced).
6. You can react to the interests of the students in real time.
7. This structure gives you an opportunity to skip worn out paths and reorder the material drastically.
8. Students will help you prove things as you are working through them on the board.
9. I found that I was doing more examples this way. For example, we discussed the blowing up of the cone over a conic along a line through the vertex to show that blowups mayn’t be what you think they are from nice pictures in books.
10. You can skip over annoying verification without the students noticing. You may think this is a drawback, but more and more I think this is a useful and necessary evil of teaching algebraic geometry.

What didn’t work so well?

1. You don’t always remember the slickest proof of every result you talk about.
2. The order of the arguments isn’t always the right one.
3. Students may never be sure that they have completely grokked all of the material up to a certain point. The material is not presented in a linear fashion. They’ll have to do a lot of work themselves.

The last point is perhaps the most serious drawback. However, IMHO it is impossible to teach algebraic geometry and truly cover all the details needed to get a sound theory. It is necessary for those who intend to work in the field or intend to use AG in a serious manner to sit down and work/read through a good deal of the material by themselves.

Would I recommend teaching a graduate course in Algebraic Geometry this way? Actually, no. First of all, if you are teaching algebraic geometry for the first time, I recommend choosing a good text to work with and sticking fairly close to it; or choose a topic, for example linear algebraic groups and representations, choose a good book for it, and aim the development of your algebraic geometry towards the topic (maybe the book already discusses some algebraic geometry in the first chapter). On the other hand, if you have taught algebraic geometry already multiple times, then you know why it is a difficult thing to do. You probably have already found a best possible method of teaching the course for you, and there is nothing I can say to change your mind!

Also, personally, the next time I will yet again use a completely different method!