I have some conceptual doubts on the notion of a lie group, or perhaps on the notion of topological group. A lie group is manifold equipped with a continuous group structure of multiplication and inverse mapping. My understanding is that the points of a manifold posses a intrinsic existence independent of the coordinate chart used, and so this should also be true of the group structure. Now consider the unit circle. It's generally said the the unit circle is a lie group on the basis that it is the group of all complex numbers of absolute value 1, and such is a lie group under usual complex multiplication.

Now it seems to me that there should be conceptual distinction between the "unit circle" as a purely geometric or topological object and the group of unit complex numbers, and that the unit circle as such could not be considered a group.

My reasoning is the following. The relation between the unit complex and the unit circle is realized via the parameter $\theta$ in the complex exponential. But $\theta$ is a coordinate on the unit circle. So to define the multiplication structure on the circle via the multiplication of the corresponding complex numbers appears as structure that depends on coordinates.

To define the group structure first we must chose invertible maps $f(\theta) \longrightarrow \ M $ (where $M$ is the circle manifold), and $h(\theta) \longrightarrow e^{i\theta} $. And then define multiplication in the circle by

$$ p_1*p_2 = f\bigg(h^{-1}\big(e^{if^{-1}(p_1)}e^{if^{-1}(p_2)}\big)\bigg) \ \ ; p_1,p_2 \in M $$

Which of course is the point corresponding to the sum of the $\theta$'s. Now this definition seems to the depend on the particular coordinate chart $f$. By choosing differently I could change the point in $M$ associated with $\theta = 0$ for example, thereby changing the identity element. I began reading about Lie groups thinking that there are groups that are fortuitously continuous and fortuitously coordinalizable, but it is now seeming that the coordinates are the group.

I'm therefore being forced to conclude that the group structure is not intrinsic to the manifold. The operations $G \times G \longrightarrow G$ are not defined over the manifold points but over coordinate space. I do not know how to multiply elements without first assigning coordinates. The same type of consideration seems to apply to the torus and the plane, that are so symmetric that is impossible to distinguish for example a identity element in a intrinsic manner. This consideration should not be a problem if the manifolds in question are just coordinate spaces themselves, n-uples of number such as $\mathbb{R}^n$ or embedded surfaces on it. But if we are to think of manifolds abstractly it seems to be relevant. In the case of the unit circle and the plane (considered simply as geometric objects) it does not seem possible unless we arbitrarily fix some point to be the identity.

So my question is, can the group structure be defined intrinsically (independent of coordinates) such as other properties of manifolds, like it's tangent space structure for example, or some other topological properties? If so can someone explain how or point me towards some reference? If it is not possible is this fact at all relevant or is the conceptual difference that is bugging me superfluous?

I'm not a mathematician and this is my first attempt at seriously studying group theory so apologies if I committed some simple misunderstanding.

Thanks in advance.

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    $\begingroup$ "A lie group is manifold equipped with a continuous group structure of multiplication and inverse mapping. " No, actually not. It is a smooth manifold with a group structure such that the group operations are differentiable (see for example here). It is useful to consider first easy examples, e.g., $G=SL_n(\Bbb R)$. We can multiply matrices, and this is "intrinsic". $\endgroup$ Jul 4 at 8:56
  • $\begingroup$ The group structure of a Lie group is intrinsically defined as a smooth map $G\times G \to G$. Its coordinate representation, indeed, depends on the choice of coordinates, but this does not mean that the product depends on this choice. It is similar with the fact that smooth maps $f\colon G\to \Bbb R$ are well-defined independently of coordinates.\\ For the circle, you can simply consider it as the quotient $\Bbb R/ \Bbb Z$, and the product becomes $[x] + [y] = [x+y]$, which is independent of any representative. $\endgroup$
    – Didier
    Jul 4 at 9:37
  • $\begingroup$ I agree the two structures, manifold and group should at first be considered separately. Then you show that they satisfy the definition of a Lie group (i.e., multiplication and inverse are smooth maps) $\endgroup$
    – Deane
    Jul 4 at 13:49
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    $\begingroup$ You are right that the circle manifold $M$ does not have a canonical group structure. This is because every point is indistinguishable from every other, and you at least need to pick a point $e \in M$ to play the role of group identity. It turns out this is basically all the choice you have, and if someone hands you the pair $(M, e)$ there is really only one way of defining a multiplication on the circle $M$ making $e$ the identity. I think you’ve correctly identified that choosing an $e \in M$ (or similar) is required, and charts are convenient for this, but they are not required. $\endgroup$
    – Joppy
    Jul 4 at 14:13
  • $\begingroup$ @Joppy Surely the choice of identity is not sufficient to pin down the group structure on the circle. Given one choice of group structure I can transform it into any other using a diffeomorphism of the the circle to itself, And there are many such diffeos... $\endgroup$
    – Gooch
    Jul 4 at 18:48

2 Answers 2


A manifold is first of all a set. And many of our favourite examples of manifolds are subsets of some $\Bbb{R}^n$. But, just because you’re sitting inside as a subset of $\Bbb{R}^n$, it doesn’t mean you’re automatically “using coordinates”.

I think this is a case where you’re (incorrectly) persuaded too much by the “modern philosophy” of coordinate-freeness/independence where you think that just because something is defined as a subset of $\Bbb{R}^n$, that coordinates are automatically involved, so you look for more abstract ways of saying something equivalent, i.e it’s like you’re prohibiting yourself from doing anything all in fear of “using coordinates”, even when you’re not.

Now, you’re talking about the circle specifically, so ok we want the circle to be a manifold. But which one? In your post you mention “the circle manifold $M$”, but if you don’t have a definition for $M$, then you can’t even talk about providing it a (Lie-)group structure. A definition for such an $M$ could come in various forms: either a concrete definition as a set together with a smooth maximal atlas (smooth structure), and finally a definition of the group operation. Or you could give an abstract “existence and ‘uniqueness’” definition (for example via some sort of universal property, followed by an explicit example to show the result is not vacuous).

The fastest route is of course to first of all simply define $M_1=\{z\in\Bbb{C}\,:\,|z|=1\}$. This is just a set first of all; note that just because I defined it to be a subset of $\Bbb{C}$, it doesn’t mean I’m automatically using coordinates. By the usual procedures, one can then show that $M_1$ inherits a smooth structure such that it becomes an embedded submanifold of $\Bbb{C}$. Then, I can define the group operation $\mu:M_1\times M_1\to M_1$ as $(z,w)\mapsto zw$. Now, you may object here and say that I am using complex multiplication here which is somehow “against the manifold philosophy”. But not really. To define a map $M_1\times M_1\to M_1$ means given a pair $(z,w)$ in the domain, I have to tell you what the output is. It doesn’t matter how I come up with the output, I just need to tell you which element of $M_1$ I have to send $(z,w)$ to. It turns out that the complex number $zw$ belongs to $M_1$, so I am perfectly justified in considering the map $M_1\times M_1\to M_1$, $(z,w)\mapsto zw$. There are actually an infinite number of maps I can define, but the map $\mu$ has the nice property that it is smooth (when we equip $M_1\times M_1$ with the product smooth structure) and satisfies the axioms for a group. In proving this fact, you’ll of course invoke some facts about complex numbers, but that’s totally fine! You can use anything in math as long as you’ve proved the necessary facts about it before. You also raise the comment

My reasoning is the following. The relation between the unit complex and the unit circle is realized via the parameter $\theta$ in the complex exponential. But $\theta$ is a coordinate on the unit circle.

Well, sort of, but not really. The fact that $\theta$ is a local coordinate function on $M_1$ is an a-posteriori observation, but nowhere in my definitions did I ever invoke this particular local coordinate. That being said, coordinates are not a completely bad thing either; the definition of a manifold relies on it, and you can often “glue together” information you obtain in each coordinate chart, so even if you happen to use coordinates (note we did not in our discussion so far) as long as at the end of the day you get something well-defined, that’s perfectly meaningful.

Ok, maybe you’re still not convinced. So, let’s look at another definition: let’s take as our set, the quotient set $\Bbb{R}/\Bbb{Z}$, as mentioned in the comments. With some group theory, you’ll see that you can consider this as the quotient of the abelian group $\Bbb{R}$ by the (necessarily normal) subgroup $\Bbb{Z}$, so you get a quotient group $M_2=\Bbb{R}/\Bbb{Z}$. Set-theoretically, this is a (slightly more) complicated object, as its elements are certain equivalence classes. Now, if I want to consider this as a Lie group, then I have to define what the smooth structure is, and I have to show that the group operations are smooth. You can certainly do this, I won’t spell out the details here. So, $M_2$ with this particular group structure and the particular smooth structure constitutes a Lie group.

Note that at this stage, you can prove as an exercise that the Lie groups $M_1$ and $M_2$ are isomorphic (there is a diffeomorphism $f:M_1\to M_2$ which is also a group isomorphism). This means as far as Lie groups alone are concerned, $M_1$ and $M_2$ are “about the same as can be”.

If you don’t like that definition either, then here’s an even more abstract looking definition. Consider the set $M_3=\Bbb{RP}^1$. Actually even here there are several ways of defining this set theoretically, all of them being different ways of formalizing that it is the set of “all possible lines in the plane through a fixed point (typically taken to be the origin in $\Bbb{R}^2$)”. This is such a far cry from being a subset of some $\Bbb{R}^n$. Then, one has a natural way of defining a topology and smooth structure on this set $M_3$. With this, one can prove that $M_3$ as a smooth manifold is diffeomorphic to $M_1$ and to $M_2$. Using this diffeomorphism, one can of course “transport the group structure” and make $M_3=\Bbb{RP}^1$ into a Lie group, so that in the sense of Lie groups, $M_1,M_2,M_3$ are all isomorphic.

Another thing you can do is consider $SO(2)$. This has a standard smooth structure and Lie group structure. With this, you can prove it is isomorphic to $M_1,M_2,M_3$ above.

Finally, the most silly way of doing things is to fix a set $S$ with cardinality equal to that of $\Bbb{R}$. Then, fix a bijection $f:M_1\to S$ for example. Then, by “transport of structure”, you can make $S$ into a smooth manifold and in fact a Lie group such that $f$ becomes a Lie group isomorphism. In other words, I can, if I were pathological, force the fat Cantor set to become a smooth Lie group, which is isomorphic to $S^1$. If I were to do things differently, I could force the fat cantor set to be isomorphic to any given finite-dimensional Lie group $G$. It’s just that this is a somewhat artificial and silly thing to do.

The point I’m trying to illustrate is that you should start with one definition. Once you have that definition, you can contemplate whether or not some other Lie group which may have been defined differently, is actually isomorphic to the one you started with.

Some tangential remarks.

One comment you made is true though (but based on what you write afterwards, I’m guessing you meant it in a slightly different way)

I'm therefore being forced to conclude that the group structure is not intrinsic to the manifold.

Correct, the group structure is not intrinsic to the manifold. It is an extra piece of information. It is not unique either: on a given manifold I can define two different group laws, and in fact I can define them in such a way that the resulting Lie groups are non-isomorphic.

Also, you write

So my question is, can the group structure be defined intrinsically (independent of coordinates) such as other properties of manifolds, like it's tangent space structure for example,…

Well, this isn’t quite right. The tangent spaces to a manifold don’t “come out of nowhere”. Yes, set theoretically, I can define them in a number of ways (derivations, equivalence classes of smooth curves, etc) but at some point you’re going to have to use a chart in order to relate this to the vector space structure of $\Bbb{R}^n$. For example, if you go the derivations route, then it may seem like you don’t use coordinates at all, and fair enough, but you don’t even know the dimension of the resulting vector space you constructed; it could be $0$ or even infinite-dimensional. If you go the equivalence-classes route, it’s not even clear how to define the vector space structure without using a chart and then showing chart-independence.

  • $\begingroup$ Thank you for taking the time to write such a thoughtful answer. I think I agree with all your points and your answer has been very clarifying. What I had in mind was indeed something along the lines of what is said in your third paragraph. Some characterization theorem for an abstract object that would essentially be the euclidean circle. If think my question comes down to if on the euclidean circle (plane or torus) you could define a group structure without first fixing arbitrarily a point as the identity. Calling this choice a coordinate was to much, but the answer still appears to be no. $\endgroup$ Jul 5 at 5:40
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    $\begingroup$ @ErrorPropagator One could invoke the characterization of 1-dimensional smooth manifolds: up to diffeomorphism, the only connected 1-dimensional smooth manifolds (without boundary) are $\Bbb{R}$ and $S^1$, so you could (but I’d suggest not) define $S^1$ to be the unique (up to diffeomorphism) compact connected smooth 1-dimensional manifold. And no, we obviously can’t define a group structure without first fixing what the identity is. But for $S^1$ in particular, it seems like any two topological group structures are isomorphic. $\endgroup$
    – peek-a-boo
    Jul 5 at 6:21
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    $\begingroup$ but these are pretty non-trivial (ok not crazy difficult, but still takes a bit of effort to prove) classification results, so I’d rather stay away from them for the elementary purpose of merely defining $S^1$ as a Lie group, for which the standard way is geometric, intuitive, and also simple. $\endgroup$
    – peek-a-boo
    Jul 5 at 6:23

Maps between manifolds can be considered abstractly, but if you want to specify a map you usually give it in coordinates. You say you do not know how to define multiplication without first calculating in coordinates, but this is the case for most maps of manifolds. This does not preclude them from being coordinate-independent.


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