# Challenge Math Experiment: Let's Prove Something!

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#### fresh_42

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2018 Award
Summary
Another Kurzweil.
This is an experiment. I thought of a way to bridge the gap between the usual challenge threads. Of course we could shorten the monthly period, but given that there are almost always untouched problems, more of them might not be the solution. Today we had a thread "Is math a language" by @frankin garcia and most of us will certainly think so. So if it is, then we can write an essay in this language, formerly known as proof. I have no idea how it will go, where it ends, or if it makes sense at all. I thought we could give it a try until the next load of challenges in December. We start with:

Let $G$ be a not necessarily Abelian group of square integrable smooth, real functions $f\, : \,I=[0,1]\longrightarrow [0,1]$ on the unit interval, and $\mathfrak{g}$ its real Lie algebra.

Now everybody can either add a conclusion based on all previous posts, e.g. "Since $I$ is compact, all functions ..." or add additional properties, e.g. "Assume $G$ is simple." or focus on additional perspectives, e.g. "Let us consider the center $Z(G)$ of $G$ ...". Conclusions must be proven (keep it short) or the relevant theorem must be quoted. In my example it could be e.g. Heine-Borel or Weierstrass, depending on which direction you want to go: topology or analysis. Please choose only one of these possibilities per post and do not post more than once in a row, i.e. you may continue after somebody else posted something. The projected runtime is until end of month, but it will depend on what actually will happen.

Now let's see whether we can prove something!

Edit: Group operation is $(fg)(x)=f(g(x))$ and integration Lebesgue. I also corrected the domain accordingly. Two dimensional might have been more fun, but let's start simple and see who participates.

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#### Math_QED

Homework Helper
Interesting experiment:

To make the thread self-contained: For what operation is $G$ a group? And by what Bracket is the Lie-algebra given?

#### fresh_42

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2018 Award
Yes, Lebesgue. But as I assume invertible functions, there shouldn't occur any Nullsets or other major differences between the two on a compact domain.

Let's assume that the group operation is the natural $(f\circ g)(x,y)=f(g(x,y))$ and $\mathfrak{g}$ the left-invariant smooth vector fields on $G$, i.e. the smooth sections of the tangent bundle. Multiplication is then defined by the flows along these fields.

Let $X,Y$ be two vector fields on the smooth manifold $G$ and $X(\gamma(t))$ a flow of $X$.
The Lie derivative from $Y$ along $X$ is defined by

\mathcal{L}_XY = \left. \frac{d}{dt} \right|_{t=0}(X^*(\gamma(t))Y)

Then

\mathcal{L}_XY = [X,Y] = X\circ Y - Y \circ X

#### Svein

It looks like a Hilbert space to me - with the usual inner product $<f, g> = \int f\cdot g$.

#### fresh_42

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2018 Award
It looks like a Hilbert space to me - with the usual inner product $<f, g> = \int f\cdot g$.
Not quite. As the elements form a multiplicative group, it cannot be a vector space.

#### Svein

Not quite. As the elements form a multiplicative group, it cannot be a vector space.
Why not? I can come up with two bases straight away:
1. The functions $1, x, x^{2},x^{3},x^{4},...$ form a countable base for G
2. The functions $\left[ \sin(n(2\pi (x-\frac{1}{2}))),\cos(n(2\pi (x-\frac{1}{2}))) \right]_{n=0}^{\infty}$ form another countable base

#### fresh_42

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2018 Award
Why not? I can come up with two bases straight away:
1. The functions $1, x, x^{2},x^{3},x^{4},...$ form a countable base for G
2. The functions $\left[ \sin(n(2\pi (x-\frac{1}{2}))),\cos(n(2\pi (x-\frac{1}{2}))) \right]_{n=0}^{\infty}$ form another countable base
$0 \notin G$. And a basis would mean only finitely many coefficients were different from zero, but $G$ contains power series. But as $x-x \notin G$, the question doesn't come up.

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#### Infrared

Gold Member
How are you putting a smooth structure on $G$? Is the topology on $G$ given by the compact-open topology?

#### fresh_42

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2018 Award
I only wanted to give a frame: group of smooth functions. I haven't elaborated details. Square integrability was already redundant after the requirement, that group multiplication is consecutive application. The Lie algebra multiplication of vector fields should only ensure, that the functions are the points of the manifold, and their tangents shouldn't be regarded to define $\mathfrak{g}$.

So which topology applies is already a first step. Just choose a "normal" one. The group allows the $L^2$ metric which is a natural candidate. I guess it's even dense in $L^2([0,1])$.

#### WWGD

Gold Member
What's the group operation? If you multiply you cannot have inverses unless you consider $f: f (x)\neq 0$? Or maybe you want your target to be $(0,1]$ or $[\epsilon,1]; \epsilon >0$ for compactness?

#### fresh_42

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2018 Award
I'm not sure whether it's better to end the experiment or restart it somehow. The goal is not to guess a proof or a theorem which I have in mind, the goal is to create one!

Let $$G=\left(\{\,f \in L^2([0,1];[0,1])\,|\,f \text{ is bijective and smooth }\,\}\, , \,\circ \, : \,(f,g) \longmapsto (x \longmapsto f(g(x)))\right)$$ be a multiplicative group with the induced topology from the $L^2$ inner product, norm and metric. This should be a Lie group, so we also have a Lie algebra $\mathfrak{g}$, the smooth tangent vector field of flows in $G$. I also guess that $G \subseteq L^2([0,1];[0,1])$ is an open, dense subset.

So that's it so far. Possible moves are:

1. conclusion from previously given or found facts (e.g. all $f$ have a minimum $x_f$ and a maximum $x_F$ by Weierstraß)
2. addition of conditions (e.g. we assume that $\mathfrak{g}$ is solvable, which also restricts $G$)
3. addition of objects (e.g. let $D$ the maximal simple subgroup of $G$)

and the only rule is not to post twice in a row. From here on it's just creativity.

#### Infrared

Gold Member
I'll try to get the party started.

Consider the smooth map $T:G\to\mathbb{R}$ given by $T(f)=\int_0^1 |f(x)|^2 dx$.

I'm still not totally sure what the smooth structure on $G$ is though- especially as it's infinite-dimensional, and shouldn't we also require elements of $G$ to have smooth inverses (so that $G$ is closed under inversion)?

#### fresh_42

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2018 Award
I'm still not totally sure what the smooth structure on $G$ is though- especially as it's infinite-dimensional, and shouldn't we also require elements of $G$ to have smooth inverses (so that $G$ is closed under inversion)?
Yes, but we can pretend that it is given by the word "group".

#### Math_QED

Homework Helper
Yes, but we can pretend that it is given by the word "group".
I think he means that the map $G \to G: g \mapsto g^{-1}$ is smooth, not that inverses of smooth maps itself are smooth, but the inversion map itself.

#### fresh_42

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2018 Award
I think he means that the map $G \to G: g \mapsto g^{-1}$ is smooth, not that inverses of smooth maps itself are smooth, but the inversion map itself.
Well, it can be decided mathematically: prove it or find a counterexample. But as there is so much boundary around, I wouldn't expect a counterexample.

To keep track of the problem posts in comparison to side discussions, please enter the sequence of previously relevant posts in any post on the problem. Right now we are at the prefix

(11,12)

#### fresh_42

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2018 Award
(11,12)

Consider the smooth map $T:G\to\mathbb{R}$ given by $T(f)=\int_0^1 |f(x)|^2 dx$.
This means, that $T(f)=||f||_2^2$ is the restriction of the squared $L^2-$norm of $\mathcal{A}:= L^2([0,1];\mathbb{R})$ on $G$ and $\operatorname{im}(T)=(0,1).$ $G$ is not closed in this norm topology: the sequence $f_n(x)= \dfrac{e^{x^n}-1}{e-1} \in G$ converges to $f(x)=\begin{cases}0 &,\,x<1\\1&,\,x=1\end{cases}$ which is not in $G$.

(... which is a hint that $G$ might not be a Lie group.)

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#### Couchyam

Here's a basic question: what's the smallest subset in $G$ that generates $G$? (Or even more basic: what subsets in $G$ generate all of $G$?) Here are a few simple conjectures:
$$\text{(i) Existence of (e.g. square) roots: if }g\in G,\text{ then }\exists! \sqrt{g} \text{ s.t. }\sqrt{g}\circ\sqrt{g}=g.$$
$$\text{(ii) Existence of flows: if }g\in G,\exists! \{g_\lambda\}_{\lambda\in \mathbb{R}^+}:\,g_1= g,\,g_0=\mathbb{I},\text{ and }g_a\circ g_b = g_{a+b}$$
$$\text{(iii) Convexity: if }g_1,\,g_2\in G\text{ and both }g_1\,g_2\text{ are convex, then so is }g_1\circ g_2.$$
$$\text{(iv) Noncommutativity: e.g. let }g_1(x)=x^2,\,g_2(x)=\frac{2}{\pi}\arcsin(x)$$
$$\text{(v) Piecewise linear approximation: if }g\in G,\,\epsilon> 0,\,\exists g_{\epsilon}(x)\text{ piecewise linear, such that }\|g\circ g'-g_\epsilon\circ g'\|<\epsilon\,\forall g'\in G.$$
$$\text{(vi) Same as (v) for powers of }g\in G\text{ up to some }n\in\mathbb N$$
$$\text{(vii) Interleaving properties of cusp singularities in piecewise linear functions under composition.}$$
$$\text{(viii) Approximating arbitrary p.w. linear functions with those generated from a countable subset.}$$
$$\text{(ix) Same as (viii) for finite subsets (of piecewise linear functions.)}$$
$$\text{(x) Criteria for commutativity: }g_1\text{ and }g_2\in G \text{ commute iff they belong to the same one-parameter flow, }g_\lambda,\,\lambda\in\mathbb R^+.(?)$$

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#### Couchyam

(i) Consider derivatives of $\sqrt{g}$.
(ii) After (i), uniqueness of flows follows from continuity.
(x) There are a large number of mutually commuting, distinct, one-parameter flows. (Are there more?)
(xi) The 'center' of $G$ consists of just the identity. This can be seen by considering the shift operation, $g(x)\rightarrow T(g,s)(x)=\overline{g(\overline{x-s})-g(1-s)}$ (where $\bar x\equiv x\text{ mod }1$.)

#### fresh_42

Mentor
2018 Award
(i) Consider derivatives of $\sqrt{g}$.
(ii) After (i), uniqueness of flows follows from continuity.
(x) There are a large number of mutually commuting, distinct, one-parameter flows. (Are there more?)
(xi) The 'center' of $G$ consists of just the identity. This can be seen by considering the shift operation, $g(x)\rightarrow T(g,s)(x)=\overline{g(\overline{x-s})-g(1-s)}$ (where $\bar x\equiv x\text{ mod }1$.)
That's not exactly what I meant. Choose an option, prove it and post it with a first line (11,12,16) indicating the posts which contributed to our proof so far. Brainstorming isn't meant as contribution. So your post should have been:

(11,12,16)

The center of $G$ is trivial: $Z(G)=\{\,\operatorname{id}_{[0,1]}=1\,\}.$
Proof: Say $f\in Z(G)$ and $g(x):= \ldots$
Or you can still prove or disprove whether $G$ is actually a Lie group.

#### WWGD

Gold Member
I'll try to get the party started.

Consider the smooth map $T:G\to\mathbb{R}$ given by $T(f)=\int_0^1 |f(x)|^2 dx$.

I'm still not totally sure what the smooth structure on $G$ is though- especially as it's infinite-dimensional, and shouldn't we also require elements of $G$ to have smooth inverses (so that $G$ is closed under inversion)?
Maybe a Banach manifold? I dont know if they can be Lie groups?

#### fresh_42

Mentor
2018 Award
Maybe a Banach manifold? I dont know if they can be Lie groups?
Addition is a problem within $[0,1]$ which is needed for the multiplication $(fg)(x)=f(g(x))$.

All my attempts to find an example $||f^{-1}-g^{-1}||_2 > \varepsilon$ given $||f-g||_2 < \delta$ were in vain. If two functions are close, then $f^{-1}$ and $g^{-1}$ were also close. But I don't have a proof; or a counterexample. So the Lie question is still open. Other topological features except "not closed" are also still open.

#### Couchyam

All my attempts to find an example $||f^{-1}-g^{-1}||_2 > \varepsilon$ given $||f-g||_2 < \delta$ were in vain. If two functions are close, then $f^{-1}$ and $g^{-1}$ were also close. But I don't have a proof; or a counterexample. So the Lie question is still open. Other topological features except "not closed" are also still open.
I might try using the reflection property of inverse functions, and the fact that $x<1\Rightarrow x^2<x$ (so the $L^2$ norm is bounded by the $L^1$ norm in this case.) It could be there exists a more natural metric, however (one that is invariant with respect to the Lie group multiplication.)

#### fresh_42

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2018 Award
Another interesting subject are involutions. But it's not my turn.

#### Couchyam

(11, 12, 16, 21)

Prop:
The function $d(f,g) \equiv \frac{1}{2}(\|x-f^{-1}\circ g(x)\|_2+\|x-g^{-1}\circ f(x)\|_2)$ is invariant under left multiplication.
Proof:
$d(af,ag) = \frac{1}{2}(\|x-f^{-1}\circ (a^{-1}\circ a)\circ g(x)\|_2+\|x-g^{-1}\circ (a^{-1}\circ a)\circ f(x)\|_2)=d(f,g).$

(The above function is just the 'average' of the distances to $g$ from the perspective of $f$ and vice versa.)

#### fresh_42

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2018 Award
(11, 12, 16, 24)

The functions $1$ and $x\longmapsto 1-x$ are clearly involutions and since $\left(f\iota f^{-1}\right)=f \iota^2 f^{-1}=1$, the involutions form a normal, infinite subgroup of $G$. Of course $f\iota f^{-1} \stackrel{i.g.}{\neq}f^{-1}\iota f\,.$