# Before we define the group operation and label the elements, how are continuous Lie groups distinguishable?

Mandatory disclaimer: I'm not a mathematician I'm a physics student.

I am learning about Lie groups and have hit a bit of a wall here. I'm going to specifically be talking about the Lie groups I'm required to know about, $$U(1)$$, $$SO(2)$$, $$SO(3)$$ and $$SU(2)$$, mostly because I know anything I say will have countless counterexamples and at the moment I'm not worried about anything outside of this scope.

So to my understanding these Lie groups can be defined as a 4-tuple, $$(G,\cdot,\tau,\mathscr{A})$$, a continuous set of elements, $$G$$, the group operation, $$\cdot$$, with all its axioms, and a topology/atlas to give it manifold structure. Now at this point, it seems like all of the Lie groups listed above are completely indistinguishable, we haven't labelled any of the elements of the set, we haven't defined the group operation and we haven't specified the composition of the topology/atlas. My first question is whether this is true, that at the purely abstract level these groups are indistinguishable right now.

I then understand that we define a representation of these groups to be a map from the elements of the Lie group itself to some subset of the general linear group:

$$\pi:G\rightarrow GL(n;\Bbb{C}).$$

It is only at this point that the Lie groups are associated with matrices, this is usually the point at which physics texts pick up the idea of Lie groups, without too much reference to this "representation map", I assume this is why Lie groups are referred to as "the matrix groups" sometimes in physics texts.

It makes sense to me that at this point we can distinguish between the Lie groups, because we make the requirement that the matrices representing, say, $$SU(2)$$ are unitary and defined over the complex numbers, and the matrices representing $$SO(2)$$ are orthogonal and defined over the real numbers. I understand that there are multiple different representations for each group, exactly what they are isn't something I'm too concerned about right now. My second question would then be, is the reason that all Lie group representations are not isomorphic to each other because there does not exist a mapping that preserves the group operation? It seems like since these Lie groups all contain a continuous set of elements there could exist a 1-to-1 mapping between them. If this second question requires a significant amount of mathematics it will probably be lost on me but I thought I'd ask.

Any help appreciated.

• Groups (in the usual sense) are a pair $(G, \cdot)$ of a set $G$ and an operation $\cdot$. Sometimes we can already distinguish groups because the sets $G$ might be different sizes, but if the sets are the same size we will have to use data about the operation $\cdot$ to distinguish groups. An example is the two groups with four elements: only one of them has an element of order 4. Apr 29, 2020 at 13:08
• For a Lie group, we can use the manifold information as well to distinguish groups. Lie groups have a dimension (as a real manifold), and isomorphic Lie groups must in particular be isomorphic manifolds, so the dimension of the manifold could be used to distinguish them. Or other topological properties, for example we can distinguish the circle group from the group $(\mathbb{R}, +)$, both Lie groups of dimension 1, because one is compact and one is not. If you have two Lie groups which are isomorphic as manifolds, it might be quite tricky to tell them apart. Apr 29, 2020 at 13:10
• By the way, you can get around having to label the elements by considering what it means to be an isomorphism of Lie groups. An isomorphism $(G, \cdot) \to (H, \circ)$ of groups is a bijection $f \colon G \to H$ such that $f(g_1 \cdot g_2) = f(g_1) \circ f(g_2)$ for all $g_1, g_2 \in G$, and we say groups are isomorphic if there is an isomorphism between them. This allows us to say which groups are isomorphic without worrying about labels. The only difference in the definition for Lie groups is that $f$ must be an isomorphism of manifolds, meaning a smooth map, with smooth inverse. Apr 29, 2020 at 13:22

To answer your first question, it is entirely inaccurate to say that $$U(1)$$, $$SO(2)$$, $$SO(3)$$ and $$SU(2)$$ are "completely indistinguishable", that "we haven't labelled any elements of the set", that "we haven't defined the group operation", and that "we haven't specified the composition of the topology/atlas".

For example, with regard to $$SO(3)$$:

• Its elements are the $$3 \times 3$$ matrices $$M$$ with real number entries such that $$M M^{T}$$ is the identity matrix and such that the determinant of $$M$$ equals $$1$$.
• Its group operation is matrix multiplication.
• Its topology is the subspace topology defined by the obvious embedding $$M \to \mathbb R^9$$, under which the matrix
• $$M = \begin{pmatrix} M_{1,1} & M_{1,2} & M_{1,3} \\ M_{2,1} & M_{2,2} & M_{2,3} \\ M_{3,1} & M_{3,2} & M_{3,3}\end{pmatrix}$$ is mapped to the $$9$$ tuple $$(M_{1,1} , M_{1,2} , M_{1,3} , M_{2,1} , M_{2,2} , M_{2,3} , M_{3,1} , M_{3,2} , M_{3,3}) \in \mathbb R_9$$

I could go on about its atlas as well, which is more technical, but I think I've made my point already.

Now, to say a little more about "distinguishability", the key issue to a mathematician is the concept of "isomorphism". One could ask: does there exist an isomorphism $$f : U(1) \to SO(2)$$? If so, then one would say that the Lie groups $$U(1)$$ and $$SO(2)$$ are "isomorphic".

But to a mathematician, before this can even make sense, the concept of isomorphism must be defined: given two Lie groups $$G,H$$, an isomorphism is a smooth bijection $$f : G \to H$$ such that $$f(gg') = f(g) f(g')$$ for all $$g,g' \in G$$.

As it turns out, $$U(1)$$ and $$SO(2)$$ are indeed isomorphic. The proof requires one to write down a formula for an isomorphism $$f : U(1) \to SO(2)$$. Each element of $$U(1)$$ is a $$1 \times 1$$ matrix $$(z)$$ consisting of a complex number such that $$|z|=1$$. Letting $$z = x+iy$$, we define $$f(z) = \begin{pmatrix} x & y \\ -y & x \end{pmatrix}$$ Now there's some work to do, i.e. that $$f$$ is a smooth bijection and that $$f(zw) = f(z) f(w)$$, but that can be done and in the end one has proved that $$U(1)$$ and $$SO(2)$$ are isomorphic. Perhaps one might take this as evidence that these two Lie groups are "indistinguishable", although to a mathematician that's terminology to avoid; from a formal point of view I would stick with saying they are "isomorphic".

But on the other hand, it turns out that $$SO(2)$$ and $$SO(3)$$ are not isomorphic. This requires one to prove a negative: there does not exist an isomorphism $$f : SO(2) \to SO(3)$$. The proof is an argument by contradiction, based upon a theorem of differential topology: if two Lie groups are isomorphic then they have the same dimension; more generally, if two smooth manifolds are diffemorphic then they have the same dimension. Now one calculates dimensions: the dimension of $$SO(2)$$ equals $$1$$ and the dimension of $$SO(3)$$ equals $$3$$. So they aren't isomorphic.

• I guess my follow up question would be at what point does the concept of a Lie group fork off into the different Lie groups? Just to illustrate what I mean, if I make myself a group out of a set $G$ containing 3 elements and an operation $\cdot$, $(G,\cdot)$, right now these elements and the operation could be anything, they are just "3 elements that form a group under $\cdot$", before I decide exactly what the elements and the operation are, all possible choices are "indistinguishable" because I haven't labelled/defined anything. Apr 29, 2020 at 18:23
• Well, I may not be from Missouri, but, I would say: show me the 3 elements, and show me the operation; until you've done that, you don't have a group. Apr 29, 2020 at 18:25
• Maybe that's where my problem is, so the object that is just a set, an operation, a topology and an atlas is not "a Lie group" until we've actually defined what the objects and operations are? Apr 29, 2020 at 18:28
• I was under the impression that you could "strip" off the labels and definitions to get a kind of template for an algebraic object like a group. Which is why it made me uncomfortable that we define Lie groups to be matrix groups, because I wanted to think of the underlying object as being an unlabelled collection of objects (set, operation, topology etc.) that we then assign meaning to by choosing the elements of the set etc. to be something concrete. Apr 29, 2020 at 18:31
• You might be thinking about the algebraic geometry concept of a "group scheme"? For example, $SL_n$ is a template which you can apply to any ring $R$ and you get a group $SL_n(R)$... but for this example, or with any group scheme, once you've applied it to a ring $R$ the resulting object $SL_n(R)$ is an actual concrete group with a concrete operation: $n \times n$ matrices with entries in $R$, with operation being matrix multiplication. Apr 29, 2020 at 22:26

How are these Lie groups distinguishable? One of the standard methods is to consider the associated Lie algebra, i.e., an associated vector space with a Lie bracket. Then it is mostly linear algebra to show that the Lie algebras are not isomorphic, and by general theory, so aren't the Lie groups you have written down. For example, often their dimension is already different (which coincides with the vector space dimension of the Lie algebra). Similar remarks concern the representations.

• So a Lie group cannot be defined without a Lie algebra? Apr 29, 2020 at 12:48
• Of course it can. What I mean is that we can easily distinguish your given matrix groups by linear algebra. Apr 29, 2020 at 13:00
• Is that not only true once we've chosen a representation? Before we've chosen to associate the elements of the Lie group with a set of matrices is there any way to distinguish between them if it is possible to define the Lie groups without their algebras? Apr 29, 2020 at 13:03
• Again, a group like $SO(3)$ is already defined by matrices, namely those $3\times 3$-matrices $A$ which satisfy $AA^I=I$ and $\det(A)=1$. No need to chose a representation. It's dimension is $3$, so it cannot be isomorphic to $SO(2)$. Apr 29, 2020 at 13:06