text:
\begin{document}

\title[Invariant tubular neighborhoods and Yang--Mills theory]{Invariant tubular neighborhoods in infinite-dimensional Riemannian geometry, with applications to Yang--Mills theory}

\author{Daniel Ramras}
\address{Department of Mathematical Sciences,
New Mexico State University, Las Cruces, New Mexico 88003-8001, USA}
\email{ramras@nmsu.edu}

\thanks{This work was partially supported by
    NSF grants DMS-0353640 (RTG), DMS-0804553, and DMS-0968766.}

\subjclass[2000]{Primary 58B20, Secondary 53C07, 58D27}

\begin{abstract}
We show that if $G$ is an arbitrary group acting isometrically on an a (possibly infinite dimensional) Riemannian manifold, then every $G$--invariant submanifold with locally trivial normal bundle admits a $G$--invariant total tubular neighborhood.  These results apply, in particular, to the Morse strata of the Yang-Mills functional over a closed surface.  The resulting neighborhoods play an important role in calculations of gauge-equivariant cohomology for moduli spaces of flat connections over non-orientable surfaces (Baird \cite{Baird-rank3}, Ho--Liu \cite{Ho-Liu-anti-perfect}, Ho--Liu--Ramras \cite{Ho-Liu-Ramras}).
\end{abstract}

\maketitle{}

\section{Introduction}
Consider a (possibly infinite-dimensional) smooth manifold $X$ acted on by a group $G$, and a smooth, $G$--invariant submanifold $Y\subset X$, which need not be closed.  The question we address is: does $Y$ have a $G$--invariant tubular neighborhood inside $X$?  For finite-dimensional manifolds and compact groups, it is well-known that the answer is yes (see Bredon \cite{Bredon-transf}, for example).  The proof for compact groups involves averaging over the group, and the problem appears to be much more difficult for non-compact groups.  A number of authors have considered specific infinite-dimensional versions of this problem: Ebin \cite{Ebin} constructed invariant tubular neighborhoods for orbits of the diffeomorphism group acting on the space of Riemannian metrics on a manifold; Kondracki and Rogulski \cite{KR} handled the case of orbits of the gauge group acting on the space of connections on a principal bundle; and recently A. Teleman \cite{Teleman-harmonic} gave an explicit construction
of a gauge-invariant tubular neighborhood for the space of connections $A$ on a Euclidean bundle satisfying
$\mathrm{dim} (\ker (d_A)) = 1$.

We will show that invariant tubular neighborhoods exist whenever $X$ is a Riemannian manifold, $G$ acts by isometries, and the orthogonal complement of $TY$ in $TX|_Y$ is a smooth subbundle (Proposition~\ref{tubular} and Theorem \ref{total-tubular}).  This last condition is automatically satisfied in the finite dimensional setting, but it is important to note that it does not come for free in the infinite-dimensional setting.
Our motivation for this work comes from Yang--Mills theory, and we will show that these conditions are all satisfied in the case of the Morse strata of the Yang-Mills functional over a (possibly non-orientable) surface.

Mathematical applications of Yang--Mills theory began with Atiyah and Bott \cite{A-B}, who
established a recursive formula for the gauge-equivariant cohomology of
the space of semi-stable holomorphic structures on a Hermitian bundle $E$ over a
Riemann surface.  They achieved this by treating the Yang--Mills functional as a gauge-equivariant Morse function on the infinite-dimensional space $\A(E)$ of connections on $E$.
In ordinary Morse theory on a manifold $M$, each critical point corresponds to a cell in $M$, whose dimension is the index of that critical point.  The zero-dimensional (co)homology of the critical point  contributes to the (co)homology of $M$, with a shift in dimension corresponding to the index.  Atiyah and Bott showed that the Morse strata of the Yang--Mills functional can be ordered, starting with the central (or semi-stable) stratum, so that each initial segment of the ordering forms an open set in $\A(E)$ (see Ramras \cite{Ramras-YM} for a detailed discussion).  Adding the strata $C_1, C_2, \ldots$ in order, the Thom isomorphism theorem shows that each stratum contributes its cohomology with a shift in dimension, given by the codimension of the stratum: letting $\C_n = \cup_{i = 1}^n C_i$, we have a long exact sequence
\begin{equation}\label{Thom}
 \cdots \to H^*_{\G(E)} (C_n, C_{n-1}) \isom H^{*-\mathrm{codim} (\C_{n})}_{\G(E)} \left(\C_{n}\right) \to H^*_{\G(E)} (\C_n)
\to H^*_{\G(E)} (\C_{n-1}) \to \cdots
\end{equation}
As discussed in Section~\ref{YM}, the $\G(E)$--invariant tubular neighborhoods constructed in this paper provide one method for establishing these Thom isomorphisms.  (Atiyah and Bott proved that the Yang--Mills functional is ``equivariantly perfect," meaning that the boundary maps in the sequence (\ref{Thom}) are all zero.  This leads to the recursive formula mentioned above.)

Ho and Liu \cite{Ho-Liu-non-orient} extended Yang--Mills theory to bundles $E$ over non-orientable surfaces, where the Yang--Mills
stratification is no longer ``equivariantly perfect."
However, the Thom isomorphisms (\ref{Thom}), together
with the orientability result of Ho, Liu, and Ramras \cite{Ho-Liu-Ramras},
establish equivariant Morse \emph{inequalities} for the space of flat connections on $E$ (see the introduction to \cite{Ho-Liu-Ramras}).
Moreover, these Thom isomorphisms
have been used by Tom Baird \cite{Baird-rank3} to produce
formulas for the $U(3)$--equivariant Poincar\'{e} series of $\Hom(\pi_1
\Sigma, U(3))$ (where $\Sigma$ is a non-orientable surface),
establishing the ``anti-perfection'' conjecture of Ho and Liu \cite{Ho-Liu-anti-perfect}.

We note that over a Riemann surface, an alternate approach to the Thom isomorphism in (\ref{Thom}) would be to use the fact that after modding out the (based, complex) gauge group, one obtains a smooth algebraic variety, and results of Shatz \cite[Section 4]{Shatz} show that the image of each stratum is a smooth subvariety.  This algebraic approach involves various technicalities, which we will not attempt to resolve here.

This paper is organized as follows.  In Section~\ref{inv}, we present our construction of invariant tubular neighborhoods, and in Section~\ref{YM} we use results of Daskalopoulos to show that our construction applies to the Morse strata of the Yang--Mills functional over a surface.  In the last section, we compare our construction of tubular neighborhoods to other constructions in the literature.

\vspace{.1in}

\noindent {\bf Acknowledgements:} I thank Robert Lipshitz and Chiu-Chu Melissa Liu for helpful conversations.  In particular, the construction of tubular neighborhoods presented here owes much to conversations with Lipshitz and Liu.  I also thank Chris Woodward for a helpful discussion regarding the connection with the work of Shatz.

\newpage

\section{Invariant Tubular Neighborhoods}$\label{inv}$

For basic definitions and terminology regarding infinite dimensional manifolds, as well as for many of the necessary constructions in this section, we follow Lang \cite{Lang-dg}.  By smooth we will always mean $C^{\infty}$.

Throughout this section, $X$ will denote a (possibly infinite-dimensional) Riemannian manifold, and $G$ will denote a group acting on $X$ by smooth isometries.  In more detail, this means $X$ is modeled on a Hilbert space $\Hilb$ (real or complex), and the tangent bundle $T(X)$ has a smoothly varying fiber-wise inner product (which we assume to be Hermitian in the complex case).  We will assume these inner products induce the correct topology on $T_x (X)$, or in other words, we assume that the Riemannian metric on $X$ is \emph{strong}.  We will view the action of $G$ as a homomorphism $G\to \Isom(X)$, where $\Isom(X)$ denotes the group of smooth isometries.  For the purposes of our arguments, both $G$ and $\Isom(X)$ can be viewed as \emph{discrete} groups.

Let $Y\subset X$ be a (locally closed), $G$--invariant submanifold; then the tangent space $T_y (Y)$ is a closed subspace of the Hilbert space
$T_y (X)$ and hence has an orthogonal complement $N_y (Y)$.  The normal bundle $N(Y)\subset T(X)|_Y$ is then a well-defined subset of $T(X)|_Y$, and since $G$ preserves the metric, it acts on $N(Y)$.  However, $N(Y)$ need not be a smooth subbundle in general.  For the construction of tubular neighborhoods given by Lang \cite[IV.5]{Lang-dg}, this is irrelevant: Lang obtains a tubular neighborhood diffeomorphic to the quotient bundle $T(X)|_Y/T(Y)$ by using a splitting of the projection $T(X)|_Y/T(Y)$, which always exists if $X$ admits partitions of unity \cite[III, Proposition 5.2]{Lang-dg}.

In this section, we show that if $N(Y)$ is a smooth subbundle of $T(X)|_Y$, then $Y$ admits a $G$--invariant total tubular neighborhood $\tau(Y)$ (Theorem~\ref{total-tubular}).  The neighborhoods $\tau(Y)$ are homeomorphic to the normal bundle $N(Y)$, but may not be diffeomorphic to $N(Y)$.  Since we are mainly interested in Thom isomorphisms, a homeomorphism is sufficient.  (If $G = \{1\}$, or if $Y$ consists of a single orbit of $G$, then we do obtain a diffeomorphism $\tau(Y) \homeo N(Y)$.)

\begin{remark} The condition that $N(Y)\subset T(X)|_Y$ is a smooth subbundle could be replaced by the condition that there exists a splitting $s$ of the map $\pi$ in the exact sequence
$$0 \maps T(Y) \maps T(X)|_Y \srm{\pi} T(X)|_Y/T(Y)\maps 0$$
whose image in $T(X)|_Y$ is invariant under $G$; our proof goes through equally well in this case, with $\mathrm{Im} (s)$ playing the role of $N(Y)$.
\end{remark}

\subsection{The tangent bundle as a Riemannian manifold}

Associated to the metric on $X$ we have the metric spray $F: TX \to T(TX)$, which is equivariant with respect to the natural actions of the isometry group $\Isom(X)$.  The spray $F$ induces an exponential map $f\co \D \to X$ (Lang \cite[VII.7]{Lang-dg}), whose domain $\D$ is an open subset of $T(X)$ containing the zero section.  It follows from the construction of $f$ that $\D$ is invariant under $\Isom(X)$ and $f$ is $\Isom(X)$--equivariant.

In addition to the exponential map $f$, we will also need a $\Isom(X)$--invariant Riemannian metric on the tangent bundle $TX$, considered as a smooth manifold in its own right.  This Riemannian metric on $T(TX)$ can be defined by explicit local formulae as in Do Carmo \cite[Chapter 3, Exercise 2]{DoCarmo}.  Here we give a more global perspective, from which it is clear that the action of $\Isom(X)$ on $T(TX)$ preserves this metric.

Recall that there is a natural exact sequence of vector bundles over $TX$ of the form
\begin{equation} \label{exact-seq} 0 \maps \pi_{TX}^* (TX) \maps T(TX) \srm{\alpha} (\pi_{TX})^* TX\maps 0,
\end{equation}
where $\pi_{TX} \co TX \to X$ is the structure map for the tangent bundle $TX$, and $\alpha$ is defined via the commutative diagram
$$\xymatrix{ T(TX) \ar@/^/[rrd]^{D \pi_{TX}} \ar@/_/[ddr]_{\pi_{T(TX)}} \ar@{.>}[dr]^\alpha\\
		& (\pi_{TX})^* TX \ar[r] \ar[d] & TX \ar[d]^{\pi_{TX}} \\
		& TX \ar[r]^{\pi_{TX}} & X.
		}
$$
The map $\alpha$ is surjective, because in a local chart $U \subset X$, this diagram becomes (see Lang  \cite[p. 96]{Lang-dg})
$$\xymatrix{ (U\cross \Hilb) \cross (\Hilb\cross \Hilb) \ar@/^/[rrrd]^{(\pi_1, \pi_3)} \ar@/_/[ddr]_{(\pi_1, \pi_2)}
					\ar@{.>}[dr]|-{(\pi_1, \pi_2, \pi_3)}\\
		& U \cross \Hilb \cross \Hilb \ar[rr]^{(\pi_1, \pi_3)} \ar[d]^{(\pi_1, \pi_2)} & & U \cross \Hilb \ar[d]^{\pi_1} \\
		& U \cross \Hilb \ar[rr]^{\pi_1} & & U.
		}
$$
To identify the kernel of $\alpha$, consider the map
$$\pi_{TX}^* (TX) = TX \oplus TX \srm{\phi} T(TX)$$
sending a pair of vectors $(v, w)\in T_x X \oplus T_x X$ to the vector in $T_v (T_x X)$ represented by the curve $t\mapsto v+tw$ in $T_x X \subset TX$.  This defines a map of vector bundles over $TX$, and
from the local representation given above, one sees that $\phi$ is a smooth, injective bundle map whose image is precisely $\ker (\alpha)$.

By naturality, the maps in the exact sequence (\ref{exact-seq}) are $\Isom(X)$--equivariant.  The next step will be to construct an $\Isom(X)$--equivariant splitting of $\alpha$.

Locally, the spray $F$ has the form
$$(x, v) \goesto (x, v, v, f_x (v))$$
where $f_x \co T_x (X)\to \Hilb$ is homogeneous of degree two, meaning that $f_x (tv) = t^2 f_x (tv)$ for any $v\in T_x (X)$ and any scalar $t$ \cite[IV, Proposition 3.2]{Lang-dg}.
We now recall a well-known lemma (see, for example, \cite[I.3]{Lang-dg}), whose proof we include for completeness.

\begin{lemma} If $q\co E\to F$ is a smooth map between Banach spaces and $q$ is homogeneous of degree $p>0$ (meaning that $q(tx) = t^p q(x)$ for all $x\in E$ and all scalars $t$), then $q(x) = \frac{1}{p!} \left(D^p q(0)\right)(x,x,\ldots, x)$, where $D^p q$ is the $p^\textrm{th}$derivative of $q$, considered as a map from $E$ in the space of multilinear functions $E^p \to F$.
\end{lemma}
\begin{proof}  The proof is by induction on $p$.
Differentiating the equation $q(tx) = t^p q(x)$ with respect to $t$ gives
\begin{equation}\label{diff} \left(Dq(tx)\right) (x) = pt^{p-1} q(x).\end{equation}
Note that when $p=1$, the result follows by setting $t=0$, completing the base case.
Setting $t=1$ in (\ref{diff}), we have
\begin{equation} \label{q} q(x) =  \frac{1}{p} \left(Dq(x)\right)(x).  \end{equation}

Now, $Dq \co E\to L(E, F)$, is homogeneous of degree $p-1$: indeed, for any vector $v\in E$ we can differentiate the equation
$$q(t(x + r v)) = t^p q(x + r v)$$
with respect to $r$, obtaining
$$\left(Dq(t(x+ r v))\right) (tv) = t^p \left(Dq(x+ r v) \right)(v);$$
setting $r = 0$ and using linearity of $Dq(tx)$, one finds that
$$\left(Dq(tx)\right) (v) = t^{p-1} \left(Dq (x)\right)(v)$$
for any scalar $t$ and any $x,v\in E$.

We can now apply the induction hypothesis to the degree $p-1$ homogeneous function $Dq$, yielding
$Dq(x) =  \frac{1}{(p-1)!} \left(D^p q(0)\right) (x,x,\ldots, x)$.  Plugging this into (\ref{q}) completes the proof.
\end{proof}

From the lemma, it follows that the quadratic function $f_x$ can be written in the form
$$f_x (v) = \frac{1}{2} \left(D^2 f_x (0)\right)(v,v),$$
meaning that $f$ is the quadratic part of the symmetric bilinear form $\frac 1 2 D^2 f_x (0)$.  Writing this bilinear form as $B(x, v, w) = \frac 1 2 \left(D^2 f_x (0)\right) (v,w)$, we obtain a splitting $s\co (\pi_{TX})^* TX = TX\oplus TX \to T(TX)$ of the exact sequence (\ref{exact-seq}) whose local form is
$$s(x, v, w) = (x, v, w, B(x, v, w)).$$
Since the spray $F$ is $\Isom(X)$--equivariant and (by polarization) $B(x, -, -)$ is the unique symmetric bilinear form satisfying $B(x, v, v) = f_x (v)$, it follows that $s$ is itself $\Isom(X)$--equivariant.  Note also that $s$ is a map of vector bundles over $TX$.

The splitting $s$ gives us an explicit, $\Isom(X)$--equivariant direct sum decomposition of vector bundles over $TX$:
$$T(TX) \isom \pi_{TX}^* (TX) \oplus \pi_{TX}^* (TX).$$
The Riemannian metric on $X$ pulls back to an inner product on $\pi_{TX}^* (TX)$ (which is again $\Isom(X)$--invariant) and now the direct sum decomposition endows $T(TX)$ with an $\Isom(X)$--invariant inner product as well, making $TX$ into a Riemannian manifold with an isometric action of the group $\Isom(X)$.

We note that there is another description of this splitting $s$ in terms of the Levi-Civita covariant derivative $D$ \cite[VIII, Theorem 4.2]{Lang-dg}: given $(v,w)\in T_x(X) \oplus T_x (X)$ with $w = \gamma'(0)$ for some curve $\gamma\co (-\epsilon, \epsilon) \to X$, there is a unique lift of $\gamma$ to a curve $\gamma_v: (-\epsilon, \epsilon) \to TX$ such that $\gamma(0) = v$ and $\gamma_v$ is parallel to $\gamma$ with respect to $D$.  The splitting $s$ now sends $(v, w)$ to $(\gamma_v)'(0) \in T_v (TX)$.  This splitting is the same as the one given above, as can be seen by examining the definition of $D$--parallel curves.

\subsection{Construction of the tubular neighborhood}

\begin{definition}$\label{tub-def}$  A \emph{tubular neighborhood} of $Y$ in $X$ is an open neighborhood $\tau(Y)$ of $Y$ in $X$ together with a homeomorphism
$\phi\co U\to \tau(Y)$ from an open neighborhood $U\subset N(Y)$ containing the zero-section $Y\subset N(Y)$, such that $\phi$ restricts to the identity from $Y\subset N(Y)$ to $Y\subset X$.  If $\phi$ is a diffeomorphism, we call $\tau(Y)$ smooth, and if $U = N(X)$, we call $\tau$ \emph{total}.
\end{definition}

\begin{proposition}$\label{tubular}$  Let $X$ be a Riemannian manifold and let $Y\subset X$ be a (locally closed) submanifold whose normal bundle $N(Y)\subset T(X)|_Y$ is a smooth subbundle.  Assume $G$ acts on $X$ by smooth isometries, leaving $Y$ invariant.  Then there exists a $G$--invariant open neighborhood $Z$ of the zero section of $N(Y)$ with $Z\subset \D\cap N(Y)$, and the exponential map restricts to an equivariant diffeomorphism $f\co Z\to V$, where $V$ is an open neighborhood of $Y$ in $X$.  Hence $V$ is a $G$--invariant tubular neighborhood of $Y$.
\end{proposition}
\begin{proof} We identify $Y$ with the zero section of $N(Y)$, and for $g \in G$ we will let $g$ denote both the self-map it induces on $X$ and the derivative of this map.

We begin by noting that the geodesic distance $d$ associated to the natural $\Isom$--invariant Riemannian metric on $TX$ constructed above yields an $\Isom(X)$--invariant distance function (topological metric) on $TX$ (see \cite[VII.6]{Lang-dg} for a discussion of the geodesic distance, and the fact that the associated metric topology is the usual topology on $TX$).  Since $G$ acts by isometries, the distance function $d$ is $G$--invariant.
For any $W\subset T(X)$ and any $w\in W$, we write $B_{\epsilon} (w, W) = \{w'\in W| d(w, w')<\epsilon\}$.  Note that we identify $X$ with the zero section of $T(X)$, so $B_\epsilon (x, X)$ is defined.

Now fix $y\in Y$. Since the exponential map $f$ restricts to a local diffeomorphism $\D\cap N(Y)\to X$
\cite[p. 109]{Lang-dg}, we know that for some $\epsilon_y>0$, $f$ restricts to a diffeomorphism $B_{\epsilon_y} (y, N(Y))\to f\left(B_{\epsilon_y} (y, N(Y))\right)$ (with $f\left(B_{\epsilon_y} (y, N(Y))\right)$ \emph{open} in $X$).
Let $\psi\co f\left(B_{\epsilon_y} (y, N(Y))\right)\to B_{\epsilon_y} (y, N(Y))$ denote the inverse map.  Now, $f\left(B_{\epsilon_y/2} (y, N(Y))\right)$ is an open neighborhood of $y$ in $X$, hence contains $B_{\epsilon'_y}(y, X)$ for some $\epsilon'_y<\epsilon_y/4$.  Set $Z_y = \psi(B_{\epsilon'_y}(y, X))\subset N(Y)$.  Then $Z_y$ is open in $N(Y)$ and we have
\begin{equation}\label{Z_y0}
Z_y \subset B_{\epsilon_y/2} (y, N(Y))
\end{equation}
and
$f(Z_y) = B_{\epsilon'_y}(y, X)$.  In fact, for any $g\in G$,
\begin{equation}\label{Z_y}
f(g (Z_y)) = B_{\epsilon'_y} (g\cdot y, X).
\end{equation}

Define $Z_{y(\mO)}$ as above for one point $y(\mO)$ from each $G$--orbit $\mO\subset Y$.  Then we claim that $f$ is injective, and hence restricts to a diffeomorphism, on the open set
$$Z = \bigcup_{\mO \in Y/G, \,\, g\in G} g(Z_{y(\mO)}).$$
Note here that in order to conclude that $f|_Z$ is a diffeomorphism onto its image, we are using the fact that $f\left(B_{\epsilon_y} (y, N(Y))\right)$ is \emph{open} in $X$; this also shows that $f(Z)$ is open in $X$.

Say $x = f(g_1 \cdot z_1) = f(g_2 \cdot z_2)$ with $z_1\in Z_{y_1}$, $z_2\in Z_{y_2}$ and $g_1$, $g_2\in G$ (here the $y_i$ are the chosen representatives for some two orbits).  Then by (\ref{Z_y})
we have
\begin{equation}\label{est}
d(g_1\cdot y_1, g_2 \cdot y_2) \leq d(g_1\cdot y_1, x) + d(x, g_2\cdot y_2) < \epsilon'_{y_1} + \epsilon'_{y_2} < \epsilon_{y_1}/4 + \epsilon_{y_2}/4.
\end{equation}
We may assume that $\epsilon_{y_1} \geq \epsilon_{y_2}$.  Then by (\ref{Z_y0}) and (\ref{est}) we have
\begin{equation}\label{est2}
\begin{split}
d(g_1 \cdot y_1, g_2 \cdot z_2) &\leqs d(g_1\cdot y_1, g_2\cdot y_2) + d(g_2\cdot y_2, g_2\cdot z_2) \\
&< (\epsilon_{y_1}/4 + \epsilon_{y_2}/4) + \epsilon_{y_2}/2
\leqs \epsilon_{y_1}.
\end{split}
\end{equation}
The fact that $f$ is injective on $B_{\epsilon_{y_1}} (y_1, N(Y))$ implies that it is also injective on $B_{\epsilon_{y_1}} (g_1 \cdot y_1, N(Y))$.  But $g_1 \cdot z_1 \in B_{\epsilon_{y_1}} (g_1 \cdot y_1, N(Y))$ by (\ref{Z_y0}), and $g_2 \cdot z_2 \in B_{\epsilon_{y_1}} (g_1 \cdot y_1, N(Y))$ by (\ref{est2}), so we have a contradiction.
\end{proof}

Before constructing a $G$--invariant total tubular neighborhood, we need a lemma which follows from the argument
in Lang \cite[VII, Proposition 4.1]{Lang-dg}.

\begin{lemma}$\label{compression}$ Let $X$, $Y$, $Z$, and $G$ be as in Proposition~\ref{tubular}, and let $\sigma\co Y\to\bbR$ be a continuous $G$--invariant function such that for all $y\in Y$,
$\sigma(y) >0$ and
$\{v\in N_y (Y) \, | \, \abs{v} < \sigma (y)\}$ lies inside $Z$.

Then there is a homeomorphism $N(Y) \to N(Y)_{\sigma}$, where
$$N(Y)_{\sigma} = \{ v\in N(Y) \,\, |\,\, ||v||<\sigma(\pi v)\}$$
(here $\pi\co N(Y)\to Y$ is the projection, and $||\cdot||$ is the norm on $T_x (X)$).  This homeomorphism restricts to the identity on $Y\subset N(Y)$.
\end{lemma}

We can now prove our main theorem.

\begin{theorem}$\label{total-tubular}$ Let $X$ be a Riemannian manifold, and let $G$ act on $X$ by isometries, leaving the submanifold $Y\subset X$ invariant.  Assume that $N(Y)$ is a smooth subbundle of $T(X)|_Y$.  Then there exists a $G$--invariant total tubular neighborhood $\tau(Y)$ of $Y$ in $X$.  The homeomorphism
$N(Y) \stackrel{\homeo}{\maps}  \tau(Y)$ is $G$--equivariant, but need not be smooth.
\end{theorem}
\begin{proof}  We apply Lemma~\ref{compression}.  Set $\sigma(y) = \sup \{\epsilon>0 : ||v||<\epsilon \implies v\in Z\}$.
Then $\sigma(y)>0$ for any $y\in Y$, $\sigma$ is $G$--invariant, and
we have $N(Y)_\sigma \subset Z$.
Continuity of $\sigma$ follows from the assumption that the inner products defining $||\cdot||$ vary continuously over $X$, together with the fact that $Z$ is open.
\end{proof}

\begin{remark} If $G = \{1\}$ and $X$ admits partitions of unity, then Lang \cite[VII.4, Corollary 4.2]{Lang-dg} shows that one may choose $\sigma$ to be smooth, and then the homeomorphism $N(Y) \to \tau(Y)$ will be smooth as well.  If $Y$ consists of a single $G$--orbit, then $\sigma$ can be chosen to be \emph{constant}, and we obtain the same conclusion.
\end{remark}

\section{Applications to Yang--Mills Theory}$\label{YM}$

We now apply Theorem~\ref{total-tubular} to the Morse strata of the Yang--Mills functional.  Let $P\to M$ be a principal $U(n)$--bundle over a surface, and
let $\cA^{k-1}(P)$ denote the affine space of connections on $P$ lying in the Sobolev space
$L^2_{k-1}$ ($k\geqs 2 1$).  Similarly, let $\cG^k(P)$ denote the $L^2_{k-1}$ Sobolev completion of the unitary gauge group of $P$, which acts smoothly on $\cA^{k-1} (P)$ (Wehrheim \cite{Wehrheim}).  Connections on $P$ correspond precisely to Hermitian connections on the associated Hermitian bundle $P\cross_{U(n)} \bbC^n$, and $\G^k(P)$ is isomorphic to the Sobolev gauge group $\G^k (E)$.  The following lemma will allow us to apply Theorem~\ref{total-tubular}.  A proof of this lemma may be found
in Kondracki--Rogulski \cite[Section 2.3]{KR}.

\begin{lemma}\label{inv-metric}
Let $K$ be a compact connected Lie group and
let $P$ be a principal $K$--bundle over
a Riemann surface $M$ equipped with
a Riemannian metric. The metric
on $M$ and a $K$--invariant inner product
on the Lie algebra $\fk$ of $K$ induce
a Riemannian metric on $\cA^{k-1}(P)$,
the space of $L^2_{k-1}$ $K$--connections on $P$.
This Riemannian metric
is $\cG^k(P)$--invariant.
\end{lemma}

\vspace{.15in}

As discussed in the introduction, the Morse strata of the Yang--Mills functional can be ordered  $C_1\leqs C_2\leqs \cdots$ , starting with the central stratum, in such a way that the union $\C_n = \cup_{i = 1}^n C_i$ of each initial segment of the ordering is open.  Such an ordering was first discussed by Atiyah and Bott \cite{A-B}, and is described in more detail in Ramras \cite{Ramras-YM}.  We can now give our main application of Theorem~\ref{total-tubular}.

\begin{corollary}$\label{YM-tubular}$ Let $P\to M$ be a smooth principal $U(n)$--bundle over a surface $M$, and let $\leqs$ denote a linear order on the set of Morse strata as above.  Then each stratum $S$ has a $\G(E)$--invariant tubular neighborhood lying inside the open set $\C_n$, and there are Thom isomorphisms in gauge equivariant cohomology
$$H^*_{\G(E)} (C_n, C_{n-1}) \isom H^{*-\mathrm{codim} (\C_{n})}_{\G(E)} \left(\C_{n}\right).$$
If $M$ is orientable, or a non-orientable surface whose double cover has genus greater than 1, these isomorphisms hold with integer coefficients; in general they hold with $\bbZ/2$--coefficients.
\end{corollary}

\begin{proof} Since $\cA^{k-1} (P)$ is an affine space modeled on the vector space of $L^2_{k-1}$ sections of $\mathrm{ad} P \otimes T^*M$, this space is a Hilbert manifold, as are the open subsets $\bigcup_{i\leqs j} C_i$.  When $M$ is orientable, the Morse strata for the Yang--Mills functional on $P$ are locally closed submanifolds of $\cA^{k-1} (P)$ (Daskalopoulos \cite{Dask}) and are invariant under $\cG^k(P)$.  So existence of the desired neighborhoods will follow by applying Theorem~\ref{total-tubular} and Lemma~\ref{inv-metric} to the submanifolds $C_j \subset \bigcup_{i\leqs j} C_i$, once we show that the normal bundle $N(C_j)\subset T(\A^{k-1} (P))|_{C_j}$ is a smooth subbundle.  Since the kernel of a smooth, surjective map of vector bundles is a smooth subbundle of the domain (see, for example, Lang \cite[III.3]{Lang-dg}), it will suffice to show that the orthogonal projections $T_A (\A^{k-1} (P)) \to T_A (\C_j)$ vary smoothly with $A\in \C_j$.

To understand these orthogonal projections, we follow Daskalopoulos' proof that the strata are locally closed submanifolds \cite[Proposition 3.5]{Dask}.  Recall \cite[Section 5]{A-B} that connections on $P$ correspond bijectively with holomorphic structures on the associated vector bundle $\Ad\, P = P\cross_{U(n)} \bbC^n$, where we view a holomorphic structure as an operator on sections
$$d'': L^2_{k-1} (\Ad\, P) \to L^2_{k-1} (T^* M'' \otimes \Ad\, P).$$
Here $T^* M''$ is the space of $(0,1)$--forms on the complex manifold $M$.  Sections in the kernel of $d''$ are then \emph{holomorphic}.  We will call $\Ad\, P$, together with a $d''$ operator, a holomorphic bundle, which we denote by $E$.  Note that by adjointness, $d''$ induces a corresponding operator on the endomorphism bundle $\End (\Ad\, P) = \End\, E$.

Given a critical holomorphic bundle $E\in C_j$ let $S$ denote the orthogonal complement of the space of $L^2_{k-1}$ holomorphic endomorphisms of $E$ inside the space of all smooth endomorphisms (that is, $S$ is the kernel of the $d''$ operator associated to $\End\, E$).
Daskalopoulos constructs a map
$$f\co S \cross H^1 (M, \End \, E) \maps \A^{k-1} (P)$$
and shows that $f$ is an isomorphism near zero, meaning that
$f\co U_0 \maps U_E$
is an isomorphism for some neighborhood $U_0$ of $0$ in the Hilbert space $S \cross H^1 (M, \End \, E)$ and some neighborhood $U_E$ of $E$.  Daskalopoulos then shows that
$$f^{-1}(U_E \cap C_j) = U_0 \cap \left(S \cross H^1 (M, \End' \, E) \right),$$
so that $f$ provides a chart for $C_j$ near $E$.
Here $\End' \, E$ is the sheaf of holomorphic endomorphisms preserving the Harder-Narasimhan filtration on $E$.  (The notation $\End' \, E$ comes from Atiyah--Bott \cite[Section 7]{A-B}; Daskalopoulos uses the notation $UT(E, *) = \End' \, E$, since these endomorphisms correspond, in a local ordered basis respecting the Harder-Narasimhan filtration, to upper triangular matrices.)

This chart provides an isomorphism between the tangent bundle to $U_E \cap C_j$ and the trivial bundle
$$\left( U_0 \cap \left(S \cross H^1 (M, \End' \, E)\right) \right) \cross (S\cross H^1 (M, \End' \, E)).$$
We need to understand the orthogonal projections
$$S\cross H^1 (M, \End \, E) \maps S\cross H^1 (M, \End' \, E)$$ with respect to the \emph{pullback}, along $f$, of the metric on $\A^{k-1} (P)$ to the various tangent spaces $S\cross H^1 (M, \End \, E)$ (although these tangent spaces are all isomorphic to one another, the pullback metric will vary depending on the behaviour of the map $f$).   Observe that this orthogonal projection is simply the identity on $S$, plus an orthogonal projection $H^1 (M, \End \, E) \to  H^1 (M, \End' \, E)$.  These cohomology groups are finite dimensional, and since the metrics vary smoothly, so do the orthogonal projections.

 Daskalopoulos produces the manifold structure at a general point $E\in C_j$ (as opposed to critical points) using a gauge transformation sending $E$ into a neighborhood of the critical set on which the manifold structure has already been established.  So local triviality of $N(C_j)$ at arbitrary points of $E$ follows from the result near the critical set.  This completes the proof of local triviality, in the case of an orientable surface.

When $M$ is non-orientable, we let $\wt{P}\to \wt{M}$ denote the pullback of $P$ to the orientable double cover $\wt{M} \to M$.  There is a natural embedding $\cG^k (P)\injects \cG^k(\wt{P})$, making the natural embedding
$\cA^{k-1} (P)\injects \cA^{k-1} (\wt{P})$ equivariant.  In both cases, the image of the embedding is the set of fixed points of an involution $\tau$ arising from the deck transformation on $\wt{M}$ (see Ho--Liu \cite{Ho-Liu-non-orient}).  The Morse strata in $\cA^{k-1} (\wt{P})$ are, by definition, just the intersections of Morse strata in $\cA^{k-1} (\wt{P})$ with $\cA^{k-1} (P)$, and hence are locally closed submanifolds invariant under $\cG^k (P) = \cG^k (\wt{P})^{\tau}$.  By Lemma~\ref{inv-metric}, we obtain a gauge-invariant metric on $\cA^{k-1} (\wt{P})$ from a choice of a metric on $\wt{M}$ and a $U(n)$--invariant inner product on $\fu (n)$.  We will choose our metric on $\wt{M}$ to be invariant under the deck transformation, so that the resulting metric on $\A^{k-1} (\wt{P})$ is invariant under $\tau$.
Now, the $\cG^k (\wt{P})$--invariant metric on $\cA^{k-1}(\wt{P})$ restricts to a $\cG^k (P) = \cG^k (\wt{P})^{\tau}$--invariant metric on $\cA^{k-1} (\wt{P}) = \cA^{k-1}(P)^{\tau}$, and it remains to check that the normal bundles are smooth subbundles $T(\A^{k-1} (P))$.

Given a Morse stratum $C \subset \cA^{k-1} (P)$, it will suffice to show that the orthogonal projection $Q_C \co T \cA^{k-1} (P)\to TC$ is a smooth bundle map, since the kernel of this projection is precisely the normal bundle to $C$.  By definition, $C = \wt{C} \cap \cA^{k-1} (P)$ for some Morse stratum $\wt{C} \subset \cA^{k-1} (\wt{P})$.  Let $Q_{\wt{C}} \co T \cA^{k-1} (\wt{P}) \to T \wt{C}$ denote the orthogonal projection, which we have shown is a smooth bundle map.  Since the involution $\tau$ is an isometry, it follows that $Q_C$ is simply the restriction of the $Q_{\wt{C}}$ to $TC = (T\wt{C})^\tau \subset T\wt{C}$ (in general, one may check that if $W$ is a closed subspace of a Hilbert space $\Hilb$ and $\tau\co \Hilb \to \Hilb$ is a linear isometry, then the orthogonal projection $\Hilb^\tau \to W^\tau$ is the restriction of the orthogonal projection $\Hilb\to W$).  Hence $Q_C$ is a smooth bundle map as well.

To produce the desired Thom isomorphisms, note that over a Riemann surface, the normal bundles to the Yang--Mills strata are naturally complex vector bundles (Atiyah--Bott \cite[Section 7]{A-B}).  For any $G$--equivariant complex vector bundle $V\to X$, the homotopy orbit bundle $V_{hG}\to X_{hG}$ is still a complex vector bundle, hence orientable.
Letting $\nu_n$ denote a $\G(E)$--invariant tubular neighborhood of $C_n$ inside $\C_n$, we may excise the complement of $(\nu_n)_{h\G(E)}$ in $(\C_n)_{h\G(E)}$ and apply the ordinary Thom isomorphism for the bundle $(\nu_n)_{h\G(E)}\to (\C_n)_{h\G(E)}$.

The same argument works over a non-orientable surface, since it is proven in Ho--Liu--Ramras \cite{Ho-Liu-Ramras} that the normal bundles to the Yang--Mills strata (and the corresponding homotopy orbit bundles) are orientable (real) vector bundles (so long as the double cover of the surface has genus greater than 1).
\end{proof}

\section{Other constructions of tubular neighborhoods}$\label{background}$

\subsection{General constructions}

In order to put our construction in context, we briefly discuss existing general constructions of tubular neighborhoods for for submanifolds $Y\subset X$.
There are many treatments of this topic in the literature, and the basic tool is always the exponential map, which is a local diffeomorphism from an open neighborhood $\D \subset N(Y)$ containing the zero section $Y\subset N(Y)$ to an open neighborhood of $Y$ in $X$.
Since the proof of Godement's lemma \cite{Godement} first appeared, most arguments \cite{Bredon-transf, Brocker-Janich, GP, Hirsch, Lang-dg, MO, Mukherjee} rely on Godemont's idea produce a smaller neighborhood $Z\subset \D$ on which the exponential map is injective, hence a diffeomorphism (in \cite{BG, Milnor-collected-III, Spivak-I}, $Y$ is assumed to be compact, and a simplified version of Godement's lemma is used).  After this point, one argues that $N(Y)$ can be compressed into $Z$.  In order to construct invariant neighborhoods for compact groups, Bredon uses integration over the group to make this compression $G$--equivariant.  Surprisingly, we have not found the non-equivariant version of our argument, making use of a distance function on $T(X)$, elsewhere in the literature.

\subsection{Relationship with the work of Kondracki--Rogulski, Ebin, and Teleman}

Kondracki and Rogulski \cite{KR} provided a construction of gauge-invariant tubular neighborhoods of the gauge orbits inside the space $\cA^{k-1} (P)$ of (Sobolev) connections on a smooth principle $K$--bundle $P$ over a compact manifold $M$, where $K$ is a compact Lie group.  Following Ebin \cite{Ebin}, Kondracki and Rogulski utilize a weak gauge-invariant metric on the space of connections, and show that with respect to this weak metric, the orthogonal complement to $T(\G^k (P) \cdot A)$ inside $T(\cA^{k-1} (P))$ is a smooth subbundle (\cite[Lemma 3.3.1]{KR}).  However, their argument also shows that the orthogonal complement with respect to the strong metric is  a smooth subbundle.  Hence Theorem \ref{total-tubular} applies to this situation.

Ebin \cite{Ebin} considers the space of Riemannian metrics on a fixed finite-dimensional manifold $M$, equipped with the action of the diffeomorphism group of $M$.  After passing to appropriate Sobolev completions of these objects, Ebin constructs diffeomorphism-invariant tubular neighborhoods for the orbits of the action.  His arguments make use of a weak Riemannian metric on the space of metrics, and it is only with respect to this weak metric that Ebin proves the normal bundles to orbits are smooth subbundles.  Ebin's arguments are somewhat subtler than those of Kondracki--Rogulski, and they do not directly apply to the strong Riemannian metric.  Hence our results do not immediately apply in this situation.  We note that Ebin uses his results in the Sobolev setting to derive results in the smooth setting.

A. Teleman \cite{Teleman-harmonic} gives an explicit construction
of a gauge-invariant tubular neighborhood for the space of connections $A$ on a Euclidean bundle satisfying the condition
$\mathrm{dim} (\ker (d_A)) = 1$.  Teleman's methods are quite different from ours, and it is unclear whether our results apply to this situation.


TITLE: Proof of Catalan numbers on a circle
QUESTION [6 upvotes]: Question: Letting 2n be the number of points on a circle, prove that the number of ways to join these points, with non-intersecting lines, into pairs is equal to the Catalan numbers.
I'm having troubles understanding how to give a solid proof for this. When I think intuitively it makes sense. Any ideas on how to start the recurrence?

REPLY [6 votes]: First of all, I need to remark that the current formulation of the question doesn't seem to be quite right, since the fact that you're working on a circle isn't really used. It also isn't very clear what you mean by "joining into pairs.
However, if I read this question in the following way, it seems to work: We choose $2n$ arbitrary points on a circle. How many ways are there of drawing $n$ lines between $2$ points, such that there are no intersections between these lines inside the circle?
How to prove this? It is easiest to follow what's going on by drawing a few examples and keeping these nearby while reading this. We will denote the number of possibilities given $n$ as $Q_n$.
Now suppose we have $n$ fixed, as well as $2n$ points on a circle. We will fix one point $p_1$ and then give names clockwise: the first point one encounters when walking along the circle, starting from $p_1$, is called $p_2$, etc. The first thing now is to consider to which points $p_1$ may be connected. We remark that if we draw a line between $p_1$ and $p_i$, this divides the remaining points in the sets $X_1=\{p_2,\ldots,p_{i-1}\}$ and $X_2=\{p_{i+1},\ldots,p_{2n}\}$, each corresponding to the points on one side of the line. Since no two lines may intersect within the circle, this imposes a restriction on the possible points $p_1$ may be connected to. Namely, this needs be done in such a way that both sets $X_i$ have an even number of points. That is, we may connect the point $p_1$ to each point $p_{2j}$ with $1\leq j \leq n$. This can easily be seen in a drawing.
Now, what happens when we join up the point $p_1$ with one such point? We have a set $X_1$ of $2n_1$ points lying on a circle that need to be pairwise connected in such a way that no lines intersect, while the same holds for the $2n_2$ points in $X_2$. We therefore notice that if we connect the point $p_1$ to $p_i$, the remaining number of possible line configurations becomes $Q_{n_1}Q_{n_2}$.
The result can now be obtained by adding the numbers of possible configurations corresponding to each line from $p_1$ to some $p_i$. What we get, is that $Q_n = Q_{n-1} + Q_1Q_{n-2}+...+Q_{n-2}Q_1+Q_{n-1}$. It is best to first write it this particular way to avoid any problems `in the middle' when $n$ is odd. If we define $Q_0$ as $1$ (one could argue that there is exactly one way not to draw any lines inside a circle at all), we can express every $Q_n$ by the recurrence relation
$Q_0=1,\qquad Q_n=\sum_{j=0}^{n-1}Q_jQ_{(n-1)-j}.$
I don't know how Catalan numbers have been defined in your case, but this recurrence relation describes them. If you don't know this yet, there must be plenty of sources out there to prove it for you.


TITLE: Vector spaces and graded vector spaces
QUESTION [0 upvotes]: Let $k$ be a field and $G$ a group. Consider $G-\mathrm{vect}_k$ the category of $G$-graded vector spaces (compare nLab).
I have the following proposition to prove:
For a field $k$ and a finite group $G$ the following holds for the categories $\mathrm{vect}_k$ and $G-\mathrm{vect}_k$:
$$(G-\mathrm{vect}_k)_0 \cong \mathrm{vect}_k$$
I suppose the equivalence is meant to be as monoidal categories, as both have the structure of a monoidal category with the normal vector space tensor product.
I understand the basic idea behind this, as vector spaces in $\mathrm{vect}$ have no grading and in every component every vector space can be considered.
But I have problems with formalizing this. Can anyone maybe help me to find a start formulating this?

REPLY [1 votes]: If I'm correct in my interpretation that $(G-\mathrm{vect}_k)_0$ is the full subcategory of $G-\mathrm{vect}_k$ on objects that are concentrated in degree $e$, that is objects $V$ such that  $\forall x, (x=e \lor V_x=0)$, then the claim is relatively easy.
Indeed, if you have two of these, $V,W$ then a map $V\to W$ is a collection of maps $f_x: V_x\to W_x$. But for all $x$ except $e$, $V_x=W_x= 0 $ so $f_x=0$: the only interesting one is $f_e$. Similarly, since we know all the other $V_x$'s (they're $0$ !) we only care about $V_e$. This shows intuitively why essentially this is the same as $\mathrm{vect}_k$.
Formally, you can define a functor $(G-\mathrm{vect}_k)_0\to \mathrm{vect}_k$ via $V\mapsto V_e$, $f:V\to W\mapsto f_e :V_e\to W_e$, check that it is fully faithful and essentially surjective (in fact, depending on how you define things, it's even literally surjective and injective, but this depends on the specific details of implementation); so it is an equivalence (you can even define a quasi-inverse quite easily !)


TITLE: Changing equation to x equals
QUESTION [0 upvotes]: Im currently stuck on this equation I need to modify to be in terms of x
$$y=-x^2+4$$
I got something like this which looks wrong
$$x = -\sqrt{y+4}$$
First you would subtract the 4 from both sides. Do you then divide by -1 then square root it?

REPLY [2 votes]: Original equation, $y=-(x^2)+4$;
Minus $4$ from both sides, $y-4=-x^2;$
Multiply $(-1)$ to both sides, $x^2=-(y-4)$;
That gives you, $x^2=4-y$;
Square root it, $x=\pm\sqrt {4-y}$
Final ans is $x=\pm\sqrt {4-y}$.


TITLE: Conceptual question about Osillations
QUESTION [1 upvotes]: I am confused about simple harmonic motion. I understand that it follows a sine wave but is it possible to explain why it does?

REPLY [1 votes]: A particle is said to be executing simple harmonic motion if it satisfies the following differential equation equation:
$$\frac{d^2\phi}{dt^2} = - \omega^2\phi \tag{1}$$
The $\sin$ wave is merely a solution to the above differential equation.
$$\phi = \phi_o \sin (\omega t + c) \tag{2}$$
You can verify it by substituting the function $(2)$ in equation $(1)$.
If a quantity is of the form $(2)$, then we say that the quantity is varying harmonically. However, this is not the only solution to the $(1)$ differential equation.
$$\phi = \phi_o \cos (\omega t + c)$$
$$\phi = \phi_{o1} \sin(\omega t) + \phi_{o2} \cos (\omega t)$$
are also the solutions for the $(1)$ differential equation.


TITLE: Variance of Chi Square Distribution as the Sum of Unit Normal Random Variables
QUESTION [2 upvotes]: Okay, so I am interested if there is a way to derive the variance for a Chi-Square distribution using the property that it is the sum of independent unit normal distributions squared.
For example, if $X$ is a Chi-Square random variable with $n$ degrees of freedom, it has the distribution: $\displaystyle\operatorname{}\left(\sum_{i=1}^n Z_i^2\right)$ where $Z$ is Normal$(0,1)$
I know that $Var(X)= E(X^{2}) -[E(X)]^{2}$
To start finding $E(X)$ I begin with the fact that each $Z$ has $E(Z)=0$ and $Var(Z)=1$. This implies that $E(Z^2)=1$ since $Var(Z)=E(Z^{2})-[E(Z)]^2$
Since $X = \displaystyle\operatorname{}\left(\sum_{i=1}^n Z_i^2\right)$ then $E(X)=\displaystyle\operatorname{}\left(\sum_{i=1}^n 1\right)=n$
I am lost on what the next step would be. I have this start, but don't where to go next. Any thoughts on how to find $E(X^{2})$?
$\begin{align}
X^2 =& (\sum_{i=1}^nZ_i^2)^2\\
=& \sum_{i=1}^nZ_i^4+\sum_{i \neq j}^nZ_i^2Z_j^2
\end{align}$
I can't see any way this sum not getting nasty. Note: I can solve this using integration of the PDF for the Chi-Square distribution, but I was wondering if there is any way to do it using the property that Chi-Squared is sum of Squared Normal.

REPLY [2 votes]: Since $X_1^2,\ldots,X_n^2$ are independent, you have
$$
\operatorname{var}(X_1^2+\cdots+X_n^2) = n \operatorname{var}(X_1^2).
$$
And this is $\operatorname{var}(X_1^2) = \operatorname{E}((X_1^2)^2) - (\operatorname{E}(X_1^2))^2$. If you know that $\operatorname{E}(X_1^2) =1,$ then it remains only to find $\operatorname{E}(X_1^4).$
\begin{align}
\operatorname{E}(X_1^4) & = \int_{-\infty}^\infty x^4 \varphi(x)\, dx \text{ where $\varphi$ is the standard normal density} \\[10pt]
& = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty x^4 e^{-x^2/2} \,dx \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty x^4 e^{-x^2/2} \,dx\text{ by symmetry} \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty x^3 e^{-x^2/2} \Big( x\,dx\Big) \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty (2u)^{3/2} e^{-u} \, du = \frac 4 {\sqrt{\pi}} \int_0^\infty u^{3/2} e^{-u} \,du \\[10pt]
& = \frac 4 {\sqrt\pi} \Gamma\left(\frac 5 2 \right) = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \Gamma\left( \frac 3 2 \right) = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \Gamma\left( \frac 1 2 \right) \\[10pt]
& = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \sqrt \pi \\[10pt]
& = 3.
\end{align}


TITLE: Derivative with respect to entries of a matrix
QUESTION [3 upvotes]: What is the derivative of this matrix expression with respect to $\theta_k$
\begin{equation}
\begin{aligned}
\mathcal{J}(X, \theta) &= {\bf trace}\left( XX^TP(\theta)^{-1} \right) +{\bf trace}\left( (Y-H(\theta)X)(Y-H(\theta)X)^T \Sigma^{-1} \right)\\
& = X^TP(\theta)^{-1}X + (Y-H(\theta)X)^T \Sigma^{-1} (Y-H(\theta)X)^T
\end{aligned}
\end{equation}
$X$ and $Y$ are vectors
$\theta$ is a vector with entries $\theta_k$
$P(\theta)$ and $H(\theta)$ are matrices constructed using some or all of the entries of $\theta$ and possibly other constants.
The matrix $\Sigma$ is an invertible known constant matrix
All vectors and matrices have compatible dimensions.
I tried to use The Matrix Cookbook to calculate this derivative. Here is my result:
\begin{equation}
\begin{aligned}
\frac{\partial \mathcal{J}(X,\theta)}{\partial \theta_k }  =&  - {\bf trace} \left( X X^T P(\theta)^{-1} \frac{\partial P(\theta)}{\partial \theta_k}P(\theta)^{-1}   \right) \\
& - 2\; {\bf trace} \left(\frac{\partial H(\theta)}{\partial \theta_k} X Y^T \Sigma_e^{-1}\right)\\
&+ 2\;  {\bf trace} \left(\frac{\partial H(\theta)}{\partial \theta_k} \Sigma_e^{-1} H(\theta) X X^T\right)
\end{aligned}
\end{equation}
Is this result correct? Can you explain if there is a mistake?
Also I would like to know if there is a better way to write this derivative.

REPLY [1 votes]: Define the vector $z = (Hx-y)$.  Then rewrite the function in terms of the Frobenius (:) product and find its differential
$$\eqalign{
 {\mathcal J} &= \Sigma^{-1}:zz^T + xx^T:P^{-1}  \cr\cr
 d{\mathcal J} &= \Sigma^{-1}:d(zz^T) + xx^T:dP^{-1} \cr
   &= \Sigma^{-1}:(dz\,z^T+z\,dz^T) - xx^T:P^{-1}\,dP\,P^{-1} \cr
   &= (\Sigma^{-1}+\Sigma^{-T}):dz\,z^T - P^{-T}xx^TP^{-T}:dP \cr
   &= (\Sigma^{-1}+\Sigma^{-T})z:dH\,x - P^{-T}xx^TP^{-T}:dP \cr
   &= (\Sigma^{-1}+\Sigma^{-T})(Hx-y)x^T:dH - P^{-T}xx^TP^{-T}:dP \cr
}$$
Now substitute $d_{\theta_k} \rightarrow d$ to obtain the desired derivative
$$\eqalign{
 \frac{\partial {\mathcal J}}{\partial \theta_k} &= (\Sigma^{-1}+\Sigma^{-T})(Hx-y)x^T:\Big(\frac{\partial H}{\partial \theta_k}\Big) - P^{-T}xx^TP^{-T}:\Big(\frac{\partial P}{\partial \theta_k}\Big) \cr
}$$
If you are uncomfortable with the Frobenius product, you can replace it with the equivalent trace expression
$$ A:B = {\rm tr}(A^TB)$$


TITLE: Class number measuring the failure of unique factorization
QUESTION [27 upvotes]: The statement that the class number measures the failure of the ring of integers to be a ufd is very common in books. ufd iff class number is 1. This inspires the following question:
Is there a quantitative statement relating the class number of a number field to the failure of unique factorization in the maximal order - other than $h = 1$ iff $R$ is a ufd?
In what sense does a maximal order of class number 3 "fail more" to be a ufd than a maximal order of class number 2?
Is it true that an integer in a field of greater class number will have more distinct representations as the product of irreducible elements than an integer in a field with smaller class number?

REPLY [4 votes]: I know I'm late to the party, but it seems to me the other answers don't directly answer the last part of the question:

Is it true that an integer in a field of greater class number will have more distinct representations as the product of irreducible elements than an integer in a field with smaller class number?

No.  Of course it's not necessarily true that a given integer in two fields $K_1 \cap K_2$ will have more distinct factorizations in $K_1$ than $K_2$ if $h(K_1) \ge h(K_2)$.  But even in simple situations where you could hope for this to be true, the opposite behavior can happen.
In this paper, I gave a combinatorial expression for the number $\eta(x)$ of distinct factorizations of $x \in \mathcal O_K$ ($K$ a number field) in terms of the ideal classes of $\mathfrak p_i$ and the exponents $e_i$, where $x \mathcal O_K$ has prime ideal factorization $\prod \mathfrak p_i^{e_i}$.  There is also a precise description of the structure of distinct factorizations, which makes it clear that the structure of the distinct factorizations can get more complicated with more complicated class groups.  (In particular, you can use this description to get Carlitz's theorem and related results.)
But to return to your question, here is an example where you can describe $\eta(x)$ simply.  Suppose $K$ is quadratic, $x=p$ is a rational prime and $\mathfrak p$ is the prime ideal of $\mathcal O_K$ above $p$, then for any $n \in \mathbb N$,
$$ \eta(p^n) = \begin{cases}
1 & \mathfrak p \text{ principal or $p$ ramifies} \\
\lfloor \frac nm \rfloor + 1 & \text{else},
\end{cases} $$
where $m$ is the order of $\mathfrak p$ in the class group.
In particular, suppose $K_1$ and $K_2$ are quadratic fields with class numbers 3 and 2 respectively, and $p$ is rational prime unramified in both $K_1$ and $K_2$.  Let $\mathfrak p_1$ and $\mathfrak p_2$ be primes of $K_1$ and $K_2$ above $p$.  Then $p^n$ (for any $n > 3$) has more distinct factorizations in $K_2$ (a class number 2 field) than $K_1$ (a class number 3 field) if and only if $\mathfrak p_2$ is not principal.


TITLE: Positive integers $n$ such that $\sum_{k=1}^{n} k!$ divides $(n + 1)!$
QUESTION [4 upvotes]: I need help writing a proof by induction for this problem.
Recall that $n! = 1 \times 2 \times 3 \times \ldots \times n$. Find all positive integers $n$ such that $1! + 2! + \ldots + n!$ divides $(n + 1)!$.
Here I have what I have already tried:
http://openstudy.com/study#/updates/527aa321e4b0dc82d54350ff
Here I have what I have already tried.

REPLY [0 votes]: First, we know that $1!+2!+3!+....+n!$ is odd. Since half the numbers before $(n+1)!$ are even, we have that the highest power of two that divides $(n+1)!$ is $\ge 2^{\lfloor \frac{n+1}{2} \rfloor}$ .
Therefore, if we have $(1!+2!+3!+....+n!)k=(n+1)!$, for some $k$, then since the RHS is multiple of $2^{\lfloor \frac{n+1}{2} \rfloor}$,  $k$ must also be one, and we have:
$$k(1!+2!+3!+....+n!)\ge 2^{\lfloor \frac{n+1}{2} \rfloor}n!$$
It is then easy to argue by induction that  $2^{\lfloor \frac{n+1}{2} \rfloor}> n+1$ for $n\ge5$.
And solutions are $n=1,2$


TITLE: convergence and concentration inequalities
QUESTION [2 upvotes]: I am trying to solve the following problem:
Let $\lambda>0$ and $(X_n, n\geq 1)$ be a sequence of i.i.d. r.v. with $X_i\sim \mathcal{E}(\lambda)$.
Define as $Y_n= e^{X_n}, S_n = \sum_{j=1}^{n} Y_i$
a) For what values of $\lambda>0$ does there exist a number $\mu$ s.t. $\lim_{n\to +\infty} S_n/n \to \mu $  a.s.?
b) For what values of $\lambda>0$ the sequence of r.v. $$Z_n = \frac{S_n - n\mu}{\sqrt(n)}$$ converges in distribution to a limiting r.v. $Z$.
c) For what values of $\lambda>0$ there exists an $s_0>0$ such that $$\Lambda(s)=\log(\mathbb{E}(e^{sY_1}))<+\infty, \forall |s|\leq s_0$$
Here is my reasoning:
For a): the sequence of $Y_n$ is i.i.d. and the cdf of $Y_i$, $$F_{Y_i}(t)=F_x(\ln(t))=1-e^{\lambda \ln(t)} = 1-t^{-\lambda}, t>0$$
Being the seqence iid the sequence $S_n/n$ by the strong law of large numbers converges a.s. to $\mathbb{E}(Y_1)=\frac{\lambda}{1-\lambda}(e^{t(1-\lambda)})\big |^{+\infty}_0$ which is finite only if $\lambda>1$. Is that correct?
For b): using the Central Limit Theorem we have that $$\frac{Z_n}{\sigma} \to^d Z \sim \mathcal{N}(0,1)$$ where $\sigma = \sqrt(\operatorname{Var}(Y_i))$ thus, $Z_n \to^d \sigma Z \sim \mathcal{N}(0,\sigma^2)$ and $\sigma^2 =  \operatorname{Var}(Y_i)$ which is finite as long as $\mathbb{E}(Y_i^2)<+\infty$ which happens if $\lambda>2$.
So, we can say that $\operatorname{Var}(Y_i)= \frac{\lambda}{(\lambda-2)(\lambda-1)^2}$ and the converging distribution is $$Z' \sim \mathcal{N}(0,\operatorname{Var}(Y_i))$$ is that correct?
I am not looking for (only) error in calculation but maninly in conceptual errors I might be making...
For c) I am not really sure on how to proceed, I tried to compute $\mathbb{E}(e^{sY_i})$ but I came across the integral
$$\lambda\int_{0}^{+\infty} e^{sy}y^{-\lambda-2}$$
which I think should be solved by reiterated integration by parts until $\frac{\partial y^{-\lambda-2}}{\partial{y}}$ reaches $0$ is this the way to go?
Or else, should I compute $\mathbb{E}(e^{s\dot s \dot X_1})$.
I also tried to bound $\log(\mathbb{E}(e^{sY_1}))\leq \mathbb{E}(sY_1)$ using Jensen's inequality, but the last term seems to be always divergent and thus I cannot distinguish much. Any suggestions?

REPLY [2 votes]: There is no $\lambda >0$ such that $Ee^{sY_1} <\infty$ for $|s|$ sufficiently small. In fact $se^{x}-\lambda x \to \infty$ as $x \to \infty$ whenever s$>0$.
As far as a) and b) are concerned you have proved one way implications. It can be shown that $S_n /n \to \infty$ a.s. in the case   of non-negative random variables with infinite mean. Using this it follows that you answer to a) works in the opposite direction also: if $\lambda \leq 1$ then $S_n/n$ does not converge a.s. to a finite limit. Whether $Z_n /n$ can converge in distribution to a limiting variable even when the variance is not finite is a harder question.


TITLE: How to prove a function is not onto?
QUESTION [7 upvotes]: Let $f : Z\to Z$ be the function defined by $f(x) = 3x + 1$. Prove that $f $ is not onto, using a proof by contradiction. (Choose an integer $n$, and then prove ($\forall m \in Z$)($f(m) ≠ n$) by contradiction.)
so far I have:
\begin{gather*}
    1.\quad \text{Let}\quad y  = -1     \qquad   \text{assumption}\\
    2. \quad \text{Let}\quad f(x) = -1       \qquad \text{hypothesis}\\
    3. \quad 3x+1 = -1        \qquad \text{Definition of $f (2)$}\\
    4. \quad 3x  = -2            \qquad \text{Algebra}\\
    5. \quad x = -2/3          \qquad \text{Algebra}
\end{gather*}
What else is needed? I don't know where to go from here?

REPLY [0 votes]: y=3x+1 is not onto because x=(y-1)/3 which does not belong to Z rather belong to R. hence we can say there does not exist any integer x corresponding to any integer in codomain y for which f(x)=y hence f is not onto


TITLE: If the product of two matrices is invertible, are each matrices in the product invertible?
QUESTION [0 upvotes]: True or flase?
  If the product of two matrices is invertible then each matrices in the product is invertible.

It was said to be true but I didn't understood the argument:
a matrix A
\begin{pmatrix}
1 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 1
\end{pmatrix}
is non-invertible, its determinant equals $0$, furthermore it is equal to the product of diagonal coefficients of $A$, $A$ not being triangular.
Okay but $A*A$ isn't invertible neither, it gives:
\begin{pmatrix}
3 & 2 & 3\\ 2 & 2 & 2\\ 3 & 2 & 3
\end{pmatrix}
So I don't undertand the point, it should have demonstrate that $A*A$ was invertible and that $A$ wasn't.

REPLY [2 votes]: We can say that also without using determinants. Let $f_A$ and $f_B$ be the linear applications associated with our matrices and let $g=f_{AB}=f_A\circ f_B$. If $AB$ is invertible, $g$ is a isomorphism, hence it is both injective and surjective, hence $f_A$ is surjective and $f_B$ is injective.
But that is the same as stating that both $A$ and $B$ are invertible matrices.


TITLE: Combinations/Permutations intuition question
QUESTION [1 upvotes]: I have a relatively question that I am struggling with that is an intuition question and I'm hoping someone can help.
I start with the basic and simple question, how many ways can you get 3 heads EXACTLY with 5 flips. The math behind this simple: You have 5 slots for the first heads, 4 for the second, 3 for the 3rd = 5*4*3= 60. However, we have to divide this by 6 because order doesn't matter. While I somewhat understand I am struggling with the intuition so my question is.
What does the 60 represent? 60 represents if the order matters correct, but what larger set is 60 a subset of? Like in simple english what set is a 60 a subset of?

REPLY [0 votes]: To make the picture simpler, take exactly TWO heads in $5$ flips.
Then you can place
 - 1st H in position $1$, and the 2nd in one of the remaining $4$
 - 1st H in position $2$, and the 2nd in one of the remaining $3$
 -  ...
that is in
$4+3+2+1=10=5/2*4={5 \choose 2}=5*4/2$
ways.
If the problem was in how many ways you can place $1$ black, $1$ red and $3$ white marbles in a row, then this is $5*4=20$.
Now, if the red marble becomes black, you have to divide the above by $2=2!$,
and you get a situation which is analogous to the two heads above.


TITLE: Problem with 2nd order differential equation
QUESTION [1 upvotes]: I have the following differential equation: $$\frac{d^2 y}{dx^2}+\frac{1}{2} x \frac{dy}{dx}-y=0.$$ I have tried to multiply by $e^{x^2/4}$ to obtain $$\frac{d}{dx}\left(e^{\frac{x^2}{4}}\frac{dy}{dx}\right)=ye^{\frac{x^2}{4}}$$ but I don't know how to proceed from here, or even if this is the best first step to take. I can't integrate this because of the right hand side, but because of the $x/2$ term multiplying the middle term of the differential equation, I thought that doing this multiplication would have been useful.
Any help would be appreciated.

REPLY [2 votes]: Considering the equation $$\frac{d^2 y}{dx^2}+\frac{1}{2} x \frac{dy}{dx}-y=0.$$ multiplying instead by $e^{-\frac{x^2}{4}}$, you should arrive to $$\frac{d}{dx}\left(e^{-\frac{x^2}{4}}\frac{dy}{dx}\right)=y\,e^{-\frac{x^2}{4}}$$ the solution of which being related to Hermite_polynomials.
Quoting the Wilipedia page, "the probabilists' Hermite polynomials are solutions of the differential equation $$\frac{d}{dx}\left(e^{-\frac{x^2}{2}}\frac{du}{dx}\right)+\lambda\,e^{-\frac{x^2}{2}}u=0$$ where $\lambda$ is a constant.
So, applied to your case, the solution should be $$y=c_1 e^{-\frac{x^2}{4}} H_{-3}\left(\frac{x}{2}\right)+c_2
   \left(\frac{x^2}{2}+1\right)$$
Edit
Expanded, the solution write $$y= c_1 \left(\sqrt{\pi } \left(x^2+2\right)
   \text{erfc}\left(\frac{x}{2}\right)-2 e^{-\frac{x^2}{4}} x\right)+ c_2 \left(x^2+2\right)$$


TITLE: Determinant of a block matrix, blocks are transformations of the same matrix
QUESTION [3 upvotes]: Let's say that $\Phi, U_{11},U_{12},U_{21},U_{22}$ are $n\times n$ matrices with coefficients in $\mathbb{C}$. Consider the block matrix:
$$
M = \begin{bmatrix}
U_{11}\Phi & U_{12}\Phi \\
U_{21}\Phi & U_{22}\Phi
\end{bmatrix}
$$
where $U_{ij}\Phi$ is the usual matrix product. Can anything be deduced about the determinant of this matrix in terms of the determinant of $\Phi$?

REPLY [4 votes]: Observe that
$M=\begin{pmatrix}U_{11}\Phi & U_{12}\Phi \\ U_{21}\Phi & U_{22}\Phi\end{pmatrix}=\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\begin{pmatrix}\Phi & 0_{n\times n} \\ 0_{n\times n} & \Phi\end{pmatrix}$.
Then by property $\det(AB)=\det(A)\det(B)$, the determinant of $M$ is
$\det(M)=\det\left(\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\right)\det\left(\begin{pmatrix}\Phi & 0_{n\times n} \\ 0_{n\times n} & \Phi\end{pmatrix}\right)=\det\left(\begin{pmatrix}U_{11} & U_{12} \\ U_{21} & U_{22}\end{pmatrix}\right)[\det(\Phi)]^2$,
where the last step comes from the fact that the determinant of a block diagonal matrix is the product of the determinants of the diagonal blocks of such matrix (see General expression for determinant of a block-diagonal matrix).


text:
\section{Framework and some results from Parts~1 and~2}\label{main-one-two}

\subsection*{The spherically symmetric Einstein-Maxwell-scalar field system with a cosmological constant}

Consider a spherically symmetric spacetime with metric
\begin{equation}\label{metric}
g=-\Omega^2(u,v)\,dudv+r^2(u,v)\,\sigma_{\mathbb{S}^2},
\end{equation}
where $\sigma_{\mathbb{S}^2}$ is the round metric on the 2-sphere.
The Einstein-Maxwell-scalar field system with a cosmological constant $\Lambda$ and total electric charge $4\pi e$ reduces to the following system of equations:
the wave equation for $r$,
\begin{equation}\label{wave_r}
\partial_u\partial_vr=\frac{\Omega^2}{2}\frac{1}{r^2}\left(\vlinha\right),
\end{equation}
the wave equation for the massless scalar field $\phi$,
\begin{equation}\label{wave_phi}
\partial_u\partial_v\phi=-\,\frac{\partial_ur\,\partial_v\phi+\partial_vr\,\partial_u\phi}{r},
\end{equation}
the Raychaudhuri equation in the $u$ direction,
\begin{equation}\label{r_uu}
\partial_u\left(\frac{\partial_ur}{\Omega^2}\right)=-r\frac{(\partial_u\phi)^2}{\Omega^2},
\end{equation}
the Raychaudhuri equation in the $v$ direction,
\begin{equation}\label{r_vv}
\partial_v\left(\frac{\partial_vr}{\Omega^2}\right)=-r\frac{(\partial_v\phi)^2}{\Omega^2},
\end{equation}
and the wave equation for $\ln\Omega$,
 \begin{equation}\label{wave_Omega}
\partial_v\partial_u\ln\Omega=-\partial_u\phi\,\partial_v\phi-\,\frac{\Omega^2e^2}{2r^4}+\frac{\Omega^2}{4r^2}+\frac{\partial_ur\,\partial_vr}{r^2}.
\end{equation}

\subsection*{The first order system}

Given $r$, $\phi$ and $\Omega$, solutions of the Einstein equations,
let
\begin{equation}\label{nu_0}
\nu:=\partial_u r,
\end{equation}
\begin{equation}\label{lambda_0}
\lambda:=\partial_v r,
\end{equation}
\begin{equation}\label{bar_rafaeli}
\omega:=\frac{e^2}{2r}+\frac{r}{2}-\frac{\Lambda}{6}r^3+\frac{2r}{\Omega^2}\nu\lambda,
\end{equation}
\begin{equation}\label{mu_0}
\mu:=\truemu,
\end{equation}
\begin{equation}\label{theta}
\theta:=r\partial_v\phi,
\end{equation}
\begin{equation}\label{zeta}
\zeta:=r\partial_u\phi
\end{equation}
and
\begin{equation}\label{kappa_0}
 \kappa:=\frac{\lambda}{1-\mu}.
\end{equation}
Notice that we may rewrite~\eqref{bar_rafaeli} as
\begin{equation}\label{omega_sq}
\Omega^2=-\,\frac{4\nu\lambda}{1-\mu}=-4\nu\kappa.
\end{equation}
The Einstein equations imply the first order system for $(r,\nu,\lambda,\varpi,\theta,\zeta,\kappa)$
\begin{eqnarray}
 \partial_ur&=&\nu\label{r_u},\\
 \partial_vr&=&\lambda\label{r_v},\\
 \partial_u\lambda&=&\nu\kappa\partial_r(1-\mu)\label{lambda_u},\\
 \partial_v\nu&=&\nu\kappa\partial_r(1-\mu),\label{nu_v}\\
 \partial_u\omega&=&\frac 12(1-\mu)\left(\frac\zeta\nu\right)^2\nu,\label{omega_u}\\
 \partial_v\omega&=&\frac 12\frac{\theta^2}{\kappa},\label{omega_v}\\
 \partial_u\theta&=&-\,\frac{\zeta\lambda}{r},\label{theta_u}\\
 \partial_v\zeta&=&-\,\frac{\theta\nu}{r},\label{zeta_v}\\
 \partial_u\kappa&=&\kappa\nu\frac 1r\left(\frac{\zeta}{\nu}\right)^2,\label{kappa_u}
\end{eqnarray}
with the restriction
\begin{equation}\label{kappa_at_u}
\lambda=\kappa(1-\mu).
\end{equation}
Under appropriate regularity conditions for the initial data, the system of first order PDE~\eqref{r_u}$-$\eqref{kappa_at_u} also implies the Einstein equations.

\subsection*{Reissner-Nordstr\"{o}m initial data}

We take the initial data on the $v$ axis to be the data on the event horizon of a subextremal Reissner-Nordstr\"{o}m solution with mass $M$.
So, we choose initial data as follows:
\begin{equation}\label{RN1}
\left\{
\begin{array}{lclcl}
 r(u,0)&=&r_0(u)&=&r_+-u,\\
 \nu(u,0)&=&\nu_0(u)&=&-1,\\
 \zeta(u,0)&=&\zeta_0(u),
\end{array}
\right.\qquad{\rm for}\ u\in[0,U],
\end{equation}
and
\begin{equation}\label{RN2}
\left\{
\begin{array}{lclcl}
 \lambda(0,v)&=&\lambda_0(v)&=&0,\\
 \varpi(0,v)&=&\omega_0(v)&=&M,\\
 \theta(0,v)&=&\theta_0(v)&=&0,\\
 \kappa(0,v)&=&\kappa_0(v)&=&1,
\end{array}
\right.\qquad{\rm for}\ v\in[0,\infty[.
\end{equation}
Here $r_+>0$ is the radius of the event horizon. We assume that $\zeta_0$ is continuous and $\zeta_0(0)=0$.
We will denote $M$ by $\varpi_0$.

\begin{Lem}\label{sign}
Suppose that $(r,\nu,\lambda,\omega,\theta,\zeta,\kappa)$ is a solution of the characteristic initial value problem\/~\eqref{r_u}$-$\eqref{kappa_at_u}, with initial
conditions\/~\eqref{RN1} and\/~\eqref{RN2}.
Then:
\begin{itemize}
\item $\kappa$ is positive;
\item $\nu$ is negative;
\item $\lambda$ is negative on ${\cal P}\setminus\{0\}\times[0,\infty[$;
\item $1 - \mu$ is negative on ${\cal P}\setminus\{0\}\times[0,\infty[$;
\item $r$ is decreasing with both $u$ and $v$;
\item $\omega$ is nondecreasing with both $u$ and $v$.
\end{itemize}
\end{Lem}

\subsection*{Well posedness of the first order system and stability of the radius at the Cauchy horizon}

\begin{Thm}\label{r-stability}
Consider the characteristic initial value problem\/~\eqref{r_u}$-$\eqref{kappa_at_u}
 with initial data\/~\eqref{RN1}$-$\eqref{RN2}. Assume $\zeta_0$ is continuous and $\zeta_0(0)=0$. Then,
the problem has a unique solution defined on a maximal past set $\mysigma$. Moreover,
there exists $U>0$ such that
$\mysigma$ contains $[0,\myumax]\times[0,\infty[$,
$$
\inf_{[0,\myumax]\times[0,\infty[}r>0
$$
and
\begin{equation}\label{r-menos}
\lim_{u\searrow 0}r(u,\infty)=r_-.
\end{equation}
\end{Thm}
Here $r_->0$ is the radius of the Cauchy horizon of the Reissner-Nordstr\"om reference solution and $$r(u, \infty) = \lim_{v \to \infty} r(u,v)$$
(which exists and is decreasing). Similarly, we also define $$\varpi(u, \infty) = \lim_{v \to \infty} \varpi(u,v).$$

Theorem~\ref{r-stability} implies that the spacetime is extendible across the Cauchy horizon with a $C^0$ metric.

\subsection*{Two effects of any nonzero field}

\begin{Thm}\label{rmenos}
 Suppose that there exists a positive sequence $(u_n)$ converging to~0 such that $\zeta_0(u_n)\neq 0$.
 Then $r(u,\infty)<r_-$ for all $u>0$.
\end{Thm}

\begin{Lem}\label{l-kappa}
 Suppose that there exists a positive sequence $(u_n)$ converging to~0 such that $\zeta_0(u_n)\neq 0$.
Then
\begin{equation}\label{integral-k}
\int_{0}^\infty\kappa(u,v)\,dv<\infty\ {\rm for\ all}\ u>0.
\end{equation}
\end{Lem}
This lemma implies that the affine parameter of an outgoing null geodesic is finite at the Cauchy horizon.

\subsection*{Well posedness for the backwards problem}

In Section~\ref{CH}, we will extend the solutions of Einstein's equations beyond the Cauchy horizon.
For this we will need to solve a backwards problem, already discussed in Part 1.
The initial conditions will be prescribed as follows:
$$
({\rm I}_u)\qquad\left\{
\begin{array}{lcl}
 r(u,0)&=&r_0(u),\\
 \nu(u,0)&=&\nu_0(u),\\
 \zeta(u,0)&=&\zeta_0(u),
\end{array}
\right.\qquad{\rm for}\ u\in\left]0,U\right],
$$
$$
({\rm I}^v)\qquad\left\{
\begin{array}{lcl}
 \lambda(U,v)&=&\lambda_0(v),\\
 \omega(U,v)&=&\omega_0(v),\\
 \theta(U,v)&=&\theta_0(v),\\
 \kappa(U,v)&=&\kappa_0(v),
\end{array}
\right.\qquad{\rm for}\ v\in[0,V],
$$
We let $$\tilde{r}_0(v)=r_0(U)+\int_0^v\lambda_0(v')\,dv',$$ for $v\in[0,V]$.
We assume the regularity conditions:
\begin{align*}
\text{(h1)} \qquad & \text{the functions } \nu_0, \zeta_0, \lambda_0, \theta_0 \text{ and } \kappa_0 \text{ are continuous, and} \\
& \text{the functions } r_0 \text{ and } \omega_0 \text{ are continuously differentiable.}
\end{align*}
We assume the sign conditions:
$$
\hspace{-4cm}
\text{(h2)} \qquad
\left\{
\begin{array}{ll}
r_0(u)>0&{\rm for}\ u\in\left]0,U\right],\\
\tilde{r}_0(v)>0&{\rm for}\ v\in[0,V],\\
\nu_0(u)<0&{\rm for}\ u\in\left]0,U\right],\\
\kappa_0(v)>0&{\rm for}\ v\in[0,V].
\end{array}\right.
$$
We assume the three compatibility conditions:
\begin{align}
& r_0'=\nu_0,\label{initial_rp}\\
\text{(h3)} \qquad & \omega_0'=\frac 12\frac{\theta_0^2}{\kappa_0},\label{initial_v}\\
& \lambda_0=\kappa_0\left(\muuz\right).\label{kappa_at_zero} \hspace{3cm}
\end{align}

\begin{Thm}\label{existence-alt}
The characteristic initial value problem
with initial conditions\/~{\rm (I$_u$)} and\/~{\rm (I$^v$)} satisfying\/ {\rm (h1)}$-${\rm (h3)}
has a unique solution defined on a maximal reflected past set $\mathcal{R}$ containing a neighborhood of $]0,U]\times\{0\}\cup\{U\}\times[0,V]$.
\end{Thm}

\subsection*{Retrieving the Einstein equations from the first order system}

Henceforth, consider the additional regularity condition
\[
\text{(h4)} \qquad \nu_0, \kappa_0 \text{ and } \lambda_0 \text{ are continuously differentiable.} \hspace{2cm}
\]

\begin{Lem}\label{regular}
Suppose that $(r,\nu,\lambda,\omega,\theta,\zeta,\kappa)$ is the solution of the characteristic initial value problem, or the backwards problem,
with initial data satisfying\/ {\rm (h1)} to\/ {\rm (h4)}.
Then the function $r$ is $C^2$, and $\kappa$ is $C^1$.
\end{Lem}

\begin{Prop}\label{einstein-2}
 The functions $r$, $\phi$ and $\Omega$ satisfy~\eqref{wave_r}, \eqref{wave_phi}, \eqref{r_uu} and~\eqref{r_vv}.
\end{Prop}

\begin{Prop}\label{einstein-3}
The first order system~\eqref{r_u}$-$\eqref{kappa_at_u} implies \eqref{wave_Omega}.
Since equations~\eqref{wave_r}$-$\eqref{r_vv} imply the first order system,
equations~\eqref{wave_r}$-$\eqref{r_vv} also imply~\eqref{wave_Omega}.
\end{Prop}

\subsection*{The partition of spacetime into four regions}
In Part~2,
we divide $[0,U]\times[0,\infty[$ into four disjoint regions, separated by three curves, $\cg_{\ckrp}$, $\cg_{\ckrm}$ and $\gam$, where different estimates can be obtained (see Appendix~\ref{Costa}).
Next we explain how these curves are constructed.

\begin{center}
\begin{turn}{45}
\begin{psfrags}
\psfrag{u}{{\tiny $u$}}
\psfrag{v}{{\tiny $v$}}
\psfrag{x}{{\tiny $\!\!\!\cg_{\ckrp}$}}
\psfrag{y}{{\tiny $\!\!\cg_{\ckrm}$}}
\psfrag{z}{{\tiny $\gam$}}
\psfrag{e}{{\tiny $\!\!\!\!\!U$}}
\includegraphics[scale=.7]{fig22b.eps}
\end{psfrags}
\end{turn}
\end{center}

\subsection*{The curves $\cg_{\ckr}$}

We denote by $\cg_{\ckr}$ the level sets of the radius function $r^{-1}(\ckr)$.
These are spacelike curves and consequently may be parameterized by
$$v\mapsto (\uckr(v),v)\ \quad {\rm or}\ \quad u\mapsto (u,\vckr(u)).$$
We choose $\ckrp$ and $\ckrm$ sufficiently close to $r_+$ and $r_-$ with $r_-<\ckrm<\ckrp<r_+$.

\subsection*{The curve $\gam=\gamma_{\ckrm,\beta}$}
 Given $\ckrm$ as before and $\beta>0$, we define $\gamma$ to be the curve parametrized by
$$
u\mapsto\big(u,(1+\beta)\vgr(u))=:(u,v_{\gamma}(u)),\ {\rm for}\ u\in[0,U].
$$

\subsection*{The choice of $\beta$ so that $r$ and $\varpi$ are controlled}
Choose any $\beta$ such that
\begin{equation}\label{beta}
0<\beta<{\textstyle\frac 12\left(\sqrt{1-8\frac{\partial_r(1-\mu)(r_+,\varpi_0)}{\partial_r(1-\mu)(r_-,\varpi_0)}}-1\right)}.
\end{equation}
Let $0<\eps<\eps_0$. Then there exists $U_\eps>0$ such that
\begin{equation}\label{espana}
r(u,v)\geq r_--\eps\qquad{\rm and}\qquad\varpi(u,v)\leq\varpi_0+\eps
\end{equation}
for $(u,v)\in J^-(\gam)\cap J^+(\cg_{\ckrm})$ and $0<u\leq U_\eps$, provided that the parameters $\ckrp$, $\eps_0$ and $\delta$ are chosen so that
\begin{equation}\label{success}
\textstyle
\beta<\frac 12\left(
\sqrt{(1+\delta)^2-8
\frac{\mbox{\tiny $(\frac{\ckrp}{r_+})^{\hat\delta^2}$}\min_{r\in[\ckrp,r_+]}\partial_r(1-\mu)(r,\varpi_0)}{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)}}-(1+\delta)
\right).
\end{equation}
Here $\hat\delta$ is a bound for $\bigl|\frac\zeta\nu\bigr|$ in $J^-(\ckrp)$.
Suppose that there exist positive constants $C$ and $s$ such that
$|\zeta_0(u)|\leq Cu^s$. Then, instead of choosing $\beta$ according to~\eqref{beta} we may choose
\begin{equation}\label{beta-s}
0<\beta<{\textstyle\frac 12\left(\sqrt{1-8\frac{(1+s)\partial_r(1-\mu)(r_+,\varpi_0)}{\partial_r(1-\mu)(r_-,\varpi_0)}}-1\right)}.
\end{equation}
In this case, \eqref{success} should be replaced by
\begin{equation}\label{success-s}
\textstyle
\beta<\frac 12\left(
\sqrt{(1+\delta)^2-8
\frac{\mbox{\tiny $[(\frac{\ckrp}{r_+})^{\hat\delta^2}$}+s]\min_{r\in[\ckrp,r_+]}\partial_r(1-\mu)(r,\varpi_0)}{\partial_r(1-\mu)(r_--\eps_0,\varpi_0)}}-(1+\delta)
\right).
\end{equation}

config: arxiv


TITLE: Fermat's Little Theorem and Polynomial Congruence Relation
QUESTION [5 upvotes]: According to this Wikipedia article, we know that an integer $n\; (\geq 2)$ is prime if and only if the polynomial congruence relation
$$
(x - a)^n \equiv (x^n - a) \pmod{n}
$$
holds for all integers $a$ coprime to $n$ (or even just for some such integer $a$, in particular for $a = 1$).
Then the Fermat Little Theorem says, $n$ is a probable prime if
$$
a^n \equiv a \pmod{n}
$$
Inside the article it said $x$ should never be substituted by a number, so If we do so, then for primality testing of a number like $211$ we would have something like this as an example :
$$
(2 + 3)^{211} \equiv 2^{211} + 3 \pmod{n}
$$
So can't we use the above equation for some reduction in modular exponentiation?

REPLY [6 votes]: If you have $n=211$, you can use it, but you have to know your $n$ is prime. And then you would have $$(2+3)^{211} \equiv 2^{211} + 3 \pmod{211}$$
For a prime $p$ it is always true that:
$$ (x+y)^p \equiv x^p + y^p\pmod{p} $$
But if you are looking for the explanation of the test, you should not do that. The part that says you shouldn't substitute $x$ means, that you should be looking at the polynomial $(x-a)^n$ which will be determined by its coefficients, i.e. if $f=(x-a)^n$ then it can be written also as a sum $f=\sum_{i=0}^{n} a_{i}x^{n}$ (if you expand the product), and you should be checking if $a_i \equiv 0 \pmod{n}$ for all $1<i<n$ and $a_0\equiv a \pmod{n}$.


TITLE: Any reparametrisation of a regular curve is regular
QUESTION [2 upvotes]: So I'm having a little trouble algebraically showing this is true, the hint is that it is an exercise of the chain rule.
From definition, a parametrised curve $\tilde\gamma : J \rightarrow \mathbb{R}^n$ is a reparametrisation of $\gamma: I \rightarrow \mathbb{R}^n$ if $\exists$ a diffeomorphism $\phi : J \rightarrow I $  such that $\tilde\gamma(\tilde t) = \gamma \circ \phi (\tilde t) \forall \tilde t \in J$.
I feel like the result is obvious, I just don't know quite how to prove it algebraically.
I'm sure differentiating both sides would be a good start but past that I'm not sure.

REPLY [0 votes]: Okay. So the definition of a regular curve, $\mu:I\rightarrow\mathbb{R}^n$ is that $\forall t\in I:\mu'(t)\neq 0$. Assume that $\gamma$ is regular. Now.. a diffeomorphism is regular because it's a smooth bijection defined on an open set with an smooth inverse.  With that in mind observe that:
$\tilde{\gamma}'(\tilde{t})=(\gamma(\phi(\tilde{t})))'=\gamma'(\phi(\tilde{t}))\phi'(\tilde{t})$
But since both $\gamma$ and $\phi$ are regular, we know that $\gamma'(\phi(\tilde{t}))\phi'(\tilde{t})\neq 0$ for all $t\in J$.


TITLE: Link between average lifespan and half time in radioactivity
QUESTION [0 upvotes]: I have learnt that in radioactivity $λ=\ln2/t_{1/2}$, thus showing the link between λ and radioactive half time. I have also learnt how to apply it in calculations (such as $R={R_0}e^{-λt}$) etc.
Now I just learnt that average life-span of radioactive substances is calculated as $1/λ$.
This leads me to the following two questions:

What does $λ$ actually stand for? My book teaches how to use it, but never defines the meaning beyond stating that it is the "constant of radioactive decay".
Is there a way we can mathematically show why $1/λ$ is equal to average life-span, perhaps linking it back to $λ=\ln2/t_{1/2}$?

REPLY [2 votes]: Say we have a sample with a bunch of the original stuff, and you can see when each of them decays. One of them will take $t_1$, another $t_2$, another...
If we calculate the average of those, we get a time called 'average life', usually denoted by the greek letter $\tau$:
$\tau = \frac{\sum_i t_i}{N}$
If the sample is big enough, the number of decays we see in a time interval $dt$ is (*)
$ N\frac{dt}{\tau} $,
or, in terms of change in the number of undecayed nuclei,
$ dN=-N \frac{dt}{\tau} $.
If you solve the differential equation, you get:
$ \frac{dN}{dt} = -\frac{N}{\tau} \Rightarrow N(t)=N_0 e^{-t/\tau} $
Which, comparing to yours,
$ N(t)=N_0 e^{-t/\tau}=N_0 e^{-\lambda t} \Rightarrow \lambda = \frac{1}{\tau} $
(*) I know this is overly simplified, but you would probably have to work from Poisson statistics up to really justify this.
Edit: to better answer question 1, let's have a look at the activity by derivating the number of particles not yet disintegrated, $ N(t) $:
$ A(t) = - \frac{dN}{dt} = - \frac{d}{dt} \left( N_0 e^{-\lambda t} \right) = \lambda N_0 e^{-\lambda t} = \lambda N(t) $
So $ \lambda $, besides being the reciprocal of the average life, also links activity and number of radioactive particles.