Category: math

The martial art of giving talks

Published January 5, 2012 by lievenlb

Last fall, Matilde Marcolli gave a course at CalTech entitled Oral Presentation: The (Martial) Art of Giving Talks. The purpose of this course was to teach students “how to effectively communicate their work in seminars and conferences and how to defend it from criticism from the audience”.

The lecture notes contain basic information on the different types of talks and how to prepare them. But they really shine when it comes to spotting the badasses in the public and how to respond to their interference. She identifies 5 badsass-types : the empreror and the hierophant (see below), the chariot (the one with a literal mind, asking continuously for details), the fool (the one who happens to sit in the talk but doesn’t belong there) and the magician (the quick smartass).

I’ll just quote here the description of, and most effective strategy against, the first two badass-types. Please have a look at the whole paper, it is a good read!

“The Emperor is the typical figure of power and authority in a given field. It refers to those people who have a tendency to think that the whole field is their own private property, and in particular that only what they do in the field is important, that the work of all others is derivative and that in any case they are not being quoted enough. These are typically pathological narcissists, so one needs to take this into account in interacting with them.
The trouble of having The Emperor in your audience is that he (it is rarely she) can very easily disrupt your presentation completely, by continuous interruptions, by running his own commentary while you are trying to stay focused on delivering your talk and by distracting the rest of the audience.
The Emperor is by far one of the most dangerous encounters you can make in the wilderness of the conference rooms.”

Counter-measure : “Keep in mind that the Emperor is a pathological narcissist: part of the reason why he keeps interrupting your talk is because he cannot stand the fact that, during those fifty minutes, the attention of the audience is focused on you and not on him. His continuous interruptions and complaints are a way to try to divert the attention of the audience back to him and away from you. That your talk gets disrupted in the process, he could not care the less.
A good way to try to avoid the worst case scenario is to make sure (if you know in advance you may be having the Emperor in the audience) that you arrange in your talk to make frequent references to him and his work. In this way, he will hopefully feel that his need to be at the center of attention is sufficiently satisfied that he can let you continue with your talk. Effectiveness: high.”

“The Hierophant represents a priestly figure. What this refers to here is the type of character who feels entitled to represent (and defend) a certain “orthodoxy”, a certain school of thought, or a certain group of people within the field.
Typically the hierophants are the minions and lackeys of the Emperor, his entourage and fan club, those who think that the Emperor represents the only and true orthodoxy in the field and that anything that is done in a different way should be opposed and suppressed.
These characters are generally less disruptive than the Emperor himself, as they are really only fighting you on someone else’s behalf. Nonetheless, they can sometime manage to seriously disrupt your presentation.”

Counter-measure : “This is essentially the same advise as in the case of the Emperor. To an objection that substantially is of the form: “This is not the right way to do things because this is not what what we do (= what the Emperor does)”, which is what you expect to hear from the Hierophant, you can reply along lines such as: “There is also another approach to this problem, developed by the Emperor and his school, which is a very interesting approach that gave nice and important results. However, this is not what I am talking about today: I am talking here about a different approach, and I will be focusing only on the specific features of this other approach…”
Something along these lines would recognize “their” work without having to make any concession on their approach being the only game in town.
Effectiveness: high (unless the Emperor is also present and is delegating to his hierophants the task of attacking you: in that case they won’t give up so easily and the effectiveness of this line of defense becomes medium/low).”

Lambda-rings for formula-phobics

Published February 5, 2010 by lievenlb

In 1956, Alexander Grothendieck (middle) introduced $\lambda $-rings in an algebraic-geometric context to be commutative rings A equipped with a bunch of operations $\lambda^i $ (for all numbers $i \in \mathbb{N}_+ $) satisfying a list of rather obscure identities. From the easier ones, such as

$\lambda^0(x)=1, \lambda^1(x)=x, \lambda^n(x+y) = \sum_i \lambda^i(x) \lambda^{n-i}(y) $

to those expressing $\lambda^n(x.y) $ and $\lambda^m(\lambda^n(x)) $ via specific universal polynomials. An attempt to capture the essence of $\lambda $-rings without formulas?

Lenstra’s elegant construction of the 1-power series rings $~(\Lambda(A),\oplus,\otimes) $ requires only one identity to remember

$~(1-at)^{-1} \otimes (1-bt)^{-1} = (1-abt)^{-1} $.

Still, one can use it to show the existence of ringmorphisms $\gamma_n~:~\Lambda(A) \rightarrow A $, for all numbers $n \in \mathbb{N}_+ $. Consider the formal ‘logarithmic derivative’

$\gamma = \frac{t u(t)’}{u(t)} = \sum_{i=1}^\infty \gamma_i(u(t))t^i~:~\Lambda(A) \rightarrow A[[t]] $

where $u(t)’ $ is the usual formal derivative of a power series. As this derivative satisfies the chain rule, we have

$\gamma(u(t) \oplus v(t)) = \frac{t (u(t)v(t))’}{u(t)v(t)} = \frac{t(u(t)’v(t)+u(t)v(t)’}{u(t)v(t))} = \frac{tu(t)’}{u(t)} + \frac{tv(t)’}{v(t)} = \gamma(u(t)) + \gamma(v(t)) $

and so all the maps $\gamma_n~:~\Lambda(A) \rightarrow A $ are additive. To show that they are also multiplicative, it suffices by functoriality to verify this on the special 1-series $~(1-at)^{-1} $ for all $a \in A $. But,

$\gamma((1-at)^{-1}) = \frac{t \frac{a}{(1-at)^2}}{(1-at)} = \frac{at}{(1-at)} = at + a^2t^2 + a^3t^3+\ldots $

That is, $\gamma_n((1-at)^{-1}) = a^n $ and Lenstra’s identity implies that $\gamma_n $ is indeed multiplicative! A first attempt :

hassle-free definition 1 : a commutative ring $A $ is a $\lambda $-ring if and only if there is a ringmorphism $s_A~:~A \rightarrow \Lambda(A) $ splitting $\gamma_1 $, that is, such that $\gamma_1 \circ s_A = id_A $.

In particular, a $\lambda $-ring comes equipped with a multiplicative set of ring-endomorphisms $s_n = \gamma_n \circ s_A~:~A \rightarrow A $ satisfying $s_m \circ s_m = s_{mn} $. One can then define a $\lambda $-ringmorphism to be a ringmorphism commuting with these endo-morphisms.

The motivation being that $\lambda $-rings are known to form a subcategory of commutative rings for which the 1-power series functor is the right adjoint to the functor forgetting the $\lambda $-structure. In particular, if $A $ is a $\lambda $-ring, we have a ringmorphism $A \rightarrow \Lambda(A) $ corresponding to the identity morphism.

But then, what is the connection to the usual one involving all the operations $\lambda^i $? Well, one ought to recover those from $s_A(a) = (1-\lambda^1(a)t+\lambda^2(a)t^2-\lambda^3(a)t^3+…)^{-1} $.

For $s_A $ to be a ringmorphism will require identities among the $\lambda^i $. I hope an expert will correct me on this one, but I’d guess we won’t yet obtain all identities required. By the very definition of an adjoint we must have that $s_A $ is a morphism of $\lambda $-rings, and, this would require defining a $\lambda $-ring structure on $\Lambda(A) $, that is a ringmorphism $s_{AH}~:~\Lambda(A) \rightarrow \Lambda(\Lambda(A)) $, the so called Artin-Hasse exponential, to which I’d like to return later.

For now, we can define a multiplicative set of ring-endomorphisms $f_n~:~\Lambda(A) \rightarrow \Lambda(A) $ from requiring that $f_n((1-at)^{-1}) = (1-a^nt)^{-1} $ for all $a \in A $. Another try?

hassle-free definition 2 : $A $ is a $\lambda $-ring if and only if there is splitting $s_A $ to $\gamma_1 $ satisfying the compatibility relations $f_n \circ s_A = s_A \circ s_n $.

But even then, checking that a map $s_A~:~A \rightarrow \Lambda(A) $ is a ringmorphism is as hard as verifying the lists of identities among the $\lambda^i $. Fortunately, we get such a ringmorphism for free in the important case when A is of ‘characteristic zero’, that is, has no additive torsion. Then, a ringmorphism $A \rightarrow \Lambda(A) $ exists whenever we have a multiplicative set of ring endomorphisms $F_n~:~A \rightarrow A $ for all $n \in \mathbb{N}_+ $ such that for every prime number $p $ the morphism $F_p $ is a lift of the Frobenius, that is, $F_p(a) \in a^p + pA $.

Perhaps this captures the essence of $\lambda $-rings best (without the risk of getting an headache) : in characteristic zero, they are the (commutative) rings having a multiplicative set of endomorphisms, generated by lifts of the Frobenius maps.

2 Comments

Grothendieck’s functor of points

Published September 29, 2009 by lievenlb

A comment-thread well worth following while on vacation was Algebraic Geometry without Prime Ideals at the Secret Blogging Seminar. Peter Woit became lyric about it :

My nomination for the all-time highest quality discussion ever held in a blog comment section goes to the comments on this posting at Secret Blogging Seminar, where several of the best (relatively)-young algebraic geometers in the business discuss the foundations of the subject and how it should be taught.

I follow far too few comment-sections to make such a definite statement, but found the contributions by James Borger and David Ben-Zvi of exceptional high quality. They made a case for using Grothendieck’s ‘functor of points’ approach in teaching algebraic geometry instead of the ‘usual’ approach via prime spectra and their structure sheaves.

The text below was written on december 15th of last year, but never posted. As far as I recall it was meant to be part two of the ‘Brave New Geometries’-series starting with the Mumford’s treasure map post. Anyway, it may perhaps serve someone unfamiliar with Grothendieck’s functorial approach to make the first few timid steps in that directions.

Allyn Jackson’s beautiful account of Grothendieck’s life “Comme Appele du Neant, part II” (the first part of the paper can be found here) contains this gem :

“One striking characteristic of Grothendieck’s
mode of thinking is that it seemed to rely so little
on examples. This can be seen in the legend of the
so-called “Grothendieck prime”.

In a mathematical
conversation, someone suggested to Grothendieck
that they should consider a particular prime number.
“You mean an actual number?” Grothendieck
asked. The other person replied, yes, an actual
prime number. Grothendieck suggested, “All right,
take 57.”

But Grothendieck must have known that 57 is not
prime, right? Absolutely not, said David Mumford
of Brown University. “He doesn’t think concretely.””

We have seen before how Mumford’s doodles allow us to depict all ‘points’ of the affine scheme $\mathbf{spec}(\mathbb{Z}[x]) $, that is, all prime ideals of the integral polynomial ring $\mathbb{Z}[x] $.
Perhaps not too surprising, in view of the above story, Alexander Grothendieck pushed the view that one should consider all ideals, rather than just the primes. He achieved this by associating the ‘functor of points’ to an affine scheme.

Consider an arbitrary affine integral scheme $X $ with coordinate ring $\mathbb{Z}[X] = \mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k) $, then any ringmorphism
$\phi~:~\mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k) \rightarrow R $
is determined by an n-tuple of elements $~(r_1,\ldots,r_n) = (\phi(t_1),\ldots,\phi(t_n)) $ from $R $ which must satisfy the polynomial relations $f_i(r_1,\ldots,r_n)=0 $. Thus, Grothendieck argued, one can consider $~(r_1,\ldots,r_n) $ an an ‘$R $-point’ of $X $ and all such tuples form a set $h_X(R) $ called the set of $R $-points of $X $. But then we have a functor

$h_X~:~\mathbf{commutative rings} \rightarrow \mathbf{sets} \qquad R \mapsto h_X(R)=Rings(\mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k),R) $

So, what is this mysterious functor in the special case of interest to us, that is when $X = \mathbf{spec}(\mathbb{Z}[x]) $?
Well, in that case there are no relations to be satisfied so any ringmorphism $\mathbb{Z}[x] \rightarrow R $ is fully determined by the image of $x $ which can be any element $r \in R $. That is, $Ring(\mathbb{Z}[x],R) = R $ and therefore Grothendieck’s functor of points
$h_{\mathbf{spec}(\mathbb{Z}[x]} $ is nothing but the forgetful functor.

But, surely the forgetful functor cannot give us interesting extra information on Mumford’s drawing?
Well, have a look at the slightly extended drawing below :

What are these ‘smudgy’ lines and ‘spiky’ points? Well, before we come to those let us consider the easier case of identifying the $R $-points in case $R $ is a domain. Then, for any $r \in R $, the inverse image of the zero prime ideal of $R $ under the ringmap $\phi_r~:~\mathbb{Z}[x] \rightarrow R $ must be a prime ideal of $\mathbb{Z}[x] $, that is, something visible in Mumford’s drawing. Let’s consider a few easy cases :

For starters, what are the $\mathbb{Z} $-points of $\mathbf{spec}(\mathbb{Z}[x]) $? Any natural number $n \in \mathbb{Z} $ determines the surjective ringmorphism $\phi_n~:~\mathbb{Z}[x] \rightarrow \mathbb{Z} $ identifying $\mathbb{Z} $ with the quotient $\mathbb{Z}[x]/(x-n) $, identifying the ‘arithmetic line’ $\mathbf{spec}(\mathbb{Z}) = { (2),(3),(5),\ldots,(p),\ldots, (0) } $ with the horizontal line in $\mathbf{spec}(\mathbb{Z}[x]) $ corresponding to the principal ideal $~(x-n) $ (such as the indicated line $~(x) $).

When $\mathbb{Q} $ are the rational numbers, then $\lambda = \frac{m}{n} $ with $m,n $ coprime integers, in which case we have $\phi_{\lambda}^{-1}(0) = (nx-m) $, hence we get again an horizontal line in $\mathbf{spec}(\mathbb{Z}[x]) $. For $ \overline{\mathbb{Q}} $, the algebraic closure of $\mathbb{Q} $ we have for any $\lambda $ that $\phi_{\lambda}^{-1}(0) = (f(x)) $ where $f(x) $ is a minimal integral polynomial for which $\lambda $ is a root.
But what happens when $K = \mathbb{C} $ and $\lambda $ is a trancendental number? Well, in that case the ringmorphism $\phi_{\lambda}~:~\mathbb{Z}[x] \rightarrow \mathbb{C} $ is injective and therefore $\phi_{\lambda}^{-1}(0) = (0) $ so we get the whole arithmetic plane!

In the case of a finite field $\mathbb{F}_{p^n} $ we have seen that there are ‘fat’ points in the arithmetic plane, corresponding to maximal ideals $~(p,f(x)) $ (with $f(x) $ a polynomial of degree $n $ which remains irreducible over $\mathbb{F}_p $), having $\mathbb{F}_{p^n} $ as their residue field. But these are not the only $\mathbb{F}_{p^n} $-points. For, take any element $\lambda \in \mathbb{F}_{p^n} $, then the map $\phi_{\lambda} $ takes $\mathbb{Z}[x] $ to the subfield of $\mathbb{F}_{p^n} $ generated by $\lambda $. That is, the $\mathbb{F}_{p^n} $-points of $\mathbf{spec}(\mathbb{Z}[x]) $ consists of all fat points with residue field $\mathbb{F}_{p^n} $, together with slightly slimmer points having as their residue field $\mathbb{F}_{p^m} $ where $m $ is a divisor of $n $. In all, there are precisely $p^n $ (that is, the number of elements of $\mathbb{F}_{p^n} $) such points, as could be expected.

Things become quickly more interesting when we consider $R $-points for rings containing nilpotent elements.