How to investigate a transformation that might form a Lie group?

[LieToMe]

I would like to investigate a function that sends $f(x)$ to $f(x) - \frac{1}{c}f(x^{c})$, or a function $g$ such that $g(f(x)) = f(x) - \frac{1}{c}f(x^{c}).$ Since symmetries produced by groups are used in physics, I thought there might be someone here who could help explain what the process is.

What I would like to know is that for at least some analytic function $f(x)$ that this transformation can form a Lie group in the constant $c$, or that translating $c$ to $c + t$ keeps the solution manifold invariant. I would imagine I need to show continuity in some way. Then, how would I show such a manifold is smooth? How would I take a Lie derivative? Is there a convenient way I can satisfy the group axioms? A lot of online text is too scattered and hastily explained to give me concise insight into how to proceed.

It might be that instead of looking at a function $g$, I should just start with $f(x) - \frac{1}{c}f(x^{c})$ and suppose there exists an operator $L$ (not necessarily linear) which sends $f(x) - \frac{1}{c}f(x^{c})$ to $f(x) - \frac{1}{c+t}f(x^{c+t})$. I'm not really sure, I don't have formal training in Lie groups so I don't know what the setup should be.

 

[fresh_42]

What lives where and is what? Since you said "Lie group in $c$" I assume a one parameter group in $c$. But what is $f$ in that scenario? A parameter, a variable? As I see it, you have a family of functions
$$
g(c)(f)=f-\frac{1}{c}f\circ p_c
$$
where $p_c$ maps $x$ to $x^c$. This looks as if $g(c)$ acts on wherever $f$ is taken from. Is it all analytic functions or a certain one? Maybe you can get a local Lie group, but for a global Lie group, one would like to have a functional equation for $g$, i.e. $g(c+t)=g(c) * h(t)$ for some operation and function $h.$

 

[LieToMe]

What lives where and is what? Since you said "Lie group in $c$" I assume a one parameter group in $c$. But what is $f$ in that scenario? A parameter, a variable? As I see it, you have a family of functions
$$
g(c)(f)=f-\frac{1}{c}f\circ p_c
$$
where $p_c$ maps $x$ to $x^c$. This looks as if $g(c)$ acts on wherever $f$ is taken from. Is it all analytic functions or a certain one? Maybe you can get a local Lie group, but for a global Lie group, one would like to have a functional equation for $g$, i.e. $g(c+t)=g(c) * h(t)$ for some operation and function $h.$

I don't have formal training in Lie groups so you probably know as much as me in how to go about this if you haven't seen it before.

The bottom line is that: I think for certain analytic functions, the span of all possible $c \in \mathbb{R}$ of $f(x) - \frac{1}{C}f(x^{C})$ generates a smooth invariant manifold, that as you continuously change $C,$ you get a continuous family of functions under some kind of transformation that forms a Lie group where functions of a certain class are mapped to other functions of the same class in that manifold. So the question then is: how do I set this up these transformations to show they are a Lie group?

 

[Office_Shredder]

Is the set of functions f important here?

In order to have a lie group, you need a group multiplication. Normally you would expect something like if $g_c(f) = f(x) - \frac{1}{c} (x^c)$ then you have $g_c g_d = g_{cd}$ or something, so you pick up the group structure of multiplication of real (non-zero) numbers. But that doesn't seem to hold here.

 

[LieToMe]

Is the set of functions f important here?

Not necesserily, but to be safe I like to start with analytic functions in case there are special properties derived by differentiating or Taylor expansions or unexpected restrictions.

In order to have a lie group, you need a group multiplication. Normally you would expect something like if $g_c(f) = f(x) - \frac{1}{c} (x^c)$ then you have $g_c g_d = g_{cd}$ or something, so you pick up the group structure of multiplication of real (non-zero) numbers. But that doesn't seem to hold here.

So you mean a group operation in the parameter C, that's what I was having trouble with. However, I thought back to Lie symmetries in differential equations. Despite not knowing how the solution depends on a generalized constant in the general case of any symmetric ODE, you can still somehow make deductions about the solution manifold being invariant under the action of a Lie group that shifts the constants. As you might have encountered, you can have expressions like $\frac{e^t}{e^t+C}$ as an invariant solution to an ODE problem, yet despite that, there is somehow a way to argue there is a Lie group in C that produces a smooth manifold. So, I don't think the Lie group property is dependent on the specific algebra, it only seems it's dependent on whether a translation or scaling in the parameter of the group preserves the manifold.

 

[fresh_42]

So you mean a group operation in the parameter C, that's what I was having trouble with.

Yes, that's a crucial point. $\frac{1}{c}f(x^c)$ may have good multiplicative properties, if $f$ has, but adding $f(x)$ makes it somehow linear, and both at the same time doesn't fit well. So maybe you can construct a local Lie group with $f(x)$ as neutral element. Here is an example for a local Lie group:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ (1st section).

 

[LieToMe]

Yes, that's a crucial point. $\frac{1}{c}f(x^c)$ may have good multiplicative properties, if $f$ has, but adding $f(x)$ makes it somehow linear, and both at the same time doesn't fit well. So maybe you can construct a local Lie group with $f(x)$ as neutral element. Here is an example for a local Lie group:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ (1st section).

Thanks, I will look at it. The mapping is linear in $f$, at least those for which $x^C$ is well defined. It probably is not linear in $C$ without a really weird change of coordinates in $C$ and $f$, but then again, one of the reasons for Lie groups is to find whether such a change of coordinates exists. Even though the algebra might be weird or involve functions that aren't even known, it doesn't necessarily mean such mappings don't exist.

 

[fresh_42]

Oh, the mappings do exist. Of course you can define a function $g_c(f)=f-\frac{1}{f}f\circ p_c$ on $C^\infty(\mathbb{K})$ with $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$. The question is why and what for. And in order to make $\{g_c\,|\,c\in \mathbb{K}^*\}$ a one parameter Lie group. you need a smooth multiplication $g_c \ast g_d = g_e$ as mentioned in post #4. $e$ doesn't have to be $c\cdot d$ or $c+d$, and $\ast$ can be defined anyhow, but yet somehow. That's why I linked the article. It is an example with an uncommon multiplication.

 

[LieToMe]

Oh, the mappings do exist. Of course you can define a function $g_c(f)=f-\frac{1}{f}f\circ p_c$ on $C^\infty(\mathbb{K})$ with $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$. The question is why and what for. And in order to make $\{g_c\,|\,c\in \mathbb{K}^*\}$ a one parameter Lie group. you need a smooth multiplication $g_c \ast g_d = g_e$ as mentioned in post #4. $e$ doesn't have to be $c\cdot d$ or $c+d$, and $\ast$ can be defined anyhow, but yet somehow. That's why I linked the article. It is an example with an uncommon multiplication.

Okay, so you're saying there is a way to prove that there does exist a group operation that is linear in C?

As for "why", I am trying to abstract certain algebraic structures to see whether they are preserved in a differential field of non-trivial functions. There is a taylor series for the inverse of f(x) = x^n - x - 1 for instance, but it is obviously very awkward and cumbersome to have to work with a Taylor expansion in every use. In a certain sense I want to see whether such a structure can implicitly be preserved to show whether a transformation admits an particular algebraic field extension without having to try and directly generalize the solution to all polynomial equations.

 

[fresh_42]

Okay, so you're saying there is a way to prove that there does exist a group operation that is linear in C?

No. There may be one, but I don't see it. I only showed you possible ways out of the problem.

 

[LieToMe]

Okay, well maybe it makes sense to test it out on something then. Let's take the case that $f(x) = x$. For general $C$, we need to restrict the domain to $(0,a)$ and perhaps even to restrict $C$ to some compact subset of $\mathbb{R}.$ In this case, we know such a function is analytic and invertible over this domain. Whatever the solution is to $y = x - \frac{1}{C}x^{C}$ in the general case has a weird and probably unknown structure, but it still exists, so it might be possible to solve for $C$, then add a constant $b$ to $C$ to it that keeps it within the compact subset, then uninvert it. We don't know what such functions look like, but they should exist.

 

[fresh_42]

Say $g(c)(x):=x-\frac{1}{c}x^c$ and we define $(g_c\ast g_d)(x):=x-\frac{1}{cd}x^{cd}$. Then $g_c\cdot g_1 = g_c$ and $(g_c)^{-1}=g_{c^{-1}}$. I haven't checked associativity, but this could be isomorphic to the Lie group $(\mathbb{K}^*,\cdot).$ Another $f$, however, complicates the situation.

 

[LieToMe]

Nice, that's a good jumping off point, I'll try to play around with it. Assuming associativity, is it a problem that $C$ and $x$ are restricted to compact subsets? That's not going to interfere with any of the group axioms that need to be satisfied is it?

 

[fresh_42]

$x$ is irrelevant, as it is only some variable. If $c$ is from a compact set, then you get only a local group, since $1/c$ is needed for the inverse.

 

[LieToMe]

$x$ is irrelevant, as it is only some variable. If $c$ is from a compact set, then you get only a local group, since $1/c$ is needed for the inverse.

Sorry I think I meant compact and continuous. So you're saying if $C \in (0,\infty)$ then that would be enough for an entire group?

 

[fresh_42]

That would work, but it isn't compact. For compactness you could choose e.g. $[2/3\, , \,3/2]$, anything that contains $1$.

 

[LieToMe]

That would work, but it isn't compact. For compactness you could choose e.g. $[2/3\, , \,3/2]$, anything that contains $1$.

If all we need is continuity then losing compactness wouldn't be an issue right? Because $(0,\infty$) should contain every inverse element of every reciprocal in that interval $(0,1)$ where $1$ is the identity element instead of 0 under addition. Consequently, choosing $c = 1$ also produces the identity element in most known spaces of functions, being $f(x) - f(x) = 0.$

Actually you should even be able to differentiate with respect to $C$ which should imply continuity, shouldn't it?

 

[strangerep]

[...] $y = x - \frac{1}{C}x^{C}$ [...]

Well, I'm totally baffled why you don't simply apply the ordinary group axioms and see whether they hold. This should be easy (or am I missing something??)

E.g., consider: $$x'(c) ~:=~ x - \frac{x^c}{c} ~.$$ The first question is "what value of $c$ corresponds to the identity transformation?". Let's see... we want:
$$x'(c) ~=~ x - \frac{x^c}{c} ~=~ x ~,~~~ \Rightarrow~~cx - x^c ~=~ cx ~,~~~~ \Rightarrow~~ x^c = 0 ~.$$Not a good start, not good at all. Those transformations are restricted to $x=0$ if you demand that they form a group.

A little more generally, consider:
$$f'(x,c) ~:=~ f(x) - \frac{f(x^c)}{c} ~.$$ What value $c_0$ makes $f'(x,c_0) = f(x)$? Well,
$$f(x) - \frac{f(x^{c_0})}{c_0} ~=~ f(x),~ \Rightarrow~~c_0 f(x) - f(x^{c_0}) ~=~ c_0 f(x),~ \Rightarrow~ f(x^{c_0}) = 0.$$ That doesn't look much better.

If then you say: ok, maybe we don't need a full group. Maybe a semigroup would be enough? (A semigroup has the associative and transitive multiplication like a group, but does not necessarily have an identity element, and maybe not even inverses.)

In that case, you have to investigate transitivity. A 1-parameter Lie semigroup is necessarily commutative: meaning that, e.g., (denoting the transformation by $T$...),
$$T(c_1)\Big(T(c_2) f \Big)~:=~ T(c_2)\Big(T(c_1) f \Big)~~~~~ (1).$$For what values of $c_1,c_2$ in some open set in ${\mathbb R}$ (or even ${\mathbb R}$) does eq(1) above hold? [Exercise.]

 

[LieToMe]

If I'm not mistaken, the Lie group operation varies $c$, it is not itself the result of applying actions. The group operation maps transformations to other transformations, so the group is a subset of bounded linear operators, which makes sense because a lot of useful Lie groups tend to generalize linear transformations. If your group operation is multiplication in $c$, then the identity element is 1.

The group action on functions on the other hand is from the space of linear operators acting on a family of functions which sends every element in that set to another function in the family of functions. Do you at least agree there is continuity in $c$ for at least some subset of $\mathbb{R}$?

 

[strangerep]

The group action on functions on the other hand is from the space of linear operators acting on a family of functions which sends every element in that set to another function in the family of functions.

It does not need to be "from the space of linear operators". A group action on a function is not necessarily linear. (The group of fractional linear transformations is an example.)

Do you at least agree there is continuity in $c$ for at least some subset of $\mathbb{R}$?

Continuity in $c$ here is irrelevant if your set of transformations does not [non-trivially] form a group. (I already showed that it doesn't, so what's the point?)

 

[LieToMe]

It does not need to be "from the space of linear operators". A group action on a function is not necessarily linear.

I didn't say it was necessarily linear in the general case, only that many useful Lie groups like the special linear group generalize properties of linear operations. In this particular case we happen to have a set of operators that are linear when acting on functions.

Continuity in here is irrelevant if your set of transformations does not [non-trivially] form a group. (I already showed that it doesn't, so what's the point?)

But it's especially relevant to the concept of Lie groups as part of their definition. In the event the group operation forms a group between a set of mappings, we then need continuity in the operation. For the most part it seems we've already found local Lie groups, the only thing at this point is adjusting the range of the one parameter to generate a smooth manifold in $f$ while preserving the identity axiom, which isn't a guarantee if we choose the bounds incorrectly nor if we place insufficient restrictions on $f$.

For many functions it is likely the case that the best we can do is create local Lie groups, but for some functions there could still be a global Lie group.

 

[strangerep]

[...] it seems we've already found local Lie groups, [...]

I don't see that. For clarity, please (re-)define explicitly the "local Lie group" you think you have found.

 

[LieToMe]

I don't see that. For clarity, please (re-)define explicitly the "local Lie group" you think you have found.

For that you're welcome to check the discussion before your first comment in this thread to see the process of how we may already be arriving at a local Lie group given invertibility restrictions in $f$ and restrictions in $c$.

For that point you were trying to make about for all $c$-transformations acting on $f$ that it implies $f(x^c)$ is identically 0, that is merely a special case of $f$. Take $f(x) = \ln(x),$ it follows then $\ln(x) - \frac{1}{c}\ln(x^c) = \ln(x) - \ln(x) \equiv 0 \ \ \ \forall x \in (0, \infty), \ c \in (0, \infty)$.

 

[fresh_42]

For that you're welcome to check the discussion before your first comment in this thread to see the process of how we may already be arriving at a local Lie group given invertibility restrictions in $f$ and restrictions in $c$.

For that point you were trying to make about for all $c$-transformations acting on $f$ that it implies $f(x^c)$ is identically 0, that is merely a special case of $f$. Take $f(x) = \ln(x),$ it follows then $\ln(x) - \frac{1}{c}\ln(x^c) = \ln(x) - \ln(x) \equiv 0 \ \ \ \forall x \in (0, \infty), \ c \in (0, \infty)$.

If you mean the multiplication $(x-\frac{1}{c}x^c)\ast (x-\frac{1}{d}x^d)=(x-\frac{1}{cd}x^{cd})$ then this is isomorphic to $(\mathbb{K}^*,\cdot)$ which is a global one dimensional Lie group. This holds true even in case we only allow positive parameters. It's basically only a complicated way to write the multiplicative group of the scalar field, and presumably not very helpful.

 

[LieToMe]

If you mean the multiplication $(x-\frac{1}{c}x^c)\ast (x-\frac{1}{d}x^d)=(x-\frac{1}{cd}x^{cd})$ then this is isomorphic to $(\mathbb{K}^*,\cdot)$ which is a global one dimensional Lie group. This holds true even in case we only allow positive parameters. It's basically only a complicated way to write the multiplicative group of the scalar field, and presumably not very helpful.

Thank you for clarifying. For the simple case of $f(x) = x$ I can see what you're saying that it might not be helpful. What is interesting to consider however is how such a structure might be preserved in other functions or across a function space. For instance, what if one had $f(x) = c_0e^x$? If such a structure is preserved in some way then perhaps it has a relation to symmetries of an ODE.

Also, what is the proper name the group you're saying it is isomorphic to? When I look here https://en.wikipedia.org/wiki/Table_of_Lie_groups I don't see it.

 

[fresh_42]

$\mathbb{R}^+\, , \,\mathbb{R}^\times\, , \,\mathbb{C}^\times$

 

[LieToMe]

Thanks, that makes more sense. We know at least one function maps to the identity function of most function spaces, being the 0 function. Analytic functions unfortunately do not form a normable space in general, so to analyze this, perhaps what needs to be considered is $C^{1}[a,b]$ where $[a,b] \subset (0,\infty)$ which would then mean a local Lei group needs to be considered. Maybe there is a space of globally monotonic functions or invertible functions but I have not seen it. I suppose since I don't see a gaurantee of uniqueness there could be multiple functions that map to 0, but I thought there was a property of linear operators in spectral theory about uniqueness of the identity element.

 

[LieToMe]

Actually there was one property I forgot to check for linearity $G[a f(x)] = a f + \frac{1}{c}a f(x^c) = a G[f]$ where $a$ is the constant function $h(x) = a$.

With group-ness established, how can we then identify the corresponding Lie algebra and Lie derivative?