## 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:
$$flongmapstovarphiequiv fcirchat{zeta}~,$$
where
$$hat{zeta} x,=,x^{-1}$$
is the inversion operation on $$G$$.

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

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?)

PS.

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:

begin{align*} 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}). end{align*}

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, $$n times n$$)
$$A = begin{bmatrix} 0 & 4.5 & 6 & 5.5 \ 4.5 & 0 & 4 & 3 \ 6 & 4 & 0 & 5 \ 5.5 & 3 & 5 & 0 \ end{bmatrix}$$

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

In Wolfram Mathematica I tried this way with mapping `Ordering`:

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

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.

## 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 $$0, such that $$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 $$0, $$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} end{align*}.

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?

## 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:

$$(a,b)mapsto(a’,b’)=left(frac{a+b}{2},quadpmsqrt{ab}right)$$

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

$$S$$ 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

$$(a,b)={(a,b),(a,-b),(-b,a),(-b,-a),(-a,-b),(-a,b),(b,-a),(b,a)};$$

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

$$(a,b)mapstoleft(frac{a+b}{2},;sqrt{ab}right),;left(frac{a-b}{2},;isqrt{ab}right)$$

$$(a,b)mapstoleft(a+sqrt{a^2-b^2},;a-sqrt{a^2-b^2}right),;left(b+isqrt{a^2-b^2},;b-isqrt{a^2-b^2}right).$$

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
E