User abdelmalek abdesselam - MathOverflow

Answer by Abdelmalek Abdesselam for A basis of the symmetric power consisting of powers

2013-05-01T16:55:15Z

I would look up a book on the calculus of finite differences in a multivariate setting. The claim here is to show that for any multi-index $\alpha=(\alpha_1,\ldots,\alpha_n)$ of length $k$ one can express the multiple derivative at zero $$ \left(\frac{\partial}{\partial t}\right)^\alpha \ (t_1x_1+\cdots+ t_n x_n)^k $$ as a linear combination of finite difference analogous expressions which should only involve the evaluation of $(t_1x_1+\cdots+ t_n x_n)^k$ at integer points $(t_1,\ldots,t_n)$ with nonnegative coordinates adding up to $k$. This is the same as the above candidate basis considered by you and Peter. I don't know if there exists a multivariate analogue of the Newton series. If so then this would immediately imply the wanted statement.

Edit: Apparently there is such a formula due to Lascoux and Schutzenberger, see Theorem 9.6.1 page 148 in the book "Symmetric functions and combinatorial operators on polynomials" by Alain Lascoux. Another source on the web is here. It also has the required property here which is that the number of finite differences taken is the same as the degree of the multiplying Schubert polynomial.

Edit: @Jesko you're right it is a bit more complicated than what I said. Also, the Lascoux-Schutzenberger formula might not be the simplest to use here.

First note that expressions $$ \prod_{i=1}^{n-1} (x_i-x_n)^{\beta_i}\ \times\ (kx_n)^{k-|\beta|}\ , $$ where $\beta$ ranges over multiindices with $n-1$ components and length $|\beta|\le k$, form a basis. Now you get the latter as derivatives $$ \left(\frac{\partial}{\partial t}\right)^{\beta} \ \left(t_1x_1+\cdots+ t_{n-1} x_{n-1}+\left(k-\sum_{i=1}^{n-1}t_i\right)x_n\right)^k $$ at $t=0$.

Call $f(t_1,\ldots,t_{n-1})$ the polynomial function to be hit with derivatives. One has a multivariate Newton expansion for it: $$ f(t)=\sum_{m} (t-a)^m \partial^m f(a_{11},a_{21},\ldots,a_{n-1,1}) $$ as follows. Here $a$ stands for a matrix of indeterminates $(a_{ij})$ with $1\le i\le n-1$ and $1\le j\le d$, with $d$ high enough. Let $\partial_{ij}$ denote the divided difference operator acting on functions of these indeterminates as $$ \partial_{ij} g=\frac{1}{a_{i,j+1}-a_{ij}}\left( g({\rm argument\ with\ }a_{i,j+1}\ {\rm and}\ a_{ij}\ {\rm exchanged})- g \right)\ . $$ The notation $m=(m_1,\ldots,m_{n-1})$ is for a multiindex with nonnegative entries. We also write the corresponding operator $$ \partial^m = \prod_{i=1}^{n-1} \left(\partial_{i, m_i} \cdots\partial_{i,2}\partial_{i,1}\right) $$ noting that finite difference operators concerning different groups of variables commute. Finally $$ (t-a)^m=\prod_{i=1}^{n-1} \left((t_i-a_{i,m_i})\cdots(t_i-a_{i,2})(t_i-a_{i,1})\right)\ . $$ The formula basically amounts to applying Newton's univariate formula in each coordinate direction separately. Now use this with the choice $a_{i,j}=j-1$, then take the beta derivative in the $t$'s and that should be it.

Answer by Abdelmalek Abdesselam for Old books still used

2012-12-28T16:33:39Z

If one needs to use tools from classical invariant theory or elimination theory then some books that come to mind are:

Grace and Young "The Algebra of Invariants", 1903.
Elliott "An Introduction to The Algebra of Quantics", 1913.
Salmon "Lessons Introductory to The Modern Higher Algebra", 1876.
Faa di Bruno "Théorie Des Formes Binaires", 1876.
Faa di Bruno "Théorie Générale de l'Elimination", 1859.

and there are quite a few more.

For Salmon's book, the 4th edition of 1885 might be best. Indeed, as I learned from a paper by Macauley, it has a discussion (on p. 87) of Cayley's very general formula for the multivariate resultant as the determinant of a complex (see the book by Gelfand, Kapranov and Zelevinsky for a modern account and a reprint of Cayley's paper).

Is there a quantum Hermite reciprocity?

2010-07-28T17:00:27Z

It is well known that there is an isomorphism of $SL_2=SL(V)$ representations $$ Sym^n(Sym^m(V))\simeq Sym^m(Sym^n(V)) $$ called Hermite reciprocity (discovered in 1854). My question is: Is there anything like this isomorphism for $U_q(sl_2)$, at least for generic $q$?

Answer by Abdelmalek Abdesselam for Decomposition into irreducibles of symmetric powers of irreps.

2013-05-30T00:16:31Z

For $\mathfrak{s}\mathfrak{l}_n$ and $\rho$ a symmetric power you can have a look at this article by Bedratyuk published in Lin. Multlin. Algebra. It has a formula for the multiplicity of the trivial representation but it is quite scary.

Answer by Abdelmalek Abdesselam for What is the "fundamental theorem of invariant theory" ?

2013-05-14T14:37:03Z

Normally, the first fundamental theorem of invariant theory (due to Cayley and Clebsch in the mid 19th century) says that all invariants can be obtained as contractions of elementary tensors like the epsilon expression in your question. See my answer to MO 121715 for an example of how that works. Coincidentally, the latter is not far from your question since there are also two Lie groups acting. However, if you want more help, you need to follow Jeff's advice and formulate your question with more mathematical precision.

Answer by Abdelmalek Abdesselam for Generate a higher degree symmetric polynomial from an existing one

2013-05-01T13:35:13Z

I don't know if the operation has a name in the context of the classical theory of symmetric functions. However, in mathematical physics this is essentially what is called a creation operator in a Boson Fock space. See, e.g., Reed and Simon "Methods of Modern Mathemtatical Physics" vol 2, page 209, 1975 edition.

Answer by Abdelmalek Abdesselam for An expression with an alternating trilinear form, written in terms of the determinant and a symmetric bilinear form

2013-04-25T13:54:17Z

I think this is specific to the $M_2(\mathbb{C})$ situation, or rather $\mathfrak{s}\mathfrak{l}_2(\mathbb{C})$. The identity boils down to $det(A)det(B)=det(AB)$ for $3\times 3$ matrices. It's a simple calculation using Pauli matrices.

Partial linearization near a hyperbolic fixed point--Classical scattering

2013-03-19T20:58:51Z

I am currently reading the famous article "Universal Properties of Maps on an Interval" by Collet, Eckmann and Lanford related to the Feigenbaum-Coullet-Tresser universality. I am in particular interested in Theorem 6.3 page 236 in that article. See the article for the precise statement, but very roughly the theorem says:

Consider a transformation $T$ on an infinite dimensional space which has a hyperbolic fixed point with one unstable direction with eigenvalue $\delta$ and a codimension one stable manifold. Then there is change of coordinates to a new system (x,y) where:

the stable manifold is given by $y=0$
the unstable manifold is given by $x=0$
the transformation T takes the form $$ (x,y)\longmapsto (M(x,y), \delta \ y) $$ in this new coordinate system.

In other words, this realizes a linearization in the unstable direction only.

I would like to know about similar/related theorems, follow-ups, improvements, etc. that exist in the literature.

Using keyword searches etc. has been quite disappointing and I can definitely use the help of people with expertise in the area. For instance, I did not know the above paper contained such a theorem until a chance discussion with one of the authors.

Edit with some context:

The method used in the CEL article is as follows. They first do some prep work in order to have a coordinate system $(x,y)$ satisfying the first two properties, i.e., such that the stable and unstable manifolds are straight. Then they construct the partial conjugation as $$ z(x,y)=\lim_{n\rightarrow \infty} \delta^{-n} y_n(x,y) $$ where $y_n$ denotes the $y$ coordinate of the $n$-th iterate of the point $(x,y)$ by a suitable cut-off modifcation of $T$.

This is very similar to the construction of wave operators in scattering theory.

The reason I am interested in this is because in recent joint work (see this paper) we proved the following:

Assume $T$ is analytic and has a hyperbolic fixed point $v_{*}$ with only one expanding direction with eigenvalue $\delta$. Then $$ \Psi(v,w)=\lim_{n\rightarrow \infty} T^n(v+\delta^{-n}w) $$ exists and is analytic (jointly in $w$ and the component of $v$ along the stable tangent space used in the analytic parametrization of the stable manifold). Here $v$ belongs to the stable manifold and $w$ is arbitrary but not too big. This function $\Psi$ is not a true linearization, not even a partial one such as the $z$ function of CEL but it shares some of that flavor. Namely, it satisfies the properties:

$T\circ\Psi(v,w)=\Psi(v,\delta\ w)$.
$\Psi$ takes its values in the unstable manifold.
$\Psi(v,w)=\Psi(v_{*},L_{v}(w))$ where $L_v$ is a $v$-dependent linear map onto the unstable tangent space.

The $\Psi$ function can be seen as the $w$ directional derivative of the $z$ function on the stable manifold. It is a "true linearization" on the unstable manifold only. I would like to know if similar results exist in the literature.

Answer by Abdelmalek Abdesselam for real symmetric matrix has real eigenvalues - elementary proof

2013-02-27T21:41:39Z

This is quite an interesting question, perhaps a research problem. I think an elementary answer should be a high school algebra answer in the sense of Abhyankar and it would have to be in the spirit of what follows. But first a little story.

I was teaching linear algebra and had just covered eigenvalues and characteristic polynomials but was not yet at the chapter on the spectral theorem for real symmetric matrices. I was looking for problems to assign for my students as homework in the textbook we were using. One of the exercises was to show that a real matrix $$ A=\left[ \begin{array}{cc} \alpha & \beta \\ \beta & \gamma \end{array} \right] $$ only had real eigenvalues. Not too hard. Write the characteristic polynomial $$ \chi(\lambda)=det(\lambda I-A)=\lambda^2-(\alpha+\gamma)\lambda+\alpha\gamma-\beta^2 $$ then its discriminant is $$ \Delta=(\alpha+\gamma)^2-4(\alpha\gamma-\beta^2)=(\alpha+\gamma)^2+4\beta^2\ge 0\ . $$ Hence two real roots.

The next problem in the book was to do the same for $$ A=\left[ \begin{array}{ccc} \alpha & \beta & \gamma\\ \beta & \delta & \varepsilon \\ \gamma & \varepsilon & \zeta \end{array} \right] $$ and (silly me) I also assigned it...

Here is the solution in the 3X3 case. All roots are real if the discriminant (for a binary cubic) is nonnegative. The discriminant of the characteristic polynomial is $$ \Delta = (\delta \varepsilon ^{2} + \delta \zeta ^{2} - \zeta \delta ^{2} - \zeta \varepsilon ^{2} + \zeta \alpha ^{2} + \zeta \gamma ^{2} - \alpha \gamma ^{2} - \alpha \zeta ^{2} + \alpha \beta ^{2} + \alpha \delta ^{2} - \delta \alpha ^{2} - \delta \beta ^{2})^{2} \\ \mbox{} + 14(\delta \gamma \varepsilon - \beta \varepsilon ^{2} + \beta \gamma ^{2} - \alpha \gamma \varepsilon )^{2} \\ \mbox{} + 2(\delta \alpha \gamma + \delta \beta \varepsilon + \delta \gamma \zeta - \gamma \delta ^{2} - \gamma \varepsilon ^{2} + \gamma ^{3} - \alpha \beta \varepsilon - \alpha \gamma \zeta )^{2} \\ \mbox{} + 2(\delta \beta \gamma + \delta \varepsilon \zeta - \varepsilon ^{3} + \varepsilon \alpha ^{2} + \varepsilon \gamma ^{2} - \alpha \beta \gamma - \alpha \delta \varepsilon - \alpha \varepsilon \zeta )^{2} \\ \mbox{} + 2(\zeta \alpha \beta + \zeta \beta \delta + \zeta \gamma \varepsilon - \beta \varepsilon ^{2} - \beta \zeta ^{2} + \beta ^{3} - \delta \alpha \beta - \alpha \gamma \varepsilon )^{2} \\ \mbox{} + 14(\zeta \beta \varepsilon - \gamma \varepsilon ^{2} + \gamma \beta ^{2} - \alpha \beta \varepsilon )^{2} \\ \mbox{} + 2(\zeta \beta \gamma + \delta \varepsilon \zeta - \varepsilon ^{3} + \varepsilon \alpha ^{2} + \varepsilon \beta ^{2} - \alpha \beta \gamma - \alpha \delta \varepsilon - \alpha \varepsilon \zeta )^{2} \\ \mbox{} + 14(\varepsilon \beta ^{2} + \zeta \beta \gamma - \delta \beta \gamma - \varepsilon \gamma ^{2})^{2} \\ \mbox{} + 2(\zeta \alpha \beta + \zeta \beta \delta + \zeta \gamma \varepsilon - \beta \gamma ^{2} - \beta \zeta ^{2} + \beta ^{3} - \delta \alpha \beta - \delta \gamma \varepsilon )^{2} \\ \mbox{} + 2(\alpha \gamma \zeta + \zeta \beta \varepsilon - \gamma ^{3} + \gamma \beta ^{2} + \gamma \delta ^{2} - \delta \alpha \gamma - \delta \beta \varepsilon - \delta \gamma \zeta )^{2}\ . $$

This formula comes from a paper by Ilyushechkin in Mat. Zametki, 51, 16-23, 1992.

I suspect the elementary answer should be as follows. First find a list of invariants or covariants of binary forms $C_1,C_2,\ldots$ such that a form with real coefficients has only real roots iff these covariants are nonnegative. Apply this to the characteristic polynomial of a general real symmetric matrix and show that you get sums of squares. I suppose these covariants, via Sturm's sequence type arguments, should correspond to subresultants or rather subdiscriminants. This seems also related to Part 2) of Godsil's answer.

Answer by Abdelmalek Abdesselam for Modern developments in finite-dimensional linear algebra

2013-02-28T14:11:30Z

Just putting the references asked for by Timur:

J. M. Landsberg, "The border rank of the multiplication of $ 2\times 2$ matrices is seven", J. American Math. Soc. 19 (2006), 447-459.
J. M. Landsberg and G. Ottaviani "New lower bounds for the border rank of matrix multiplication", 2011 preprint.

Answer by Abdelmalek Abdesselam for What mathematical treatment is there on the renormalization group flow in a space of Lagrangians?

2011-04-26T22:41:00Z

The answer to your question depends on whether you are interested in the perturbative RG or the nonperturbative one.

Typically one starts with a Gaussian measure $d\mu_{0,\infty}$ on a space of fields $\phi$ given by a covariance $$C_{0,\infty}(x,y)=\int \phi(x)\phi(y)\ d\mu_{0,\infty}(\phi)\ . $$ For instance one can take for the covariance $\frac{1}{\xi^2}$ in Fourier space. Then one introduces a UV regularization at length scale $l$ by multiplying for instance by $\exp(-l^2 \xi^2)$ which cuts-off momenta $\xi$ which are larger than $l^{-1}$. This defines $$ C_{l,\infty}(x,y)=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} \frac{e^{-l^2\xi^2}}{\xi^2} e^{i\xi(x-y)}\ d^d\xi\ . $$ The RG is used in order to study quantities of the form $$ \int e^{-V(\phi)}\ d\mu_{l,\infty}(\phi)\ . $$ The idea is to use a ``rescaling to unit lattice'', i.e., a scaling change of variable so one has an integral as before with $l=1$ (with a different $V$ that I will still call $V$ to keep notations simple). Then one uses a decomposition of Gaussian measures $$ \int e^{-V(\phi)}\ d\mu_{1,\infty}(\phi) =\int \int e^{-V(\psi+\zeta)} d\mu_{1,L}(\zeta)d\mu_{L,\infty}(\psi) $$ where $d\mu_{1,L}$ is the Gaussian measure corresponding to the covariance $C_{1,L}=C_{1,\infty}-C_{L,\infty}$ and $L$ is some number $>1$. If one defines the constant $[\phi]=\frac{d-2}{2}$, called the scaling dimension of the field, then the law of the field $\psi(x)$ is the same as that of $\phi_L(x)=L^{-[\phi]}\phi(L^{-1}x)$ where $\phi$ is sampled according to the original measure $d\mu_{1,\infty}$. Hence $$ \int e^{-V(\phi)}\ d\mu_{1,\infty}(\phi) =\int \left(\int e^{-V(\phi_L+\zeta)} d\mu_{1,L}(\zeta)\right)d\mu_{1,\infty}(\phi) $$ $$ =\int e^{-V'(\phi)}\ d\mu_{1,\infty}(\phi) $$ where $$ V'(\phi)=-\log\left( \int e^{-V(\phi_L+\zeta)} d\mu_{1,L}(\zeta) \right)\ . $$ The renormalization group transformation on the space of Lagrangians is the map $V\rightarrow V'$. One can also do this infinitesimally by taking $L\rightarrow 1$, in which case one talks about an RG flow rather than a transformation. In the perturbative RG one writes the dynamical variable $V$ which is a complicated functional of the field as a formal power series in some variable which you can think of as Planck's constant. In the nonperturbative RG one essentially wants to use analysis to control the sum of this series. There are rigorous ways to study both RGs. The perturbative one is of course much simpler.

What you will find in Costello's book is only the perturbative RG. He does treat Yang-Mills in flat space using the Batalin-Vilkovisky formalism, which is quite remarkable for an introductory book. For the curved case, see the paper http://arxiv.org/abs/0705.3340 by S. Hollands which appeared in J. Math. Phys.

If you would be happy learning about the RG flow on $\phi^4$ instead of Yang-Mills, then much simpler perfectly rigorous presentations are available:

the book "Renormalization: an introduction" by Manfred Salmhofer, Springer, 1999.
the review article http://arxiv.org/abs/hep-th/0208211 by Volkhard Mueller which appeared in Rev. Math. Phys.

As for the rigorous nonperturbative RG, the Park City lectures by Brydges mentioned by jc is definitely the best place to start. The issue here is that for Bosons one cannot really take the log in the definition of $V'$. This is called the large field problem, and one algebraic way around it is to use a so-called polymer representation. All this is explained by Brydges. Another nice introduction to the nonperturbative RG for Bosons is a set of (preliminary) lecture notes by Antti Kupiainen (you can find them on Google if you search for "Introduction to the renormalization group kupiainen").

For Fermions, taking the log is not a problem and good mathematical presentation can be found, e.g., in:

the book "Non-perturbative renormalization" by Vieri Mastropietro, World Sci. 2008.
the book "Renormalization group" by Giuseppe Benfatto and Giovanni Gallavotti, Princeton University Press, 1995.
the book "Fermionic Functional Integrals and the Renormalization Group" by Joel Feldman, Horst Knoerrer and Eugene Trubowitz, AMS-CRM, 2002, also available at http://www.math.ubc.ca/~feldman/papers/aisen-all.pdf

Also, for the nonperturbative RG there is another approach which is closer to BPHZ renormalization. It is presented in the book by Vincent Rivasseau http://www.cpht.polytechnique.fr/cpth/rivass/articles/book.ps

If you would like a very short account of the kind of theorems one would like to prove in the nonperturbative RG setting you can also look up my recent Oberwolfach extended abstract: http://arxiv.org/abs/1104.2937

Edit: An in depth study of the rigorous nonperturbative RG is in the paper I just posted on arXiv: Rigorous quantum field theory functional integrals over the p-adics I: anomalous dimensions, with A. Chandra and G. Guadagni.

Answer by Abdelmalek Abdesselam for Invariant polynomials for a product of algebraic groups

2013-02-13T17:33:01Z

It is best to think of these invariants in terms of pictures but it is hard for me to draw one here. You are asking about invariant polynomials in variables $X_{a,b}$ with $1\le a\le n$ and $1\le b\le m$. The invariants are all complete contractions of $X$'s with the following four pieces: $\delta_{a,a'}$ the $n\times n$ Kronecker delta for the $a$'s; $\rho_{b,b'}$ the $m\times m$ Kronecker delta for the $b$'s; $\epsilon_{a_1,\ldots,a_n}$ the completely antisymmetric tensor in the $a$'s normalized so that $\epsilon_{1,2,\ldots,n}=1$ and finally $\eta_{b_1,\ldots,b_m}$ which is the same thing for the $b$'s.

The rule is you only contract $a$'s with $a$'s and $b$'s with $b$'s. Now you get a lot of graphs in this way. If you only use the Kronecker delta, i.e., $\delta$'s and $\rho$'s all you can make are cycles: basically traces of powers of $X^T X$. These must be the invariants in Kac's table.

It turns out that if we now also include the $\epsilon$'s and $\eta$'s then we can get rid of them almost completely. For instance if you have two antisymmetric pieces of the same kind, e.g., two $\epsilon$'s, then it is easy to see that you can trade them for an antisymmetrized sum only involving $\delta$'s. This is nothing more than the formula $det(AB)=det(A)det(B)$. So one is reduced to the case of only one $\epsilon$ and one $\eta$.

If you have an $\epsilon$ and say no $\eta$ then you must have a path of $X^TX$'s starting from a leg of that $\epsilon$ and returning to another leg. This gives zero because it is a contraction of something symmetric against something antisymmetric.

For the remaining case:by the same leg to leg path argument it is easy to see that starting from a leg of the unique $\epsilon$ one must arrive (after an odd number of $X$ steps) to a leg of the unique $\eta$ and vice versa. This can only happen if $n=m$ and the invariant we are considering is in fact $det(X)$. Your list is missing that one if the dimensions are odd.

In fact even if the dimension is even it is impossible to reduce this determinant to a polynomial in the invariants from Kac's list. Take a diagonal $n\times n$ matrix $X$ with eigenvalues $\lambda_1,\ldots,\lambda_n$. From the basic theory of symmetric polynomials it is easy to see that the even power sums are not enough to get the determinant.

Conclusion:

1) The generators are $tr((X^T X)^{p})$ for $p=1,\ldots,m$ when $n>m$ and they are algebraically independent.

2) If $n=m$ one also needs to add $det(X)$. There is a single relation coming from the trace expansion of $(det(X))^2=det(X^T X)$.

Edit: I just realized the conclusion is correct but the argument is a bit more complicated. Indeed in the case with one $\epsilon$ and one $\eta$ one has $n=m$ and several odd paths realizing the connections between the legs of $\epsilon$ and those of $\eta$. The problem is that these paths can have different lengths. But these lengths are all odd and therefore $\ge 1$. One can use the identity

$$\sum_{a_1,\ldots,a_n=1}^{n} \epsilon_{a_1,\ldots,a_n}\ X_{a_1,b_1}\cdots X_{a_n,b_n} =det(X)\times \eta_{b_1,\ldots,b_n}$$

to peal off the first layer of $X$'s growing from the $\epsilon$ and thus reduce to the situation with at least two $\eta$'s.

Edit2: For the statement about no other relation than $(det(X))^2=\ldots$ this can be done easily with symmetric functions of the eigenvalues. Basically one has to show there are no nonzero polynomials $P$ and $Q$ in the even power sums $p_2,p_4,\ldots$ such that $Q\ e_n=P$ where $e_n$ is the determinant,i.e., the $n$-th elementary symmetric function. Then write the expansion of $e_n$ in terms of power sums which contains $p_1^n$ and compare both sides as polynomials in $p_1$ with coefficients in the ring of polynomials in $p_2,p_3,\ldots,p_n$ .

Answer by Abdelmalek Abdesselam for Mathematicians whose works were criticized by contemporaries but became widely accepted later

2013-02-12T13:38:27Z

Galois maybe?

Also, a famous example is Hilbert's work on invariant theory. I don't know if there is truth in the "theology and not mathematics" story regarding Hilbert's first paper with the basis theorem, but in any case it took a while before this new way of doing algebra became accepted.

Answer by Abdelmalek Abdesselam for "Classical" consequences of Bezout's theorem in dimensions $>2$

2013-02-11T14:51:00Z

A small remark: Bezout's theorem is not just about plane curves but includes the higher-dimensional version. See: General Theory of Algebraic Equations. The French original is freely available here.

Answer by Abdelmalek Abdesselam for What, precisely, does Klein's Erlangen Program state?

2013-01-28T18:23:51Z

A small complement:

It is good to have a look at what the contemporaries thought about Klein's program and how it articulated with the big deal of the time: classical invariant theory. In particular Franz Meyer's review mentioned in my answer to MO96140 is a useful reference in this regard (Klein's program is mentioned on pages 19, 42 and 45).

At the opposite extreme of the time axis, one can note that Klein's program continues to be a source of inspiration for mathematicians, see e.g. this article posted today on arXiv by Freed and Hopkins.

Answer by Abdelmalek Abdesselam for What should be offered in undergraduate mathematics that's currently not (or isn't usually)?

2011-05-11T16:58:52Z

Computing with Feynman diagrams in zero dimension, i.e., a graphical calculus for tensor contractions. It is very elementary yet can lead quickly into rather deep mathematics. It would make later studies in say mathematical physics or low dimensional topology much more congenial. Possible applications could be

redoing a good portion of linear algebra see e.g.: http://arxiv.org/abs/0910.1362
doing some basic representation theory following the formalism in the book by Cvitanovic: http://birdtracks.eu/
projective geometry on the line and on the plane and some elimination theory, Bezout's theorem is very easy to understand in this language.
computer graphics in the spirit of J. F. Blinn see, e.g., the account given in: http://www-m10.ma.tum.de/foswiki/pub/Lehrstuhl/PublikationenJRG/52_TensorDiagrams.pdf
since many answers in this post are about asymptotics, another application is to compute the higher order terms of the Laplace method.
combinatorial enumeration, e.g., a proof and examples of application of Lagrange inversion, explicit forms of the implicit function theorem, etc. etc.

Answer by Abdelmalek Abdesselam for New grand projects in contemporary math

2013-01-02T15:37:10Z

Work on the mathematical foundations of quantum field theory. See for instance the recent review by Michael Douglas: "Foundations of quantum field theory" in Proc. Symp. Pure Math. vol 85, 2012. See also this recent Stony Brook conference on the subject.

Answer by Abdelmalek Abdesselam for Invariants for the exceptional complex simple Lie algebra $F_4$

2012-05-09T22:56:10Z

This is more of an extended comment/enlargement of the original question/wild speculation than an answer. Due to my poor recollection of reading, a while back, Klein's Erlangen program and Franz Meyer's review on classical invariant theory, I thought Jose's Q 1 was an easy consequence of a 19th's century theorem which would say (disclaimer: what follows is meta-math):

`Let $G$ be a group acting on some vector space $V$. Let $A$ denote a collection of tensors in some tensor powers of $V$ and its dual (such as $Q$ and $C$ above). Let $H$ be the subgroup of $G$ defined as the stabilizer of $A$. Suppose we know the first fundamental theorem for $G$: namely a description of invariants of $G$ using contractions of a finite set of elementary pieces. Then one can get the FFT of $H$ simply by adjoining the pieces in the collection $A$.'

Suppose for instance one looks at a polynomial $P({\mathbf{F}})$ where $\mathbf{F}$ is some generic $V$-tensor. Suppose $P(g{\mathbf{F}})=P(\mathbf{F})$ for all $g$ in $H$. Then the previous meta-theorem is the statement that there exists a polynomial $\Phi(\mathbf{F},\mathbf{A})$ involving this time a generic tensor or collection of tensors $\mathbf{A}$ of the same format as $A$, such that:

1) $\Phi$ is invariant by the full group $G$, i.e., $\Phi(g\mathbf{F},g\mathbf{A})=\Phi(\mathbf{F},\mathbf{A})$ for all $g$ in $G$.

2) the specialization of $\Phi(\mathbf{F},\mathbf{A})$ to $\mathbf{A}:=A$ is the original subgroup invariant $P(\mathbf{F})$.

If this were true then the FFT for $G=GL(V)$ would be the mother of all FFT's. The statement works for going from $GL(n)$ to $SL(n)$ and taking $A$ to be the epsilon Levi-civita tensor. For othogonal and symplectic groups one takes for $A$ a symmetric or antisymmetric form. But it works also if $G=SL(2)$ and $A=e_1=(1, 0)^T$ so that $H$ is the group of upper triangular matrices (I guess for $n>2$ one would need $A=e_1, e_1\wedge e_2,\ldots$). I thought the meta-theorem was mentioned in Franz Meyer's review, but in fact the only case he explicitly talks about is the last one. He calls that the study of "peninvariants" which is the same as seminvariants sources of covariants of binary forms etc.

Now here is a meta-meta-proof of the meta-theorem:

Let $\Phi(\mathbf{F},\mathbf{A}):=P(g\mathbf{F})$ where $g$ is some element of $G$ such that $\mathbf{A}=g^{-1}A$.

This is well defined since $g_1^{-1}A=g_2^{-1}A$ implies by definition of $H$ that $g_2 g_1^{-1}\in H$. Therefore $P(g_2\mathbf{F})=P((g_2 g_1^{-1})g_1\mathbf{F})= P(g_1\mathbf{F})$ by $H$-invariance of $P$.

This is a $G$-invariant because by construction, if $\mathbf{A}=g^{-1}A$ then $g'\mathbf{A}=(g g'^{-1})^{-1}A$ and so $\Phi(g'\mathbf{F},g'\mathbf{A})=P((gg'^{-1})g'\mathbf{F}) =P(g\mathbf{F})=\Phi(\mathbf{F},\mathbf{A})$.

Now of course there is a catch: one seems to need the $G$-orbit of $A$ to essentially (Zariski dense?) be all of the space where $\mathbf{A}$ belongs. This holds in the previous instances of the meta-theorem. Now when $A=C,Q$ as in Jose's problem, I am less sure. So I am not as convinced now by what I said in my comment above that Jose's $\Phi$ should be expressible in terms of $Q,C,\nu$ alone. Although, Bruce seems to be able to show this, so stay tuned.

In fact one does not seem to need the orbit of $A$ to fill everything, but one needs the existence of a $G$-invariant extension of the $\Phi$ I defined to the full space of $\mathbf{A}$. An idea to do this is to find some extension of $\Phi$ not necessarily invariant and then average it over $G$. For instance if $G=SL(n)$, one needs to take the Haar measure average over $SU(n)$. Usually if one does this the risk is to get zero, but this cannot happen here since the restriction of the average to $A$ should be the original $P$ which must be nonzero, otherwise the problem of expressing $P$ is moot. This sounds too good to be true, in particular in view of Nagata's counterexample for finite generation of rings of invariants. A specialist of algebraic groups would be better able to delineate the boundaries of what is solid math and what is pure fancy in the above arguments.

Since Jose's question 1 really is: is there a FFT for $F_4$? I briefly searched the literature and found this recent paper by Bruno Blind. It has some good references, in particular by Gerald Schwarz who solved this problem for $G_2$, and several papers by Iltyakov (who proves things about $F_4$ but uses notations and definitions I do not understand).

By the way, another paper by Schwarz in AIF is about binary cubics. This seems to be related to the instance of the meta-theorem where $G=SL(4)$ and $H$ is an imbedded $SL(2)$ using the 3rd symmetric power map. The $A$ in this case, I guess would be the twisted cubic curve. In general, one would need a hypersurface given by the Veronese imbedding (Hesse's transfer principle referred to in Klein's program). Classics studied the kind of things addressed in Schwarz's AIF article under the heading of "invariant types" see the book by Grace and Young Ch. XV and XVI.

The above setup also makes sense if $H$ is finite. I wonder if one can prove the fundamental theorem of symmetric polynomials (in a very complicated way) along these lines. I was toying with $A=x_1\cdots x_n$ or $x_1^p+\cdots+x_n^p$ for some well chosen power $p$ but one needs to get rid of a torus or roots of unity as well as solve the extension problem.

Update: I recently came across this article by Alexander Schrijver "Tensor subalgebras and first fundamental theorems in invariant theory". J. Algebra 319 (2008), no. 3, 1305–1319. It is related to what I said above since it deduces the FFT for classical groups from that of $GL(n)$.

Answer by Abdelmalek Abdesselam for The Unreasonable Effectiveness of Physics in Mathematics. Why ? What/how to catch?

2012-12-17T13:38:36Z

It seems to me the answer to: "Do physicists have some tools/ideas/techniques which allow them to make insights, which are not seen for mathematicians?" is indeed yes. Not only such a tool exists but, in my opinion, it is also unique: functional integrals. Predictions based on that tool are what mathematicians have a hard time reproducing and justifying using well-established rigorous theories. Imagine a world where we still would not know how to define an ordinary integral rigorously, e.g., via Riemann sums, but where physicists, engineers etc. use them on a daily basis and with great success. This would be quite similar to the situation today with the heuristic theory of functional integrals developed by physicists. Also, there is an area of mathematics which aims at constructing and studying these objects rigorously: constructive quantum field theory or rigorous renormalization group theory.

Answer by Abdelmalek Abdesselam for Test functions with small support and nonnegative Fourier transform

2012-10-28T15:31:33Z

I didn't think about the non Abelian case, but for a commutative group say like $\mathbb{R}$ you can do the following. Take a function with support in $\frac{1}{2} U$ and symmetric with respect to the origin. Its convolution with itself answers your question.

Answer by Abdelmalek Abdesselam for Computing the relations in invariant algebra

2012-10-24T12:24:51Z

See the book "Invariant Theory of Finite Groups" by Neusel and Smith, as well as this other book by Derksen and Kemper.

Answer by Abdelmalek Abdesselam for 2D Ising model partition function expansion

2012-10-18T13:58:41Z

I don't think Onsager's solution will help you in the presence of a magnetic field. In the low temperature regime there are rigorous expansions for pretty much all the quantities of interest. These go under the names of low temperature cluster expansions, contour expansions, Pirogov-Sinai theory (the most general theory for this kind of things). You can find presentations of these tools especially in the archetypal example of the Ising model in just about any book on Gibbs measures: e.g. the books by Malyshev and Minlos, Preston, Georgii, Sinai, etc. A good place to start is the lecture notes by Velenik. If you do not read French then look up the references in these lecture notes.

Answer by Abdelmalek Abdesselam for Fourier transform of $e^{it|\xi|^{\alpha}}$

2012-10-12T12:26:26Z

For $t=i$ there is a formula involving an integral of a Bessel function, so I doubt there is a simple closed formula for $K_{\alpha}$ in general. You can find the formula I mentioned in the first page of the article "Some theorems on stable processes" by Blumenthal and Getoor. Also see this related MO question.

Answer by Abdelmalek Abdesselam for which homogeneous polynomials split into linear factors?

2012-10-10T23:29:36Z

1) This is the Chow variety of degree $n$ zero cycles in $\mathbb{P}^{d-1}$.

2) Yes, this collection of polynomials can be bundled together into the Brill form or covariant.

3) Rather explicit descriptions of the Brill equations can be found in the book by Gelfand, Kapranov and Zelevinsky on resultants. There is also a paper by Rota and Stein. But first check out Emmanuel Briand's page and in particular the articles "Covariants decomposing on totally decomposable forms" and "Brill's equations for the subvariety of factorizable forms" and if you read French (or German) the translation of the original article by Gordan (respectively the article itself).

As an aside, analogues of the Brill equations for the variety of forms which are powers of forms of degree dividing $n$ have been given recently in my paper with Chipalkatti "On Hilbert covariants".

Answer by Abdelmalek Abdesselam for Tensor powers of the standard representation

2012-10-08T21:14:18Z

The problem has been solved in the reference indicated in the comment by Vasu Vineet, namely: "Combinatorial Operators for Kronecker Powers of Representations of Sn" by Alain Goupil and Cedric Chauve. However, one cannot say that the formulas in Propositions 1 and 2 of this paper are "nice".

Answer by Abdelmalek Abdesselam for Reference request: a conjecture of Rota on positive functions of a random variable

2012-09-21T14:42:46Z

I think your reformulation of the conjecture in the Rota-Shen paper is correct. Also your counterexample is correct. So I guess the conjecture was not stated properly in that article. Perhaps one should restrict to translation invariant polynomials only. By that I mean polynomials in the moments which remain invariant if we change the basic random variable $X$ to $X+c$ where $c$ is some constant, the first example being the variance $m_2-m_1^2$. I think one can make the conjecture more precise by asking the polynomial to be a positive linear combination of evaluations of squares of products of differences of umbral letters.

It is hard to know what Rota's motivation was. It is not clear to me if for him this conjecture was a matter of probability theory or of (real) classical invariant theory. The Hankel determinants which determine moment sequences and motivate this conjecture are essentially the catalecticants of binary forms and their umbral expression with squared Vandermondes is the standard representation of these catalecticants using the classical symbolic method. My feeling is Rota's motivation might be to understand a big problem in classical invariant theory which is how does one know that the evaluation of an umbral polynomial is nonzero (see this very interesting article by Alexandersson and Shapiro).

Also for the proper formulation of the conjecture, I suggest contacting J. Shen. Her latest article, "Least-squares halftoning via human vision system and Markov gradient descent (LS-MGD): algorithm and analysis", SIAM Rev. 51 (2009), no. 3, 567–589, has a contact email address for her. Another person who works in this area and might know what exactly Rota's conjecture was is Elvira Di Nardo who gave lectures at the SLC 67 on cumulants and umbral calculus.

Answer by Abdelmalek Abdesselam for The discriminant for the plane cubic curve

2012-09-18T23:28:16Z

For a less abstract and more computational approach you can for instance look at Example 5.48 in the book "Introduction a la resolution des systemes polynomiaux" by Elkadi and Mourrain (in French). It gives the resultant of three conics as the determinant of an explicit matrix. Apply this to the partial derivatives of your cubic and you will get the discriminant. The method goes back to Sylvester. Also, if you have JSTOR access and don't mind reading rather old fashioned algebra, you can look at this paper by Morley.

Answer by Abdelmalek Abdesselam for Invariants of group action: SL_n acts simultaneously on m symmetric matrices

2012-09-17T14:18:58Z

This is part of classical invariant theory: the study of joint invariants of several quadratic forms. As far as I know these rings of invariants are only known in a few special cases. There has been work by Turnbull and Todd for the $n=3$ case. A recent paper on the subject which contains pointers to the classical literature is ``La théorie des invariants des formes quadratiques ternaires revisitée'' by Bruno Blind.

Answer by Abdelmalek Abdesselam for Faa di Bruno's formula for inverse functions ?

2012-05-31T16:20:22Z

You should be able to get a formula, first by reducing to the case where f(0)=0 and the evaluation of the derivatives (for both f and its inverse) is at 0. Then, work formally by replacing f by its Taylor-MacLaurin series at 0. The problem then becomes that of the reversion of power series. It has been done in many places and typically involves summing over trees.

Shimura-Taniyama-Weil VS Grothendieck's dessins

2012-05-10T22:37:00Z

When listening to the beautiful lectures by Gilles Schaeffer at the SLC68, the following (perhaps crazy) question occurred to me: did anyone attempt (succeed?) to combinatorially prove modularity of elliptic curves using dessins d'enfant?

Of course I am not talking about a combinatorial proof of the general result due to Wiles, Taylor, Breuil, Conrad and Diamond. If such a thing existed, everyone and their dog would have heard about it. I am interested in learning about combinatorial proofs, if any, even for very modest examples. As I do not know anything about the subject, references to the relevant literature would be appreciated.

This question can be broken down into the following three:

1) Can one tell `by looking at a dessin' if the corresponding curve is defined over $\mathbb{Q}$? If this is too hard, can one construct an explicit collection of dessins which catches all elliptic curves defined over $\mathbb{Q}$?

2) Does one know explicit dessins for all modular curves?

3) Let $X\rightarrow\mathbb{P}^1$ and $Y\rightarrow\mathbb{P}^1$ be two coverings given by dessins. Is there some sufficient criteria for the existence of a cover $X\rightarrow Y$?

Crazy addendum to a crazy question:

Can one `count' $H_{X,Y}$ the number of covers in question 3)? Again, I am talking about examples.

Comment by Abdelmalek Abdesselam

2013-05-27T18:39:30Z

Contrary to the situation over the reals, the delta function (up to multiplicative constant) is the only distribution supported at the origin. There is no notion of derivatives in the sense of distributions of \delta and therefore there is no good notion of local derivative. One can define fractional derivatives via Fourier but these are nonlocal. This issue is also related to the classification of homogeneous distributions and the appearance of poles with respect to the homogeneity parameter.

Comment by Abdelmalek Abdesselam

2013-05-08T19:10:58Z

@Jesko: see my last edit.

Comment by Abdelmalek Abdesselam

2013-04-25T17:44:02Z

Not sure this a good example. For some field theories constructed rigorously the function is still analytic even though the Taylor series has zero radius of convergence. The issue is the location of the point around which the Taylor expansion is made: in the middle of the domain of analyticity (ordinary summation which requires positive radius of convergence) versus on the boundary (Borel summation).

Comment by Abdelmalek Abdesselam

2013-04-24T00:03:48Z

@Qiaochu: nice. Thanks.

Comment by Abdelmalek Abdesselam

2013-04-19T21:05:50Z

then it's called a circulant determinant.

Comment by Abdelmalek Abdesselam

2013-04-17T23:29:16Z

Of course. But I think the issue here rather is about published mathematical papers telling insights and intuitions yet without the gory details.

Comment by Abdelmalek Abdesselam

2013-04-15T16:49:36Z

it's done on page 181 of Itzykson-Zuber 1980 edition.

Comment by Abdelmalek Abdesselam

2013-04-03T16:14:38Z

@Steve: You need to give more details, otherwise it is not very respectful of people who would make the effort of answering your question. Is mu a real valued positive measure? if so, is the sigma algebra that of Borel sets etc.

Comment by Abdelmalek Abdesselam

2013-03-28T22:51:36Z

This is very pretty.

Comment by Abdelmalek Abdesselam

2013-03-20T16:27:33Z

@Adam: Thanks I will look at Sec 6.

Comment by Abdelmalek Abdesselam

2013-03-20T13:38:36Z

@Adam: I was looking at this paper yesterday but it is quite long and I have yet to see where the kind of theorem I am interested in is embedded.

Comment by Abdelmalek Abdesselam

2013-03-20T00:52:01Z

@Peter: I would not say "strikingly". My question is quite unsurprisingly related to this kind of questions simply because it explores extensions of normal form theory in the direction of weaker statements than e.g. full linearizations.

Comment by Abdelmalek Abdesselam

2013-03-18T20:38:18Z

@Piotr: degree bounds are not easy either. For binary form CIT there is an old one by Jordan see: www.math.lsa.umich.edu/~hderksen/preprints/bound.ps I don't know if it has been improved recently.

Comment by Abdelmalek Abdesselam

2013-03-04T20:51:59Z

@anon: indeed, there is a lot more in that article than what we now know as the Hilbert Basis Thm: the Hilbert polynomial, the Hilbert Syzygy Thm, the application to rings of invariants via averaging over SU(n) done by way of Cayley's omega operator...

Comment by Abdelmalek Abdesselam

2013-03-04T14:41:29Z

you can define being Gaussian by saying the moments are given by the Isserlis-Wick's theorem or the log-moment generating function is quadratic.