Math Experiment: Let's Prove Something

[fresh_42]

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.

 

[member 587159]

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]

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
\begin{equation}
\mathcal{L}_XY = \left. \frac{d}{dt} \right|_{t=0}(X^*(\gamma(t))Y)
\end{equation}
Then
\begin{equation}
\mathcal{L}_XY = [X,Y] = X\circ Y - Y \circ X
\end{equation}

 

[Svein]

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

 

[fresh_42]

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]

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.

 

[Infrared]

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

 

[fresh_42]

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]

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]

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]

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]

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".

 

[member 587159]

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]

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]

(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.)

 

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

 

[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]

(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]

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 don't know if they can be Lie groups?

 

[fresh_42]

Maybe a Banach manifold? I don't 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]

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]

(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 $f\iota f^{-1} \stackrel{i.g.}{\neq}f^{-1}\iota f\,.$

 

[sysprog]

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.

To me a Kurzweil is a keyboard musical instrument -- I would like to ask that a constraint be added to this venture, and to ask also (by "also", I mean 'too'; not 'thus' ) that it be adopted as retroactive; viz, the constraint that all terms or words used be absent of names of persons, and instead be named after by-reason-intellegible characteristics, so that. for example, the term 'non-Abelian' does not occur -- I don't have an agenda against crediting Prof. Abel, who contributed so much of great worth; however, one can understand the term 'non-commutative' without having to learn much about who discovered or invented what.

 

[fresh_42]

Abelian is a standard to refer to commutativity in abstract algebra. It shouldn't be a big thing on a website where you can hardly find a post without a Lagrangian or a Hamiltonian.

 

[sysprog]

Abelian is a standard to refer to commutativity in abstract algebra. It shouldn't be a big thing on a website where you can hardly find a post without a Lagrangian or a Hamiltonian.

It's a pet peeve of mine -- I detest the practice of naming things after persons -- I think that it's better to name things after their properties.

 

[fresh_42]

It's a pet peeve of mine -- I detest the practice of naming things after persons -- I think that it's better to name things after their properties.

.
I prefer to use what is standard. E.g. when you look for "commutative group" on Wikipedia, you end up here:
https://en.wikipedia.org/wiki/Abelian_group. And all the other examples, where a name shortens a variety of properties: Artinian or Noetherian rings, Lie groups, the many algebras named after persons? This is a senseless discussion.

 

[sysprog]

.
I prefer to use what is standard. E.g. when you look for "commutative group" on Wikipedia, you end up here:
https://en.wikipedia.org/wiki/Abelian_group. And all the other examples, where a name shortens a variety of properties: Artinian or Noetherian rings, Lie groups, the many algebras named after persons? This is a senseless discussion.

Standards -- so many of them to choose from -- in system(s) programming, we often shorten names of things by using acronyms -- not as often by naming things after persons -- I disagree with your contention regarding this discussion being "senseless", and would prefer, for example, along with my prior example, to see the term 'planar', where and when reasonable, used in preference to 'Euclidian', at least whenever planar is all that is actually meant.

 

[fresh_42]

Most people associate planar with something two dimensional whereas Euclidean is not limited. As this is off-topic here, I suggest to start a thread in general discussion about it. There is certainly more to say about this topic.

 

[sysprog]

Oh and by the way, especially for anyone else reading this, an acknowledgment to for and of @fresh_42: I'm pretty sure that he's very extremely much better at math than I am.

 

[sysprog]

. . . and he's not the only one here on PF of whom that's true . . .

 

[fresh_42]

Oh and by the way, especially for anyone else reading this, an acknowledgment to for and of @fresh_42: I'm pretty sure that he's very extremely much better at math than I am.

Thanks, but I already have forgotten many things. Ask @Infrared and @Math_QED , they catch all my carelessness and inaccuracies. We are a quite luxurious website, because errors don't go unseen. I love this about PF and it can't be said too often.

 

[fresh_42]

Resume where we are:

(1) Let $G$ be the group of square integrable smooth, real functions $f\, : \,I=[0,1]\longrightarrow [0,1]$.

(11) $G \subseteq L^2([0,1],\mathbb{R})$ and $(f\circ g)(x)=f(g(x))$

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

(16) 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 $G$.

(24) The function $d(f,g) \equiv \dfrac{1}{2}\left(\|x-f^{-1}\circ g(x)\|_2+\|x-g^{-1}\circ f(x)\|_2\right)$ is invariant under left multiplication.

(25) $x \longmapsto 1-x$ is an involution of $G$. Involutions form a normal, infinite subgroup of $G$.

Copy template for continuation: (1, 11, 12, 16, 24, 25)