TITLE: Asymptotic complexity of sum of poly-logarithmic functions
QUESTION [3 upvotes]: I'm trying to figure out what's the asymptotic complexity for the following sums:
$$\sum_{k=1}^n lg^s k$$
$$\sum_{k=1}^n k^rlg^s k$$
s and r are positive constants.
I think i should be using integrals for these, but given this: https://en.wikipedia.org/wiki/List_of_integrals_of_logarithmic_functions, I just can't find a way to solve these.
Any help will be appreciated,

REPLY [1 votes]: Using integrals which needed to be guessed looked overly complicated for this problem, i asked a friend at work and we believe that we came up with an answer that does not involve integrals:
$\sum_{k=1}^nlg^sk = \sum_{k=1}^{\lfloor n/2\rfloor}lg^sk+\sum_{k=\lfloor n/2\rfloor}^nlg^sk\ge$
$\sum_{k=1}^{\lfloor n/2\rfloor}0+\sum_{k=\lfloor n/2\rfloor}^nlg^s(n/2)=$
$0+n*lg^s(n/2)=\Omega(n*lg^s(n))$
I know the last part is true, but i don't really know how to prove it...
A similar trick can be applied for the upper bound.


TITLE: Is $P(A \mid B) = P(A \mid B,C) + P(A \mid B , C^c)$?
QUESTION [0 upvotes]: I am confused with these statements:
\begin{align}
P(A\mid B) &= P(A\mid B \cap \Omega)\\
&= P(A\mid B \cap (C \cup C^c) \\
&= P(A\mid B \cap C) + P(A\mid B \cap C^c) \quad\text{because $C$  and $C^c$ are disjoint} \\
&= P(A\mid B,C) + P(A\mid B ,C^c)
\end{align}
But according to LOTP
$$P(A\mid B) =  P(A\mid B,C) P(C\mid B) + P(A\mid B,C^c) P(C^c\mid B)$$
Can someone tell me where I am wrong?
Or is it that $P(A\mid B,C) \neq P(A\mid B \cap C)$?

REPLY [1 votes]: The wrong assumption is
$$
P(A \mid B\cap(C\cup C^c) ) = P(A\mid B\cap C) + P(A\mid B\cap C^c).
$$
It is true that $B\cap C$ and $B\cap C^c$ are disjoint, but this does not imply the identity above. Why do you think it would?
Note that $P(X\mid Y) = \frac{P(X\cap Y)}{P(Y)}$ and we have
$$
\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}
\qquad\text{but}\qquad
\frac{a}{c+c'} \neq \frac{a}{c} + \frac{a}{c'}.
$$


TITLE: Matrix exponential of any matrix
QUESTION [2 upvotes]: I want to be able to compute the exponential of any square matrix $A\in M_n(\mathbb{R})$.
I know that:

If $A$ is diagonalizable, then we can write $A=PDP^{-1}$, where $D=\left(\begin{array}{ccc}
\lambda_{1} & \cdots & 0\\
\vdots & \ddots & \vdots\\
0 & \cdots & \lambda_{n}
\end{array}\right)$
and it is easy to show that $e^A=Pe^DP^{-1}$.
If we can write $A$ in the form $$\left(\begin{array}{ccccc}
\lambda_{1} & 1 & \cdots & 0 & 0\\
 & \lambda_{2} & \ddots & 0 & 0\\
 &  & \ddots & 1 & \vdots\\
\vdots &  &  & \lambda_{n-1} & 1\\
0 & \cdots &  &  & \lambda_{n}
\end{array}\right)
 $$
then  we can write $A=D+N$ where $D$ is diagonal and $N$ is nilpotent. If $DN=ND$, then $e^A=e^{D+N}=e^De^N$, and this shouldn't be too hard to compute.

Now, what if both cases fail? For example, what if $A$ is not diagonalizable and, furthermore, the decomposition $D,N$ is such that $DN\ne ND$? Or, even worse, if there is no such  decomposition?
For example: $A=\left(\begin{array}{cc}
2 & 1\\
0 & 1
\end{array}\right)$ is diagonalizable, so we can use the first case above. But we can't use the second, since $A=\left(\begin{array}{cc}
2 & 0\\
0 & 1
\end{array}\right)+\left(\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right)$ but these matrices do not commute. If $A$ was not diagonalizable, I was in a bad situation.

REPLY [1 votes]: It is well known that we can write any matrix in Jordan normal form by a change of basis. This means it is a block-diagonal matrix, with block-entries the 'Jordan blocks' (not sure if that's standard terminology):
$$ J(\lambda, n) = \begin{bmatrix} \lambda & 1 & & \\ & \lambda &1 &\\ &&\ddots &1 \\ &&&\lambda \end{bmatrix} = \lambda \cdot I + N,$$
where $N$ is the matrix with ones at all entries immediately above the diagonal.
These blocks, then, can be split as indicated above, and $\lambda I \cdot N = N \cdot \lambda I$, for $I$ commutes with all matrices. Since each block on the diagonal of the Jordan normal form commutes, we can split the entire matrix in a diagonal matrix and a nilpotent matrix. Since each block commutes, the 'entire thing' commutes as well.
So, to summarise: we can find a Jordan normal form: $A = U \cdot J \cdot U^{-1}$, and this $J$ can be split in a diagonal and a nilpotent matrix.

REPLY [1 votes]: The Jordan-Chevalley decomposition guaranties that any matrix can be expressed in the form $A=PJP^{-1}$ with the entries of $P$ and $J$ in $\mathbb{C}$ ( $J$ is the Jordan form of the linear transformation) , and $J=D+N$ with $DN=ND$.
If you  are interested in the computational ways to find the exponential of a matrix, you can see here.
(Note that in your example the matrix is diagonalizable, so the exponential is easy calculated) .


TITLE: A ring which has many distinct left inverses
QUESTION [2 upvotes]: I'm trying to find a ring which has an element that has many(more than two) distinct left inverses.
I first tried the (square) matrices but then realized that if a matrix is a left unit, then it has determinant greater than 0, thus it is also a right unit and has unique inverse.
And then next I tried to do it with functions with usual addition and composition, and I found this example(in the first answer, by user23211): Kaplansky's theorem of infinitely many right inverses in monoids?
But I realized that the space of functions does not form a ring since the distribution law fails.
To make distribution law hold, we can only have homomorphisms; while the elements in the above example are not.
I also have seen that we can construct a similar examples with infinite vector spaces, but they are not rings.
Is there any ring like this?

REPLY [3 votes]: Consider a countably infinite dimensional vector space $V$ and let $R$ be the endomorphism ring of $V$.
Suppose the vector space has a basis $\{v_k:k\in\mathbb{N}\}$; define the endomorphism $f$ by
$$
f(v_k)=v_{k+1}
$$
and, for every $v\in V$, the endomorphism $g_v$ by
$$
g_v(v_0)=v,\qquad g_v(v_{k+1})=v_k
$$
Then, for every $k\in\mathbb{N}$,
$$
g_v(f(v_k))=v_k
$$
so $g_v$ is a left inverse of $f$ in $R$.

It is an interesting exercise proving that, if the element $a\in R$ ($R$ any ring) has two distinct left inverses, then it has infinitely many.


TITLE: Geometric series for values between 0 and 1
QUESTION [0 upvotes]: I am given that geometric series is defined as the following $1-x+x^2-x^3+x^4$ for values in range $0<x<1$.
I am also told expected value can be calculated by using the following equation: $1/1+x..$
Then i am told i can keep calculating the series until  i am in the error range of the expected value.
Error = |currentSumofGeoSeries+expectedValue|
Given x = 0.5 and error tolerance =  0.001

Term 1 1 0.333333
Term 2 0.5 0.166667
Term 3 0.75 0.0833333
Term 4 0.625 0.0416667
Term 5 0.6875 0.0208333
Term 6 0.65625 0.0104167
Term 7 0.671875 0.00520833
Term 8 0.664063 0.00260417
Term 9 0.667969 0.00130208
Term 10 0.666016 0.000651042

I am just confused as to how term 1 ...term 2 are calculated. how did they get 1 and 0.333 for term 1? please advice

REPLY [1 votes]: The series is given by
$$\sum_{k=1}^\infty (-x)^k \to \frac1{1+x} \text{ iff } |x| < 1$$
Your problem is that the last value of the output is the error of the limit $\frac1{1+\frac12} = \frac23$ compared to the truncated sum.
The output is formatted like this:
$$\text{Term } k \qquad \sum_{j=0}^k (-\frac12)^j \qquad \Bigg| \sum_{j=0}^k (-\frac12)^j - \underbrace{\sum_{j=0}^\infty (-\frac12)^j}_{= \frac23} \Bigg|$$

As per request: calculation of Terms $1$ and $2$:
I denote
$$S_n := \sum_{k=0}^{n-1} (-\frac12)^k$$
The output of your program then gives $S_n$ and $\epsilon_n := |S_n - \frac23|$.
$$S_1 = \sum_{k=0}^{1-1} (-\frac12)^k = (-\frac12)^0 = 1$$
so
$$\epsilon_1 = |1-\frac23| = \frac13$$
. Now for term $2$:
$$S_2 = \sum_{k=0}^{2-1} (-\frac12)^k = (-\frac12)^0 + (-\frac12)^1 = 1 - \frac12 = \frac12$$
so that
$$\epsilon_2 = |\frac12- \frac23| = |-\frac16| = \frac16$$


TITLE: Integrated luminosity in a Monte Carlo Simulation
QUESTION [0 upvotes]: Suppose we have a Monte Carlo simulation comprising momenta values. How do we calculate the integrated luminosity? There is no time and all values are discrete so I don't understand how I should integrate it.

REPLY [1 votes]: Simulation is a way to model a certain physics process. One can generate as many events of this process ($N_{signal}$) as it is needed. Knowing the number of events in your given sample, and the cross-section of this process (either theoretical or experimentally measured in the past) you can convert the number of events into the integrated luminosity by simply doing $\mathcal{L} = N_{signal}/\sigma_{process}$. Note that if the process involves a decay of a particle A into a certain final state, you also need to take into account both production cross-section of the particle A and the branching fraction of its decay to this final state. In addition, detector imperfections lead to detection efficiency being lower than 1. Therefore, you need to know the number of generated events rather than the number of events reconstructed by the detector in your simulation.
However in reality, the concept of luminosity is rarely used when working with simulation, but rather used when describing the amount of experimental data.


TITLE: Different methods for measuring intensity of IR wavelengths
QUESTION [1 upvotes]: I am proposing a teaching experiment at my university for undergraduates to learn about black body radiation and part of it will involve measuring intensities of different near - mid IR wavelengths. This is where the problem lies.
As far as I am aware there are two methods I could propose to do this:

Using a spectrometer that can measure in the required range. This equipment is however very expensive and so is probably a no go.
Using IR filters and a suitable photodiode to measure intensities. The problem here however is that I do not know of any filters that will only allow small IR bandwidths through. I can only find filters that block out optical wavelengths and allow all IR to pass through.

Are there any other methods I could use or am I going to have to bite the bullet and propose buying a £20,000 for this lab?

REPLY [0 votes]: Building your own spectrometer as suggested by @ostrichCamel is not a bad idea to save cash because a grating spectrometer is rather simple, although it would take some effort to get right.  Another option is these filters, which are still a little pricy, but I think they might be just what you need.
Another issue is that you’ll need either a hot lamp source or have to shell out for a cooled detector (TEC or LN$_2$).  This is necessary because the detector is constantly emitting black body radiation at its own temp. Thus, a 300 K detector will detect no net light from a 300 K black body source. So you need to either heat up the source or cool the detector.  If you plan to use an incandescent lamp, probably the glass will block the emission in the mid-IR range, so that’s another issue.
You could probably do this decently for 1000-2000 $/£.  But if your department will agree to get you a spectrometer, I think it could pay dividends in many other ways as well, since it is such a versatile tool.
Good luck!


TITLE: Specificity of domain of multivariable function
QUESTION [0 upvotes]: Say we have the function $$h(x,y) = \dfrac{1}{\sqrt{x-y^2}}$$
Is $\text{dom} \ h = \{(x,y) \in \mathbb{R}^2 : x > y^2\}$? WolframAlpha puts the additional restriction that $x > 0$, but I assume this is already implied in $x > y^2$, since $y^2$ is bounded below by $0$.

REPLY [1 votes]: Your domain is correct. It is likely that WolframAlpha is being redundant in its domain specification because its algorithm find the constraints $x > 0$ and $x > y^2$ separately, and reports both without realizing that $x > y^2$ implies the former.


text:
\begin{document}

\vspace*{0.5in}
\begin{center}
   {\LARGE Tangency quantum cohomology \par}
   \vskip 1.5em
   {\large \lineskip .5em
      Joachim Kock\footnote{
      Supported by the National Science Research Council of Denmark.

      Date: 10th of June 2000.

      Current address: Dept.\ of Mathematics, \ Royal Institute of Technology, \
      100 44 Stockholm, Sweden.

      E-mail: {\tt kock@math.kth.se}}
   \par}
   \vskip 1em
   {\normalsize Universidade Federal de Pernambuco\\
   Recife, Brazil}
\end{center}
\par
\vskip 1.5em

\begin{abstract}
   Let $X$ be a smooth projective variety.  Using modified psi classes on the
   stack of genus zero stable maps to $X$, a new associative quantum product is
   constructed on the cohomology space of $X$.  When $X$ is a homogeneous
   variety, this structure encodes the characteristic numbers of rational
   curves in $X$, and specialises to the usual quantum product upon resetting
   the parameters corresponding to the modified psi classes.  For $X=\P^2$, the
   product is equivalent to that of the contact cohomology of
   Ernstr\"om-Kennedy.
\end{abstract}

\section*{Introduction}

Let $X$ be a smooth projective variety over the complex numbers.  The
Gromov-Witten invariants of $X$ are constructed by pulling back cohomology
classes to the stack of stable maps to $X$ and integrating over the virtual
fundamental class (cf.~Behrend-Manin~\cite{Behrend-Manin},
Behrend~\cite{Behrend}, Li-Tian~\cite{Li-Tian}).  Let $\Phi$ denote the
generating function for the genus zero Gromov-Witten invariants.  The quantum
product is defined on the cohomology space of $X$ by taking the third
derivatives of $\Phi$ as structure constants,
$$
\basis_i * \basis_j \df \sum_{e,f} \Phi_{ije} \; g^{ef} \; \basis_f,
$$
cf.~Kontsevich-Manin~\cite{KM:9402}.  The associativity of this product is
equivalent to the fact that $\Phi$ satisfies the WDVV equations
\begin{eqnarray*}
	\sum_{e,f} \Phi_{ije} \; g^{ef} \; \Phi_{fkl}
	&=& \pm
\sum_{e,f} \Phi_{jke} \; g^{ef} \; \Phi_{fil}.
\end{eqnarray*}
A feature of this structure is that when $X$ is a homogeneous variety, the
Gromov-Witten invariants are solutions to enumerative problems of rational
curves in $X$ subject to incidence conditions, and then the WDVV equations
provide relations among the numbers.

It is straightforward to generalise the construction by including psi classes in
the integrals, to get a larger quantum product: instead of using just the
primary Gromov-Witten potential as above, one uses the whole gravitational
Gromov-Witten potential.  This product is associative too, due to the
(generalised) WDVV equations (Witten~\cite{Witten}).  This larger product has
not brought about as much interest as the first one, for two reasons: the
gravitational descendants are not directly interpretable in enumerative
geometry, and the equations corresponding to associativity are ``perpendicular
to the descent'' in the sense that each equation relates only invariants with
the same psi classes as factors.

\bigskip

The present work introduces a new extension of the quantum product, the {\em
tangency quantum product}, and proves its associativity.  When $X$ is a
homogeneous variety, this structure encodes all the characteristic numbers of
rational curves in $X$.

The construction is based on a different potential $\Gamma$, called the {\em
tangency quantum potential}, which incorporates one level of {\em modified psi
classes}.  Modified psi classes are boundary corrections of the tautological psi
classes motivated from enumerative geometry: tangency conditions are easily
expressible in terms of modified psi classes.  They were first discovered and
studied by R.~Pandharipande (unpublished \cite{Pand:mail}) who used them to
establish a topological recursion for the characteristic numbers of rational
curves in $\P^2$, and later, independently, by myself~\cite{_9902}.  (This work
is subsumed in \mbox{Graber-Kock-Pandharipande~\cite{Grab-Kock-Pand}}).

In contrast to the tautological psi classes, the modified ones restrict to the
boundary in a non-trivial way, giving rise to certain {\em diagonal classes}.
Therefore the splitting lemma (\ref{split2}), which is the engine in the proof
of the associativity, is complicated by correction terms.  This mess is handled
by the introduction of a new ``metric'' $\gamma_{ef}$ on the cohomology space of
$X$.  It is a deformation of the Poincar\'e metric $g_{ef}$ in much the same way
as the tangency quantum potential is a deformation of the Gromov-Witten
potential. Amazingly it encodes all the combinatorics of the diagonal
corrections in the splitting lemma, leading to certain ``deformed'' WDVV
equations.  These in turn are equivalent to associativity of the {\em tangency
quantum product}, defined as
$$
\basis_i * \basis_j \df \basis_i \cp \basis_j  +
\sum_{e,f} \Gamma_{ije} \, \gamma^{ef} \, \basis_f.
$$
(This deformed metric also clarifies the topological recursion, as explained in
\cite{Grab-Kock-Pand}.)

The last section is devoted to a technical detail: since modified psi classes
are not defined in degree zero, the same goes for the tangency potential.  The
missing degree zero part is now constructed separately to account for the
classical part of the new product.  This construction shows that the new product
is in fact ``integrable'' in the sense that its structure constants (with
respect to $\gamma$) are the third derivatives of a single function.  With this
last piece of data, the results of the paper can be summarised in saying that
the new structure is that of a formal Frobenius (super)manifold (over the power
series ring in the new variables).

Acknowledgements: this work is a part of my PhD thesis~\cite{_thesis}, and I am
indebted to my advisor Israel Vainsencher.  This particular part was carried out
while I was visiting the California Institute of Technology; I am thankful to
Caltech for its hospitality, and in particular to Rahul Pandharipande for
encouraging my research, and for many stimulating conversations.  I
have also benefited from conversations with Marzia Polito and correspondence
with Gary Kennedy.

\section{First enumerative descendants}

\begin{blanko}{Set-up.}
	Throughout we work over the field of complex numbers.  Let $X$ denote a
	smooth projective variety, and let $\basis_0,\ldots,\basis_r$ denote the
	elements of a homogeneous basis of the cohomology space $H\upperstar (X,\Q)$.
	As in Manin's book~\cite{Manin} we consider $H\upperstar (X,\Q)$ as a linear
	supermanifold.

	Let $\ov M_{0,S}(X,\beta)$ denote the moduli stack of Kontsevich stable maps
	of genus zero whose direct image in $X$ is of class $\beta\in H_2^+(X,\Z)$,
	and whose marking set is $S=\{p_1,\ldots,p_n\}$.

	For each mark $p_i$, let $\nu_i : \ov M_{0,S}(X,\beta) \pil X$ denote the
	evaluation morphism that sends the class of a map $\mu$ to $\mu(p_i)$;
	pull-backs along evaluation morphisms of cohomology classes of $X$ are
	referred to as {\em evaluation classes}.  The reader is referred to
	Fulton-Pandharipande~\cite{FP-notes} for definitions and basic properties of
	stable maps, Gromov-Witten invariants and quantum cohomology.
\end{blanko}

\begin{blanko}{Modified psi classes (cf.~\cite{Grab-Kock-Pand}).}
	Let $\pi_0 : \ov M_{0,S\cup\{p_0\}}(X,\beta) \pil \ov M_{0,S}(X,\beta)$
	denote the forgetful morphism that forgets the extra mark $p_0$.  Together
	with the sections $\sigma_i$ corresponding to each of the marks in $S$, it
	constitutes the universal family.  For each mark there is a psi class
	defined as $\psiclass_i \df c_1(\sigma_i\upperstar \kan_{\pi_0})$, where
	$\kan_{\pi_0}$ is the relative dualising sheaf of $\pi_0$.  When $\beta\neq
	0$, the {\em modified psi class} is $\mpsiclass_i \df\hat\pi_i\upperstar
	\psiclass_i$, where $\hat\pi_i : \ov M_{0,S}(X,\beta) \pil \ov
	M_{0,\{p_i\}}(X,\beta)$ is the forgetful morphism that forgets all marks but
	$p_i$.

	The importance of modified psi classes comes from enumerative geometry: for
	example, if $X$ is a homogeneous variety and $Z\subset X$ is a very ample
	hypersurface of degree $\coh z\in H^2(X,\Q)$, then the cycle of maps tangent
	to $Z$ at $p_i$ is of class $\nu_i\upperstar (\coh z) \cp ( \mpsiclass_i +
	\nu_i\upperstar (\coh z))$.  It then follows from a transversality argument
	that the characteristic numbers of rational curves are top products of
	evaluation classes and modified psi classes.  For this reason, these top
	products (integrated over the virtual fundamental class of $\ov
	M_{0,S}(X,\beta)$),
   $$
   \Brac{\mtau_{m_1}(\coh z_1)\cdots \mtau_{m_n}(\coh z_n)}\!_{\beta} \df
\int \ov\psiclass{}_1^{m_1}\cp\nu_1\upperstar (\coh z_1) \cp\cdots \cp
\ov\psiclass{}_n^{m_n}\cp\nu_n\upperstar (\coh z_n)
\ \cap [\ov M_{0,S}(X,\beta) ]^{\text{virt}}
   $$
	are called {\em enumerative descendants}.

The enumerative descendants that appear in the characteristic number problem are
only the {\em first} enumerative descendants, i.e.\ those made up of factors of
type $\mtau_0(\coh x)$ and $\mtau_1(\coh y)$, for $\coh x, \coh y \in
H\upperstar (X,\Q)$.  For simplicity, we will work in the fixed basis
$\basis_0,\ldots,\basis_r$ and employ the following notation for the
corresponding integrals:
\begin{equation}\label{tauab}
\brac{\mbtau{0}{\fat a} \ \mbtau{1}{\fat b}}\!_{\beta} \df
\Brac{ \smallprod{k=0}{r} (\mtau_0(\basis_k))^{a_k}
\smallprod{\ell=0}{r} (\mtau_1(\basis_\ell))^{b_\ell}}\!_{\beta},
\end{equation}
where $\fat{a}=(a_0,\ldots,a_r)$ and $\fat{b}=(b_0,\ldots,b_r)$ are vectors of
non-negative integers.  The integral is zero unless $\sum_k
\brok{1}{2}\deg(\basis_k) (a_k+b_k+1) = \vdim_{\C} \ov M_{0,S}(X,\beta)$, where
$S$ is of cardinality $n= \sum (a_k + b_k)$.
\end{blanko}

\begin{blanko}{Diagonal classes (cf.~\cite{Grab-Kock-Pand}).}\label{diagonal}
	The {\em diagonal class} $\deltaclass_{ij} \in H^2(\ov M_{0,S}(X,\beta), \Q)$
	is defined as the sum of all boundary divisors having $p_i$ and $p_j$
	together on a contracting twig.  It can also be described as the pull-back
	from $\ov M_{0,\{p_i,p_j\}}(X,\beta)$ of the divisor $D_{ij}$ having both
	marks on a contracting twig; from this description it is clear that diagonal
	classes are invariant under pull-back along forgetful morphisms.

   Let $\pi_0 : \ov M_{0,S\cup\{p_0\}}(X,\beta) \pil \ov M_{0,S}(X,\beta)$
   denote the morphism that forgets $p_0$.  Then
   \begin{equation}\label{push-delta}
      \pi_0{}\lowerstar \deltaclass_{0i} = 1,
      \end{equation}
   (the virtual fundamental class of $\ov M_{0,S}(X,\beta)$).

   The diagonal classes owe their name to  the following properties.
   \begin{eqnarray}
      \deltaclass_{ij}\deltaclass_{ik} & = &
      \deltaclass_{ij}\deltaclass_{jk}\notag\\
      -\deltaclass_{ij}^2 \; = \;\deltaclass_{ij}\mpsiclass_i  & = &
      \deltaclass_{ij}\mpsiclass_j\notag\\
      \deltaclass_{ij}\nu_i\upperstar (\coh z)
      & = & \deltaclass_{ij}\nu_j\upperstar (\coh z),\label{swap}
   \end{eqnarray}
   for $\coh z\in H\upperstar (X,\Q)$.
\label{D_ij}
	The swapping property also holds when $\deltaclass_{ij}$ is replaced with
	$D_{ij}$.
\end{blanko}

\begin{blanko}{Key formula.}
   Let $D = D(S', \beta' \mid S'', \beta'')$ denote the boundary divisor which
   is the image of the morphism
$$
\rho_D : \ov M_{0,S' \cup \{x'\}}(X,\beta')
\times_X \ov M_{0,S'' \cup \{x''\}}(X,\beta'')
\langpil \ov M_{0,S}(X,\beta)
$$
	consisting in gluing together the two maps at $x'$ and $x''$.  We will assume
	that $S'$ and $S''$ are non-empty and that $\beta'$ and $\beta''$ are
	non-zero.  Then $\rho_D$ is birational onto $D$.  For short, let $\ov M'$ and
	$\ov M''$ denote the factors of the fibred product above, and let
   $$
   \jmath_D : \ov M' \times_X \ov M'' \langpil \ov M' \times \ov M''
   $$
   denote the inclusion in the cartesian product.

   Now there is the following formula for restricting a modified psi class to the
   boundary (assuming $p_i \in S'$).
   \begin{equation} \label{restr-mpsi}
   \rho_D\upperstar \ov \psiclass_i =
\jmath_D\upperstar\big( \mpsiclass_i+\deltaclass_{ix'}\big).
   \end{equation}
\end{blanko}

\begin{lemma}\label{split2}\textsc{Splitting lemma for first enumerative
	descendants.} Let $D= D(S',\beta' | S'',\beta'')$ be a boundary divisor with
	$\beta'$ and $\beta''$ non-zero.  Let $\coh x$ and $\coh y$ be generic even
	elements of $H\upperstar (X,\Q)$, and let there be given a product
	$\mtau_{0}(\coh x)^a\,\mtau_1(\coh y)^b$ such that the marks
	corresponding to $a' \leq a$ and $b' \leq b$ belong to $S'$, and the
	remaining marks, corresponding to $a'' \leq a$ and $b'' \leq b$, belong to
	$S''$.  Then \setlength{\multlinegap}{0pt}
   \begin{multline*}

      \brac{ D \cdot
   \mtau_{0}(\coh x)^a\,\mtau_1(\coh y)^b}\!_\beta =
   \sum_{p,q} \sum_{{s}',{s}''}
   \smallbinom{{b}'}{{s}'}\smallbinom{{b}''}{{s}''}
   \brac{\mtau_{0}(\coh x)^{a'}\,\mtau_1(\coh y)^{b'-s'} \,
   \mtau_0(\coh y^{{s}'}\!\cp \basis_p)}\!_{\beta'}
   \\ g^{pq} \
   \brac{\mtau_0(\basis_q\cp\coh y^{{s}''})
   \,\mtau_{0}(\coh x)^{a''}\,\mtau_1(\coh y)^{b''-s''}}\!_{\beta''}.
   \end{multline*}

	{\rm The outer sum is over the splitting indices $p$ and $q$ running from $0$
	to $r$, and the inner sum is over all ${s}'$ and ${s}''$: the symbol
	$\mtau_{1}(\coh y)^{ b'- s'}$ makes sense only for ${s}'\leq{b}'$, but since
	there is a binomial coefficient $\binom{{b}'}{{s}'}$ in front of it, which is
	zero unless ${s}'\leq{b}'$ we will allow any ${s}'$.  Ditto for ${s}''$.  }
   \end{lemma}
\begin{dem}
   There are four ingredients in the proof of this formula: the first is of
   course the splitting axiom of Gromov-Witten theory
   (cf.~Kontsevich-Manin~\cite{KM:9402}), which accounts for the overall shape
   of the formula.  Second, formula~(\ref{restr-mpsi}) tells how each factor
   $\mtau_1(\coh y)$ (say realised at mark $p_i$ as $\mpsiclass_i \cp
   \nu_i\upperstar (\coh y)$) restricts to give $(\mpsiclass_i +
   \deltaclass_{ix})\cp\nu_i\upperstar (\coh y)$ on the twig containing $p_i$ ---
   here $x$ denotes the gluing mark of that twig.  Now (on each twig) expand the
   product of all these restrictions into a sum over $s$ (which accounts
   for the binomial coefficients).  Third, apply formula~(\ref{swap}) to write $
   \deltaclass_{ix}\cp \nu_i\upperstar (\coh y) = \deltaclass_{ix}\cp
   \nu_{x}\upperstar (\coh y)$, with the effect of accumulating evaluation classes
   on the gluing mark.  Finally, by equation~(\ref{push-delta}) we can push down
   each term involving a diagonal class $\deltaclass_{ix}$ along the forgetful
   morphism forgetting $p_i$; the effect is simply deleting $\deltaclass_{ix}$.
\end{dem}

\newcommand{\ring}{\Lambda}

\begin{blanko}{The tangency quantum potential }\label{Gamma}
	$\Gamma$ shall now be defined as the generating function for the first
	enumerative descendants.
	Let $\coh x= \sum x_i \basis_i$ and $\coh y = \sum y_i \basis_i$ be
	generic even elements of $H\upperstar (X,\Q)$, and set
\begin{eqnarray}\label{Gammadef}
\Gamma(\coh{x},\coh{y})
&\df& \sum_{\beta>0} q^\beta
\brac{ \exp( \mtau_0(\coh x) + \mtau_1(\coh y))}\!_\beta \\
&=& \sum_{\beta>0} q^\beta \sum_{a,b}
\brac{\frac{\mtau_0(\coh x)^a}{a!} \frac{\mtau_1(\coh y)^b}{b!}}\!_\beta \notag .
\end{eqnarray}
The inner sum is over {\em all} non-negative integers $a$ and $b$ --- this is
meaningful since there is no $\beta=0$ term in the outer sum.  (The degree-zero
case is considered and included below, in \ref{beta0} and \ref{interpret0}.)
The coefficients $q^\beta$ are necessary only to ensure formal convergence; they
belong to the Novikov ring $\ring$ (as defined in Getzler~\cite{Getzler:9612}),
which will be our coefficient ring when treating the tangency quantum potential.

For convenience we identify $\coh x$ and $\coh y$ with their coordinate vectors
$\fat x = (x_0,\ldots,x_r)$ and $\fat y = (y_0,\ldots,y_r)$ with respect to the
basis $\basis_0,\ldots,\basis_r$.  The potential thus belongs to the
power series ring $\ring[[\fat x, \fat
y]]=\ring[[x_0,\ldots,x_r,y_0,\ldots,y_r]]$, and expands to
\begin{equation}\label{ybxa}
\Gamma(\fat{x},\fat{y})  = \sum_{\beta>0}q^\beta\sum_{\fat{a},\fat{b}}
\frac{\fat{y}^{\fat{b}}}{\fat{b}!}\frac{\fat{x}^{\fat{a}}}{\fat{a}!}
\brac{\mbtau{0}{\fat a} \ \mbtau{1}{\fat b}}\!_{\beta}.
\end{equation}
Here
the inner sum is over all pairs of vectors
$\fat{a}=(a_0,\ldots,a_r)$ and $\fat{b}=(b_0,\ldots,b_r)$ of non-negative
integers, and we employ multi-index notation,
e.g. $\fat a!= a_0!\cdots a_r!$.  For the formal variables $\fat x$ and $\fat y$,
the multi-index notation is reverse, in order to preserve the signs arising from
odd variables, e.g.\
$\fat x^{\fat a} = x_r^{a_r}\cdots x_0^{a_0}$

The variables $\fat{x}$ are the usual formal variables from quantum cohomology,
so when $\fat{y}$ is set to zero, $\Gamma$ reduces to the usual (quantum part of
the) genus-zero Gromov-Witten potential.

The viewpoint of (\ref{ybxa}) is advantageous for the sake of extracting the
invariants, as well as for checking the validity of certain formal operations on
the potential.  This task is safely left to the reader and henceforth only the
more compact notation of (\ref{Gammadef}) is used.
\end{blanko}

\section{Deformation of the Poincar\'e metric}

While the usual quantum potential is based on the Poincar\'e metric constants
$g_{ij} = \otrace{\basis_i\cp\basis_j}$, the tangency quantum potential relates
more naturally to a deformation of them, a certain ``metric'' with
values in $\Q[[\fat{y}]]=\Q[[y_0,\ldots,y_r]]$.

\begin{blanko}{The classical product in the Poincar\'e metric.}
	The two important structures on $H=H\upperstar (X,\Q)$ are the intersection
	product $\cp$, and the trace map $\otrace{} : H \pil \Q$ which is just
	integration over the fundamental class of $X$.  Set
	$g_{ij}=\otrace{\basis_i\cp\basis_j}$ and $g_{ijk} = \otrace{\basis_i \cp
	\basis_j \cp \basis_k}$.  Let $(g^{ij})$ be the inverse matrix to $(g_{ij})$.
	It is used to raise indices as needed; in particular, with $g_{ij}^k = \sum_e
	g_{ije}\,g^{ek}$, we have the multiplication formula
\begin{equation}\label{gijk}
\basis_i \cp \basis_j = \smallsum{k}{} g_{ij}^k \; \basis_k.
\end{equation}
\end{blanko}

\begin{blanko}{Intersection polynomials of $X$.}
	For a generic even element $\coh y \in H$, (identified with its coordinates
	with respect to $\basis_0,\ldots,\basis_r$ as in \ref{Gamma}), let
	$\phi(\coh{y}) \in \Q[[\fat{y}]]$ be the generating function for the
	integrals on $X$,
	\begin{eqnarray*}
		\phi(\coh y) \;\df\; \otrace{\exp(\coh y)}
		& = & \sum_{n\geq 0} \frac{1}{n!}\otrace{\coh y^n} .

	\end{eqnarray*}

	Note that the expansion is $\exp(y_0)$ times a polynomial
	in $y_1,\ldots,y_r$.

   Now set
$$
   \phi_{ij} \df \frac{\d^2}{\d y_i \d y_j} \, \phi
	= \otrace{\exp(\coh y)\cp\basis_i\cp\basis_j}
$$
and use the matrix $(g^{ef})$ to raise indices, putting
\begin{equation}\label{gandphi}
	 \phi^i{}_j \df \smallsum{e}{} g^{ie} \, \phi_{ej} ,
	   \quad\text{ }\quad
	 \phi_j{}^i{} \df \smallsum{e}{}  \phi_{je}\, g^{ei} ,
	   \quad\text{ and }\quad
	 \phi^{ij} =\sum_{e,f} g^{ie} \, \phi_{ef} \, g^{fj}.
\end{equation}

The two first entities appear in the following multiplication formula,

\begin{eqnarray}
	\exp(\coh y) \cp \basis_p \label{exptophi}
	&=& \smallsum{e,f}{} \, \basis_e \, g^{ef} \, \otrace{\basis_f \cp
\exp(\coh y) \cp \basis_p}  \\
	 & = & \smallsum{e}{} \, \basis_e \, \phi^e{}_p(\coh y),\notag
\end{eqnarray}

and similarly
$
\basis_q \cp \exp(\coh y) = \smallsum{f}{} \, \phi_q{}^f(\coh y) \, \basis_f .
$
(These are identities in $H[[\fat y]]$.)
\end{blanko}

\begin{lemma}
   \textsc{Sum formula.} Let $\coh y'$ and $\coh y''$ be generic even elements
   of $H$. Then
   $$
    \phi(\coh y'+\coh y'') =
  \sum_{e,f} \phi_e(\coh y') \, g^{ef} \, \phi_f(\coh y'').
   $$
\end{lemma}
\begin{dem}
   Write down the exponential series for $X\times X = X'\times X''$,
   $$
   \exp\big( \smallsum{}{} y'_i \basis_i' + \smallsum{}{} y''_i \basis_i''\big)
   $$
	and integrate over the diagonal $\Delta$.  Computing this integral using the
	isomorphism $X\simeq \Delta$ yields the left hand side.  On the other hand,
	computing the integral using the K\"unneth formula for $\Delta$ and the
	projection formula yields the right hand side.
\end{dem}
This sum formula (and its derivatives) ensures that the expected
index-raising rules hold.  For example, the useful formula
\begin{equation}\label{phiijk}
   \phi^i{}_k{}^j = \smallsum{\ell}{} \, g_{k\ell}^i \, \phi^{\ell j}
\end{equation}
amounts to $\phi^i{}_k{}^j( \coh 0 + \coh y) =
	\smallsum{\ell,m}{}  \phi^i{}_{k\ell}(\coh 0)\,  g^{\ell m} \,
	\phi_m{}^j(\coh y)
$.

\begin{blanko}{The deformed metric.}\label{defmet}
   Instead of using the integral $\otrace{} : H \pil \Q$, the new metric is based
   on the linear map $\ytrace{ } :H \pil \Q[[\fat{y}]]$ defined as
\begin{eqnarray*}
	\ytrace{\coh z} & \df & \otrace{ \exp(-2\coh y) \cp \coh z } .

\end{eqnarray*}
   It is thought of as a deformation of $\otrace{}$ since we recover this
   map upon setting $\coh y = \coh 0$.
   Now define the new metric $(\gamma_{ij})$ by
   $$
   \gamma_{ij} \df \gamma_{ij}(\coh y) \df
   \ytrace{\basis_i \cp \basis_j} = \phi_{ij}(-2\coh{y}),
   $$
	and adopt the obvious notation $\gamma_{ijk} = \ytrace{\basis_i \cp \basis_j\cp
	\basis_k} = \phi_{ijk}(-2\coh{y})$.  Let $(\gamma^{ij})$ denote the
	inverse matrix to $(\gamma_{ij})$.  Then it follows readily from the sum
	formula above that
   \begin{equation}  \label{gamma-def}
      \gamma^{ij} = \phi^{ij}(2\coh{y})  = \sum_{e,f}
      \phi^i{}_e \, g^{ef} \, \phi_f{}^j.
   \end{equation}
   The sum formula also yields
	$\sum_{e} \gamma_{ije} \, \gamma^{ef}
	= \sum_e \phi_{ije}(-2\coh{y})\; \phi^{ef}(2\coh{y}) =
   \phi_{ij}{}^f(\coh{0}) = g_{ij}^f$, so the intersection product can be
   written
   $$
   \basis_i \cp \basis_j = \sum_{e,f} \gamma_{ije} \, \gamma^{ef} \, \basis_f .
   $$
\end{blanko}

\begin{eks}\label{phiP2}
   For $\P^2$ (with $h\df c_1(\OO(1))$ and basis $\basis_i\df h^i$), we get
   $$
   (\gamma^{ij}) = \exp(2y_0)\begin{pmatrix}
   \phantom{0}0\phantom{0} & \phantom{0}0\phantom{0}& \phantom{0}1\phantom{0} \\
   0 & 1 & 2y_1 \\
   1 & 2y_1 & 2y_1^2 + 2y_2
   \end{pmatrix}.
   $$
   This matrix (with $y_0$ and $y_2$ set to zero) was first written down in the
   pioneer
   article~\cite{diFrancesco-Itzykson} of Di~Francesco and Itzykson.
\end{eks}

\section{WDVV equations and the tangency quantum product}

\begin{blanko}{Notation.}
	Let lower indices denote partial derivatives, e.g.
	$$
	\Gamma_{ij} \df \Gamma_{x_i x_j} \df \frac{\d^2}{\d x_i \d x_j}\Gamma =
	\sum_{\beta>0} q^\beta
	\brac{ \exp( \mtau_0(\coh x) + \mtau_1(\coh y))\cdot\mtau_0(\basis_i)\,
	\mtau_0(\basis_j)}\!_\beta ,
	$$
	and set
\begin{equation}  \label{Gammaij}
   \Gamma_{(ij)} = \Gamma_{(x_i x_j)} \df \sum_{k=0}^r g_{ij}^k \ \Gamma_{x_k},
\end{equation}
--- the ``directional derivative with respect to the product
$\basis_i\cp \basis_j = \sum g_{ij}^k \basis_k$''.
\end{blanko}

\begin{satz}\label{WDVV}
	The following form of the WDVV equations holds for the tangency quan\-tum
	potential.
\begin{eqnarray*}
	\Gamma_{(ij)k\ell} + \Gamma_{ij(k\ell)} +
	\sum_{e,f}\; \Gamma_{ije} \ \gamma^{ef} \ \Gamma_{fk\ell} & = & \pm\big(
   \Gamma_{(jk)i\ell} + \Gamma_{jk(i\ell)} +
	\sum_{e,f}\; \Gamma_{jke} \ \gamma^{ef} \ \Gamma_{fi\ell}\big),
\end{eqnarray*}
where $\pm$ denotes the sign $(-1)^{\deg \basis_i(\deg \basis_j +\deg\basis_k)}$.
\end{satz}
\begin{obs}
   In the special case $X=\P^2$, it turns out the only non-trivial relation is
   the one with $i=j=1$ and $k=\ell=2$.  Since $\Gamma_0 = 0$ by the string
	equation (cf.~\cite{Grab-Kock-Pand}), the relation then reads
$$
\Gamma_{222} = \exp(2y_0)\big(\Gamma_{112}^2 - \Gamma_{111}\Gamma_{122} +
2y_1 ( \Gamma_{122}\Gamma_{112} - \Gamma_{111}\Gamma_{222} )
+ (2y_1^2 + 2y_2)( \Gamma_{122}^2 - \Gamma_{112}\Gamma_{222})\big).
$$
This equation was first found by L.~Ernstr\"om and G.~Kennedy~\cite{EK2},
cf.~Remark~\ref{EKremark} below, while the special case of $y_0=y_2=0$ goes back
to Di~Francesco-Itzykson~\cite{diFrancesco-Itzykson}.  Setting also $y_1=0$ we
are back to the celebrated formula of M.~Kontsevich~\cite{KM:9402}.
\end{obs}
\begin{dem}
	The proof follows the line of arguments of the proof of the WDVV equations
	for the usual Gromov-Witten potential, cf.~Kontsevich-Manin~\cite{KM:9402}.
	The novelty is the splitting lemma for enumerative descendants and the
	appearance of the deformed metric.

	For fixed $\beta>0$ and $a,b\geq 0$, consider the moduli stack $\ov
	M_{0,\{p_1,p_2,p_3,p_4\}\cup S}(X,\beta)$ where $S$ is a marking set of
	cardinality $a+b$.  Consider the product
	$$
	\mtau_0(\basis_i)\mtau_0(\basis_j)\mtau_0(\basis_k)\mtau_0(\basis_\ell) \
	\frac{\mtau_0(\coh x)^a}{a!}\frac{\mtau_1(\coh y)^b}{b!}
	$$
	where the first four classes correspond to the marks $p_1,p_2,p_3,p_4$.
	Now integrate the product over each side of the
	fundamental equivalence
   \begin{equation}\label{equiv}
   (p_1, p_2 \mid p_3, p_4) = ( p_2, p_3 \mid p_1, p_4),
   \end{equation}
	where $(p_1, p_2 \mid p_3, p_4)$ denotes the sum of all boundary divisors
	having $p_1$ and $p_2$ on one twig and $p_3$ and $p_4$ on the other.  Summing
	up these equations over all $a,b\geq 0$ and over all $\beta>0$ (as in the
	definition of $\Gamma$) we'll get the desired equation.  Let us treat the
	left hand side.  On the right hand side of the equation the arguments are the
	same; only it is necessary initially to permute the four special factors,
	which accounts for the sign.

	On the left hand side of the equation, let us first consider the contribution
	from the trivial degree partitions, say $\beta'=0$.  Then the only possible
	distribution of the marks giving contribution is when all the spare marks
	fall on the right hand twig, which leaves us with the single boundary divisor
	$D_{12}$.  Now according to \ref{D_ij}, the effect of multiplication with
	this divisor is to merge the two classes $\mtau_0(\basis_i)$ and
	$\mtau_0(\basis_j)$, so in the end we get
   $$
   \brac{\mtau_0(\basis_i\cp \basis_j)\mtau_0(\basis_k)\mtau_0(\basis_\ell)
\frac{\mtau_0(\coh x)^a}{a!}\frac{\mtau_1(\coh y)^b}{b!}}\!_{\beta} .
$$
	Similarly, the case $\beta''=0$ gives
	$\brac{\mtau_0(\basis_i)\mtau_0(\basis_j)\mtau_0(\basis_k\cp \basis_\ell)
	\frac{\mtau_0(\coh x)^a}{a!}\frac{\mtau_1(\coh y)^b}{b!}}\!_{\beta}$.
	Summing over $\beta>0$ and $a,b\geq 0$ gives exactly the two linear terms on
	the left hand side of the promised equation, cf.~(\ref{Gammaij}).

   Now for those boundary divisors in the linear equivalence corresponding to
   strictly positive degree partitions.  To each irreducible component on the
   left hand side of (\ref{equiv}), we apply the Splitting Lemma~\ref{split2},
   getting all together
	   \begin{multline*}

   \bigsum{}{}\sum_{p,q}
   \brac{\frac{\mtau_{0}(\coh x)^{a'}}{a'!}\frac{\mtau_1(\coh
   y)^{b'}}{b'!}\mtau_0(\basis_i)\mtau_0(\basis_j) \,
   \mtau_0(\frac{\coh y^{c'}}{c'!}\!\cp \basis_p)}\!_{\beta'}
   \\ g^{pq} \
   \brac{\mtau_0(\basis_q\cp\frac{\coh y^{c''}}{c''!})
   \,\mtau_0(\basis_k)\mtau_0(\basis_{\ell})\frac{\mtau_{0}(\coh x)^{a''}}{a''!}
	\frac{\mtau_1(\coh y)^{b''}}{b''!}}\!_{\beta''}.
   \end{multline*}

where the big outer sum is over all $a'+a''=a$ and all $b'+c' + b''+c''= b$.
(The $b'$ corresponds to what was called $b'-s'$ in the splitting lemma.)
Now sum over all $a$ and $b$ getting
   \begin{multline*}

   \sum_{p,q}
   \brac{\exp(\mtau_{0}(\coh x)+\mtau_1(\coh
   y))\cdot\mtau_0(\basis_i)\mtau_0(\basis_j) \,
   \mtau_0(\exp(\coh y)\!\cp \basis_p)}\!_{\beta'}
   \\ g^{pq} \
   \brac{\mtau_0(\basis_q\cp\exp(\coh y))
   \,\mtau_0(\basis_k)\mtau_0(\basis_{\ell})\cdot\exp(\mtau_{0}(\coh x)+
	\mtau_1(\coh y))}\!_{\beta''}.
   \end{multline*}
	Next, use equation (\ref{exptophi}) to get rid of $\exp(\coh y)$,
	and sum over all $\beta>0$ as in the definition of $\Gamma$, arriving at
$$
\underset{e,f}{\sum_{p,q}} \ \Gamma_{ije} \, \phi^e{}_p \, g^{pq} \,
\phi_q{}^f \, \Gamma_{fk\ell} .
$$

By equation~(\ref{gamma-def}),  this is just the quadratic term of the left hand
side of the desired equation.

\end{dem}

\begin{blanko}{Topological recursion relation.}
	The above deformed WDVV equations alone are not sufficient to determine all
	the first enumerative descendants from the primary Gromov-Witten invariants.
	But topological recursion is available also for the enumerative descendants
	cf.~\cite{Grab-Kock-Pand}.  In the present set-up, that topological recursion
	relation takes the following pleasant form.
   $$
   \Gamma_{y_i x_j x_k } = \Gamma_{x_i(x_j x_k)} - \Gamma_{(x_i x_j) x_k}
   - \Gamma_{(x_i x_k) x_j} +
   \sum_{e,f} \; \Gamma_{x_i x_e} \ \gamma^{ef} \ \Gamma_{x_f x_j x_k}.
   $$
The shape of this equation stems from the boundary expression of the modified psi
class, $\mpsiclass_i = (p_i\mid p_j, p_k) - \boundary_i$, where $\boundary_i$ is
the sum of all boundary divisors such that $p_i$ is on a contracting twig.
The deformed metric enters in the quadratic terms for the same reason as in the
proof above.

\end{blanko}

\begin{blanko}{The tangency quantum product.}\label{tprod}
	The {\em tangency quantum product} is the $\ring[[\fat{x},\fat{y}]]$-bilinear
	product on $H\upperstar (X,\Q) \tensor_\Q \ring[[\fat{x},\fat{y}]]$ defined
	by the rule
$$
\basis_i * \basis_j \df \basis_i \cp \basis_j +
\sum_{e,f} \; \Gamma_{ije} \ \gamma^{ef} \ \basis_f.
$$
\end{blanko}
Observe that this product specialises to the usual quantum product upon setting
the formal variables $\fat{y}$ to zero.  Clearly the product is supercommutative.

\begin{satz}\label{ass}
   The tangency quantum product is associative.
\end{satz}
\begin{dem}
   This is a straightforward consequence of Theorem~\ref{WDVV};  it amounts to
   checking the associativity relations on the generators, using the definition
   of the product.  The only subtle point in the verification is the identity
   $$
      \sum_{e,f} \;\Gamma_{ije} \ \gamma^{ef} \ (\basis_f \cp \basis_k)
   = \sum_{\ell,m}\;\Gamma_{ij(k\ell)} \ \gamma^{\ell m} \ \basis_m ,
$$
which follows from the properties of the structure constants $g_{ij}^k$
listed in (\ref{gijk}), (\ref{phiijk}), and (\ref{Gammaij}).
\end{dem}

\begin{obs}\label{EKremark}
	In the special case $X=\P^2$, this product (or a simple change of coordinates
	of it) was previously constructed via ad hoc methods by L.~Ernstr\"om and
	G.~Kennedy~\cite{EK2}, who also gave a tour de force proof of its
	associativity.  Their construction relies on the space of stable lifts, and
	seems to be peculiar to the projective plane.
\end{obs}

\bigskip

\section{Integrability}

\label{integrability}

\begin{blanko}{The classical potential} for $X$ is the generating function for
the triple top products
\begin{eqnarray*}
\otrace{ \frac{\coh x^3}{3!}}
&=&\sum_{i,j,k} \frac{x_k x_j x_i}{6}
\otrace{ \basis_i \cp \basis_j \cp \basis_k }.
\end{eqnarray*}

By construction, its third derivatives are just $g_{ijk}$, the structure
constants for the cup multiplication (in the Poincar\'e metric).  In usual
quantum cohomology, this potential is reinterpreted as the $\beta=0$ part of the
Gromov-Witten potential: a quantum correction (the $\beta>0$ part) is added to
the classical potential, in such a way that the third derivatives of this sum
are the structure constants of a new associative product, the quantum product.
\end{blanko}

\begin{blanko}{``The tangency classical potential''.}\label{tcp}
The tangency quantum product relates to $\ytrace{}$ exactly as the usual
quantum product relates to $\otrace{}$. So let us introduce a potential
\begin{eqnarray*}
\Phi^0(\coh{x},\coh{y}) \df \ytrace{\frac{\coh x^3}{3!}}
= \sum_{i,j,k} \frac{x_k x_j x_i}{6} \ytrace{\basis_i
\cp \basis_j \cp \basis_k} ,
\end{eqnarray*}
which to fit into the picture could be called the tangency classical potential,
although it is neither classical, nor has anything particular to do with
tangency.  By construction, its third derivatives are $\gamma_{ijk}$, the
structure constants of the cup multiplication, but this time in the deformed
metric:
$$
\basis_i \cp \basis_j = \sum_{e,f} \Phi^0_{ije} \ \gamma^{ef} \ \basis_f .
$$

\end{blanko}

\begin{blanko}{The tangency potential, including $\beta=0$.}\label{beta0}
   Introducing the potential
$$
\Phi(\coh{x},\coh{y}) \df \Phi^{0}(\coh{x},\coh{y}) +
\Gamma(\coh{x},\coh{y}),
$$
whose third derivatives are $\Phi_{ijk} = \gamma_{ijk} + \Gamma_{ijk}$,
the tangency quantum product can be written
$$
\basis_i * \basis_j =
\sum_{e,f} \Phi_{ije} \ \gamma^{ef} \ \basis_f,
$$
and the  WDVV equation~\ref{WDVV} then takes the usual form
   $$
   \sum_{e,f}\; \Phi_{ije} \
   \gamma^{ef} \ \Phi_{fk\ell}
   =
   \pm \sum_{e,f}\; \Phi_{jke} \
   \gamma^{ef} \ \Phi_{fi\ell}.
   $$
\end{blanko}

\begin{blanko}{Interpretation of the degree-zero term.}
   \label{interpret0}
   Writing down this potential $\Phi=\Phi^0 + \Gamma$ calls for an
   interpretation of $\Phi^0$ in terms of some top products on the degree-zero
   spaces.  Unfortunately there is no way to define the modified psi class on
   $\ov M_{0,n}$, if we want it to satisfy the two rules: (i) it should be
   compatible with pull-back along forgetful morphisms, and (ii) it should
   satisfy the push-down formula $\pi\lowerstar \mpsiclass = -2$, independent of
   the number of marks, to give the dilaton equation for modified psi classes
   (cf.~\cite{Grab-Kock-Pand}). Basically this is impossible because the
   1-pointed spaces do not exist in degree zero.

   The best one can do is to define the class on a fixed 4-pointed space and
   then pull it back to the hierarchy lying over this space, but this definition
   depends on the choice of the three extra marks.

So take a moduli space $\ov M_{0,S\cup \{q_1,q_2,q_3\}}(X,0)$ with three
distinguished marks.  For each of the other marks $p_s\in S$, define the
modified psi class $\mpsiclass_s$ as the pull-back from $\ov
M_{0,\{p_s,q_1,q_2,q_3\}}\simeq \P^1$ of the class of degree $-2$.  One easily
checks that this is equivalent to defining $\mpsiclass_s \df \psiclass_s -
\boundary_s$, where $\boundary_s$ is the sum of all boundary divisors such that
$p_s$ is on a (contracting) twig together with at most one of the distinguished
marks.  (This description is then
compatible with the boundary description of the modified psi class in the
$\beta>0$ case (cf.~\cite{Grab-Kock-Pand}), since in that case there are no
distinguished marks.)

In this setting, define the invariant
$$
\brac{\frac{\mtau_0(\coh x)^a}{a!}\frac{\mtau_1(\coh y)^b}{b!}\,
\mtau_0(\basis_i)\mtau_0(\basis_j)\mtau_0(\basis_k)}\!_0
$$
in the obvious way, with the last three classes corresponding to the three
distinguished marks.  Identifying $\ov M_{0,S\cup\{q_1,q_2,q_3\}}(X,0)$ with
$\ov M_{0,S\cup\{q_1,q_2,q_3\}} \times X$, all the evaluation morphisms are just
the projection $p$ to $X$, so the integrand has a factor $p\upperstar(\frac{\coh
x^a}{a!}\cp\frac{\coh y^b}{b!} \cp\basis_i\cp\basis_j\cp\basis_k)$.  The
remaining factors are modified psi classes from $\ov
M_{0,S\cup\{q_1,q_2,q_3\}}$; since each of them is alone on its mark, we can
push them down one by one, arriving at a factor $(-2)^b$.  The remaining
integral $\int 1$ is zero for dimension reasons, unless we have come down to
just $\ov M_{0,\{q_1,q_2,q_3\}}$, which means $a=0$.  We conclude
$$
\brac{\frac{\mtau_1(\coh x)^b}{b!}\,
\mtau_0(\basis_i)\mtau_0(\basis_j)\mtau_0(\basis_k)}\!_0
=
\otrace{ \frac{(-2\coh y)^b}{b!} \cp \basis_i\cp \basis_j
\cp \basis_k }.
$$
Summing over all $b$ (and $a$) we get
$$
\otrace{ \exp(-2\coh y) \cp \basis_i \cp \basis_j \cp \basis_k}
= \ytrace{\basis_i \cp \basis_j \cp \basis_k}
= \gamma_{ijk}
$$
showing that at least the third derivatives of $\Phi^0$ have an
interpretation as top products on degree-0 moduli spaces, and after all
it is the third derivatives that really matter.
\end{blanko}

Once we know that the structure constants of the tangency quantum product are
third derivatives of the single potential $\Phi$, we are in position to
given an interpretation in terms of Frobenius manifolds.

\begin{blanko}{Formal Frobenius manifolds.}
	For convenience let us recall (from Manin~\cite{Manin}, Ch.~III) the
	definition of a formal Frobenius manifold (over $k$).  Let $k$ be a
	supercommutative $\Q$-algebra.  Let $H$ be a free $k$-module of finite rank,
	with generators $\basis_0,\ldots,\basis_r$, and let $g : H \tensor H \pil k$
	denote an even symmetric non-degenerate bilinear pairing.  Let $K=k[[H^t]]$
	be the completed symmetric algebra of the dual module $H^t$.  In other words,
	if

	$\coh x = \sum x_i \basis_i$ is a generic even element of $H$

	then $K=k[[x_0,\ldots,x_r]]$.  Now the structure of a formal Frobenius
	manifold on $(H,g)$ is given by an even potential $\Phi\in K$ (defined up to
	quadratic terms) satisfying WDVV. In other words, the multiplication
	$\basis_i * \basis_j \df \sum_{e,f} \Phi_{ije}\; g^{ef}\; \basis_f$ makes
	$H\tensor_k K$ into an associative supercommutative $K$-algebra.
\end{blanko}

\smallskip

With $H\df H\upperstar (X,\ring)$, the results of this section readily imply:

\begin{satz}
	The cohomology $\ring[[\fat{y}]]$-module $H[[\fat{y}]]$ with bilinear
	non-degenerate pairing $\gamma : H[[\fat{y}]] \tensor H[[\fat{y}]] \pil
	\ring[[\fat{y}]]$, equipped with the tangency quantum potential $\Phi \in
	\ring[[\fat{x},\fat{y}]]$ constitutes a formal Frobenius manifold over
	$\ring[[\fat{y}]]$.  \qed
\end{satz}

In fact, this formal Frobenius manifold is a deformation over $\Q[[\fat{y}]]$
of the formal Frobenius manifold of usual quantum cohomology.

While the underlying space is trivially deformed under
this deformation, the metric and the potential vary non-trivially.

config: arxiv


TITLE: What is the definition of polynilpotent Lie algebras?
QUESTION [0 upvotes]: I am looking for definition of polynilpotent Lie algebras. Is there any equivalent concept for that?

REPLY [1 votes]: Usually poly-P means that there is a subnormal series (or normal series; here it doesn't matter) in which all subquotients have Property P.
When P is "abelian" or "solvable", clearly poly-P is the same as solvable. Since nilpotent is between solvable and abelian, poly-nilpotent would just be the same as solvable.
So the definition has no interest. Unless one counts the number of steps. Then being $n$-step poly-nilpotent, i.e. polynilpotent with a subnormal series of length $n$ has a reasonable meaning. For instance for $n=2$ it is called meta-nilpotent.
Over a field, every solvable finite-dimensional Lie algebra is nilpotent-by-abelian, hence 2-step polynilpotent.
In infinite dimension one can construct solvable Lie algebra of arbitrary large minimal step of polynilpotency.


text:
\begin{document}

\title{Graphical patterns in quadratic residues}

\author{I. Jim\'enez Calvo}
\address{C.S.I.C., C/Serrano 144, 28006--Madrid, Spain.}
\email{ijcalvo@terra.es}

\begin{abstract}
Plots of quadratic residues display some visual features that are analyzed mathematically in this paper.
The graphical patterns of quadratic residues modulo $m$ depend only on the residues of $m$ modulo the
lowest positive integers.
\end{abstract}

\maketitle

\section{Introduction}
The quadratic residue (QR) of an integer $x$ modulo $m$ is the remainder $a$
of the division of $x^2$ by $m$. It is expressed by the congruence,
$$
x^2 \equiv a \pmod m,
$$
where the residue class $a$ can be represented by an integer from 0 to $m-1$.
Figure 1 shows a typical plot of QRs where a rather chaotic distribution can be appreciated.
As $x$ and $m-x$ have the same QR, then only those QRs corresponding to $0 \le x < m/2$ are represented.
It is not surprising that QRs modulo a big integer of unknown factorization are used in cryptography
and random sequence generators \cite{BBS}. Nevertheless, they always show a remarkable
presence of points that seem to draw multiple ``acute parabolas'' that spread along the graphic.
A closer look confirms that the acute parabolas have a quite regular distribution, for values of $x$
near simple fractions of the modulus in a quantity related to the denominator of the fraction,
so that a lower number of
best defined acute parabolas correspond to a simpler fraction .  Not only
QRs, but also $n$-power residues spread as integer points of sets of $n$-degree
polynomials for $x$ values near a fraction of the modulus as it is stated in \cite{JS}. In that paper,
it was shown that some of those ``acute parabolas'' can be used to obtain low quadratic residues
used in the factorization of large integers. In fact, those parabolas correspond to the
Montgomery polynomials
of the MPQS ({\it Multipolynomial Quadratic Sieve}) \cite{Pomerance,Silverman}  and
HMPQS ({\it Hypercube Multipolynomial Quadratic Sieve}) \cite{Peralta}  factorization algorithms.

A detailed treatment of the presence and regular disposition of the ``acute parabolas'' is
done in Section 2. Nevertheless, such features can be explained in a rather informal way with
a fairly straightforward reasoning. Let be an integer $x$ and a modulus $m$ and consider the
difference between the square of $x$ and its nearby integers, i.e.
$\Delta=(x+i)^2-x^2=2ix+i^2$, when $x$ is the closest integer to a fraction $a/b$ of the modulus.
We consider small $i$ values and neglect the term $i^2$. A second approximation is done replacing
$x$ by $(a/b)m$, so $\Delta \approx  (2ia/b) m$.
If $b$ is odd, the residue of the fraction $(2ia/b) m$ modulo $m$ is another fraction
$ (a'/b) m$ with $a' \equiv 2ia \pmod b$ because $2ia=a'+kb$ for some integer $k$.
When $b$ is even, we can note that $\Delta \approx (ia/b') m$ with $b'=b/2$
and its residue modulo $m$ is the fraction $ (a'/b') m$ with $a' \equiv ia \pmod {b'}$.
Therefore, the QRs for values near $x$ are spaced approximately by multiples of $m/b'$.
The term $i^2$, which is not considered in this approximation, accounts for building the parabolas at
regular intervals.

Functions involving QRs have been used to generate computer graphics. The function
$f(x,y) = x^2+y^2 \pmod m$ shows concentric circles with centres at coordinates that are made up of simple fractions of the modulus. They are actually, truncated paraboloids present
in this bi-dimensional case (see fig.~2). Another type of patterns can be observed sometimes in QRs
plots besides the ``acute parabolas''.  For some moduli, it may happen that the outlines of broader
diffused ``ghost parabolas'' materialize, as can be observed in fig.~1 and fig.~3.
These kind of patterns are analyzed in Section 3. To conclude, we can state that the graphical
patterns of QRs mod $m$ plots depend only on the residues of $m$ modulo the lower positive integers.

\begin{figure}
\centerline{\psfig{figure=crypt1.eps,height=8cm}}
\caption{Quadratic residues mod 20171.}
\end{figure}

\begin{figure}
\centerline{\psfig{figure={x2y2.eps},height=8cm}}
\caption{Function $f(x,y)=(x^2+y^2)$ mod 415.}
\end{figure}

\section{The geometry of the ``acute parabolas''}
The congruence relation is a relation of equivalence between integers. Nevertheless, a similar relation
between rationals can be established:
\begin{definition}
Two rationals, $s$ and $t$, are Q-congruent modulo an integer $m$ when $(s-t)/m$ is an integer
and then we can write
$$
   s \equiv_Q t \pmod m\quad \mbox{and}\quad s = t + km,
$$
for some integer $k$.
\end{definition}
$Q$-congruence relation is reflexive, symmetric and transitive as well as the congruence relation
between integers. Moreover, two congruent integers are also $Q$-congruent. Therefore, the notation
``$\equiv$'' between rationals must be read as ``$\equiv_Q$'' and when the relation
is between integers, it can be read in both ways.

Let $m$ be a modulus and $a/b$ an irreducible fraction. We define the integer $x_0$ and its
quadratic residue $r_0$ as
\begin{equation}
   x_0=\round{ \frac a b m },\quad x_0^2 \equiv r_0 \pmod m \; ,
\end{equation}
where $\round{x}=\lfloor x+1/2\rfloor$, that is, the nearest integer to $x$.

\begin{proposition}
$r_0$ is the nearest integer to the rational $\beta (m/b^2)$ where \\
$$
\beta \equiv a^2m - 2a\alpha \pmod {b^2} \quad {with} \quad \alpha \equiv am \pmod b, \:|\alpha| \leq b/2.
$$
\end{proposition}
\begin{proof}
We note that $x_0=(am-\alpha)/ b$ from the definition of the integer $\alpha$ and then,
$$
  x_0^2=\frac {a^2m^2-2a \alpha m} {b^2} +  \frac {\alpha^2} {b^2} =
\left({\frac {a^2m-2a\alpha-\beta } {b^2}}\right)m+ \beta \frac m {b^2}+
 \frac {\alpha^2} {b^2} \; .
$$
With the definition of $\beta$, the term $(a^2m-2a\alpha-\beta)/b^2$ becomes an integer. Taking
into account that $|\alpha| \leq b/2$, we can then write
\begin{equation}\label{r_0}
  r_0 \equiv \beta \frac m {b^2} +\frac {\alpha^2} {b^2} \pmod m,
  \quad 0 \leq \frac {\alpha^2} {b^2}  \leq 1/4 \;.
\end{equation}
\end{proof}

We now calculate the QRs of integers near $x_0$. From the considerations explained in
the introduction, it becomes natural to consider lattice points defined by
\begin{equation}\label{lattice}
x=x_0+i+jb', \quad {\rm with} \quad b'=\left\{
 \begin{array}{lc}
    b, & {\rm if\:{\it b}\: odd}\\
    b/2, & {\rm if\:{\it b}\: even}
 \end{array} \right. ,\quad |i| \le b'/2, \quad j \in \mathbb{Z}_m.
\end{equation}
Expanding the square of the trinomial, we have
$$
x^2 = (x_0+i)^2+j^2{b'}^2+2(x_0+i)jb'.
$$
Taking into account that $x_0=(am-\alpha)/ b$ and with $r$ and $r_i$ as the quadratic
residues of $x$ and $x_0+i$ respectively, we can write
\begin{equation}\label{r}
r \equiv {b'}^2j^2+2\left(b'i- \frac \alpha c\right)j+r_i \pmod m \;,\quad
c=\frac b {b'} = \left\{
\begin{array}{ll}
    1, & {\rm if\:{\it b}\: odd.}\\
    2, & {\rm if\:{\it b}\: even.}
 \end{array} \right.
\end{equation}

\begin{proposition}
The quadratic residues of integers near $(a/b) m$ split as integer points of
$b$ parabolas if $b$ is odd, or $b/2$ parabolas if $b$ is even, with their
vertices placed at $x_v=(a/b) m$ and equidistant in between.
\end{proposition}
\begin{proof}
It is clear that (\ref{r}) is a parabola under variable $j$ for each one of the $b'$ values
of $i$ in (\ref{lattice}). Those parabolas will be named as $P_i(j)$.
The coordinates at the vertex are given when the derivative of (\ref{r}) with respect to $j$ becomes null.
Since $r_i$ is fixed for each $i$ value they are given by
$$
j_v= \frac \alpha {c{b'}^2}-\frac i {b'}, \quad
P_i(j_v)=r_i-\left(i-\frac \alpha b\right)^2  \;.
$$
Substituting the value of $j_v$ in (\ref{lattice}), and taking into account that $x_0=(am-\alpha)/b$,
the coordinate at the vertex is $x_v=(a/b) m$ which is independent from $i$ value.
Then, all parabolas related to the fraction $a/b$ have vertices at the same $x$ value.

Since $(x_0+i)^2=x_0^2+2ix_0+ i^2$, then
$r_i \equiv r_0 + 2i(am - \alpha)/b + i^2  \pmod m$.
Replacing $r_i$ and simplifying we obtain,
$$
P_i(j_v) \equiv r_0+2ia \frac m b- \frac {\alpha^2} {b^2} \pmod m\;.
$$
Taking the value of $r_0$ from (\ref{r_0}), we obtain,
$$
P_i(j_v) \equiv \beta \frac m {b^2} + 2ai \frac m b \pmod m\;.
$$
Let be
$$
a' \equiv \frac 2 c ia \pmod {b'},
$$
with $c$ as defined in (\ref{r}).  Then
\begin{equation}\label{minimos}
P_i(j_v) \equiv \beta \frac m {b^2} + a'\frac m {b'} \pmod m\;.
\end{equation}
As much as $i$ runs over $\Z_{b'}$ and $\gcd(a,b')=1$, $a'$ takes all values in $\Z_{b'}$.
From the above equation, it follows that vertices of $P_i(j)$
lie exactly at multiples of $m/b^2$ and are separated by exact gaps of $m/b'$.
This concludes the proof.
\end{proof}

\section{Patterns}
Propositions 1 and 2 give account of the number and position of the ``acute parabolas''.
Near $x=(a/b) m$, $b$ parabolas (or $b/2$ if $b$ is even) appear and all are equidistant in
$\Z_m$. Figure 2 shows a representation in gray levels of the function $f(x,y)=(x^2+y^2) \pmod m$.
It is remarkable the symmetry of the graphic because it displays all the range of variables in $Z_m$.
There is only one paraboloid at each node
related to fractions 0/1, 1/2 and 1/1. That corresponds to centre, corner and the midpoint of the sides
of fig.~2.
Such paraboloids have a sharp profile form and they are surrounded by rings due to truncations.
Two paraboloids are present at each node related to 1/4 and 3/4 producing more fuzzy patterns.
Other patterns alike can be observed for more complex fractions.

At first glance, when plots like fig.~1 are observed, the presence of ``acute parabolas'' and
their positions can be recognized.
Since $b= b'c$, equation (\ref{minimos}) can be rewritten as
$$
P_i(j_v) \equiv \frac \beta {cb} \frac m {b'} + a'\frac m {b'} \pmod m.
$$
Let $\beta' \equiv \beta \pmod {cb}$, then
$$
P_i(j_v)  \equiv \frac {\beta' +sbc} {bc} \frac m {b'} + a' \frac m {b'} \pmod m,
$$
for some integer $s$.

Taking $k=s+a'$, we have
\begin{equation}\label{vertices}
P_i(j_v)  \equiv \beta' \frac m {b^2} + k \frac m {b'} \pmod m,\quad  k \in \Z_{b'}.
\end{equation}

If the set of parabolas $P_i$ corresponding to the fraction $a/b$ is considered as a whole, without paying
attention to the position of any single parabola in the set, the above equation shows that their
position is defined only by the value of $\beta'$. Proposition 1 shows that $\beta$
(and also $\beta'$) depends only on $m$ for each fraction $a/b$. Since the sets of parabolas
corresponding to the simplest fractions are those that are relevant to the eye,
it follows that the general appearance
of a plot only depends on  the values
of $m\; {\rm mod}\; {cb}$ for the lowest values of $b$, that is for $cb=4,3,8,5,12, \cdots$.

Let us consider the integer
$$
\Lambda(n)= 2 \,{\rm LCM}(2,3,4\cdots n).
$$
For example, let $\Lambda=\Lambda(9)=2^4\cdot3^2\cdot5\cdot7=5040$. Plots of QRs modulo integers that are congruent
modulo $\Lambda$ will have the parabolas corresponding to fractions with denominator equal or
less than 9  in the same position. Moreover, since $cb$ divides $\Lambda$
for $b=10,12,14,15,16,18 \cdots$, the parabolas related to fractions with these denominators will
also have the same position.
As an example, Figures 3(a) and 3(b) have the same appearance, but for their density,
because the modules are $4\Lambda+19$ and $5\Lambda+19$ respectively and then
$\beta\; {\rm mod}\; {cb}$ do not change when $b\le 9$ or $b=10,12,14,15,16,18 \cdots$.
Furthermore, we can prove that the vertices are rational points of a bundle of lines.
\begin{proposition}
Let the integers $\Lambda$, $m$ and $s$ be such that $s \equiv m \pmod \Lambda$ and let ${\mathcal B}$ be
the set of positive integers $b$ such that $b|\Lambda$ if $b$ is odd or $2b | \Lambda$ if $b$ is even.
The vertices of the parabolas defined in Proposition 2 associated to fractions $a/b$ with $b \in {\mathcal B}$
are rational points of the bundle of lines of the form
$$
   Y \equiv 2nX-sX^2 \pmod 1,\quad n=0,\pm1,\pm2, \cdots,\quad X=\frac x m,\;Y=\frac {x^2 \pmod m} m.
$$
\end{proposition}
\begin{proof}
The relative position of the vertices depends on the value of $\beta$ mod  $cb$ as it was justified above.
When $b$ is odd, $c=1$ and the expression of $\beta$  in Proposition 1 can reduce to
$\beta \equiv -a^2m \pmod b$. If $b$ is even, $c=2$. We put $\alpha = am -kb$, for an integer $k$
such as $|\alpha| \le b/2$.
Then, $\beta \equiv a^2m-2a(am-kb) \pmod {b^2}$, that reduced to mod $2b$ is also
$\beta \equiv -a^2m \pmod {2b}$. We conclude that $\beta \equiv -a^2m \pmod {cb}$ for any parity of $b$.
Note that, from the definition of $\Lambda$
and $s$, we have that when $b\in{\mathcal B}$, $s \equiv m \pmod {cb}$ and then
$\beta \equiv -sa^2 \pmod {cb}$.
From Proposition 2 and equation (\ref{vertices}), it follows that the vertices of the parabolas are at
$$
X_v=x_v/m=a/b,\quad Y_v=P_i(j_v)/m \equiv (\beta\;  {\rm mod}\; cb)\frac 1 {b^2} + k \frac 1 {b'} \equiv
-s \frac {a^2} {b^2} + \frac {ck} b \pmod 1.
$$
Note that $ck \equiv 2na \pmod b$ can always be solvable in $n$ for any value of $k\in \Z_{b'}$
and for any parity of $b$. Note also that $n$ can take any value with the appropriate value of
$k\in \Z_{b'}$. Then, we have that
$Y_v=-s \frac {a^2} {b^2} + 2n \frac a b = -sX_v^2 + 2nX_v$ for some $n$.
\end{proof}

When $s=0$ the bundle of quadratics degenerates in straight lines( fig.~3(c)) that are outlined due to
the accumulation of points near the vertices. As $s$ has a higher absolute value,
the straight lines bend into parabolas as can be seen in fig.~3 (d) and 3(e).
When $s$ has absolute value as high as in fig~3(f), the lines of the bundle have such a high slope
that they do not allow to notice the ``ghost parabolas''.

To conclude, we have shown that Proposition 1, 2 and 3 explain the main visual features of QR plots
and that those features are related solely with the residues of $m$ modulo the lower positive integers.

\begin{figure}
\centerline{\psfig{figure={ghost.eps},height=20cm}}
\caption{Plots of quadratic residues, $X=x/m$, $Y=(x^2\; {\rm mod}\; m)/m$, $\Lambda=5040$.}
\end{figure}


TITLE: How exactly does "classical physics is continuous" posit "spacetime is a set with a certain topology"?
QUESTION [1 upvotes]: Context:
In Lecture 1 of the Central Lecture Course, Professor Schuller begins with the so-called "physical key definition" of space-time:

Spacetime is a four-dimensional topological manifold with a smooth
  atlas carrying a torsion-free connection compatible with the
  Lorentzian metric and a time orientation satisfying the Einstein
  equations.
At the coarsest level, "spacetime" is a "set". Not enough to talk about
  continuity of maps.  Why do we as physicists want to talk about continuity of maps? Well, in classical physics we have the idea that curves do not "jump".
Weakest structure that can be established on a set which allows a definition of continuity of maps is called a topology.

Not all the first terms in the first paragraph are clear to me yet. However, this question thread is solely based on what I understood (and didn't understand) from the first lecture.

As far as I understand:

If $M$ is a set, a topology $\mathcal{O}$ is a subset $\mathcal{O}\subseteq \mathcal{P}(M)$ satisfying
(i) $\phi\in \mathcal{O}$
(ii) $u,v\in\mathcal{O}\implies u\cap v \in \mathcal{O}$
(iii) $u_{\alpha}\in \mathcal{O} \implies (\bigcup_{\alpha\in A}u_{\alpha}) \in \mathcal{O}$ where $\alpha$ comes from an arbitrary (not necessarily countable) index set.
We can define "continuous maps" between two sets $M$ (equipped with a topology $\mathcal{O}_M$) and $N$ (equipped with a topology $\mathcal{O}_N$).
A map $f:(M,\mathcal{O}_M)\to (N,\mathcal{O}_N)$ is called continuous if

$$\forall \ V  \in \mathcal{O}_N: \text{preim}_f(V) \in \mathcal{O}_M$$
Now, the problem is that I'm not being able to relate "there cannot be jumps in classical physics" to "spacetime is a set with a certain topology".

Questions:

What exactly is meant by "there cannot be jumps in classical physics"? Are they simply referring to the fact that particles in classical mechanics follow "continuous trajectories"? Or do are they referring to something more fundamental to classical mechanics? By the way, I'm also not completely sure that "trajectories are continuous" is an absolute necessity in classical mechanics.
To define a continuous map (in this context), we will need two sets equipped with their own topologies. If spacetime is considered to be a set with its own topology, what is the other set (to which continuous maps can be defined)? (As of now) it sort of seems meaningless to me when they say that spacetime is continuous in general relativity without mentioning what the other set (equipped with a certain topology) is.
If the answer to point (1) is "yes", then how exactly is the fact that "trajectories in classical mechanics are continuous" related to the fact that "spacetime is continuous", and in turn how is that related to the fact that "continuous maps can be defined between two sets equipped with their own topologies". Basically, I'm not being able to connect these three notions. At least, not mathematically.
How are we certain that the weakest structure on a set, which allows for continuity of maps is a topology? Is it a mathematical fact? If yes, could someone point me out a proof and/or a reference which discusses this?

REPLY [3 votes]: What follows is a very rough sketch which I hope may help.  See the end for some pointers.
The trick here is that, as you say, to define continuous maps you need two sets with topologies.  And we then choose one of these sets to be a set we understand very well with a topology we understand very well: the set $\mathbb{R}^n$ with the usual topology.
The usual topology on $\mathbb{R}^n$
The easiest way to define the usual topology on $\mathbb{R}^n$ is to do it backwards:

for any $x = (x_1,\ldots,x_n), y = (y_1,\ldots,y_n)\in\mathbb{R}^n$ define $d(x,y) = \sqrt{(x_1 - y_1)^2 + \cdots + (x_n - y_n)^2}$ -- this is the normal euclidean distance function;
for any point $p\in\mathbb{R}^n$, define an open neighbourhood of radius $r > 0$ from $p$ as $N_r(p) = \left\{q\in\mathbb{R}^n: d(p,q) < r\right\}$
now define the open sets of the topology as $\emptyset$, all the $N_r(p)$ for all $p$, their unions & finite intersections and $\mathbb{R}^n$ itself.

It's pretty easy to show that this:

is a topology;
is the one you'd expect for $\mathbb{R}$ ($d$-neighbourhoods are just open intervals);
is not very sensitive to the details of $d$ -- the topology is not dependent on the particular distance function and many distance functions will give the same topology.

It's also possible to show that for this topology the topological definition of continuity is exactly the same as the traditional $\epsilon$-$\delta$ one.  So the continuous mappings using this topology between, say $\mathbb{R}$ and $\mathbb{R}$ are just what we would hope they would be, and all our intuitions of what it means for functions on the reals to be continuous work.
Topological manifolds
So, then the trick is that we define a topological manifold to be a set $M$ with a topology, and the the requirement that each point of $M$ has an open neighbourhood which has a 1-1, continuous, map onto an open neighbourhood of $\mathbb{R}^n$.
So long as we use the usual topology on $\mathbb{R}^n$ this really means that $M$ is 'locally like' $\mathbb{R}^n$ with the usual topology.  And this in turn means that our intuitions about continuous functions on the reals also work in $M$, at least locally.
Note that there is no requirement for there to be a single, global, 1-1, continuous map between $M$ and $\mathbb{R}^n$, and there often is not (for instance the 2-sphere has no such single map).  But thanks to the nature of open sets in the usual topology the maps will end up with overlaps (or it's possible to define them so that there are always overlaps by shifting them around in obvious ways).
An example: the unit circle
This is a simple example where you can easily visualise what's going on.  The unit circle, which I'll call $S$, is $S = \{(x,y)\in\mathbb{R}^2: \sqrt{x^2 + y^2} = 1\}$.  This is a one-dimensional manifold, so it wants to be mapped onto (open sets of) $\mathbb{R}$.
The first obvious trick is to say that $(x, y) = (\cos\theta, \sin\theta), \theta\in\mathbb{R}$.  This makes it easy to talk about the topology we want on $S$: it's just the usual topology on $\mathbb{R}$: for some $\theta_0$ the neighbourhoods of it are $(\theta_0 - \epsilon, \theta_0 + \epsilon), \epsilon > 0, \epsilon < 2\pi$. Note I am now using $(a,b)$ to mean $\{x: x > a, x < b\}$ & similarly for $[a,b]$, $(a, b]$ & so on, rather than to mean a tuple, and note also the $\epsilon < 2\pi$ condition which avoids us looping around $S$.
So, we now need some 1-1, continuous mappings between $S$ & open subsets of $\mathbb{R}$.  It's tempting to just pick a single one, $m(\theta) = \theta$.  This is continuous but it's not 1-1, because $\theta + 2n\pi, n\in\mathbb{N}$ is the same point as $\theta$.  Well, we could limit the range of $\theta$ to be $(0, 2\pi)$ but this interval doesn't include $0$ and so does not map the whole of $S$.  We can map the whole of $S$ by having $\theta\in[0, 2\pi)$, but this isn't an open set of $\mathbb{R}$ in the usual topology.
So, it turns out we need at least two mappings.  There's a lot of freedom in choosing them, but for instance

$m_1(\theta) = \theta, \theta\in(0, 3\pi/2)$
$m_2(\theta) = \theta, \theta\in(\pi, 5\pi/2)$

Note that these overlap: in $(pi, 3\pi/2)$, $m_1(\theta) = m_2(\theta)$, and in $(0, \pi/2)$, $m_1(\theta) = m_2(\theta - 2\pi)$.  And between them they map all the points of $S$ in a 1-1, continuous way, to open subsets of $\mathbb{R}$.
It should be obvious that we need more than one mapping: although $S$ locally looks like $\mathbb{R}$, it doesn't look globally like $\mathbb{R}$ at all.  And that's part of the point of the whole topological manifold thing: we want to be able to talk about objects which locally are like $\mathbb{R}^n$ but which globally may not be like it at all.
A note on overlaps.  Above I said that 'open sets in the usual topology always end up with overlaps'.  What this means is easy to see by considering the intersection of two open intervals of $\mathbb{R}$: $(a,b) \cap (c,d)$.  This set is either empty, if $c \ge b$, or it contains more than one point, if $c < b$.  It never contains just a single point.  This is a property of the usual topology on $\mathbb{R}$: it's not true for all topologies, but it is true for the usual topology.  As an example it is not true for the discrete topology where all sets are open, because then $\{x\} \cap \{x\}$ is a single point, and $\{x\}$ is open in this topology: that's why the discrete topology is useless.  I don't know the name of this condition: it might be the Hausdorff condition or an implication of it, but I'm not sure.
So long as we stick to a topology where this is true then we can be sure that if we have maps for all of a manifold, then those maps will have overlaps, and this means that we can work our way between them using the overlaps.  And that means that there are no 'bad places' in the manifold where continuity breaks down: if there are going to be such bad places then we need to cut them out of the manifold somehow (such bad places are singularities, of course).
The questions

What is meant by 'there cannot be jumps' is that, for instance, trajectories of particles in the manifold are continuous, which translates directly to trajectories in the images of the maps between the manifold and $\mathbb{R}^n$ being continuous in the way we expect from basic analysis.  There are, in fact, no jumps.
You don't in fact need two sets, or rather the two sets can be copies of each other.  However the trick sketched above of defining a manifold by requiring there to be well-behaved mappings between it and $\mathbb{R}^n$ means that you always do, in fact, have two sets, and you can bootstrap things like continuity from elementary notions of continuity on the reals.
The manifold is continuous because it allows these continous mappings.  If it was not continous then these mappings would not exist at all.
I think this is backwards: defining a topology on a set defines what 'continous' means: if you define a different topology (there are various more-or-less pathological ones such as the discrete topology (all sets are open) and the trivial topology (only the whole set and the empty set are open)) you will get a different notion of continuity.

Pointers

I have always liked Geometrical methods of mathematical physics by Schutz, which gives a much better version of the sketch I've given here.  It was the first book I read which made me understand what topology actually is and why it's useful.
Analysis, manifolds and physics by Choquet-Bruhat, deWitt-Morette with Dillard-Bleick is a much more serious approach which starts with a fairly fierce overview of topology and analysis.  I used to use this as a reference.

There are certainly more modern references and certainly other equally old and perhaps better references: these are just the books I used before about 1990.


TITLE: Boundary conditions in minimizing Dirichlet energy for an image processing problem.
QUESTION [1 upvotes]: Suppose
$$\mathcal{L} =\mathcal{L}(x,y,u,u_x,u_y) = \frac{1}{2} \lVert \nabla u \rVert^2$$
and I want to find $u$ such that the functional
$$
E(u)=\int_{\Omega} \mathcal{L}dxdy
$$
is minimized, computing the variational derivative leads me to
$$
\frac{\delta E}{\delta u} = \int_{\partial \Omega}(u_y - u_x)h d\gamma -\int_{\Omega} \Delta u h dxdy
$$
therefore the $u$ function I'm looking for is given by
$$
\left\{
\begin{array}{ll}
\left(\frac{\partial}{\partial y} - \frac{\partial}{\partial x}\right) u = 0 & u (x,y) \in \partial \Omega \\
\Delta u = 0 & (x,y) \in \text{Int}(\Omega)
\end{array}
\right.
$$
Though the question might be silly, I wonder how the operator
$$
A =\frac{\partial}{\partial y} - \frac{\partial}{\partial x}
$$
might be discretized, my attempt would be using be applying the Sobel masks in pixels at $\partial \Omega$ and adding the results, would this be correct? Or is that operator something know that I'm missing? I thought at the beginning it was the divergence operator, but I was wrong since $u$ is a real valued function.
Update (Full derivation of the integral)
We have
$$
\frac{E(u + \alpha h) - E(u)}{\alpha} = \frac{1}{\alpha}\int_{\Omega} \mathcal{L}(x,y,u+\alpha h,u_x+\alpha h_x, u_y+\alpha h_y) -\mathcal{L}(x,y,u,u_x, u_y)dxdy = \\ \int_{\Omega} \frac{\partial \mathcal{L}}{\partial u}h + \frac{\partial \mathcal{L}}{\partial u_x}h_x + \frac{\partial \mathcal{L}}{\partial u_y}h_y dxdy = \int_{\Omega} \frac{\partial \mathcal{L}}{\partial u}h dxdy + \int_{\Omega}\frac{\partial \mathcal{L}}{\partial u_x}h_x + \frac{\partial \mathcal{L}}{\partial u_y}h_y dxdy.
$$
Consider the integral
$$
\int_{\Omega}\frac{\partial \mathcal{L}}{\partial u_x}h_x + \frac{\partial \mathcal{L}}{\partial u_y}h_y dxdy
$$
observing
$$
\frac{\partial \mathcal{L}}{\partial u_x}h_x = \frac{d}{dx}\left(\frac{\partial \mathcal{L}}{\partial u_x}h\right) - \frac{d}{dx} \frac{\partial \mathcal{L}}{\partial u_x} h
$$
and likewise for $\frac{\partial \mathcal{L}}{\partial u_y}h_y$ we have
$$
\int_{\Omega}\frac{\partial \mathcal{L}}{\partial u_x}h_x + \frac{\partial \mathcal{L}}{\partial u_y}h_y dxdy = \int_{\partial \Omega} \left(\frac{\partial \mathcal{L}}{\partial u_y} - \frac{\partial \mathcal{L}}{\partial u_x}\right)h d\gamma - \int_{\Omega} \left( \frac{d}{dx} \frac{\partial \mathcal{L}}{\partial u_x} + \frac{d}{dy} \frac{\partial \mathcal{L}}{\partial u_y}\right)h dxdy
$$
Taking the limit for $\alpha \to 0$ yields
$$
\frac{\delta E}{\delta u} = \int_{\Omega} \left(\frac{\partial \mathcal{L}}{\partial u} - \frac{d}{dx} \frac{\partial \mathcal{L}}{\partial u_x} - \frac{d}{dy} \frac{\partial \mathcal{L}}{\partial u_y}\right)h dxdy + \int_{\partial \Omega} \left(\frac{\partial \mathcal{L}}{\partial u_y} - \frac{\partial \mathcal{L}}{\partial u_x}\right)h d\gamma
$$
And explicit computation of the partial derivatives of $\mathcal{L}$ yields to expression above.

REPLY [2 votes]: The variation should be:
$$
\delta\int_\Omega \frac12|\boldsymbol{\nabla} u|^2 = -\int_\Omega\Delta u\ \delta u +
\int_{\partial\Omega}\boldsymbol{\nabla} u\cdot\boldsymbol n\ \delta u
$$
So the second equation is
$$
\left.\boldsymbol{n}\cdot\boldsymbol{\nabla} u = 0 \right|_{\partial\Omega},
$$
where $\boldsymbol n$ is a normal vector to $\partial\Omega$. Or in other words that gradient flow $\boldsymbol{\nabla} u$ should be perpendicular to the boundary $\partial\Omega$.
Discretization of this depends on many things. If boundary is simple (rectangular), you can just set one of the sobel values to zero. If the boundary is complex, you might try to detect pixels of boundaries, calculate the tangent there and write linear equations on sobel values.
Edit
Consider your integral
$$
I = \int_\Omega\left(\frac{\partial\mathcal L}{\partial u_x}h_x+\frac{\partial\mathcal L}{\partial u_y}h_y\right)dx\,dy =
\int_\Omega (F\,h_x+G\,h_y)\,dx\,dy =\\
 \int_\Omega\left(\frac\partial{\partial x}(F h)-F_xh+\frac\partial{\partial y}(G h)-G_yh\right)dx\,dy = \\
 -\int_\Omega h\left(F_x+G_y\right)dx\,dy +  \int_\Omega\left(\frac\partial{\partial x}(F h)+\frac\partial{\partial y}(G h) \right)dx\,dy.
$$
Green's theorem for second integral gives
$$
\int_{\partial\Omega}-Ghdx+Fhdy = \int_{\partial\Omega}h(F\,dn_x + G\,dn_y).
$$
where $\boldsymbol{dn} = (dy, -dx)$


TITLE: Irreducibility of $8x^3 -6x +1$ in $\mathbb{Q}[x]$
QUESTION [2 upvotes]: I want to use Gauss' Theorem for that. So I have to show that $f(x) = 8x^3 -6x +1$ is irreducible in $\mathbb{Z}[x]$ which is equivalent to it not having any roots in $\mathbb{Z}$.
Since any integral root would have to divide $1$, one only has to check $\pm 1$. But $f(\pm 1) \ne 0$. So $f$ is irreducible in $\mathbb{Z}[x]$ and then, by Gauss, irreducible in $\mathbb{Q}[x]$.
Is that correct?

REPLY [4 votes]: Set $u=2x$ and rewrite the polynomial as $u^3-3u+1$. The original polynomial  is reducible over $\mathbf Q$ if and only if this polynomial is. Having degree $3$,  it is reducible if and only if it has a rational root. Now the rational root theorem says this root can be only $\pm 1$. Furthermore, this root must be positive, hence you only have to test $1$, which is not a root. Hence the original polynomial is irreducible.

REPLY [3 votes]: As almagest pointed out in comments, your argument would seemingly apply to the reducible polynomial $4x^2-4x+1$ as well.  For non-monic polynomials, you need to check for rational roots, whose denominators divide the lead coefficient.  So in your case, you would need to evaluate $f(x)$ for $x=\pm1$, $\pm1/2$, $\pm1/4$, and $\pm1/8$.
For a different approach, note that
$$f(x+1)=8(x+1)^3-6(x+1)+1=8x^3+24x^2+18x+3$$
This allows you to use Eisenstein's criterion (with the prime $p=3$) to show the polynomial is irreducible.


TITLE: Does an asymmetric top return to its initial position &orientation, while rotating freely in space? (contradiction in Landau&Lifshitz mechanics book)
QUESTION [1 upvotes]: The Poinsot's ellipsoid tells us that the angular momentum ($M$) (in the rotating frame) should be lying on the intersection curve between the conservation of momentum sphere radius $|M|$: $M_1^2 + M_2^2 + M_3^2=|M|^2$
and  the conservation of energy ($E$) ellipsoid:
$$1 = M_1^2/(2E I_1) + M_2^2/(2E I_2) + M_3^2/(2E I_3)\, .$$
This intersection curve is closed so the motion of $M$ relative to the body should be periodical, and as written in the Landau & Lifshitz mechanics book page 117:

First of all, we may note that, since the paths are closed, the motion of the vector M relative to the top must be periodic; during one period the vector M describes some conical surface and return to its original position.

and because $M$ is fixed (unchanged) in the world frame, there the body moves, but it will return to its original starting orientation (and place as the center of mass stationery) after some time. as the relative motion $M$ and the body is periodical.
But then on page 120 it was written:

This incommensurability has the result that the top does not at any time return exactly to its original position.

Additionally on page 119, regarding the angular velocity function of time, it was written:

After a time T the vector W returns to its original position relative to the axes of the top. The top itself, however, does not return to its original position relative to the fixed system of co-ordinates

So is there a mistake in this book or where do I wrong? assume no initial translational motion.

REPLY [1 votes]: There is no mistake in the book. To really understand the motion of a freely rotating top (Euler's top), you have to look at the full system of differential equations that describe the motion of the top.
Let us denote by capital letters $\vec{X}$ the coordinates with respect to the rotating frame (body-fixed frame) and the  by lower-case letters $\vec{x}$ the coordinates with respect to the inertial world-fixed frame (the world frame is an inertial frame of reference attached to the center of the top, so the center of the top is stationary). Knowing the position(orientation) of the top at any moment of time $t$ in the world space means that you know the $3 \times 3$ rotational matrix (also called orthogonal matrix) $U = U(t)$ that transforms the coordinates $\vec{X}$ of any point on the top to its coordinates in $\vec{x}$ in the world frame:
$$\vec{x}(t) = U(t)\, \vec{X}$$ Notice, in the body fixed frame the coordinates $\vec{X}$ of a point do not change with time, but in the world frame the coordinates $\vec{x}$ of the same point change with time, because the body is reorienting constantly.
The velocity of our arbitrarily chosen point is
$$\frac{d\vec{x}}{dt} = \frac{dU}{dt}\, \vec{X}$$
which combined with $\,\vec{X} = U^T\, \vec{x}\,$, where $U^T$ is the matrix $U$ transposed, which in the case of a rotational matrix is the inverse matrix, i.e. $U^T = U^{-1}$, gives the expression
$$\frac{d\vec{x}}{dt} =\frac{dU}{dt}\, \left(U^T \,\vec{x}\right) = \left(\frac{dU}{dt}\,U^T \right)\vec{x}$$ As it turns out, in the case of rotational matrices, there always exists a unique vector $\vec{\omega} = \vec{\omega}(t)$ such that
$$\frac{d\vec{x}}{dt} = \left(\frac{dU}{dt}\,U^T \right)\vec{x} = \vec{\omega} \times \vec{x}$$ This $\vec{\omega}$ is the angular velocity with respect to the inertial world frame. If we apply the inverse rotation matrix $U^T$ from the left to the latter equation we obtain
$$\vec{V} = U^T\,\frac{d\vec{x}}{dt} = U^T \Big(\vec{\omega} \times \vec{x}\Big) =
 \big( U^T\vec{\omega} \big)\times \big( U^T\vec{x}\big) = \vec{\Omega} \times \vec{X}$$ where $\vec{V}$ are the coordinates of the velocity vector $\frac{d\vec{x}}{dt}$ in the body-fixed frame and $\vec{\Omega} = U^T\, \vec{\omega}$ are the coordinates of the angular velocity in the body-fixed frame. Another way of writing the latter chain of equations is
$$\vec{V} = U^T\,\frac{d\vec{x}}{dt} =  U^T\, \left(\frac{dU}{dt}\, \vec{X}\right) = \left(U^T\,\frac{dU}{dt}\right)\vec{X} = \vec{\Omega} \times \vec{X}$$ i.e.
$$  U^T\,\frac{dU}{dt} = \Big(\vec{\Omega}\times \cdot\Big)$$ where by $\Big(\vec{\Omega}\times \cdot\Big)$ I denote the skew-symmetric matrix that acts as follows:
$$\Big(\vec{\Omega}\times \cdot\Big) \vec{X} = \vec{\Omega} \times \vec{X}$$ Let us denote by $J$ the $3 \times 3$ inertia matrix calculated in the body-fixed frame. In the body-fixed frame, $J$ is a constant matrix, i.e. it doesn't change with time because the body is fixed in that frame and doesn't change its orientation. In the inertial world frame however, the inertia tensor constantly changes with time because the body (and thus its mass distribution) changes orientation. So in the inertial world frame, the inertia matrix is
$$j = j(t) = U(t)\, J\, U(t)^T$$ The angular momentum in world frame coordinates is
$$\vec{m} = j\,\vec{\omega}$$ and by the conservation of angular momentum law:
$$\frac{d \vec{m}}{dt} = \vec{0}$$
Now, the ultimate goal of the rigid-body dynamics is to obtain the rotation matrix $U = U(t)$ as a function of time. This matrix, via $\vec{x} = U\,\vec{X}\,$, gives you the position $\vec{x}$ in the world frame of every point $\vec{X}$ from the body at any moment of time $t$. The equation $\frac{d \vec{m}}{dt} = \vec{0}$ is the right equation to start from, but just knowing the angular momentum in the world frame doesn't immediately provide us with a way to find the rotation matrix $U(t)$. So the strategy is to transform the angular momentum equation into an equation for the angular velocity $\vec{\Omega}$ in the body fixed frame, solve it, find $\vec{\Omega} = \vec{\Omega(t)}$, and then solve the equation $U^T\, \frac{dU}{dt} = \big(\vec{\Omega} \times \cdot\big)$. Why in the body-fixed frame, you may ask. Because in the world frame the inertia matrix is complicated time-dependent matrix, while in the body-fixed frame it is a constant matrix, which can be even diagonal if the body frame is chosen appropriately.
The coordinates of the angular momentum in the body-fixed frame are
\begin{align}
\vec{M} &= U^T\, \vec{m} = U^T\, (j\, \vec{\omega}) \\
&= U^T \Big( U\, J\, U^T \, \vec{\omega}\Big) = U^T \Big( U\, J\, U^T \, U\, \vec{\Omega}\Big) = \big(U^T\, U\big)J\big(U^T\, U\big) \vec{\Omega}\\
&= J\, \vec{\Omega}
\end{align}
Thus
\begin{align}
\vec{0} &= \frac{d \vec{m}}{dt} = \frac{d}{dt}\big(U\, \vec{M}\big) \\
&=\frac{d}{dt} \big(U\, J\, \vec{\Omega}\big) = \frac{dU}{dt}\, J\,\vec{\Omega} + U\, \frac{d}{dt} \,\big( J\vec{\Omega}\big)
\end{align}
If you multiply the latter equation by the orthogonal matrix $U^T$ from the right, you get
\begin{align}
\vec{0} &= U^{T}\,\vec{0} = U^T\left( \frac{dU}{dt}\, J\,\vec{\Omega} + U\, \frac{d}{dt} \,\big( J\vec{\Omega}\big)\, \right) \\
&= \left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega} + \big(U^T\,U\big)\, \frac{d}{dt} \,\big( J\vec{\Omega}\big) \\
&= \left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega} + \frac{d}{dt} \,\big( J\vec{\Omega}\big)
\end{align}
The latter equation is
\begin{align}
\vec{0} &= \left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega}\, +\,  \frac{d}{dt} \,\big( J\vec{\Omega}\big) \\
& = \frac{d}{dt} \,\big( J\vec{\Omega}\big)  \, + \, \left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega}
\end{align}  and if we recall that  $$\left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega} = \vec{\Omega} \times \big( J\vec{\Omega}\big)$$
The conservation of angular momentum equation becomes the following equation for the angular velocity in the body-fixed frame:
$$\frac{d}{dt}  \big( J\vec{\Omega}\big) \, = \, - \,\, \vec{\Omega} \times \big( J\vec{\Omega}\big)$$
Thus, the full system of equation, that determines the rotation $U = U(t)$ that we really care about is
\begin{align}
J \, &\frac{d\vec{\Omega}}{dt} = \big( J\vec{\Omega}\big) \times \vec{\Omega}\\
&\frac{d}{dt} U = U\, \big(\vec{\Omega} \times \cdot \big)
\end{align} Using the conservation laws, which say that the magnitude of $\vec{M} = J\,\vec{\Omega}$ is conserved (because the vector $\vec{m}$ itself   is constant in the world frame, so its representation $\vec{M}$ in the body-fixed frame should have a constant length, but not a constant direction) and that the energy is conserved, we arrive at the fact that the solutions of the first set of equations $J \, \frac{d\vec{\Omega}}{dt} = \big( J\vec{\Omega}\big) \times \vec{\Omega}$ always lie on the pair of ellipsoids
\begin{align}
&(J\,\vec{\Omega} \, \cdot \, J\,\vec{\Omega}) = c_1 \,\,\, \text{(conservation of momentum)}\\
&(J\,\vec{\Omega} \, \cdot \, \vec{\Omega}) = c_2 \,\,\, \text{(conservation of energy)}\\
\end{align} So any solution $\vec{\Omega} = \vec{\Omega}(t)$ lies on a closed curve obtained from the intersection of the two ellipsoids and is therefore periodic, i.e. there is a time period $T>0$ such that  $\vec{\Omega}(t+T) = \vec{\Omega}(t)$.  But only because $\vec{\Omega}(t)$ is periodic doens't mean that the solution to the second set of equations
$$\frac{d}{dt} U = U\, \Big(\vec{\Omega}(t) \times \cdot \Big)$$ will be periodic. Think of the simpler scalar equation $$\frac{du}{dt} = w(t) u$$ where $w(t) = 1 + \cos(t)$. The function $w(t)$ is clearly $2\pi-$periodic, but the full solution is $$u(t) = u_0\, e^{t + \sin(t)}$$ and it is clearly not periodic. What happens in the case of the rigid body is that
the angular velocity can be decomposed as follows
$$\vec{\Omega}(t) = |\vec{\Omega}(t)| \,  \left(\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}\right)$$ where the unit vector $\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$ determines the instantaneous axis of rotation of the body at time $t$ while the magnitute $|\vec{\Omega}(t)|$ is the angular speed, i.e. it represents the instantaneous angle of rotation of the solid body around the instantaneous  axis  $\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$. Now, after time $T$, the axis  of rotation $\frac{\vec{\Omega}(t + T)}{|\vec{\Omega}(t + T)|}$ comes back exactly to $\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$, but the angle increments of rotation of the body around this axis, which are represented by the angular speed $|\vec{\Omega}(t)|$ from time $t$ to time $t+T$, generally do not exactly sum up to  one full rotation around the axes $\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$. In other words $$U(t + T) \neq U(t)$$ and this discrepancy between the period of  $\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$ and the accumulation of rotational increments $|\vec{\Omega}(t)|$ is what the authors probably mean by "incommensurability". The actual evolution of the top is in most cases quasi-periodic, i.e. at time $t+T$ the top's orientation in space returns very close to where it was at time $t$ but not exactly.
Description of the angular velocity in the inertial frame. I should add here a description of the angular velocity in the inertial frame. The conservation of energy yields the equation
$$\big(j\,\vec{\omega}(t) \cdot \vec{\omega}(t)\big) = c_2$$ but since the angular momentum $j\,\vec{\omega}(t) = \vec{m}_0$ in the inertial world frame is constant for all $t$, we obtain the equation of a plane
$$\big(\vec{m}_0 \cdot \vec{\omega}(t)\big) = c_2$$ Therefore, in the inertial world frame, the angular velocity $\vec{\omega}(t)$ traverses a curve lying on the fixed plane $\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$, where the plane is perpendicular to the angular momentum $\vec{m}_0 $. At the same time, in the body-fixed frame, if we look at the ellipsoid $\big(J\,\vec{X} \cdot\vec{X} \big) = c_2$, we know that the angular velocity at any time $t$ lies on it, i.e.
$\big(J\,\vec{\Omega} \cdot \vec{\Omega} \big) = c_2$. The equation of the ellipsoid's tangent plane at the point with position vector $\vec{\Omega}$ is given by the equation $\big(J\,\vec{\Omega} \cdot  \vec{X} \big) = c_2$. Let us apply the change of coordinates $\vec{x} = U \, \vec{X}$, which takes us from the body-fixed frame at fixed time $t$ to the inertial world frame. It's inverse transformation is $\vec{X} = U^T \, \vec{x}$ and the body-fixed ellipsoid $\big(J\,\vec{X} \cdot\vec{X} \big) = c_2$ with the tangent plane $\big(J\,\vec{\Omega} \cdot\vec{X} \big) = c_2$ is transformed into the world ellipsoid $\Big((U\,J \,U^T)\,\vec{x} \cdot\vec{x} \Big) = c_2$ with the corresponding tangent plane $\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$ ($ $ because $\vec{m}_0 = U\,J\,\vec{\Omega}\,\,$).
Consequently, if we let time $t$ run, the angular velocity  $\vec{\omega}(t)$ in the inertial world-frame traverses a curve always lying on the fixed plane $\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$ with that plane being perpendicular to the fixed angular momentum $\vec{m}_0 $ and tangent to the moving ellipsoid $\Big(\big(U(t)\,J \,U^T(t)\big)\,\vec{x} \, \cdot\,\vec{x} \Big) = c_2$ at the tip of the vector $\vec{\omega}(t)$. Consequently, the trajectory of $\vec{\omega}(t)$ can be seen as the planar curve traversed by the point of contact $\vec{\omega}(t)$ of the ellipsoid $\Big(\big(U(t)\,J \,U^T(t)\big)\,\vec{x} \,\cdot \,\vec{x} \Big) = c_2$, rolling without slipping on the plane $\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$ so that the center of the ellipsoid is always fixed at the origin.
When the rolling ellipsoid has different principal axes, the curve $\vec{\omega}(t)$ is in general not closed, and thus the angular velocity is not periodic, because $$\vec{\omega}(t) = U(t)\, \vec{\Omega}(t)$$ and as discussed earlier $\vec{\Omega}(t)$ is periodic, but $U(t)$ in general is not, hence neither is $\vec{\omega}(t)$. In some cases though, when the parameters and the constants are chosen correctly, the time-evolution of the angular velocity $\vec{\omega}(t)$ will be periodic and the rotation matrix $U=U(t)$ will also be periodic.
In the case of an inertial ellipsoid with two equal principal axes, however, we can see that the angular velocity $\vec{\Omega}(t)$ in the body-frame moves along a circle and so its length $|\vec{\Omega}(t)| = c_3$ is constant. But length is independent quantity from Cartesian frames. Therefore, in the world frame $|\vec{\omega}(t)| = |\vec{\Omega}(t)| = c_3$ so $\vec{\omega}(t)$ also has a constant length, which means that $\vec{\omega}(t)$ points to a point that always lies on the sphere $|\vec{x}| = c_3$. Therefore, as time runs, the trajectory of  $\vec{\omega}(t)$ always lies on the plane $\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$ and on the sphere $|\vec{x}| = c_3$, so the said trajectory must be a circle (because it is the intersection of a sphere and a plane). Therefore, both $\vec{\Omega}(t)$ and $\vec{\omega}(t)$ are periodic, but in general they will have different periods, i.e. there will be two different numbers $T_0 > 0$ and $T_1 > 0$ such that $\vec{\Omega}(t + T_0) = \vec{\Omega}(t)$ and $\vec{\omega}(t + T_1) = \vec{\omega}(t)$. In general, these periods will be "incommensurable" in the snece that there is no positive rational number $\frac{m}{n}$ such that $T_1 = \frac{m}{n}\, T_0$. Then, the rotation matrix $U(t)$ will not be periodic and the symmetric top will not be returning periodically to its original orientation in space. However, in the rare occasions when there is a positive rational number $\frac{m}{n}$ such that $T_1 = \frac{m}{n}\, T_0$, the two representations of the angular velocity will have a common period $T = n\, T_1 = m\, T_0$ and then the rotation matrix $U(t)$ will be periodic $U(t+T) = U(t)$ and the symmetric top will be returning periodically to its original orientation in space.