text:
\chapter{Introduction}
\label{intro}

One of the fundamental questions in harmonic analysis is the classification of the unitary dual of a locally compact group.
An important class of locally compact groups is the $F$-points of reductive groups over a local field $F$.
We will only consider $p$-adic groups in this paper.
The history of this problem goes back at least to \cite{MR0023246} and \cite{MR0046370}.\footnote{Note that these papers were
published well before \cite{MR0093558}, which is usually considered the beginning of the representation theory of reductive $p$-adic groups.}

The classification of the unitary dual for general linear groups in the $p$-adic case was solved in \cite{MR870688}.
(The archimedean case can be solved along the same lines -- see \cite{TadicBonn1985} and \cite{MR2537046}.)
An important input is Bernstein's result \cite{MR748505} on irreducibility of unitary parabolic induction.
Subsequently, the classification of the unitary dual of inner forms of the general linear group was obtained
(\cite{MR2055385} and \cite{MR2492994}; see also \cite{MR1040995}).
A unified and simplified proof was recently discovered by E. Lapid and A. M\'inguez \cite{MR3573961} (see also \cite{MR3269346}).
Their point of departure is the reducibility points for parabolic induction of cuspidal representations of general linear groups,
which in the split case are always $\pm1$.
This also gives a uniform and simplified approach for the classification of the admissible dual
(see \cite[Appendix]{MR3573961} and \cite{MR3456591}).

Little is known in general about the classification of the unitary dual in case of (other) classical groups, except for some important subclasses
of unitary representations, such as generic representation \cite{MR2046512} or spherical representations \cite{MR2767523}.

A natural question is whether one can attack the unitarizability problem for the classical groups
using the cuspidal reducibility points as the starting point (as is the case for the general linear groups).
Such an approach was proposed in \cite{MR3969882}.
The main goal of this paper is carry out this approach in the corank (at most) 3 case, namely to classify,
irreducible unitarizable subquotients of representations Ind$_P^G(\tau)$, where $G$ is a classical group over a $p$-adic field of
characteristic zero, $P$ is a parabolic subgroup of $G$ of corank (at most) 3 and $\tau$ is an irreducible cuspidal
representation of a Levi factor $M$ of $P$.

The first order of business is to understand cuspidal reducibility for classical groups.
This was done by C. M\oe glin in terms of the local Langlands correspondence, which is a consequence of
Arthur's endoscopic classification (\cite{MR3135650}, which relies on \cite{MR3823813} and \cite{MR3823814} among other things).
The bottom line is that the points of cuspidal reducibilities are half-integers
\[
0,\pm\tfrac12,1,\pm\tfrac32,\pm2, \dots.
\]
(All half-integers can occur.) This is considerably more complicated than in the case of general linear groups
(and their inner forms) where there is a single cuspidal reducibility point (up to $\pm$).

Another important difference is that the parabolic induction of irreducible unitarizable representations is not irreducible in general.

For any reductive $p$-adic group $G$, denote by $\tilde G$ the set of all equivalence classes of its irreducible smooth representations
(the admissible dual of $G$) and by $\hat G$ the subset of the unitarizable classes (the unitary dual of $G$).
The unitarizability problem is the determination of the subset $\hat G$ of $\tilde G$.
It breaks down naturally into two parts: construction and exhaustion.
The exhaustion is usually achieved by showing that all the classes in $\tilde G$ other than the ones constructed in the first step
are non-unitarizable.
We may call such an approach to the exhaustion ``a proof by elimination''.

In the construction of the representations of $\hat G$, arguably the hardest part is the construction of representations that are isolated
in the natural topology of $\hat G$.
(Explicating the reducibility of unitary parabolic induction and the complementary series are other difficult problems.)
Our expectation is that at least for split classical $p$-adic groups, all isolated representations are of Arthur type
(and consequently, they occur as local constituents of automorphic representations in the discrete spectrum).
This is known to be the case for spherical representations by \cite{MR2767523}.
Here, M\oe glin's results on the structure of Arthur packets provide a powerful tool for the construction of isolated representations.

Regarding exhaustion, proof by elimination (which is used in this paper) is unfeasible in the higher rank case
(as this paper clearly indicates for rank 3).
Namely, it requires a very detailed knowledge of the structure of the representations in $\tilde G\backslash \hat G$.
Thus, especially in higher rank (where $\tilde G\backslash \hat G$ is much larger than $\hat G$)
we are spending most of our effort on the ``wrong'' class of representations.

Unfortunately, the prospect of finding a direct approach to the exhaustion problem for classical groups does not seem
to be on the horizon at the moment.
In fact, so far, the only successful direct approach to the exhaustion problem for reductive groups
in higher rank seems to be that of \cite{MR870688}, \cite{TadicBonn1985} for the case of general linear groups
(see also \cite{MR2537046} and \cite{MR1181278}).
The statement (if not the proof) of the classification of the unitary dual in this case is rather simple
(see Theorem \ref{ud-gl} below).\footnote{There is also D. Vogan's classification of unitary duals of
$\GL(n,\C), \GL(n,\R)$ and $\GL(n,\mathbb{H})$ (Theorem 6.18 of \cite{MR827363}).
One can find at the end of the seventh section of \cite{MR2684298} remarks
about relation between our approach and that of Vogan. We shall quote here only a part which indicates the main difference
between these two approaches: ``Vogan's classification is conceptually very different from Tadi\'c's classification.
It has its own merits, but the final result is quite difficult to state and to understand, since it uses sophisticated concepts
and techniques of the theory of real reductive groups.''}
This was arguably surprising at the time, although it should be noted that the first lists of candidates for the unitary duals
of the closely related groups $SL(n,\C)$, which go back to 1947 in \cite{MR0023246}, albeit incomplete, were very simple
(and not far from the actual unitary duals).
It took almost four decades to get a direct approach to the exhaustion in the case of general linear groups
(thereby fulfilling the vision of I. M. Gelfand and M. A. Naimark).

Although this paper is about unitarizability, most of it deals with non-uni\-tarizability because of the
exhaustion proof by elimination. A very small part of the admissible dual is unitarizable, and its unitarizability,
excluding only the representations \eqref{intro-dist}, is very natural to expect (and not too hard to prove).
In the analysis of the non-unitarizability of representations, the most delicate ones are those whose $\GL$-support
is contained in a segment of cuspidal representations which contains the reducibility point, and which are not fully induced
(non-unitarizability of the other representations is obtained by deformation to these representations or reducing to the
non-unitarizability in the case of general linear groups).
In \S\ref{sec: unit3} we settle the non-unitarizability in the most delicate cases, save for a few exceptions.
The remaining five cases are dealt with separately in \S\ref{basic 1},\ref{basic 0}.

In order to show non-unitarizability of $\pi$ in the most delicate case we consider the parabolic induction $\Pi$ of $\pi$ tensored
with a suitable irreducible unitarizable representation $\tau$ of a general linear group.
We show that the length $\ell$ of $\Pi$ is larger than the multiplicity $m$ of $\tau\otimes\pi$ in the Jacquet module of $\Pi$.
This implies that $\Pi$ cannot be semisimple, let alone unitarizable. Hence, $\pi$ cannot be unitarizable.
(In one case, we will also use the fact that if $\Pi$ is semisimple and $\ell$ is equal to $m$, then each copy of $\tau\otimes\pi$
must occur as a direct summand in the Jacquet module of $\Pi$.)

We already noted that in the construction of new irreducible unitarizable representations,
the most difficult cases are the isolated representations.
The simplest examples of representations that are often (though not always) isolated in the unitary dual are the square-integrable ones
(whose unitarizability is obvious) and their dual representations
(whose unitarizability is not obvious, except for the trivial representation which is isolated by \cite{MR0209390} if the split rank of
the simple group is not one).
In the case of $p$-adic general linear groups, the first case of an isolated representation (modulo the center) not of this type is for
$\GL(9)$.\footnote{In general, these representations are $u(\delta(\rho,m),n)$ where $m,n>2$ -- see below for notation.}
In the case of classical groups, such examples first occur in corank 3, when the reducibility point is $>1$.
These are the representations \eqref{intro-dist} below whose unitarizability
was proved by M\oe glin. (The smallest group which accommodates such a representation is the split $SO(11,F)$.)

We shall now briefly describe some parts of the strategy in \cite{MR3969882} proposed to handle the unitarizability problem
in the case of classical $p$-adic groups.
We first note that it is easy to reduce to the case of representations
supported on real twists of selfcontragredient irreducible cuspidal representations of general linear groups and irreducible cuspidal
representations of classical groups (see \S\ref{reduction-u-w} for more details).\footnote{In the case of unitary groups we need to
consider $F'/F$-contragredients, whose definition is recalled in the second chapter of the paper.
For simplicity, in the introduction we only consider symplectic and orthogonal groups.}
Because of this, we shall consider in the sequel only such representations, which we call weakly real representations.

Jantzen decomposition attaches to an irreducible (weakly real) representation $\pi$, irreducible representations
supported on single cuspidal lines
\[
\pi\ra (\pi_1,\dots,\pi_k)
\]
(see \cite{MR1481814} or \cite[\S8]{MR3969882} for more details).
In chapter \ref{mixed-case} we conjecture that this decomposition preserves unitarizability in both directions,
i.e., that $\pi$ is unitarizable if and only if all $\pi_i$ are unitarizable.
(See \cite{MR3969882} for some very limited support for this conjecture.)
If true, this would reduce the general case to unitarizability pertaining to the cases of a single cuspidal lines
(and single cuspidal reducibilities).

Consider now an irreducible representation $\pi$ of a classical group that is supported on a cuspidal line
$X_\rho$ along a selfcontragredient irreducible cuspidal representation $\rho$ of a general linear group,
and an irreducible cuspidal representation $\s$ of a classical group.
To the pair $\rho,\s$ corresponds a unique non-negative reducibility exponent $\alpha_{\rho,\s}\in\tfrac12\Z$.
The next question is whether the unitarizability of $\pi$ can be described in terms of the reducibility exponent
$\alpha_{\rho,\s}$ alone. (See \cite{MR3969882} for a precise formulation.)
If true, the unitarizability problem for classical $p$-adic groups
would amount to the determination of a certain (complicated) combinatorial data.

We shall below use standard Bernstein-Zelevinsky notation $\times$ for parabolic induction in the setting of general linear groups,
and its natural extension $\rtimes$ to the setting of classical groups (see \S\ref{notation} for details).
By $|\ \ |_F$ is denoted the normalized absolute value on $F$, and by $\nu$ the character $g\mapsto |\det(g)_F|$ of $\GL(n,F)$.

The main goal of this paper is to classify, following the above proposed strategy,
the irreducible unitarizable (weakly real) subquotients of the representations
\begin{equation} \label{k-times}
\theta_1\times\dots\times\theta_k\rtimes\s, \quad k\leq 3,
\end{equation}
where $\theta_i$, $1\leq i\leq k$, and $\s$ are irreducible cuspidal representations of general linear groups and of
a classical group respectively.
In particular, we completely classify the unitary dual of classical $p$-adic groups of the split rank (at most) three.
This gives some very limited support for the possibility of the above approach to the unitarizability to work in general.

In the last chapter of the paper we prove that the Jantzen decomposition preserves unitarizability in both directions for the cases that we consider in this paper.
More precisely, we prove the following

\begin{proposition} \label{intro-red}
Let $\pi$ be a weakly real irreducible subquotient of $\theta_1\times\dots\times\theta_k\rtimes\s$, where $\theta_i$ are irreducible
cuspidal representations of general linear groups and $k\leq 3$. Then, $\pi$ is unitarizable if and only if all $\pi_i$
in the Jantzen decomposition of $\pi$ are unitarizable.
\end{proposition}

For a general connected reductive group $G$ over $F$ there is a natural involution $\pi\mapsto D_G(\pi)$ established in \cite{MR1285969}
and \cite{MR1471867}, which carries an irreducible representations of $G$ to an irreducible representation of $G$,
up to a sign. It is modeled after Deligne--Lusztig duality (in the context of groups over finite fields)
and will be henceforth referred to as \ASS involution, or duality\footnote{This involution is also called Zelevinsky involution, or Aubert involution, or Aubert-Schneider-Stuhler involution.}
(see also \cite{MR3769724}).
Take $\e_\pi\in\{\pm1\}$ such that $\e_\pi D_G(\pi)$ is a representation.
We denote then $\e_\pi D_G(\pi)$ by $\pi^t$ and call it the \ASS involution of $\pi$.

Now we shall describe unitarizability in the case of corank up to three.
We shall express the classification of irreducible subquotients in the shortest way.

By Proposition \ref{intro-red}, it is enough to consider representations supported a on single cuspidal line.
It means that we fix an irreducible selfcontragredient cuspidal representation $\rho$ of a general linear group
and an irreducible cuspidal representation
$\s$ of classical group.
Then, there exist a unique non-negative $\alpha_{\rho,\s}\in\tfrac12\Z$ such that
$$
\nu^{\alpha_{\rho,\s}}\rho\rtimes\s
$$
is reducible.
Then, to simplify notation, we denote
$$
\alpha:=\alpha_{\rho,\s}.
$$
Suppose that $\alpha>0$. Let $k$ be a non-negative integer. Define the generalized Steinberg representation
$$
\delta([\alpha,\alpha+k]^{(\rho)};\s)=\soc(
\nu^{k+\alpha}\rho\times \nu^{k-1+\alpha}\rho\times \dots \times \nu^{\alpha}\rho\rtimes\s)
$$
where $\soc$ is the socle.
These representations are irreducible and square-integrable.
(For $k=0$ we simply write $\delta([\alpha]^{(\rho)};\s)$.)

Denote
$$
\R^k_{++}=\{(x_1,\dots,k_k)\in\R^k; 0\leq x_1\leq \dots\leq x_k\}.
$$
It is well-known that for each representation $
\pi:=\nu^{x_1}\rho\times \dots \times \nu^{x_k}\rho\rtimes\s
$, $x_i\in
\R$, there exists $(x_1',\dots,x_k')\in \R^k_{++}$ such that $\pi$ and $\nu^{x_1'}\rho\times \dots \times \nu^{x_k'}\rho\rtimes\s$ have the same composition series.
Now we describe irreducible unitarizable subquotients of representationsn\eqref{k-times}. For $k=1$ the answer is:

\begin{remark}
Unitarizability in the corank 1 is very simply to describe:
an irreducible subquotient $\pi$ of $\nu^x\rho\rtimes\s$, $x\in\R_{\geq0}$, is unitarizable $\iff$ $x\leq \alpha$.
\end{remark}

The following answer to the unitarizability problem in the corank 2
is more or less already well known (although we were unable to find a complete reference in the literature).

\begin{proposition}
The irreducible unitarizable subquotients of $\nu^{x_1}\rho\times\nu^{x_1}\rho\rtimes\sigma$, $(x_1,x_2)\in\R^2_{++}$, are the following.
\begin{enumerate}
\item $(\alpha>1)$ All irreducible subquotients when $x_1+1\leq x_2\leq \alpha$.
\item $(\alpha\ne\frac12)$ All irreducible subquotients when $x_1+x_2\leq 1$.
\item $(\alpha=\tfrac12)$ All irreducible subquotients when $x_2\le\tfrac12$.
\item $(\alpha>0)$ The representations $\delta([\alpha,\alpha+1];\s)$ and $\delta([\alpha,\alpha+1];\s)^t$.
\end{enumerate}
\end{proposition}

For the answer of unitarizability for corank 3 we need additional notation.

Suppose $\alpha>1$. Denote by
\begin{equation} \label{intro-dist}
L([\alpha]^{(\rho)}, [\alpha-1]^{(\rho)};\delta([\alpha]^{(\rho)};\s))
\end{equation}
the Langlands quotient of $\nu^{\alpha}\rho\times \nu^{\alpha-1}\rho\rtimes\delta([\alpha]^{(\rho)};\s)$.
(This representation is invariant under the \ASS involution.)

For any positive integer $n$, the representation
$$
\delta(\rho,n)=\soc(\nu^{\frac{n-1}2}\rho \times \nu^{\frac{n-3}2}\rho \times\dots\times \nu^{-\frac{n-1}2}\rho)
$$
is irreducible and square-integrable modulo center.

For $\alpha=\tfrac12$, the representation $\delta(\rho,2)\rtimes\s$ contains a unique irreducible subquotient
which is not a subquotient of $[\tfrac12]^{(\rho)}\rtimes\delta ([\tfrac12]^{(\rho)};\s).$
We denote it by
$$
\delta([-\tfrac12,\tfrac12]^{(\rho)}_-;\s).
$$
Let $\alpha=0$. For any positive integer $k$, the representation
$
\nu^{\frac k2}\delta(\rho,k+1)\rtimes\s
$
has precisely two irreducible subrepresentations, and they are both square-integrable. We denote them by
$$
\delta([0,k]^{(\rho)}_+;\s)\quad \text{and}\quad \delta([0,k]^{(\rho)}_-;\s).
$$

The unitarizability statement for corank 3, which is the main result of the paper, is the following

\begin{theorem} \label{thm: mainintro}
The irreducible unitarizable subquotients of $\nu^{x_1}\rho\times \nu^{x_2}\rho\times \nu^{x_3}\rho\rtimes \sigma$
where $(x_1,x_2,x_3)\in\R^3_{++}$ are the following.
\begin{enumerate}
\item $(\alpha\ge1)$ All irreducible subquotients when $\mathbf{x}=(x_1,x_2,x_3)$ lies in the closure of one of the domains
\begin{gather*}
x_2+x_3<1, \\
x_1+x_2<1,\, x_3-x_2>1,\, x_3<\alpha,\ \ \ (\alpha>1),\\
x_1+x_2<1,\, x_1+x_3>1,\, x_3-x_1<1,\, x_3<\alpha,\\
x_2-x_1>1,\, x_3-x_2>1,\, x_3<\alpha,\ \ \ (\alpha>2).
\end{gather*}
\item $(\alpha=\tfrac12)$ All irreducible subquotients when $x_3\le\tfrac12$.
\item $(\alpha=0$) All irreducible subquotients when $x_1=0$, $x_2+x_3\le1$.
\item $(\alpha>0)$ The representations $\delta([\alpha,\alpha+2];\s)$ and $\delta([\alpha,\alpha+2];\s)^t$.
\item $(\alpha=0)$ The representations $\delta([0,2]_\pm;\s)$ and $\delta([0,2]_\pm;\s)^t$.
\item $(\alpha>0)$ The complementary series $[x]\rtimes \delta([\alpha,\alpha+1];\s)$ and
$[x]\rtimes \delta([\alpha,\alpha+1];\s)^t$ for $0\le x<\abs{\alpha-1}$ (if $\alpha\ne1$)
and its irreducible subquotients for $x=\alpha-1$.
\item $(\alpha=0)$ The complementary series (including subquotients at the ends)
\[
[x]\rtimes\delta([0,1]_\pm;\s),\ [x]\rtimes \delta([0,1]_\pm;\s)^t,\ \ 0\le x\le1
\]
\item $(\alpha>1)$ The representation $L([\alpha-1],[\alpha];\delta([\alpha];\s))$.
\item $(\alpha=\tfrac12)$ The complementary series (including subquotients at the ends)
\begin{gather*}
\nu^x\delta(\rho,2)\rtimes\delta([\tfrac12];\s),\
\nu^x\delta(\rho,2)^t\rtimes\delta([\tfrac12];\s)^t,
\ \ 0\le x\le1
\\
\nu^x\rho\rtimes\delta([-\tfrac12,\tfrac12]_-;\s),\
\nu^x\rho\rtimes\delta([-\tfrac12,\tfrac12]_-;\s)^t,\ \ 0\le x\le\tfrac32
\\
\nu^x\delta(\rho,3)\rtimes\s,\ \nu^x\delta(\rho,3)^t\rtimes\s,\ \ 0\le x\le\tfrac12.
\end{gather*}
\end{enumerate}
\end{theorem}

Note that Theorem \ref{thm: mainintro} directly implies that the
\ASS involution preserves unitarizability in the cases at hand,
a fact that is expected to hold in general (see \cite{MR794086} and \cite{MR1010153} for some steps in that direction).
Theorem \ref{thm: mainintro} also implies that all isolated representations are automorphic. (In \S\ref{sec: conjectures}
we conjecture this to hold in general.)

Furthermore, note that in the above classifications only the reducibility point $\alpha$ plays a role in determining the exponents of the representations that are unitarizable
(not $\rho$ and $\s$ themselves).

The motivation for writing this paper came from a discussion with C. M\oe glin at the
Simons Symposium on Geometric Aspects of the Trace Formula in Schloss Elmau in Germany (2016).
The paper \cite{MR3573961} of E. Lapid and A. M\'inguez was also a strong motivation for us to try to understand unitarizability based
only on the reducibility points between irreducible cuspidal representations, at least at corank 3.
Some of the results of this paper were presented in a minicourse at the Special Trimester on
Representation Theory of Reductive Groups Over
Local Fields and Applications to Automorphic forms, which was held at the Weizmann Institute in spring 2017.
We are thankful to the Simons Foundation and the Weizmann Institute. Discussions with
M. Hanzer, I. Mati\'c and A. Moy were helpful during the writing of this paper.

Especially we would like to thank C. M\oe glin and E. Lapid.
C. M\oe glin wrote the appendix with
the proof that the representation \eqref{intro-dist}
is in an Arthur packet, which proves the unitarizability of that representation.
Thanks to E. Lapid and his huge help, this paper is much easier to understand.
In particular, he helped us to change the organization of the paper, and considerably simplified the exhaustion part
by adding a geometric argument.
The final work on this paper took place at Weizmann Institute in winter 2020, and we are very thankful to the institute
for its hospitality and the very pleasant and relaxed working atmosphere.

The contents of the paper are the following.
Chapter \ref{notation} introduces notation and recalls known results that we use throughout the paper.
In chapters \ref{CC-corank12} -- \ref{CC-0} the unitarizability is solved for representations supported on
a segment of cuspidal representations that contains the reducibility point.
The heart of the paper is chapter \ref{sec: unit3} where we solve the cases $\alpha>1$.
In chapters \ref{basic 1} and \ref{CC-0} we address the remaining cases $\alpha=1,\tfrac12,0$ which require additional work.
All these cases are completely new.
In chapters \ref{rem-on-uni}, \ref{corank-3} and \ref{mixed-case} we complete the solution of the unitarizability in corank $\le 3$.
Appendix \ref{appendix-M} by C. M\oe glen shows the unitarizability of the isolated representations $L([\alpha-1],[\alpha];\delta([\alpha];\s))$
(for $\alpha>1$). Appendix \ref{App-JM} provides formulas for the Jacquet modules of these representations.

\endinput


TITLE: Is Cosine the only function satisfying $ f'(x)= f(x+\frac{\pi}{2})$?
QUESTION [6 upvotes]: Basically most of us know that $\frac {\textrm{d}}{\textrm{d}x} \cos x = -\sin x$ . Also $ \cos (x+\frac{\pi}{2})=-\sin x$
That makes $$ \frac {\textrm{d}}{\textrm{d}x} \cos x = \cos (x+\frac{\pi}{2}) $$ So, out of curiosity I wondered what other functions could have this property. It's pretty interesting that by translating the graph of a function by some vector we get the graph of its derivative.
Therefore, here's my attempt:  I tried to find the defining property of this function. Here we go:
$$ f'(x)= f(x+\frac{\pi}{2}) $$ That makes
$$ f''(x)= \left( f'(x)\right)'= f(x+\frac{\pi}{2})'=f'(x+\frac{\pi}{2})=f(x+\pi)$$
Generally $$f^n(x)=f(x+n\frac{\pi}{2})$$
With $f^n(x) $ being the $n$-th derivative of $f$.
 And I'm stuck at this step. What makes cosine adapt to this property is that $ \cos (x+\pi)=-\cos x$ which allows it to be the solution of the following differential equation:
$$f''(x)+f(x)=0$$
Though I don't want to admit that $f(x+\pi)=-f(x) $  So, could you please help me? Isn't there any way to determine all the functions that satisfy the previously acknowledged property? Are they infinite?

REPLY [1 votes]: We have:
$$\sin(x+\pi/2)=-\sin(\pi/2-x)=\cos x=(\sin x)'$$
Actually, $\cos(x+a)$ for any $a$ will work.
Applying Fourier transform rules, your original equation becomes:
$$i\omega \hat f(\omega)=e^{i\omega \pi/2}\hat f(\omega)$$
So, when $i\omega \neq e^{i\omega\pi/2}$, $\hat f$ must be zero. The set of solutions to $i\omega =e^{i\omega\pi/2}$ is discrete, and contains only $-1$ and $1$ on the real line. Are there complex solution? I don't know.
The result is that $\hat f$ must be a linear combination of delta functions at the roots $\omega_1,\omega_2,...$, an thus that $f$ is a linear combination of $e^{i\omega_k x}$, The cases $\omega=1,-1$ give $\sin x$ and $\cos x$.
(This corresponds to the comment given above, where $\frac{\ln \lambda}{\lambda}=\frac{\pi}{2}$, by setting $\lambda =i\omega$.)


TITLE: Why are eigen-things better/more useful than non-eigen-things?
QUESTION [2 upvotes]: An eigenvector is a vector that doesn't change direction when a transformation is applied.  So in the case of, say, an energy (Hamiltonian) eigenstate, it's a state that doesn't change 'direction' (not exactly sure what to map that to).  But why is that more special/important/useful than any of the other states?
Is it because you can potentially 'construct' any of the other states with the eigenstates, so that's all you need to worry about?

REPLY [3 votes]: You can write down quantum states as $| \psi \rangle = \sum_k c_k | \phi_k \rangle$ for any basis $| \phi_k \rangle$. However, if you choose basis which is not an eigenbasis of Hamiltonian the coefficients will change with time and it would be extremely difficult to keep track of them. Therefore such expansion is not very useful. To sum up: every basis is fundamentally as good as any other. We usually use particular bases, such as eigenbasis of Hamiltonian because then it's easier to solve real physical problems.


TITLE: Does diamond dust conduct heat as well as diamond?
QUESTION [6 upvotes]: Diamond is one of the best thermal conductors you can get. If the diamond is crushed into dust and spread out over a flat surface, but still held fairly compact (for instance in a small petri dish), would it still conduct as well as before?

REPLY [0 votes]: It is highly unlikely that diamond dust would be anywhere near in thermal conductivity, what a solid diamond exhibits.
The thermal conductivity of diamond, is extremely dependent on the quality of the diamond, both as to crystal defects, and also impurity doping.
At room Temperature, a typical Type I diamond was measured, giving a thermal conductivity of about 10 Watts per deg C per cm; at least double what silver or copper give.   But the conductivity, typically peaks (for all diamonds) at much lower Temperatures, about 100 kelvins, where it is 3 to 4 times the 300 K value.  Conductivity is best for Type IIa diamonds, with a 100 k value of around 100 W/deg/cm.
Measured conductivity values seem to indicate that the simple isotropic homogenious model of heat flow is not followed in diamond.
The very highest quality (white) gem diamonds, are usually type IIa.   To the best of my recollection, I believe the Cullinan diamond was type IIa.  (largest ever found).


TITLE: Prove by induction that $1+2^{2^n}+2^{2^{n+1}}$ is divisible by $7$
QUESTION [2 upvotes]: As said in the title you have to show by induction that 1+$2^{2^n}+2^{2^{n+1}}$ is divisible by 7. So you start with n=0, that gives 1+2+4=7. So the start is shown.
Let 1+$2^{2^n}+2^{2^{n+1}}$ be divisble by 7 for a n.
I've tried several attempts but I ended up in a mess.
For example the latest attempt:

We have $1+2^{2^{n+1}}+2^{2^{n+2}}$
Adding $2^{2^n}-2^{2^n}$ gives you
$1+2^{2^{n+1}}+2^{2^n}+2^{2^{n+2}}-2^{2^n}$
As the first three summands resemble our Assumption, only $2^{2^{n+2}}-2^{2^n}$ needs to be proved as a multiple of seven.

But I am stuck at trying to show this.
As stated that might be a incorrect approach by myself, so I'm not really sure whether or not that is going in the correct direction. In the end I'd really appreciate some help on this question!

REPLY [1 votes]: Starting with $n = 0$, we have $1+2^{2^0}+2^{2^{0+1}} = 7$, which is clearly divisible by $7$. Now suppose inductively that $n \ge 1$ and argument holds for all $n$. Then for $n+1$, we have
$$1+2^{2^{n+1}}+2^{2^{n+2}} = 1+2^{2\cdot2^n}+2^{2\cdot2^{n+1}} = 1+(2^{2^n})^2+(2^{2^{n+1}})^2$$
Now and and subtract $(2^{2^n})^2$, we have
$$1+2(2^{2^n})^2+(2^{2^{n+1}})^2-(2^{2^n})^2 = [1+(2^{2^n})^2]^2-(2^{2^n})^2$$ $$ = [1+(2^{2^n})^2+2^{2^n}]\cdot[1+(2^{2^n})^2-2^{2^n}]$$
$$ = [1+2^{2^{n+1}}+2^{2^n}]\cdot[1+2^{2^{n+1}}-2^{2^n}]$$
Here, notice that by inductive hyphothesis, $7|(1+2^{2^{n+1}}+2^{2^n})$. Therefore $7$ divides the whole expression and argument holds for $n+1$. Therefore by induction, argument holds for all $n$.


TITLE: How to know whether a chain is a boundary?
QUESTION [0 upvotes]: Well,
Suppose $c$ is a $k$-chain in some open set $U\subset \mathbb{R}^n$, Spivak says in his Calculus on Manifolds that $\partial c =0$ doesn't imply that $c$ is the boundary of some $k+1$ chain.
With differential forms, we know that if $\omega$ is some $k$ form in the open set $U$, then $d\omega = 0$ doesn't imply that there is some $k-1$ form $\eta$ with $\omega=d\eta$. However, if we add the hypothesis that $U$ is contractible then $\eta$ exists.
Is there some hypothesis that when added to $U$ grants that $\partial c = 0$ imply $c$ being the boundary of some other chain $r$?
Thanks very much in advance!

REPLY [1 votes]: The hypothesis is that the $k$th homology group of $U$ is zero. I'm not sure what coefficients you are using for your chains, so I will assume you allow only integer coefficients here (but the story works for coefficients in other abelian groups).
Let $Z_k$ denotes the set of closed $k$-chains in $U$, i.e. those $k$-chains $c$ such that $\partial c=0$, and let $B_k$ denote the set of boundary $k$-chains, i.e. those $k$-chains $c$ such that $c = \partial a$ for some $(k+1)$-chain $a$. Each of $B_k$ and $Z_k$ is an abelian subgroup of the group of all $k$-chains.
Since $\partial^2=0$, we have $B_k\subset Z_k$. The $k$th homology group of $U$ is defined to be the quotient group
$$H_k(U,\mathbb{Z}) = Z_k/B_k.$$
It can be shown that this is a homotopy invariant, i.e. if $U$ and $V$ are homotopic then they have isomorphic homology groups. In particular, this is true if $U$ and $V$ are homeomorphic.
By definition, we have $H_k(U,\mathbb{Z})=0$ if and only if $Z_k=B_k$, which is exactly the condition that every $k$-chain $c$ with $\partial c=0$ is the boundary of some $(k+1)$-chain.
One topological condition to guarantee that this happens is if $U$ is contractible (homotopic to a single point). In this case, $U$ has the same homology groups as a point, which are all zero (except for $k=0$).
Regarding the result you state about differential forms, there is a map from the $k$th homotopy group to the $k$th homology group,
$$\pi_k(U)\to H_k(U,\mathbb{Z}),$$ called the Hurewicz map. Part of the Hurewicz theorem says that this map is surjective for $k=1$. So if $U$ is simply connected, which is equivalent to saying that its first homotopy group $\pi_1(U)$ is trivial, then the Hurewicz theorem implies that $H_1(U,\mathbb{Z})$ is also trivial.


TITLE: Fluid pressure in Neumann's triangle for triple junctions
QUESTION [1 upvotes]: Consider a point of contact where three fluid phases meet. The  Neumann's triangle rule  for this case is usually written as,
\begin{equation}
\frac{\sigma_{\alpha\beta}}{\sin \gamma} = \frac{\sigma_{\beta\gamma}}{\sin \alpha}  = \frac{\sigma_{\gamma\alpha}}{\sin \beta}
\end{equation}
where $\sigma_{\alpha\beta}$ is the surface tension between two fluids and the angle $\gamma$ shows the angle created by phase $\gamma$ at the triple contact point. The angles naturally satisfy $\alpha + \beta +\gamma = 2\pi$.
In the case of diffuse interface modelling, the surface tension effects on the fluid momentum balance (Navier-Stokes equations) are expressed as,
\begin{equation}
\frac{\partial \rho\mathbf{u}}{\partial t} + \nabla\cdot\nabla\cdot(\rho\mathbf{u}\mathbf{u}) = -\nabla p + \nabla\cdot S -\sum_{\alpha<\beta}\sigma_{\alpha\beta}\kappa_{\alpha\beta}\nabla\varphi_{\alpha\beta} \label{Eq:NV}
\end{equation}
where $S$ is the shear stress tensor, $\kappa_{\alpha\beta}$ is  curvature, $\nabla\varphi_{\alpha\beta}$ is gradient of the phase indicator function of the interface. In equilibrium, when the fluid velocity is zero $\mathbf{u} = 0$, we can write,
\begin{equation}
-\nabla p -\sum_{\alpha<\beta}\sigma_{\alpha\beta}\kappa_{\alpha\beta}\nabla\varphi_{\alpha\beta} = 0
\end{equation}
I have some questions regarding this:
1) I understand the Lami's law for force balance. But, in order to extend the Lami's law we must specify how the surface tension is converted to an equivalent force. Specifically, we should mention which length scale is divided with surface tension $\sigma_{\alpha\beta}$ etc so that a force term is generated. Also why this length scale is same for all the three interfaces?
2) From the two phase fluid case, I know that the pressure difference across the interface is related to the surface tension as $\Delta p = \sigma\kappa$, where $\kappa$ is the curvature. Why we do not consider the pressure drop across different interfaces while considering the equilibrium situation of three phase contact? Surely there will a component of the pressure drop $\Delta p_{\alpha\beta}$ in a direction tangent to the interface  for $\alpha\gamma$ and $\beta\gamma$ interfaces. What am I missing here?
3) The Neumann's triangle relation is equivalent,
\begin{equation}
\sigma_{\alpha\beta}\hat{\tau}_{\alpha\beta} + \sigma_{\beta\gamma}\hat{\tau}_{\beta\gamma} + \sigma_{\gamma\alpha}\hat{\tau}_{\gamma\alpha} = 0
\end{equation}
where $\hat{\tau}_{\gamma\alpha}$ is a unit vector along tangential direction to $\gamma\alpha$ interface. How do the tangential components come into play when in the Navier-Stokes equation (see Eq.~(\ref{Eq:NV})) we include surface tension forces in a direction normal to the interface?
Any hint or justification will be highly appreciated.

REPLY [1 votes]: some hints here:

The Lami law is two-dimensional, while the point of contact of three fluids is actually a line. The "unit length" is this third dimension, the interface line, that rarely appears in the drawings. All drawings i found represent only a 2D slice. Strictly speaking, the tension at the point (line) is a force per unit length.
At the interface point we disregard the curvature, since its contribution is small; the dominant contribution is the tension along the surface. Note also that difference in pressure has negligible contribution at the point of contact (arbitrary small surface).
Locally, the surface tension is much greater than the force due to pressure difference.
I do not know how to write Navier-Stokes... I will say as before, at the point you have forces only tangential, nothing normal to the interface


TITLE: What is the Laurent Series of $\cos{\frac{1}{\sqrt{z}}}$ at point $z=0$?
QUESTION [1 upvotes]: The function $\cos{\frac{1}{\sqrt{z}}}$ is analytic at any neighborhood of $z=0$, but I find it hard to get the Laurent Series of the function. Any help?

REPLY [2 votes]: $$\begin{align}
\cos z = \sum_{n=0}^\infty \frac{(-1)^n z^{2n}}{(2n)!}
\\
\cos \frac{1}{\sqrt{z}} = \sum_{n=0}^\infty \frac{(-1)^n z^{-n}}{(2n)!}
\end{align}$$
what's wrong with that?