TITLE: Finding the Matrix of a linear map $T$.
QUESTION [1 upvotes]: Let $T\colon \mathcal M_{22}(\Bbb R) \to \mathcal M_{22}(\Bbb R)$  be  defined by:
$
T\left(\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\right) = \begin{bmatrix} 2c & a+ c\\
b-2c & d\\
\end{bmatrix}
$
a) Find the eigenvalues of $T$.
Sort of stuck on this. Is there a standard basis $\mathcal B$ that I can use to find $[T]_{\mathcal B}$? I've really never dealt with this sort of problem. Any sort of hints or help? :O

REPLY [2 votes]: The canonical basis is $$\left\{\begin{pmatrix}1&0\\0&0\end{pmatrix}\;\begin{pmatrix} 0&1\\0&0\end{pmatrix}\;\begin{pmatrix} 0&0\\1&0\end{pmatrix}\;\begin{pmatrix} 0&0\\0&1\end{pmatrix}\right\}$$
In general, the canonical basis for the space $\Bbb R^{n\times m}={\bf M}_{n\times m}(\Bbb R)$ is the set of $n\cdot m$ matrices $\{E^{\ell k}:1\leq \ell\leq n,1\leq k\leq m\}$, defined by $$(E^{\ell k})_{ij}=\delta_{\ell i}\delta_{kj}$$ where $$\delta_{ab}=\begin{cases} 1&a=b\cr 0&\text{ otherwise }\end{cases}$$
that is, the matrix with a $1$ at the entry $(\ell,k)$ and $0$ elsewhere. As you're being suggested, maybe you feel more comfortable working in $\Bbb R^4$ where we can think of your transformation as $$(a,b,c,d)\mapsto (2c,a+c,b-2c,d)$$ under the canonical basis $\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}$. Then you can roll it back to $\Bbb R^{2\times 2}$.

REPLY [0 votes]: There's an isomorphism between $\mathcal{M}_{22}$ and the vector space of $4$-tuples that simply unfolds a matrix. Then you can use the same standard basis you're used to, which simply corresponds to the matrices with a single nonzero entry set to $1$.


TITLE: When is the Hodge diamond concentrated in $H^{n,n}$'s?
QUESTION [2 upvotes]: Let $X$ be a smooth projective complex algebraic variety. The Hodge decomposition tells us that $H^n(X, \mathbf C) = \oplus H^{p,q}$.
Here is my question:

For what kind of $X$ is $H^{2n}(X) = H^{n,n}$ for all $n$ (I don't
  care about the odd cohomology groups $H^{2n+1}$)? Has this condition been studied in the literature?

Or a slightly more relaxed question: given some $X$, can one construct a related variety $X'$ so that the property holds for $X'$? Or, maybe, give bounds on $dim H^{2n} - dim H^{n,n}$ etc.
Thank you!

REPLY [5 votes]: A more restrictive condition, which probably has been studied, is to look at varieties with Hodge level $\le 1$. This means that $h^{pq}=0$ unless $|p-q|\le 1$. If you assume the general Hodge conjecture, then this is equivalent to the motive being a summand of $\bigoplus C_i(n_i)$ for some smooth projective curves $C_i$. If just want examples, they are easy enough to find. Take your favourite  variety with cohomology generated by algebraic cycles (e.g. a toric variety, a flag variety…) and then optionally take the product with a curve.
I am a bit skeptical about the last two questions. If $X$ is a degree $d$ surface, then $\dim H^2 - h^{11}$ grows with $d$. I am not sure what you would take for $X'$ in such a case.


TITLE: Show that $a_{n+1} =a_n + \frac{1}{3^n}$ defines a Cauchy sequence
QUESTION [2 upvotes]: I just need help.
Let ${a_n}$ be the sequence defined by the relations $a_1 = 1$ and $a_{n+1} =a_n + \frac{1}{3^n}$ for $n = 1, 2,....$ Prove that ${a_n}$ is a Cauchy sequence using definition.

REPLY [3 votes]: It's easy to see that
$$a_n = \sum_{k=0}^{n}\frac{1}{3^k}$$
The RHS is the $n$'th partial sum of a convergent geometric series (with limit $3/2$), so $a_n$ is a convergent sequence, hence a Cauchy sequence.

Edit after the OP added "using definition":
Let $m > n$. Then
$$\begin{aligned}
a_m - a_n &= \sum_{k=n+1}^{m}\frac{1}{3^k} \\
&= \frac{1}{3^{n+1}}\sum_{k=0}^{m-n-1}\frac{1}{3^k} \\
&= \frac{1}{3^{n+1}}\frac{1 - 1/3^{m-n}}{1 - 1/3} \\
&= \frac{3}{2}\frac{1}{3^{n+1}}\left(1 - \frac{1}{3^{m-n}}\right) \\
\end{aligned}$$
Observe that
$$0 \leq 1 - \frac{1}{3^{m-n}} \leq 1$$
Therefore,
$$0 \leq a_m - a_n \leq \frac{3}{2} \frac{1}{3^{n+1}} = \frac{1}{2}\frac{1}{3^n}$$
Given $\epsilon > 0$, the RHS will be less than $\epsilon$ whenever
$$3^n > \frac{1}{2\epsilon}$$


TITLE: General relativity as a gauge theory of the Poincaré algebra
QUESTION [8 upvotes]: Let the Poincaré algebra be given without any factors of i as
$[P_\mu,P_\nu]=0$,
$[M_{\rho \sigma},P_\mu]=\eta_{\sigma\mu}P_\rho-\eta_{\rho\mu}P_\sigma$,
$[M_{\mu\nu},M_{\rho\sigma}]=\eta_{\nu\rho}M_{\mu\sigma}+\eta_{\mu\sigma}M_{\nu\rho}-\eta_{\mu\rho}M_{\nu\sigma}-\eta_{\mu\sigma}M_{\nu\rho}$,
where $P$ are the generators of the translational symmetries of the Poincaré group, $M$ are the generators of the Lorentz rotations and boosts, and $\eta$ is the spacetime metric 1.
The author of this thesis 2 writes on pages 20 and 21, that infinitesimal transformations take the form,
$\delta \bullet = \xi^A P_A$
where recall that the $P_A$ translation generator is given by $\partial_\mu$. This holds for the bullet denoting an arbitrary field.
When discussing the first step in gauging the Poincare algebra — assigning a gauge field to each generator — the author states that the transformation rules for these gauge fields can be derived from the above expression and the structure constants of the algebra.
From here he states that we can "define a connection $A_\mu$ that takes values in the adjoint of the gauge group."
$A_\mu = E_\mu^A P_A +\frac{1}{2} \Omega_\mu{}^{AB} M_{AB}$
where $E_\mu^A$ and $\Omega_\mu{}^{AB}$ will eventually be realized as the vielbein and spin connection.
Where did this come from? Can anyone provide me some more clear motivation for constructing such an object? My only idea is comparing it to the following expression for the global Poincaré transformation acting on a scalar field,
$\delta(a,\lambda)\phi(x)=[a^\mu \partial_\mu - \frac{1}{2} \lambda^{\mu\nu}M_{\mu\nu}]\phi(x)=[a^\mu P_\mu - \frac{1}{2} \lambda^{\mu\nu}M_{\mu\nu}]\phi(x)$
where it seems the construction of $A_\mu$ was done by considering the "vielbein and spin connection" as parameters for the Poincaré and Lorentz symmetry transformations.
I have addition questions after this is resolved, but I will either edit the question later to include more, or post another question.
1 D. Freedman and A. Van Proeyen, Supergravity, Textbook (2012).
2 T. Zojer, Non-relativistic supergravity in three space-time dimensions, Van Swinderen Institute for Particle Physics and Gravity at Rijksuniversiteit Groningen, PhD Thesis (2016).

REPLY [5 votes]: Main idea to introduce gauge field for every generator, is to provide invariance under some group of transformation, in your case under group of diffeomorphism and local Lorentz transformations (local version of global Poincaré group). This logic is very similar to gauge invariance, where we introduce gauge field and covariant derivative to provide invariance under local gauge transformations. But now gauge group is Poincaré group.
1) Gauge symmetry with generators $T^t$:
$$
\nabla_m = \partial_m - i A_m^t T^t
$$
Poincaré group with generators $P_a$ and $M_{ab}$:
$$
\nabla_m = \partial_m -i e_m^{\;a}P_a -\frac{i}{2}\omega_m^{\;\;\;cd}M_{cd}
$$
2) Now we can directly calculate commutator:
$$
[\nabla_m, \nabla_n] = -i R_{mn}^{\;\;\;a}P_a -\frac{i}{2}R_{mn}^{\;\;\;ab}M_{ab}
$$
$R_{mn}^{\;\;\;a}$ is curvature (or field strength), that corresponds to $P_a$, $R_{mn}^{\;\;\;ab}$ curvature for $M_{ab}$. They expressed only in terms $e_m^{\;\;a}$ and $\omega_m^{\;\;cd}$. They will be identified with tetrad and spin connection.
3) Gauge transformations for $e_m^{\;\;a}$ and $\omega_m^{\;\;cd}$ follow from requirement:
$$
\phi^\prime = h \phi
$$
$$
(\nabla_m\phi)^\prime = \nabla_m^\prime\phi^\prime = h \nabla_m\phi
\;\;\;\;\;\Rightarrow \;\;\;\;\; \nabla_m^\prime = h \nabla_m h^{-1}
$$
$$
h = e^{i a^a P_a + i\omega^{ab}M_{ab}}
$$
Here $\phi$ is some field, $h$ is element of Poincaré group.
From here you can found laws of transformation for $e_m^{\;\;a}$ and $\omega_m^{\;\;cd}$. From this transformations you can argue, that this fields transform as tetrad and spin connection.
Note, that curvatures transforms homogeneously $R^\prime = hRh^{-1}$.
4) In gravity, spin connection is expressed in terms of tetrad. So we need covariantly constraint geometry:
$$
R_{mn}^{\;\;\;a} =0
$$
From this equation one can express $\omega_m^{\;\;\;cd}$ in terms of $e_m^{\;a}$.
One can also check vielbein postulate:
$$
\nabla_n e_m^{\;\;a} = 0
$$
5) From tetrads one can construct metric:
$$ g_{mn}(x) = e_m^{\;a}(x) e_n^{\;b}(x) \eta_{ab}$$
Indeces $a,b, \dots$ are Lorentz indeces, and transforms onle under Lorentz transformations.
Indeces $m,n, \dots$ are world indeces, and transforms only under diffeomorphism transformations.
Using tetrad one can convert Lorentz indeces to curved indeces:
$$
\xi_m = e_m^{\;\;a}\xi_a
$$
From Riemannian tensor $R_{mn}^{\;\;\;ab}$ one can construct standard gravity actions.
Please, Ask questions, I will try to clarify answer..!


TITLE: Locus of a point on triangle side
QUESTION [1 upvotes]: Let $P$ be a point on line segment $XY$. On the perpendicular line to $XY$ through $P$ lies point $Z$. Prove that for any position of $Z$ there is a unique point $Q$ on line segment $XZ$ for which $\angle ZYQ = \angle ZXY $. The perpendicular line on $XZ$ through $Q$ then always passes through the same (fixed) point $R$ on line segment $XY$, independent of the position of $Z$.
From geogebra it seems that
$$\angle ZSQ = \angle ZYQ =\angle ZXY  $$ and
$$\angle YZP = \angle YQS$$
$S$ is the intersection of $ZP$ and $QR$.
Holding $Z$ and shifting $Q$ on $XZ$ does not affect $\angle ZSQ$.

REPLY [0 votes]: Solution
It's easy to have that the circumcircle of $\triangle QXY$ touches $ZY$ at $Y$,and $Z,P,Q,R$ are cyclic.
Hence, $$ZY^2=ZX \cdot ZQ=ZX \cdot (ZX+XQ)=ZX^2+ZX \cdot XQ=ZX^2+PX \cdot XR.$$
This shows that $$XR=\frac{ZY^2-ZX^2}{PX}=\frac{PY^2-PX^2}{PX}=\frac{PY^2}{PX}-PX.$$
Notice that $PX,PY$ are fixed length. Then the length of $XR$ is a constant. As a result, $R$ is a fixed point independent of $Z.$


TITLE: Amount of "lucky" tickets
QUESTION [0 upvotes]: In Russia (at least in my town) we have bus tickets with 6 digits from 0 to 9. A ticket is considered "lucky" if the sum of its first 3 digits is equal to the sum of the last 3.
Is there any formula to calculate the amount of lucky tickets? I know I can just use brute force and calculate it on a computer with a for-loop, but I want more elegant solution.

REPLY [1 votes]: One simplification is to compute the number of three digit combinations that have each sum.  There is one combination that sums to $0$ (assuming leading zeros are allowed), three that sum to $1$, and so on.  This is easy to do with a spreadsheet.  It takes advantage of the fact that the sum of $258$ is the same as $456$ and so on. There are $75$ ways to sum to $13$ The number of lucky tickets is the sum of the squares of these numbers because there are $3^2=9$ tickets that have both the first and last sums $1$ and $75^2=5625$ tickets that have both the first and last sums $13$.  I find a total of $55252$ lucky tickets.


TITLE: Degree of a field extension for three fields
QUESTION [2 upvotes]: Given field extension $L/K$ and fields $E_1,E_2$ with $$(1)\ K\subset E_1\subset L,\ [E_1:K]=n_1$$ $$(2)\ K\subset E_2\subset L,\ [E_2:K]=n_2.$$ If $\gcd(n_1,n_2)=1$ then $K=E_1\cap E_2$.

Proof: Since $E_1\cap E_2\subset E_1,E_2$ we can write the following:$$(i)\ K\subset E_1\cap E_2\subset E_1,\ n_1=[E_1:K]=[E_1:E_1\cap E_2][E_1\cap E_2:K]$$ $$(ii)\ K\subset E_1\cap E_2\subset E_2,\ n_2=[E_2:K]=[E_2:E_1\cap E_2][E_1\cap E_2:K].$$ Since $\gcd(n_1,n_2)=1$ then $$(a)\ n_1\ |\ [E_1:E_1\cap E_2]$$ $$(b)\ n_2\ |\ [E_2:E_1\cap E_2].$$ Hence $E_1\cap E_2=K$.

REPLY [1 votes]: Almost! You have the right idea, but you didn't finish the proof all the way. You have the following:
$$n_1 \mid [E_1 : E_1 \cap E_2] \mid [E_1 : K]=n_1$$
Since $n_1$ is both a divisor and factor of $[E_1 : E_1 \cap E_2]$, we have:
$$[E_1 : E_1 \cap E_2]=n_1 \implies [E_1 : E_2 \cap E_2]=[E_1 : K]$$
Thus, by substitution:
$$[E_1 : K]=[E_1 : E_1 \cap E_2][E_1 \cap E_2 : K] \implies [E_1 : K]=[E_1 : K][E_1 \cap E_2 : K]$$
Now, divide both sides by $[E_1 : K]$:
$$[E_1 \cap E_2 : K]=1 \implies E_1 \cap E_2 = K$$
I'm pretty sure you already knew all of this, but you have to write it all out so that your teacher knows what you are thinking.


text:
\chapter{Introduction}

\section{Elementary knot theory}\lbl{sec:ElemKnotTheory}
We begin by recalling two facts from elementary knot theory.  These
simple statements have deep consequences for Lie algebras and
Vassiliev invariants.  The two facts can be summarized by the catch
phrases ``$1+1=2$'' and ``$n\cdot 0 = 0$.''

\begin{itemize}
\item ``$1+1=2$.'' This refers to a fact in ``abacus
arithmetic.''  On an abacus, the number $1$ is naturally represented
by a single bead on a wire, as in Figure~\ref{fig:One}, which we think
of as a tangle.  The fact that $1+1 = 2$ then becomes the equality of
the two tangles in Figure~\ref{fig:OneOneTwo}.  On the left side of the
figure, ``$1+1$'', the two beads are well separated, as for connect
sum of links or multiplication of tangles; on the right side, ``$2$'',
we instead start with a single bead and double it, so the two beads
are very close together.

In other terms, the connected sum of two Hopf links is the same as
doubling one component of a single Hopf link, as in
Figure~\ref{fig:HopfSum}.
\begin{figure}[hbtp]
\begin{center}
  \subfigure[The link ``$\One$.'']
   {\makebox[.22\linewidth]{\figscaled{One}{0.4}\lbl{fig:One}}}
\hfil\subfigure[``$\One + \One = \Two$.'']
   {\makebox[.35\linewidth]{\figscaled{OneOneTwo}{0.33}\lbl{fig:OneOneTwo}}}
\hfil\subfigure[An alternate version of ``$\One + \One = \Two$.'']
   {\makebox[.35\linewidth]{\figscaled{HopfSum}{0.36}\lbl{fig:HopfSum}}}
\end{center}
\caption{Elementary knot theory, part 1}\lbl{fig:Abacus}
\end{figure}

\item ``$n\cdot 0 = 0.$''  In the spirit of abacus
arithmetic, $0$ is represented as just a single vertical strand.  We
prefer to close it off, yielding the knot in Figure~\ref{fig:Zero}.
The knot $n\cdot 0$ is then this knot repeated $n$ times, as in
Figure~\ref{fig:nZero}.

The two knots are clearly the same, up to framing.

\begin{figure}[hbtp]
\begin{center}
  \subfigure[The knot ``$\Zero$.'']
   {\makebox[.3\linewidth]{\figscaled{Zero}{0.25}\lbl{fig:Zero}}}
\qquad\subfigure[The knot ``$n\cdot\Zero$,'' shown here for $n=3$.]
   {\makebox[.3\linewidth]{\figscaled{nZero}{0.25}\lbl{fig:nZero}}}
\end{center}
\caption{Elementary knot theory, part 2}
\end{figure}
\end{itemize}

\section{The Duflo isomorphism} \lbl{sec:intro-Duflo} The first of
these equations, ``$\One+\One=\Two$'', is related with the
{\em Duflo isomorphism}.  The Duflo isomorphism is an
algebra isomorphism between the invariant part of the symmetric
algebra and the center of the universal enveloping algebra for any Lie
algebra $\frakg$.  (The Poincaré-Birkhoff-Witt theorem gives a vector
space isomorphism.)  This isomorphism was first described for
semi-simple Lie algebras by Harish-Chandra,
and for all Lie algebras by Duflo~\cite{Duflo:Operateurs}.

Let us review briefly the Duflo isomorphism.
Every Lie algebra $\frakg$ has two associated algebras: the universal
enveloping algebra $U(\frakg)$, generated by $\frakg$ with relations
$xy - yx = [x, y]$, and the symmetric algebra $S(\frakg)$, generated
by $\frakg$ with relations $xy - yx = 0$.  There is a natural map
between the two,
\[
\chi: S(\frakg) \longrightarrow U(\frakg),
\]
given by taking a monomial $x_1\dots x_n$ in $S(\frakg)$ and averaging
over the product (in $U(\frakg)$) of the $x_i$ in all possible
orders. By the Poincaré-Birkhoff-Witt (PBW) theorem, $\chi$ is an
isomorphism of vector spaces and $\frakg$-modules.  Since $S(\frakg)$
is abelian and $U(\frakg)$ is not, $\chi$ is clearly not an algebra
isomorphism.  Even restricting to the invariant subspaces on both
sides,
\[
\chi: S(\frakg)^\frakg \longrightarrow U(\frakg)^\frakg\cong Z(\frakg),
\]
$\chi$ is still not an isomorphism of algebras.  The following theorem
gives a modification that is an algebra isomorphism.

\begin{theorem}[Duflo~\cite{Duflo:Operateurs}]\lbl{thm:duflo} For any
finite-dimensional Lie algebra $\frakg$, the map
\[
\Upsilon: S(\frakg)^\frakg \longrightarrow Z(U(\frakg))
\]
is an algebra isomorphism, where
\begin{equation*}
\begin{split}
\Upsilon &= \chi \circ \partial_{j^{\frac12}} \\
j^{\frac12}(x) &=
 \det\nolimits^{\frac12}\left(
  \frac{\sinh(\frac{1}{2} \ad x)}
       {\frac{1}{2} \ad x}\right)
\end{split}
\end{equation*}
\end{theorem}

The notation of $\partial_{j^{\frac12}}$ means to consider
$j^{\frac12}(x)$ as a power series on $\frakg$ and plug in the
(commuting) vector fields $\partial/\partial x^*$ on $\frakg^*$.
(Note that for $x^*\in\frakg^*$, $\partial/\partial x^*$ transforms
like an element of $\frakg$).  The result is an infinite-order
differential operator on $\frakg^*$, which we can then apply to a
polynomial on $\frakg^*$ ($\equiv$ an element of $S(\frakg)$).
$j^\frac{1}{2}(x)$ is an important function in the theory of Lie
algebras.  Its square, $j(x)$, is the Jacobian of the exponential
mapping from $\frakg$ to the Lie group $G$.

\section{Wheels} The bridges between the knot theory of
Section~\ref{sec:ElemKnotTheory} and the seemingly quite disparate Lie
algebra theory of Section~\ref{sec:intro-Duflo} are a certain spaces
of uni-trivalent diagrams modulo local relations.  (``Uni-trivalent''
means that the vertices have valence 1 or 3; the 1-valent vertices are
called the ``legs'' of the diagram.)  On one hand, diagrams give
elements of $U(\frakg)$ or $S(\frakg)$ for every metrized Lie algebra
$\frakg$ in a uniform way, as we will see in
Chapter~\ref{chap:liealg}; on the other hand, they occur naturally in
the study of finite type invariants of knots, as we will see in
Chapter~\ref{chap:vassiliev}.  Like the associative algebras
associated to Lie algebras, these diagrams appear in two different
varietes: $\cA$, in which the legs have a linear order, as in
Figure~\ref{fig:A-examp}, and $\cB$, in which the legs are unordered,
as in Figure~\ref{fig:B-examp}.  As for Lie algebras, they each have a
natural algebra structure (concatenation and disjoint union,
respectively); and there is an isomorphism $\chi:\cB\rightarrow
\cA$ between the two (averaging over all possible orders of the legs).

\begin{figure}[tbp]
\begin{center}
  \subfigure[A sample element of $\cA$.]
   {\makebox[.3\linewidth]{\figscaled{A-examp}{0.3}\lbl{fig:A-examp}}}
\qquad\subfigure[A sample element of $\cB$.]
   {\makebox[.3\linewidth]{\figscaled{B-examp}{0.25}\lbl{fig:B-examp}}}
\end{center}
\caption{Example web diagrams}
\end{figure}
There is one element of the algebra $\cB$ that will be particularly
important for us: the ``wheels'' element of the title.  It is the
diagrammatic analogue of the function $j^\frac12$ above.
\begin{equation} \lbl{eq:OmegaDef}
  \Omega=\exp \sum_{n=1}^\infty b_{2n}\omega_{2n} \in \cB.
\end{equation}
where
\begin{itemize}
\item The `modified Bernoulli numbers' $b_{2n}$ are defined by the power
series expansion
\begin{equation} \lbl{eq:MBNDefinition}
  \sum_{n=0}^\infty b_{2n}x^{2n} = \frac{1}{2}\log\frac{\sinh x/2}{x/2}.
\end{equation}
These numbers are related to the usual Bernoulli numbers $B_{2n} =
4n\cdot(2n)!\cdot b_{2n}$ and to the values of the Riemann
$\zeta$-function on the even integers. The first three modified
Bernoulli numbers are $b_2=1/48$, $b_4=-1/5760$, and $b_6=1/362880$.

\item The `$2n$-wheel' $\omega_{2n}$ is the degree $2n$ web diagram
made of a $2n$-gon with $2n$ legs:
\begin{equation}\label{eq:omega-definition}
  \omega_2=\mathcenter{\figscaled{2wheel}{0.5}},\quad
  \omega_4=\mathcenter{\figscaled{4wheel}{0.5}},\quad
  \omega_6=\mathcenter{\figscaled{6wheel}{0.5}},\quad\ldots.
\end{equation}
\end{itemize}

Our main theorem is written in terms of $\partial_\Omega$, the
operation of applying $\Omega$ as a differential operator, which takes
a diagram $D$ and
attaches some of its to $\Omega$.  (See
Section~\ref{sec:diagram-differential} for the precise definition.)

\begin{theorem}[Wheeling; joint with D.~Bar-Natan]\lbl{thm:wheeling}
The map $\Upsilon = \chi \circ \partial_\Omega: \cB \rightarrow \cA$
is an algebra isomorphism.  \end{theorem}

(All the notation above, including the definitions of $\cA$ and $\cB$,
is explained in Chapter~\ref{chap:liealg}.  The proof of the theorem
is in Chapter~\ref{chap:wheeling}.)

Although Theorem~\ref{thm:wheeling} was motivated by Lie algebra
considerations when it was first
conjectured~\cite{BGRT:WheelsWheeling, Deligne:letter}, the proof we
will give, based on the equation ``$1+1=2$'' from
Section~\ref{sec:ElemKnotTheory}, is entirely independent of Lie
algebras and is natural from the point of view of knot theory.  In
particular, we obtain a new proof of Theorem~\ref{thm:duflo} for
metrized Lie algebras, with some advantages over the original proofs
by Harish-Chandra, Duflo, and Cartan: our proof does not require any
detailed analysis of Lie algebras, and so works in other contexts in
which there is a Jacobi relation.  For instance, our proof works for
super Lie algebras with no modification.

Theorem~\ref{thm:wheeling} has already seen several applications.  In
Chapter~\ref{chap:wheels} we will use it to compute the Kontsevich
integral of the unknot, using our second elementary knot theory
identity ``$n\cdot 0 = 0$''.

\begin{theorem}[Wheels; joint with T.~Le]\label{thm:wheels}
The Kontsevich integral of the unknot is
\[
Z(\bigcircle) = \Omega \in \cB.
\]
\end{theorem}

We also compute the Kontsevich integral of the
Hopf link $\Hopf$; which is intimately related to the
map $\Upsilon$ above.

\section{Related work} Theorem~\ref{thm:wheeling} was first
conjectured by Deligne~\cite{Deligne:letter} and Bar-Natan,
Garoufalidis, Rozansky, and Thurston~\cite{BGRT:WheelsWheeling}, who
also conjectured Theorem~\ref{thm:wheels}.

Further computations for a sizeable class of knots, links, and
3-manifolds (including all torus knots and Seifert-fiber homology
spheres) have been done by Bar-Natan and
Lawrence~\cite{Bar-NatanLawrence:RationalSurgery}.  Hitchen and
Sawon~\cite{HitchinSawon:CurvatureHyperKahler} have used
Theorem~\ref{thm:wheeling} to prove an identity expressing the $L^2$
norm of the curvature tensor of a hyperk\"ahler manifold in terms of
Pontryagin classes.  And in a future paper~\cite{Thurston:Torus} I
will show how to write simple formulas for the action of $\sltwo(\ZZ)$
on the vector space associated to a torus in the perturbative TQFT of
Murakami and Ohtsuki~\cite{MO:TQFTUniversal}.

These applications suggest that Theorem~\ref{thm:wheeling} is a
fundamental fact.  Another sign of fundamental facts is that they have
many proofs.  Besides the earlier work of Harish-Chandra, Duflo, and
Cartan mentioned above, there are two other recent proofs of
Theorem~\ref{thm:wheeling}. One is due to
Kontsevich~\cite[Section~8]{Kontsevich:DeformationQuantization}, as
expanded by~\cite{ADS:KontsevichQuantization}.  Kontsevich's proof is
already at a diagrammatic level, similar to the one in this thesis,
although it is more general: it works for all Lie algebras, not just
metrized ones.  His proof again uses a transcendental integral,
similar in spirit to the one in Chapter~\ref{chap:kontsevich}.
Another proof is due to Alekseev and
Meinrenken~\cite{AlekseevMeinrenken:NonCommutativeWeil}.  The Alekseev
and Meinrenken paper is not written in diagrammatic language, but
seems to extend to the diagrammatic context without problems.  Their
proof does not involve transcendtal integrals: the only integral in
their proof is in the proof of the Poincaré lemma (the homology of
$\RR^n$ is trivial in dimension $>\,0$).

\section{Plan of the thesis} The first few chapters are standard
introductory material, included here to make this thesis as
self-contained as possible.  In Chapter~\ref{chap:liealg}, we review
the diagrammatics of Lie algebras and define the space $\cA$ and
$\cB$.  In Chapter~\ref{chap:vassiliev} we review the definition of
Vassiliev invariants, again arriving at the same spaces $\cA$ and
$\cB$; in addition, a few more spaces are defined in
Section~\ref{sec:diagrams2}.  The exposition is related to the
standard exposition~\cite{BarNatan:OnVassiliev} and the work of
Goussarov~\cite{Goussarov:KnottedGraphs, Goussarov:3Manifolds} and
Habiro~\cite{Habiro:Claspers}, but is different from both.
Chapter~\ref{chap:kontsevich} is a review of the Kontsevich integral
and its main properties, notably the behaviour under connect sum and
cabling.

After this introductory material, we give the proof of
Theorem~\ref{thm:wheeling} in Chapter~\ref{chap:wheeling}, modulo
fixing the coefficients $b_{2n}$, which is done in
Appendix~\ref{app:coefficients}.  In Chapter~\ref{chap:wheels}, we
prove Theorem~\ref{thm:wheels}, as well as similar results for the
Hopf link.

config: arxiv


TITLE: Algerbraic manipulation
QUESTION [0 upvotes]: I'm sure this will be simple for someone, how do you go from the line I've highlighted (with the oversized arrow) to the next line. I can't see it.

Thanks
Mike

REPLY [0 votes]: We have that $$I = \int (\frac {e^{2x}+1 - e^{2x}}{e^{2x}+1}) \mathrm {d}x $$ $$= \int (\frac {e^{2x}+1}{e^{2x}+1} - \frac {e^{2x}}{e^{2x}+1}) \mathrm {d}x $$ $$= \int (1 - \frac {e^{2x}}{e^{2x}+1}) \mathrm {d}x $$
The fact that we can rearrange the terms and cancel out common terms is used here. Hope it helps.


TITLE: Why do people prefer cosine to sine when speaking of harmonic oscillation?
QUESTION [9 upvotes]: In almost all of the physics textbooks I have ever read, the author will write the oscillating function as

$$x(t)=\cos\left(\omega t+\phi\right)$$

My question is that,  is there any practical or historical  reason why we should prefer $\cos$ to $\sin$ here? One possible explanation I can think of is that, to trigger a harmonic oscillation movement, we usually push the mass (to the maximum displacement) from the balance point at the initial moment, for which the cosine function will be neater to use than sine ($\phi=0$). But is it really the case?

REPLY [1 votes]: Another reason is to use the initial position as a parameter.
Compare $z=z_0\lambda^t$  and $h=h_0\cos \omega t$
You can't do the same with $\sin$ - at best you can say perhaps $d=d_{max}\sin \omega t$
By the way, this argument fails once you introduce a phase shift!


TITLE: Relative velocity of two cars
QUESTION [0 upvotes]: Two cars A and B are at $100\,\text{m}$ and $200\,\text{m}$ distance from the origin $O$ at time $t = 0$.
They start simultaneously with velocities $10\,\text{m/s}$ and $5\,\text{m/s}$ respectively. A will overtake B at a distance of $x$ from $O$.
The answer given: $300\,\text{m}$.
What I have tried so far: I tried to calculate the relative velocity of car A and then to find out the time and the distance but I got stuck. And each time I try, I go with a different approach, so, really, I haven't a clue even  on how to proceed to solve this problem.

REPLY [1 votes]: Probably there are many ways to approach the problem, The simplest, the first that comes to my mind is : find the time it takes.
Think of the time it takes A to cover/catch up the difference in space using the difference of speed.
P.S I hope this hint is not against the rule, if it is, just tell me and I delete my answer.


TITLE: how many subsets in a set?
QUESTION [0 upvotes]: I know this is a pretty basic question, but I fail to understand the notion behind it.
Consider the following:

If $A=\left\{a,b,c\right\}$, how many subsets can be created from $A$?

It is simple to write all possibilities and see it is $2^3=8$, but I don't understand why this is idea behind the formula.

REPLY [3 votes]: Let's say $A$ has $n$ elements (in your example $n=3$). To determine a subset of $A$, we go through each element of $A$ and decide whether to put it in our subset or to omit it from our subset. This creates two possible options for each element of $A$, which yields $2^n$ possible combinations once we go through all elements of $A$. Hence, $A$ has $2^n$ subsets.


TITLE: Stronger version of finite additivity of Legesgue measure
QUESTION [3 upvotes]: Let $m_*(A)$ denote the Lebesgue outer measure, and when $A$ is measurable, let $m(A)=m_*(A)$ be the Lebesgue measure.
Let $U=\{E_1,\cdots,E_N\}$ be a finite collection of pairwise disjoint Lebesgue measurable sets (of $\mathbb{R}^n$), and denote $E:=\bigcup_{i=1}^{N} E_i$. We know that (finite additivity):
$$ m(E)=\sum_{i=1}^{N}m(E_i)\tag{1}$$
If we add another set $A$, which may not be measurable, is it true that
$$m_*(A\cap E)=\sum_{i=1}^{N}m_*(A\cap E_i)\tag{$*$}$$
(1) is a special case of ($*$) when $A=E$. Anyone have any idea on proving  ($*$)?
I thought $m_*(A\cap E)=m_*(A)-m_*(A\setminus E)$ may be useful but I don't know how to proceed.
Note that the definition of a set $E$ is measurable is that $m_*(A)=m_*(A\setminus E)+m_*(A\cap E)$ for all $A$.

REPLY [1 votes]: Consider the case $U = \{E_1, E_2\}$. You can easily get the general statement by induction. Because $E_1$  is measurable, and using  $A \cap (E_1 \cup E_2) $ as a "test" set, we have:
\begin{align}
m_*(A \cap (E_1 \cup E_2) \cap E_1 ) &= m_*(A \cap (E_1 \cup E_2) )   - m_*(A \cap (E_1 \cup E_2) \setminus E_1 )  \\
&=m_*(A \cap (E_1 \cup E_2) ) - m_*(A \cap E_2 ).
\end{align}
On the other hand $m_*(A \cap (E_1 \cup E_2) \cap E_1 ) = m_*(A \cap E_1)$.


TITLE: Can Quarks and Gluons Escape from Protons and Neutrons?
QUESTION [0 upvotes]: An answer to a related question described protons and neutrons as made up of a "sea of quarks, anti-quarks, gluons, ...." with a net makeup to provide either a unit charge or no net charge.  If so, the question arises as to whether it is possible to "boil" a quark or other particle out of a neutron.  If so, I would like to know what is the theoretical energy to remove a quark or other particle from a neutron.  Such a value would be useful in a model on which I am working.
And, is it reasonable to consider a proton or neutron as a "container", perhaps one from which escape is not possible?

REPLY [2 votes]: Quarks and gluons are subject to “color confinement”: the strong interaction is so strong that only color-singlet objects are independently observable systems. At low energies, the relevant degrees of freedom are the baryons (with three valence quarks) and the mesons (whose valence structure is a quark-antiquark pair).  The strong-interaction decays of excited baryons are generally via meson emission.
At very high energy, the vacuum undergoes a phase transition to a “quark-gluon plasma,” in which the baryon-meson degrees of freedom disappear and the quark-gluon degrees of freedom are better. But you can’t extract a single quark from a QGP; it is a collective phenomenon, more like a liquid. A QGP cools by condensing into baryons, antibaryons, and mesons.
Whenever I give this little spiel, a heavy-quark person appears to remind me that the bottom and top quarks live so briefly that they don’t really hadronize. So there is a sense in which you can talk about “a bottom quark,” where if you had the same conversation about charm you would eventually need to switch to “a charmed meson.”


TITLE: If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
QUESTION [5 upvotes]: If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$

My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}$
By squaring both sides, we get,
$\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}$
$\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
So, now, we have to prove that,
$-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0$
I cannot do this. Please help!

REPLY [0 votes]: My Solution:: Given $$\displaystyle \frac{\sin^4 x}{a}+\frac{\cos^4 x}{b} = \frac{1}{a+b}.$$
Now using the Cauchy-Schwarz inequality we get
$$\displaystyle \frac{(\sin^2 x)^2}{a}+\frac{(\cos^2 x)}{b}\geq \frac{\left(\sin^2 x+\cos^2 x\right)^2}{a+b} = \frac{1}{a+b}$$
and equality holds when $$\displaystyle \frac{\sin^2 x}{a} = \frac{\cos^2 x}{b}.$$
Now using ratio and proportion, we get $$\displaystyle \frac{\sin^2 x}{a} = \frac{\cos^2 x}{b}=\frac{\sin^2 x+\cos^2 x}{a+b}=\frac{1}{a+b}.$$
So $$\displaystyle \sin^2 x = \frac{a}{a+b}$$ and $$\displaystyle \cos^2 x = \frac{b}{a+b}.$$ So we get $$\displaystyle \frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{a^3}{a^2\cdot (a+b)^3}+\frac{b^3}{b^2\cdot (a+b)^3} = \frac{1}{(a+b)^2}$$


TITLE: Find max rate of change of $f(x,y) = \sin(xy)$ at the point $(1,0)$ and in the direction in which it occurs.
QUESTION [1 upvotes]: Find max rate of change of $f(x,y) = \sin(xy)$ at the point $(1,0)$ and in the direction in which it occurs.
I did the following:
$$\nabla f = <\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}>$$
$$= <y\cos(xy), x\cos(xy)>$$
$$\nabla f_{(1,0)} = <(0)\cos(0),\cos(0)>$$
$$= <0,1>$$
This vector tells me both the direction and the magnitude in which the maximum rate of change occurs because it happens to already be a unit vector, no?
But shouldn't I somehow get a scalar for the maximum rate of change?

REPLY [1 votes]: Hint:
The magnitude of the gradient is a scalar.