TITLE: Two questions about the order type of $2^\alpha$ equipped with the lexicographical ordering
QUESTION [2 upvotes]: Let $2^\alpha=\{f\mid f\colon\alpha\to2\}$.
1, Is it provable in ZFC that for all infinite ordinals $\alpha$, $2^{\alpha+1}$ cannot be embedded into $2^\alpha$, where $2^{\alpha+1}$ and $2^\alpha$ are equipped with the lexicographical ordering?
2, We can prove in ZFC that for all infinite cardinals $\kappa$ and all ordinals $\alpha<\kappa$, $\kappa$ cannot be embedded into $2^\alpha$, where $2^\alpha$ is equipped with the lexicographical ordering.
Is this proposition provable in ZF? Or is it provable in ZF that $\omega_1$ cannot be embedded into $2^{\omega+\omega}$, where $2^{\omega+\omega}$ is equipped with the lexicographical ordering? (I have proved in ZF that $\omega_1$ cannot be embedded into $2^\alpha$ for all $\alpha<\omega+\omega$.)

REPLY [2 votes]: A friend of mine proves in ZF the proposition in Question 1 as follows.
Assume towards a contradiction that $H$ is an embedding of $2^{\alpha+1}$ into $2^\alpha$. For two different $f,g\in2^\alpha$, the branching point of $f$ and $g$, which will be denoted by $\mathrm{brp}(f,g)$, is defined to be the least ordinal $\beta<\alpha$ such that $f(\beta)\neq g(\beta)$. Let $<_l$ be the lexicographical ordering. The following two facts are easily verifiable:
(i) If $e<_lf<_lg<_lh$, then either $\mathrm{brp}(e,h)<\mathrm{brp}(e,f)$ or $\mathrm{brp}(e,h)<\mathrm{brp}(g,h)$.
(ii) If $e\leqslant_lf<_lg\leqslant_lh$, then $\mathrm{brp}(e,h)\leqslant\mathrm{brp}(f,g)$.
Let $F$ be the function on $2^{\leqslant\alpha}$ into $\alpha$ such that for all $u\in2^{\leqslant\alpha}$,
$$F(u)=\mathrm{brp}(H(u\cup\langle0\mid\mathrm{Dom}(u)\leqslant\beta\leqslant\alpha\rangle),H(u\cup\langle1\mid\mathrm{Dom}(u)\leqslant\beta\leqslant\alpha\rangle)).$$
We define now by recursion a function $G$ on $\alpha$ into $\{0,1\}$ as follows. $G(\delta)=0$, if $F(G\upharpoonright\delta)<F((G\upharpoonright\delta)\cup\{(\delta,0)\})$, and $G(\delta)=1$, otherwise. By the fact (i) above, we have that if $G(\delta)=1$ then $F(G\upharpoonright\delta)<F((G\upharpoonright\delta)\cup\{(\delta,1)\})$.
Let $T$ be the function on $\alpha+1$ into $\alpha$ such that for all $\beta<\alpha+1$, $T(\beta)=F(G\upharpoonright\beta)$. By the definition of $G$, $\ T(\delta)<T(\delta+1)$ for all $\delta<\alpha$. By the fact (ii) above, we have that for all limit ordinals $\gamma<\alpha+1$ and all $\beta<\gamma$, $T(\beta)\leqslant T(\gamma)$. Hence $T$ is an increasing function from $\alpha+1$ into $\alpha$, which is a contradiction.


TITLE: Which variables to choose in a change of variables for a PDE
QUESTION [0 upvotes]: I want to solve the following PDE:
$$\frac{\partial f}{\partial x}-3 \frac{\partial f}{\partial y}=x+y$$
with boundary condition:
$$f(x,-2 x)=\sin \left(x^{2}\right)$$
Using a change of variables. How can I determine the correct variables in such a transformation?

REPLY [0 votes]: $$\frac{\partial f}{\partial x}-3 \frac{\partial f}{\partial y}=x+y$$
System of characteristic ODEs (Charpit-Lagrange) :
$$\frac{dx}{1}=\frac{dy}{-3}=\frac{df}{x+y}$$
A first characteristic equation from solving $\frac{dx}{1}=\frac{dy}{-3}$ :
$$3x+y=c_1$$
A second characteristic equation from :
$$\frac{dx}{1}=\frac{dy}{-3}=\frac{(x)dx+(-\frac13 y)dy}{1(x)-3(-\frac13 y)}=\frac{xdx-\frac13 ydy}{x+y}=\frac{df}{x+y}$$
$$xdx-\frac13 ydy=df$$
$$f-\frac12 x^2+\frac16 y^2=c_2$$
General solution of the PDE expressed on implicit form $c_2=\Phi(c_1)$ :
$$f-\frac12 x^2+\frac16 y^2=\Phi(3x+y)$$
$\Phi$ is an arbitrary function, to be determined according to the boundary condition.
$$f(x,y)=\frac12 x^2-\frac16 y^2+\Phi(3x+y)$$
Condition :
$$f(x,-2x)=\sin(x^2)=\frac12 x^2-\frac16 (-2x)^2+\Phi(3x+(-2x))$$
$$\Phi(x)=\sin(x^2)+\frac16 x^2$$
Now the function $\Phi$ is known. We put it into the above general solution where the argument is $(3x+y)$ :
$$f(x,y)=\frac12 x^2-\frac16 y^2+\sin\big((3x+y)^2\big)+\frac16 (3x+y)^2$$
After simplification :
$$f(x,y)=2x^2+xy+\sin\big((3x+y)^2\big)$$


TITLE: Compute lambda of Poisson distribution
QUESTION [1 upvotes]: $p = \lambda e^{ -\lambda }$
If $p$ and $e$ are known. How can I calculate $\lambda$?
I tried on log on both side but it did not help. Any suggestions?

REPLY [3 votes]: You say that $e$ is known -- does that mean you're using $e$ as a variable name rather than to denote Euler's number? That would be rather confusing.
In either case, what you need is the Lambert W function. Using the defining relation
$$z=W(z)\mathrm e^{W(z)}$$
(where $\mathrm e$ as usual denotes Euler's number) and
$$-p=-\lambda\mathrm e^{-\lambda}\;,$$
you get
$$\lambda =-W(-p)\;.$$
Note that this is a) multivalued and b) only defined for $p\le1/e$.
If you really did intend to use $e$ as a variable name, we can replace it by $a$ to keep things clear and then write
$$
\begin{eqnarray}
p
&=&
\lambda a^{-\lambda}
\;,
\\
p
&=&
\lambda \mathrm e^{-\lambda\log a}
\;,
\\
-p\log a
&=&
-\lambda\log a \mathrm e^{-\lambda\log a}
\;,
\\
\lambda
&=&
-\frac{W(-p\log a)}{\log a}
\;.
\end{eqnarray}$$


TITLE: Why doesn't this argument show the Möbius bundle is trivial?
QUESTION [4 upvotes]: I wrote the following argument to prove that $S^1$ is parallelizable, that is, to show that the tangent bundle is trivial. It looks fairly reasonable to me.

Let $\tau=2\pi$.
We define a map $\varphi:S^1\times\Bbb R\to TS^1$ by
  $\varphi((e^{i\tau\theta},
 t))=(e^{i\tau\theta},t\frac{\partial}{\partial
 x^1}\Big|_{e^{i\tau\theta}})$. This map is clearly a bijection:
  injectivity follows trivially and surjectivity follows since the
  tangent space $T_{e^{i\tau\theta}}$ is one-dimensional.
It remains to be shown that it is a diffeomorphism. We will, in fact,
  show that it is the identity map on suitably chosen coordinates. For
  any point $(e^{i\tau\theta},t)\in S^1\times\Bbb R$, we can choose the
  coordinate chart such that
  $(e^{i\tau(\theta+\varepsilon)},(t+\delta))$, for sufficiently small
  $\varepsilon$ and $\delta$, is given by $(\varepsilon,\delta)$.
Similarly, for any point $(e^{i\tau\theta},t\frac{\partial}{\partial
 x^1}\Big|_{e^{i\tau\theta}})\in TS^1$, we can choose the coordinate
  chart such that
  $(e^{i\tau(\theta+\varepsilon)},(t+\delta)\frac{\partial}{\partial
 x^1}\Big|_{e^{i\tau(\theta+\varepsilon)}})$, for sufficiently small
  $\varepsilon$ and $\delta$, is given by $(\varepsilon,\delta)$.  Then,
  we have that $\widehat\varphi$ (the function with respect to these two
  coordinates) sends $(\varepsilon,\delta)\mapsto (\varepsilon,\delta)$,
  as desired.

However, I don't see where this is using the tangent bundle construction in any meaningful way. It seems that this would work just as well for any rank-1 vector bundle over the sphere.
That concerns me, because of course the conclusion is not true: for instance the Möbius strip is (diffeomorphic to) a vector bundle over the sphere. Since it is not orientable, we know that the bundle is nontrivial.
So the question is: Where does this break for a general vector bundle? Or if the argument is simply invalid, can it be fixed without much trouble?

REPLY [2 votes]: You're implicitly using the fact that a non-vanishing section (the coordinate vector field $\partial/\partial x_{1}$) exists. The tangent bundle admits such a section, while the non-trivial line bundle doesn't.


TITLE: $U\in{Mat(\mathbb{C})} , $$U$ is unitary and $U+iI$ is self adjoint, prove: $U = -iI$
QUESTION [4 upvotes]: I'm preparing for a test and this question was a real pain to do. I get a bit confused by all the terms, so I'd appreciate if you guys peek at my attempt to solve this, and see if its correct.
What I did: First of all, $U+iI$ is self adjoint, therefore its normal and unitary diagonalizable, which means there exists a unitary matrix $P$ and a diagonal matrix $D$ so: $$U+iI = PDP^{-1}$$
Now because $U+iI$ is diagonalizable it has $n$ distinct eigenvectors, and for every such vector $v_i$ there is an eigenvalue $\lambda_k$ so: $$(U+iI)v_i = {\lambda_k}v_i$$ $$Uv_i + iIv_i = {\lambda_k}v_i$$ $$Uv_i = (\lambda_k-i)v_i$$
Therefore $(\lambda_k-i)$ is an eigenvalue of $U$, but $U$ is unitary which means that $(\lambda_k-i) (\overline{\lambda_k-i}) = 1 $ and finally $\lambda_k = 0$. This occurs for all $v_i$ eigenvectors, therefore $U+iI$ has one eigenvalue and its zero. Therefore $D$ has only zeroes on its diagonal, and $U+iI = 0$ therefore $U = -iI$.
I'm not really sure if the argument I used about the eigenvectors\values is correct, if its not, would love to hear a few tips on how to proceed, thanks!

REPLY [3 votes]: This solution is correct as far as I can tell.  For a simpler solution:
$U + iI$ is self-adjoint, so it is equal to its complex-conjugate transpose.  This means that all elements on the diagonal of $U + iI$ must be real.  Thus the imaginary part of all the diagonal elements of $U$ must be $-1$.  Since $U$ is unitary, each row and column must have a vector length of 1, but since $-i$ has magnitude 1, all other terms have to be zero (adding a real part to a diagonal element or any nonzero element off the diagonal will give that row and column a length greater than one).  Therefore $U = -i I$.


TITLE: Is it possible to identify a unique square within a grid given its vertices?
QUESTION [0 upvotes]: For example, I have a grid where the x-axis goes from -1.5 to 1.5, with an interval of 0.3 (i.e. -1.5 would go to -1.2 then -0.9 and so on).
The y-axis goes from 3.0 to -3.0 with an interval of 0.3 as well.
How can I identify a unique square within this grid just by its vertices?
I tried to sum up the vertices which make up the square, but the sum of the vertices is not unique and can apply to a few other squares within the grid as well.
EDIT FOR CLARIFICATION:
Is there a way to use vertices like these to help me identify the square in the graph? I am designing a game where I want to be able to check if the box is currently occupied. I only have access to the vertices surrounding the box.

REPLY [1 votes]: You can find the center of a square by finding the mid-point of it's diagonal:
$$(h,k) = \left( \frac {x_2+x_1} {2}, \frac {y_2+y_1} {2} \right)$$
Where $(h,k)$ represents the center. You can then get the side length by finding the distance between any two adjacent vertices:
$$\text{side length (r)} = \sqrt{(x_2-x_1) + (y_2-y_1)}$$
And then you can use this formula to identify the square:
$$\left|(x - h) + (y - k)  \right|+\left| (x - h)  - (y - k)  \right|=r$$


TITLE: Thermal center of a set
QUESTION [2 upvotes]: The current Wikipedia article on the heat equation includes a gif which shows how the solution to the heat equation evolves over time if the initial data is the indicator function of a domain.

Suppose $w(t,x)$ is the solution to the heat equation $w_t=\Delta w$ satisfying $w(0,x)=\mathbb{1}_A(x)$ for a nice set $A$. Suppose further that the function $w(t,x)$ (as a function of $x$) attains a unique maximum at $x=a(t)$ for each $t$ in some interval $(0,\varepsilon)$, and we can define the thermal center of $A$ to be the limit $a=\lim\limits_{t\to0^+} a(t)$. (I made up this concept.)
Can this be given an integral formula, or integral characterization? Or some other nice and interesting characterization? For example, maybe it happens to be the center of mass of $A$, or something. Or, what is it for, say, a triangle?

REPLY [1 votes]: TL;DR In any dimension the position(s) of the thermal center(s) belong to the set
$$\mathbf{x}_{th}=\text{argmax}_x \min_{r\in \partial A} d(x,r)$$
Here is a sketch of a proof in $d=2$. (To extend the idea to arbitrary $d$ one just needs to replace $C_r^x\to (S^{d-1})_r^x$ and make appropriate adjustments to the measures and estimates, but I don't think this precludes the proof from going through).  Assume that the set $A$ is compact. Define the density of neighbors of a point $x$ at distance $r$ from that point to be
$$\rho(r;x)=\frac{dN}{dr}(r;x)=r\mu(C^x_r\cap A)$$
where $C_r^x $ is a circle of radius $r$ centered at $x$ and $\mu(C_r^x)=2\pi$. This function in a certain sense measures how many neighbors the point $x$ has a certain distance away. Note that by definition
$$\int_0^{\max _{r\in A}d(x,r)}\rho(r;x)dr=\mu(A)$$
EXAMPLE: For a circle of radius $R$ centered at the origin, the density of neighbors can be computed to be
$$\rho(r;x)=\begin{Bmatrix}2\pi r&r\in (0,R-|x|)\\ 2r\arccos\left(\frac{|x|^2+r^2-R^2}{2|x|r}\right)&~~~~~~~~~r\in (R-|x|,R+|x|)\end{Bmatrix}$$
Critically, notice here that the neighbor density for any two points in the interior of $A$ is the same up until some radius $L=\min\{\min_{r\in A}d(x_1,A), \min_{r\in A}d(x_2,A)\}$. The reason why this is important is because the idea is that in the limit $t\to 0^+$ the difference in value between two points is determined by the difference in the local environment of the two points. In layman words, the furthest away you are from the boundary, the more points you have around you and therefore you get a bigger contribution in the limit.
With this notation established, notice now that the solution to the initial value problem posed in the question is given by
$$w(x,t)=\frac{1}{4\pi t}\int_{0}^{\max _{r\in A}d(x,r)}e^{-\delta^2/4t}\rho(\delta;x)d\delta$$
Finally, define $r_0=\max_x\min_{r\in \partial A}d(x,r)$. For an arbitrary point $x$, define $r_x=\min_{r\in \partial A}d(x,r)$ and $R_x=\max\{\max_{r\in \partial A}d(x,r), \max_{r\in \partial A}d(\mathbf{x}_{th},r)\}$. It should be clear that $r_x\leq r_0$. Consider the difference
$$4\pi t(w(\mathbf{x}_{th},t)-w(x,t))=\left(\int_{r_x}^{r_0}+ \int_{r_0}^{R_x}\right)e^{-\delta^2/4t}(\rho(\delta;\mathbf{x}_{th})-\rho(\delta;x))d\delta:= I_1+I_2$$
One can show easily that $I_1\geq 0$ and hence establish the bounds
$$C(x) e^{-r_0^2/4t}\leq I_1 \leq 4\pi t(e^{-r_0^2/4t}- e^{-r_x^2/4t})\\|I_2|\leq 4\pi t (e^{-r_0^2/4t}-e^{-R_x^2/4t})$$
for $C(x):=\int_{r_x}^{r_0}(2\pi \delta- \rho(\delta;x))d\delta$ independent of $t$. Thus, in the limit $t\to 0^+$ it can be shown that $|I_2|/I_1\to 0$, and it can also be shown that for a small enough time
$$ t < A\epsilon C(x)~,~ \epsilon<\min\left(1, \frac{W^2-r_0^2}{4}\right)$$
where $A$ a positive constant, and $W$ denotes the radius of the smallest circle containing $A$.  Hence, it must be true that $I_1+I_2>(1-\epsilon)I_1>0$. It follows immediately that
$$\forall t<\epsilon(x):= AC(x)\epsilon~~,~~ w(x,t)\leq w(\mathbf{x}_{th},t)$$
By definition, it is true that the function $C(x)$ is always bounded and non-negative on compact sets and zero only when $x\in \mathbf{x}_{th}$. If we restrict $x$ to only consider $x\notin \mathbf{x}_{th}$, by the estimate above it is seen that for any of those $x$, there exists a finite time interval for which the last statement holds, which allows us to conclude that the thermal set defined above contains the maximum in the limit $t\to 0$.
NOTE: Obviously when $\mathbf{x}_{th}$ contains only one element, then that is the unique thermal center. When it doesn't, the thermal center is determined by subleading contributions to the integral above. (My intuition is that thermal centers can only be non-unique when the set possesses a symmetry group that allows for the existence of multiple members of the thermal set with the same environment. In any other case, there must be a unique thermal center.)
COMMENT 1: You can think about the last step in the proof as follows. For any small enough finite non-zero time $t=T$ it guarantees that there is a set of points in the set $x\notin \mathbf{x}_{th}$ for which $w(x,T)<w(\mathbf{x}_{th},T)$ (the points for which the condition $C(x)>T/A\epsilon$ is fulfilled). As $T\to 0$, the set shrinks until it becomes the empty set.
COMMENT 2: There exist sets $A$ where $\mathbf{x}_{th}$ can be relatively large. An example is given by the set $$A=C_R^{(-L,0)}\cup ([-L,L]\times[-R,R])\cup C_R^{(L,0)}$$
By construction it admits the thermal set
$$\mathbf{x}_{th}=[-L,L]\times\{0\}$$
However, it appears that the thermal center is the origin, which can be justified by inspecting $\rho(r;x\in \mathbf{x}_{th})$.


TITLE: What does it mean for commutator of position and momentum to be equal to the metric tensor?
QUESTION [0 upvotes]: I was reading: https://arxiv.org/abs/hep-th/9505152
And got confused by the first line stating
$$ [P^{\mu}, X^{\nu}] = ig^{\mu\nu} \tag{1.1}.$$
I want to believe that this a generalization of the statement
$$ XP - PX = i\hbar\mathbb{1}.$$
But I can't seem to guess what is meant by $g^{\mu \nu}$. I have seen this type of expression before as an expression for the metric tensor, wich makes me wonder if the expression $[P^{\mu}, X^{\nu}]$ is some kind of tensor product but again I've never heard of a out product of momentum and position being equal to the metric before, so that doesn't sit well with me.
How do I make sense of this?

REPLY [1 votes]: My foreword is that IMHO that expression is wrong (EDIT: I'm too bold sometimes, let's say incomplete), as the right hand side is clearly symmetric in the interchange $\mu\leftrightarrow\nu$, while the left hand side is not, unless we impose by assumption that kind of symmetry, see below for more detail.
Nevertheless, let's forget it for a moment and focus on the purpose of the paragraph.
Basically, the paragraph wants to explain why Quantum Field Theory is the way it is, and not otherwise, in particular why do we have fields as operators and not anymore position $\vec x$, which has been underemployed as a simple parameter, together with time.
Indeed, Quantum Mechanics is a nonrelativistic theory, meaning that time and position are not dealt with on the same level, for example the first is a parameter while the second is an operator.
Reunite QM with Special or General relativity would mean that every 3-vector like position $\vec X$ and momentum $\vec P$ must be promoted to a 4-vector $X^\mu$ and $P^\mu$, and the commutation relations (which are the core of QM) must be extended as well.
The most natural extension would then be to take spatial momentum $\vec P$ and the Hamiltonian $H$ and build the 4-vector operator $(H,\vec P)$, and similarly for the position, to take time $t$, promoting it to an operator $T$, and build a four vector $(T,\vec X)$.
The next step is to generalize the commutation relations:
$$[X^\mu,P^\nu]=?$$
As $X^\mu$ and $P^\mu$ are 4-vectors, i.e. $(1,0)$ contravariant tensors, their product $X^\mu P^\mu$ will be a $(2,0)$ contravariant tensor. If, as we said at the beginning, we assume that the amount of noncommutativity which exists between some direction $X^\mu$ and a momentum direction $P^\nu$, has no physical reason to be expected different from the amount of noncommutativity between some direction $X^\nu$ and a momentum direction $P^\mu$, then we conclude that $[X^\mu,P^\nu]$ must be a symmetric entity, so on the right hand side of $[X^\mu,P^\nu]$ there must lie some symmetric $(2,0)$ tensor.
In a generic spacetime which is endowed with a metric $g_{\mu\nu}$ and its beloved inverse $g^{\mu\nu}$, the best candidate to represent our commutation relations is exactly $g^{\mu\nu}$:
$$[X^\mu,P^\nu]=c\cdot g^{\mu\nu}\mathbb I$$
where $\mathbb I$ is the identity operator in the Hilbert space where the operators act.
The constant $c$ is determined considering that anyway $[X^\mu,P^\nu]$ is anti-Hermitian, and in case of flat spacetime the spatial commutation relations should yield the canonical $[X^i,P^j]=i\hbar\delta^{ij}$, so finally
$$[X^\mu,P^\nu]=i\hbar \cdot g^{\mu\nu}\mathbb I$$
This fact could be seen in a clearer way, without the need of handwavy arguments, if we consider the definition of $P^\mu$ as the generator of infinitesimal translations:
$$x^\mu\rightarrow x^\mu+dx^\mu$$
in terms of the position eigenbasis $|x^\mu\rangle$, we postulate the existence of an operator $\mathcal J(dx^\mu)$ such that:
$$\mathcal J(dx^\mu)|x^\mu\rangle=|x^\mu+dx^\mu\rangle$$
Following closely the same steps that you can find, for example, in Sakurai, and that lead to $[X^i,P^j]=i\hbar \delta^{ij}$, you will define an operator infinitesimally close to the identity, keeping only its first order in $dx^\mu$:
$$\mathcal J(dx^\mu)=\mathbb I-G(dx^\mu) +o(dx^\mu{}^2)$$
As $\mathcal J(dx^\mu)$ has not any free index, it is to be considered a scalar, so the operator $G(dx^\mu)$ must involve a combination of $P^\mu$ and $dx^\mu$ of this kind, i.e. a scalar product:
$$G(dx^\mu)=\frac{i P^\mu g_{\mu\nu}dx^\nu}{\hbar}$$
from which the generalization involving the (inverse of the) metric tensor.
Now, everything I said so far is completely useless, because the remainder of the paragraph explains that defining a $T$ operator violates a fundamental physical principle: the boundedness from below of the energy (and its quantization).
So, this way has to be discarded, and this is the main reason why in QFT, instead of having a 4-vector position operator, we have 4 parameters $(\vec x,t)$, i.e. $\vec x$ has been downgraded from operator to c-number.
Hope this helps


TITLE: Prove R/J is noetherian
QUESTION [0 upvotes]: I have a noetherian ring $R$ and $J\triangleleft R$. Then $R/J$ is noetherian?

I suppose $I\triangleleft R/J$ and $\tilde{I}=\{r\in R:r+J \in I\}$.
$\tilde{I}\triangleleft R$, $R$ is noetherian so exist $a_1,....,a_n \in \tilde{I}: \tilde{I}=Ra_1+\cdots+Ra_n$
Can I say $I=\tilde{I}/J$?

REPLY [0 votes]: All ascending chains of ideals in $R/J$ correspond to ascending chains of ideals in $R$ and so must stop.


text:
\begin{document}

\title{\textbf{On the some properties of circulant matrix with third order
linear recurrent sequence}}
\author{Arzu Coskun and Necati Taskara\thanks{
e mail: \ \textit{arzucoskun58@gmail.com, ntaskara@selcuk.edu.tr}} \\

Department of Mathematics, Science Faculty, \\
Selcuk University, Campus, 42075, Konya, Turkey}
\maketitle

\begin{abstract}
In this paper, firstly, we give the some fundamental properties of Van Der
Laan numbers. After, we define the circulant matrices $C(Z)$ which entries
is third order linear recurrent sequence. In addition, we compute
eigenvalues, spectral norm and determinant of this matrix. Consequently, by
using properties of this sequence, we obtain the eigenvalues, norms and
determinants of circulant matrices with Cordonnier, Perrin and Van Der Laan
numbers.

\textit{Keywords}: Third order linear recurrent sequence, Cordonnier
numbers, Perrin numbers, Van Der Laan numbers, circulant matrix, norm,
determinant.
\end{abstract}

\section{Introduction}

Shannon, Horadam and Anderson \cite{4} defined respectively Cordonnier
sequence $\left\{ P_{n}\right\} _{n\in

\mathbb{N}

}$\ as
\begin{equation}
P_{n+3}=P_{n+1}+P_{n},\ \ \ \ P_{0}=1,~P_{1}=1,\ P_{2}=1,  \label{1}
\end{equation}
Perrin sequence $\left\{ Q_{n}\right\} _{n\in

\mathbb{N}

}$\ as

\begin{equation}
Q_{n+3}=Q_{n+1}+Q_{n},\ \ \ \ Q_{0}=3,~Q_{1}=0,\ Q_{2}=2,  \label{2}
\end{equation}
and Van Der Laan sequence $\left\{ R_{n}\right\} _{n\in

\mathbb{N}

}$\ as

\begin{equation}
R_{n+3}=R_{n+1}+R_{n},\ \ \ \ R_{0}=0,~R_{1}=1,\ R_{2}=0.  \label{3}
\end{equation}

The fact that the Cordonnier numbers and Perrin numbers are a linear
combination of $\alpha ^{n}$, $\beta ^{n}$ and $\gamma ^{n}$, that is,
\begin{equation}
P_{n}=\frac{\alpha ^{n+4}}{\left( \alpha -\beta \right) \left( \alpha
-\gamma \right) }+\frac{\beta ^{n+4}}{\left( \beta -\alpha \right) \left(
\beta -\gamma \right) }+\frac{\gamma ^{n+4}}{\left( \gamma -\alpha \right)
\left( \gamma -\beta \right) },  \label{1.2}
\end{equation}

\begin{equation}
Q_{n}=\alpha ^{n}+\beta ^{n}+\gamma ^{n}.  \label{2.2}
\end{equation}

Hence, the relations are hold
\begin{equation}
\alpha +\beta +\gamma =0,\;\alpha \beta \gamma =1,\ \alpha \beta +\beta
\gamma +\alpha \gamma =-1,  \label{1.1}
\end{equation}
where $\alpha $, $\beta $ and $\gamma $ are roots of equations (\ref{1}), (
\ref{2}), (\ref{3}).\qquad

Elia \cite{12}, gave third order linear recurrent sequence$\left\{
T_{0},T_{1},T_{2},...\right\} $ defined by the recurrence

\begin{equation}
T_{n+3}=pT_{n+2}+qT_{n+1}+rT_{n},\ \ \ \ T_{0}=a,~T_{1}=b,\ T_{2}=c.
\label{5}
\end{equation}
Also, he studied Tribonacci cubic form and solve the integer representation
problem for the Tribonacci cubic form.

Recently, it has been widely studied the some properties of the circulant
matrices with special numbers. For instance, in \cite{6}, they defined the
generalized $k$-Horadam numbers and computed the spectral norm, eigenvalues
and the determinant of circulant matrix with this numbers. Solak \cite{7}
defined \ the $n\times n$ circulant matrices $A=\left[ a_{ij}\right] $ and $
B=\left[ b_{ij}\right] $, where $a_{ij}\equiv F_{\left( \func{mod}\left(
j-i,n\right) \right) }$ and $b_{ij}\equiv L_{\left( \func{mod}\left(
j-i,n\right) \right) }$. Additionally, he investigated the upper and lower
bounds of the matrices $A$ and $B$, respectively. In \cite{5}, Ipek obtained
the spectral norms of circulant matrices $A=\left[ a_{ij}\right] $ and $B=
\left[ b_{ij}\right] $, where $a_{ij}\equiv F_{\left( \func{mod}\left(
j-i,n\right) \right) }$ and $b_{ij}\equiv L_{\left( \func{mod}\left(
j-i,n\right) \right) }$. Shen and Cen, in \cite{8,11}, have found upper and
lower bounds for the spectral norms of $r$-circulant matrices and obtained
some bounds for the spectral norms of Kronecker and Hadamard products of
these matrices. Also, they gave the determinants and inverses of circulant
matrices with Fibonacci and Lucas numbers. In \cite{3}, it has been studied
the norms, eigenvalues and determinants of some matrices related to
different numbers. Yazlik \cite{9} obtained upper and lower bounds for the
spectral norm of an $r$-circulant matrices $H=C_{r}\left(
H_{k,0},H_{k,1},H_{k,2},...,H_{k,n-1}\right) $ whose entries are the
generalized $k$-Horadam numbers. Additionally, he find new formulas to
calculate the eigenvalues and determinant of the matrix $H$. In \cite{10},
they defined the circulant matrices with Jacobsthal and Jacobsthal-Lucas
numbers and computed the determinants and inverses of these matrices.

In the light of the above studies, in here, we present some properties of
the Van Der Laan sequence as Binet formula, sum. After, we find eigenvalues,
spectral norm and determinant of circulant matrix with the third order
sequence. Finally, we give eigenvalues, norms and determinants of circulant
matrices with Cordonnier, Perrin and Van Der Laan numbers via properties of
circulant matrix with this third order sequence.

Now, we give some preliminaries about circulant matrix and the spectral norm
of a matrix.

The circulant matrix $C=\left[ c_{ij}\right] \in M_{n,n}\left(

\mathbb{C}

\right) $ is defined by the form
\begin{equation*}
c_{ij}=\{
\begin{array}{c}
c_{j-i},\ \ \ \ \ \ \ j\geq i \\
c_{n+j-i},\ \ \ j<i
\end{array}
\end{equation*}

The spectral norm of $A$, for a matrix $A=\left[ a_{i,j}\right] \in M_{m,n}(

\mathbb{C}

)$, is given by
\begin{equation*}
\left\Vert A\right\Vert _{2}=\sqrt{\max\limits_{1\leq i\leq n}\lambda
_{i}\left( A^{\ast }A\right) },
\end{equation*}
where $A^{\ast }$ is the conjugate transpose of matrix $A$.

\begin{lemma}
\bigskip \lbrack 2] Let $A=circ(a_{0},a_{1},\cdots ,a_{n-1})$ be a $n\times
n $ circulant matrix. Then we have
\begin{equation*}
\lambda _{j}\left( A\right) =\sum_{k=0}^{n-1}a_{k}w^{-jk},
\end{equation*}
where $w=e^{\frac{2\pi i}{n}},\ i=\sqrt{-1},\ j=0,1,\ldots n-1.$
\end{lemma}

\begin{lemma}
\lbrack 1] Let $A$ be an $n\times n$ matrix with eigenvalues $\lambda
_{1},\lambda _{2},\ldots ,\lambda _{n}.$ Then, $A$ is a normal matrix if and
only if the eigenvalues of $AA^{\ast }\ $are $\left\vert \lambda
_{1}\right\vert ^{2},\left\vert \lambda _{2}\right\vert ^{2},\ldots
,\left\vert \lambda _{n}\right\vert ^{2},$ where $A^{\ast }$ is the
conjugate transpose of the matrix $A$.
\end{lemma}

\section{Circulant Matrices with Third Order Linear Sequences}

Firstly, since some of results of this paper concern about the spectral
norm, eigenvalues and determinant of the circulant matrix entried by the Van
Der Laan numbers, we need to introduce some properties of this sequence.

The Binet formula of Van Der Laan sequence, for every $n\in

\mathbb{N}

$, is obtained as
\begin{equation}
R_{n}=\frac{\alpha ^{n}}{\left( \alpha -\beta \right) \left( \alpha -\gamma
\right) }+\frac{\beta ^{n}}{\left( \beta -\alpha \right) \left( \beta
-\gamma \right) }+\frac{\gamma ^{n}}{\left( \gamma -\alpha \right) \left(
\gamma -\beta \right) },  \label{3.2}
\end{equation}
\ \ \ \ \qquad\ where $\alpha $, $\beta $ and $\gamma $ are roots of
equation (\ref{3}).

Also, for $n\geq 1$, we have
\begin{equation*}
\overset{n}{\underset{k=0}{\dsum }}R_{k}=R_{n+5}-1,
\end{equation*}
\begin{equation*}
\overset{n}{\underset{k=0}{\dsum }}
R_{k}^{2}=R_{n+2}^{2}-R_{n-1}^{2}-R_{n-3}^{2}+1.
\end{equation*}

If we take $p=0,q=1,r=1$ in (\ref{5}), we obtain as

\begin{equation}
Z_{n+3}=Z_{n+1}+Z_{n},  \label{4}
\end{equation}
where initial conditions $Z_{0}=a,~Z_{1}=b,\ Z_{2}=c$ . Also, the Binet
formula of $Z_{n}$ sequence, for every $n\in

\mathbb{N}

$, is obtained as

\begin{equation}
Z_{n}=\frac{\left( \alpha ^{2}-1\right) a+\alpha b+c}{\left( \alpha -\beta
\right) \left( \alpha -\gamma \right) }\alpha ^{n}+\frac{\left( \beta
^{2}-1\right) a+\beta b+c}{\left( \beta -\alpha \right) \left( \beta -\gamma
\right) }\beta ^{n}+\frac{\left( \gamma ^{2}-1\right) a+\gamma b+c}{\left(
\gamma -\alpha \right) \left( \gamma -\beta \right) }\gamma ^{n},
\label{4.2}
\end{equation}
where $\alpha $, $\beta $ and $\gamma $ are roots of equation $x^{3}-x-1=0.$

\begin{proposition}
For $n\geq 1$, the following relations are hold:
\begin{equation*}
\overset{n}{\underset{k=0}{\dsum }}Z_{k}=Z_{n+5}-Z_{4},
\end{equation*}
\begin{equation*}
\overset{n}{\underset{k=0}{\dsum }}
Z_{k}^{2}=Z_{n+2}^{2}-Z_{n-1}^{2}-Z_{n-3}^{2}+T
\end{equation*}
where $T=2a\left( a-c\right) -\left( b+c\right) ^{2}$.
\end{proposition}

We formulate eigenvalues, spectral norms and determinants of the circulant
matrices with the third order linear sequences. In order to do that, we can
define \textit{the circulant matrix} as follows.

\begin{definition}
$n\times n$ circulant matrix with third order linear sequence entries is
defined by
\begin{equation}
C(Z)=\left[
\begin{array}{ccccc}
Z_{0} & Z_{1} & Z_{2} & \cdots & Z_{n-1} \\
Z_{n-1} & Z_{0} & Z_{1} & \cdots & Z_{n-2} \\
Z_{n-2} & Z_{n-1} & Z_{0} & \cdots & Z_{n-3} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
Z_{1} & Z_{2} & Z_{3} & \cdots & Z_{0}
\end{array}
\right] .  \label{1.3}
\end{equation}
\end{definition}

The following theorem gives us the eigenvalues of circulant matrix with
third order linear recurrent sequence.

\begin{theorem}
Let $C(Z)=circ(Z_{0},Z_{1},...,Z_{n-1})$ be circulant matrix. Then the
eigenvalues of $C(Z)\ $are
\begin{equation*}
\lambda _{j}\left( C(Z)\right) =\frac{Z_{n}-a+\left( Z_{n+1}-b\right)
w^{-j}+\left( Z_{n-1}-c+a\right) w^{-2j}}{w^{-3j}+w^{-2j}-1},
\end{equation*}
where $w=e^{\frac{2\pi i}{n}},\ i=\sqrt{-1},\ j=0,1,\ldots ,n-1$.
\end{theorem}

\begin{proof}
By taking $X=\left( \alpha -\beta \right) \left( \alpha -\gamma \right) $,$\
Y=\left( \beta -\alpha \right) \left( \beta -\gamma \right) $,$\ W=\left(
\gamma -\alpha \right) \left( \gamma -\beta \right) ,\ $from Lemma 1, we have
\begin{eqnarray*}
\lambda _{j}\left( C(Z)\right) &=&\sum_{k=0}^{n-1}Z_{k}w^{-jk} \\
&=&\sum_{k=0}^{n-1}\left(
\begin{array}{c}
\frac{\left( \alpha ^{2}-1\right) a+\alpha b+c}{X}\alpha ^{k}+\frac{\left(
\beta ^{2}-1\right) a+\beta b+c}{Y}\beta ^{k} \\
+\frac{\left( \gamma ^{2}-1\right) a+\gamma b+c}{W}\gamma ^{k}
\end{array}
\right) w^{-jk} \\
&=&
\begin{array}{c}
\frac{\left( \alpha ^{2}-1\right) a+\alpha b+c}{X}\left( \frac{\left( \alpha
w^{-j}\right) ^{n}-1}{\alpha w^{-j}-1}\right) +\frac{\left( \beta
^{2}-1\right) a+\beta b+c}{Y}\left( \frac{\left( \beta w^{-j}\right) ^{n}-1}{
\beta w^{-j}-1}\right) \\
+\frac{\left( \gamma ^{2}-1\right) a+\gamma b+c}{W}\left( \frac{\left(
\gamma w^{-j}\right) ^{n}-1}{\gamma w^{-j}-1}\right)
\end{array}
\\
&=&
\begin{array}{c}
\frac{\left( \alpha ^{n}-1\right) \left[ \left( \alpha ^{2}-1\right)
a+\alpha b+c\right] }{X\left( \alpha w^{-j}-1\right) } \\
+\frac{\left( \beta ^{n}-1\right) \left[ \left( \beta ^{2}-1\right) a+\beta
b+c\right] }{Y\left( \beta w^{-j}-1\right) }+\frac{\left( \gamma
^{n}-1\right) \left[ \left( \gamma ^{2}-1\right) a+\gamma b+c\right] }{
W\left( \gamma w^{-j}-1\right) }
\end{array}
.
\end{eqnarray*}
By considering (\ref{1.1}) and (\ref{4.2}), we obtain
\begin{equation*}
\lambda _{j}\left( C(Z)\right) =\frac{Z_{n}-a+\left( Z_{n+1}-b\right)
w^{-j}+\left( Z_{n-1}-c+a\right) w^{-2j}}{w^{-3j}+w^{-2j}-1}
\end{equation*}
which is desired.
\end{proof}

By using Theorem 5, we obtain eigenvalues of circulant matrices with
Cordonnier, Perrin and Van Der Laan sequences as follows.

\begin{corollary}
$i)$ Let $C(P)=circ(P_{0},P_{1},...,P_{n-1})$ be circulant matrix. If we
take $a=b=c=1$ in Theorem 5, then the eigenvalues of $C(P)\ $are
\begin{equation*}
\lambda _{j}\left( C(P)\right) =\frac{P_{n}-1+\left( P_{n+1}-1\right)
w^{-j}+P_{n-1}w^{-2j}}{w^{-3j}+w^{-2j}-1},
\end{equation*}
where $P_{n}$ is the $n$th Cordonnier number and $w=e^{\frac{2\pi i}{n}},\ i=
\sqrt{-1},\ j=0,1,\ldots ,n-1$.
\end{corollary}

$ii)$Let $C(Q)$ be circulant matrix with Perrin numbers. If we take $
a=3,b=0,c=2$, the eigenvalues of $C(Q)\ $are
\begin{equation*}
\lambda _{j}\left( C(Q)\right) =\frac{Q_{n}-3+Q_{n+1}w^{-j}+\left(
Q_{n-1}+1\right) w^{-2j}}{w^{-3j}+w^{-2j}-1},
\end{equation*}
where $Q_{n}$ is the $n$th Perrin number and $w=e^{\frac{2\pi i}{n}},\ i=
\sqrt{-1},\ j=0,1,\ldots ,n-1$.

$iii)$Let $C(R)=circ(R_{0},R_{1},...,R_{n-1})$ be circulant matrix. If we
take $a=c=0,b=1$ in theorem 5 the eigenvalues of $C(R)\ $are
\begin{equation*}
\lambda _{j}\left( C(R)\right) =\frac{R_{n}+\left( R_{n+1}-1\right)
w^{-j}+R_{n-1}w^{-2j}}{w^{-3j}+w^{-2j}-1},
\end{equation*}
where $R_{n}$ is the $n$th Van Der Laan number and $w=e^{\frac{2\pi i}{n}},\
i=\sqrt{-1},\ j=0,1,\ldots ,n-1.$

Because matrix $C(Z)$ is normal matrix, we can write $\lambda _{j}\left(
C(Z)C(Z)^{\ast }\right) =\left\vert \lambda _{j}\left( C(Z)\right)
\right\vert ^{2}$. Also, since the matrices $C(P)$, $C(Q)$ and $C(R)$ are
normal matrices, we can write $\lambda _{j}\left( C(P)C(P)^{\ast }\right)
=\left\vert \lambda _{j}\left( C(P)\right) \right\vert ^{2}$, $\lambda
_{j}\left( C(Q)C(Q)^{\ast }\right) =\left\vert \lambda _{j}\left(
C(Q)\right) \right\vert ^{2}$ and $\lambda _{j}\left( C(R)C(R)^{\ast
}\right) =\left\vert \lambda _{j}\left( C(R)\right) \right\vert ^{2}$. Then,
we have the following theorem and corollary that deal with spectral norm.

\begin{theorem}
Let $C(Z)$ be an $n\times n$ circulant matrix with the $Z_{n}$ entries.
Then, we have
\end{theorem}

\begin{equation*}
\left\Vert C(Z)\right\Vert _{2}=Z_{n+4}-Z_{4}.
\end{equation*}

\begin{proof}
From Lemma 2, we can write
\begin{equation*}
\left\Vert C(Z)\right\Vert _{2}=\sqrt{\left( \max\limits_{0\leq j\leq
n-1}\left\vert \lambda _{j}\left( C(Z)\right) \right\vert ^{2}\right) }.
\end{equation*}
In this last equality, for $j=0$, $\lambda _{0}$ \ becomes \ the maximum
eigenvalue. Thus, $\left\Vert C(Z)\right\Vert _{2}=\left\vert \lambda
_{0}\left( C(Z)\right) \right\vert $. Also, from the Theorem 5, we clearly
obtain
\begin{equation*}
\left\Vert C(Z)\right\Vert _{2}=Z_{n+4}-Z_{4}.
\end{equation*}
Hence proof is completed.
\end{proof}

For special cases of $a,b,c$ in Theorem 7, we obtain following corollary for
spectral norms of $C(P),C(Q)$ and $C(R)$.

\begin{corollary}
$i)$ Let $C(P)$ be an $n\times n$ circulant matrix with the Cordonnier
numbers entries.  If we take $a=b=c=1$ in (\ref{4}),we obtain Cordonnier
sequence. And so, from Theorem 7, we obtain
\begin{equation*}
\left\Vert C(P)\right\Vert _{2}=P_{n+4}-2
\end{equation*}
where $P_{n}$ is $n$th Cordonnier number.
\end{corollary}

$ii)$ Similarly to $i)$, if we take $a=3,b=0,c=2$ in (\ref{4}), we obtain
for $C(Q)$ matrix
\begin{equation*}
\left\Vert C(Q)\right\Vert _{2}=Q_{n+4}-2.
\end{equation*}
where $Q_{n}$ is $n$th Perrin number.

$iii)$ Similarly,  if we take $a=c=0,b=1$ in (\ref{4}), we have
\begin{equation*}
\left\Vert C(R)\right\Vert _{2}=R_{n+4}-1.
\end{equation*}
where $R_{n}$ is $n$th Van Der Laan number.

Let $A=\left[ a_{ij}\right] $ be an $n\times n$ matrix. Since row norm and
column norm of the $A$ matrix are $\left\Vert A\right\Vert _{\infty }=
\underset{1\leq i\leq n}{\max }\left( \underset{j=1}{\overset{n}{\dsum }}
\left\vert a_{ij}\right\vert \right) $ and $\left\Vert A\right\Vert _{1}=
\underset{1\leq j\leq n}{\max }\left( \underset{i=1}{\overset{n}{\dsum }}
\left\vert a_{ij}\right\vert \right) $, it can be easily seen that $
\left\Vert C(Z)\right\Vert _{2}=\left\Vert C(Z)\right\Vert _{1}=\left\Vert
C(Z)\right\Vert _{\infty }.$ Also, $\left\Vert C(P)\right\Vert
_{2}=\left\Vert C(P)\right\Vert _{1}=\left\Vert C(P)\right\Vert _{\infty },$
$\left\Vert C(Q)\right\Vert _{2}=\left\Vert C(Q)\right\Vert _{1}=\left\Vert
C(Q)\right\Vert _{\infty }$ and $\left\Vert C(R)\right\Vert _{2}=\left\Vert
C(R)\right\Vert _{1}=\left\Vert C(R)\right\Vert _{\infty }.$

The following theorem gives us the determinant of $C(Z)$.

\begin{theorem}
The determinant \ of \ the matrix $C(Z)=circ\left( Z_{1},Z_{2},\ldots
,Z_{n}\right) $ is written by
\begin{equation*}
\det (C(Z))=\frac{\left( Z_{n}-a\right) ^{n}\left(
1-K^{n}-L^{n}+K^{n}L^{n}\right) }{\left( -1\right) ^{n}\left(
Q_{-n}-Q_{n}\right) },
\end{equation*}
where $K=\dfrac{b-Z_{n+1}-\sqrt{\left( Z_{n+1}-b\right) ^{2}-4\left(
Z_{n}-a\right) \left( Z_{n-1}-c+a\right) }}{2\left( Z_{n}-a\right) }$,
\end{theorem}

$L=\dfrac{b-Z_{n+1}+\sqrt{\left( Z_{n+1}-b\right) ^{2}-4\left(
Z_{n}-a\right) \left( Z_{n-1}-c+a\right) }}{2\left( Z_{n}-a\right) }$.

\begin{proof}
From Theorem 5, we have
\begin{eqnarray*}
\det (C(Z)) &=&\prod\limits_{j=0}^{n-1}\lambda _{j}\left( C(Z)\right) \\
&=&\prod\limits_{j=0}^{n-1}\frac{Z_{n}-a+\left( Z_{n+1}-b\right)
w^{-j}+\left( Z_{n-1}-c+a\right) w^{-2j}}{w^{-3j}+w^{-2j}-1}.
\end{eqnarray*}
By considering the equality
\begin{equation*}
\prod\limits_{k=0}^{n-1}\left( x-yw^{-k}+zw^{-2k}\right) =x^{n}\left(
1-\left( \frac{y-\sqrt{y^{2}-4xz}}{2x}\right) ^{n}-\left( \frac{y+\sqrt{
y^{2}-4xz}}{2x}\right) ^{n}+\left( \frac{z}{x}\right) ^{n}\right) ,
\end{equation*}
\ we have
\begin{equation*}
\det (C(Z))=\frac{\left( Z_{n}-a\right) ^{n}\left(
1-K^{n}-L^{n}+K^{n}L^{n}\right) }{\left( -1\right) ^{n}\left(
Q_{-n}-Q_{n}\right) }.
\end{equation*}
Hence proof is completed.
\end{proof}

By using in Theorem 9, we obtain determinants of circulant matrices with
Cordonnier, Perrin and Van Der Laan sequences as follows.

\begin{corollary}
$i)$ Let $a=b=c=1$ in Theorem 9. Then, the determinant \ of \ the matrix $
C(P)=circ\left( P_{1},P_{2},\ldots ,P_{n}\right) $ is written by
\begin{equation*}
\det (C(P))=\frac{\left( P_{n}-1\right) ^{n}\left(
1-K^{n}-L^{n}+K^{n}L^{n}\right) }{\left( -1\right) ^{n}\left(
Q_{-n}-Q_{n}\right) },
\end{equation*}
where $K=\dfrac{1-P_{n+1}-\sqrt{\left( 1-P_{n+1}\right)
^{2}-4P_{n}P_{n-1}+4P_{n-1}}}{2\left( P_{n}-1\right) }$,
\end{corollary}

$L=\dfrac{1-P_{n+1}+\sqrt{\left( 1-P_{n+1}\right) ^{2}-4P_{n}P_{n-1}+4P_{n-1}
}}{2\left( P_{n}-1\right) }$.

$ii)$ If we take $a=3,b=0,c=2$ in Theorem 9. Then, the determinant \ of \
the matrix $C(Q)=circ\left( Q_{1},Q_{2},\ldots ,Q_{n}\right) $ is written by
\begin{equation*}
\det (C(Q))=\frac{\left( Q_{n}-3\right) ^{n}\left(
1-K^{n}-L^{n}+K^{n}L^{n}\right) }{\left( -1\right) ^{n}\left(
Q_{-n}-Q_{n}\right) },
\end{equation*}
where $K=\dfrac{-Q_{n+1}-\sqrt{Q_{n+1}^{2}-4Q_{n}Q_{n-1}-4Q_{n}+12Q_{n-1}+12}
}{2\left( Q_{n}-3\right) }$ and

$L=\dfrac{-Q_{n+1}+\sqrt{Q_{n+1}^{2}-4Q_{n}Q_{n-1}-4Q_{n}+12Q_{n-1}+12}}{
2\left( Q_{n}-3\right) }$.

$iii)$Let $a=c=0,b=1$ in Theorem 9. Then, the determinant \ of \ the matrix $
C(R)=circ\left( R_{1},R_{2},\ldots ,R_{n}\right) $ is written by
\begin{equation*}
\det (C(R))=\frac{R_{n}^{n}\left( 1-K^{n}-L^{n}+K^{n}L^{n}\right) }{\left(
-1\right) ^{n}\left( Q_{-n}-Q_{n}\right) },
\end{equation*}
where $K=\dfrac{1-R_{n+1}-\sqrt{\left( 1-R_{n+1}\right) ^{2}-4R_{n}R_{n-1}}}{
2R_{n}}$,

$L=\dfrac{1-R_{n+1}+\sqrt{\left( 1-R_{n+1}\right) ^{2}-4R_{n}R_{n-1}}}{2R_{n}
}$.

config: arxiv


TITLE: uniform integrability of all conditional expectations of a fixed $L^1$ function
QUESTION [2 upvotes]: Let $Z$ be a real $L^1$ random variable on a probability space $(\Omega, \mathscr{A}.\mu).$ Why is the family of all $E[Z|  \mathscr{B}]$ uniformly integrable when $ \mathscr{B}$ ranges over the sub-sigma-algebras of $ \mathscr{A}$?  This is supposed to be trivial but even assuming $Z\geq 0$ i don't know how to control $\int_A  E[Z|  \mathscr{B}]d\mu$ when $A\notin \mathscr{B}$.

REPLY [3 votes]: Recall that because $Z$ is $L^1$, for any $\epsilon > 0$, there is a $\delta > 0$ such that $E[|Z|;A] < \epsilon$ whenever $\mu(A) < \delta$. Also we use an equivalent definition of uniform integrability that a family $\{X_i\}$ of r.v.s is u.i. if $\lim_{M\to\infty}\sup_iE[|X_i|;|X_i|>M] = 0$.
Now, for any sub-$\sigma$-algebra $\mathscr B$ of $\mathscr A$, let $X$ be a version of $E[Z|\mathscr B]$ and $Y$ be a version of $E[|Z|\mid\mathscr B]$. Then,
\begin{align*}
E[|X| ; |X|>M] &\le E[Y;Y>M]\\
&= E[|Z|; Y > M].
\end{align*}
Now, $\mu(Y > M) \le \frac{1}{M}E[Y] = \frac{1}{M}E[|Z|]$. If we choose $M$ so large that $E[|Z|] < M\delta$, then $\mu(Y>M) < \delta$, so $E[|X| ; |X|>M] < \epsilon$.


TITLE: Estimating $\sum_{p_1\cdots p_k\leq n} \frac{1}{p_1\cdots p_k}$ for various $k$
QUESTION [3 upvotes]: Let $$D_2(n) =\sum_{pq\leq n}  1,$$
and
$$F_2(n) =\sum_{pq\leq n}  \frac{1}{pq}$$
where $p,q$ are primes. Similarly define
$$D_k(n) =\sum_{p_1\cdots p_k\leq n}  1,$$
and
$$F_k(n) =\sum_{p_1\cdots p_k\leq n}  \frac{1}{p_1\cdots p_k}.$$
I believe my calculations showing
$$F_2(n)\sim \frac{(\log\log n)^2}{2}$$
and
$$D_2(n)\sim \frac{n \log\log n}{\log n}$$
are correct (I used PNT and Mertens theorems).
What about larger $k$?
Is it possible to allow $k$ to grow very slowly and still get an estimate?

REPLY [1 votes]: Define the function $F^*(n)=\underset{p_1\dots p_k\leq n}{\sum_{p_, \ldots, p_k}}\frac{1}{p_1\dots p_k}$. Then we have for each fixed $k$ the asymptotics $F_k^*(n)\sim(\log\log n)^k$. To see this note the obvious upper and lower bounds
$$
\left(\sum_{p\leq n^{1/k}}\frac{1}{p}\right)^k\leq F_k(n)\leq \left(\sum_{p\leq n}\frac{1}{p}\right)^k,
$$
and use the asymptotics $\sum_{p\leq n}\frac{1}{p}\sim\log\log n$.
To deduce an asymptotics for $F_k(n)$, note first that summands in $F_k^*$ corresponding to tuples which are not pairwise different are negligible, in fact, using a variation of the upper bound above we see that this contribution is $\mathcal{O}((\log\log n)^{k-1})$. If the primes are pairwise different, then the corresponding summand occurs $k!$-times in $F_k^*$ and only once in $F_k$, thus $F_k(n)\sim\frac{1}{k!}(\log\log n)^k$, confirming your computation.
As $k$ grows, we use the bound $\sum_{p\leq n}\frac{1}{p}=\log\log n +\mathcal{O}(1)$, thus $F_k^*(n)=(\log\log n)^k+\mathcal{O}(k(\log\log n)^{k-1})$. Deleting tuples which contain the same prime multiple times gets more complicated. A crude estimate would be $B_k(\log\log n)^{k-1}<k^k(\log\log k)^{k-1}$, where $B_k$ denotes Bell numbers. In this way you get a uniform bound almost up to $k=\log\log\log n$. If you need better estimates, you would have to consider partitions more carefully, which is doable, but some work.


TITLE: Need verification for the entropy equation in statistical thermodynamics
QUESTION [1 upvotes]: The relationship between entropy $S$, the total number of particles $N$, the available energy levels $E_j$ and a yet to be defined parameter $\beta$ is:
$$S(\beta)=k_B \cdot N \cdot \ln\bigg(\sum_{j=1}^n[e^{\beta E_j}]\bigg) - k_B\cdot \beta \sum_{j=1}^n\bigg[\frac{N}{\sum_{i=1}^n [e^{\beta E_i}]}\cdot e^{\beta E_j}\cdot E_j\bigg]$$
The partition function $Z$ and the total energy $U$ as functions of $\beta$ are:
$$\sum_{j=1}^n[e^{\beta E_j}] = Z(\beta) \tag{1}$$
$$\sum_{j=1}^n\bigg[\frac{N}{\sum_{i=1}^n [e^{\beta E_i}]}\cdot e^{\beta E_j}\cdot E_j\bigg]=U(\beta) \tag{2}$$
I need to first derive $\frac{dS}{d\beta}$ and then substitute parameters within $\frac{dS}{d\beta}$ with $\frac{dZ}{d\beta}$ and $\frac{dU}{d\beta}$ if there are any. I came up with the following:
$$\frac{dS}{d\beta}=k_B\cdot \beta \cdot \bigg(\frac{d\big(\ln(Z)\big)}{d\beta}\cdot U - \frac{dU}{d\beta}\bigg)$$
However, I am not sure if this derivation is correct and would really appreciate a verification on this.

REPLY [1 votes]: $$S(\beta) = k_B \cdot N \cdot \ln(Z(\beta)) - k_B\cdot \beta \cdot U(\beta).$$
$$U(\beta) = \sum_{j=1}^n\bigg[\frac{N}{\sum_{i=1}^n [e^{\beta E_i}]}\cdot e^{\beta E_j}\cdot E_j\bigg] = \frac{N}{\sum_{i=1}^n [e^{\beta E_i}]}\,\sum_{j=0}^{n} e^{\beta\,E_j}\cdot E_j = \frac{N}{Z(\beta)} \sum_{j=0}^{n} e^{\beta\,E_j}\cdot E_j$$
$$\frac{dZ}{d\beta} = \sum_{j=0}^{n} e^{\beta\,E_j}\cdot E_j = \frac{U(\beta)\cdot Z(\beta)}{N}$$
$$\frac{dS}{d\beta}=k_B\cdot \left( N\cdot\frac{d\left(\ln(Z)\right)}{d\beta} - U - \beta \cdot \frac{dU}{d\beta}\right)$$
$$\frac{dS}{d\beta}=k_B\cdot \left( \frac{N}{Z}\cdot\frac{dZ}{d\beta}-U-\beta\cdot\frac{dU}{d\beta} \right)$$
$$\frac{dS}{d\beta}= k_B\cdot \left( \frac{N}{Z}\cdot\frac{U\cdot Z}{N}-U-\beta\cdot\frac{dU}{d\beta} \right) = -\beta\cdot k_B\cdot \frac{dU}{d\beta}$$
Your first term vanished somewhere in my solution. I cannot tell where either of us did a mistake without comparing solutions.

Part II
Counting in degeneracy:
$$S=k_B\bigg(N \cdot \ln(Z(\beta)) - \beta U(\beta) + \ln(\frac{N}{Z(\beta)}) \cdot N - \frac{N^2}{Z(\beta)}\bigg)$$
which has the last two terms in addition to the non-degenerate solution. So let's just solve for these two (using the fact that $\ln(a/b)=\ln(a)-\ln(b)$):
$$\frac{d}{d\beta}\left( k_B\cdot\left[ N \cdot \ln(\frac{N}{Z(\beta)}) - \frac{N^2}{Z(\beta)} \right] \right) = k_B\cdot\left(
- N \cdot\frac{1}{Z}\frac{dZ}{d\beta}
+ \frac{N^2}{Z^2}\cdot\frac{dZ}{d\beta}
\right)$$
$$ = k_B \cdot\frac{dZ}{d\beta}\cdot\frac{N}{Z}\cdot\left( \frac{N}{Z} - 1 \right) = k_B \cdot\frac{U\cdot Z}{N}\cdot\frac{N}{Z}\cdot\left( \frac{N}{Z} - 1 \right) $$
$$ = k_B \cdot U(\beta) \cdot\left( \frac{N}{Z(\beta)} - 1 \right) $$
Therefore, overall derivative considering degeneracy is:
$$\frac{dS}{d\beta} = k_B\cdot\left( - \beta\cdot\frac{dU}{d\beta} + \frac{U(\beta)\cdot N}{Z(\beta)} - U(\beta) \right)$$
So, no, unless you have a partition function of constant value $N$, factoring in degeneracy changes the outcome.


TITLE: When (or what is the meaning of) $I \propto E^2$?
QUESTION [5 upvotes]: For a monochromatic plane wave: $$\mathbf E = \mathbf E _0e^{i(\mathbf k \cdot \mathbf r -\omega t)},\qquad \mathbf H = \dfrac{\mathbf B}{\mu _0}= \mathbf H_0e^{i(\mathbf k \cdot \mathbf r-\omega t)},$$
it's straightforward to show, using Maxwell's equations, that the Poynting vector $\mathbf S = \mathbf E \times \mathbf H$ is given (in MKS units) by$$\mathbf S = \varepsilon _0cE_0^2 \cos ^2 (\mathbf k \cdot \mathbf r - \omega t)\mathbf n,$$
where $\mathbf n = \mathbf k /k$ is the direction of propagation of the wave, so that the average energy flux in that direction is $$I= \overline {\mathbf S \cdot \mathbf n}=\lim _{T\to \infty} \intop ^T \mathbf S \cdot \mathbf n \text dt\propto E_0^2.$$
This relation is usually (compare, for example 1) assumed as valid. That is, the irradiance is defined (or assumed to be) proportional to the average of the square of the electric field (1 puts directly $I =  \overline {\mathbf E \cdot \mathbf E ^*}$) even when the field is the superposition of two plane harmonic waves.
Now, if the two waves (say $\mathbf E _0i, \mathbf k _i$ for $i=1,2$) share the same direction of propagation, I see no problem in the definition. But when the two directions are different, it's easy to see that the magnitude of the Poynting vector isn't strictly proportional to $(\mathbf E_1+\mathbf E_2)^2$. Moreover the direction of $\mathbf S$ is a quite complicated function of $\mathbf n _1 ,\mathbf n _2, \mathbf E _1, \mathbf E _2$ (in the case of two linearly polarized waves one has: $$\frac{1}{\varepsilon _0 c}\mathbf S = E_1 ^2 \mathbf n _1 + E_2 ^2 \mathbf n _2 + (\mathbf n _1 +\mathbf n _2)\mathbf E _1 \cdot \mathbf E _2 - (\mathbf n _1 \cdot \mathbf E _2 ) \mathbf E _1 - (\mathbf n _2 \cdot \mathbf E _1) \mathbf E _2.$$
As an extreme example, if one has two waves travelling in opposite direction, with $\mathbf k _2 = -\mathbf k _1$ and $\mathbf E _1 = \mathbf E _2,$ the flux of energy is clearly zero, while $I=4E^2$ according to the above definitions.
On the other hand, from the above expression, it is seen that if $\mathbf n _2 = \mathbf n _1 +\delta \mathbf n$, the difference $\mathbf S - \varepsilon _0 c E^2\mathbf n$ is $O(\delta \mathbf n).$
To sum up, what is the meaning of the relation $I \propto E^2$? In particular is it

An approximation of the exact law (see above). Or
Is it a definition? In this case, why is it a useful one?

Thank you in advance.

1 Grant R. Fowles, “Introduction to modern optics”.

REPLY [3 votes]: The author confused things in section 3.1.
In macroscopic EM theory of radiation (radiometry), technically irradiance at a given point of a plane is defined as time average of the normal component of the Poynting vector (normal to the plane):
$$
I = \overline{\mathbf S \cdot \mathbf n}
$$
As you have shown, this is function of $|\mathbf E|^2$ only in some cases, like single plane wave; it also seems reasonable for isotropic thermal radiation. For general EM field, however, irradiance cannot be expressed as function of $|\mathbf E|^2$ only.


TITLE: Anyons as particles?
QUESTION [24 upvotes]: I'm trying to understand the basics of anyons physics. I understand there is neither a Fock space they live in (because Fock is just the space of (anti-)symmetrized tensor product state, see e.g. Wikipedia), nor a (pseudo / fictitious) commutation relation for them, as discussed at this Phys.SE post. But still I've a few questions regarding their statistics:

Can we associate a creation / destruction operator to an anyon mode ? Does it make sense to talk about mode of anyons?
Is there a general occupation function like Fermi-Dirac or Bose-Einstein for fermions or bosons ? Is it model dependent, i.e. does it depend on the type of anyon ? Does it make sense to discuss number of anyons?
What is the ground state of anyons, like a Fermi sea or a Bose-Einstein condensate for fermions or bosons ? Does it make sense to talk about a ground state of a gas of anyons?

I believe this bunch of questions can all be contracted to

Does it make sense to talk of the anyons as particles ?

Because in principle a particle exists independently of the Fock space construction, isn't it ? We could still construct a space of the tensor product of non (anti-)symmetrised states.
I realised that a perhaps better approach on the question would be:

To which extend is the anyon statistic a (quantum) statistic ?

provided the two other quantum statistics I know (Bose and Fermi) provide a ground state, an occupation number, and some second-quantised version of operators.
Post-scriptum : This Phys.SE question is partially related to mine.

REPLY [1 votes]: Let me partially address the question as to whether annoys and be considered particles or not.
As you have rightful pointed out the anyonic nature of the braiding relation means that a conventional Fock representation as creation and annihilation operators acting on a vacuum is difficult to use.
One should however consider anyons as proper particles in the sense that they have (usually) a finite energy gap to the ground state out of which they are created and also has a finite lifetime.
An easy example here is the creation anyons in the quantum Hall effect at $\frac13$ filling for instance. Here pairs of anyons can be created out of the background quantum hall fluid. These excitations have fractional electric changes (this is related to them being anyons) but cost a finite energy and have a finite life time.
Anyons can also be created in the Quantum Hall system when the number of flux quantum $N_\phi$ compared to the number of electrons $N_e$ is not exactly $N_e/N_\phi=\frac13$. Essentially for each extra magnetic flux an anyonic quasi-hole is created and these are stable as they make up the ground state at that filling fraction.


TITLE: How do I expand the Hermitian conjugate in the Lagrangian?
QUESTION [0 upvotes]: I am trying to expand the following equation:
$$\mathcal{L} = \partial_\mu \phi^\dagger_1\partial^\mu \phi_1 - m^2_1 \phi_1^\dagger \phi_1 - \lambda (\phi^\dagger _1 \phi_1)^2\tag{1}$$
using the following:
$$\phi_1 \to \phi'_1 = e^{i\theta_1}\phi_1\tag{2}$$
but I am not sure how to simplify it.
My take on it is that the Hermitian conjugate transforms equation (2) into:
$$\phi_1^\dagger \to (\phi'_1)^{\dagger} = (e^{i\theta_1}\phi_1)^\dagger = \phi_1^\dagger e^{-i\theta} \tag{3} $$
and when trying to apply this to the first part of equation (1),  $\partial_\mu \phi^\dagger_1\partial^\mu \phi_1 $, I think this leads to
$$\partial_\mu \phi^\dagger_1\partial^\mu \phi_1 = e^{i\theta} \partial_\mu \phi_1 \partial^\mu \phi_1^\dagger e^{-i\theta} \tag{4}$$
I think that this could be further simplified but I am not sure how, or even if what I have done is actually correct...any help would be appreciated.

REPLY [3 votes]: It should be $\partial_\mu(\phi_1^\dagger e^{-i\theta})\partial^\mu(e^{i\theta}\phi_1)$, which by the product rule is a sum of four terms which for spacetime-constant $\theta$ simplifies to just one of them, $\partial_\mu\phi_1^\dagger\partial^\mu\phi_1$. For $\theta$ varying across spacetime, the result is more complicated, motivating gauge covariant derivatives.
Edit: as requested, I'll show how the local symmetry works. Define $D_\mu\phi_1:=\partial_\mu\phi_1+iqA_\mu\phi_1$. We don't just replace of $\phi_1$ with $\phi_1^\prime:=e^{i\theta}\phi_1$; we also replace $A_\mu$ with whatever value of $A_\mu^\prime$ obtains$$D_\mu^\prime\phi_1^\prime=e^{i\theta}D_\mu\phi_1\implies (D_\mu^\prime\phi_1^\prime)^\dagger D^{\mu\prime}\phi_1^\prime=(D_\mu\phi_1)^\dagger e^{-i\theta}e^{i\theta}D^\mu\phi_1=(D_\mu\phi_1)^\dagger D^\mu\phi_1.$$So$$\begin{align}0&=(\partial_\mu+iqA_\mu^\prime)(e^{i\theta}\phi_1)-e^{i\theta}(\partial_\mu+iqA_\mu)\phi_1\\&=\partial_\mu(e^{i\theta}\phi_1)+iq(A_\mu^\prime-A_\mu)e^{i\theta}\phi_1-e^{i\theta}\partial_\mu\phi_1\\&=iq(A_\mu^\prime-A_\mu+\tfrac1q\partial_\mu\theta)e^{i\theta}\phi_1,\\A_\mu^\prime&=A_\mu-\tfrac1q\partial_\mu\theta.\end{align}$$


text:
\begin{document}

\maketitle

\vskip.15in

The natural  sum operation for symplectic manifolds is defined
by gluing  along
codimension two submanifolds.  Specifically, let $X$ be a  symplectic
$2n$-manifold with a symplectic $(2n-2)$-submanifold  $V$.  Given a similar
pair $(Y,\overline{V})$ with a
symplectic identification $V=\overline{V}$ and a complex anti-linear
isomorphism between the normal bundles of $V$
and $\overline{V}$, we can form the symplectic sum
$Z=X\#_{V=\overline{V}}Y$.  This note announces a general formula
for computing the Gromov-Witten invariants of the  sum $Z$ in terms of relative Gromov-Witten
invariants of
$(X,V)$ and $(Y,\overline{V})$.

Section 1 is a review of the GW invariants for symplectic manifolds and the
associated
invariants, which we call $TW$ invariants, that count reducible curves.
The corresponding relative
invariants of a symplectic pair $(X,V)$ are defined in section 2.  The  sum
formula is stated in a
special case in section 3, and in general as Theorem 4.1.  The last section
presents two applications: a
short derivation of the Caporaso-Harris formula [CH], and new proof that
the rational enumerative
invariants of the rational elliptic surface are given by the ``modular
form'' (\ref{BL}).

 Related results, involving symplectic sums along contact manifolds,  are
being developed by Li and Ruan
[LR]  and by Eliashberg and Hofer.

\vskip.3in

\setcounter{equation}{0}
\section{Symplectic Invariants}
\medskip

The  moduli space of $(J,\nu)$-holomorphic
maps from  genus $g$ curves  with $n$ marked points representing a class
$A$ in the free part of
$H_2(X;\Z)$  has a compactification $\overline{\M}_{g,n}(X,A)$.  This
comes with a map
\begin{equation}
\overline{\M}_{g,n}(X,A)\ \longrightarrow \ \overline{\M}_{g,n}\ti
X^n\ti H_2^{\mbox{\scriptsize {free}}}(X;\Z),
\label{modulimap1}
\end{equation}
where the first factor is the ``stabilization'' map $st$ to the
Deligne-Mumford moduli space (defined by collapsing all unstable components
of the domain curve), the second factor records the images of the marked
points, and the last one
keeps  track of the homology class $A$ of the image.  After
perturbation (cf. [LT]), the image defines a rational homology class,
\begin{equation}
\left[\overline{\M}_{g,n}(X,A)\right] \ \in\
H_*(\overline{\M}_{g,n};\Q)\otimes H_*(X^n;\Q)\otimes
H_*(H_2^{\mbox{\scriptsize {free}}}(X;\Z);\Q)
\label{modulimap2}
\end{equation}
for each choice $g,n, A$. The last term in (\ref{modulimap2}) can be
identified with the rational group ring
$RH_2(X)$ of $H_2^{\mbox{\scriptsize{free}}}(X;\Z)$, that is, with finite
sums $\sum c_A
t_A$ over $A\in H_2^{\mbox{\scriptsize{free}}}(X;\Z)$ where $c_A\in\Q$ and
the $t_A$ are
variables satisfying  $t_At_{B}=t_{A+B}$.

 After summing on $A$ and dualizing,
(\ref{modulimap2}) defines a map
\begin{equation}
GW_{g,n,X}:\, H^*(\overline{\M}_{g,n})\otimes H^*(X^n) \ \to \ RH_2(X).
\label{1.3}
\end{equation}
(This is well-defined for $2g+n\geq 3$, and we will always assume  that it
has been extended to
all $g,n$ by the stabilization procedure of [IP1] \S 3).

 Finally, we sum over $n$ and $g$ by setting  $\overline{\M} =
\bigcup_{g,n}\overline{\M}_{g,n}$,    letting ${\Bbb T}(X)$ denote the
total tensor algebra
${\Bbb T}(H^*(X))$ on the rational cohomology of $X$, and introducing a
variable  $\lambda$ to keep track of
the euler class.  The total Gromov-Witten invariant  are then the map
\bear
GW_X: H^*(\overline{\M})\otimes  {\Bbb T}(X)\ra RH_2(X)[\la].
\label{defGW}
\eear
which is linear in the second factor and defined by the Laurent expansion
\begin{equation}
GW_{X}\ =\ \sum_{g,n,A} \frac{1}{n!}\,GW_{g,n,A,X}\ t_A\ \la^{2g-2}.
\label{defn.7}
\end{equation}
The diagonal action of the symmetric group on
$\overline{\M}_{g,n}\ti X^n$ leaves $GW_X$ invariant up to sign, and if
$\kappa\in H^*(\overline{\M}_{g,n})$
then
$GW_X(\kappa,
\alpha)$ vanishes unless $\alpha$ is a tensor of length $n$.

We can recover the familiar geometric interpretation of these invariants
by  evaluating on cohomology classes.  Given $\kappa\in
H^*(\overline{\M};\Q)$ and  a
vector $\alpha=(\alpha_1,\dots, \alpha_n)$ of rational cohomology classes
in $X$ of length $|\alpha|=n$,
fix a generic $(J,\nu)$ and generic
geometric representatives $K$ and $A_i$ of the Poincar\'{e} duals of
$\kappa$ and of the $\alpha_i$. Then
$GW_{g,n,A,X}(\kappa,\alpha)$ counts (with orientation) the number of genus
$g$ $(J,\nu)$-holomorphic maps
$f:C\to X$ with $C\in K$ and  $f(x_i)\in A_i$ for each marked point $x_i$.  By the usual dimension counts,
this vanishes   unless
$$
\deg \kappa+\sum\deg \alpha_i- 2 |\alpha|=(\mbox{dim}\, X-6)(1-g)-2K_X\cdot A
$$
 where $K_X$ is the canonical class of $X$.

It is sometimes useful to also incorporate the `$\psi$ classes'.   There
are canonical oriented real 2-plane bundles ${\cal L}_i$ over
$\overline{\M}_{g,n}(X,A)$
whose fiber at each map $f$ is the  cotangent space to the domain curve
at the $i^{\mbox{\tiny th}}$ marked point.  Let $\psi_i$ be the euler class of
${\cal L}_i$, and for each vector $D=(d_1,\dots d_n)$ of non-negative
integers let $\psi_D=\psi_1^{d_1}\cup\dots \cup\psi_n^{d_n}$.  Replacing
the lefthand side of (\ref{modulimap2}) by the pushforward of the cap product
$\psi_D\cap \left[\overline{\M}_{g,n}(X,A)\right]$
and again dualizing gives invariants
\begin{equation}
GW_{g,n,D,X}:\, H^*(\overline{\M}_{g,n})\otimes H^*(X^n) \ \to \ RH_2(X)
\label{1.45}
\end{equation}
which agree with (\ref{1.3}) when $D$ is the zero vector.  These invariants
can be included in
(\ref{defGW}); the series (\ref{defn.7}) then has additional variables that keep
track of the vector $D$. To keep the notation manageable we will not do this
explicitly, although these
invariants will reappear at the end of section 4. (The similar classes
$\phi_i$ defined by the
 cotangent space of the stabilized domain are already part of (\ref{defGW})).

\bigskip

The  invariant (\ref{defGW}) counts
$(J, \nu)$-holomorphic maps from {\it connected}  domains.  It is often
more natural to work with domains
with more components, as  Taubes  does in [T] (see also [IP1]).  Let
$\wt{\M}_{\chi,n}$
be the space of all compact Riemann surfaces of euler characteristic $\chi$
with any number of components, of any possible genus, and with $n$ ordered
points distributed in all possible ways. More precisely, let ${\cal P}_n$ be
the set of all ordered partitions of the set $\{x_1,\dots x_n\}$. Each
$\pi=(\pi_1,\dots,\pi_{\ell})\in{\cal P}_n$ records which marked points go
on each component. Then
\best
\wt \M_{\chi,n}=\ma\bigsqcup_{\pi\in {\cal P}_n}\;
\ma\bigsqcup_{g_i}\  \overline{\M}_{g_1,\pi_1}\ti \dots \ti
 \overline{\M}_{g_l,\pi_l}
\eest
where $\overline{\M}_{g_i,\pi_i}$ is the Deligne-Mumford  space with $n_i$
marked points labeled by the set $\pi_i$ and where the second sum is over
all $g_i$ with
$\sum(2-2g_i)=\chi$.  Define the  ``Taubes-Witten'' invariant
$$
TW_X: H^*(\wt{\M})\otimes {\Bbb T}(X) \ \to \ RH_2(X)[\la]
$$
by
\begin{equation}
TW_X\ =\ e^{GW_X}.
\label{defnTW}
\end{equation}
As before, when this is expanded as a Laurent series
$$
TW_X(\kappa, \alpha)\ =\ \sum {1\over n!}\;TW_{X,A,\chi,n}(\kappa,\alpha)\
\la^{-\chi}\,t_A
$$
the coefficients count the number of curves (not necessarily connected)
with euler characteristic $\chi$ representing $A$ satisfying the
constraints $(\kappa,\alpha)$.

\medskip

\rem \ \ Specifically, if $\al=\al_1\otimes\dots \otimes \al_n$, then for
each partition $\pi=(\pi_1,\dots \pi_\ell)$
let $\al_{\pi_i}$ be the product of $\al_j$ for all $j\in \pi_i$. If
$\kappa=\kappa_1\otimes \dots \otimes \kappa_\ell$, then:
\best
TW_{X, n}(\kappa,\alpha)\ &=&\;\sum_{\pi\in{\cal P}_n}
{\ep(\pi) \over \ell! }{ n \choose n_1, \dots, n_\ell}
 GW_{X,n_1}(\kappa_{1},\al_{\pi_1})
\otimes  \dots \otimes GW_{X, n_\ell}(\kappa_{\ell},\alpha_{\pi_\ell})
\eest
where $\ep_\pi=\pm 1$ depending on the sign of the permutation
$(\pi_1,\dots, \pi_\ell)$ and the degrees of $\al$.
 Note that  $A$ and $\chi$ add when one
take disjoint unions, so the variables $t_A$ and $\la$ multiply.

\vskip.4in

\setcounter{equation}{0}
\section{Relative Invariants}
\bigskip

The invariants of Section 1 can be extended to  invariants for $(X,\w)$
relative to a codimension
two symplectic submanifold
$V$  (a related invariant has been defined by Li-Ruan [LR]). This requires addressing three major issues:
``rim tori'',  higher order contact between the
holomorphic curves and $V$, and   ``chambers''.

\bigskip

Let ${\cal J}^V$ be the space of pairs $(J,\nu)$ where $J$ is an almost
complex structure on $X$
compatible with the symplectic form and for which $V$ is a holomorphic
submanifold, and $\nu$ is a
perturbation (as in [RT]) such that $\nu|_V$ takes values in the tangent
space to $V$. A
$(J,\nu)$-holomorphic map into $X$  is {\it $V$-regular} if the inverse image
of $V$ is a finite set of
points.  For each $(g,n)$, the $V$-regular maps form an open subset
$\M_{g,n}^V(X)$ of the moduli space
 of maps, and this is a manifold for   generic $(J,\nu)\in{\cal J}^V$. We
will write this moduli space as a
disjoint union of strata and compactify each strata separately.

\medskip

The strata are  labeled by sequences $s=(s_1, s_2,\dots, s_\ell)$ of
integers with
$s_i\geq 1$.  Let ${\cal S}$ be the
set of all such sequences and define the degree, length, and order of
$s\in{\cal S}$ by
$$
\mbox{deg }s =\sum s_i \qquad \ell(s) = \ell, \qquad |s|=s_1 s_2\cdots s_\ell.
$$
Associated to each $s$ is the moduli space
$$
\M_{g,n,s}^V(X)\,\subset\ \M_{g,n+\ell(s)}^V(X)
$$
consisting of those maps $f$ such that $f^{-1}(V)$ is exactly the points
$x_i$, $n+1\leq i \leq n+\ell(s)$, each with multiplicity $s_i$. Then
$\M_{g,n,s}^V(X)$ defines a stratum of
$\M_{g,n}^V(X)$ via the projection that forgets the last
$\ell(s)$ points, and  $\M_{g,n}^V(X)$ is the disjoint union of these strata.

\medskip

 To compactify  $\M_{g,n,s}^V(X)$, we introduce the ``symplectic
compactification'' $\widehat{X}$ of $X\setminus V$.  This is a compact,
symplectic manifold  whose
boundary is the circle bundle $S$ of the normal bundle to $V$ in $X$
together with a projection
$\pi:\widehat{X}\to X$ which is the projection $S\to V$ on the boundary and
is a  symplectomorphism in the
interior. Let $V^*$ be $V$ with the discrete topology, and let $S^*$  be
$S$ topologized as the disjoint
union of its fiber circles.  Then
$\pi:S^*\to V^*$ and the inclusions $V^*\subset X$ and
$S^*\subset \hat{X}$ are continuous.  The long exact sequence of the pair
$(\hat{X},S^*)$ gives maps
\bear
H_2(\hat{X})  \longrightarrow H_2(\hat{X},S^*)
\overset{\rho}\longrightarrow H_1(S^*).
\label{2.1}
\eear

 Each $V$-regular map  lifts to  a map into $\hat{X}$; this determines an
element of $H_2(\hat{X},S^*)$
that projects under $\rho$ to  an element of the symmetric
product $\mbox{Sym}\,V$.  Similarly, each $f\in\M_{g,n,s}^V(X)$ lifts to
$\hat{X}$, and evaluation at the
last $\ell(s)$ marked points gives a map into  the space
$$
{\cal S}V\ =\ \bigcup_{s\in{\cal S}}\ V_s
$$
where each $V_s$ is a copy of $V^{\ell(s)}$.    We give $SV$ the
topology of that makes this a
disjoint union.  Restricting (\ref{2.1}) to $\mbox{Sym}\,V$ and pulling
back by the projection $
{\cal S}V \to \mbox{Sym}\,V$,  then gives a space  ${\cal
H}_X^V=H_2(\hat{X},S^*)
\ti_{\footnotesize{\mbox{Sym $V$}}} {\cal S}V$ with an evaluation map $\ep$
into ${\cal S}V$:
\begin{equation}\begin{array}{clcl}
H_2(\hat{X}) & \longrightarrow &{\cal H}_X^V & \\
& & \Big\downarrow\ep &  \\
& &  {\cal S}V. &
\end{array}
\label{HSVproj}
\end{equation}
 We topologize ${\cal H}_X^V$ so that $\ep$ is a covering map. With this
setup, lifting maps into
$\hat{X}$ gives a diagram
\begin{equation}\begin{array}{clccc}
\M_{g,n,s}^V(X) & \overset{\phi}\to & {\cal H}_X^V& \overset{\ep}\to &{\cal
S}V \\
\Big\downarrow & & \Big\downarrow & &\Big\downarrow\\
\M_{g,n}^V(X) &  \to &  H_2(\hat{X},S^*) & \overset{\rho}\to & \mbox{Sym}\, V.\\
\end{array}
\label{liftmap}
\end{equation}
Note that the intersection divisor with $V$ can be recovered by applying
$\rho$, and that the image of $\phi$ implicitly describes that class of the
lift in $H_2(\hat{X},S)$.
However, these  data cannot be separated because the covering
(\ref{HSVproj}) is not, in general, trivial.

\medskip

\begin{rem}
The covering (\ref{HSVproj}) can be partially trivialized.  Let ${\cal
R}_X$ be the kernel of the
projection $H_2(\hat{X})\to H_2(X)$.  Elements of ${\cal R}_X$ are called {\it rim tori} because they can
be represented by tori  of the form  $\alpha\ti \mu$ where $\alpha$ is  a
curve in $S$ and $\mu$ is a
fiber of $S\to V$.   Projecting 2-cycles in  $(\hat{X},S^*)$ into $X$
defines a map $\pi_*:{\cal H}_X^V\to
H_2(X)$.  Then (\ref{HSVproj}) can be rewritten as a covering map
\begin{equation}
\begin{array}{clc}
{\cal R}_X & \longrightarrow &{\cal H}_X^V \\
& & \Big\downarrow\\
& &   H_2(X)\ti {\cal S}V.
\end{array}
\label{Hfibration2}\nonumber
\end{equation}
\end{rem}

\medskip

\begin{lemma}
 For generic $(J,\nu)$, the image of $\M_{g,n,s}^V(X)$ under $\phi$  has
compact closure in ${\cal
H}_X^V$ for each $s\in {\cal S}$,  and these images define  elements of
$H_*({\cal H}_X^V;\Q)$.
\label{prop1}
\end{lemma}

The relative GW invariant of the pair $(X,V)$ is then obtained by
repeating the discussion leading from (\ref{modulimap2}) to (\ref{defGW}).
Writing ${\Bbb T}(H^*(X,V))$ as
${\Bbb T}(X,V) $, the relative  invariant is thus  a map
$$
GW_X^V: H^*(\overline{\M})\otimes  {\Bbb T}(X,V) \ \longrightarrow\
 H_*({\cal H}_X^V)[\la]
$$
with an expansion
\bear
GW_X^V =\ \sum_{g,n} {1\over n!}\; GW_{g,n,X}^V\ \la^{2g-2}.
\label{defnRelInvt2}
\eear
 Formula (\ref{defnTW}) extends this to a relative Taubes-Witten invariant
\begin{equation}
TW_X^V:  H^*(\overline{\M})\otimes  {\Bbb T}(X,V)  \to    H_*({\cal
H}_X^V)[\la].
\label{defn5}
\end{equation}
When $V$ is the empty set ${\cal H}_X^V$ is $H_2(X)$ and the relative
invariant takes values
in $RH_2(X)$.  The relative and absolute invariants are then equal:
$GW_X^{\emptyset}=GW_X$.
\smallskip

The following theorem shows that these relative
invariants are well-defined, although they depend on a `chamber structure'
in ${\cal J}^V$. Let ${\cal W}$ be the subset  of ${\cal J}^V$ consisting
of all those
$(J,\nu)$ which admit a $(J,\nu)$-holomorphic map $C\subset V$ such that
the restriction of the linearization $D_C$ to the normal bundle
to $V$ has non-trivial cokernel.  This `wall' ${\cal W}$ has codimension
one and depends only on the 1-jet of
$(J,\nu)$ along $V$.  The wall separates ${\cal
J}^V\setminus{\cal W}$ into components called  the chambers of ${\cal J}^V$.

\begin{theorem}
\begin{tabular}[t]{l}
 $GW^V$ and $TW^V$ are constant for generic $(J,\nu)$ in each chamber of
${\cal J}^V$.
\end{tabular}
\label{thm2.3}
\end{theorem}

\medskip

 The geometric interpretation of the relative invariant
(\ref{defnRelInvt2}) is similar to that of
(\ref{defn.7}).
Given $\kappa\in H^*(\overline{\M})$, a vector
$\alpha=(\alpha_1,\dots, \alpha_n)$ of classes in $H^*(X,V)$, and a
$\gamma\in H^*({\cal H}_X^V)$, fix generic
geometric representatives $K\subset \overline{\M}_{g,n}$,
$\Gamma\subset {\cal H}_X^V$ and $A_i\subset X$ of their Poincar\'{e}
duals. Then the evaluation pairing
$\gamma \cdot GW_{g,X}^V(\kappa,\alpha)$ counts   the oriented number of
genus $g$ $(J,\nu)$-holomorphic maps $f:C \to X$ with $C\in K$,
$\phi(f)\in \Gamma$, and $f(x_i)\in A_i$ for each marked point $x_i$.

Note that the condition $\phi(f)\in \Gamma$ constrains both the homology
class $A$ of the map and the boundary values of the curve.  In the special
case when there are no rim
tori,  these homology and the boundary value constraints can be fully separated.

\medskip

\begin{ex} When there are no rim tori, ${\cal H}_X^V$ is the subset of
$H_2(X)  \ti \bigcup_s V^{\ell(s)}$
consisting of pairs $(A,s)$ with $\deg s=A\cdot V$.  The homology of
${\cal H}_X^V$ is the corresponding
subalgebra of $RH_2(X)\otimes {\Bbb {CT}}$ where ${\Bbb {CT}}$ is the
``contact tensor algebra'' of $V$:
\best
{\Bbb {CT}}(V)\,=\, {\Bbb {T}}(H_*(V)\ti {\Bbb N}).
\eest
The relative invariants are then maps
\best
 H^*(\overline{\M})\otimes {\Bbb T}(X,V) \ra  {\Bbb {CT}}(V)\otimes RH_2(X)[\la]
\eest
and have Laurent expansions like (\ref{defn.7}) with coefficients in ${\Bbb
{CT}}(V)$.  In fact, a basis for
the contact algebra is given by  elements of the form
$C_{s,\gamma}=C_{s_1, \gamma_1}\otimes\dots \otimes  C_{s_\ell, \gamma_\ell}$,
where $\gamma_i$ are elements of $H_*(V;\Q)$ and $s_i\ge 1$ are integers.
Let $\{C^*_{s,\gamma}\}$ denote the
dual basis.  With $\kappa$ and $\alpha$ as above, we can expand
\bear
 \sum_{s,\gamma}  GW_{g,A,X}^V(\kappa,\alpha; C_{s,\gamma})\ \
C^*_{s,\gamma}\ t_A\ \la^{2g-2}
\label{2.last}
\eear
where the coefficients count the oriented number of genus $g$
$(J,\nu)$-holomorphic
maps $f:C \to X$ with $C\in K$, $f(x_i)\in A_i$, and having a contact of
order $s_j$ with $V$ along fixed representatives $\Gamma_j$ of the
Poincar\'{e} duals of the $\gamma_j$.
\label{ex2.3}
\end{ex}

\vskip.4in

\setcounter{equation}{0}
\section{The Convolution}
\bigskip

 Having defined the relative invariants, we now turn to the problem of
describing how they behave under
the symplectic  sum operation. The goal is to give a formula
expressing the absolute invariants of a
symplectic sum $Z=X\#_{V=\overline{V}} Y$ in terms of the relative
invariants of $X$ and
$Y$.  This turns out to be most
natural when one works with the $TW$ invariants, and when one works with
relative invariants throughout.

\bigskip

Let $X$ be a symplectic manifold with two disjoint symplectic  submanifolds
$U$ and $V$ with real
codimension two.  Suppose that $V$ is symplectically identified with a
submanifold of
similar triple  $(Y,\overline{V},W)$ and that the normal bundles of
$V\subset X$ and
$\overline{V}\subset Y$ have opposite chern
classes.  Let $(Z,U,W)$ be the resulting symplectic  sum.  We will choose
$(J,\nu)$ on the $X$ and $Y$
sides so that the chambers along $V$ and $\overline{V}$  match when the sum
is formed.  In  this section we
will define a ``convolution'' operation  and establish a formula of the  form
\bear
TW_X^{U\cup V}\, *\, TW_Y^{ \overline{V}\cup W}\, = \, TW_Z^{U\cup W}
\label{3.1}
\eear
under the assumption that all curves contributing to the invariants are
$V$-regular (this condition will be
eliminated in the next section).

\medskip

The convolution operation is assembled from several pieces.   The first
ingredient is a gluing map defined
on Deligne-Mumford spaces.  Given stable curves $C_1$ and
$C_2$ (not necessarily connected) with euler
characteristics $\chi_i$ and $n_i+\ell$ marked points, we can construct a
new curve by identifying the last
$\ell$ marked points of
$C_1$ with the last $\ell$ marked points of $C_2$, and then
forgetting the marking of these new
double points.  This  defines an attaching  map
\best
\xi_\ell:\wt{\M}_{\chi_1,n_1+\ell}\ti \wt{\M}_{\chi_2,n_2+\ell}\ \longrightarrow
\ \wt{\M}_{\chi_1+\chi_2-2\ell,n_1+n_2}
\eest
whose image is a subvariety of complex codimension $\ell$.  Taking the union
over all
$\chi_1, \chi_2, n_1$ and $n_2$ gives a  map
$\xi_\ell:\wt{\M}\ti\wt{\M}\to\wt{\M}$ for each $\ell$.

The second ingredient describes how maps are glued along $V$.  Consider the
evaluation map
$$
\ev: \M^V(X) \ti \M^{\overline{V}}(Y)  \ \overset{\phi}\longrightarrow \
{\cal H}_X^V
\ti {\cal H}_Y^{\overline{V}}
\ \overset{\ep}\longrightarrow\  {\cal S}V\ti {\cal S}V.
$$
defined by the top row of (\ref{liftmap})  on each factor;
this evaluation map
 records the intersection points with $V$, together with their
multiplicities.  The diagonal of
${\cal S}V\ti {\cal S}V$ is the union of components
$$
{\bold \Delta}_s\ \subset \ {\cal S}V\ti {\cal S}V.
$$
labeled by sequences $s$.   With this notation, we can state our first main
result.  It is a `gluing
theorem' for families of pseudo-holomorphic maps which is proved by PDE
methods [IP2].

\begin{theorem}  For each sequence $s$, there is an
${|s|\over \ell(s)!}$-fold cover of $ev^{-1}({\bold \Delta}_s)$,
denoted \newline $\M^V(X) \ti_{ev_s} \M^{\overline{V}}(Y)$, and a commutative
diagram
$$
\begin{array}{ccc}
\M^V(X) \ti_{ev_s} \M^{\overline{V}}(Y) & \longrightarrow &
\M(Z)\\
\Big\downarrow\vcenter{\rlap{$st$}} &  & \Big\downarrow
\vcenter{\rlap{$st$}}  \\
\wt{\M}\ti \wt{\M} &
\overset{\xi_{\ell(s)}}\longrightarrow &  \wt{\M}
\end{array}
$$
where the top arrow is a diffeomorphism.
\label{gluing thm}
\end{theorem}

\medskip

The next step is to pass to homology.  We can glue a curve in $X$ to a
curve in $Y$ provided the two
curves meet $V$ at the same points with the same
multiplicity; this defines a map
\bear
g:{\cal H}_X^{U,V}\ti_{\ep} {\cal H}_Y^{\overline{V},W}\ra {\cal H}_Z^{U,W}.
\label{gmap}
\eear
\non Combining this with the map $\xi_\ell$ above gives the convolution
operator that describes how homology
classes of maps combine in the gluing operation for the symplectic sum.

\begin{defn}\ \ The convolution operator
$$
*: \ H_*(\wt{\M}\ti {\cal H}_X^{U,V};\Q[\la])
\otimes H_*(\wt{\M}\ti {\cal H}_Y^{\overline{V},W};\Q[\la])
\ \longrightarrow\  H_*(\wt{\M}\ti {\cal H}_Z^{U,W};\Q[\la])
$$
is given  by
\begin{equation}
(\kappa\otimes h)\,*\,(\kappa'\otimes h') \ =\
\sum_{s} {|s|\over \ell(s)!} \
\left(\xi_{\ell(s)}\right)_*(\kappa\otimes \kappa')  \
g_*\left[ \left. h\ti h'\right|_{ \ep^{-1}\left(\bold
{\Delta}_{s}\right)}\right]\ \la^{2\ell(s)}.
\label{wteddiagonal1}
\end{equation}
\end{defn}

\medskip

 To conform with this definition, we will regard the
$TW$
invariant of
$X$ as a map
$$
TW_X^{U,V}:    {\Bbb T}(X,V)  \to   H_*(\wt{\M}\ti {\cal H}_Z^{U,V};\Q[\la])
$$
 by dualizing the first term in (\ref{defn5}). Finally, we say that a
constraint $\alpha\in H^*(Z)$ {\it separates} as $\alpha_X+\alpha_Y$ if its
Poincar\'{e} dual
$PD(\alpha)$ is the image of $(PD(\alpha_X),PD(\alpha_Y))$ under the
Mayer-Vietoris map $H_*(X\setminus
V)\oplus H_*(Y\setminus \overline{V})\to H_*(Z)$. These generate a subalgebra
${\Bbb T}_s(Z)$ of ${\Bbb T}(Z)$,  namely the span
of tensors of the form
\begin{equation}
\label{separates}
\alpha= \alpha_X+\alpha_Y
\end{equation}
with  $\alpha_X\in
{\Bbb T}(X,V)$ and
$\alpha_Y\in {\Bbb T}(Y,\overline{V})$.  Then, when $\alpha$ has this form
and assuming that all
curves contributing to the
invariants are $V$-regular,  (\ref{3.1}) holds in the sense that
\bear
TW_Z^{U, W}(\alpha)\, = \,TW_X^{U,V}(\alpha_X)\, *\, TW_Y^{
\overline{V}, W}(\alpha_Y).
\label{3.5}
\eear

\bigskip

\begin{ex} {\bf (The S-matrix)}\ \  Starting from the normal bundle $N_V$
of $V$ in $X$, we can form the
$\P^1$ bundle ${\Bbb F}=\P(N_V\oplus\cx)$ over $V$.  In  ${\Bbb F}$, the
zero section $V$   and the infinity section $\overline{V}$ are disjoint
symplectic submanifolds.  The
symplectic  sum of $(X,U,V)$ and $({\Bbb F}, \overline{V},V)$  is a
symplectic deformation of $(X,U,V)$,
so has the same  $TW$ invariant. The convolution then defines a operation
$$
 \ H_*(\wt{\M}\ti {\cal H}_X^{U,V};\Q[\la])
\otimes H_*(\wt{\M}\ti {\cal H}_{{\Bbb F}}^{\overline{V},V};\Q[\la])
\ \longrightarrow\  H_*(\wt{\M}\ti {\cal H}_X^{U,V};\Q[\la]).
$$
Thus for each choice of constraints $\alpha\in {\Bbb T}({\Bbb F},V\cup
\overline{V})$, the
$TW$ invariant of ${\Bbb F}$ relative to its zero and infinity section
defines an ``S-matrix''
\bear
\label{3.Endo}
TW_{{\Bbb F}}^{\overline{V},V}(\alpha)\,\in\, \mbox{End}\
\left(H_*(\wt{\M}\ti {\cal H}_X^{U,V};\Q[\la])  \right)
\eear
which describes how families of curves on $X$
are modified --- ``scattered''--- as they pass through a neck modeled on
$({\Bbb F}, \overline{V},V)$  containing the constraints $\alpha$.
\label{ex3.1}
\end{ex}

\bigskip

The identity  in the endomorphism algebra (\ref{3.Endo}) is always realized
as  the convolution
by the element
\best
{\Bbb I}\in H_*(\wt{\M}\ti {\cal H}_{{\Bbb F}}^{\overline{V},V};\Q[\la])
\eest
corresponding to that part of $TW(1)$ coming from the maps whose domain is
unstable and whose
image is a multiple cover of a fiber of ${\Bbb F}$.  As in physics, it is
convenient to write the $S$-matrix
(\ref{3.Endo}) as ${\Bbb I}+ R^{\overline{V},V}$; the inverse of the
$S$-matrix is then given by the series
\bear
(TW_{\Bbb F}^{\overline{V},V})^{-1}={\Bbb I}\ -\  R^{\overline{V},V} \ +
\  R^{\overline{V},V}*  R^{\overline{V},V}\ -\
R^{\overline{V},V}* R^{\overline{V},V}*R^{\overline{V},V}\ +\ \dots.
\label{3.inverseSmatrix}
\eear

\bigskip

\begin{ex}
When $V=\P^1$,   ${\Bbb F}\to V$ is one of the rational ruled surfaces
with its standard symplectic and holomorphic structure.  If we wish to
count all pseudo-holomorphic maps,
without constraints on the genus
or the induced complex structure, the relevant S-matrix is  the relative
$TW$ invariant  with
$(\kappa,\alpha)= (1,1)\in H^0(\overline{\M})\ti H^0(X)$.  This  case
works out
neatly:  a dimension count shows that  $TW_{{\Bbb
F}}^{\overline{V},V}(1,1)={\Bbb I}$.
\label{ex3.2}
\end{ex}

\vskip.4in

\setcounter{equation}{0}
\section{The Sum Formula}
\bigskip

The full  formula for the invariants of a symplectic sum is obtained from
(\ref{3.5}) by a `stretching the
neck' argument.   The  sum $X\#_VY$ can be viewed as assembled from
$n+1$ symplectic sums, where the
middle $n$ pieces are copies of the ruled space ${\Bbb F}$ associated to
$V$. We again pick  $(J,\nu)$ on the pieces of this sum so that the
chambers match along each
cut.  The  pigeon-hole principle implies that, for each $A\in H_2(Z)$, we
can take $n$ sufficiently large so
that  all pseudo-holomorphic  $A$-curves in $Z$ limit to $V$-regular curves
in the $n+1$-fold
 sum.  Thus the
curves in $Z$ decompose into curves in $X$ joined to curves in $Y$ by a
chain of curves in the
intermediate copies of ${\Bbb F}$. Applying formula (\ref{3.5})  along each
cut yields the general sum formula.

\medskip

\begin{theorem}  Let $(Z,U,W)$ be the symplectic  sum of
$(X,U,V)$ and  $(Y,\overline{V},W)$ along $V=\ov V$.  Suppose that
$\alpha\in{\Bbb T}(Z)$
separates as in
(\ref{separates}).  Then the relative TW invariant of $Z$ is given
in terms of the  invariants of $(X,U,V)$ and $(Y,\overline{V},W)$ by
$$
TW_Z^{U,W}(\al)\ =\   TW_X^{U,V}(\al_X) \, * \,
(TW_{\Bbb F}^{\overline{V},V})^{-1}\, * \, TW_Y^{\overline{V},W}(\al_Y)
$$
where  the middle term is the  inverse of the $S$-matrix
(\ref{3.inverseSmatrix}).
\label{mainthm}
\end{theorem}

One interesting special case is when $U$ and $W$ are empty and we put no
constraints on the
complex structure of the curves.  Theorem \ref{mainthm} then gives a
formula expressing the absolute
invariant of $Z$ in terms of the relative invariants of $X$ and $Y$.

In this case the algebra of the formula simplifies:  we drop all references
to $\wt{\M}$ and, because $U$ and
$W$ are  empty, $H_*({\cal H}_Z^{U,W})$ is simply
$RH_2(Z)$.  The convolution (\ref{wteddiagonal1}) can then be thought of as
a bilinear pairing

$$
\langle\ ,\ \rangle:\ H_*({\cal H}_X^V;\Q[\la])
\otimes H_*({\cal H}_Y^{\overline{V}};\Q[\la])
\longrightarrow  RH_2(Z)[\la].
$$

\medskip

\non In this case $g:{\cal H}_X^{V}\ti_{\ep} {\cal H}_Y^{\overline{V}}\ra
RH_2(Z)$, so for each sequence $s$,
$\ep^{-1}\left({\bold \Delta}_s\right)$ is the union of components ${\bold
\Delta}_{A,s}=\ep^{-1}\left({\bold \Delta}_s\right) \cap g^{-1}(A)$,  and
the formula for the bilinear
pairing becomes
$$
\langle  h\,, \, h'  \rangle
\ =\  \sum_{s} {|s|\over \ell(s)!}  \sum_{A\in H_2(Z)}\ g_*[{\bold
\Delta}_{A,s}\cap (h\ti h')]\ t_A \
\la^{2\ell(s)}.
$$
The $S$-matrix is  the operator  obtained by letting
$TW_{{\Bbb F}}^{\overline{V},V}(1,1)\,\in\, H_*({\cal H}_X^{U,V};\Q[\la])
$ act  by  convolution; let $S_{\overline{V},V}$ denote its inverse.

\medskip

\begin{cor}  Let $Z$ be the symplectic  sum of $(X,V)$ and
$(Y,\overline{V})$ and suppose that $\alpha\in{\Bbb T}(Z)$ separates as in
(\ref{separates}). Then
$$
TW_Z(\al)\ =\ \left\langle\, TW_X^V(\al_X),\ S_{\overline{V},V}\,
\cdot\,TW_Y^{\overline{V}}(\al_Y)
\,\right\rangle.
$$
\label{cor4.2}
\end{cor}

Another simplification occurs when there are no rim tori in $X$ and $Y$,
and therefore in $Z$.
Then  the
relative invariants have an expansion of the form (\ref{2.last}).  For
simplicity we take $U$ and $W$ both
empty.  Then ${\cal H}_Z^{U,W}$ is simply  $H_2(Z)$ and the map $g$ of
(\ref{gmap}) is the
identification ${\cal S}V\ti_\rho {\cal S}V$ with ${\cal S}V$ together with the addition
$H_2(X)\ti H_2(Y)\to H_2(Z)$.  The
key part of the convolution (\ref{wteddiagonal1}) is then given by the cap
product with the Poincar\'{e}
dual of the diagonal:
$$
g_*\left[ \left.h\ti h'\right|_{\bold {\Delta}_{s}}\right]\ =\
\sum_{B\in H_2(X)}\sum_{ C\in H_2(Y)} {\mbox{PD}}\left(\bold
{\Delta}_{s}\right)\cap (h\ti h')\ t_B t_C.
$$
We can then `split the diagonal' by fixing a basis $\{h^{p}\}$ of
$H^*({\cal S}V)$
and writing
$$
{\mbox{PD}}\left(\bold {\Delta}_{s}\right)\ =\
\sum_{p,q} Q^V_{p,q}\ h^p\ti h^q= \
\sum_{p} \ h^p\ti h_p
$$
where $Q^V_{p,q}$ is the intersection matrix of the Poincar\'{e}
dual basis, and $\{h_p\}$ is the dual basis with respect to $Q^V$.  In fact,
fix a basis $\{\gamma^{i}\}$ of $H_*(V)$, and let
$\{\gamma_{i}\}$ be
the dual basis of $H_*(V)$ with respect to the intersection form.
 Then the $\{h^{p}\}$ can be taken to be the
basis $\{C^*_{s,\gamma^I}=C^*_{s,\gamma^1}\otimes\dots \otimes
C^*_{s_\ell,\gamma^\ell}\}$ of  Example 2.4.
The convolution then has the more explicit form

$$
(\kappa\otimes h)\,*\,(\kappa'\otimes h') \ =\
\sum_{B,C,s} {|s|\over \ell(s)!} \
\left(\xi_{\ell(s)}\right)_*(\kappa\otimes \kappa')  \
 \sum_{I} C^*_{s,\gamma^I}(h)\,  C^*_{s,\gamma_J}(h') \
t_{B+C}\la^{2\ell(s)}.
$$

Finally, we can also include the $\psi$-classes (\ref{1.45}) to obtain a
sum formula for the
numbers
$$
TW_{\chi,A,D,X}(\kappa,\alpha)
$$
in terms of the relative invariants (also numbers)
$TW^V_{\chi,A,D,X}(\kappa,\alpha ;
C^*_{s,\gamma})$.    For this, we will take $\kappa=1$ and  pair  each
$\psi$ class with an $\alpha$.
This case is usually written in terms of the classes $\tau_d(\alpha)=
\psi^d_i\cup ev_i^*(\alpha)$, so we adopt the notation
$$
TW^V_{\chi,A,X}\left(\tau_D(\alpha) ;
 s,\gamma\right)\ :=\ TW^V_{\chi,A,D,X}(1,\alpha ; C^*_{s,\gamma}).
$$
Using dual bases $\{\gamma^I,\gamma_I\}$ as above, we let
$TW_{\chi, A}(s_1,\gamma^{I};s_2,\gamma_{J})$ denote the relative invariant
of ${\Bbb F}$
satisfying the contact constraints $ C^*_{s_1,\gamma^{I}}$ along
$\overline{V}$ and $C^*_{s_2,\gamma_{J}}$ along $V$.    Then the inverse
$S$-matrix is the alternating
series
\bear
S^{\overline{V},V}_{\chi, A}(s_1,\gamma_{I}; s_2,\gamma^{J})
 & = & {\Bbb I}(s_1,\gamma_{I}; s_2,\gamma^{J})\ -\
          TW^{\overline{V},V}_{\chi, A}( s_1,\gamma_{I}; s_2,\gamma^{J})\\
& + &\hskip-.2in
  \sum_{A_1+A_2=A\atop\chi_1+\chi_2-2\ell(s_3)=\chi} \ma\sum_{s_3,
I_3}{|s_3|\over \ell(s_3)!}\
TW^{\overline{V},V}_{\chi_1, A_1}(s_1,\gamma_I; s_3,\gamma^{I_3})\,
TW^{\overline{V},V}_{\chi_2, A_2}(s_3,\gamma_{I_3}; s_2,\gamma^{J})-\dots
\nonumber
\label{4.longprop}
\eear

\begin{theorem} Suppose there are no rim tori in $(X,V)$ and
$(Y,\overline{V})$ and that the constraint $\al$ separates as in
(\ref{separates}). Then the number
$$
TW_{\chi,A,Z}(\tau_D(\alpha))
$$
is given by
\best
\sum  \  \ \frac{|s_1|\, |s_2|}{\ell(s_1)!\, \ell(s_2)!}\ \
TW^V_{\chi_1,A_1,X}(\tau_{D}(\alpha_X);s_1,\gamma^I)\
S^{\overline{V},V}_{\chi_0,A_0}(s_1,\gamma_{I};s_2,\gamma^{J})
\ TW^{\overline{V}}_{\chi_2,A_2,Y}(s_2,\gamma_J;\phantom{i^I}
\tau_{D}(\alpha_Y))
\label{easymainformula}
\eest
where the sum is over all (a) decompositions $A=A_1+A_0+A_2$ and
$\chi=\chi_1+\chi_0+\chi_2-2\ell(s_1)-2\ell(s_2)$, (b)
pairs of sequences $s_1,s_2\in{\cal S}$ of degree $\deg s_i=
A_i\cdot V$, and (c) pairs of multiindices $I,J$  enumerating  bases
$\{\gamma^I,\gamma_I\}$ of
$H^*(V^{\ell(s_i)})$ dual by the intersection form.
\label{2easymainformula}
\end{theorem}

In the case where $TW_{V,\overline{V}}(1,1)={\Bbb I}$ only the leading term
of (\ref{4.longprop}) is present, and
this sum formula   simplifies to
$$
\sum_{A=A_1+A_2\atop
\chi_1+\chi_2-2\ell(s)=\chi}\sum_{s,I}
 \frac{|s|}{l(s)!}\
TW^V_{\chi_1,A_1,X}\left(\tau_{D_X}(\alpha_X); s,\gamma^I\right)
\ TW^V_{\chi_2,A_2,Y}\left(s,\gamma_I ;\tau_{D_Y}(\alpha_Y)\right).
$$

\vskip.4in

\setcounter{equation}{0}
\section{Applications}
\bigskip

This section presents two consequences of Theorem \ref{mainthm}:  the
Caporaso-Harris recursion formula for
the number of curves in $\P^2$, and the  formula (related to the
``Yau-Zaslow conjecture'') for the  number
of curves in the rational elliptic surface.  While these formulas
are both known, the proofs outlined here are considerably easier
and more transparent than the existing ones.

\bigskip

We begin by examining the relative invariants for the pair $(\P^2,L)$ (the
manifold $\hat{X}$ of section 2
is then the  4-ball, and in that sense we are considering the  relative
invariant
of $(B^4,S^3)$).  This pair can
be written as a symplectic  sum
$$
(\P^2,L)\ma\#_{L=E} ({\Bbb F}_1,E,L)\ = \ (\P^2,L)
$$
where $({\Bbb F}_1,E,L)$ is the ruled surface with euler class one with
its zero section $L$ and its infinity section $E$.  We will obtain a
recursive formula for the relative invariants of $(\P^2,L)$ by moving one point
constraint $p$ to the ${\Bbb F}$ side, and applying the sum formula.

 Following [CH], let
$N^{d,\delta}(\al,\b)$ be the number of degree $d$ curves in $\P^2$  with
$\delta$ double points with a
contact with $L$ of order $k$ at $\al_k$ fixed  points, and at $\b_k$
moving points, and passing through the
appropriate number of fixed generic points off $L$.   Similarly, let
$N^{a,b,\delta}_{\Bbb
F}(\al',\b';\; p\; ; \al,\b)$ be the number of curves in ${\Bbb F}$
representing $aL+bF$
that have contact described by
$(\al', \b')$ along $E$, $(\al, \b)$ along $L$, and pass through  $p\in {\Bbb
F}$.  By elaborating on Example 2.4, one can relate these numbers $N^{d,\delta}$
 to the relative $TW$ invariants.   Theorem \ref{mainthm} then becomes
\bear
 N^{d,\delta}(\al,\b)=\ma\sum   |\al'|\,|\b'|\  \
 N^{d',\delta'}(\al',\b') \ \,
N^{d-d',d,\delta''}_{\Bbb F}(\b',\al';\; p\; ; \al,\b)
\label{5.CH1}
\eear
(the $S$-matrix is the identity by Example  3.4).  Furthermore, a
dimension count shows that   there  are exactly two types of curves that
contribute to the
last term in (\ref{5.CH1}):  either (i)  several rational covers of the
fiber, one of them passing
through the point $p$, or (ii) several rational covers of the fiber and a
rational curve in the class $L+a F$ passing through $p$ and having all
contact points with $E$ and $L$ fixed.  In each situation there is a unique
such curve. In the case (i) nothing changes, except that a contact of some
order $k$
at a moving point becomes a contact of order $k$ at the fixed point $p_0$
where the fiber through $p$ hits $E$; in case (ii) the degree of the curve
drops by 1, and the curve looses some of the contact conditions at fixed
points and gains some at moving points, so $\al\ge \al'$ and  $\b'\ge \b$.
Thus the product in (\ref{5.CH1}) becomes a
sum of two terms.  The resulting formula has the form
\best
\ N^{d,\delta}(\al,\b)= \ma \sum k
N^{d,\delta'}(\al-\ep_k,\b+\ep_k) +
\ma \sum |\b'-\b|\cdot {\al\choose \al'} {\b'\choose b}
\cdot N^{d-1,\delta'}(\al',\b')
\eest
and is exactly the formula of Caporaso-Harris ([CH]  Theorem 1.1).

\smallskip

The same method, applied to the sum
${\Bbb F}_n\#{\Bbb F}_n={\Bbb F}_n$ yields the recursion formula of Vakil [V].

\bigskip

We next consider the  invariants of the rational elliptic surface
$E$.  Let $f$ and $s$ denote, respectively, the homology class of the fiber
and the section of the
elliptic fibration $E\to\P^1$.  We will consider the invariants
$GW_{g,s+df}$, which count the number of
connected genus $g$ curves in the class $s+df$ through $g$ generic points.
These define  power
series
$$
F_g(t)=\ma\sum_{d\ge 0} GW_{g,s+df}(p^g)\  t^{d}\,t_s
$$
where $t=t_f$. Recently,  G\"{o}ttsche  and Pandharipande [GP] obtained
recursive relations for the
coefficients of $F_0$,  and Bryan-Leung [BL]  proved the  closed-form expression
\bear
F_0(t)=t_s\, \left(\ma\prod_{d}{1 \over 1-t^d}\right)^{12}.
\label{BL}
\eear
(this formula was very strongly motivated by the work of Yau-Zaslow [YZ]).   We will prove this by relating
it to  the similar series of elliptic ($g=1$)  invariants
\best
H(t)&=&\ma\sum_{d\ge 0}\ GW_{1,s+df}(\tau_1(f^*)) \  t^{d}\,t_s
\eest
where $f^*\in H^2(E)$ is the Poincar\'{e} dual of the fiber class. The
``topological recursion relation''
for $g=1$ (TRR)  easily implies that
\bear
H(t)\ =\ \frac{1}{12} \left(t\,F_0' -F_0\right) +F_0 G
\label{rel1}
\eear
where $G=\sum\sigma(d)t^d$ is the generating function for
$\sigma(d)=\sum_{a|d}a$.   On the other hand, we
can write $E=E\#_{F}T^2\ti S^2$, and compute $H$ by moving the
constraint onto the $T^2\ti S^2$
side and using Theorem \ref{mainthm}. For this situation, the only possible
splitting is a genus 0
curve on $E$ side representing $s+d_1F$ and a genus 1 curve on the
$T^2\ti S^2$ side. But the genus 1 invariant of $T^2\ti S^2$
is computable from TRR, and  comes out to be $2(G(t)-1/24)$. Therefore,
$$
H(t)\ =\ 2 F_0\l(G-{1\over 24}\r).
$$
Equating this with (\ref{rel1}), we see that $F_0$ satisfies the ODE
\best
t\, F_0'=12\,G F_0
\eest
with $F_0(0)=GW_{0,s}t_s = t_s$.  Integrating and rearranging  gives (\ref{BL}).

\vskip.4in

\small

\medskip


TITLE: Conjugation of Matrices and Conjugation of Complex Numbers
QUESTION [4 upvotes]: Are conjugation of matrices and conjugation of complex numbers related?
What I mean is that if $A$ is an $n \times n$ matrix then the conjugation of $A$ by an invertible $n \times n$ matrix $C$ is given by $CAC^{-1}$. On the other hand, if $a + bi$ is a complex number then it's conjugate is $a-bi$. These two operations don't really seem to have anything to do with one another but if they're unrelated why is the same term used to describe the operation?

REPLY [1 votes]: I know this is super old but there's a good answer that nobody gave here. Both operations are automorphisms. That is, the sum of two conjugates is the conjugate of the sum, and the conjugate of the product is the product of the conjugates. This holds in both cases.


TITLE: simple algebraic problem (related to graph theory)
QUESTION [2 upvotes]: Given $\binom{n}{r} p^{\binom{r}{2}} = 1$, I want to obtain an expression for $r$.
In particular, applying Stirling approximation $$n! \sim \sqrt{2 \pi n} (n/e)^n$$ We see that
$$\binom{n}{r} p^{\binom{r}{2}} \sim (2 \pi)^{-1/2} n^{n+1/2} (n-r)^{-n+r-1/2} r^{-r-1/2} p^{r(r-1)/2}$$
So, my aim is to isolate $r$ in:
$$(2 \pi)^{-1/2} n^{n+1/2} (n-r)^{-n+r-1/2} r^{-r-1/2} p^{r(r-1)/2} = 1$$
Setting $b = 1/p$, I know the answer I should get is:
$$r = 2 \log_b n - 2 \log_b \log_b n + 2 \log_b\left(\frac{e}{2}\right) + 1 + o(1)$$
However... I dont know how can I do it. I would be grateful if someone can help me!
(By the way, this is part of Bollobas' proof for the chromatic number of $G(n,p)$ with $p$ constant)
Thanks!

REPLY [2 votes]: Firstly, we have $0<r\le n$. As a direct result of Stirling's approximation, we have
$$\ln n!=n\ln n-n+\frac12\ln n+O(1)\tag1$$
Take logarithm at $\binom nrp^{\binom r2}=1$, we have
$$\ln\binom nr+\binom r2\ln p=0$$
For $0<p<1$, let $c=-\ln p$, and $b=1/p$, we have $c>0$, therefore $L=\frac c2r(r-1)$
where
$$L=\ln\binom nr$$
We apply a rough approximation at first:
$$L=n\ln n-r\ln r-(n-r)\ln(n-r)+O(n)$$
therefore $L=O(n\log n)$, which results in $r=O(\sqrt{n\log n})=o(n)$.
Now observe more closely
$$L=(n\ln n-n)-(r\ln r-r)-((n-r)\ln(n-r)-(n-r))+O(\log n)$$
and $\ln(n-r)=\ln n+\ln(1-r/n)$, thus
$$L=r\ln n-r\ln r-(n-r)\ln(1-r/n)+O(\log n)$$
therefore $\frac c2(r-1)=O(\log n)$ and $r=O(\log n)$.
Now we apply approximation (1), more exact than the proceding approximations:
\begin{multline}
L=\left(n\ln n-n+\frac12\ln n\right)-\left(r\ln r-r+\frac12\ln r\right)\\
-\left((n-r)\ln(n-r)-(n-r)+\frac 12\ln(n-r)\right)+O(1)
\end{multline}
Simplify it, we get
$$L=r\ln n-r\ln r-(n-r)\ln(1-r/n)-\frac12\ln r-\frac12\ln(1-r/n)+O(1)$$
Notice that $\ln(1-r/n)=-r/n+O(r/n)^2=O(1)$, and we have
$$L=r\ln n-r\ln r+r-\frac12\ln r+O(1)$$
At first, go roughly
$$L=r(\ln n+O(\log r))=r(\ln n+O(\log\log n))$$
thus
$$\frac c2(r-1)=\ln n+O(\log\log n)\implies r=\frac{2\ln n}c+O(\log\log n)$$
then go exactly
$$L=r\ln n-r\ln r+r+O(\log\log n)$$
thus
$$\frac c2(r-1)=\ln n-\ln r+1+O\left(\frac{\log\log n}{\log n}\right)$$
where
\begin{align}
\ln r
&=\ln\left(\frac{2\ln n}c+O(\log log n)\right)\\
&=\ln\left(\frac{2\ln n}c\right)+\ln\left(1+O\left(\frac{\log\log n}{\log n}\right)\right)\\
&=\ln\ln n+\ln 2-\ln c+O\left(\frac{\log\log n}{\log n}\right)
\end{align}
Finally, we have
\begin{align}
r&=\frac{2\ln n}c-\frac{2\ln\ln n}c-\frac{2\ln 2}c+\frac{2\ln c}c+\frac 2c+1+O\left(\frac{\log\log n}{\log n}\right)\\
&=2\log_b c-2\log_b\log_b c+2\log_b\frac e2+1+O\left(\frac{\log\log n}{\log n}\right)
\end{align}


TITLE: Identity of Polynomials in positive charcteristic
QUESTION [5 upvotes]: In positive charcteristic $p$, we know that for every field element $x\in\mathbb{F}_{p}$ we get $x^p = x$.
Then I think (and I might be wrong, but I don't see how) monomials of the form $t^{p^i}\in\mathbb{F}_p[t]$ for arbitrary $i\in \mathbb{N}$ are all the same, since the functions $p_0(t)=t$, $p_1(t)=t^p$, $p_2(t)=t^{p^2}$ and so on are all actually the same functions. Not? I mean certainly they are equal on all elements, that is
$$p_i(x)= p_j(x)$$ for all $x\in\mathbb{F}_p$ and $i,j\in\mathbb{N}_0$. But in textbooks on finite function field and such, they are treated as if they where different. But this seems to contradict the (poinwise) definition of a function in terms of evaluation on elements.
So would be great to get some clarification here.

REPLY [6 votes]: There are indeed two different notions of a polynomial over a field $K$: the first is that of formal expressions of the form $\sum_{i=0}^k a_iX^i$ where $a_i \in K$, and the second is that of functions $f : K \to K$ where there exist $a_i \in K$ such that for all $x \in K$ we have
$$
f(x) = \sum_{i = 0}^k a_ix^k.
$$
If we write $K[X]$ for the formal expressions, and $\mathrm{Pol}(K)$ for the functions, then there is a mapping $K[X] \to \mathrm{Pol}(K)$ given by interpreting the formal expression as a function. In fact, it is a homomorphism whose kernel is precisely those polynomial expressions which evaluate as the zero function.
Note that if $K$ is infinite, then the mapping $K[X] \to \mathrm{Pol}(K)$ is a bijection, and we tend to identify the two sets without too much worry.
When $K$ is finite, identifying the sets makes little sense: the set $\mathrm{Pol}(K)$ is a finite set -- in fact the set of all functions $K \to K$ -- that does not catch the structure we find interesting about polynomials over $K$. For a concrete example, $\mathrm{Pol}(\mathbb F_2)$ only has four elements, and they can be written (as polynomial expressions) as $0, 1, X, X - 1$. There are no polynomials of degree at least $2$ to form any splitting fields!


TITLE: Do ordered lists contain more information than unordered lists?
QUESTION [2 upvotes]: Suppose you have a set of 4 elements {A,B,C,D}, but without specifying their order. Now, if you want to specify which precedes which, that would mean you need to provide more information.
But at the same time, writing down the ordered list {B,C,A,D} doesn't take more text. That's probably because order is implicit when writing down things, even when I say that {A,B,C,D} is unordered, the way I write it down has order, I just tell you to ignore it.
So which is true? Does order carry additional information, and if so, why don't ordered lists take more text to describe than unordered ones?

REPLY [6 votes]: Ordered lists do contain more information.
Precisely, you can order $n$ elements in $n!$ different ways, so say you got $4$ elements ${A,B,C,D}$ with no order defined. When writing them down in a particular order, the order itself defines a number from $0$ to $4! = 23$, so you basically stored $log_2 24$ additional bits.
Now when you write down unordered elements, you'll of course need to choose some order (which is "waste" information in that moment). But if you choose a particular order, you can use this to store additional information in the same message. This is actually a neat trick in steganography, i.e. sending hidden messages, in data where order is of no importance (HTML attribues are great!).
On the other hand, if order does matter, you cannot afterwards rearrange the items to store extra information or the original order is lost.