text:
\section{Minimizers}\label{sec:minz}

In this section we use our Theorems \ref{thm:main}, \ref{thm:codim} to prove the partial regularity Theorem \ref{thm:partial-reg} for codimension-one area-minimizing free-boundary currents in a manifold $N$.  We will use the Nash embedding theorem and reduce our problem to one in Euclidean space, as in Section \ref{sec:codim}.  We prove some technical Lemmas, establishing appropriate non-concentration (\ref{lem:minz-non-conc}) and compactness (\ref{lem:minz-compact}) for almost-area-minimizing currents.  We then classify low-dimensional minimizing cones with free-boundary (Lemmas \ref{lem:minz-planes}, \ref{lem:low-dim-cones}), and use a standard-dimension reducing argument to get our partial regularity bound.

First, we establish some notation.  Given an open set $U \subset \R^{n+1+k}$, we let $\cI_n(U)$ be the set of integer-multiplicity rectifiable $n$-currents in $U$ with locally finite mass.  Given $T \in \cI_n(U)$, we will write $||T||$ for the mass measure, and $\spt T \subset U$ for the support of $T$.  Note that associated to every such $T$, there is an integral varifold which we will often denote by $T$ also, such that $\mu_T = ||T||$.  Given an open set $E \subset U$, we write $[E]$ for the $(n+1)$-current in $U$ obtained by integrating the standard orientation over $E$.  We say $E$ is a set of locally finite perimeter in $U$ if $\del[E] \in \cI_n(U)$.

Given a domain $\Omega \subset \R^{n+1+k}$, and $T \in \cI_n(U)$, we say $T$ is minimizing with free-boundary in $\Omega$ if $T = T\llcorner \sint\Omega$, and satisfies
\begin{equation}\label{eqn:T-minz}
||T||(W) \leq ||T+S||(W)
\end{equation}
for all $W \subset\subset U$, and all $S \in \cI_n(U)$ satisfying $\spt S \subset \overline{\Omega} \cap W$, $\del S \llcorner \sint\Omega = 0$.

Given $A \geq 0$, $\alpha \geq 0$, $\delta > 0$, we say $T$ is $(A, \alpha, \delta)$-almost-minimizing with free-boundary in $\Omega$ if $T = T \llcorner \sint\Omega$, and given $W$, $S$ as above then $\diam(W) \leq \delta$ and we have the inequality
\[
||T||(W) \leq ||T+S||(W) + A r^{n+\alpha}, \quad \diam(W) \leq \min\{ \delta,  r \} .
\]
instead of \eqref{eqn:T-minz}.

Let $\OmegaRef = \Omega_0 \times \R^{m+k}$ be a polyhedral cone domain in $\R^{n+1+k}$, and $\Omega \in \cD_\eps(\OmegaRef)$.  Recall that $\Omega$ is closed in $B_1$, $\overline{\sint\Omega} = \Omega$, and we have the natural stratification
\[
\del_0\Omega \subset \del_1\Omega \subset \ldots \subset \del_{n+1+k}\Omega = \Omega,
\]
where $\del_i\Omega$ consists of the points near which $\Omega$ is diffeomorphic to a polyhedral cone domain having at most $i$ dimensions of translational symmetry.

Given $T \in \cI_n(B_1)$ with $T = T \llcorner \sint\Omega$, we define $\reg T$ to be the set of points $x \in \spt T \subset \Omega$ with the following property: there is an $r > 0$, a $C^2$ diffeomorphism $\phi : B_r(x) \to B_r(x)$, a $V \in \cP_{T_x\Omega}$, and a $C^{1,\alpha}$ function $u : B_{2r}(x) \cap (x + V) \to V^\perp$, so that
\[
\phi(\spt T \cap B_r(x)) = \graph_{x + V}(u) \cap B_r(x).
\]
Note that this implies $\del_0\Omega \cap \reg T = \emptyset$ (but also recall that $\del_i\Omega = \emptyset$ for $i < m+k$).  Note further that if $x \in \reg T$, then the tangent plane of $T$ at $x$ lies in $\cP_{T_x\Omega}$.  We define $\sing T = \spt T \setminus \reg T$.

The global domains we are interested in are locally polyhedral domains, defined precisely here.
\begin{definition}\label{def:domain}
Let $N^{n+1}$ be a complete manifold.  A \emph{locally polyhedral domain} is a closed domain $\Omega \subset N$ with non-empty interior which is locally diffeomorphic to some polyhedral cone domain at every point.  Precisely, for every $x \in \Omega$ there is a radius $r > 0$, a polyhedral cone domain $\OmegaRef$, and a diffeomorphism $\phi : B_1(0^{n+1}) \to B_r(x \in N)$ so that $\phi(B_1 \cap \OmegaRef) = \Omega \cap B_r(x)$ and $D\phi|_0$ is an isometry.  We say $\Omega$ is $C^k$ is the associated diffeomorphism is $C^k$.

In the special case when every model polyhedral cone $\OmegaRef$ takes the form $L(\R^i \times [0, \infty)^{n-i})$, for some $i$ and linear isomorphism $L$, then $\Omega$ is said to be a \emph{domain with corners}.  We say $\Omega$ has dihedral angles $\leq \pi/2$ (resp. $=\pi/2$) if every model $\OmegaRef$ as above has dihedral angles $\leq \pi/2$ (resp. $=\pi/2$).  As before, we may call $\Omega$ with dihedral angles $\leq \pi/2$ \emph{non-obtuse}.
\end{definition}

\vspace{5mm}

The following two Lemmas prove the required compactness and closure theorems for our (almost-)area minimizing currents.
\begin{lemma}[Boundary non-concentration]\label{lem:minz-non-conc}
Let $\OmegaRef$ be a polyhedral cone domain in $\R^{n+1+k}$.  Let $\Omega \in \cD_\eps(\OmegaRef)$, and $T \in \cI_n(B_1)$ be $(A, 0, 1)$-almost-minimizing with free-boundary in $\Omega$, such that $\del T \llcorner \sint\Omega = 0$.

Then for $\eps \leq \eps(\OmegaRef)$, there is a continuous function $\eta(\tau) : [0, 1] \to [0,1]$ depending only on $\theta, \OmegaRef$, satisfying $\eta(0) = 0$, such that
\[
||T||(B_\tau(\del\Omega) \cap B_\theta) \leq ||T||(B_1) \eta(\tau) + 2 A.
\]
\end{lemma}

\begin{remark}
It should in fact be true for minimizers that $||\del T||(B_\theta) \leq C(\theta, \OmegaRef, ||T||(B_1))$ (c.f. \cite{Gr}).  However the above weaker statement is easier to prove in our more singular setting setting, and suffices for our purposes.
\end{remark}

\begin{proof}
Fix any unit vector $\tau \in \sint\Omega$, and let $H$ be the half-space with outer normal given by $-\tau$.  Then for $\eps(\OmegaRef)$ sufficiently small $\Omega \subset H$, and we can write $\del\Omega$ as a Lipschitz graph over $\del H$.  Given $R \geq 2$, $\eps(\OmegaRef)$ small, and $x \in B_1 \cap \Omega$, there is a unique $\zeta(x) \in \del\Omega$ such that $\zeta(x) - R\tau = \lambda_x (x - R \tau)$ for some $\lambda_x \geq 1$.  Provided we fix $R(\theta, \OmegaRef)$ to be sufficiently large, then $\zeta$ is a Lipschitz function satisfying
\begin{equation}\label{eqn:T-non-conc-1}
\zeta(B_{\theta}) \subset B_{(1+\theta)/2}, \quad |x - \zeta(x)| \leq c(\OmegaRef) d(x, \del\Omega), \quad ||D\zeta||_{L^\infty(B_1)} \leq c(\OmegaRef) .
\end{equation}

Since $T = T \llcorner \sint\Omega$, let us view $T \in \cI_n(\sint\Omega \cap B_1)$, in which case $\del T = 0$.  Define $f(x) = d(x, \del\Omega) + g(|x|)$, where $g = 0$ on $B_\theta$, $g$ is increasing, $g = 1$ on $B_{(1+\theta)/2}$, and $|Dg| \leq 4/(1-\theta)$.  Then for all $h \leq 1$, $T \llcorner \{ f < h \} \in \cI_n(\sint\Omega \cap B_{(1+\theta)/2})$.

Let $T_h = \langle T, f, h\rangle$ be the slice (see \cite[section 28]{simon:gmt}) of $T$ at $f = h$, defined for a.e. $h$, and let $m(h) = ||T \llcorner \{ f < h\}||(B_1)$.  By the coarea formula, we have
\begin{equation}\label{eqn:T-non-conc-2}
||T_h||(B_1) \leq c(\theta) m'
\end{equation}
for a.e. $h < 1$.  On the other hand, again for a.e. $h$, by slicing we have
\[
T_h = \del (T \llcorner \{ f < t\}).
\]
Therefore if we define
\[
F(t, x) = t x + (1-t) \zeta(x), \quad t \in [0, 1], x \in B_\theta,
\]
then $\del F_\sharp ([0, 1] \times T_h) \llcorner \sint\Omega = T_h$, $\spt F_\sharp ([0, 1] \times T_h) \subset \Omega \cap B_{(1+\theta)/2}$, and so by area comparison and the homotopy formula (\cite[26.22]{simon:gmt}) we have
\begin{equation}\label{eqn:T-non-conc-3}
m(h) \leq ||F_\sharp T_h||(B_1) + A \leq c(\OmegaRef) h ||T_h||(B_1) + A,
\end{equation}
having used \eqref{eqn:T-non-conc-1}, and the fact that $|x - \zeta(x)| \leq c d(x, \del\Omega) \leq c f(x)$.  Together, \eqref{eqn:T-non-conc-2} and \eqref{eqn:T-non-conc-3} imply that for a.e. $h< 1$ we have
\[
m(h) \leq c(\OmegaRef, \theta) h m' + A,
\]
and therefore since $m$ is increasing there is an $\alpha > 0$ so that $h^{-\alpha} (m(h) - A)$ is increasing for $h \in (0, 1)$.  Letting $\eta(h) = h^\alpha$ proves the Lemma.
\end{proof}

\begin{lemma}\label{lem:minz-compact}
Let $\Omega = \Omega_0^l \times \R^{m+k}$ be a polyhedral cone domain in $\R^{n+1+k}$.  Let $\eps_i \to 0$, $A_i \to 0$, $\Omega_i \in \cD_{\eps_i}(\Omega)$, and $T_i \in \cI_n(B_1)$.  Suppose $T_i$ is $(A_i, 0, 1)$-area-minimizing with free-boundary in $\Omega_i$, and satisfies
\begin{equation}\label{eqn:compactness-hyp1}
\del T_i \llcorner \sint\Omega = 0, \quad \sup_i ||T_i||(B_1) < \infty.
\end{equation}
Then after passing to a subsequence, there is a $T \in \cI_n(B_1)$ which is area-minimizing with free-boundary in $\Omega$, such that $T_i \to T$ as currents in $B_1$ and $||T_i|| \to ||T||$ as Radon measures on $B_1$.

If the $T_i$ as varifolds lie in $\cIVT(\Omega_i, B_1)$ with $||H^{tan}_{T_i}||_{L^\infty(B_1)} \to 0$, then $T \in \cIVT(\Omega, B_1)$, $T_i \to T$ as varifolds, $||H^{tan}_T||_{L^\infty(B_1)} = 0$.

Write $\Omega_i = \Phi_i(\Psi_i(\Omega)) \cap B_1$, and let $N_i = \Phi_i(\Psi_i(\Omega_0 \times \R^m \times \{0^k\})) \cap B_1$.  Suppose there are relatively open sets $E_i \subset N_i \cap B_1$ so that $T_i = \del[E_i] \llcorner \sint\Omega_i$.  Then $T$ is multiplicity-one $||T||$-a.e. and for every $x \in \reg T \cap B_1$, there is a neighborhood $B_r(x)$ such that $\sing T_i \cap B_r(x) = \emptyset$ for all $i >> 1$.
\end{lemma}

\begin{proof}
By assumption we have $|\Phi_i - \mathrm{id}|_{C^2(B_2)} \to 0$, $|\Psi_i - \mathrm{Id}| \to 0$.  We can find $C^2$ functions $F_i : B_2 \to B_2$ functions satisfying
\[
\Omega_i = F_i(\Omega) \cap B_1, \quad |F_i - \mathrm{id}|_{C^2(B_2)} \to 0,
\]
such that for every $U \subset\subset \sint\Omega \cap B_1$, we have $F_i|_U = \mathrm{id}$ for all $i>>1$.

By a diagonalization argument, after passing to a subsequence, we can find a $T \in \cI_n(\sint\Omega)$ so that $T_i(\omega) \to T(\omega)$ for any smooth $n$-form compactly supported in $B_1 \cap \sint\Omega$.  We have $||T||(\sint\Omega \cap B_1) < \infty$, so extend $T$ to be an element of $\cI_n(B_1)$ by restriction $T := T \llcorner \sint\Omega$.  Because of our definition of $T$, and convergence $T_i \to T$ as currents on compact subsets of $\sint\Omega \cap B_1$, we can also assume that the $||T_i||$ limit to some Radon measure on $B_1$, which is $\geq ||T||$.

By slicing theory and \eqref{eqn:compactness-hyp1}, and after passing to a further subsequence, we can find $d_j \to 0$, $r_k \to 1$ so that if $D_{jk} = B_{r_k} \cap \{ d_{\del\Omega} > d_j\}$, then
\begin{equation}\label{eqn:compactness-1}
||\del (T_i \llcorner D_{jk} )||(B_1) \leq C_{jk}
\end{equation}
for $C_{jk}$ independent of $i$.  Since, for any fixed $j, k$, $T_i \llcorner D_{jk} \to T \llcorner D_{jk}$, we have $\del (T_i \llcorner D_{jk}) \to \del (T \llcorner D_{jk})$ and hence
\begin{equation}\label{eqn:compactness-2}
||\del (T \llcorner D_{jk} )||(B_1) \leq C_{jk}
\end{equation}
also.

By Lemma \ref{lem:minz-non-conc}, we have
\begin{equation}\label{eqn:compactness-4}
||T_i ||(B_{r_k} \setminus D_{jk}) \leq \eta_k(d_j) + 2A_i
\end{equation}
for some continuous function $\eta_k$ independent of $i, j$, satisfying $\eta_k(0) = 0$.  Lower semi-continuity and our hypothesis $A_i \to 0$ implies
\begin{equation}\label{eqn:compactness-5}
||T||(B_{r_k} \setminus D_{jk}) \leq \eta_k(d_j).
\end{equation}
Inequalities \eqref{eqn:compactness-4} and \eqref{eqn:compactness-5} imply that $T_i \to T$ as currents in $B_1$.

Fix $S \in \cI_n(B_1)$ with $\spt S \subset \Omega \cap B_1$., and $\del S \llcorner \sint\Omega = 0$.  We wish to prove that
\[
||T||(B_1) \leq ||T+S||(B_1).
\]
Since $T = T\llcorner \sint\Omega$, there is no loss in assuming $S = S \llcorner \sint\Omega$ also.  Define $S_i = (F_i)_\sharp S$, so that for $i >> 1$: $\spt S_i \subset \Omega_i \cap B_1$, $\del S_i \llcorner \sint \Omega_i = 0$, $S_i = S_i\llcorner \sint\Omega_i$ and $||S - S_i||(B_1) \to 0$.

Fix a $k$ such that $\spt S \subset B_{r_k}$ for all $i$ large, and fix any $j$ arbitrary.  By \eqref{eqn:compactness-1}, \eqref{eqn:compactness-2}, slicing theory and the deformation theorem (see e.g. \cite[Theorem 7.2.4]{simon:gmt}), we can find $P_i \in \cI_{n+1}(B_1)$, $R_i \in \cI_n(B_1)$ such that
\[
(T_i - T) \llcorner D_{jk} = \del P_i + R_i
\]
and
\[
\spt P_i, \spt R_i \subset \overline{D_{j+1,k}}, \quad ||P_i||(B_1) + ||R_i||(B_1) + ||\del P_i||(\del B_{r_k}) \to 0.
\]
Since $T_i$ is $(A_i, 0, 1)$-almost-minimizing with free-boundary in $\Omega_i$, and $\del P_i$ is an admissible competitor, we have
\[
||T_i||(B_r) \leq ||T_i + \del P_i + S_i||(B_r) + A_i \quad \forall r_k < r < 1,
\]
and hence taking $r \to r_k$, we get
\begin{align}
||T_i||(B_{r_k})
&\leq ||T_i + \del P_i + S_i||(B_{r_k}) + ||\del P_i||(\del B_{r_k}) + A_i \\
&\leq \left( ||T \llcorner D_{jk} + S|| + ||S - S_i|| + ||R_i|| \right)(B_{r_k}) + ||\del P_i||(\del B_{r_k}) + A_i
\end{align}
Let $i \to \infty$, and then by lower-semi-continuity we get
\begin{align}
||T||(B_{r_k})
&\leq ||T \llcorner D_{jk} + S||(B_{r_k}) \\
&\leq ||T +S||(B_{r_k}) + ||T ||(B_{r_k} \setminus D_{jk}) \\
&\leq ||T+S||(B_{r_k}) + \eta_k(d_j) \label{eqn:compactness-3}
\end{align}
Since \eqref{eqn:compactness-3} holds for every $j$ ($k$ fixed), we can take $j \to \infty$ to get
\[
||T||(B_{r_k}) \leq ||T+S||(B_{r_k}),
\]
which proves that $T$ is minimizing.  If we apply the same argument to $S = 0$, then we get $||T||(B_{r_k}) = \lim_i||T_i||(B_{r_k})$, which by lower-semi-continuity implies $||T_i|| \to ||T||$.  This proves the first part of the Lemma.

\vspace{5mm}

Abusing notation, let us write $T_i$, $T$ for the underlying varifolds associated to the $n$-currents, and assume as in our hypothesis that $T_i \in \cIVT(\Omega_i, B_1)$.  By Theorem \ref{thm:first-var}, the $T_i$ have uniformly bounded first variation, and hence after passing to a further subsequence we can assume $T_i$ converge as varifolds in $B_1$.  Since $T_i$, $T$ are both integral varifolds, and $||T_i|| \to ||T||$, then we must have $T_i \to T$ as varifolds.

Fix any $X \in C^1_c(B_1)$ tangential to $\Omega$.  Let $X_i(x) = DF_i|_{F_i^{-1}(x)} X(F_i^{-1}(x))$.  Then $X_i \in C^1_c(B_1)$, and $X_i$ is tangential to $\Omega_i$, and $|X_i - X|_{C^0(B_1)} \to 0$.  Pick $\theta < 1$ such that $\spt X_i, \spt X \subset B_\theta$.  We have
\[
|\delta T_i(X)| \leq C(\Omega)|X|_{C^0(B_1)} ||H^{tan}_{T_i}||_{L^\infty(B_1)} ||T_i||(B_\theta) + C(\Omega) |X_i - X|_{C^0(B_1)} ||\delta T_i||(B_\theta),
\]
and hence $\delta T(X) = 0$.  This implies $T \in \cIVT(\Omega, B_1)$, and $H^{tan}_T = 0$.

\vspace{5mm}

Suppose $T_i = \del[E_i] \llcorner \sint\Omega_i$ for some relatively open sets $E_i \subset N_i \cap B_1$.  Then, after passing to a further subsequence as necessary, we can assume $[E_i] \to E \subset \Omega_0 \times \R^m \times \{0^k\}$, and hence $T = \del[E] \llcorner \Omega_0 \times \R^m$ has multiplicity-one $||T||$-a.e.  If $x \in \reg T$, then at sufficiently small scales $T \llcorner B_r(x)$ is varifold-close to a multiplicity-one plane in $\cP_{T_x\Omega}$, and hence the $T_i \llcorner B_r(x)$ lie close to this plane also.  For $r$ sufficiently small, and $i$ sufficiently large, we deduce by Theorems \ref{thm:main}, \ref{thm:codim} that $\sing T_i \cap B_r(x) = \emptyset$.
\end{proof}

\vspace{5mm}

We now work towards classifying low-dimensional tangent cones.
\begin{lemma}\label{lem:minz-planes}
Let $\Omega^{n+1}$ be a $0$-symmetric polyhedral cone domain.  If $n \geq 2$ assume $\Omega$ is non-obtuse.  Then $\Omega$ is a domain-with-corners, and given any $T \in \cI_n(\R^{n+1})$ a free-boundary minimizing cone in $\Omega$, such that $\spt T$ is contained in a plane, then $T = 0$.
\end{lemma}

It is plausible Lemma \ref{lem:minz-planes} fails in higher dimensions when $\Omega$ does not satisfy the dihedral angle condition.
\begin{example}\label{ex:possible-minz-plane}
Let $\Omega^3$ be the intersection of the half-spaces with outer normals given by
\[
(1, 0, 1), \quad (-1, 0, 1), \quad (0, \eps, 1).
\]
Then when $\eps > 0$ is small-ish, it may be the case that the plane $y = 0$ is not minimizing with free boundary in $\Omega$.
\end{example}

\begin{proof}[Proof of Lemma \ref{lem:minz-planes}]
By the constancy theorem $T = k[P \cap \sint\Omega]$ for some integer $k \geq 0$, and some oriented plane $P$.

We perform induction on $n$. First assume $n = 1$: $\Omega = W^2$ is a $2$-dimensional wedge with interior angle $\beta < \pi$, and $P \cap W$ is a ray.  If we write $\del W = L_1 \cup L_2$ where $L_i$ are the two rays meeting at angle $\beta$, then after relabeling as necessary the rays $P$, $L_1$ meet at some angle $< \pi/2$.  Take any point $q\in P$, and let $q_1=\proj_{L_1}q$. Then $|qq_1|<|oq|$, so $P$ is not length-minimizing. We therefore must have $k=0$ and hence $T=0$.

Assume the statement holds for all positive integers less than $n$. Since $\Omega^{n+1}$ is $0$-symmetric and non-obtuse, by \cite[Theorem 1.1]{Coxter1934discrete}, $\Omega$ is simplicial. In other words, there exists $n+1$ half spaces $H_1,\cdots,H_{n+1}$ such that $\Omega=\cap_{i=1}^{n+1}H_i$, and if $F_i=\partial\Omega\cap \partial H_i$, then each $F_i$ and $F_j$ meet along some $(n-1)$-dimensional set.  Let $\nu_i$ denote the outer unit normal vector of $\partial H_i$. Since $\Omega$ is non-obtuse, $\nu_i\cdot \nu_j\le 0$, and
\[\Omega=\{x\in \R^n: x\cdot \nu_i\le 0, i=1,\cdots,n+1\}.\]

Suppose, for the sake of contradiction, that $k\ne 0$. We first observe that $\spt T\cap \partial_{n-1}\Omega=\{0\}$. Otherwise, let $q\in \spt T\cap \partial_{n-1}\Omega$. Then $Tan_q T=T'\times \R^{n+1-j}$, $T_q\Omega=\Omega'^j\times \R^{n+1-j}$ for some $2\le j\le n$, $\Omega'$ is $0$-symmetric, and $T'$ is free-boundary minimizing in $\Omega'$, contradicting the induction hypothesis.

Therefore $\spt T\cap \partial \Omega$ only in the smooth part of $\partial \Omega$, and thus $\spt T$ meets $\partial \Omega$ orthogonally. Denote $\Omega_0=P\cap \Omega$. Note that $\Omega_0$ is a $0$-symmetric $n$-dimensional polyhedral cone. Thus by dimension counting, $P$ intersects exactly $n$ faces, say $F_1,\cdots,F_n$ orthogonally. After a rotation if necessary, we may assume that $P=\{x_{n+1}=0\}$, and $e_n\in P\cap \sint \Omega$. Since $e_n\in \sint\Omega\subset H_{n+1}$, we have $e_{n}\cdot \nu_{n+1}\le 0$. On the other hand, regarded as a convex polyhedron in $P\cong \R^n$, $\Omega_0$ has outer unit normals $\nu_1,\cdots, \nu_n$, and $e_n\in \sint(\Omega_0)$. Therefore there exists a linear combination
\[e_n=\sum_{j=1}^n \lambda_j \nu_j\]
with $\lambda_j\le 0$. Hence
\[e_n\cdot \nu_{n+1}=\sum_{j=1}^n \lambda_j \nu_j\cdot \nu_{n+1}\ge 0.\]

As a result, $e_n\cdot \nu_{n+1}=0$, and hence $e_n\in \partial H_{n+1}$. This implies that $P\cap F_{n+1}\ne \{0\}$, contradiction.
\end{proof}

\begin{lemma}\label{lem:low-dim-cones}
Let $T \in \cI_n(\R^{n+1})$ be a dilation-invariant minimizing current with free-boundary in a polyhedral cone domain $\Omega^{n+1}$.  Then $T$ is entirely regular if:
\begin{enumerate}
\item $n = 1$;
\item $n = 2$ and $\Omega$ is non-obtuse;
\item $n \leq 6$ and $\Omega = [0, \infty)^l \times \R^{n+1-l}$.
\end{enumerate}
\end{lemma}

\begin{proof}
Suppose $n \leq 6$ and $\Omega = [0, \infty)^l \times \R^{n+1-l}$.  When $l = 0$ then $T$ is minimizing without boundary in $\R^{n+1}$, and hence $T$ is planar by Simons' theorem \cite{Simons}.  For general $l$, we can assume by induction that $\sing T \subset \{0\}$.  Otherwise, if $x \in \sing T \setminus \{0\}$, then by \eqref{eqn:lower-reg} and Lemma \ref{lem:minz-compact} we would be able to find a non-zero singular tangent cone $T' \times [\R]$ at $x$ which is minimizing with free-boundary in some $[0, \infty)^{l-1} \times \R^{n-l}$.

Now by reflection we can obtain a cone $\tilde T$ in $\R^{n+1}$ which is $C^{1,\alpha}$ away from $0$, and smooth and stable away from finitely-many $(n-2)$-planes.  By standard interior elliptic regularity and a cutoff argument it follows that $\tilde T$ is smooth and stable on all of $\R^{n+1} \setminus \{0\}$, and hence by Simons' theorem $\tilde T$ is planar.

Suppose $n = 1$.  By the previous characterization, we can WLOG assume $\Omega = W^2$ is a non-obtuse wedge.  Then $T$ is a cone over finitely-many points, and hence is a union of minimizing rays (with possible multiplicity) in $W^2$.  But then by Lemma \ref{lem:minz-planes} $T = 0$.  This completes the $n = 1$ case.

Suppose $n = 2$ and $\Omega$ is non-obtuse.  We claim that $\spt T \cap (\del_1\Omega \setminus \{0\}) = \emptyset$.  Otherwise, if there were an $x \in \spt T \cap (\del_1\Omega \setminus \{0\})$ then by \eqref{eqn:lower-reg}, Lemma \ref{lem:minz-compact} (and a rotation as necessary) we could obtain a non-zero tangent cone $T' \times [\R]$ in some cone $W^2 \times \R$, where $W^2$ is a wedge.  But $T'$ is minimizing in $W^2$, and hence by our $n = 1$ case we would have $T' = 0$, which is a contradiction.

Therefore $\spt T \setminus \{0\}$ meets $\del\Omega$ only where $\del\Omega$ is planar, and hence by reflection and interior regularity $\spt T$ is smooth away from $0$.  Moreover, by the free-boundary condition $\spt T$ satisfies the usual stability inequality
\[
\int_{\spt T} |A|^2 \zeta^2 \leq \int_{\spt T} |\nabla \zeta|^2 \quad \forall \zeta \in C^1_c(\R^{n+1} \setminus \{0\}),
\]
and $\del_n |A| = 0$ along $\spt T \cap \del\Omega \setminus \{0\}$.  (Here $A$ is the second fundamental form of $\spt T$.)  It then follows by the usual proof of Simons' theorem that $\spt T$ is a finite union of disjoint free-boundary planes in $\sint\Omega$.

As before we can assume WLOG that $\Omega$ is either $W^2 \times \R$ or $0$-symmetric.  If $\Omega = W^2 \times \R$ then since $\spt T$ is planar and $\spt T \cap (\{0\}\times \R) \subset \{0\}$ we must have $\spt T = W^2 \times \{0\}$, and hence $T$ is regular.  If $\Omega$ is $0$-symmetric then by Lemma \ref{lem:minz-planes} we must have $T = 0$.
\end{proof}

\begin{comment}
\begin{lemma}\label{lem:cone-is-plane}
Let $\Omega^{n+1}$ be a polyhedral cone domain, for $1 \leq n \leq 6$.  Suppose $T \in \cI_n(\R^{n+1})$ is a free-boundary minimizing cone in $\Omega$, such that $\sing T \subset \{0\}$ and $\del T \llcorner \sint\Omega = 0$.  Suppose $\reg T \setminus \{0\}$ extends in $C^{1,1}$ to $\Omega \setminus (\{0\}\cup \del_{n-3}\Omega)$, and admits the bound
\begin{equation}\label{eqn:cone-plane-hyp}
||A(x)||_{L^\infty(\Omega \cap U \setminus B_\delta(\Omega_{n-3}\Omega)))} \leq c(\OmegaRef, K) \delta^{\alpha-1} \quad \forall \delta > 0
\end{equation}
for any $U \subset \subset \R^{n+1} \setminus \{0\}$.  Here $A(x)$ is the second fundamental form of $\reg T$, $\alpha > 0$.

Then $\spt T$ is a finite union of planes $P_i$, which are disjoint in $\sint\Omega$.
\end{lemma}

\begin{remark}
The bound \eqref{eqn:cone-plane-hyp} arises naturally from an estimate like \eqref{eqn:higher-main-concl}.
\end{remark}

\begin{proof}
The case when $n = 1$ is standard, following from the fact that $T \llcorner \del B_1$ is a finite union of points.  Suppose now $n \geq 2$.  Let $|A|$ be the norm of the second fundamental form of $\reg T \llcorner \sint\Omega$.  Since $\del_n\Omega$ is piecewise-flat, by reflection $A$ extends smoothly to $\del_n\Omega \setminus \del_{n-1}\Omega$, and satisfies $\nu_{\Omega} \cdot \nabla |A_\Sigma| = 0$ here.  The stability inequality becomes
\[
\int |A|^2 \zeta^2 d\mu_T \leq \int |\nabla \zeta|^2 d\mu_T
\]
for all $\zeta \in C^1_c(\R^{n+1} \setminus \del_{n-1}\Omega)$.  Replacing $\zeta$ with $|A| \zeta$ like in the usual proof of Simons' theorem, we obtain
\begin{equation}\label{eqn:flat-cone-1}
2 \int |x|^{-2} |A|^2 \zeta^2 d\mu_T \leq \int |A|^2 |\nabla \zeta|^2 \quad \forall \zeta \in C^1_c(\R^{n+1} \setminus \del_{n-1}\Omega).
\end{equation}

Fix any $\zeta \in C^1_c(\R^{n+1} \setminus (\{0\} \cup \del_{n-3}\Omega))$.  Let $d_i : \R^{n+1} \to \R$ be the signed distance function to $\del_i \Omega$ (which is positive inside $\sint\Omega$).  By our hypothesis that $T$ is regular away from $0$, we have
\[
\mu_T(\{ x \in \spt \zeta : d_i \leq r \}) \leq C r^{n-i+1}
\]
for all $r > 0$ sufficiently small, and $C$ independent of $r$.  Let $d_{i,\sigma}$ be a standard mollifcation of $d_i$, with a positive, radially symmetric mollifier.  Take $\rho > 0$.  For $\sigma = \sigma(\rho, \zeta)$ sufficiently small, we have
\[
\frac{1}{2} d_i \leq d_{i,\sigma} \leq 2d_i, \quad |D d_{i,\sigma}| \leq 1
\]
for $x \in \spt \zeta \cap \{ \rho^2/2 \leq d_i(x) \leq 2\rho \}$.  Define $\phi \equiv \phi_{\rho, \zeta} : \Sigma \to \R$ be the usual Lipschitz log-cutoff function:
\begin{equation}\label{eqn:flat-cone-2}
\phi(x) = \min \{ 1, \frac{1-\log(d_{n-1,\sigma}(x)/\rho^2)_+}{\log(1/\rho)} \}
\end{equation}
with the choice of $\sigma$ as above.  Let $N$ be the smallest such that $2^N \geq 1/\rho$.  Then we have for $\rho$ small:
\begin{align}
\int \zeta^2 |\nabla \phi|^2 d\mu_T
&\leq C(\zeta) \int_{\spt (\zeta \phi)} \frac{1}{d_{n-1}(x)^2 \log^2 \rho} d\mu_T \\
&\leq C \sum_{i=0}^N \frac{\mu_T \{ x \in \spt \zeta : d_{n-1}(x) \leq 2^{i+1} \rho^2 \}}{4^i \rho^4 \log^2 \rho}  \\
&\leq C \sum_{i=0}^N \frac{4^{i+1} \rho^4}{4^i \rho^4 \log^2 \rho} \\
&\leq -C/\log(\rho) ,
\end{align}
for $C$ independent of $\rho$.  Similarly, for $\rho$ sufficiently small we have $|\nabla\phi| \leq |\nabla \phi|^2$, and so we have the same estimate for $\int_{\spt \zeta} |\nabla \phi| d\mu_T$.

By hypothesis $|A|$ extends to a locally bounded function on $\del\Omega \setminus \del_{n-3}\Omega$, and hence $|A|$ is uniformly bounded on $\spt \zeta$.  Therefore if we replace $\zeta$ with $\zeta\phi$ in \eqref{eqn:flat-cone-1}, we can estimate
\begin{equation}\label{eqn:flat-cone-3}
2\int |x|^{-2} |A|^2 \zeta^2 \phi^2 d\mu_T \leq \int |A|^2 |\nabla \zeta|^2 \phi^2 d\mu_T + C/\log\rho,
\end{equation}
for $C$ independent of $\rho$.  By the dominated convergence theorem we deduce that \eqref{eqn:flat-cone-1} holds all $\zeta \in C^1_c(\R^{n+1} \setminus (\{0\}\cup \del_{n-3}\Omega))$.

Now fix a $\zeta \in C^1_c(\R^{n+1} \setminus \{0\})$.  The hypothesis \eqref{eqn:cone-plane-hyp} implies that there are constants $C = C(\zeta, T)$, $\alpha = \alpha(\Omega)$, so that
\[
|A(x)| \leq C d_{n-3}(x)^{\alpha-1}, \quad \text{ for } \mu_T-a.e. x \in \spt \zeta \cap \spt T \setminus \del_{n-3}\Omega.
\]
Let $\phi$ be as in \eqref{eqn:flat-cone-2}, except with $d_{n-3}$ in place of $d_{n-1}$.  Then, as before, we can compute
\begin{align}
\int_{\spt \zeta} |A|^2 |\nabla \phi|^2 d\mu_T
&\leq C \sum_{i=0}^N \frac{ \mu_T \{ x \in \spt \zeta : d_{n-3}(x) \leq 2^{i+1} \rho^2 \}}{(2^{i+1} \rho^2)^{4-2\alpha} \log^2 \rho} \\
&\leq C \sum_{i=0}^N \frac{ (2^i \rho^2)^{2\alpha}}{\log^2\rho } \\
&\leq -C / \log \rho,
\end{align}
for $C$ independent of $\rho$.  Similarly, the same estimate holds for $\int_{\spt \zeta} |A|^2 |\nabla \phi| d\mu_T$.  We can therefore replace $\zeta$ with $\zeta \phi$ to get \eqref{eqn:flat-cone-3} (with a different constant $C$), which implies as before that \eqref{eqn:flat-cone-1} holds for all $\zeta \in C^1_c(\R^{n+1} \setminus \{0\}$).

As in the usual proof of Simons' theorem (\cite{Simons}, \cite{Sc-Si-Ya}), we can plug in $\zeta = \zeta(|x|)$ into \eqref{eqn:flat-cone-1}, to deduce that $|A| \equiv 0$, and hence $\spt T$ is a finite union of planes.  Disjointnesss of the planes in $\sint\Omega$ follows directly from standard interior regularity.
\end{proof}

\begin{lemma}\label{lem:minz-higher-reg-cone}
Let $\Omega^{n+1}$ be a non-obtuse polyhedral cone domain for $1 \leq n \leq 6$.  Let $T^n$ be a free-boundary minimizing cone in $\Omega^{n+1}$, with $\del T \llcorner \sint\Omega = 0$.  Then $\sing T \subset \del_{n-2}\Omega$.  In fact if $n  \leq 2$ then $\sing T = \emptyset$.

If the dihedral angles of $\Omega$ are $=\pi/2$, or more generally if $\Omega = W^2 \times [0, \infty)^{l-2} \times \R^m$, then $\sing T = \emptyset$.
\end{lemma}

\begin{proof}
We induct on $n$.  If $n = 1$, then $T$ consists of a finite union of rays in $\Omega$, and hence by Lemma \ref{lem:minz-planes} $T = 0$.  So Lemma \ref{lem:minz-higher-reg-cone} vacuously holds in this case.  We can therefore assume by inductive hypothesis that Lemma \ref{lem:minz-higher-reg-cone} holds for $n-1$ in place of $n$.

By interior regularity and reflection, we already know that $\sing T \subset \del_{n-1}\Omega$.  Given $x \in \spt T \cap \del\Omega \setminus (\del_{n-2}\Omega \cup \{0\})$, then by Lemma \ref{lem:minz-compact}, Lemma \ref{lem:monotonicity}, Corollary \ref{cor:lower-reg}, we can find a sequence of $r_i \to 0$ so that (after a rotation as necessary necessary), we get convergence
\[
(\eta_{x, r_i})_\sharp T \to T' \times [\R]
\]
as both currents and varifolds, and $\Omega_{x, r_i} \to \Omega' \times \R$ in the local Hausdorff distance.  Here $T' \in \cI_{n-1}(\R^n)$ is a non-zero free-boundary area-minimizing cone in $\Omega' \subset \R^n$.  Since $x \in \del\Omega \setminus \del_{n-2}\Omega$, we have $0 \in \del\Omega' \setminus \del_{n-3}\Omega'$, and hence by inductive hypothesis $0 \in \reg T'$.  Our regularity Theorem \ref{thm:main} then implies that $x \in \reg T$ also.

If $0 \in \del_{n-2}\Omega$ then we are done.  Otherwise, suppose $0 \in \del_{n-1}\Omega$.  Then $\Omega = W^2\times \R^{n-1}$ for some wedge $W$ having interior angle $\leq \pi/2$, and $\sing T \subset \{0\}$.  At each point of $\spt T \cap \del\Omega \setminus \del_{n-1}\Omega$, standard interior regularity (possibly after reflection) implies $\spt T$ is smooth nearby.  For each point $x\in \spt T \cap \del_{n-2}\Omega \setminus \{0\}$, then after a rotation we can write
\[
(\spt T - x)\cap B_r(0) = \{ y + u(y) : y \in (W^2 \times \R^{n-2}) \cap B_r(0) \},
\]
where $u$ is a $C^{1,\alpha}$ function.  Theorem \ref{thm:higher-main} then implies that $u$ is in fact $C^{2,\alpha}$.  Therefore we can apply Lemma \ref{lem:cone-is-plane} to deduce $\spt T$ is a union of free-boundary planes $\{V_i\}_{i=1}^N$, which are disjoint in $\sint\Omega$.

It suffices to show that each $V_i \in \cP_{\Omega}$, since then it will follow by disjointness that $N = 1$, and hence $0 \in \reg T$.  Otherwise, if $V_i \not\in \cP_{\Omega}$, then it must follow that $\{0\} \times \R^{n-1} \subset V_i$, and hence $\sing T \supset \{0\}\times \R^{n-1}$.  However this contradicts our earlier conclusion, that $\sing T \subset \{0\}$.  This proves $\sing T \subset \del_{n-2}\Omega$.

If $n \leq 2$, then $\Omega$ is either $W^2 \times \R$ or $0$-symmetric.  We just showed $\sing T = \emptyset$ in the first case.  In the second case, then the above argument implies that $\spt T$ is a union of free-boundary planes which are disjoint in $\sint\Omega$.  But then by Lemma \ref{lem:minz-planes} we must have $T = 0$.

If the dihedral angles are all exactly $= \pi/2$, then we can in fact prove that $\spt T$ coincides with a single plane $V \in \cP_{\Omega}$.  This vacuously holds when $n = 1$, and so we can assume it holds with $n-1$ in place of $n$ by inductive hypothesis.  By the same argument as above, we must have $\sing T \subset \{0\}$.  By reflection, we can obtain a minimal cone $T'$ in $\R^{n+1}$ which is smooth and stable away from $0$, and hence by Simons' theorem $\spt T'$ is a plane.

When $\Omega = W^2 \times [0, \infty)^{l-2}\times \R^m$, then by a similar argument after reflections we reduce to the case $\Omega = W^2 \times \R^{n-1}$, in which case $\sing T = \emptyset$.
\end{proof}\end{comment}

\vspace{5mm}

Using the previous results, and the Nash embedding theorem, partial regularity is now a standard argument.

\begin{proof}[Proof of Theorem \ref{thm:partial-reg}]
Since $N$ is a complete, Riemannian $C^3$ $(n+1)$-manifold we can by isometrically embed it in some $\R^{n+1+k}$ space (for $k = k(n)$).  WLOG assume $0 \in \del\Omega \cap \spt T$, $T_0N = \R^{n+1} \times \{0^k\}$, and by dilating $N$ as necessary we can assume that the map $\exp_{T^\perp N}(x, v)$ that takes $x \in N$, $v \in T^\perp_x N$ to $x + v$ is a diffeomorphism onto its image for $x \in B_2$, $|v| \leq 2$, and $\exp_{T^\perp N}( B_2, B_2) \supset B_1$.  Let $\Omega' = \exp_{T^\perp N}(\Omega, B_2) \cap B_1$, and then $\Omega' \in \cD_\eps(T_0\Omega \times \R^k)$, where $\eps$ can be made arbitrarily small by dilating $N$.

Suppose $T$ is area-minimizing with free-boundary in $\Omega$.  Then $T$ is $(A, 1, 1)$-almost-area-minimizing in $B_1$ with free-boundary in $\Omega'$, where $A$ depends only on the curvature of $N$.  Since $T$ has zero tangential mean curvature in $\Omega \subset N$, then $T$ has bounded tangential mean curvature in $\Omega' \subset B_1$ (again controlled by the curvature of $N$).  By standard codimension-one theory we can reduce to the case when $T = \del[E] \llcorner \sint\Omega'$, for $E \subset N$ relatively open.

Suppose $T = \del[E] \llcorner \sint\Omega$ is an isoperimetric region.  Then $T$ is $(A, 1, \delta)$-almost-area-minimizing in $N$ with free-boundary in $\Omega$ for some constants $A, \delta$ (depending on the volume $|E|$; see \cite[Example 21.3]{maggi_2012}), and hence as above $T$ is $(A', 1, \delta/2)$-almost-area-minimizing in $B_1$ with free-boundary in $\Omega'$.  Similarly, since $T$ has bounded tangential mean curvature in $\Omega \subset N$, then $T$ has bounded tangential mean curvature in $\Omega' \subset B_1$ also.

In either case, if we consider a sequence of dilations $T_i = (\eta_{0, r_i})_\sharp T$ then by \eqref{eqn:lower-reg}, \eqref{eqn:monotonicity} and Lemma \ref{lem:minz-compact}, the $T_i$ will subsequentially converge as currents and varifolds to some non-zero area-minimizing cone $T''$ with free-boundary in $\Omega''$, where $\Omega'' = T_0\Omega \subset T_0N$.  Moreover, if $x \in \reg T''$, then $\sing T_i \cap B_r(x) = \emptyset$ for all $i$ large.

Lemma \ref{lem:low-dim-cones} implies that:
\begin{enumerate}
\item if $T''$ is $(n-1)$-symmetric then $T''$ is regular;
\item if $T''$ is $(n-2)$-symmetric and $\Omega$ is non-obtuse, then $T''$ is regular;
\item if $T''$ is $(n-6)$-symmetric and $\Omega$ has dihedral angles $=\pi/2$, then $T''$ is regular.
\end{enumerate}
The partial regularity of Theorem \ref{thm:partial-reg} then follows from a standard dimension reducing argument.

\end{proof}

\begin{comment}
\begin{prop}\label{prop:isotropic-minz}
Suppose $E \subset B_1$ is a set of locally finite perimeter in $B_1$, and write $T = \del[E] \llcorner int\Omega \cap B_1$.  Suppose $E$ minimizes the functional
\[
\cF(E) = \int_{E \cap B_1 \cap int\Omega \Omega } g d\leb^{n+1} + \int_{int\Omega} h d\mu_{T},
\]
for some fixed functions $g, h \in C^1_c(B_1)$ satisfying $h \geq h_0 > 0$.

If we write $M$ for the varifold associated to $T$, then $M \in \cIVT_n(\Omega, B_1)$, and $T$ is $(c_0, 1, \delta)$-almost-minimizing in $\Omega$.
\end{prop}

\begin{proof}
DO ME!
\end{proof}
\end{comment}


TITLE: Matching one variable and its slope on a surface in 3-dimensions
QUESTION [1 upvotes]: I have the following equation (also viewable and more nicely formatted at the bottom of page 5 of this PDF):
$$z = \frac{2500}{ \left(\frac{ \left (\frac{X^{2}}{(Y - 4.5) + 0.954038 - 0.0558178X + 0.000733819X^{2}}\right )}{0.109757 + 0.00792236X + 0.000197693X^{2}}\right )} + 4.5$$
That translates to this surface (with si = z, age = x, and height = y):
$\hskip2in$
My problem is that I want to find si (hereafter Site Index) with only height and the slope of height. Conceptually I have been able to make inroads to doing this in two dimensions if I model the individual SI curves as a polynomial. I also understand a little bit spatially what it is I want to accomplish, which is demonstrated in the below image.
On the left of this image is the original publication showing discrete Site Index curves and I have highlighted holding height constant at 30 (purple), 70 (orange), and 130 (yellow). On the right are the same constant heights in three dimensions. In two dimensional space, for a given Site Index curve, I can obviously find the age with ease. My problem is that I actually have a given constant height and the change in height (slope) and need to match those up with the right "curve".
To do this I think I need to take the second derivative of the original equation (first to get down to the two dimensional space on the left, and a second time to match up the slope) but I am now pretty stuck. I have been looking at the right image a long time and some how have a block - I cannot figure out with respect to what variable I would be differentiating(?), and if this is a partial derivative problem or not?

REPLY [1 votes]: I am going to assume that by "change in height", you mean $m:=\frac{\partial y}{\partial x}$, when $y$ is considered as a function of $x$ and $z$ (so $z$ is held constant as the derivative with $x$ is being taken). Now let's simplify the notation considerably, by introducing two new functions in $x$, and replacing the nasty constants by tidy letters: Let $a = 2500, b= 4.5$, and let $$p = −4.5+0.954038−0.0558178x+0.000733819x^2\\q = 0.109757+0.00792236x+0.000197693x^2$$
Then your expression is $$z - b= \frac a{\left[\frac {\left(\frac {x^2}{y + p}\right)}q\right]} = \frac {aq(y+p)}{x^2}$$ or
$$y + p= \frac {x^2(z-b)}{aq} $$
Hence $$m + p' = \frac{\partial y}{\partial x} + p' = \frac{z-b}a\frac {2xq - x^2q'}{q^2}$$
Solving for $z$ again, $$z -b = \frac{aq^2(m+p')}{2xq - x^2q'}$$
where $p'$ and $q'$ are the derivatives of $p$ and $q$ with respect to $x$.


TITLE: Why does induction prove equality?
QUESTION [4 upvotes]: I know how to prove equality for my homework assignment, but I can't understand why it actually proves that equality. Could you please explain?

REPLY [3 votes]: Assume that all $\emptyset \neq B \subset \mathbb{N}$ has the smallest element $n_0$. Let $X =\{ n \in \mathbb{N}: P(n) \ \mbox{is true}\}$($P$ of property). Hence if
\begin{equation}
(1) \quad P(n_0) \quad \mbox{is true.}
\end{equation}
and
\begin{equation}
(2) \quad P(n) \Rightarrow P(n+1)
\end{equation}
Then $X \supset \{n \in \mathbb{N}:n \ge n_0\}$. In fact, by (1) $n_0 \in X$. If $X\neq\{n \in \mathbb{N}:n \ge n_0\}$ there exist $n_1$ the smallest element of $\mathbb{N} - X$ because $\emptyset \neq X \subset \mathbb{N}$. Then $n_{1} - 1 \in X$ and by (2) we have $n_1 \in X$. Contradiction.
Consider now $P(n)$ is true if
\begin{equation}
(3) \quad  \sum_{k=1}^{n}k = \dfrac{n(n+1)}{2}.
\end{equation}
 As $P(1)$ is true and (2) holds, $X=\mathbb{N}$ and (3) hold for all $n \in \mathbb{N}$.


TITLE: Time evolution of a quantum system
QUESTION [1 upvotes]: A quantum system has Hamiltonian $H$ with normalised eigenstates $\psi_n$ and corresponding energies $E_n$ ($n = 1,2,3...$). A linear operator $Q$ is defined by its action on these states:
  $$ Q\psi_1 = \psi_2 $$
  $$ Q\psi_2 = \psi_1 $$
  $$ Q\psi_n = 0, n>2 $$
  Show that $Q$ has eigenvalues 1 and -1 and find the corresponding normalised eigenstates $\zeta_1$ and $\zeta_2$, in terms of energy eigenstates. Calculate $\langle H \rangle$ in each of the states $\zeta_1$ and $\zeta_2$.
A measurement of $Q$ is made at time=0, and the result 1 is obtained. The system is then left undisturbed for a time $t$, at which instant another measurement of $Q$ is made. What is the probability that the result will again be 1? Show that the probability is 0 if the measurement is made after a time $T = \pi \hbar/(E_2 - E_1)$, assuming $E_2 - E_1> 0$.

I found
$$ \zeta_1 = (\psi_1 + \psi_2)/\sqrt{2} $$
$$ \zeta_2 = (\psi_1 - \psi_2)/\sqrt{2} $$
and $\langle H \rangle = (E_1 + E_2)/2$ for both.
I have trouble doing the second part.
Doesnt the system collapse into $ \zeta_1$ given we know this is the state at time =0?
so probability will be 1?

REPLY [2 votes]: Doesnt the system collapse into ζ1 given we know this is the state at time t=0?

True, but after the measurement at $t_0$ the system follows the unitary dynamics given by $H$. As you can easily check, $\zeta_1$ is not an eigenstate of the Hamiltonian, iff $E_1\neq E_2$. Hence the system will not remain in the state $\zeta_1$ during time-evolution.
If you were to perform another measurement directly after the first one, the outcome would be the same. Here and generally this is not true for arbitrary times between measurements.
As a hint, I remind you that the time-evolution of an eigenstate of the Hamiltonian looks like
$$ U(t,t_0)\psi_n = \exp\left[-i(t-t_0)E_n / \hbar \right] \psi_n $$


TITLE: Floating point division resulting in a value exceeding 1 but should be equal to 1...maybe!
QUESTION [2 upvotes]: I am computing the apparent magnitude of comets and minor planets using data from the Minor Planet Center.  The formula I use has a division and that result is passed to the arccosine function (in Python).
Occasionally the result of the division is slightly greater than 1, for example
6.272370848320445 / 6.2723697617591405 = 1.0000001732297912
I assume the result is not actually greater than 1; rather the result is an artefact of floating point arithmetic, particularly given the number of zeros after the decimal point.
If my assumption is correct, I want to catch this sort of situation and treat the result as if the division yielded exactly 1 (and so can safely be passed to arccosine).
If I make an arbitrary choice of say five decimal points and truncate the division result, then if the result is 1 I'm good, otherwise, drop the result.
I'm not sure if what I'm proposing is correct mathematically, so looking for guidance please.  I have assumed this issue arises due to floating point arithmetic, but if that assumption is wrong, then I'm not sure what else I can do.  It is possible the source data from the Minor Planet Center contains spurious data which is causing the actual error (and thus I shouldn't be trying to work around it but just let it fail).
The formula can be found here (using the H,G model).

REPLY [5 votes]: I looked at the link you mention in your question and I don't
have time to study all of the details. However, the expression
beta = acos((rp*rp + rho*rho - rsn*rsn)/(2*rp*rho));

seems like an instance of
Law of cosines.
This can be rewritten as
beta = acos(1 -(rsn+rho-rp)(rsn-rho+rp)/(2*rp*rho));

The problem in computing this is that rsn+rho-rp or
rsn-rho+rp loses precision if rsn and |rho-rp|
are of comparable value and may become negative. This
happens when the triangle is close to degenerate where
one of the angles becomes very close to zero. In these
cases you can use the haversine
idea and use the formula
beta = 2*asin(sqrt(max(0,(rsn+rho-rp)(rsn-rho+rp)/(4*rp*rho)));

or the approximation when the angle is close to zero
beta = sqrt(max(0,(rsn+rho-rp)(rsn-rho+rp)/(rp*rho)));

where the max(0,.) ensures that we don't attempt to
take the square root of a negative number.
Note that the acos(x) formula is great when $x$ is
close to zero, but the asin(x) formula is better when
$x$ is close to one in absolute value.


TITLE: Prove that $\{x + y \cdot \sqrt{3} | x, y \in \mathbb{Z} \}$ is a ring (or not)
QUESTION [1 upvotes]: How to prove that $(R, +, \cdot)$ is a ring (or not), where $R = \{x + y \cdot \sqrt{3} | x, y \in \mathbb{Z} \}$?
Update. Is this proof correct?
$(R, +)$ is an abelian group:

Closure: $a, b \in R \implies (a + b) \in R$.
Associativity: $a, b, c \in R \implies (a + b) + c = a + (b + c)$.
Identity: $0 \in R$.
Inverse: $-(x + y \sqrt{3}) \in R$.
Commutativity: $a, b \in R \implies a + b = b + a$.

$(R, \cdot)$ is a monoid:

Closure: $(x_1 + y_1 \sqrt{3}) \cdot (x_2 + y_2 \sqrt{3}) = x + y \sqrt{3}$, where $x = x_1 x_2 + 3 y_1  y_2 \in \mathbb{Z}$ and $y = x_1 y_2 + x_2 y_1 \in \mathbb{Z}$.
Identity: $1 \in R$.

Multiplication distributes over addition:
$a, b, c \in R \implies$

Left distributivity: $a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)$
Right distributivity: $(b + c) ⋅ a = (b ⋅ a) + (c ⋅ a)$

REPLY [1 votes]: Hint $\ $ Employ the subring test to prove that $\,S\,$ is a subring of some known larger ring (e.g. $\,\Bbb R),\,$ i.e. show $\,1\in S\,$ and $\,S\,$ is closed under  subtraction and multiplication (by $\,\sqrt 3^2 = 3\in S)\,$
Alternatively, show that $\,S\,$ is the image of $\,\Bbb Z[x]\,$ in $\,\Bbb R\,$ under the evaluation hom $\,x\mapsto \sqrt 3.$


TITLE: Discrete subgroup of topological group
QUESTION [1 upvotes]: I want to find textbook which contains the next proposition(?).
I think it is true, but I can't find proof of that.
Please teach me a textbook in which the next proposition(?) are proved.
Proposition(?) (Characterization of discrete subgroup)
Let $G$ be a topological group and e be identity element of G.
For a subgroup $H \subset G\ $, The following are equivalent.
(1) Relative topology of H is discrete.
(2) There exists an open neighborhood $U$ of $\ e\ \ $such that$\ \ H \bigcap U = \left\{ 0 \right\}$.
(3) There exists an open neighborhood $V$ of $\ e \ \ $such that$\ \ \ \forall x \in G ,\ \#(H \bigcap xV) \leq 1\ $ where$\ \#\ $means cardinality of a set.
(4) When we give the discrete topology to H,$\ \ $action of $H$ to $G\ $:$\ H \times G \rightarrow G$,$\ \ h \mapsto h \cdot g\ $ is properly discontinuous.

REPLY [1 votes]: Here are some Hints for the exercise.
For (1) and (2)- notice the $e$ is open in the relative topology so $e= H\cap U$
For (2) and (3) - given $V$ as in (3) take $U=V$ and given $U$ as in (2) take $V$ to be a symmetric open set s.t $VV\subset U$.
For (4): (Notice that the term "properly discontinuous" usually assumes $G$ is locally compact. )
If $H$ is discrete then the intersection of $H$ with a compact set is finite, this implies the action is  "properly discontinuous" since for any compact set $K\subset G$ then $KK^{-1}$ is compact as well. so #$\{h\ : \ hK\cap K\}=$#$\{h\ : \ h\in KK^{-1} \}<\infty$.
In the other direction, if we assume $H$ is not disreate then there is a accumulation point $x$- thus we will not be able to find a $V$ satisfying (3)


TITLE: Concerning the weak* closed convex hull of a subset of states of a C*-algebra
QUESTION [1 upvotes]: In Dixmier's "C*-Algebras", lemma 3.4.1 states the following: if $A$ is a unital C*-algebra, $S(A)$ the set of states, and $Q\subseteq S(A)$ a discerning subset, then $\overline{\operatorname{conv}}^{w*}Q = S(A)$. Here a subset $Q$ of $S(A)$ is called discerning if whenever $x\in A$ is self-adjoint, and $f(x)\ge 0$ for all $f\in Q$, then $x$ is positive.
I'm having trouble following the proof. Let $x\in A$ be self-adjoint, and $\alpha\in\mathbb{R}$ be such that $f(x)\ge \alpha$ for all $f\in Q$. Then $x - \alpha\cdot 1\ge 0$, so $f(x)\ge \alpha$ for all $f\in S(A)$. Apparently, this is enough to conclude that $S(A)\subseteq\overline{\operatorname{conv}}^{w*}Q$, but I fail to see why. (The opposite inclusion is easy to see.)
The proof is so short on details that I imagine I'm missing something simple, but I've come back to this a couple times now and still can't see the correct thought process.

REPLY [1 votes]: Ah, the key is the geometric Hahn-Banach. Suppose we had some $f\in S(A)$, but not $C:=\overline{\operatorname{conv}}^{w*}Q$. Assuming the latter set is absorbing in at least one place (which I haven't checked yet), we can exhibit $x\in A$ and $\alpha\in\mathbb{R}$ such that $\Re f(x)\ge \alpha \ge \Re g(x)$ for all $g\in C$. By passing to $\Re x = (x + x^*)/2$, we can exhibit a hermitian $x$ such that $f(x)\ge \alpha\ge g(x)$ for all $g\in C$ (since $f(\Re x) = \Re f(x)$).
By assumption that $Q$ is discerning, $g(\alpha\cdot1 - x)\ge 0$, so that $\alpha\cdot1 - x$ is positive, and so $h(\alpha\cdot1 - x)\ge 0$ for all $h\in S(A)$. Yet $f(\alpha\cdot1 - x)\le 0$, yielding a contradiction.
Also, I should note that this lemma occurs in a slightly different form in Brown and Ozawa (also labelled lemma 3.4.1).


TITLE: Let $x>0$ and let $n_0$ be the largest integer such that $n_0\leq x$...(Help proving the existence of $n_0$)
QUESTION [1 upvotes]: I'm reading Rudin's book (PMA) and it says Let $x>0$ and let $n_0$ be the largest integer such that $n_0\leq x$ (Note that the existence of $n_0$ depends on the arquimedean property of $\Bbb R$). My question is how can I prove that using the arquimedean property of $\Bbb R$?

REPLY [3 votes]: Given $x>0$, by the Archimedean property there is a positive integer $n$ with $x < n$. Let $A$ be the set of all positive integers $n$ with $x<n$. Since $A$ is a nonempty subset of the natural numbers, and the natural numbers are well-ordered, $A$ has a smallest element, say $m_0$. Then $n_0 = m_0 - 1$ must satisfy $n_0 \leq x$, and it is also the largest integer with $n_0 \leq x$ because any integer larger than $n_0$ must lie in $A$.
In short, the Archimedean property guarantees that the set $A$ is nonempty.

REPLY [1 votes]: Split two cases. $ x\le 1 $ and $x \gt 1$. First case is obvious. The second one is disposed as follows.
Note that given $x \in \Bbb R$ by the Archimedian Property there exists $n \in \Bbb N$ such that $ n \gt x $. Then the set $ A =  \{ n \in \Bbb N \ | \ n \gt x\} $ is non-empty.
Then by the Well-Ordering Principle $A $ has a least element $ n' $. |et $ n_{\circ} = n' - 1 $
Now prove that $n_{\circ}$ is indeed the smallest integer such that $n_{\circ} \le x$. Assume not and reach a contradiction.