Sunday, 31 December 2006

oc.optimization control - how to estimate a polyhedron(convex hull) classifier from data sample

Given a set of points $XinRe^D$, they have labels $Yin${$-1,+1$}. I would like to separate the data labeled +1 and the data labeled -1 by a polyhedron.



$min_w sum_i xi_i + frac{1}{2}|w|_2^2$



subject to: $ xi_i > max_{j=1}^K[1-(w_j^Tx_i+b_j)]$, for $y_i=+1$



and $ xi_i > min_{j=1}^K[1+(w_j^Tx_i+b_j)]$, for $y_i=-1$
and $ xi_i > 0 $, for all $i$.



Where K is the number of faces of the polyhedron, i represents each sample, j represents each face of the polyhedron. I assume that all positive data go inside the polyhedron while negative data are outside. Following the max-margin principle, we let the distance of the point to the face offset by a margin 1.



Optimizing with the first constraint is straightforward. But the second one seems difficult.



Is there anyway to optimize them in a fast way to the optimal?

ag.algebraic geometry - Endofunctors of CRing which give schemes when composed with schemes?

Here is a nontrivial example I like. Let $W:mathrm{Rings}tomathrm{Rings}$ denote the Witt vector functor of some fixed finite length. (You can consider the $p$-typical Witt vectors, for some prime $p$, but everything works with the other standard flavors.) Then the functor $W_*(-)=-circ W$ is an endofunctor of the category of functors $mathrm{Rings}tomathrm{Sets}$, and it takes schemes to schemes. The scheme $W_*(X)=Xcirc W$ is the so-called arithmetic jet space of $X$, extensively studied by Buium in the case of $p$-adic formal schemes. The fiber over $p$ is the Greenberg transform.



There is a standard method for proving $W_*$ takes schemes to schemes (or rather for proving almost that), though it probably doesn't work for every functor $F:mathrm{Rings}tomathrm{Rings}$ such that $F_*$ takes schemes to schemes. First you show that the category of sheaves of sets on the category of affine schemes w.r.t. the etale topology is stable under $W_*$. This is true since $W$ takes etale covers of rings to the same and also takes cocartesian squares of etale rings maps to the same. (These properties of $W$ are not obvious.) Then you show $W_*$ takes sheaf epimorphisms to the same, and etale maps of sheaves to the same. (These properties are much easier.) Therefore any etale equivalence relation on an affine scheme is sent to an etale equivalence relation on an affine scheme, and the quotient of the first is sent to the quotient of the second. Therefore the category of quasi-compact quasi-separated algebraic spaces is stable under $W_*$. I have no doubt you could find reasonable abstract properties on the endofunctor $W$ of Rings that allow this argument to go through.



Showing $W_*$ takes schemes to schemes is a bit subtler. You have to deal with quasi-compactness issues (as a right adjoint, $W_*$ doesn't necessarily behave well w.r.t. disjoint unions) and also the fact that it's harder to tell whether a functor is represented by a scheme than by an algebraic space. But as I said, in the Witt vector example above, it is true.



Presumably the same argument works, and is much easier, for the functor $F$ defined by $F(R)=R[t]/(t^{n+1})$. Then $F_*(X)$ should be the usual jet space functor of length $n$. The case $n=1$ should give the total space of the tangent bundle, at least when $X$ is smooth.



Edit: I'm reminded below that this example is just a particular case of the representability of the Weil restriction of scalars for a finite flat map $Ato B$. There you consider the endofunctor of the category of $A$-algebras given by $F(R)=Botimes_A R$. In particular, it's reasonable to view $W_*$ as a generalized Weil resitrction of scalars.

Saturday, 30 December 2006

cv.complex variables - Mode of convergence of a power series

An example, due to Fejér, appears in Hille's Analytic function theory on page 122 of the second edition, volume 1.



Erdős wrote a paper showing the existence of classes of examples with many zero coefficients, and in that paper he states that Hardy was the first to find an example of what you're looking for. However, Erdős refers to Landau's book for evidence, and I don't read German, so someone more capable and willing may check that out.



Edit: I initially misnamed the author and book containing the example, as pointed out by Harald Hanche-Olsen.



Update: Coincidentally, while skimming through Sasane's Algebras of holomorphic functions and control theory, I came upon another reference to Hardy's example. This time the reference was to Dienes's The Taylor series. I checked, and sure enough there is Hardy's example in the chapter "The Taylor series on its circle of convergence." I scanned it, along with a page containing a result used in the example. Here it is. (The example starts two thirds of the way down page 464, and the cited "105.IV" is the "IV" that appears on page 441.)

soft question - Math Vs Social Science

I will suppose you have an algebraic taste, and that you want to develop theory rather than analyze data.



A starting point is the observation that relationships in a social group can be modeled by a semigroup. Among the leaders of this area, developed around the 70's, you can find Harrison White. He completed a PhD in theoretical physics at MIT, then a PhD in sociology at Princeton. Currently, he is at Columbia. A serious trouble of this approach (from my point of view) is that relationships change over time, so we do not have a stable algebraic structure. There are ways around it... If you want to follow this lead, then I recommend you the book Algebraic Models for Social Networks by Philippa Pattison.



Another observation is that social sciences are full of relationships we don't measure with numbers. So, a categorical-functorial approach has been developed. In that sense Lorrain and White wrote an enjoyable article:
http://www.tandfonline.com/doi/pdf/10.1080/0022250X.1971.9989788
I would like to know how to use some category theory in this context...



I must say many sociologists dislike this approach a lot, because they consider it reductionist. I am afraid research in this area may fall in no man's land i.e neither mathematicians nor sociologists read it. However, there are some groups trying to develop this field. In particular, UC Irvine has the Institute for Mathematical Behavioral Sciences: http://www.imbs.uci.edu/. I also recomend to look up for Santa fe institute.



On other hand, a big problem for social sciences is measurement. Suppose, you develop a social theory (for example [1]). To validate the model you need to test it with reality. This is a big issue in social sciences, because you cannot take people inside a box, isolated other variables, and measure a specific behavior. Plus, real life events have many variables to account for. However, these days Google, Facebook, NSA, etc, are collecting a lot of social generated data; and there is not good theory for this sort of data. The void is being filled by computer scientists; for example Carnegie Mellon has a PhD program at its Center for Computational Analysis of Social and Organizational Systems. I attended their main conference some time ago...It impressed me their focus on extract, analyze, and represent relational data rather than understand it (in the sense of constructing theory). That perspective can be very useful, but I dislike it. Hopefully, social generated data can be used one day for constructing theories. (Noam Chomsky argued for developing theory here:
http://www.theatlantic.com/technology/archive/2012/11/noam-chomsky-on-where-artificial-intelligence-went-wrong/261637/).



[1] "A New Model of Wage Determination and Wage Inequality." Rationality and Society, 2009 by Guillermina Jasso

Friday, 29 December 2006

set theory - Inaccessible cardinals and Andrew Wiles's proof

I'm writing a new community wiki answer because it seems to me the consensus in the comments is that the accepted answer doesn't really tell the right story, and since this is something that pops up all the time it'd be good to have a single place to point people without making them read through all the comments. Please improve my answer.



In the most naive sense Wiles proof does depend on existence of Grothendieck universes (and thus on existence of inaccessible cardinals). By this I mean if you took every reference in Wiles proof and read the first published proof of that fact you'd certainly end up somewhere in SGA where, due to Grothendieck's love of generalization, you'd find universes popping up.



However, this certainly doesn't mean the proof really uses universes. It's widely believed (though for some people this belief may not come from much direct evidence) that in any practical situation you don't actually need universes. However, there are some concrete situations (BCnrd mentions some involving sheafification on the crystalline site) where it's not necessarily known how to eliminate the use of universes.



As a result, in order to figure out if Wiles's proof uses universes, or whether it's relatively easy to avoid them, you'd need to either read the proofs yourself or find someone who was both deeply familiar with the details of the proof, and someone who cares a lot about details. One person who comes quickly to mind is BCnrd. BCnrd was one of the mathematicians who proved the Modularity Theorem, which showed that all elliptic curves over $mathbb{Q}$ are modular. This is a strengthening of Taylor and Wiles' result, which applied only to semi-stable elliptic curves, and the proof involved understanding and building on Taylor and Wiles' work. BCnrd is also famous for his attention to detail and for consulting underlying foundational sources; he is the author of a book dedicated to simplifying and correcting the presentation of Grothendieck duality in Hartshorne's book Residues and Duality.



As explained in the comments to Pete's answer, BCnrd says there's really no issue at all in Wiles's proof. All of the specific things that Wiles uses stay far away from any of the difficult issues where you might be worried about needing to invoke universes.

soft question - Dimension Leaps

Here is a closely related pair of examples from operator theory, von Neumann's inequality and the theory of unitary dilations of contractions on Hilbert space, where things work for 1 or 2 variables but not for 3 or more.



In one variable, von Neumann's inequality says that if $T$ is an operator on a (complex) Hilbert space $H$ with $|T|leq1$ and $p$ is in $mathbb{C}[z]$, then $|p(T)|leqsup{|p(z)|:|z|=1}$. Szőkefalvi-Nagy's dilation theorem says that (with the same assumptions on $T$) there is a unitary operator $U$ on a Hilbert space $K$ containing $H$ such that if $P:Kto H$ denotes orthogonal projection of $K$ onto $H$, then $T^n=PU^n|_H$ for each positive integer $n$.



These results extend to two commuting variables, as Ando proved in 1963. If $T_1$ and $T_2$ are commuting contractions on $H$, Ando's theorem says that there are commuting unitary operators $U_1$ and $U_2$ on a Hilbert space $K$ containing $H$ such that if $P:Kto H$ denotes orthogonal projection of $K$ onto $H$, then $T_1^{n_1}T_2^{n_2}=PU_1^{n_1}U_2^{n_2}|_H$ for each pair of nonnegative integers $n_1$ and $n_2$. This extension of Sz.-Nagy's theorem has the extension of von Neumann's inequality as a corollary: If $T_1$ and $T_2$ are commuting contractions on a Hilbert space and $p$ is in $mathbb{C}[z_1,z_2]$, then $|p(T_1,T_2)|leqsup{|p(z_1,z_2)|:|z_1|=|z_2|=1}$.



Things aren't so nice in 3 (or more) variables. Parrott showed in 1970 that 3 or more commuting contractions need not have commuting unitary dilations. Even worse, the analogues of von Neumann's inequality don't hold for $n$-tuples of commuting contractions when $ngeq3$. Some have considered the problem of quantifying how badly the inequalities can fail. Let $K_n$ denote the infimum of the set of those positive constants $K$ such that if $T_1,ldots,T_n$ are commuting contractions and $p$ is in $mathbb{C}[z_1,ldots,z_n]$, then $|p(T_1,ldots,T_n)|leq Kcdotsup{|p(z_1,ldots,z_n)|:|z_1|=cdots=|z_n|=1}$. So von Neumann's inequality says that $K_1=1$, and Ando's Theorem yields $K_2=1$. It is known in general that $K_ngeqfrac{sqrt{n}}{11}$. When $n>2$, it is not known whether $K_nltinfty$.



See Paulsen's book (2002) for more. On page 69 he writes:




The fact that von Neumann’s inequality holds for two commuting contractions
but not three or more is still the source of many surprising results and
intriguing questions. Many deep results about analytic functions come
from this dichotomy. For example, Agler [used] Ando’s theorem to deduce an
analogue of the classical Nevanlinna–Pick interpolation formula
for analytic functions on the bidisk. Because of the failure of a von
Neumann inequality for three or more commuting contractions, the analogous
formula for the tridisk is known to be false, and the problem of finding the
correct analogue of the Nevanlinna–Pick formula for polydisks
in three or more variables remains open.


Wednesday, 27 December 2006

nt.number theory - Quadratic forms that evaluate to zero mod p only when their input is zero.

Let $Q$ be a quadratic form in $n$ variables with integer coefficients. Let us say that $Q$ has the "special property" mod $p$, if the relation $Q(x_1,...,x_n)=0$ (mod $p$) implies that $(x_1,...,x_n)=(0,...,0)$ (mod $p$). (There must be a name for this property, but I don't know it, which is why I'm calling it "the special property".) Let us say that $Q$ is "special infinitely often" if there are infinitely many primes $p$ such that $Q$ has the special property mod $p$. For example, the one-variable quadratic form $Q(x)=x^2$ is special infinitely often. Another simple example is that $Q(x,y)=x^2+y^2$ is special infinitely often because it is special mod $p$ whenever -1 is a quadratic non-residue mod $p$. In fact, any two-variable quadratic form that is non-degenerate over the rationals is special infinitely often because $Q(x,y)=ax^2+bxy+cy^2$ is special whenever $p$ is odd and the discriminant $b^2-4ac$ is a quadratic non-residue mod $p$.



I've been trying to find a quadratic form in more than two variables that is special infinitely often, but I'm doubtful that such a thing exists. As far as I know, each of the following statements could either be valid or invalid. (Although obviously [c] implies [b] implies [a].)



[a] For every integer quadratic form $Q$ in three or more variables, the set of $p$ such that $Q$ has the special property mod $p$ is finite.



[b] The set of primes $p$ such that there exists an integer quadratic form $Q$ in three or more variables having the special property mod $p$ is finite.



[c] The set of primes $p$ such that there exists an integer quadratic form $Q$ in three or more variables having the special property mod $p$ is empty.



Can anyone help resolve any of these questions?

rt.representation theory - Innocent question on tensor products of modular representations

Jim has already given the correct answer "no".



Here is a hopefully instructive example. Let $k$ be an alg. closed field of pos. char $p$, and let $G = SL_2(k)$. Write $V = k^2$ for the "natural" 2-dimensional representation of $G$ say with basis $e_1,e_2$. Let $W = S^pV$ be the $p$-th symmetric power of $V$. Then $W$ contains a 2 dimensional submodule $A$ spanned by the $p$-th powers $e_1^p$ and $e_2^p$; the module $A$ is isom. to the "first Frobenius twist" of $V$.



It is an exercise to check that there is no $G$-stable complement to $A$ in $W$; i.e. the SES
$$0 to A to W to W/A to 0$$
is not split.
Thus
$W$ is not completely reducible. Evidently there is a surjective mapping
$V^{otimes p} to W$, thus also the $p$-th tensor power $V^{otimes p}$ is not completely
reducible.



But $V$ is a simple (hence completely reducible) $G$-module; thus tensor powers of a completely reducible module are not in general completely reducible. In fact,
the $(p-1)$-th tensor power $V^{otimes p-1}$ is completely reducible;
arguing as before, one sees that $V otimes (V^{otimes p-1})$ is not completely
reducible; thus in general the tensor product of two completely reducible modules
is not completely reducible.



I gave some further remarks about semisimplicity of tensor products in an answer to
this question.

The homotopy cofiber of the smash product of two maps of spectra

It is a standard fact that smashing with a fixed spectrum $Z$ preserves cofiber sequences. So if I have a cofiber sequence $$X xrightarrow{f} Y rightarrow C_f$$ then there is also a cofiber sequence $$Z wedge X rightarrow Z wedge Y rightarrow Z wedge C_f$$



If more generally I have a map $Z xrightarrow{g} W$, is there any formula for the cofiber of the map $$Z wedge X xrightarrow{g wedge f} W wedge Y$$ in terms of $C_f$ and $C_g$? (The above discussion corresponding to $g = mathrm{id}_Z$).

ag.algebraic geometry - Invariant Polynomials under a Group Action (hidden GIT)

The actions of $S_n$ and $mathbb Z_n$ differ in the sense that in the first case the quotient is smooth (it is again $mathbb C^n$) while in the second case it is singular. This is why in the fist case we have a nice presentation, but in the second not really. For example, the number of generators of the quotient can not be less than the dimension of Zariski tangent space to the singularity at zero of $mathbb C^n/mathbb Z_n$.



Still in principle the presentation can be provided by toric geometry (http://www.cs.amherst.edu/~dac/toric.html) because the quotient is the toric singularity. For example, in your case of $mathbb C^3/mathbb Z_3$ let us change the coordinates so that $mathbb Z_3$ is acting as $w_0to w_0$, $w_1to mu w_1$, $w_2to mu^2 w_2$ (here $mu^3=1$). Then you can write the minimal set of four generators:



$w_0, w_1^3, w_2^3, w_1w_2$, and one obvious relation $(w_1^3w_2^3)=(w_1w_2)^3$



The case $mathbb C^n/mathbb Z_n$ for $n>3$ will be more involved, but the idea is the same roughly. First you chose the coordinates on $mathbb C^n$ $w for which the action is diagonal. Then pick the minimal set of monomials (in these new coordinates) that are invariant under the action, and generate the whole set of invariant monomials (of positive degree).



Consider one more case $n=4$, and chose the coordinates $w_i$, so that $Z_4$ is acting as $w_ito mu^iw_i$, $mu^4=1$. The number of generators is $7$ this time:



$w_0, w_1^4, w_3^4, w_2^2, w_1w_3, w_1^2w_2, w_3^2w_2$

Monday, 25 December 2006

sp.spectral theory - Is there a name for this differential operator and/or its corresponding spectrum?

Let $mathcal{M}$ be a real, compact, orientable manifold and let $X$ be a vector field on $mathcal{M}$. Consider the functional



$$E(f) = int_{mathcal{M}} X_p(f)^2 dV$$



where $X_p(f)$ is the directional (Lie) derivative of $f$ along $X$ at the point $p$ and $dV$ is a volume form on $mathcal{M}$ -- this functional essentially measures the total amount of change in $f$ along $X$ over all of $mathcal{M}$ in the $L^2$ sense. Then $delta E(f)$ is a differential operator whose eigenspectrum



$$delta E(f) = lambda f$$



(for $lambda in mathbb{R}$) yields the critical points of $E$ over the set of functions with unit norm. Is there an established name for this operator (or functional) and/or its corresponding eigenspectrum?



The prototype for this operator is Dirichlet's energy



$$int_{mathcal{M}} ||nabla f||^2 dV$$



which has as its (unit-norm) critical points the Laplacian eigenspectrum



$$nabla^2 f = lambda f,$$



the main difference being that Dirichlet's energy measures the total gradient, i.e., the change in all directions, rather than just the change along a particular direction at each point.

Saturday, 23 December 2006

ag.algebraic geometry - Detecting etale maps on reduced points

Suppose I have a morphism of schemes for which I know the relative cotangent complex is trivial, and the map on reduced subschemes is an isomorphism. Is the map an isomorphism?
More generally, given a morphism of schemes with zero relative cotangent complex, which is of finite presentation on the reduced points. Is the map of finite presentation, and thus etale?



(Maybe a better way to phrase this is - what's the reference for these statements?
are they in SGA or in Illusie somewhere?)

genetics - Why was it so hard to decode the corn genome?

Why was decoding the genome so significant?
Because decoding the genome gives us a complete picture of the genetic makeup of an organism.



What made it so difficult?
The repetitive sequences as mentioned by other people is a main problem. Imaging you are trying to complete a big puzzle with millions of pieces. Each piece represents a sequencing reads that we get from the experiment. If there are repetitive sequences in the genome, basically the pieces that representing these regions would look very similar (or even entirely identical). Undoubtedly, this would make the puzzle very difficult to finish. To give a scale, the maize genome is ~2.3 billion base pairs and each sequencing reads is <1,000 base pairs.

Friday, 22 December 2006

ca.analysis and odes - Asymptotics of Hermite and hypergeometric function

I am looking for the asymptotics of the following integral




$int_{mathbb{R}} H_m^2(x) {rm e}^{-2 alpha^2 x^2} {rm d} x = 2^{m-1/2} alpha^{-2m -1} (1-2alpha^2)^m Gamma(m+1/2) ~ _2F_1left(-m,m,1/2-m,frac{alpha^2}{2alpha^2-1}right)$




where $H_m$ is the $m^{rm th}$ Hermite polynomial (orthogonal under the weight ${rm e}^{-x^2}$), and $_2F_1$ is the hypergeometric function.



I found this formula from p. 803 of "Table of Integrals, Series, and Products" by Gradshteyn-Ryzhik. However, I have idea about the asymptotics of the $_2F_1$ term. Can anyone enlighten me on the asymptotics of




$_2F_1left(-m,m,1/2-m,betaright)$




when $m$ is large? In fact I tried mathematica and it seems $_2F_1left(-m,m,1/2-m,betaright) sim |4 beta|^m$. Any reference on this issue?



Now given the above asymptotics is true, observe that the norm of $H_m$ under the weight ${rm e}^{-2 alpha^2 x^2}$ has the same exponent for all $alpha$, including the original weight ($alpha^2 = 1/2$). Is this a common phenomenon for orthogonal polynomials?

pr.probability - Is every probability space a factor space of the Haar Measure on some group?

It is possible to find the following: A compact abelian group G with Haar measure $mu_G$, a subset $Ssubseteq G$ of full outer Haar measure and a measurable function $fcolon Sto X$ with $mu_P(E)=mu_S(f^{-1}(E))$ for measurable $Esubseteq P$. In fact, as you mention, G can be taken to be a large enough product of the circle group.



Here, I am implicitly referring to the sigma algebra $mathcal{B}(S)equiv{Scap Ecolon Einmathcal{B}(G)}$ and $mu_S$ is the restriction of the Haar measure to $mathcal{B}(S)$, $mu_S(Scap E)=mu_G(E)$. Ideally, we would like to enlarge S so that it is actually of full measure, then it could be enlarged to all of G. I'm not sure if this is possible though (and suspect that it is not possible in ZFC). The problem is that if $fcolon Pto Q$ is a measure preserving map of probability spaces then $f(P)$ will be of full outer measure, but need not be measurable. If $f^{-1}colonmathcal{B}_Qtomathcal{B}_P$ is an onto map of their sigma algebras then P and Q are "almost isomorphic" probability spaces. If, however, $f(P)$ is not measurable then f does not have a right inverse (even up to zero probability sets), unless we restrict to the subset $f(P)subseteq Q$.



In the following, I write 2={0,1}, so that, for a set I, 2^I is the set of {0,1}-valued functions from I.
Letting $pi_icolon2^Ito{0,1}$ be the projection onto the i'th coordinate, the sets of the form $pi_i^{-1}(S)$ generate a sigma algebra, which I will denote by $mathcal{E}_I$.



First, we can reduce the problem to that of measures on $2^I$.




Step 1: Given a collection ${A_i}_{iin I}$ of sets generating the sigma algebra on P, construct a probability measure $mu_I$ on $2^I$ and a measure preserving map $fcolon Pto 2^I$ such that $f^{-1}colonmathcal{E}_Itomathcal{B}_P$ is onto.




Then, f will have a right inverse $gcolon f(P)to P$ which is automatically measurable and measure preserving.
To construct f, define $f(p)in 2^I$ by $f(p)(i)=1$ if $pin A_i$ and =0 otherwise. It can be checked that $mu_I(S)=mu(f^{-1}(S))$ for $Sinmathcal{E}_I$ satisfies the required properties.



Now, I will use $G^I=(mathbb{R/Z})^I$, which is a compact abelian group with Haar measure $mu_{G^I}$, which is the product of the uniform measure on the circle $G=mathbb{R/Z}$.




Step 2: Construct a measure preserving map $fcolon G^Ito 2^I$.




Once this map is constructed, putting it together with step 1 gives what I claimed.



For any $Jsubseteq I$, use $pi_Jcolon 2^Ito2^J$ and $rhocolon G^Ito G^J$ to denote restriction to J. Also use $mu_J(S)=mu_I(pi_J^{-1}(S))$ for the induced measure on $2^J$.
Zorn's lemma guarantees the existence of a subset $Jsubset I$ and measure preserving map $fcolon G_Jto 2^J$ which is maximal in the following sense: for $Jsubseteq Ksubseteq I$ and measure preserving $gcolon G^Kto2^K$ with $pi_Jcirc g=fcircrho_J$ then K=J.



Then, J=I and we have constructed the required map. If not, choose any $kin Isetminus J$, $K=Jcup{k}$ and define $h_kcolon G^Kto 2$ as follows (I let $A_k=pi^{-1}_k(1)subseteq2^I$).



$$
h_k(x)=begin{cases}
1,&textrm{if }0le x_klemu_I(A_kmidmathcal{E}_J)_{f(x)}\\
0,&textrm{otherwise}.
end{cases}
$$



Defining $gcolon G^Kto2^K$ by $g(x)_i=f(x)_i$ for $iin J$ and $g(x)_k=h_k(x)$ gives a measure preserving map extending f, and contradicting the maximality of J.

Thursday, 21 December 2006

nt.number theory - adding an n-th root to Q_p

If $n$ is prime to $p$, then ${mathbb Q}_p(a^{1/n})$ is unramified if $n | v_p(a)$,
and is tamely ramified otherwise.
To see this, we note that we may first of all divide $a$ by powers of $p^n$, and so assume
that $0 leq v_p(a) < n.$



If in fact $v_p(a)=0$, i.e. $a$ is a unit, then the extension is unramified, and the ring
of integers is equal to ${mathbb Z}_p[a^{1/n}]$ (by Hensel's lemma, since $x^n - a$
is then a separable equation mod $p$).



Otherwise, if $0 < v_p(a) < n,$ we get a tamely ramified extension (essentially by the
definition of tamely ramified).



If $p | n$ then the situation is a little more complicated. For example, if $n = p$
and $0 < v_p(a) < p,$ then the extension is wildy ramified.



If $a$ is a unit, then we may write $a = zeta u,$ where $zeta$ is a $(p-1)$st root of 1
and $u equiv 1 bmod p,$ and since $zeta^p = zeta,$ we see that
${mathbb Q}_p(a^{1/p}) = {mathbb Q}_p(u^{1/p}).$
Now (supposing that $p$ is odd, for simplicity) if $u equiv 1 bmod p^2,$ then $u$ is
in fact a $p$th power in ${mathbb Q}_p,$ and so the extension is trivial. On
the other hand, if $u equiv 1 bmod p,$ but not mod $p^2$, then the extension is
wildy ramified of degree $p$, with ring of integers equal to ${mathbb Z}_p[u^{1/p}].$



To see this last claim, note that if $X^p - u = 0,$ and we write $Y = X - 1$,
then $(Y + 1)^p - u = 0,$ i.e. $Y^p + pY^{p-1} + cdots + p Y + (u-1) = 0,$
and so $Y$ satisfies an Eisenstein polynomial of degree $p$. This implies that the extension is wildly ramified of degree $p$, that $Y$ is a uniformizer in the extension,
and that the ring of integers is equal to ${mathbb Z}_p[Y] = {mathbb Z}_p[u^{1/p}].$



Added in response to Keith Conrad's comments below: As Keith points out, the extension
${mathbb Q}_p(a^{1/n})$ is not really well-defined unless ${mathbb Q}_p$ contains the
$n$th roots of $1$, or equivalently, if $n$ divides $p-1$ (or 2 if $p = 2$).



But note e.g. if $p$ does not divide $n$, then adding the $n$th roots of unity gives
an unramified extension of ${mathbb Q}_p(a^{1/n})$, and so the ramification behaviour
is independent of the choice of $n$th root, while in the case when $n = p$ also treated above,
adjoining the $p$th roots of unity is a tamely ramified extension of ${mathbb Q}_p$,
so the claims regarding wild ramification are independent of the choice of $p$th root.

st.statistics - Where can i find 'getting started' resources for statistical prediction

I wanted to learn prediction, forecasting etc. I also have time series data on millions of online videos. I would like to test out prediction algorithms etc on this data set, for eg. Linear Prediction, Kalman filter.



Are there any good resources out there to get me started on those?



Edit: Let me rephrase the question. If you were given such a time series data set with a million videos with some number of attributes to each, what steps/path path would you take to come up with a decent view prediction scheme? For example, factor analysis or PCA etc first.

Tuesday, 19 December 2006

ag.algebraic geometry - How does one find vanishing algebraic cycles?

I have a question, related to what I asked before.
Let's consider a smooth hyperplane section $X$ of a smooth projective variety $Y$ over $mathbb C$.
According to Weak Lefschetz theorem, cohomology groups of $X$ coincide with those of $Y$ in all dimensions except for the middle one. In the middle dimensions the pull-back $i^*: H^d(Y) to H^d(X)$ is injective, but not surjective, and the "new" cycles on $X$ are called vanishing cycles. (The reason for such a name is that these "new" cycles will vanish when we approach singular fibers on the Lefschetz pencil.)
Vanishing cycles also can be described as the ones that live in the kernel of $i_*: H^d(X) to H^{d+2}(Y)$.



Let's consider the case when $X$ is even-dimensional, so that we can hope that the vanishing cycles are algebraic.



For example, for a smooth even-dimensional quadric in $mathbb {CP}^n$ there exist one vanishing cycle - it is a difference $[E_1] - [E_2]$ of two maximal linear subspaces from different classes.



Another example I thought about is a smooth cubic surface in $mathbb {CP}^3$, vanishing cycles here are generated by differences of pairs of lines $[l_1] - [l_2]$ lying on the cubic.



Now I wanted to ask, what are other examples people have in mind?
I'm interested in the case, when vanishing cycles actually are algebraic.



Is there a general method to describe such cycles in concrete situations (like MG(3,6))?



Thanks



EDIT: Probably winter break is not a best time to start a bounty...

biochemistry - Decreasing the alcohol proof and faster in hangover, why?

I would argue that your central point is intrinsically flawed. Your claim is that starting at a high alcohol percentage, then gradually decreasing alcohol content, gets you intoxicated faster. @nico is correct that you will get drunk faster this way (more alcohol sooner rather than later, so this is trivial). If we allow only one variable to change, the alcohol content (and therefore the water content), then the first drink is the most potent, and each drink afterwards is getting progressively more dilute, then this implies you are able to rehydrate better the more you drink. This argues exactly opposite to your claim.



Since you offered up the example of vodka to wine, this introduces two new important aspects: fermentation impurities and sugar content. Vodka is distilled and relatively pure, so it offers little in addition to alcohol and therefore easy to process by the liver. Scotchs and whiskeys are low sugar, higher impurities, and so would be a bit more troubling to filter and process. As you move down to wine, impurities rise and sugar content is much higher. This added sugar load dehydrates faster as well as causes a cascade of hormonal changes that are more immediately troubling than the alcohol. Beer would generally be the extreme end of the impurity and sugar continuum.



Since you bring up the USSR and the KGB, I imagine that if they did stick to this regimen of interrogation, it could be more effective by power of suggestion. The victim believes they are more intoxicated than they are, and perhaps more likely to offer up their secrets.

Monday, 18 December 2006

evolution - Alternatives to fittest-win and Moran processes as simple mathematical models of selection

When modeling selective sweeps as a micro-building block in models of macroevolution (not to be confused with misuses of this in creationist arguments), I use the fittest-win model of selection as a first approximation, or Moran process model when I want a more reasonable approximation.



In the fittest win model the probability for a mutant of fitness r to invade a host population of fitness 1 is 100% if r > 1 and 0 otherwise. In the Moran process model, the mutant of fitness r invades with probability $frac{1 - r^{-1}}{1 - r^{-n}}$ for a finite population. Alternative in the limit as n goes to infinite, a mutant with r > 1 invades with probability $1 - frac{1}{r}$ and 0 otherwise.



In general I am interested in simple models of selection of a single (or small concentration of) mutant invading an asexual host population of fitness 1 (with fitness constant and independent of frequency). Are there other common mathematical models for selection of this type?

ag.algebraic geometry - Is a 'generic' variety nonsingular? Or singular?

Executive summary: If you look at the whole Hilbert scheme associated to a given polynomial, the locus of points corresponding to nonsingular (which I take to mean smooth) subschemes can sometimes be very small in terms of dimension and number of irreducible components. So in this sense, most subschemes are singular.



Details: The Hilbert scheme $operatorname{Hilb}^P_{mathbf{P}^n}$ associated to a given Hilbert polynomial $P$ is connected (a theorem of Hartshorne), but in general it has many irreducible components, each with its own generic point. Thus there are several different "generic" closed subschemes with the same Hilbert polynomial, each a member of a different family.



The locus of points in the Hilbert scheme corresponding to smooth (=nonsingular) subschemes of $mathbf{P}^n$ is a Zariski open subset, which implies that it is Zariski dense in the union of the components that it meets, but there are often other components of the Hilbert scheme all of whose points correspond to singular subschemes.



Because the Hilbert scheme need not have a single generic point, one might ask: How many of these generic points parametrize singular subschemes, and what are the dimensions of the corresponding components of the Hilbert scheme?



As a case study, consider the Hilbert scheme $H_{d,n}$ of $d$ points in $mathbf{P}^n$, i.e., the case where $P$ is the constant polynomial $d$. Points of $H_{d,n}$ over a field $k$ correspond to $0$-dimensional subschemes $X subseteq mathbf{P}^n$ of length $d$, or in other words, such that $dim_k Gamma(X,mathcal{O}_X) = d$. Each smooth $X$ with this Hilbert polynomial is a disjoint union of $d$ distinct points. These smooth $X$'s correspond to points of an irreducible subscheme of $H_{d,n}$, and the closure of this irreducible subscheme is a $dn$-dimensional irreducible component $R_{d,n}$ of $H_{d,n}$. Sometimes $H_{d,n}=R_{d,n}$, which means that every $X$ is smoothable. But for each fixed $n ge 3$, Iarrobino observed that $dim H_{d,n}$ grows much faster than $dim R_{d,n}$ as $d to infty$. (He proved this by writing down large families of $0$-dimensional subschemes, like $operatorname{Spec} (k[x_1,ldots,x_n]/mathfrak{m}^r)/V$, where $mathfrak{m}=(x_1,ldots,x_n)$ and $V$ ranges over subspaces of a fixed dimension in $mathfrak{m}^{r-1}/mathfrak{m}^r$.) This shows that $H_{d,n}$ is not irreducible for such $d$ and $n$, and that the ``bad'' components all of whose points parametrize singular subschemes can have much larger dimension than the one component in which a dense open subset of points parametrize smooth subschemes. With a little more work, one can show that the number of irreducible components of $H_{d,n}$ can be arbitrarily large (and as already remarked, the components themselves can have larger dimension than $R_{n,d}$). So in this sense, one could say that for $n ge 3$, most $0$-dimensional subschemes in $mathbf{P}^n$ are singular.



For more details about $H_{d,n}$, including explicit examples of nonsmoothable $0$-dimensional schemes, see the following articles and the references cited therein:



The moduli space of commutative algebras of finite rank



Hilbert schemes of 8 points



(Warning: my notation $H_{d,n}$ is different from the notation of those articles.)

Friday, 15 December 2006

Haar measure on a quotient, References for.

I remember reading Weil's "Basic Number Theory" and giving up after a while. Now I find myself thinking of it(thanks to some comments by Ben Linowitz).



Right from the very beginning, Weil uses the fact that when you have a locally compact topolgocal group $G$ and a locally compact subgroup $H$, in addition to the Haar measures on $G$ and $H$, there exists a "Haar measure" on the coset space $G/H$, with some properties.



For instance, the upper half plane $mathbb H$ is the quotient $SL_2(mathbb R)/SO_2(mathbb R)$ and the usual measure there which gives rise to the usual hyperbolic metric, is arising in this way.



I originally assumed this theorem and went ahead(but not much) with that book.



I want to have a reference for the above theorem. A reference which is not written by Weil. I find him very hard to penetrate. This should exclude Bourbaki's "Integration", as I supppose it would be heavily influenced by him, and thus is a horrible book(note to Harry: this is personal opinion; spare me the brickbats).



I had originally seen the construction of Haar measure on H. Royden's "Real Analysis", in which he is not considering any quotients.

Simplicial Covering Map

In Rezk's paper "A model for the homotopy theory of homotopy theory" numerous references to simplicial covering maps are made. It's first appearance being at the bottom of page 8. Unfortunately no definition is provided in the paper and I was wondering if there is a purely combinatorial definition for this concept or whether we have to pass to the geometric realization.



Maps of simplicial sets already match cells of the same dimension (roughly speaking), but it is the evenly covered concept that requires some work (I imagine).



Any help would be appreciated.

pr.probability - devise a joint distribution of $alpha$ and $beta$

I am not entirely certain what your question is. It might be



(i) Is it always possible to find a joint distribution of $(alpha, beta)$ for any prescribed distributions of $alpha, beta$ and $alpha / beta$ ?



(ii) Is it possible to find/calculate a joint distribution from the three distributions when you know the joint distribution exists e.g. because these are observations in an experiment?



(i) is not possible in general. Set $alpha = exp(X)$ and $beta = exp(-Y)$ then $log(alpha / beta) = X + Y$.
Now let $X$ and $Y$ be uniform on $[0,1]$ and choose a distribution for $X+Y$ so that $P(X+Y < 0.5) = 1$. This means $P(X > 0.5) = 0$ a contradiction to uniform. A way to visualize this might be looking at mass distributions on the square $[0,1]times[0,1]$. Prescribing the margins (here uniform) is a restriction on the projections to the axes (i.e. $0times[0,1]$ and $[0,1]times 0$) and the remaining freedom is distributing the mass in the square.



(ii) Looks more like statistics than probability. There are a number of ways of coming up with a joint distribution. But you would need to specify more context to find a reasonable approach.

fa.functional analysis - Isomorphisms of Banach Spaces

Indeed, $ell_1$ provides a strong counterexample. As noted by Matt, the spaces C(X), where X is countable and compact, provide nonisomorphic Banach spaces whose duals are isomorphic to $ell_1$. If X is countable and compact (and Hausdorff, of course!) then X is homeomorphic to a closed ordinal interval [0, a] (equipped with its natural order topology) for some countable ordinal a; this result is known as the Mazurkiewicz-Sierpinski theorem. About 50 years ago Bessaga and Pelczynski showed that if a and b are infinite countable ordinals and a < b, then C([0, a]) is isomorphic to C([0, b]) if and only if b < a^*w*, where w denotes the first infinite ordinal. Thus C([0, w]) is in fact isomorphic to C([0, w^2]), contrary to Gerald's assertion above. Moreover, the least infinite ordinal b such that C([0, b]) is not isomorphic to C([0, w]) is b = w^w. Combining the Mazurkiewicz-Sierpinski result with the Bessaga=Pelczynski result, one has that each space C(X), where X is countable and compact, is isomorphic to C([0, w^(w^*a*)]) for a unique countable ordinal a. In particular, there are uncountably many nonisomorphic Banach spaces whose duals are isometrically isomorphic to $ell_1$, namely the spaces C([0, w^(w^*a*)]), where a varies over the set of countable ordinals.



$ell_1$ has other preduals too. For example, a 1972 paper from the Israel Journal of Mathematics by Yoav Benyamini and Joram Lindenstrauss exhibits a Banach space E whose dual is isometrically isomorphic to $ell_1$, but E is not isomorphic to any space of the form C(X).



Although Gerald's assertion that the Cantor-Bendixson rank distinguishes between spaces of continuous functions on countable compact spaces is not correct, there is a rank/index that does distinguish between the isomorphism classes of these spaces, namely the Szlenk index. In fact the Szlenk index may be used to prove the 'only if' direction of the Bessaga-Pelczysnki result.



I strongly recommend Chapter 2 of the book Biorthogonal Systems in Banach Spaces by Hajek et al if you want to read about the Szlenk index and a proof of the Bessaga-Pelczynski result. That book also contains a sketch proof of the Mazurkiewicz-Sierpinski theorem; for a full account of that theorem I recommend Section 8 of Semadeni's classic book Banach spaces of continuous functions.



If you want a different example, I think the James-tree space JT would do. JT is separable but its dual is (at least I think so - it is worth checking!) isometrically isomorphic to the dual of the direct sum $*JT oplus H*$, where H is Hilbert space of dimension equal to the cardinality of the continuum. In particular, $*JT oplus H*$ cannot be isomorphic to JT because they have different density characters, but their duals are isometrically isomorphic. If you want to check the details of this example consult Chapter 13 of the Albiac and Kalton book recommended above by Matthew.

Thursday, 14 December 2006

human biology - Do white Australians have a distinct look?

I've heard from many people working in tourism or similar industries that, white Australians can be recognized as Australian solely by their facial features. Being Australian myself I've heard variants of the claim quite often and lean towards the claim being at least, a possibility.



While it's true that people tend to see patterns when there are none or incorrectly attribute, why they correctly recognize something. I don't necessarily think that is the case here. Australians don't tend to dress in a distinguishing way when overseas, and people have recognized them before hearing them speak. Given the extent to which I hear this claim made, I can't completely dismiss it.





This article from an Australian newspaper in 1943 interviews an artist who when asked if Australians have a racial type, says he thinks that it was generally possible to distinguish an Australian [from other Caucasians] due to differences in the nose and differences in the jaw.



Googling for terms like "look Australian" returns examples of the claims strewn around the internet, such as this forum thread and this blog.



Also of note is the art project to show the average face of Sydney, of which the following quote is relevant:




In many cases, however, the likeness is so strong that it's possible
to guess the nationality just by taking a cursory glance at the photo.






I'm fairly ignorant when it comes to biology, although there are a few reasons I think this might be possible. The (relatively) small gene pool from when Australia was colonized could have led to features from a small group being inherited in a large population.



There seems to be some evidence that environment can play a large role in influencing genetics more so than ethnicity. I can't find the paper investigating that, although will update this question if I am able to.



It seems possible to me that facial features belonging to the original colonists could have spread through subsequent populations, becoming a defining characteristic for some Australians.



I don't think most immigrants are relevant to this question as they will not have inherited distinguishing features (if they exist), nor will they or their parents/grandparents have been able to be influenced by the environment - if that is even a relevant factor.





Do a significant number of white Australians, excluding 1st or 2nd generation immigrants, have distinct facial features unique or generally only found in Australia?



Is it reasonable that some white Australians could be recognized as Australian, going only by facial features?

Wednesday, 13 December 2006

Complex and Elementary Proofs in Number Theory

The Prime Number Theorem was originally proved using methods in complex analysis. Erdos and Selberg gave an elementary proof of the Prime Number Theorem. Here, "elementary" means no use of complex function theory.



Is it possible that any theorem in number theory can be proved without use of the complex numbers?



On the one hand, it seems a lot of the theorems using in analytic number theory are about the distributions of primes. Since the Prime Number Theorem has an elementary proof, this might suggest that elementary proofs exist in other cases.



On the other hand, the distribution of primes is intimately related to the zeros of the Riemann Zeta function. Perhaps the proofs of other statements in analytic number theory require more direct references to the Riemann Zeta function.



This topic is more of a fascination for me, as I am not a number theorist. I would be interested if there are other examples of elementary proofs of theorems originally proved with complex analytic methods.

What is an intuitive view of adjoints? (version 2: functional analysis)

I've accepted Harald Hanche-Olsen's answer to this question. However, I feel that it needs a little expansion to explain exactly why. The short reason is that his answer triggered a connection in my brain that meant that a few idiosyncrasies of Hilbert spaces suddenly became clear. I'd like to explain them. I should add that I fully expected this question to be one of those where there wasn't a "right" answer.



The key point was that his answer made me think about the rank one case where everything is much simpler. As he said, a vector in $H$ "is" a linear map $mathbb{C} to H$, $t mapsto t eta$. Similarly, a functional on $H$ is a linear map $H to mathbb{C}$. Riesz rep says that such a functional is of the form $xi mapsto langle eta, xi rangle$. Put them together and you find that rank one operators $H to H$ are of the form $xi mapsto langle eta, xirangle zeta$. Or in "bra-ket" notation, $|xirangle mapsto |zetarangle langle eta | xi rangle$. So the operator is $|zeta rangle langle eta|$.



Next is to think about adjoints. What's the adjoint of this operator? Simple: swap everything around. So $|zeta rangle langle eta|^star = |eta rangle langle zeta|$. From this we can deduce that "self-adjoint" means "$eta = zeta$", "skew-adjoint" means "$eta = izeta$", whilst "normal" is a little more complicated.



However, this doesn't feel very insightful. The "click" was to think about $|zeta rangle langle eta|$ and $|eta rangle langle zeta|$ and try to work out what the swap operation was doing. The block, for me, on why this is purely an algebraic thing and not at all geometric is that the representation of functionals as "inner-product-with-vector" is one of those "it just works" theorems, with little insight as to why it works (yes, yes, I know why, I just don't know why). But if one removes Riesz Representation altogether then one is simply working in a Banach space and adjoints don't exist. So I needed to remove the conclusion of Riesz Rep, but leave enough of it to still ensure that I know I'm in a Hilbert space.



The key step in Riesz Rep is the complementary subspace property: that each subspace has a complement (turns out that this characterises Hilbert spaces, as proved by Lindenstrauss and Tzafriri in 1971). This is much more intuitive for me: it's geometric.



So when we think of a rank one linear transformation on $H$, we have two obvious subspaces: the image and the kernel. The image has dimension $1$, the kernel has codimension $1$. The complementary subspace property then implies that each of these has a complement, the complement of the image being of codimension $1$ and of the kernel being dimension $1$. So up to scalar, these specify another linear transformation. And that linear transformation is the adjoint. Then everything else falls into place.



So the key, for me, is that the fact that $ker A^star = (operatorname{Im} A)^perp$ and vice versa is not a happy consequence of adjoints but rather the almost defining property.



That "almost" rescues me a little from having missed this before. The phrase "up to scalar" that I highlighted above is important. To get the true adjoint one has to do a little more work to show that there is an operator that satisfies this "almost defining" property and that the map $T mapsto T^star$ is "nice" - it's easier to go the other way around and start with the adjoint and work backwards.



Or one could do something a little different. Instead of saying "this works up to scalars, can we fix that?" we should say "this works up to scalars, so let's ignore scalars" and, being Good Geometers and Topologists we know how to do that! Work with projective spaces, and more generally with Grassmannians.



Concentrating again on rank one operators, we have that their images lie in $mathbb{P}H$ and their kernels in $mathbb{P}^{infty - 1}H$ (closed codimension 1 subspaces), so up to scalars, a rank one operator is a point in $mathbb{P}H times mathbb{P}^{infty -1}H$. The crucial fact that the closest point property gives is that $mathbb{P}H cong mathbb{P}^{infty - 1}H$ (Riesz Rep is then the trivial observation that $mathbb{P}^{infty - 1}X cong mathbb{P}X$ for any normed vector space $X$). So really a rank one operator is a point in $mathbb{P}H times mathbb{P}H$. Adjunction is then the obvious $mathbb{Z}/2$ action on this.



It's not difficult to extend this to any finite rank operator, as Harald says. The question is to make the leap to infinite. At this point, Harald says:




Admittedly, from here to infinite rank is something of a stretch, but I think it might help anyhow.




which just goes to show that he doesn't know his own brilliance. The amazing thing is that there is no stretch! In a Hilbert space we have the approximation property which says that every continuous linear operator is the weak limit of a sequence of finite rank operators. Since adjunction is a weak property - in that you know that you have the adjoint if a load of evaluations say that you do - if I have a weakly convergent sequence of operators then the adjoints weakly converge to the adjoint! And once I have the adjoint, it is the adjoint no matter how bizarrely I found it.



In conclusion, it was Harald Hanche-Olsen's answer that triggered this in my brain so he gets the "accepted" tag. It may not be the "right" answer, but it was the answer that taught me a heck of a lot about something that I thought I knew a lot about already.



(PS Obviously, this answer is community wiki so voting for it gains me no rep and merely says that you agree with or like what I've said. However, if you do like this answer enough to vote for it, you should vote for Harald's answer as well since that sparked all of this one.)

gt.geometric topology - Necessary and sufficient criteria for a surface to cover a surface

I don't know the reference for this question, but I am pretty sure that it should follow from some known statement. Anyway let me give the answer in the case when S has negative Euler characteristic, orientable and CONNECTED. At least you can compare with your own answer. Denote by p(S) the number of punctures.



1) chi(S')/chi(S)=d with $d$ positive integer



2) $p(S)le p(S') le p(S)d$.



3) $p(S)d-p(S')$ should be even.



Moreover, in the case Genus(S)=0 you have an additional condition



$p(S)d-p(S')>d-2$. This condition assures that S' is connected.



I think that these conditions are necessary and sufficient. It is obvious that 1), 2) are necessary. Condition 3) comes from the fact, that the permutation corresponding to going around all punctures on S is a commutator, so it should be even.



In order to show that these conditions are sufficient, you could use the old result of Ore that tells that every even permutation is a commutator of two permutations.
Oystein Ore. Some remarks on commutators. Proc. Amer. Math. Soc., 2:307–314, 1951. Let me prove that conditions are sufficient in the case when genus of S is two or more.



Sketch of a proof. We want to show that there exists a collection of permutation in $S_d$,
$s_1,...,s_{2g}, t_1,...,t_p$ ($p=p(S)$) that act transitively on ${1,...,d}$ such that
$s_1s_2s_1^{-1}s_2^{-1}...=t_1...t_p$, where $t_1,...,t_p$ are given permutations with the product in the alternating group $A_d$. Then we chose $s_1$ and $s_2$ in such a way that $s_1s_2s_1^{-1}s_2^{-1}=t_1...t_p$ (Ore result) and take $s_3=s_4$ - cycles of length $d$, while all other permutations $s_i$ should be trivial. Clearly the action on ${1,...d}$ is transitive. Now the existence of a cover follows by standard arguments.



If you manage to make the proof very short it is worth to put it in the article, or at least give a hint. Otherwise, indeed, as Pete said it would be nice to find a reference.

Monday, 11 December 2006

biochemistry - How is RNAse contamination in RNA based experiments prevented?

You need to be careful about everything that comes in contact with your samples. Every spatula, beaker and every magnetic stir bar need to be RNAse-free.



For metal- and glassware we usually put everything into a drying closet at 200-250 °C for a few hours. That should get rid of RNAses pretty efficiently.



The RNA chemicals should be seperated from other chemicals if not everyone in the lab is working with RNA. They shouldn't be weighed using spatulas, but just poured into an RNAse-free container to avoid contaminating the storage container.



Use disposable plastic wherever possible, you should be able to perform most stuff in falcons and Eppendorf tubes.



Be careful with performing e.g. a DNA-prep at the same bench as the RNA work. The first buffer for the DNA-prep contains RNAse.



Avoid keeping your Eppendorf tubes or falcons open for long, that should reduce the potential for contamination.



We've had pretty good experience with just using water from a Millipore system and autoclave it once, that seems to get rid of the RNAses. I know that this is not a sure thing as the RNAses can survive autoclaving, but it seems to work well enough.

Sunday, 10 December 2006

oc.optimization control - optimize with respect to domain shape

Let $Gamma$ be the set of all closed $C^2$ curves in the plane which enclose unit area and let $Omega$ be the set of all subsets of $mathbb{R}^2$ that are enclosed by some curve in $Gamma$. Now let $f: mathbb{R}^2rightarrowmathbb{R}$ be a real-valued function on the plane. How can we find the set $omegainOmega$ with boundary $gammainGamma$ such that
$
int_omega f
$
is minimized?



A related question (with a more physics-style interpretation): Let $Omega$ and $Gamma$ be as before. Let $u(vec{x})$ be the real-valued function on the plane that solves the PDE
$$
Delta u(vec{x}) = 0 text{for } vec{x}inomega
$$



$$
u(vec{x}) = 1 text{for } vec{x}in partialomega,
$$



where $omega$ is some set in $Omega$. How can we find the $omega$ that minimizes the quantity $int_{gamma}|du/dn|^2$ where $gamma$ is the boundary of $omega$ with unit outward normal $n$ ?



I'm more interested in finding out which branch of mathematics studies questions like this and what concepts/tools are important to approach questions like this. Any references or suggestions to similar problems are greatly appreciated. (A friend suggested I tag this as geometric measure theory, but I don't know how appropriate that is)

Saturday, 9 December 2006

homotopy theory - A Model Structure on Symmetric Monoidal Categories

One basic problem is that the category of symmetric monoidal categories isn't complete. Its completion, in a basic sense, is the category of multicategories, on which it seems reasonable to conjecture there is a model category structure whose homotopy category "is" the connective part of stable homotopy -- we hope to prove this soon. See Elmendorf and Mandell, "Permutative categories, multicategories, and algebraic K-theory", which just appeared in Algebraic and Geometric Topology.

dg.differential geometry - Definition of a complex structure on a vector bundle

For any antiholomorphic Diffeomorphism $fcolon Sto S$ we get a canonical identification $f^starbar K=K,$ $ K $ and $bar K$ being the canonical and anticanonical bundle of the Riemann surface. A holomorphic structure on a complex vectorbundle $E$ is the same as an complex operator
$DcolonGamma(E)toGamma(bar KE)$ satisfying the (Cauchy Riemann) Leibnitz rule. (holomorphic sections are exactly the one in the kernel of D). (For higher dimensions this is not true.)



Now, $f^* E$ has a natural complex structure (it's just i).
Therefore one gets an anti-holomorphic structure $bar DcolonGamma(f^star E)toGamma(Kf^* E)$ satisfying the antiholomorphic Cauchy Riemann equation.
But the complex conjugate bundle $bar E$ also has a anti-holomorphic structure, since $overline{bar K E}=Kbar E.$ Therefore, $f^* bar E$ has a natural holomorphic structure.



These two holomorphic structures are not isomorphic in general:
In the case of a line bundle $L=E$ of degree $0$ one might see this as follows: every holomorphic structure $D$ gives rise to an unique unitary flat connection $nabla$ such that
$D=1/2(nabla+i*nabla).$ Then the anti-holomorphic structure on $bar L$ is given by $1/2(nabla-i*nabla)$ and, the unitary flat connection corresponding to the holomorphic structure on $f^* L$ is the connection $f^* nabla.$ But this connection is not gauge equivalent to $nabla$ in general: For example, on a square torus with f given by $zmapsto bar z$ the connection $d+c idx$ is not gauge equivalent to $d+ci dy$ for $cin Rsetminus 2pi Z.$

Wednesday, 6 December 2006

biochemistry - How does water buffer a sudden drop in temperature?

From pure biophysical viewpoint the question



"Why does water buffer sudden temperature changes?" can be answered in the following way:



Water has relatively high specific heat capacity. This is the measure of the energy required to raise the temperature of one kilogram of a substance by one kelvin without a change of state occurring. "Relatively high" means that water can absorb more energy without noticeable (for the cell) increase in temperature.



Here are some heat capacity values from the above-linked Wikipedia article:




Animal (incl. human) tissue 3500 Jkg-1
Water at 25 °C liquid 4181.3 Jkg-1
Methanol liquid 2597 Jkg-1
Ethanol liquid 2440 Jkg-1
Paraffin wax solid 2500 Jkg-1
Graphite solid 710 Jkg-1




Water has relatively high thermal conductivity. This means that the absorbed heat is quickly distributed over the complete cell volume, leveling down the focal temperature increases in cell. Here are again some values for comparison:




Water 0.6 Wm-1K-1
Wood 0.2 Wm-1K-1
Paper 0.05 Wm-1K-1
Glycerol 0.3 Wm-1K-1




The follow-up questions:



"Why does water have relatively high heat capacity?" and "Why does water have relatively high thermal conductances?" are not within the scope of this site. However, here exactly comes your explanation with hydrogen bonds in play (from Wikipedia):




Hydrogen-containing polar molecules like ethanol, ammonia, and water
have powerful, intermolecular hydrogen bonds when in their liquid
phase. These bonds provide another place where heat may be stored as
potential energy of vibration, even at comparatively low temperatures.
Hydrogen bonds account for the fact that liquid water stores nearly
the theoretical limit of 3 R per mole of atoms, even at relatively low
temperatures (i.e. near the freezing point of water).


mathematics education - Teaching and students

A little bit more information about your background and situation would be helpful. Are you: a graduate student, a post-doc, a tenure-track professor? Are you teaching at a university? Are you teaching undergraduate or graduate courses? (I will assume that by "analysis" you mean something which is at the advanced undergraduate level, at least.)



I think the first reaction that most mathematicians will have to your question is "Oh, she should learn the material better." This is natural because part of being a mathematician is a constant desire to understand mathematics both more broadly and deeply than you presently do. So let's acknowledge that learning some more analysis couldn't be a bad idea. But it's not clear that that's the best answer to your qustion. Let's move on to other considerations.



First, it's a little unfortunate that you "have to teach" somewhat advanced courses in a discipline that you feel is far away from your training and current interests. Are there not other faculty who are more qualified to teach analysis courses than a young, inexperienced algebraist? If the answer is no, then that's no fault of yours, and it is somewhat valiant of you to be willing to pitch in outside of your core expertise.



If you are feeling worried by a lack of expertise, one strategy (not infallible, but worth considering) is to be open and honest with your students about this. Say at the beginning of the course that you are outside of your core knowledge but that's okay -- you do have plenty of training in learning mathematics in real time and solving problems. Tell them that you may not know the answer to each question they ask on the fly, but you are nevertheless more than capable of teaching the course and that at the end your knowledge will have increased just like theirs.



The prospect of teaching a course outside of your area of expertise becomes much more exciting and positive if you really do look on it as a learning opportunity. You have to trade in the mental image of the instructor as an omniscient sage whose job is to transport some platonically perfect mathematics from her head to those of the students for the image of a scholar who is always learning more, rethinking and reorganizing what she already knows, and presenting this material to the students and interacting with them in a temporal way.



In fact I was in a situation which sounds similar to yours: I am by training an arithmetic geometer (which is a kind of algebraist, I think) and as a postdoc less than two years out of my PhD I taught a second semester undergraduate analysis class. The big difference between my story and yours is that I wasn't required to do this -- on the contrary I campaigned quite actively (even a bit pushily, because it seemed necessary) to be able to teach an advanced class in any discipline, rather than the multivariable calculus / linear algebra that it seemed like I would otherwise get stuck with. I also had plenty of time to prepare for the course in advance, so I went to the library and thumbed through many different texts to get an idea of what the possibilities were. I had a fantastic time teaching the course, and my understanding of real analysis is much stronger now than it was before. (This was also the first course for which I typed up rather extensive lecture notes: scroll down to Real Analysis II on http://www.math.uga.edu/~pete/expositions.html to see them.)



The final thing I want to emphasize is that not being able to answer a question on the fly is not necessarily a sign of inadequate expertise: it may, on the contrary, be a very positive sign that the students are thinking in unexpected and novel ways. I remember my high school physics and calculus teacher (i.e., the same person taught me both): he was very adept at engaging the brightest students in deeper issues and getting them to think about problems that were a level or more above what was officially happening in the course. Often several students would stay with him after class and try to figure out some tough issue with him on the spot. It was great, because he was treating us almost as though we were equals -- which we most certainly were not; he knew far more than all of us put together -- and we felt like we were really involved. Often we would ask him tough questions in class, questions that he didn't know the answer to on the spot, and he would usually think about them out loud and try to work them out, sometimes unsuccessfully. On rare occasions he would even get a little stuck in some derivation in his lecture and have to quit and move on to the next topic. He took great pride though in coming back the next day with a fantastic explanation. He was, of course, the best math and science teacher I ever had, and by the way he had a PhD in theoretical physics, so he was not in any reasonable sense underqualified to be teaching those classes!



If you think about it, getting asked a question that you don't know the answer to immediately (and if you don't know it immediately, there's no shame in not giving the answer in that class period -- how much time are you willing to spare standing silently and thinking before you give up at least temporarily? not very much time at all, I hope -- any question which has given you such pause is well over the heads of the majority of the class, who are just waiting for you to get on with it) is just about the most positive experience you can have in a classroom. One of the reasons that I can't get excited about teaching freshman calculus is that I know from long experience that anything that I find remotely interesting is going to be difficult (and worse, boring) for 95% of the students. The number of interesting questions that I get when teaching calculus is, very sadly, about one every two or three courses. By the way, I don't recall ever getting stumped in that real analysis class that I taught. I sincerely wish I had been -- it would have meant that the students were engaging with the material at a much higher level. By way of contrast, for each of the three days of the course I am currently teaching I have received at least one question that really caused me to think and in my answers say things that I thought were true but wasn't completely sure. (Sample question: let $p$ and $p'$ be distinct prime numbers. Are the rational numbers equipped with the $p$-adic topology and the rational numbers equipped with the $p'$-adic topology homeomorphic? Probably, right? But I wasn't and still am not completely sure.) This is fantastic -- I can't wait for tomorrow. By the way, the title of this course is (graduate) Number Theory II, i.e., I am supposed to be an expert on this.

anatomy - What is the morphological difference between Leydig cell in human and pig?

I have studied all available (via University library) literature on the Leydig cells and I think your teacher might have this article summarizing the morphological studies on Leydig cells in different animal models in mind.



To put it straight, the most common animal models for studying Leydig cells are rats, mice and pigs.



The development of Leydig cells in rats was shown to be bi-phasic: development of the Leydig cell population in fetus and its degradation while transforming into the mature cells. It has been believed for quite long that this is also true for human Leydig cells. However, some studies performed on pigs and monkeys delivered an evidence for an additional phase of cell degradation sometimes between the fetal and neonatal development shifts. These data are summarized on the following Figure (taken from the referenced paper):



Leydig cell development flowchart



So, by comparing the Leydig cells of rats and humans from the same fetus age or shortly after the birth (neonatal time) you might see the degradation of Leydig cells -- they decrease in size and many of them eventually die -- compared to normal state of these cells in rats.

ag.algebraic geometry - Is there something like Čech cohomology for p-adic varieties?

The first comment to make is that Cech theory is really extremely general, and can be set up to compute the cohomology of any complex of abelian sheaves on any site (provided you have coverings that are cohomologically trivial). This is explained at least somewhat in SGA4, Expose 5 and EGA III, Chap 0, section 12.



I think you should be working with the rigid analytic space attached to $X$, and not with the $mathbf{Q}_p$-points of $X$, and the latter really has no good topology on it besides the totally disconnected one induced from the topology on $mathbf{Q}_p$.



Let's assume that $X$ has a model
$mathcal{X}$ over $mathbf{Z}_p$ that is smooth and proper and write $widehat{mathcal{X}}$
for the formal completion of $mathcal{X}$ along its closed fiber. Then the (Berthelot) generic fiber $widehat{mathcal{X}}^{rig}$ of $widehat{mathcal{X}}$ is a rigid analytic space that is canonically identified with the rigid analytification of $X$ (using properness here). Moreover, one has a "specialization morphism" of ringed sites
$$sp:X^{an}simeq widehat{mathcal{X}}^{rig}rightarrow widehat{mathcal{X}}$$
with the property that for any (Zariski) locally closed subset $W$ of the target, the inverse image $sp^{-1}(W)$ is an admissible open of the rigid space $X^{an}$ (called the open tube over W). In this way, coverings of the special fiber by locally closed subsets give coverings of the rigid generic fiber by admissible opens, and you can use Cech theory with these coverings and or your favorite spectral sequence to compute sheaf cohomology in the rigid analytic world. Again using properness, by rigid GAGA this cohomology agrees with usual (Zariski) cohomology on the scheme $X$ (provided your sheaf is a coherent sheaf of $mathcal{O}_X$-modules, say).



This idea of computing cohomology using admissible coverings of the associated rigid space is a really important one as it allows you to use the geometry of the special fiber. It occurs (allowing $mathcal{X}$ to have semistable reduction) in the work of Gross on companion forms, of Coleman on $mathcal{L}$-invariants and most prominently in Iovita-Coleman (see their article on "Frobenius and Monodromy operators"). This latter article might be a good place to start.



I would also highly recommend the articles of Berthelot:



http://perso.univ-rennes1.fr/pierre.berthelot/publis/Cohomologie_Rigide_I.pdf



http://perso.univ-rennes1.fr/pierre.berthelot/publis/Finitude.pdf



I'd also suggest the AWS 2007 notes by Brian Conrad for learning about rigid geometry, which seems generally quite pertinent to your situation.
For etale cohomology of rigid spaces, you might want to look at the article of Berkovich, though this would require learning about his analytic spaces.



In any case, I hope this is a good start.

molecular biology - Can I elute my GFP-tagged protein off anti-GFP antibody using a peptide?

I found a paper that describes a method to purify GFP fusion proteins using affinity chromatography: "Purification of GFP fusion proteins with high purity and yield by monoclonal antibody-coupled affinity column chromatography" (1). Unfortunately I don't have access to the full text, so I can't check how exactly they elute the GFP fusion protein from the column. They describe a specific monoclonal antibody they use, there should be information on how to obtain it in the paper.



Abstract:




GFP has often been used as a marker of gene expression, protein
localization in living and fixed tissues as well as for protein
targeting in intact cells and organisms. Monitoring foreign protein
expression via GFP fusion is also very appealing for bioprocess
applications. Many cells, including bacterial, fungal, plant, insect
and mammalian cells, can express recombinant GFP (rGFP) efficiently.
Several methods and procedures have been developed to purify the rGFP
or recombinant proteins fused with GFP tag. However, most current GFP
purification methods are limited by poor yields and low purity. In the
current study, we developed an improved purification method, utilizing
a FMU-GFP.5 monoclonal antibody (mAb) to GFP together with a
mAb-coupled affinity chromatography column. The method resulted in a
sample that was highly pure (more than 97% homogeneity) and had a
sample yield of about 90%. Moreover, the GFP epitope permitted the
isolation of almost all the active recombinant target proteins fused
with GFP, directly and easily, from the crude cellular sources. Our
data suggests this method is more efficient than any currently
available method for purification of GFP protein.




(1) Zhuang R. et al., Purification of GFP fusion proteins with high purity and yield by monoclonal antibody-coupled affinity column chromatography. Protein Expr Purif 2008, 59(1), 138-43

Tuesday, 5 December 2006

st.statistics - Estimate population size based on repeated observation

I take the bus to work every day. Every bus has a serial number, but unlike in the German Tank Problem, I don't know if they are numbered uniformly $1...n$.



Suppose the first $k$ buses are all different, but on day $k+1$ I take one I've been on before. What is the best estimate for the total number of buses?

ag.algebraic geometry - Obstruction bundle for spaces with Kuranishi structure

Here's a view of the symplectic side of the bridge.



The Kuranishi model (see Donaldson-Kronheimer, The geometry of four-manifolds, ch. 4) goes like this. You're interested in a (moduli) space $M$ cut out as $psi^{-1}(0)$, for some smooth but nonlinear map of Banach spaces, $psi colon (E,0) to (F,0)$ such that $delta:=D_0psi$ is a Fredholm operator (finite dim kernel and cokernel). That means that $delta$ is "almost" an isomorphism, and Kuranishi's principle is that one can construct a non-linear map $kappa colon ker(delta) to mathrm{coker}(delta)$, such that $kappa(0)=0$ and $D_0 kappa =0$, and (locally near $0$) a homeomorphism $M to kappa^{-1}(0)$. This gives a finite-dimensional model of $M$. "Kuranishi structures" are a formalism in which one can say that $M$ is everywhere-locally given as the zeros of maps like $kappa$.



In the case of genus 0 GW theory, $M$ is the moduli space of (say) parametrized pseudo-holomorphic maps from $S^2$ to an almost complex manifold $X$; $psi$ is a non-linear Cauchy-Riemann operator, and, for a pseudo-holomorphic map $uin M$, $D_u psi$ is a linearized C-R operator - the $(0,1)$-part of a covariant derivative acting on sections of $u^ast TX$. Its kernel can be identified with the holomorphic sections $H^0(S^2,u^ast TX)$ of the holomorphic structure on the vector bundle $u^ast TX$ defined by the C-R operator. Its cokernel is isomorphic to $H^1(S^2,u^ast TX)$. If you're lucky, you have a Zariski-smooth moduli space $M$ whose Zariski tangent space at $u$ is $ker D_upsi$. In this case, one could take $kappa=0$, and then the spaces $mathrm{coker} (D_upsi)$ form a vector bundle $Obs to M$, which is what symplectic geometers usually call the obstruction bundle. One can now try to divide everything by $Aut(S^2)$, and quite possibly get into orbi-mathematics.



In the integrable case, still with non-singular but excess-dimensional $M$, I could write $Obs$ as $R^1pi_* Phi^*T_X$, where $T_X$ is the holomorphic tangent sheaf, $Phi$ the evaluation map $mathbb{P}^1times M to X$, and $pi$ the projection to $M$. At this point, if what I've said is accurate, you and other algebraic geometers out there are better placed than me to answer your question. Is it your $(E_{-1})^vee$?



Of course, you didn't really want to assume $M$ smooth. A place where these deformation theories are compared in generality is Siebert's 1998 paper Algebraic and symplectic Gromov-Witten invariants coincide.

Monday, 4 December 2006

algorithms - Thousands of rays intersections with Triangles in 3D space

This is a basic question in ray tracing. Do a Google search on "ray tracing acceleration data structures," or pick up a copy of the PBRT book by Pharr & Humphreys. Classic examples of acceleration structures include the kd-tree and bounding volume hierarchy (BVH). Although kd-trees are theoretically optimal in many situations, actual performance depends largely on the distribution of triangle sizes and locations. You also need to take into account the complexity of building the structure -- building a kd-tree nominally takes O(n log n) time in the number of triangles, but it is difficult to make incremental updates (e.g., for dynamic geometry). A more modern acceleration structure is the BIH (bounding interval hierarchy), which is useful for dynamic scenes.



(Also, a resource more specific to your problem than stackoverflow.net are the forums on ompf.org)

soft question - How to select a journal?

Many shall find this criterion worthless, but personally I can't help taking the production quality of the journal into account (even if it is not of primary importance).



Quality paper and ink improve the reading comfort, and some journals (some of Elsevier particularly) have pages with pale ink, or thin paper that let the verso appear. On the other end of the scope are journals like Acta mathematica of Publications mathématiques de l'IHES with nice paper and fonts.



Another aspect of this criterion is the processing quality. To compare two very different experiences, in my first article incredible mistakes have been added after the proofs (expressions like $n/2$ replaced by $n^2$) while the AMS journals do an amazing job, showing you exactly what they changed in your paper (corrected spellings e.g.) and sometimes asking for confirmation. This part you cannot judge before being published in the journal, or discussing with colleagues.

senescence - Why isn't the p16-INK4a gene involved in apoptosis expressed in heart or liver tissues?

p16-INK4a is a part of a very important checkpoint mechanism. It's the "bad guy" in the context of aging because it induces senescence, and too much senescence leads to aging-related tissue degradation.



But senescence is important. It's one of the responses cells take when something goes wrong-- DNA damage, viral infection, telomere depletion, that sort of thing. Senescent cells have stopped proliferating. We have a word for cells that don't stop proliferating, and that word is "cancer". So, p16-INK4a is a major tumor suppressor. A universal p16-INK4a knockout would have a much harder time shutting down the proliferation of cells that had undergone DNA damage, and would therefore be much more prone to cancer. You'd have very young-looking tissues filled with tumors.



So, the headline question is why INK4a is not expressed in the heart or liver if it's so important. This is speculation on my part, but I think it's because those tissues are especially prone to being damaged by fibrosis, and a build-up of senescent cells would lead to increased fibrosis. Senescence is just one possible response to DNA damage, though. Another is apoptosis. If the senescence-induction pathways aren't active in heart and lung tissue, I'd expect the apoptosis-induction pathways to be pretty active.

st.statistics - Statistical test comparing two relative frequencies

Assume they are four independent beta random variables $X_i$, with means $mu_i$.



Note that the density functions would depend on the observed samples.



We could then test the hypothesis that $mu_1mu_2>mu_3mu_4$ by setting up a quadruple integral of the joint density function over the set



${ (x_1,x_2,x_3,x_4)mid x_1x_2>x_3x_4 }$.



If this integral is small, we reject the hypothesis $mu_1mu_2>mu_3mu_4$.

Sunday, 3 December 2006

Does bioluminescence occur in humans too?

From the article you linked:




virtually all living organisms emit extremely weak light, spontaneously without external photoexcitation. This biophoton emission is categorized in different phenomena of light emission from bioluminescence, and is believed to be a by-product of biochemical reactions in which excited molecules are produced from bioenergetic processes that involves active oxygen species




They reference these two works. The first is a 1988 review from Popp et al. (1988):



Biophoton emission - Experientia 44:543–600. (sorry, I cannot find the link to a full text...)



Fritz-Albert Popp is the biophysicist who first developed the biophoton theory.



The second work they reference is by the same first author:



In vivo imaging of spontaneous ultraweak photon emission from a rat's brain correlated with cerebral energy metabolism and oxidative stress. - Kobayashi M, et al.



Finally, a search in Pubmed reveals various other articles by different authors studying different species.

kt.k theory homology - Hilbert $C^*$-modules and approximate units

No, there is not always such an approximate unit. (This will be easier to formulate in terms of left modules, and with inner products linear in the first entry. The warning seems necessary due to the common convention in C*-module theory to do the opposite.)



Example



Let $A=B(mathbb{C}^2)$ (linear operators), $E=mathbb{C}^2$, with the module action given by operators acting on vectors (on the left) and the $A$-valued inner product of $x$ and $y$ in $E$ given by $<x,y>_A(z)=<z,y>_mathbb{C}x$. Then $<x,y>_A$ has rank at most one for all $x$ and $y$, so no such approximate identity exists. $square$




Any finite dimensional Hilbert space with dimension at least 2 would give a slight modification of this example. Or, let $E=H$ be a separable, infinite dimensional Hilbert space, and let $A=mathcal{K}(H)$ be the algebra of compact operators on $H$. (Added: Note that fullness follows from the fact that the span of the range of the inner product is the set of finite rank operators.) Or, given a C*-algebra $B$, one could form $H_B=Boplus Boplusldots$ with its usual right $B$-module structure and consider the analogous construction with $A=mathcal{K}(H_B)$ and $E=H_B$. If $B$ is $sigma$-unital, then so is $A$, but there will be no approximate identity of the desired form.



I do not have anything useful to say about formulating sufficient conditions for such an approximate identity to exist, but this simple example shows that lack of existence is common.

pr.probability - Generalized binomial coefficients and Gaussian density

I ran into an expression calculating the expected value of $exp(i t sigma)$ where $sigma$ is the total number of cycles in a uniformly chosen $S_n$ element. The expression is
$$E_n (exp(i t sigma)) = Gamma(n + exp(it)) / (Gamma(exp(it)) n!)$$
where $E_n$ denotes the expectation under the uniform distribution on $S_n$. The paper then claims that using Binet's form of Stirling approximation one can get
$$E_n (exp(it sigma)) = n^{exp(it) -1}/Gamma(exp(it)) (1 + o(1))$$



Then here comes the derivation I cannot understand:
using the last expression, they claim one gets the following central limit theorem
$$lim_{n to infty} E_n(exp(it (sigma - log n)/sqrt{log n})) = exp( -1/2 t^2)$$
for any real $t$.



I would highly appreciate anyone who can tell me why this is true. It appears to be related to some property of the Gamma function over the complex number.
The relevant paper is Shepp and Lloyd: Ordered lengths in a random permutation
John Jiang

complex geometry - Is there a "simple commutation" relation between $D^{''}$ and $delta^{'}$, with $D^{''}$ the (0,1) part of the chern connection of a vector bundle and $delta^{'}$ the adjoint of the (1,0) part?

Yes, the relation is that they anti-commute.



Let's see this very briefly (you can find it in almost all books on Kähler geometry and Hodge theory).



We want to compute $[D''_E,delta'_E]$, where $[bullet,bullet]$ is the graded commutator and let me call $omega$ (instead of $g$) the Kähler form. Then one has that
$$
[Lambda_omega,D''_E]=-idelta'_E,
$$
so that
$$
[D''_E,delta'_E]=i[D''_E,[Lambda_omega,D''_E]],
$$
where $Lambda_omega=*^{-1}L_omega*$ is the formal adjoint of the operator of degree $(1,1)$ given by $L_omegabullet=omegawedgebullet$.



Now, the (graded) Jacobi identity gives
$$
-[D''_E,[Lambda_omega,D''_E]]+[D''_E,underbrace{[D''_E,Lambda_omega]}_{=-[Lambda_omega,D''_E]}]+[Lambda_omega,underbrace{[D''_E,D''_E]}_{=0}]=0,
$$
thus $-2[D''_E,[Lambda_omega,D''_E]]=0$ and
$$
D''_Edelta'_E+delta'_ED''_E=[D''_E,delta'_E]=0.
$$

Saturday, 2 December 2006

ra.rings and algebras - Mathematical software for computing in integral group rings of discrete groups?

You can do this with GAP. The example below assumes that you have the polycyclic package installed.



First, you tell GAP which group you want to work with. Luckily, Heisenberg groups are polycyclic, and the polycyclic package provides a command to obtain them:



gap> G:=HeisenbergPcpGroup(1);
Pcp-group with orders [ 0, 0, 0 ]


Note that we could have also defined it by some other means if polycyclic was not available (e.g. as a matrix group), but this way is the most convenient. Now let's form the integral group ring:



gap> ZG:=GroupRing(Integers,G);
<free left module over Integers, and ring-with-one, with 6 generators>


The extra three generators come from the inverses of x, y and z (note that internally it calls them g1,g2,g3; it would be possible to change that with some effort, but that's beyond the scope here). Let's assign the corresponding generators of the group ring to variables x, y, z, and verify the relations you have given:



gap> x:=ZG.1;; y:=ZG.2;; z:=ZG.3;;
gap> x*z=z*x and y*z=z*y and y*x=x*y*z;
true


Here is an example of powering a group element (this works with more complicated ones, too, but I picked a small one to keep the output readable).



gap> (x+7*y)^2;
(1)*g1^2+(7)*g1*g2*g3+(7)*g1*g2+(49)*g2^2


I hope this helps.

Friday, 1 December 2006

Is there a way to define Hecke operators "inherently" as certain endomorphisms of the Jacobian?

Philip remarks that he wants to define the Hecke operator as an endomorphism of the Jacobian of "any Riemann surface $X$ such that the endomorphism ring is defined over $mathbf{Q}$", and at the same time he wants the Hecke operator to reduce to ${rm Frob}_p + p{rm Frob}_p^{-1}$. I think that for a generic Riemann surface $X$, the endomorphism ring is $mathbf{Z}$. However, ${rm Frob}_p + {rm Ver}_p$ is very unlikely to be an integer, so for a general $X$ it is highly unlikely that there exists an element of ${rm End}(X)$ that lifts ${rm Frob}_p + {rm Ver}_p$. (I'm using ${rm Ver}_p$ to denote the dual of Frobenius.)



Here is another closely related point. When $A$ is an abelian variety with good reduction at a prime $p$, there is a natural map ${rm End}(A) to {rm End}(A_{{mathbf F}_p})$. (See my remark here). I think this map is injective (consider the induced map on Tate modules at some good prime $ell$). Thus you could define the Hecke operator $T_p$ to be the unique (if it exists!) lift of ${rm Frob}_p + {rm Ver}_p$. That's intrinsic and makes no reference to any moduli space.

dg.differential geometry - PDE on manifolds

To make things coordinate-free, it is sufficient to reformulate differential equations in the form that makes use of exterior derivatives and exterior products of differential forms. Any set of differential equations can be cast into this form, the only subtlety being that it may require an infinite collection of differential forms to be introduced.
As topologist you may be aware of Sullivan's work "Infinitesimal computations in topology"
in which a sort of such equations were studied. Though to do real PDE it is necessary to work with zero-degree forms too, which he did not.
Such equations have applications in physics, e.g. you may write the equations describing black-hole without referring to any coordinates.



The typical system has the form $d W^A=F^A(W)$, where $W^A$ is a set of some forms valued in some linear spaces, $W^A$ do not necessary have the same degree, $F^A(W)$ expands only in terms of exterior products of $W^A$ with constant coefficients.



As an example, take one-forms $Omega^I$, $F^I=f^I_{JK}Omega^IOmega^K$, then $dOmega^I=f^I_{JK}Omega^IOmega^K$, the integrability for this equations implies $f^I_{JK}$ be the structure constants for some Lie algebra. The covariant constancy equations can be formulated in the same form.