European mathematics in 1927

Here’s a map of the (major) mathematical centers in Europe (in 1927), made for the Rockefeller Foundation.

Support by the Rockefeller foundation was important for European Mathematics between the two world wars. They supported the erection of the Mathematical Institute in Goettingen between 1926-1929 and creation of the Institut Henri Poincare in Paris at about the same time.

Careers of people such as Stefan Banach, Bartel van der Waerden and Andre Weil benefitted hugely from becoming fellows of the Rockefeller-funded International Educational Board in the 20ties.

The map itself shows that there were three major centers at the time: Goettingen, Paris and Rome (followed by Berlin and Oxford, at a distance).

Also the distribution by topics (the pie-charts per university) is interesting: predominantly Analysis (red) with a fair share of Geometry (yellow), Number Theory (green) and Applied Mathematics (blue). Philosophy (black) was even more important than Algebra (orange) which existed only in Goettingen (Noether, van der Waerden) and Berlin.

I’d love to see a similar map for 2014…

A larger version of the map can be found here.

There’s a corresponding map for the USA here.


the birthday of Grothendieck topologies

This is the story of the day the notion of ‘neighbourhood’ changed forever (at least in the geometric sense).

For ages a neighbourhood of a point was understood to be an open set of the topology containing that point. But on that day, it was demonstrated that the topology of choice of algebraic geometry, the Zariski topology, needed a drastic upgrade.

This ultimately led to the totally new notion of Grothendieck topologies, which aren’t topological spaces at all.

Formally, the definition of Grothendieck topologies was cooked up in the fall of 1961 when Grothendieck visited Zariski, Mike Artin and David Mumford in Harvard.

The following spring, Mike Artin ran a seminar resulting in his lecture notes on, yes, Grothendieck topologies.

But, paradigm shifts like this need a spark, ‘une bougie d’allumage’, and that moment of insight happened quite a few years earlier.

It was a sunny spring monday afternoon at the Ecole Normal Superieure. Jean-Pierre Serre was giving the first lecture in the 1958 Seminaire Claude Chevalley which that year had Chow rings as its topic.

That day, april 21st 1958, Serre was lecturing on algebraic fibre bundles:

He had run into a problem.

If a Lie group $G$ acts freely on a manifold $M$, then the set of $G$-orbits $M/G$ is again a manifold and the quotient map $\pi : M \rightarrow M/G$ is a principal $G$-fibre bundle meaning that for sufficiently small open sets $U$ of $M/G$ we have diffeomorphisms

$\pi^{-1}(U) \simeq U \times G$

that is, locally (but not globally) $M$ is just a product manifold of $G$ with another manifold and the $G$-orbits are all of the form $\{ u \} \times G$.

The corresponding situation in algebraic geometry would be this: a nice, say reductive, algebraic group $G$ acting freely on a nice, say smooth, algebraic variety $X$. In this case one can form again an orbit space $X/G$ which is again a (smooth) algebraic variety but the natural quotient map $\pi : X \rightarrow X/G$ rarely has this local product property…

The reason being that the Zariski topology on $X/G$ is way too coarse, it doesn’t have enough open sets to enforce this local product property.

(For algebraists: let $A$ be an Azumaya algebra of rank $n^2$ over $\mathbb{C}[X]$, then the representation variety $\mathbf{rep}_n(A)$ is a principal $\mathbf{PGL}_n$-bundle over $X$ but is only local trivial in the Zariski topology when $A$ is a trivial Azumaya algebra, that is, $End_{\mathbb{C}[X]}(P)$ for a rank $n$ projective module $P$ over $\mathbb{C}[X]$.)

But, Serre had come up with a solution.

He was going to study fibre bundles which were locally ‘isotrivial’, meaning that they had the required local product property but only after extending them over an unamified cover $Y \rightarrow X$ (what we now call, an etale cover) and he was able to clasify such fibre bundles by a laborious way (which we now call the first etale cohomology group).

The story goes that Grothendieck, sitting in the public, immediately saw that these etale extensions were the correct generalization of the usual (Zariski) localizations and that he could develop a cohomology theory out of them in all dimensions.

According to Colin McLarty Serre was ‘absolutely unconvinced’, since he felt he had ‘brutally forced’ the bundles to yield the $H^1$’s.

We will never known what Serre actually wrote on the blackboard on april 21st 1958.

The above scanned image tells it is an expanded version of the original talk, written up several months later after the ICM-talk by Grothendieck in Edinburgh.

By that time, Grothendieck had shown Serre that his method indeed gives cohomology in all dimensions,and convinced him that this etale cohomology was likely to be the “true cohomology needed to prove the Weil conjectures”.

What we want from a geometry over ${\mathbb{F}_1}$

From now on the main object (the irony of this terminology will become clear) of study is ${\mathbb{F}_1}$. I will first elaborate on what Kapranov-Smirnov calls the “folklore imagery”, trying to understand the motivation behind a statement like

A vector space of ${\mathbb{F}_1}$ is just a set.

Afterwards I will draw some conclusions from these observations, motivating some of the approaches outlined in Mapping ${\mathbb{F}_1}$-land.

Demystifying the ${\mathbb{F}_1}$-lore

In Projective geometry over ${\mathbb{F}_1}$ and the Gaussian binomial coefficient a nice build-up is given. I will summarise and give my own viewpoint.

We already know that ${\mathbb{F}_1}$ doesn’t exist as a field (because we need ${0\neq 1}$ by the axioms). But what if we loosen up the axioms and say the trivial ring should be taken as ${\mathbb{F}_1}$? In that case modules over this ring are in relationship with vector spaces over ${\mathbb{F}_1}$. But there are only trivial modules over the trivial ring, hence no nontrivial vector spaces. This approach obviously doesn’t work. A conclusion I like to make after vastly generalising this approach:

Setting ${\mathbb{F}_1}$ to be something doesn’t work.

The idea of looking at vector spaces looks promising though: given a field it is the most obvious structure built upon it, so we want to make sense of it over ${\mathbb{F}_1}$. We know that the cardinality of an ${n}$-dimensional vector space ${V}$ over a finite field ${\mathbb{F}_q}$ is ${q^n}$. Applying this to ${\mathbb{F}_1}$ we get a single point, for all ${n}$. So far so good, but we also know that a basis for ${V}$ consists of ${n}$ elements. The problem with this approach is that it’s too direct: we cannot construct objects over ${\mathbb{F}_1}$, we need to get there by an analogy that avoids contradictions like this. Applying this for instance to ${\mathbb{F}_1[t]}$ we see that this doesn’t yield any satisfying definition either. The conclusion after another round of generalisation:

Look at induced objects, not constructions.

The same applies to noncommutative geometry by the way. But let’s focus on ${\mathbb{F}_1}$ for the moment.

So we need a simple object over ${\mathbb{F}_1}$ where we can avoid an explicit construction, getting facts about that object only by analogy without running into contradictions. Let’s try this for ${\mathbb{P}^n/\mathbb{F}_1}$. The construction of this over an actual field ${k}$ consists of constructing ${\mathbb{A}^{n+1}/k}$ (an ${n+1}$-dimensional vector space over ${k}$) and setting ${\mathbb{P}^n/k}$ to be the set of lines through the origin.

If we take ${k=\mathbb{F}_q}$ the number of points in ${\mathbb{A}^{n+1}/\mathbb{F}_q}$ is ${q^{n+1}}$. The number of lines through the origin is ${(q^{n+1}-1)/(q-1)}$: just take any point in ${\mathbb{A}^{n+1}/\mathbb{F}_q\setminus\left\{ 0 \right\}}$, this defines a line through the origin, but there are ${q}$ points (including the origin) on this line, therefore we divide by ${q-1}$. If we write down the polynomial function counting the number of points in ${\mathbb{P}^n/\mathbb{F}_q}$ (i.e., ${q^n+q^{n-1}+\ldots+1}$) and evaluate in ${q=1}$ we see something that leads to a nontrivial object! To make this analogy really sound you have to write down what you actually want from a finite projective space and axiomatise it. But let’s rejoice for now, and conclude

The projective space ${\mathbb{P}^n/\mathbb{F}_1}$ contains ${n+1}$ points.

What we have actually done is changing the construction of ${\mathbb{P}^n/\mathbb{F}_1}$ from an algebraic-geometric viewpoint to a combinatorial-geometric viewpoint, something that is quintessential in finite geometry (and I guess I’m the angs+’er with the most background in this kind of stuff :)). If people are interested in a write-up about this, I’d be happy to provide one but I suspect my fellow seminarians are not that into finite geometry and combinatorics. The only important observation we need to make is that a line in ${\mathbb{P}^n/\mathbb{F}_1}$ contains exactly two points in this sense, as a line in ${\mathbb{P}^n/\mathbb{F}_q}$ contains ${q+1}$ points by the axioms of a combinatorial-geometric projective space.

Now taking ${\mathbb{F}_1}$-vector spaces to be sets actually makes sense: ${\mathbb{A}^{n+1}/\mathbb{F}_1}$ is an ${(n+1)}$-set, taking any point as the distinguished base point (or origin) and considering “lines” through the origin, or more appropriately ${2}$-subsets, of which we have exactly ${n}$ as we started with ${n+1}$ elements and fixed an origin. But this (inherently geometric) idea of “fixing an origin” has its downsides in what follows, where we will adjoin an origin which is more in the sense of algebra as adding a zero vector (which doesn’t exist over ${\mathbb{F}_1}$) doesn’t change the dimension or cardinality of a base for a vector space.


Now we can switch our attention to Kapranov’s and Smirnov’s unfinished paper. If Wittgenstein were an algebraic geometer interested in ${\mathbb{F}_1}$-geometry I guess he would have written a Tractatus Absoluto-Geometricus, which could have looked (in a very crude sense) like this

  1. 1 Geometry over ${\mathbb{F}_1}$ can only be understood through induced objects.
  2. 2 Vector spaces over ${\mathbb{F}_1}$ are plain sets.
    1. 2.1 Dimension equals cardinality.
    2. 2.2 ${\mathrm{GL}_n(\mathbb{F}_1)=\mathrm{S}_n}$.
    3. 2.3 ${\mathrm{SL}_n(\mathbb{F}_1)=\mathrm{A}_n}$.
    4. 2.4 ${\det\colon\mathrm{GL}_n(\mathbb{F}_1)\rightarrow\mathbb{F}_1^\times}$ is the sign homomorphism.
    5. 2.5 The Grassmannian ${\mathrm{Gr}(k,n)(\mathbb{F}_1)}$ is the set of ${k}$-subsets.
    6. 2.6 There is no harm in formally adjoining a zero vector in a ${\mathbb{F}_1}$-vector space turning it into a pointed set, just be careful with interpretations.
  3. 3 The polynomial ring ${\mathbb{F}_1[t]}$ can only be understood through its automorphisms.
    1. 3.1 Polynomial automorphisms are a generalisation of field automorphisms by evaluation at “zero”.
    2. 3.2 We have ${\mathrm{GL}_n(\mathbb{F}_1[t])\rightarrow\mathrm{GL}_n(\mathbb{F}_1)}$.
    3. 3.3 ${\mathrm{GL}_n(\mathbb{F}_1[t])=\mathrm{B}_n}$ the braid group on ${n}$ strings, by analogy of the canonical ${\mathrm{B}_n\rightarrow\mathrm{S}_n}$.
    4. 3.4 For more information I refer you to the grandmaster himself and his blog post ${\mathbb{F}_1}$ and braid groups.
  4. 4 Finite fields have finite extensions and algebraic closures and so does ${\mathbb{F}_1}$.
    1. 4.1 ${\mathbb{F}_{1^n}}$ as a vector space over ${\mathbb{F}_1}$ is the (pointed) set ${\mu_n}$ consisting of the ${n}$-th roots of unity and an adjoined zero (which will serve as the point of the pointed set).
    2. 4.2 By choosing a primitive root in ${\mu_n}$ we get a non-canonical isomorphism ${\mathrm{C}_n\cong\mu_n}$.
    3. 4.3 ${\mathbb{A}^1/\mathbb{F}_1}$ as a scheme is ${\mathrm{Spec}\,\mathbb{F}_1[t]}$.
    4. 4.4 ${\mathrm{Spec}\,\mathbb{F}_1[t]}$ describes the algebraic closure of ${\mathbb{F}_1}$.
    5. 4.5 ${\overline{\mathbb{F}_1}}$ therefore corresponds to ${\mathrm{\mu}_\infty\cup\left\{ 0 \right\}}$.
  5. 5 We can generalise linear algebra to finite extensions.
    1. 5.1 The action of ${\mathbb{F}_{1^n}}$ on ${V}$ is the action of ${\mu_n}$ on ${V\setminus\left\{ 0 \right\}}$.
    2. 5.2 ${\mathbb{F}_{1^n}}$-vector spaces are nothing but ${\mu_n}$-sets.
    3. 5.2 A ${d}$-dimensional ${\mathbb{F}_1}$-space contains ${dn}$ points.
    4. 5.4 The ${\mu_n}$-action is strictly multiplicative, there is no additive structure.
    5. 5.5 The lack of additive structure agrees with the notion of vector spaces as (pointed) sets.
    6. 5.6 A ${\mathbb{F}_{1^n}}$-basis is a set containing a representative of every orbit under the ${\mu_n}$-action.
  6. 6 We can interpret other finite objects over ${\mathbb{F}_1}$.
    1. 6.1 ${\mathbb{F}_q}$ is a ${\mathbb{F}_{1^n}}$-vector space if ${q\equiv 1\bmod n}$ because ${(\mathbb{F}_q^\times,\cdot)\cong(\mathrm{C}_{nd},\cdot)}$ for ${d=(q-1)/n}$, and therefore a ${\mathbb{F}_{1^n}}$-algebra.
    2. 6.2 This vector space structure induces the (necessarily unique) ${\mathbb{F}_{1^n}}$-vector space structure on ${\mathbb{F}_q^e}$ of dimension ${ed}$.
    3. 6.3 All development of techniques should coincide with this observation.
    4. 6.4 For a construction of exact sequences over ${\mathbb{F}_1}$ I refer you to Absolute linear algebra.

For the real Wittgenstein aficionado, my apologies for not hitting the right tone and ideas. This post mostly served as a way for me to get all my ${\mathbb{F}_1}$-folklore straight, so that I can explain to a complete outsider why stuff in the ${\mathbb{F}_1}$ is taken as what it is. If you feel any gaps present, please tell me so.

If you followed me this far you should be familiar enough with the ideas and twists of mind necessary to understanding some parts of Kapranov-Smirnov, or you have made up your mind and will take all ${\mathbb{F}_1}$-stuff to be dadaist nonsense. But before tackling Kapranov-Smirnov I suggest you read the series of posts developing the same theme as this one did but in greater depth:

  1. The ${\mathbb{F}_1}$ folklore
  2. Absolute linear algebra
  3. ${\mathbb{F}_1}$ and braid groups

For the next (and last) post of this series, the main idea will be Stuff over ${\mathbb{F}_1}$ contains multiplicative structure, not additive structure.

No bollocks, just rings

Today (one of) the goals of this blog will be neglected, and we’ll focus solely on its URL. No $\mathbb{F}_{1}$ or Riemann hypothesis, just some old-fashioned ring theory.

Misplaced Poetics

I’d like write up one of (Jacobson-) Herstein’s commutativity theorems in noncommutative ring theory. It’s a beautiful testament to the power of subdividing rings into nontrivial classes, and is filled to the brim with small, simple, but extremely elegant ideas. Part of the theorem’s appeal stems from the fact that once confronted with its statement, you’re bound to ask: who cares?! It seems so useless, and that somehow makes it all the more fun. Of course, there was a good reason for proving the theorem, but I’ll just leave you to ponder some applications. I’m in a “l’art pour l’art”-mood, hoping Poe doesn’t turn in his grave.


Without further ado, here’s the theorem I’ve been going on about:

Theorem. A ring $R$ is commutative iff for any $a,b \in R$, there exists an integer $n(a,b)>1$ such that $(ab-ba)^{n(a,b)}=ab-ba$.

Of course, your first idea, like everyone else’s, is that this can be proven by (possible) extreme amounts of formula bashing. I encourage you to try this, as I have, and get very tired (and a tad bit frustrated) after wasting the better part of a Friday morning (and wacking half a tree). If you do manage to find a proof by pure calculation, let me know! The way we’ll tackle the theorem is by a reduction argument along the following lines:

  • If $R$ is a division ring: see the next paragraph
  • If $R$ is a left primitive ring: reduce to the previous case
  • If $R$ is a semiprimitive ring: reduce to the previous case
  • If $R$ is a (general) ring: reduce to the previous case

Just lovely, isn’t it.

Thé lemma

Crucial for proving the first part of the theorem is the following lemma, which will be applied in the next paragraph.

Lemma. Let $D$ be a division ring with nonzero characteristic. If $a$ is a non-central, periodic element in $D^{*}$, then there exists an additive commutator $b$ in $D^{*}$ such that $bab^{-1}=a^{i} \neq a$, for some $i>0$.

Taking $F_{p}$ to be the prime subfield of $D$, the $F_{p}$-algebra generated by $a$ is a finite subfield of $D$, of dimension $n$ over $F_{p}$, which we’ll denote $K$. The order of this field is $p^{n}$, and $a^{p^{n}}=a$. Since $a$ isn’t central, the inner derivation $\delta$, defined by $\delta(x)=ax-xa$, is non-zero. You can easily check that it is however a $K$-linear mapping of the $K$-vector space $D$. Using the Frobenius, one sees that $\delta^{p^{n}}=\delta$. Since any element $b \in K$ satisfies $b^{p^{n}}=b$, the following equation holds true $$t^{p^{n}} -t= \prod_{b \in K}(t-b) \in K[t].$$ Substituting $\delta$ into this equation, and remembering that $\delta \neq 0$ and all mono’s are left cancellable, there has to exist a $b_{0} \in K^{*}$ such that $\delta-b_{0}$ is not a mono. This means $\delta$ has an eigenvector $d$ with eigenvalue $b_{0}$. Using the definition of $\delta$, it follows that $dad^{-1}=a-b_{0} \in K \ \backslash \{a\}$. Since $a$ and $dad^{-1}$ have the same order in the cyclic group $K^{*}$, they generate the same subgroup, and $dad^{-1}=a^{i} \neq a$. To make sure $d$ is an additive commutator, replace it by $\delta(d)$, and you’ll see that the equation still holds up.

Division rings

Supposing $R$ is a division ring which doesn’t equal its centre $F$, you can easily see that there must exist an additive commutator which isn’t central, say $a=bb’-b’b$. Taking an element $c$ in $F^{*}$, a quick calculation shows that $ca$ is an additive commutator which can’t be central. The assumptions then imply that there is a number $k$ such that $1=a^{k}=(ca)^{k}=c^{k}a^{k}$, so any element in $F^{*}$ is periodic, and $D$ has non-zero characteristic. Using the theorem in the previous paragraph on $a$, an additive commutator $y$ exists, such that the group generated by $a$ and $y$ is a finite, periodic subgroup of $D^{*}$. This means the group has to be cyclic, which contradicts the statement $yay^{-1} \neq a$, and $R$ is commutative.

The proof of the pudding …

Let’s finish the proof. Suppose $R$ is a left primitive ring. The structure theorem for left primitive rings (basically a revamped version of Jacobson & Chevalley’s density theorem), says $R$ is isomorphic to a matrix ring over a skew field $D$, or $R$ has an infinite number of subrings $R_{m}$, each one having $M_{m}(D)$ as a quotient. Supposing $m>1$, and denoting by $E_{ij}$ the matrix with a $1$ in the $ij$-place and zeroes everywhere else, the obvious identity $$E_{11} E_{12} – E_{12} E_{11}=E_{12}$$ gives us a contradiction (just raise it to a power greater than $1$). This means $R \cong D$, and we reduced it to a previous case.

If our ring is semiprimitive, the Jacobson radical, denoted $J(R)$, is zero. Given a left primitive ideal $M_{i}$, take the quotient ring $R/M_{i}$, which is obviously left primitive, and inherits the property in the theorem because it’s a surjective image of $R$. This means each one of these rings is commutative. Now $R$ can be embedded in the product of all these rings, since the Jacobson radical, which is equal to the intersection of all left primitive ideals, is zero. Thus $R$ is commutative.

For a random ring, $R/J(R)$ is semiprimitive, and thus commutative. For $a, b$ in the ring, we know there exists an $n$ such that $(ab-ba)(1-(ab-ba)^{n-1})=0$. Since $ab-ba$ sits in $J(R)$, we know $(1-(ab-ba)^{n-1})$ is invertible, and $ab-ba=0$. Voilà!

More of this fun stuff can be found in Lam’s A first course in noncommutative rings and Lectures on modules and rings.


Prep-notes dump

Here are the scans of my rough prep-notes for some of the later seminar-talks. These notes still contain mistakes, most of them were corrected during the talks. So, please, read these notes with both mercy and caution!

Hurwitz formula imples ABC : The proof of Smirnov’s argument, but modified so that one doesn’t require an $\epsilon$-term. This is known to be impossible in the number-theory case, but a possible explanation might be that not all of the Smirnov-maps $q~:~\mathsf{Spec}(\mathbb{Z}) \rightarrow \mathbb{P}^1_{\mathbb{F}_1}$ are actually covers.

Frobenius lifts and representation rings : Faithfully flat descent allows us to view torsion-free $\mathbb{Z}$-rings with a family of commuting Frobenius lifts (aka $\lambda$-rings) as algebras over the field with one element $\mathbb{F}_1$. We give several examples including the two structures on $\mathbb{Z}[x]$ and Adams operations as Frobenius lifts on representation rings $R(G)$ of finite groups. We give an example that this extra structure may separate groups having the same character table. In general this is not the case, the magic Google search term is ‘Brauer pairs’.

Big Witt vectors and Burnside rings : Because the big Witt vectors functor $W(-)$ is adjoint to the tensor-functor $- \otimes_{\mathbb{F}_1} \mathbb{Z}$ we can view the geometrical object associated to $W(A)$ as the $\mathbb{F}_1$-scheme determined by the arithmetical scheme with coordinate ring $A$. We describe the construction of $\Lambda(A)$ and describe the relation between $W(\mathbb{Z})$ and the (completion of the) Burnside ring of the infinite cyclic group.

Density theorems and the Galois-site of $\mathbb{F}_1$ : We recall standard density theorems (Frobenius, Chebotarev) in number theory and use them in combination with the Kronecker-Weber theorem to prove the result due to James Borger and Bart de Smit on the etale site of $\mathsf{Spec}(\mathbb{F}_1)$.

New geometry coming from $\mathbb{F}_1$ : This is a more speculative talk trying to determine what new features come up when we view an arithmetic scheme over $\mathbb{F}_1$. It touches on the geometric meaning of dual-coalgebras, the Habiro-structure sheaf and Habiro-topology associated to $\mathbb{P}^1_{\mathbb{Z}}$ and tries to extend these notions to more general settings. These scans are unintentionally made mysterious by the fact that the bottom part is blacked out (due to the fact they got really wet and dried horribly). In case you want more info, contact me.