TITLE: Calculating the index of an aditive subgroup.
QUESTION [1 upvotes]: Let $\mathbb{F}_{q}$ a finite field with characteristic $2$, and define over it the set $N=\{x^2+x:x\in\mathbb{F}_q\}$ show that $[\mathbb{F}_q:N]=2$.
I've already proof that $N$ is an aditive normal subgroup of $\mathbb{F}_q$ but, I don't have any idea of how to verify the value of the index using cosets, somebody can give me a hint?

REPLY [2 votes]: Hint:

Show that $f:x\mapsto x+x^2$ is a homomorphism of additive groups from $\Bbb{F}_q$ to itself.
Determine the subgroup $\operatorname{ker} f$.
The first isomorphism theorem.


TITLE: The probability of sum 5 after n time’s
QUESTION [2 upvotes]: Assuming the die is unbiased, what is the probability of getting sum of 5 after $n$ time (use only one cube at a time).
It's clear to me that more than 5 time is wrong.(Because the lowest number is $1$ so $1+1+1+1+1$ is $\max$)
I tried this way: $\frac{1}{6}+\frac{4}{6}*\frac{1}{6}…$
First time only 5 is ok,in the second time 4 possible numbers are ok(1,2,3,4) then there is one complementary number.
Is this the way solution?

REPLY [1 votes]: The number of cases corresponding to the sum $5$ for $n$ throws of a fair dice will be the coefficient of $x^5$ in the following polynomial:
$p(x)=(x^1+x^2+\ldots+x^6)^n$
$=x^n(1+x+\ldots+x^5)^n=x^n\frac{(1-x^6)^n}{(1-x)^n}=x^n(1-x^6)^n(1-x)^{-n}$
$=x^n(1-nx^6+\ldots+(-1)^nx^{6n})\left(1+{n \choose 1}x+{n+1 \choose 2}x^2+{n+2 \choose 3}x^3+{n+3 \choose 4}x^4+{n+4 \choose 5}x^5+\ldots\right)$
$=(x^n-nx^{6+n}+\ldots)\left(1+{n \choose 1}x+{n+1 \choose 2}x^2+{n+2
\choose 3}x^3+{n+3 \choose 4}x^4+{n+4 \choose 5}x^5+\ldots\right)$
As can be seen from above,

When $1 \leq n \leq 5$, the coefficient of $x^5$ in the above polynomial product will be exactly same as the coefficient of $x^{5-n}$ in the RHS polynomial $(1-x)^{-n}$, i.e., it will be ${n+5-n-1 \choose 5-n} = {4 \choose 5-n}$. Hence the requeired probability will be $\frac{{4 \choose 5-n}}{6^n}$. For example, we can list the probabilities in the following table:

$\begin{align}
\quad \quad  n \quad & prob \\
\quad \quad  1 \quad & \frac{{4 \choose 4}}{6^1}=\frac{1}{6}\\
\quad \quad 2 \quad & \frac{{4 \choose 3}}{6^2}=\frac{1}{9}\\
\quad \quad 3 \quad & \frac{{4 \choose 2}}{6^3}=\frac{1}{36}\\
\quad \quad 4 \quad & \frac{{4 \choose 1}}{6^4}=\frac{1}{324}\\
\quad \quad 5 \quad & \frac{{4 \choose 0}}{6^5}=\frac{1}{7776}\\
\end{align}$

When $n>5$, the required probability is $0$.

Hence, we have the following required probability:
$\text{Prob}(\text{sum of $5$ in $n$ die throws}) = \left\{\begin{array}{lr}
        \frac{{4 \choose 5-n}}{6^n}, & \text{for } & 1\leq n \leq 5\\
        0, & \text{ } & \text{otherwise}
        \end{array}\right\}$


text:
\section{Linearly Ordered Space is Compact iff Complete}
Tags: Linearly Ordered Spaces, Compact Spaces, Complete Lattices

\begin{theorem}
Let $T = \struct {S, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Then $T$ is a [[Definition:Compact Topological Space|compact space]] {{iff}} it is [[Definition:Complete Ordered Set|complete]].
\end{theorem}

\begin{proof}
=== Necessary Condition ===
Let $T$ be a [[Definition:Compact Topological Space|compact space]].
Let $A \subseteq S$.
{{AimForCont}} $A$ has no [[Definition:Supremum of Set|supremum]].
Consider the [[Definition:Set|sets]]:
:$P_\alpha = \set {x \in S: x \prec \alpha}$ for $\alpha \in A$
:$B_\beta = \set {x \in S: \beta \prec x}$ for $\beta$ an [[Definition:Upper Bound of Set|upper bound]] of $A$.
We have that $P_\alpha$ and $B_\beta$ [[Definition:Cover of Set|cover]] $S$ but contain no [[Definition:Finite Subcover|finite subcover]].
Thus $T$ is not a [[Definition:Compact Topological Space|compact space]].
Similarly if $A$ has no [[Definition:Infimum of Set|infimum]].
It follows that $A$ has both a [[Definition:Supremum of Set|supremum]] and [[Definition:Infimum of Set|infimum]].
As $A$ is arbitrary, it follows that $S$ is a [[Definition:Complete Ordered Set|complete ordered set]].
{{qed}}
=== Sufficient Condition ===
Let $T$ be a [[Definition:Complete Ordered Set|complete ordered set]].
Let $\UU$ be an [[Definition:Open Cover|open cover]] of $S$.
Let $a$ be the [[Definition:Infimum of Set|infimum]] of $S$.
Let $B$ be the [[Definition:Set|set]] of the [[Definition:Element|elements]] $y \in S$ for which $\hointr a y$ have a [[Definition:Finite Cover|finite cover]] from elements of $\UU$.
Let $\alpha$ be the [[Definition:Supremum of Set|supremum]] of $B$.
If $\alpha \in U \in \UU$, then $U \subseteq B$.
So unless $\alpha$ is the [[Definition:Supremum of Set|supremum]] of $S$ itself, there exists an [[Definition:Open Interval|open interval]] $\openint x y \subseteq U$ such that $\alpha \in \openint x y$.
But we have that $\alpha$ be the [[Definition:Supremum of Set|supremum]] of $B$.
Hence $\openint x y = \O$.
But that means $y \in B$ which is impossible.
Thus $B = S$.
That is, the [[Definition:Set|set]] of [[Definition:Element|elements]] of $S$ which have a [[Definition:Finite Subcover|finite subcover]] of $\UU$ is $S$ itself.
Hence $T$ is a [[Definition:Compact Topological Space|compact space]] by definition.
{{qed}}
\end{proof}

config: wiki


TITLE: Prove Convergence in Probability of $\frac{x_n}{n}$ where $x_n$ = {# of times in Bernoulli RV that success is followed by failure}
QUESTION [1 upvotes]: We have IID Bernoulli trials with $p=1/3$. Let $x_n = ${number of times in the first n trials that a success is followed by a failure}. Prove that $\frac{x_n}{n}$ converges in probability to $\frac{2}{9}$.
This is the first probability convergence problem I have come across and I am struggling with where to begin. My first that was to use the weak law of large numbers. But I am not sure how to apply it here because technically the sequence of interest is not Bernoulli.
Can you get me started?

REPLY [3 votes]: Here is an approach you can follow (there are still computations to do):
Expectations:
You can compute $E[\frac{X_n}{n}]$ very easily via the indicator function approach that you mentioned.
Since $0 \leq \frac{X_n}{n} \leq 1$ for all $n$, if $\frac{X_n}{n}$ converges to some constant $c$ with probability 1 (or even in the weaker sense "in probability") then $E[\frac{X_n}{n}]$ must also converge to $c$.  You can justify this, for example, by the Lebesgue dominated convergence theorem and/or the "bounded convergence theorem."
DTMC
If we define a Discrete Time Markov Chain (DTMC) with state space $\{SS, SF, FS, FF\}$, where the state represents the history of the past two trials, then $\lim_{n\rightarrow\infty} \frac{X_n}{n}$ is the same as the fraction of time we are in state $SF$. We know by steady state theory that this fraction of time converges to some steady state value $\pi(SF)$ with probability 1 (regardless of the initial state).
So you can either compute the steady state of this 4-state DTMC directly, or you can use the fact that steady state exists, and so $\frac{X_n}{n}\rightarrow c$ with probability 1 for some constant $c$, then compute $c$ by $c=\lim_{n\rightarrow\infty} E[\frac{X_n}{n}]$.


TITLE: Does it mean " scalar field changes in a particular direction" by that the curl of the gradient of a scalar field is zero in practice?
QUESTION [0 upvotes]: What does mean that the curl of the  gradient of a scalar field is zero in practice?

My effort: the gradient means the direction at which the magnitude of vector change maximally and the curl says how it changes its direction. So  that means a scalar field changes in the same direction at all the points of level curve. How is that wrong?

REPLY [1 votes]: It is $0$ in practice as a result of being mathematically $0$ as well, in this case :).
Calculate: If $\varphi$ is a physical scalar field (Temperature, say. When represented as a real number).
Mathematically, we have a function $\varphi :R^3 \to R$ giving a real number for each point in $R^3$. It is not a vector field, yet.
It is a scalar field.
If we assume the 2nd derivatives exists and are continous (In classical theory of fields we enjoy this mathematical property), then we have mathematically: $$\frac{\partial^2 \varphi}{\partial x_i \partial x_j} = \frac{\partial^2 \varphi}{\partial x_j \partial x_i}$$ for any $i,j \in \{1,2,3\}$ being cartesian indices as a nice choice. It is called second derivative symmetric.
Consequently, $\nabla \times (\nabla \varphi)$ is of the form $$\frac{\partial^2 \varphi}{\partial x_i \partial x_j} - \frac{\partial^2 \varphi}{\partial x_j \partial x_i}=0$$ in each of $x,y,z$ directions. Write it down as a 'determinant' to see it more clearly.


TITLE: Modern Reference for Natural Deduction Proof that Logical Extension by Definition is Conservative
QUESTION [1 upvotes]: In logical systems (I'm specifically in the context of language of first order logic) it is possible to extend a theory by adding definitions to compactify/simplify formulas and theorems. It can be proven that, under appropriate (intuitive) conditions these extensions of the language are conservative in that they don't allow for the deduction of any essentially new formulas.
I would like to find a modern reference that demonstrates this proof in a natural deduction context (especially avoiding model theory). Kleene, Introduction to Metamathematics (1974) provides a proof of this that I believe satisfies what I am looking for. However, the reference is a bit old and the specific proof I'm looking for is scattered throughout the book. I'm hoping to find another, ideally more modern, reference which is readable and presents the proof clearly.

REPLY [2 votes]: Life is a little bit harder if you want to define functions. This is something that certainly ought to appear in textbooks, but seems to get brushed aside. There is some relevant information in Mendelson's Introduction to Mathematical Logic (see "definition, possible" in the index) or Boolos, Burgess and Jeffrey's Computation and Logic (see discussion of elimination of names and function symbols). If you only need to deal with functions that are uniquely determined by their arguments, then a syntactic proof of conservativity can be given but it involves contextual transformations. Dealing with loosely defined functions requires semantic methods or some chicanery.
[Aside: my versions of the cited books are ancient hence I haven't given page or section references as they will be out of date.]


TITLE: When a manifold is a subset of a sphere?
QUESTION [0 upvotes]: Let $M \subset \mathbb{R}^3$ a smooth connected 2-dimensional manifold.
I need to show that TFAE
1) for any smooth $\gamma : [0,1] \rightarrow M$  we have $\int_{\gamma} xdx + ydy + zdz = 0$
2) $M$ is a subset of a 2-dimensional sphere of some radius.
attempt:
denote $\omega=xdx+ydy+zdz$
Assume 1) and let $\gamma : [0,1] \rightarrow M$ so
$\int_{\gamma}\omega = 0$ implies $\int_0^1 \langle \gamma (t),\dot{\gamma} (t)\rangle dt =0$.
I know from basic ODE, that such curves form circles, but I don't know how to formalize it.
For the other direction I don't have a clue.

REPLY [4 votes]: Hint: Can you write $\omega=df$ for some $f$?
Edit: Let $f=r^2/2$; Stokes' theorem says $\int_\gamma\omega=\int_{\partial\gamma} f=f(\gamma(1))-f(\gamma(0))$, which is zero iff both endpoints are at the same radius. Then mumble something about connectedness and get the quantifiers in the right order, and you're done!


TITLE: Calculating displacement in 3d projectile motion with magnus force
QUESTION [0 upvotes]: I am trying to calculate the displacement in x-z plane in 3D projectile motion of a rotating object where the force generated due to the magnus effect is applicable. The magnus force constitutes of a cross product between the unit vectors of the angular velocity ($\vec{w}$) and linear velocity ($\vec{V}$).
$$
F_M=[\rho AC_LV^2(\hat{w} \times \hat{V})]/2
$$
In the above equation, angular velocity is a variable vector while all other values are constant. The direction of the force changes throughout the projectile motion due to the changing velocity and hence using the standard equations for projectile motion would not give an accurate value for the displacement in x-z plane. In such a case, how would you calculate the displacement in x-z plane?

REPLY [0 votes]: If the projectile and the gun barrel are well designed, the angular velocity $\boldsymbol\omega(t)$ and the velocity $\mathbf{v}(t)$ of the projectile are aligned in the same direction of the axis of the projectile itself, and thus
$\mathbf{F}_M = \dfrac{1}{2} \rho A V^2 C_L \underbrace{\boldsymbol{\hat{\omega}} \times \mathbf{\hat{v}}}_{=\mathbf{0}} = \mathbf{0}$


TITLE: Trying to show a measure is inner regular on open sets
QUESTION [1 upvotes]: Background:
Let $X$ be a locally compact Hausdorff space, $C_{00}$ the collection of continuous functions on $X$ with relatively compact support.  Then let $I \in C_{00}^*$ such that $f \in C_{00}$ nonnegative implies $I(f)\geq 0$.
Define for nonnegative lower semi continuous $g$, $$J(g) = sup(I(f): f\in C_{00}, 0\leq f \leq g)$$.
For general nonnegative $g$,  $$K(g) = inf(J(f) : f\text{ lower semi continuous, }g \leq f)$$

My question:
Let $U \subset X$ be open.
Let $b < K(\chi_U)$.  Next line chooses a $f\in C_{00}, f\geq 0$ such that $I(f) > b$ and $f \leq \chi_U$.  I cannot verify that this is possible from the definitions above. How can it be done?

REPLY [1 votes]: $\chi_U$ is lower semicontinuous: Let $x \in X$ and $\epsilon > 0$. If $x \in U$, then $\chi_U$ is continuous at $x$ (as constant on the neighbourhood $U$ of $x$), so let $x \not\in U$. Then for each $y \in X$, we have $\chi_U(y) \ge \chi_U(x) - \epsilon = -\epsilon$. So $\chi_U$ is lower semicontinuous on $X \setminus U$ also.
Hence $K(\chi_U) = J(\chi_U)$ and the existence of $f$ follows from the definition of $J$.


text:
\section{Examples: Geometry and values} \label{sec:examples}

In this section we explain how to evaluate top degree classes, we
provide examples illustrating the geometry at infinity of the
compactification $\LMS$ and examples of our formulas for the normal
bundles, the Chern polynomials and the Euler characteristic.

\subsection{Evaluation of top $\xi$-powers} \label{sec:topxi}

\par
\medskip
First of all we explain how to evaluate the expression in
Theoreom~\ref{intro:ECformula}, see \cite{CoMoZadiffstrata} for many
algorithmic details. We only need to explain how to evaluate
$\int_{\ol{B}} \xi^{d}$, i.e. top powers of $\xi$ on generalized strata.
\par
Suppose that $B = \bP\Omega\cM_{g,1}(2g-2)$ is a stratum parametrizing
connected surfaces with a single zero. Then the generating series of
top $\xi$-powers is given by a simple power series inversion that
arises in the computation of Masur-Veech volumes, see \cite{SauvagetMinimal}
and \cite[Theorem~3.1]{CMSZ}, and Table~\ref{cap:xitop} for some values.
\par
Suppose that $B = \bP\Omega\cM_{g,n}(\mu)$ is a stratum parametrizing
connected surfaces of holomorphic type, i.e., with all $m_i \geq 0$
and with $n \geq 2$. Then $\int_B \xi^d = 0$ by
\cite[Proposition~3.3]{SauvagetMinimal}.
\par
It remains to explain how to  evaluate $\int_{\ol{B}} \xi^{d}$ top powers of $\xi$ on meromorphic generalized strata. First of all, we write $\xi$ with the help of Proposition~\ref{prop:Adrienrel} as a $\psi$-class and boundary strata.
The product of such objects, which are standard generators as in~\ref{eq:addgenR} of
the tautological ring, can be rewritten as a sum of standard additive generators via the algorithm explained in the proof of \autoref{thm:addgen}, more specifically in the part in which we show that $R_{fg}^\bullet(\ol{B}) = R^\bullet(\ol{B})$ is a ring. Now that we have rewritten $\xi^{d}$ in terms of standard additive generators, by \autoref{lem:evaluation}, it only remains to explain how to evaluate a top-dimensional standard generator, i.e. the top power of a $\psi$-class, on
a generalized stratum~$\ol{B}$. Since $\psi$-classes are pulled back from $ \barmoduli[g,n]$, we can use a push-pull argument and express
\[\int_{\ol{B}} \psi_i^d=\int_{ \barmoduli[g,n]} \pi_*([\ol{B}])\psi_i^d\]
where $\pi:\ol{B}\to  \barmoduli[g,n]$ is the forgetful morphism and where we used as always the abuse of notation $\psi_i=\pi^*(\psi_i)$.
\par
If we can express the class $\pi_*([\ol{B}])$ in terms of the standard generators of $\barmoduli[g,n]$, we can use the sage package {\tt admcycles} in order to obtain a number.
\par
If $\ol{B}$ is a stratum parameterizing meromorphic differentials on connected surfaces without residue conditions, the class $\pi_*([\ol{B}])$  was computed in \cite{sauvagetstrata}
and \cite{BHPSS}, and the algorithmic task can be performed again by the sage
package {\tt admcycles}, which implements the algorithm based on
the formula in~\cite{schmittDimTh} and~\cite{BHPSS}.
\par
If the stratum $\ol{B}$ is more general parametrizing differentials on disconnected surfaces and with residue conditions, we first of all use repetitively Proposition~\ref{prop:AdrienR} to write the class of $\ol{B}$ into the associated stratum without residue conditions in terms of additive generators of the stratum with not conditions.
We then reduced to the computation of the class $\pi_*([\ol{B}])$ in the case that $\ol{B}$ has no
more residue conditions, but is potentially disconnected.
If~$\ol{B}$  is  disconnected then actually $\pi_*([\ol{B}])$ is zero. In fact, since we can scale the differentials on the components independently,
the fiber dimension to a product of $\barmoduli[g_i,n_i]$ is positive and by definition of push-forward we get the zero class.
\par
\begin{figure}[h]
	$$ \begin{array}{|c|c|c|c|c|c|c|}
	\hline  &&&&&& \\ [-\halfbls]
	\mu  & (0) & (2) & (4) & (6) & (0,0,-2) & (2,-2)  \\
	[-\halfbls] &&&&&& \\
	\hline &&&&&& \\ [-\halfbls]
	\int_{\ol{B}} \xi^{\dim(B)}
	& \frac{1}{24} &  -\frac{1}{640} & \frac{-305}{580608},
	& -\frac{87983}{199065600} & 1 &  -\frac{1}{8} \\
	[-\halfbls] &&&&&& \\
	\hline  &&&&&& \\ [-\halfbls]
	\mu  & (1,1,-2) & (4,-2) & (3,1,-2) & (2,1,-3) & (5,-3) & (8,-2,-2,-2)\\
	[-\halfbls] &&&&&& \\
	\hline &&&&&& \\ [-\halfbls]
	\int_{\ol{B}} \xi^{\dim(B)}
	& 0  & - \frac{-23}{1152},
	& 0 & \frac{5}{8} &  -\frac{21}{20} & -\frac{4527}{32}     \\
	[-\halfbls] &&&&&& \\
	\hline
	\end{array}
	$$
	\captionof{table}[foo2]{Integrals of top $\xi$-powers for some
		connected strata}
	\label{cap:xitop}
\end{figure}
For the subsequent examples we present some cases where we can directly
evaluate the top $\xi$-power for meromorphic strata in genus~$0$ and
genus~$1$.
\par
\begin{prop}\label{prop:topxieasyexample}
The integrals of the top $\xi$-power are given ($a_i,k \geq 0$)
\bas
&\text{for} \quad B \= \bP\Omega_{0,n+1}(-2-\sum_{i=1}^{n} a_i,a_1,...,a_n) && \text{by} \quad
\int_B \xi_B^{n-2} \= (-1-\sum_{i=1}^n a_i)^{n-2}\,, \\
&\text{for} \quad B \=\bP\Omega_{1,2}(-k,k) && \text{by} \quad
\int_B \xi_B \= -\frac{(k-1)(k^2-1)}{24}\,, \\
&\text{for} \quad B \= \bP\Omega_{1,3}(-k-1,1,k)&& \text{by}\quad
\int_B \xi^2_B\ =\frac{(k^4-1)}{24}\,.
\eas

\end{prop}
\begin{proof}
	The first statement follows easily from Proposition~\ref{prop:Adrienrel}, which in this case implies $\xi=(-1-\sum_{i=1}^{n}a_i)\psi_1$. Indeed there cannot be any divisors which have the pole on lower level. Hence
	\[\int_B \xi^{n-2}=(-1-\sum_{i=1}^{n}a_i)^{n-2}\int_{\cM_{0,n+1}}\psi_1^{n-2}=(-1-\sum_{i=1}^{n}a_i)^{n-2}.\]
\par
The second statement follows immediately as the previous one, since again there cannot be divisors where the pole is on lower level. Hence
\bes \int_B\xi
\= -(k-1)\int_{\cM_{1,2}}\pi_*([B])\psi_1=-\frac{(k-1)(k^2-1)}{24}\,,
\ees
where we used \cite[Proposition~3.1]{ChenEED} for the computation of $\pi_*([B])$.
\par
For the proof of the last statement, notice that there can only be one non-horizontal divisor $D_3$ which has the pole on lower level (see \autoref{sec:exk1} for the full boundary description). Using Proposition~\ref{prop:Adrienrel} we find then $\xi = -k\psi_1 -D_3$, which yields
\bas
\int_{\overline{B}} \xi^2
&\= - \int_{\overline{B}} \xi (D_3 + k\psi_1)
\=  \Bigl(- 1/24 + \int_{\overline{B}} k\psi_1(D_3 + k\psi_1)\Bigr) \\
&\=\Bigl(-1/24 + k^2\int_{\cM_{1,3}}\pi_*([B]) \cdot\psi_1^2 \Bigr)=
(k^4 -1)/24
\eas
again by the computation in \cite[Proposition~3.1]{ChenEED} of the class of the class $\pi_*(B)$ of the stratum in $\cM_{1,3}$.
\end{proof}

\subsection{The minimal stratum $\bP\Omega\cM_{2,1}(2)$}

This stratum is small enough so that we can show all the level graphs,
including those with horizontal nodes and their adjacency in
Figure~\ref{cap:H2}.
\begin{figure}

\input{OmegaM_22_20200422}
\caption{Level graphs appearing in the boundary of $\Omega\cM_{2,1}(2)$. Graphs
corresponding to components of the same dimension are in the same row (divisors
in the first row, points in the bottom row). The lines connecting the graphs
symbolize degeneration.} \label{cap:H2}
\end{figure}
The picture shows the dual graphs of stable curves in the boundary
of this stratum, the top level is on top of each graph. The number in
the vertex denotes the genus, a black dot corresponds to genus zero.
The numbers associated to the legs are the orders of zero. In this
stratum, all interior edges have enhancement~$\kappa_e =1$, so the
discussion of prong-matchings is void here. There are only three graphs
without horizontal nodes, in fact $|\LG_1(B)| = 2 $ and $|\LG_2(B)| = 1$,
where $B = \bP\Omega\cM_{2,1}(2)$ as usual. Taking into also account the
entire stratum and the stack structure of the banana graphs, and using
the values of top $\xi$-powers from Section~\ref{sec:topxi}, we get
\bes
(-1)^3 \cdot \chi(B) \= 4 \cdot \frac{-1}{640} + 0 + 2 \cdot \frac{1}{24}
\cdot \frac{-1}{8} + 2 \cdot \frac{1}{2} \cdot \frac{1}{24} \cdot 1 \cdot 1
\= \frac{1}{40}
\ees
as in the table in the introduction, in accordance with the fact that
this stratum is a $6$-fold unramified cover of~$\moduli[2]$
and $\chi(\moduli[2]) = -\tfrac{1}{240}$.

\subsection{The stratum $\bP\Omega\cM_{1,3}(-k-1,1,k)$}
\label{sec:exk1}

This example illustrates the quotient stack structure at the boundary of
the smooth compactification that result from prong-matchings, i.e., from points
with $\Tw[\Gamma] \neq \sTw[\Gamma]$. We have chosen a genus-one stratum
with a simple zero, since the projection to $\moduli[1,2]$ provides an
alternative way to compute all invariants in this case. We label the
points $z_1$ (pole), $z_2$ (simple zero) and $z_3$. The boundary divisors
here are $D_{\text{h}}$ and five more types of divisors, namely there are the
divisors
\input{bd_stratumk1mrest}
where  $a,b \geq 1$ and $a+b = k+1$. Here $D_{1,a} = D_{1,k+1-a}$
and if $k$ is odd, the middle divisor $D_{1,(k+1)/2}$ has an
$\bZ/2$-involution. Moreover, there are the divisors
\input{bd_stratumk1mD4D5}
where $a'\in\{1,\dots, k-1\}$ and $b'=k-a'$. Again, $D_{5,a'} = D_{5,k-a'}$
with an involution on $D_{5,k/2}$ if $k$ is even. The local exponents
are
\bes
\ell_{1,a} \= \lcm(a,k+1-a), \quad \ell_2 = k+2, \quad   \ell_3 = 1, \quad
\ell_4\=k-1, \quad  \ell_{5,a'} = \lcm(a',k-a')\,
\ees
and the dimensions of the top level components are
\bes
N_1^\top =1, \quad N_2^\top =2, \quad N_3^\top =2, \quad N_4^\top =1,
\quad N_5^\top =2\,.
\ees
We abbreviate

$D_1 \= \frac12 \,\sum_{a=1}^{k} D_{1,a}$ and
$D_5 \= \frac12 \,\sum_{a'=1}^{k-1} D_{5,a'}$.
\par
\medskip
\paragraph{\bf The local geometry of the boundary divisors} We give
a summary of the boundary points and intersection behaviour of the
boundary divisors listed above. We start with the boundary divisors that
map to the interior of $\moduli[1,2]$ under the map to $\moduli[1,1]$
forgetting the second point. These are represented by the thin lines in
Figure~\ref{fig:bd_stratumH1kmk-1}, while thick lines are mapped to the point
at infinity of $\moduli[1,1]$.
\begin{figure}
	\includegraphics[scale=1.4]{bd_stratumH1kmk-1}
\caption{The intersection behavior of the boundary in the stratum
$\bP\Omega\cM_{1,3}(-4,1,5)$. The figure has to be considered as quotient by
the elliptic involution that interchanges $D_{1,1}$ with~$D_{1,4}$ and
$D_{5,1}$ with~$D_{5,3}$ etc.
	} \label{fig:bd_stratumH1kmk-1}
\end{figure}
The divisor $D_3$ is simply a copy of the modular curve, intersecting
once $D_{\text{h}}$.
\par
The divisor $D_2$ minus its intersection with other boundary divisors is
the union of the modular curves $X_1(d) = \bH/\Gamma_1(d)$ for all
divisors~$d>1$ of~$k+1$. The only intersections with other boundary divisors
are $\left\lceil (k+1)/2\right \rceil -1 $ intersection points
with $D_{\text{h}}$ and $\gcd(a,b)$-points  with $D_{1,a}$.
\par
The divisor $D_4$ minus its intersection with other boundary divisors is
the union of the modular curves $X_1(d) = \bH/\Gamma_1(d)$ for all
divisors~$d>1$ of~$k$. The only intersections with other boundary
divisors are $\left\lceil (k)/2\right \rceil -1 $ intersection points
with $D_{\text{h}}$ and $\gcd(a',b')$-points  with $D_{5,a'}$.
\par
The curves $D_{1,a}$ and $D_{5,a'}$ form the exceptional divisor when
realizing the level compactification $\proj\LMS[-k-1,1,k][1,3]$ as
a blowup of $\barmoduli[1,2]$ in the node of the forgetful map to
$\barmoduli[1,1]$. Without prong-matchings the curve~$D_{1,a}$
were just an $\overline{M}_{0,4}$ (with a stack structure of an involution
if $a = (k+1)/2$). The three boundary points correspond to the intersection
with $D_{5,a-1}$ and $D_{5,a}$ (respectively with $D_{\text{h}}$ and $D_{5,a}$
if $a=1$), and with $D_2$. By the formulas in Section~\ref{sec:PM}, at
the generic point of $D_{1,a}$ (and also near the intersection with $D_2$)
there are $\gcd(a,b)$ prong-matching equivalence classes. At the
intersections with $D_5$ there is just one prong-matching
equivalence class. This implies that each $D_{1,a}$ is a $\gcd(a,b)$-fold cover
of $\barmoduli[0,4]$, totally ramified over the two points of intersection
with~$D_5$.
\par
Similarly, the divisor $D_5$ is a $\gcd(a',b')$-fold cover of
$\overline{M}_{0,4}$, totally ramified over the two points that correspond
to the intersections with~$D_1$. We compute the normal bundles of the divisor
using the special geometry of this example, independently of Theorem~\ref{thm:nb}.
\par
\begin{prop} 	\label{prop:selfintHk1mk}
The self-intersection number of $D_{1,a}$ is
\bes
D_{1,a}^2 \= - \delta_a^{k+1}\, \cdot k g_{1,a}/\ell_{1,a}
\quad \text{where} \quad g_{1,a} \= \gcd(a,b)
\ees
and where $\delta^{k+1}_{a}=1/2$ if $a = (k+1)/2$
and $\delta^{k+1}_{a}=1$ otherwise. The self-intersection number of $D_{5,a}'$ is
  \bes
D_{5,a'}^2 \= - \delta_{a'}^{k}\, \cdot (k+1)\, g_{5,a'}/\ell_{5,a'}
\quad \text{where} \quad g_{5,a'} = \gcd(a',b')\,.
\ees
\end{prop}
\par
\begin{proof}
We consider the fibration $\pi:\proj\LMS[-k-1,1,k][1,3] \to \barmoduli[1,1]$
obtained from forgetting the last two marked points and take a smooth chart
of the quotient stack near the image of the curves $D_{1,a}$ and~$D_{5,a'}$. From
the intersection discussion above we deduce that the fiber over~$\infty$
in $\barmoduli[1,1]$ consists of a ring of rational curves intersecting in the
order
\bes
D_{\text{h}} -D_{1,1}-D_{5,1}-D_{1,2}-D_{5,2}-\cdots- D_{1,k-1} - D_{5,k} - D_{1,k+1}- D_{\text{h}}\,,
\ees
see again Figure~\ref{fig:bd_stratumH1kmk-1}.
We claim that the multiplicity of $D_{1,a}$ in the fiber $F = \pi^{-1}(\infty)$
is equal to $(k+1)/\gcd(a,k+1)$.  This can be deduced from the fact that $\pi|_{D_2}$
is a cover of degree $(k+1)^2 -1$ and from the order of the cusp stabilizers (see
\cite[Section~3.8]{DiaShu}, in particular the explanation around Figure~3.2) since
$D_2$ and $D_{1,a}$ intersect transversally in $\proj\LMS[-k-1,1,k][1,3]$.
Similarly, the multiplicity of $D_{5,a'}$ in this fiber is $k/\gcd(a',k)$.
Using the orbifold degree of the intersection points given in~\eqref{eq:eGamma}
and $D_{1,a}\cdot F = 0$ we find with $a' = a-1$ and $b'=b-1$ that
\bas
D_{1,a}^2 &\= - \frac{\gcd(a,b)}{k+1} \cdot \frac{abk}{\ell_{1,a}} \cdot
\Bigl(\frac{a'}{\ell_{5,a'} \gcd(a',k-a'))} +  \frac{b'}{\ell_{5,b'} \gcd(b',k-b'))}\Bigr) \\
&\= - \frac{\gcd(a,b)}{k+1} \cdot \frac{abk}{\ell_{1,a}} \cdot \frac{k+1}{ab}
\= - k \cdot g_{1,a}/\ell_{1,a}\,.
\eas
The proof of $D_{5,a'}$ is similar.
\end{proof}
\autoref{prop:selfintHk1mk} agrees with \autoref{thm:intro:nb}. Indeed,
since the dimension of the top (resp. bottom) level stratum of $D_{1,a}$
(resp. $D_{5,a'}$) is zero dimensional, we compute
\bas
D_{1,a}^2&\=\c_1(\cN_{D_{1,a}})
\=\frac{1}{\ell_{1,a}}\left(-\xi_{D_{1,a}}^\top -\cL_{D_{1,a}}^\top+\xi_{D_{1,a}}^\bot\right)\\
&\=\frac{K_{1,a}}{\ell_{1,a}^2\Aut(D_{1,a})}\xi_{B_{1,a}^\bot}=\frac{g_{1,a}}{\ell_{1,a}\Aut(D_{1,a})}\cdot (-k)
\eas
where in the last two equalities we used \autoref{lem:evaluation} about evaluating top classes and the computation of top powers of $\xi$ which can be done analogously as in the first case of \autoref{prop:topxieasyexample}. Similarly we also get
\bas
D_{5,a'}^2&\=\c_1(\cN_{D_{5,a'}})
\=\frac{1}{\ell_{5,a'}}\left(-\xi_{D_{5,a'}}^\top+\xi_{D_{5,a'}}^\bot -\cL_{D_{5,a'}}^\top\right)\\
&\=-\frac{K_{5,a'}}{\Aut(D_{5,a'})\ell_{5,a'}^2}\xi_{B_{5,a'}^\top}-\frac{1}{\ell_{5,a'}}\cL_{D_{5,a'}}^\top\\
&\=\frac{g_{5,a'}}{\Aut(D_{5,a'})\ell_{5,a'}}\cdot (-k)-\frac{1}{{\ell_{5,a'}}}\left([D_{5,a'}]\cdot([D_4]+[D_{1,a'}]+[D_{1,a'+1}])\right)\\
&\=\frac{g_{5,a'}}{\Aut(D_{5,a'})\ell_{5,a'}}\cdot (-k-1).
\eas
\smallskip
\paragraph{\bf The Euler characteristic} We give two ways proof of the
following fact.
\par
\begin{prop} The  moduli space $\bP\Omega\cM_{1,3}(-k-1,1,k)$
has Euler characteristic equal to $k(k+1)/6$.
\end{prop}
\par
\begin{proof} The first proof uses the description of $B =\bP\Omega\cM_{1,3}(-k-1,1,k)$
as the complement of $D_2$ and $D_4$ in $\moduli[1,2]$. By the above description
of $D_2$ and $D_4$ we need to compute

\bes
\sum_{\substack{d|k\\ d\not = k}} \chi(X_1(k/d))
\= \chi(\cM_{1,1}) \sum_{\substack{d|k\\ d\not = k}}[\SL_2(\bZ):\Gamma_1(k/d)]
\=-\frac{k^2-1}{12}\,,
\ees
which holds, since the rightmost sum counts the number of non-zero $k$-torsion points
in an elliptic curve. Together with $\chi(\moduli[1,2]) = -1/12$ implies the claim.
\par
The second proof evaluates Theorem~\ref{intro:ECformula}, given in the
surface case concretely by
\[
\chi(B) \= 3\xi_B^2  +\sum_{\Gamma \in \twolev }\frac{K_\Gamma \cdot N_\Gamma^\top}{|\Aut(\Gamma)|}
\left(\int_{B_\Gamma^\top}\xi_{B_\Gamma^\top}+\int_{B_\Gamma^\bot}\xi_{B_\Gamma^\bot}\right)+\sum_{\Delta \in \LG_2(B) }
\ell_{\delta_0(\Delta)}\ell_{\delta_1(\Delta)}[D_{\Delta}]\,.
\]
Using the third statement of \autoref{prop:topxieasyexample} we find
\bas
3 \int_{\overline{B}} \xi^2\=3(k^4 -1)/24.
\eas
For the divisors $D_{1,a}$ and~$D_4$
the contribution from $\xi_{B_\Gamma^\top}$ is zero and that of $\xi_{B_\Gamma^\bot}$
is non-zero, while for~$D_2$, $D_3$ and $D_{5,a'}$ the converse holds.
We evaluate in detail the contribution of that last divisor type. Its top
levels are $D_{5,a'}^\top=\bP\Omega_{0,4}(1,a'-1,b'-1,-k-1)$. Using
again \autoref{prop:topxieasyexample} we get
\bes
\sum_{\Gamma = D_{5,a} \atop a=1,\ldots k/2} \frac{K_\Gamma \cdot N_\Gamma^\top}{|\Aut(\Gamma)|}
\int_{B_\Gamma^\top}\xi_{B_\Gamma^\top}
\= \frac{1}{2}\sum_{a'=1}^{k-1} 2a'(k-a')
\cdot(-k)
\=\frac{-k^2(k^2-1)}{6}\,
\ees
Similar computations using again \autoref{prop:topxieasyexample} yield
\bas
\sum_{\Gamma = D_{1,a} \atop a=1,\ldots (k+1)/2} \frac{K_\Gamma N_\Gamma^\top}{|\Aut(\Gamma)|}
\int\xi_{B_\Gamma^\top} \= -k^2\frac{k^2+3k+2}{12}\,, \quad &2\ell_{D_3} \int\xi^\top_{D_3}
\= \frac{1}{12} \\
2\ell_{D_2} \int\xi^\top_{D_2} \= -2k(k+2)\frac{(k+1)^2-1}{24}\,,
\quad &\ell_{D_4} \int\xi^\top_{D_4} \= -(k-1)^2\frac{k^2-1}{24}
\eas
Using the evaluation result of \autoref{lem:evaluation} we finally get
\bes
\sum_{\Delta \in \LG_2(B) }\ell_{\delta_0(\Delta)}\ell_{\delta_1(\Delta)}[D_{\Delta}]
\= k(k+1)\frac{k^2+k+1}4\,.
\ees
Adding these contributions gives the claim.
\end{proof}

\subsection{Hyperelliptic components} \label{sec:hypstrata}

We recall from \cite{kozo1} that strata of holomorphic Abelian differentials
have up to three connected components, distinguished by the parity of
the spin structure and possibly hyperelliptic components. The strata
$\omoduli[g,1](2g-2)$ and $\omoduli[g,2](g-1,g-1)$ have hyperelliptic
components. Their Euler characteristics are easy to compute.
\par
\begin{prop} \label{prop:HEEuler}
The Euler characteristic of the hyperelliptic components
are
\bas
\chi(\bP\omoduli[g,1](2g-2)^{\hyp}) &\= \frac{-1}{4g(2g+1)}
\quad \text{and} \quad \\
\chi(\bP\omoduli[g,2](g-1,g-1)^\hyp)
&\= \frac{1}{(2g+1)(2g+2)}\,.
\eas
\end{prop}
\par
\begin{proof} In the first case the surfaces are double covers of surfaces
the stratum $\cQ_0(-1^{2g+1},2g-3)$ of quadratic differentials with unnumbered
points, which is isomorphic to $\moduli[0,2g+2]/S_{2g+1}$. The claim follows
from $\chi(\moduli[0,n+3]) = (-1)^n \cdot n!$, taking into account the global
hyperelliptic involution on the stratum.
\par
Double covers of surfaces the stratum $\cQ_0(-1^{2g+2},2g-2)$ of quadratic
differentials with unnumbered points produce the Abelian differentials second
case, and this stratum is isomorphic to $\moduli[0,2g+3]/S_{2g+2}$.
The extra factors~$2$ from labelling the zeros of order~$g-1$ and~$1/2$
from the global hyperelliptic involution cancel each other.
\end{proof}

\subsection{Meromorphic strata and cross-checks} \label{sec:merocross}

In this section we provide in Table~\ref{cap:EulerMero} some Euler
characteristics for meromorphic strata. We abbreviate $\chi(\mu)
= \chi(\bP\omoduli[g,n](\mu))$.
\begin{figure}[h]
$$ \begin{array}{|c|c|c|c|c|c|}
\hline  &&&&& \\ [-\halfbls]
\mu  &  (4,-2) & (3,1,-2) & (2,2,-2) & (2,1,1,-2) & (1^4,-2) \\
[-\halfbls] &&&&& \\
\hline &&&&& \\ [-\halfbls]
\chi(B)
& -\frac{19}{24}  &  \frac{28}{15} & \frac{17}{10} & -6  &  26 \\
    [-\halfbls] &&&&& \\
\hline &&&&& \\
\mu  &  (4,-1,-1) & (3,1,-1,-1) & (2,2,-1,-1) & (2,1^2,-1,-1) & (1^4,-1,-1) \\
[-\halfbls] &&&&& \\
\hline &&&&& \\ [-\halfbls]
\chi(B)
& -\frac{8}{5}  &  -4 & -4 & 14  &  -63 \\
    [-\halfbls] &&&&& \\
\hline

\end{array}
$$
\captionof{table}[foo2]{Euler characteristics of some meromorphic strata}
\label{cap:EulerMero}
\end{figure}
Moreover we provide several cross-checks for our values. First note
that the union of the strata of types $(4), (3,1), (2,2), (2,1,1)$ and
$(1^4)$ glue together to the projectivized Hodge bundle over $\barmoduli[3]$,
if all of them are taken with unmarked zeros. In fact, we read off
from Table~\ref{cap:EulerHolo} that
\bes
\chi(4) + \chi(3,1) + \frac12 \chi(2,2) + \frac12 \chi(2,1,1) +
\frac{1}{4!}\chi(1^4) \= \frac3{1008} \= \chi(\bP^2) \cdot \chi(\moduli[3])\,.
\ees
The value $\chi(4) = -55/504$ can also be retrieved from
Proposition~\ref{prop:HEEuler} and the computations of Bergvall
\cite[Table~4]{bergvall}, that gives the
cohomology of the stratum with odd spin structure $\omoduli[g,1](4)^\odd$
with $\bZ/2$-level structure. Computing the alternating sum weighted by
dimension gives~$-141120$. Since $|\Sp(6,\bZ)| = 1451520$ this checks with
\bes
\frac{-55}{504} \= \chi(4) \= \chi(4^\hyp) + \chi(4^\odd)
\= \frac{-1}{84} + \frac{-141120}{1451520}\,.
\ees
(A few other strata in $g=3$ might be cross-checked with table in
\cite{bergvall}, but one has to take into account that Bergvall glosses over
the existence of hyperelliptic curves in non-hyperelliptic strata.)
\par
Another cross-check is the Hodge bundle twisted by twice the universal
section over $\moduli[2,1]$. It decomposes into the unordered strata
$(4,-2), (3,1,-2), (2,2,-2)$, $(2,1,1,-2), (1^4,-2), (2,0), (1,1,0), (2)$
and the ordered stratum $(1,1)$, since the simple zero at the unique
marked point is distinguished. Note that $\chi(2,0) = 3 \chi(2)$ and
$\chi(1,1,0)=3 \chi(1,1)$. We can now add up the contributions
listed in Table~\ref{cap:EulerHolo} and Table~\ref{cap:EulerMero}
and find that the sum equals $ \frac{1}{40}
= \chi(\bP^2) \cdot \chi(\moduli[2,1])$. A similar cross-check can
be made for the Hodge bundle over $\moduli[2,2]$ twisted by every
section once, using the second row of Table~\ref{cap:EulerMero}.
\par

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