measure theory – Functions on a group. Functions of the inverse element.

Consider a Lie group $G$ and a Hilbert space $mathbb{V}$ equipped with a dot product $~~langle~,~rangle$.

Let ${^G{cal{L}}}$ be a space of square-integrable functions mapping $G$ to $mathbb{V}$:
{^G{cal{L}}}=left{ f:~~ Glongrightarrowmathbb{V}qquadBig{|}quad int dg langle f(g), f(g)rangle < infty right}.

We can map each $ f$ to a function of the inverse argument:
f(x)longmapstovarphi(x)equiv f(x^{-1})~,quad xin G

or, in short:
flongmapstovarphiequiv fcirchat{zeta}~,

hat{zeta} x,=,x^{-1}

is the inversion operation on $G$.

Doing this for all $f$, we obtain a new space of functions:
{cal{L}}^{G}=left{~varphi:~Glongrightarrow{mathbb{V}}~~Big{|}~~varphi= fcirchat{zeta}~,~~fin{^G{cal{L}}}

To examine if $^{G}{cal{L}}$ and ${cal{L}}^{G}$ are copies of the same functional space, we check the square-integrability of the new functions:
int dg langle varphi(g),varphi(g)rangle=int d(g^{-1}) langle varphi(g^{-1}) , varphi(g^{-1})rangle=
int d(g^{-1}) langle f(g),f(g)rangle

For unimodular groups, the measure is invariant and $d(g^{-1}) = dg $, wherefrom
int dg langle varphi(g),varphi(g)rangle=
int dg langle f(g),f(g)rangle < infty~,

so the spaces $^{G}{cal{L}}$ and ${cal{L}}^{G}$ coincide.

For what nonunimodular groups would this outcome stay valid? (Say, for compact ones? For locally compact ones?)


I understand that, by Haar’s theorem, if a group is locally compact (i.e. if its identity element has a compact neighborhood), then it admits a unique left-invariant and a unique right-invariant measure (unique — up to multiplication by a positive constant). Under the transformation $gto g^{-1}$, a left-invariant measure transforms into a right-invariant one, and vice versa. Can this fact be somehow employed here?

abstract algebra – Saracino Ex 3.17: Associative binary operation with unique right identity, left inverse implies group

Suppose $G$ is a set and $star$ is an associative binary operation on $G$ such that there is a unique right identity element and every element has a left inverse. Prove that $(G,star)$ is a group.

Here’s my work so far:
We know there is a unique $e in G$ such that $x star e = x$ for all $x in G$. We also know that for each $x in G$ there exists some $x^{-1} in G$ such that $x^{-1} star x = e$. My trouble is with establishing that $e$ is also a left identity element. If I can show that $e star x = x star e = x$ for all $x$, then I am able to prove that $x star x^{-1} = x^{-1} star x = e$ for all $x in G$ with the following argument:

x star e = x implies (x star e) star x^{-1} &= x star x^{-1} && text{Multiplying by } x^{-1} text{ on the right} \(3pt)
implies (x star e) star x^{-1} &= e && text{(Defn. of inverse)} \(3pt)
implies (e star x) star x^{-1} &= e && (text{Assuming } x star e = e star x) \(3pt)
implies e star ( x star x^{-1}) &= e && text{(Associativity)} \(3pt)
implies x star x^{-1} &= e && (e text{ is the unique right identity element}).

My attempts at showing that $e$ is a left identity have been fruitless…I know I somehow must use the fact that $e$ is a unique right identity element because the text explained that without the uniqueness condition, $(G,star)$ is not a group. Any help would be appreciated!

Map Inverse Ordering on Matrix

I have an matrix (symmetric, but it doesn’t matter, $n times n$)
$$A = begin{bmatrix}
0 & 4.5 & 6 & 5.5 \
4.5 & 0 & 4 & 3 \
6 & 4 & 0 & 5 \
5.5 & 3 & 5 & 0 \

I want to replace this matrix on matrix with ranks in each row (ascending), ex:
$$A^* = begin{bmatrix}
0 & 1 & 3 & 2 \
3 & 0 & 2 & 1 \
3 & 1 & 0 & 2 \
3 & 1 & 2 & 0 \

In Wolfram Mathematica I tried this way with mapping Ordering:

ranked = Ordering /@ Ordering /@ A - 1;

Out(1): {{0,1,3,2},{3,0,2,1},{3,1,0,2},{3,1,2,0}}

Is there a way to do it simpler? Without mapping two times ordering function.

Thanks in advance for your answers.

matrices – Pseudo inverse of a diagonal matrix

Let $A in R^{n times n}$ matrix with $k$ diagonal elements with $k< n$ and rest of the elements are zero. I am trying to find the pseudo inverse of $A + lambda I$ when $lambda$ approaches zero. Then $frac{1}{a_i + lambda}$ would be the diagonal elements for $i$ going from 1 to $k$ of the pseudo inverse and $frac{1}{lambda}$ would be the rest of the diagonal elements. If I put $lambda$ equal to zero then the pseudo inverse would a matrix with elements of $A$ matrix inverted, but there would be elements going to infinity. But that does not sound right. What is wrong in this logic?

dg.differential geometry – Are a map with constant singular values and its inverse always conjugate through isometries?

Let $U subseteq mathbb R^2$ be an open, connected, bounded subset. Fix $0<sigma_1<sigma_2$, such that $sigma_1 sigma_2=1$.

Suppose that there exist a diffeomorphism $f:U to U$ such that the singular values of $df$ equal $sigma_1,sigma_2$ everywhere on $U$.

Question: Do there exist smooth isometries $
phi_1, phi_2:U to U$
such that $phi_1 circ f circ phi_2=f^{-1}$?

We must have $phi_i in operatorname{O}(2)$ (isometries are affine), but the point here is that I want them to map $U$ into $U$.

Note that $df^{-1}=(df)^{-1}$ has singular values $sigma_1,sigma_2$, the same as $df$.

Here are two examples where this phenomena happens:

1. Affine maps on ellipses:

Let $0<a<b$, $ab=1$, and let
U=U_{a,b}=biggl{(x,y) ,biggm | , frac{x^2}{a^2} + frac{y^2}{b^2} < 1 biggr}.

Take $f(x,y)=Apmatrix{x\y}$, where $$begin{align*} & A=A(theta)= begin{pmatrix} a& 0 \ 0 & b end{pmatrix} begin{pmatrix} costheta & -sintheta \ sintheta & cos theta end{pmatrix}begin{pmatrix} 1/a& 0 \ 0 & 1/b end{pmatrix}=
begin{pmatrix} costheta & -frac ab sintheta \ frac ba sintheta & cos theta end{pmatrix}

Then $A(theta)^{-1}=A(-theta)=JA(theta) J$, where $J=begin{pmatrix} 1& 0 \ 0 & -1 end{pmatrix}$ is the reflection around the $y$ axis.

2. Non-affine maps on the disk:

Let $U=Dsetminus{0}$ where $D subseteq mathbb R^2$ is the unit disk.

$f_c: (r,theta)to (r,theta+clog r )$. Then we have $f_{c}^{-1}=f_{-c}=Jf_{c}J$.

Note that
(df_c)_{{ frac{partial}{partial r},frac{1}{r}frac{partial}{partial theta}}}=begin{pmatrix} 1 & 0 \ c & 1end{pmatrix},

so the singular values of $f_c$ are constants which depend on $c$.

Now, in general there are many local solutions to the PDE $sigma_i(df)=sigma_i$, so I don’t expect such a special relation between $f$ and its inverse in general. But I don’t have a counter-example yet.

calculus and analysis – Integration that involves inverse of a big symbolic matrix

I have a $300times300$ symbolic matrix and I want to calculate its inverse matrix. It has been two days and Mathematica is still executing Inverse(T). $T$ is the name of that matrix.

Why am I calculating inverse?

Actually, $G=Inverse(T)$ is a Green’s function. After calculating this Green’s function I want to use $NIntegrate( Tr(G.(some:other:matrices)) )$ function to calculate a 3D integral.

In MATLAB, taking an inverse was not a problem because MATLAB’s $integral3()$ function does not need to calculate the inverse of the symbolic matrix.

My question is, is there any way to use $NIntegrate()$ such that I don’t have to calculate the inverse of the symbolic matrix (the way MATLAB’s integral3() works). Or is there any other way to solve the integration that I am trying to solve?

real analysis – Inverse of stereographic projection

I am trying to compute the inverse of

$$f(theta,phi) = left(frac{costheta sinphi}{1-cosphi}, frac{sintheta sinphi}{1-cosphi}right)$$ but I lack some basic knowledge on how I can do that. Could please provide some guidance?

Thank you in advance.

sequences and series – What is the set of pairs of points in $mathbb C$ generated from one pair by the AGM iteration and its inverse?

The arithmetic mean of two complex numbers $a,b$ is $tfrac12(a+b)$, and the geometric mean is $pmsqrt{ab}$, which is two-valued. The arithmetic-geometric mean is the limit of the pair of sequences formed by iterating

$$mathbb C^2tomathbb C^2:;(a,b)mapsto(a’,b’)=left(frac{a+b}{2},;sqrt{ab}right)$$

(and choosing one of the two possible square roots at each step). See Cox’s paper The Arithmetic-Geometric Mean of Gauss.

Instead of taking one path and indexing by $mathbb N$, let’s consider the set $Ssubseteqmathbb C^2$ of all pairs of points which can be reached in this way, taking both square roots, and also iterating backwards:


$$(a’,b’)mapsto(a,b)=left(a’pmsqrt{a’^2-b’^2},quad a’mpsqrt{a’^2-b’^2}right).$$

$S$ could be indexed by some directed graph where each vertex has two successors and two predecessors. The two successors of $(a,b)$ (given by the two choices of geometric mean) share their two predecessors: $(a,b)$ itself and $(b,a)$. The two predecessors of $(a’,b’)$ share successors: $(a’,b’)$ and $(a’,-b’)$.

With that in mind, we can define equivalence classes


note that $(a,b)in S$ if and only if $(a,b)subseteq S$. Our multi-valued functions act on these by



Let’s also define $A={amidexists b:(a,b)in S}={bmidexists a:(a,b)in S}subseteqmathbb C$, and $X^*$ as the derived set (the set of accumulation points, or limit points) of a set $X$. Now for the questions.

Given a single pair $(a_0,b_0)$ to generate $S$, in general, what does $S$ or $A$ look like? Where does it accumulate?

If $a^*in A^*$, is there some $b^*$ such that $(a^*,b^*)in S^*$?

If $(a^*,b^*)in S^*$, must $a^*=pm b^*$?

If $(a^*,b^*)in S^*$, is it actually the limit of some sequence $(a_n,b_n)to(a^*,b^*)$ following the two-valued function and its inverse? that is, a sequence where, for all $ninmathbb N$, $(a_{n+1},b_{n+1})$ is one of the four values obtainable from $(a_n,b_n)mapsto(a_{n+1},b_{n+1})$?

The case $a_0=b_0=0$ is trivial: $S={(0,0)}$.

The case $a_0neq b_0=0$ is tractable:

$$S={(a_0/2^n,0),,(ia_0/2^n,ia_0/2^n)mid ninmathbb Z}$$

(I know, this is abusing set notation to have $(a,b)in S$ rather than $(a,b)subseteq S$)

$$A={0,,pm a_0/2^n,,pm ia_0/2^nmid ninmathbb Z}$$

$$S^*={(0,0)},quad A^*={0}.$$

The case $a_0=pm b_0neq0$ has the same form.

It remains to consider the general case $a_0b_0(a_0!^2-b_0!^2)neq0$.

real analysis – What is the inverse of $h(x,y) = left( |x|^{frac{1}{p-1}}text{sgn}(x), |y|^{frac{1}{p-1}}text{sgn}(y) right)$

Question; Given $pin (1,infty)$ and the mapping $h:(mathbb{R}^2, |cdot|_p) to (mathbb{R}^2, |cdot|_q) $ defined by
$$h(x,y) = left( |x|^{frac{1}{p-1}}text{sgn}(x), |y|^{frac{1}{p-1}}text{sgn}(y) right)$$
where $dfrac{1}{p} + dfrac{1}{q} = 1$ and
$$|(x,y)|_p= left(|x|^p + |y|^p right)^{1/p}.$$

I know that $h$ is a bijection function and thus I would like to find its inverse.
However, I am not able to do so.
Any hint is appreciated.

How to find the Inverse of numeric Function that was created from NDSolve

I’m a newbie to Mathematica and i do have a simple question.
I have a function H that was created numerically as solution form NDSolve (like H(R)), and i’m trying to create the Inverse function R(H) for my next step of manipulation. I was trying to use InverseFunction method, but it does not seems to work since it expects a square matrix of data.
any idea?
Thank u