Is that the definition of a lie group?

[davi2686]

I learned a lie group is a group which satisfied all the conditions of a diferentiable manifold. that is the real rigour definition or just a simplified one?

thanks

 

[wrobel]

... and the group's multiplication respects the differential structure

 

[zinq]

Suppose the Lie group is on the manifold M. Then not just the multiplication: (g, h) → gh

mult: M × M → M​

but also the operation of taking an element to its inverse: g → g^-1

inv: M → M​

are both differentiable.

And note that saying a mapping is differentiable is the normal way to express that the mapping "respects the differentiable structure" of a manifold. In the first case, this refers to the differentiable structure of the product manifold M × M.

In fact, every Lie group's operations mult and inv are not only differentiable, but infinitely differentiable, which means that in local coordinates, every partial derivative of any order, including all mixed partials, is itself differentiable with respect to all coordinates. (Even more can be said: mult and inv are even real analytic, meaning locally expressible in terms of convergent power series.)

 

[davi2686]

Thanks very much Wrobel and Zinq.

So let me see if i understand,

(1) a differentiable structure is a map of a set into itself where this map is k-differentiable,

(2) this implies that set is also a manifold (if k=infinite a differentiable manifold)

(3) and if that set also a group, so we have a lie group.

the three satatements above are right?

 

[zinq]

To answer your questions:

(1) A differentiable structure (DS) is something different. It is possible to have a DS only on a topological space that is a topological manifold (usually defined as a Hausdorff space such that every point has a neighborhood that is topologically equivalent to an open set in Euclidean space Rⁿ (where n is the same for the whole space).

Given that M is a topological manifold M as above, a differentiable structure is designed so that differentiability makes sense on this manifold. Here is what it is, precisely: It is a collection of nonempty open sets V_j of M such that their union ∪_j V_j is equal to all of M, and for each index j there is a mapping

h_j: Vj → Rⁿ

that is a homeomorphism onto its image, such that for any V_i and V_j with a nonempty intersection V_i ∩ V_j, the mapping

h_ij = h_i o h_j ^-1: h_j(V_i ∩ V_j) → h_i(V_i ∩ V_j)

(as a mapping from part of Rⁿ to part of Rⁿ) is differentiable.

A manifold together with a differentiable structure on it is a differentiable manifold.

This may seem complicated, but it is part of understanding what a Lie group is. And when you are familiar with it, it will seem simple. For more information, see these notes: http://www2.math.uu.se/~khf/Otter.pdf.


(2) As you can see from (1), yes, a differentiable structure can only be on a manifold. But: If all the maps h_ij in (1) are C^r (r times differentiable) then M is called a C_r manifold. A C_r manifold is automatically a C_k manifold for all k ≤ r. A differentiable manifold is just a C₁ manifold. (But it might be a C^r manifold for r > 1.) A smooth manifold means it is a C^r manifold for all r ≥ 1, or in other words all the maps h_ij are infinitely differentiable (C^∞). All Lie groups are in fact C^∞ by a very deep theorem due to Deane Montgomery, Andrew Gleason and others, completed about 1952. (Not that you asked, but something even stronger is true: All Lie groups are real analytic (C^ω) manifolds, which means that all the maps h_ij can be expressed by power series with real coefficients.)

(3) Yes, sort of. If a differentiable manifold has a group structure on it and the group operations of multiplication and inverse on it are continuous in its topology, then it is a Lie group. The casual phrasing of your (3) is not precise enough to include this possibility.
——————————————————————————————

To elaborate on that last sentence: Consider the circle (curve) as a topological space. It is the same cardinality as the real line. This mean that using a bijection of the circle to the real line, you could put the additive structure of the reals (as the Lie group ℝ) on the circle if you wanted to. This group structure would not make the circle into a Lie group, since its multiplication and inverse mappings would be discontinuous on the circle as a topological space.

 

[micromass]

It is possible to have a DS only on a topological space that is a topological manifold

Uh, that is not necessarily true. But it is true that things are well-behaved only on topological manifolds. A differentiable structure can be put on very general spaces however. https://arxiv.org/abs/0807.1704

 

[zinq]

Thanks, micromass. You are right. (I actually knew that but was trying to keep things simple.)

 

[fresh_42]

Uh, that is not necessarily true. But it is true that things are well-behaved only on topological manifolds. A differentiable structure can be put on very general spaces however. https://arxiv.org/abs/0807.1704

Mathematicians have a strange kind of humor: "Convenient Category". Considering axiom 3 of Chen-spaces and the following examples I'm curious to see what will be left under so much generalization.

Btw.: Couldn't we just say: "A Lie Group is a topological group with analytic group operations."?

 

[micromass]

Mathematicians have a strange kind of humor: "Convenient Category". Considering axiom 3 of Chen-spaces and the following examples I'm curious to see what will be left under so much generalization.

Surprisingly much actually...

 

[zinq]

"Btw.: Couldn't we just say: 'A Lie Group is a topological group with analytic group operations.' ?"

Yes, we could. But mathematicians (like me) usually prefer to define something initially with the fewest necessary conditions, and then state the interesting and worthwhile consequences of that definition as theorems — it is more informative that way.

Then you know that anytime you encounter the thing with its more spartan definition, you will have the thing with all its nice consequences. (It might be more difficult to know, initially, that you have something with all those nice consequences to begin with.)

 

[fresh_42]

But mathematicians (like me) usually prefer ...

Well, someone should tell him.

 

[zinq]

Please feel free to elaborate on that comment, fresh_42.

 

[davi2686]

Thanks very much to all, especially Zinq. I am think its time to read carefully your last answer to me and read the notes you had linked before playing thoughtless questions. I am really appreciate your help it helped me a LOT, thanks.
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