User najdorf - MathOverflow

Cardinal Arithmetic, foundations and constructive math

2012-11-04T01:55:51Z

This is not my area but a question occurred to me that I can not find the answer to. There is a very strong axiom of constructibility which ironically gives us highly non-constructive math (GCH is one of its implications). What would be an equally strong axiom in the opposite direction? And I mean direction in a philosophical sense, so what would be the strongest axiom that constructivists/intuitionists would approve of?

My first idea was to find the largest $\kappa$ such that $2^{\aleph_0} = \aleph_{\kappa}$ is consistent with ZF but this set is unbounded ($\kappa$ can be any finite number) and $2^{\aleph_0} < \aleph_{\omega}$. Which brings up the question, how much fundamental difference are there between CH and $2^{\aleph_0} = \aleph_{118}$ for example?

A possible mistake in Kac's "Infinite Dimensional Lie Algebras"

2012-07-27T06:18:46Z

I have a paperback 3rd edition and on page 65 you can find Proposition 5.8. My question is about part (c):

If $A$ is of indefinite type, then $$ \overline{X} = \{ h \in { \frak h_{\mathbb{R}} }| \left \langle \alpha, h \right \rangle \geq 0 \text{ for all } \alpha \in \Delta_+^{im} \}, $$ where $\overline{X}$ denotes the closure of $X$ in the metric topology of $\frak h_{\mathbb{R}}$.

Some context: $A$ is a generalized Cartan matrix, $\frak h_{\mathbb{R}}$ is a real form of the Cartan subalgebra in the Kac-Moody algebra associated to $A$ and finally $X$ is the Tits cone.

Question: if one reads the proof given by Kac, it is written for $X$ and not $\overline{X}$. Is there a mistake or am I missing something here?

Subsystems of Peano arithmetic and incompleteness theorem

2011-02-13T12:43:05Z

I think everyone is familiar with Goedel's incompleteness theorems. In particular they imply that PA (Peano arithmetic) can not prove its own consistency. Now my question is what is the largest subsystem of PA that "can" prove its own consistency? You definitely can't have the induction axiom but is that enough?

Explicit isomorphism between distributions and universal enveloping algebra

2010-10-08T09:04:46Z

The universal enveloping algebra of a Lie algebra $\mathfrak{g}$ is isomorphic to the algebra of distributions on the Lie group $G$ with support at the identity. The proof I have of this fact uses the universal property of the universal enveloping algebra and hence it is not constructive. I was wondering if there is an explicit map? For example what would happen for $\mathfrak{g} = \mathfrak{sl}(2, \mathbf{C})$ and $G = SL(2, \mathbf{C})$? For an explicit monomial in $\mathscr{U} \ \left ( \mathfrak{sl}(2, \mathbf{C}) \right )$ can we write the corresponding distribution and how it acts on functions on $SL(2, \mathbf{C})$?

Non-trivial integral forms of algebras

2011-04-14T20:34:03Z

Suppose $\mathcal{A}$ is a $\mathbf{C}$-algebra then an integral form would be a subring $\mathcal{B} \subset \mathcal{A}$ such that the canonical map $\mathcal{B} \otimes_{\mathbf{Z}} \mathbf{C} \rightarrow \mathcal{A}$ is a bijection.

For some algebras there is an obvious integral form in the following sense: there is a preferred $\mathbf{C}$-basis for $\mathcal{A}$ and the $\mathbf{Z}$-span of that basis is $\mathcal{B}$. Now my question is do we have examples where $\mathcal{B}$ is non-obvious? In other words the basis coming from $\mathcal{B}$ would look very strange for those who only work with $\mathcal{A}$. Is there such an example where $\mathcal{A}$ is commutative?

Question on Linear Operators

2011-07-21T01:15:29Z

Let $V$ be a normed infinite dimensional vector space. Let $L: V \longrightarrow V$ be a bounded linear operator. Moreover assume that $L$ is 'locally nilpotent' that is: $$ \forall v \in V \quad \exists n \in \mathbf{N}: L^n (v) = 0. $$ Now my question is if the linear operator: $$ \exp (L) = \sum_{n=0}^{\infty} \frac{L^n}{n!} $$ is bounded or not.

Embedding of algebraic surfaces

2011-01-20T12:11:39Z

If I am not mistaken there is a theorem that says any curve $C$ can be embedded in $\mathbf{P}^3$. What can be said about surfaces? Do we have a theorem like:

All surfaces can be embedded in $\mathbf{P}^{N}$ for some fixed $N$.

And if not what is the simplest counterexample? A counterexample would be an infinite sequence of surfaces: $\{S_1, S_2, \cdots \}$ and infinite sequence of strictly increasing positive integers: $\{n_1, n_2, \cdots \}$ such that $S_i$ can not be embedded in $\mathbf{P}^{n_i}$.

Answer by Najdorf for Careers advice for Ph.D.s without current postdocs or university jobs

2011-03-30T03:07:26Z

I am more or less in the same situation as you are. After careful consideration I decided that I should abandon academia for good. Because as far as I have seen your career in academia is very dependent on where you start out. If you do a postdoc in University of Nowhere the chances of getting a decent job after that will be even lower than it is now. There is a temptation to think that by hard work you can compensate for the bad start in your career, resist that temptation.

As it stands now I will be applying for finance jobs later in summer. Consider applying to big traditional banks (J P Morgan, Chase Manhattan, Wells Fargo, ...) investment banks (Goldman-Sachs, Morgan Stanley) and big hedge funds (Renaissance Technologies, D E Shaw, Citadel). If you want more info about 2.5 years ago there was a small article in Notices on the transition from academia (specifically math) to finance. The author had a Ph.D. in number theory and went on working for D E Shaw (PDF): http://www.ams.org/notices/200806/tx080600700p.pdf

Also don't worry about the state of financial firms. They have fully recovered from the effects of the great recession of 2007-2009: http://motherjones.com/kevin-drum/2011/03/chart-day-finance-back

"Resume de cours" by J. Tits

2011-03-18T13:51:20Z

I have been reading a number of papers by J. Tits (mostly written in the second half of 1980s) and in them he frequently refers to following publications of his:

Resume de cours, Annuaire du college de France, 81e annee (1980-1981), 75-86.
Resume de cours, Annuaire du college de France, 82e annee (1981-1982), 91-105.

Unfortunately I have not been able to locate these two papers (internet, library, faculty). Any help with finding these two would be immensely appreciated.

Classification of generalized Cartan matrices (GCMs)

2011-02-15T12:24:17Z

A GCM is square matrix $A = (a_{ij})$ satisfying: (1) $a_{ij} \in \mathbf{Z}$ (2) $a_{ii} = 2$ for all $i$. (3) $a_{ij} \leq 0$ for $i \neq j$. (4) $a_{ij} = 0$ iff $a_{ji} = 0$. There is a standard notion of irreducibility among GCMs and the common term for it is "indecomposable". Now if you look at most books there is a standard basic classification of GCMs: An indecomposable GCM is of 3 kinds: finite, affine and another category which can be described as all-other-GCMs-that-I-know-little-about-so-I-will-bundle-them-together. There is also another classification based on whether the GCM is "symmetrizable" or not (the finite and affine GCMs are ~~symmetric and hence~~ symmetrizable).

What I am looking for is a smart classification of the GCMs that are not finite or affine. For example here is a number of questions:

[1]

[2]

[3]

EDIT: corrected a mathematical mistake pointed out by Jim Humphreys, also improved the format as to make things more visible. Again suggested by Jim Humphreys.

What a geometer should know ...

2011-02-13T12:29:45Z

I am wondering what are the prerequisites for being a modern geometer? It seems that the amount you have to know is just huge: differential geometry, differential topology, algebraic topology, algebraic geometry, symplectic geometry, ... and then there is all kinds of overlap. So my question is the following: what should you know (from the topics I mentioned and the topics that I forgot) to call yourself a geometer (lets say by the time you get your Ph.D.)? And where are the best books/articles covering all that?

Quiver varieties and the affine Grassmannian

2011-01-24T15:05:53Z

Recently I was watching a talk: http://media.cit.utexas.edu/math-grasp/Ben_Webster.html and at the end the lecturer gave a correspondence (I was having trouble with subscripts so changed the notation a bit):

$$ \mathfrak{Q} (\lambda, \mu) \Leftrightarrow \mathfrak{M} (\lambda, \mu)$$

Between Quiver varieties and slices of affine Grassmannian. I would like to learn how this exactly works (like precise definition of both objects is a start!). If you have the time and the patience you can post a lengthy response here or you could just point me towards the right place to start reading on my own. Thank you!

A question about the affine Grassmanian

2011-01-23T02:46:26Z

For $SL(2, \mathbf{C} ((t)))$ the affine Grassmanian is defined as: $$SL(2, \mathbf{C}((t))) / SL(2, \mathbf{C} [[t]])$$ Now that is fine but $SL(2, \mathbf{C} ((t)))$ has smaller parabolic subgroups. Define: $$B = \begin{pmatrix} \mathcal{O}^{\times} & \mathcal{O} \\ t\mathcal{O} & \mathcal{O}^{\times} \end{pmatrix}$$ where $\mathcal{O} = \mathbf{C}[[t]]$. Then: $$ SL(2, \mathbf{C} [[t]]) = B \left \{ 1, \begin{pmatrix} & 1 \\ -1 & \end{pmatrix} \right \} B $$

Question: Why are people so much interested in the affine Grassmanian when it actually sits in a larger, equally well behaved projective ind-scheme (in our example above $SL(2, \mathbf{C}((t)))/ B$)? Both are flag varieties for the loop group so it seems natural to go with the full flag variety.

Quotients of unipotent groups

2010-09-13T06:55:49Z

Let $U (\mathbf{R})$ be the standard unipotent subgroup of $SL(3, \mathbf{R})$. So $U(\mathbf{R})$ is the group of 3 by 3 upper triangular matrices with 1s on the diagonal. I am interested in the quotient space $U (\mathbf{R})/ U (\mathbf{Z})$. I think it is just the cube, $[0,1]^3$, but I am having difficulty writing the actual map. The only non-obvious part is the (1,3)-entry where the addition is not straightforward. Also is the generalization that one gets the $\frac{1}{2}(n^2 - n)$ dimensional cube for the standard unipotent subgroup of $SL(n, \mathbf{R})$ correct?

@ Bill

Your answer helped clear a lot of confusion but it also created a number of new questions:

So basically $U(\mathbf{R})$ has three copies of $\mathbf{R}$ however they way they are knit together is not the simple direct sum. Is that right?
If I pick an element $u \in U$ is the group generated by $u$ isomorphic to $\mathbf{G}_a$ ?
Is the space $U (\mathbf{R})/ U (\mathbf{Z})$ compact? The Haar measure on $SL(3, \mathbf{R})$ gives a measure on $U(\mathbf{R})$. Then presumably I get a measure on $U (\mathbf{R})/ U (\mathbf{Z})$ what is the volume of the quotient with respect to this measure? Is there a place I can find these calculations?

Ultimately I am interested in the arithmetic quotient $SO(3)\backslash SL(3, \mathbf{R}) / SL(3, \mathbf{Z})$ and $U (\mathbf{R})/ U (\mathbf{Z})$ is supposed to be its 'base' and the end result should be an algebraic surface (hence a 4-dimensional real manifold I think).

Nilpotent Lie algebras and unipotent Lie groups

2010-09-11T01:35:02Z

$\mathbf{n}$ is nilpotent Lie algebra with $N$ being the corresponding algebraic Lie group. Now one neat feature of this setting is that you can take the exponential map to be identity. In other words you can define a group structure on $\mathbf{n}$ using the Campbell-Hausdorff formula. I have the following questions (they might be very easy; I just don't know):

If $\pi: \mathbf{n} \rightarrow \text{End}(V)$ is a representation of the nilpotent Lie algebra then does $x \in N$ acts as $\exp (\pi (x))$ on $V$ ?
If $\mathbf{m}$ is a subalgebra of $\mathbf{n}$ then you get a corresponding group $M \subset N$. Do we have: $N/M = \mathbf{n}/\mathbf{m}$ as sets?

EDIT: Victor's comments are to the point so in order to clear up the confusion I added that $N$ is algebraic.

Explicit equations for Schubert varieties

2010-08-26T23:49:12Z

How can one compute the Schubert variety (by compute I mean having actual polynomials that define it) for SL(n)? If this is well known forgive my ignorance and just point me to the right book/paper.

EDIT: Sorry I did not return here for quite some time. It is kind of amusing that the way I learned about Schubert varieties is not even mentioned. Here is how I learned it:

$G$ an algebraic group with Lie algebra $\mathbf{g}$.

$L(\Lambda)$ is an integrable highest weight module for $\mathbf{g}$.

For $w$, an element of the Weyl group, consider the 1-dimensional root space $L(\Lambda )_{w \cdot \Lambda}$.

Denote the vector space $U(\mathbf{b}) \bullet L(\Lambda)_{w \cdot \Lambda}$ by $E_w(\Lambda)$ (you take the 1-dimensional root space and act on it by all the raising operators plus the Cartan). Then $E_w(\Lambda) \subset L(\Lambda)$.

Now we are ready: since $L(\Lambda )_{w \cdot \Lambda}$ is 1-dimensional it becomes a single point in $\mathbf{P} \left (E_w(\Lambda) \right)$. We look at the orbit $B \bullet L(\Lambda)_{w \cdot \Lambda} \subset \mathbf{P} \left (E_w(\Lambda) \right)$. We call its closure the Schubert variety associated to $w$ and $\Lambda$ and denote it by $S_{w, \Lambda}$.

I don't know if this is a good way of computing things but in principle it should give you any Schubert variety you need.

Comment by Najdorf

2012-11-05T15:09:27Z

Also above I said that my idea of constructivism is that it should be practically computable (I mentioned MP in addition to the law of excluded middle). I have no idea if this is even a topic of discussion in constructive/intuitionist circles but if you look at it from a programming point of view it is the most natural definition.

Comment by Najdorf

2012-11-05T15:07:09Z

Alright that example clarified things a bit. But I am fine with unprovabally finite subsets of a finite set. My initial thought after reading your statement was that, there is a freakish model where a finite set had a non-finite subset (that would is stupid but from my experience you can't just call something obviously stupid and move one hence my confusion).

Comment by Najdorf

2012-11-05T09:17:31Z

Law of excluded middle is non-constructive. I have doubts about MP as well, if $A$ and $A \to B$ but it took a trillion years to get from $A$ to $B$ would that be constructive? I have realized that I use constructive in a much stronger form than anyone else here. For example, I have a class of graphs with $\chi = 5$ but the proof is classical, then I get a 5-coloring algorithm of order $O(|G|^{|G|})$. To my mind the algorithm is progress but still not constructive. In short being constructive without being computable in some sensible sense of the word does not make sense to me ...

Comment by Najdorf

2012-11-05T08:57:44Z

@Andrej: I am very sorry for the lack of clarity, I have not thought about these things and strangely enough I think I use constructive in a stronger form than anyone else here which just worsens the matters. The subset/Random Variable is not a real subset, so if the non-finite subsets of a finite set are only of that type then I am fine with the original statement. The next question is whether what I call real subset can be formalized in any sense or not ...

Comment by Najdorf

2012-11-04T23:05:32Z

@all: I call an axiom AX non-constructive if it has non-constructive consequences in everyday classical mathematics we use. I thought that was the normal understanding of the term.

Comment by Najdorf

2012-11-04T23:03:06Z

@Blass I think you are right in your first comment.

Comment by Najdorf

2012-11-04T23:01:58Z

@Andrej, @Guillaume: In those cases I would take issue with the way you define the subsets which depend on an odd black box. What you have is not a fixed subset, it can be different things in the future, it is essentially a random variable.

Comment by Najdorf

2012-11-04T11:07:08Z

@Noah, my understanding is that the model for V=L done by Godel is "constructive" but the theory itself (which is always much bigger than any single model ...) is not.

Comment by Najdorf

2012-11-04T11:02:56Z

@Ricky, well GCH implies AC ...

Comment by Najdorf

2012-11-04T11:01:31Z

Alright that was helpful but I still have some issues: (1) When you say there are models for your first list of statements, what is the base system of constructive math you are working with? (2) Throughout the whole discussion I am assuming a degree of reasonableness, an axiom stating the fact that a finite set can have non-finite subsets or that we have an embedding $\mathbf{R} \to \mathbf{N}$ are not resonable (3) I might be mistaken but I thought that intuitionists did not like Church's thesis ...

Comment by Najdorf

2012-08-14T07:54:27Z

Does Kac answer email? Actually I found what seems to be a full proof in "Introduction to Kac-Moody Algebra" by Zhexian Wan (Proposition 5.6, page 98).

Comment by Najdorf

2012-07-31T18:11:55Z

Why are all integral points of X^{\prime} (defined as points where all simple roots give us integers) dense in the metric topology? I think it would be far from dense and actually discrete.

Comment by Najdorf

2012-07-27T13:45:16Z

I used the word "mistake" because I think the statement is false and you should replace the closure with the cone. Of course I might be mistaken myself, hence "possible mistake".

Comment by Najdorf

2012-07-27T13:43:10Z

But if you read the proof he says let X^{\prime} be the right hand side and then shows that it is equal to X itself! The closure in metric topology is not mentioned even once!

Comment by Najdorf

2011-05-14T21:35:17Z

Can you give a link? I can't find it.