Rings, Modules and the Lie Bracket

  • #1
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I have been reading about Rings and Modules. I am trying reconcile my understanding with Lie groups.

Let G be a Matrix Lie group. The group acts on itself by left multiplication, i.e,

Lgh = gh where g,h ∈ G

Which corresponds to a translation by g.

Is this an example of a module over a ring where g ∈ R and h ∈ M and the scalar multiplication is interpreted as a linear map (matrix multiplication)?

If so, how does one interpret the Lie bracket [x,y] in terms rings and modules (or can one)?
 

Answers and Replies

  • #2
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11,691
I have been reading about Rings and Modules. I am trying reconcile my understanding with Lie groups.

Let G be a Matrix Lie group. The group acts on itself by left multiplication, i.e,

Lgh = gh where g,h ∈ G

Which corresponds to a translation by g.

Is this an example of a module over a ring where g ∈ R and h ∈ M and the scalar multiplication is interpreted as a linear map (matrix multiplication)?
Not really. A Lie group has in general only one operation, a ring and a module have two. In order to make a Lie group a ring and a module, one has to consider Abelian structures. This is possible, but a rather boring example of a Lie group. I'm not quite sure which conditions have necessarily to hold in order to press a Lie group into a ring, will say I haven't checked whether the examples I have in mind are necessary or only sufficient. E.g. the additive structures of ##\mathbb{C}^n## or ##\mathfrak{gl}(n)## are Lie groups and also rings, if we define ##xy=0## on ##\mathbb{C}^n## and matrix multiplication on ##\mathfrak{gl}(n)##. But their analytic structure is trivial, because it's linear. So the answer is: Making enough parts of the definitions hold trivially, then yes.
If so, how does one interpret the Lie bracket [x,y] in terms rings and modules (or can one)?
I'd first like to see how your ring and module structures are formally defined, i.e. what are the additions and scalar products in your example?

The usual way is a different one.

Start with a complex, finite dimensional associative algebra, define ##[X,Y]=XY-YX## and get a complex, finite dimensional Lie algebra. Then this Lie algebra can be written as a subalgebra of ##\mathfrak{gl}_n(\mathbb{C})## by Ado's theorem. It has thus a corresponding Lie group. But this group is in general neither a ring nor a module, unless you define all structures to be trivial: Abelian, trivial operation etc.

The other way is to consider a Lie group, which again is in general neither a ring nor a module, build it's tangent space and get it's Lie algebra.

What you do have is an operation of the Lie group ##G## on it's Lie algebra ##\mathfrak{g}## via conjugation: ##X \longmapsto gXg^{-1}## and ##\mathfrak{g}## is a non associative ring as well as a ##G-##module. If you require ##G=\mathfrak{g}## not much of a typical Lie structure would be left.
 
  • #3
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OK. I forgot about the requirement for an Abelian group structure - a matrix ring operating on module elements that are vectors would be fine because the vectors form an Abelian group. I think I may also be confusing 'generic' vector spaces associated with abstract algebra with tangent spaces associated with differential geometry. Is it fair to interpret the latter as a specific type of vector space whose single algebra operation is the Lie bracket?
 
  • #4
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OK. I forgot about the requirement for an Abelian group structure - a matrix ring operating on module elements that are vectors would be fine because the vectors form an Abelian group.
This sounds a bit confused in my opinion. You can have a matrix ring, but to be a group at the same time, you must either choose it's additive group which normally is Abelian as the one which determines the Lie structure, or its units. It depends on the case to give a statement about the latter. Matrix rings have often only diagonal matrices as units, or even multiples of ##\operatorname{id}_V##, which both are again Abelian and their Lie structure is quite simple. So to press a matrix ring into the structure of a Lie group requires a lot of restrictions. Again, without being told an example, I have only the examples I already mentioned in mind. Of course I could try and elaborate those restrictions, but it's your idea, not mine. I don't expect anything non trivial here.
I think I may also be confusing 'generic' vector spaces associated with abstract algebra with tangent spaces associated with differential geometry.
Whatever generic should be. An algebra is a vector space with a multiplication, associative or not.
We get Lie algebras as tangent spaces of Lie groups. Neither of them needs to be represented by matrices, but often they are.
If we have any associative algebra, e.g. a matrix algebra, then ##[X,Y] := XY-YX## defines a Lie algebra. The theorem of Ado says, that all complex and finite dimensional Lie algebras can be obtained in this way, i.e. isomorphic copies.
Is it fair to interpret the latter as a specific type of vector space whose single algebra operation is the Lie bracket?
In a way, yes. Tangent spaces are of course vector spaces, by construction. However, I don't see that arbitrary tangent spaces carry an algebra or Lie algebra structure except a trivial one. But if we have
  1. a topology on the object we defined the tangent space for,
  2. and an analytic structure, too, - we must differentiate for the tangents anyway -
  3. and a group structure, too,
  4. and both multiplication
  5. and inversion are smooth,
if all this is given, then we have a Lie group and a Lie algebra structure on the tangent space. The latter means that ##[X,X]=0## and ##[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0##. This product is usually called a commutator, whether it is defined by ##[X,Y]=XY-YX## or not.
Now matrix groups are a suitable example, as they can easily bring all these requirements with them. And in this case, ##[X,Y]## turns out to be ##XY-YX##.
 
  • #5
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Thank you for your reply (and patience!).

My use of matrix ring may be incorrect. I wanted it to mean that the elements of the ring are matrices and these matrices act on vectors that are elements of the module. I don't know if that would effect your answer.

Generic means a space equipped with the operations of vector addition and scalar multiplication.

The tangent space being referred to is associated with a smooth C manifold which is also a Lie group. The tangent vectors at each point are equal to the value of the vector field at the points along the integral curve. Each integral curve is a one parameter subgroup that is 'connected' to the algebra via the exponential map. Again, I don't know if this modifies your answer.

Incidentally, the thing that precipitated this discussion is definition 19 here: https://math.berkeley.edu/~reb/courses/261/4.pdf

Thanks again.
 
  • #6
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Thank you for your reply (and patience!).

My use of matrix ring may be incorrect. I wanted it to mean that the elements of the ring are matrices and these matrices act on vectors that are elements of the module. I don't know if that would effect your answer.
No. It's a perfect example of a ring and a module. It just has little to nothing to do with a Lie group and its tangent space.

If you start with a Lie group, e.g. ##G=\operatorname{GL}(V)##, then ##\mathfrak{g}=\mathfrak{gl}(V)## is a Lie algebra, the latter not the former can as in this example incidentally also be an associative matrix ring, operating on ##V##. So the group acts on ##V## as well as the tangent space, better vector fields. It is just not the same operation, although both are a matrix applied to a vector, but it doesn't make ##G## and ##\mathfrak{g}## equal in any way. As you said, you have ##(L_g(h))(v)=(gh)(v)=g(h(v))##. But for the vector fields we have ##[X,Y](v)=X(Y(v))-Y(X(v))##. The difference is, that ##G## has multiplicative inverses and ##\mathfrak{g}## has none, and ##\mathfrak{g}## is closed under addition whereas ##G## is not.
Your original question has been:
L_gh = gh where g,h ∈ G

Which corresponds to a translation by g.

Is this an example of a module over a ring where g ∈ R and h ∈ M
And the answer is no, simply because a group isn't a ring. Your elements ##g,h## were both in the group, so ##g\in R## or ##h\in M## doesn't make sense, because ##G\neq R\,(*)## and ##G\neq M\,(**)##. I only discussed the implications of such requirements ##(*),(**)## for that was what you asked, and I saw only trivial examples to make a Lie group ##G## equal to a ring ##R## or a module ##M##.
Generic means a space equipped with the operations of vector addition and scalar multiplication.
... which is a vector space. Generic isn't needed to refer to an arbitrary vector space. It suggests an additional meaning which isn't there.
The tangent space being referred to is associated with a smooth C manifold which is also a Lie group. The tangent vectors at each point are equal to the value of the vector field at the points along the integral curve. Each integral curve is a one parameter subgroup that is 'connected' to the algebra via the exponential map. Again, I don't know if this modifies your answer.
No. You just explained what the tangent space of a Lie group is, which carries a Lie algebra structure. That doesn't make a group a ring.
Incidentally, the thing that precipitated this discussion is definition 19 here: https://math.berkeley.edu/~reb/courses/261/4.pdf
The facts you stated are correct. But you've written in post #1 ##g,h \in G \ldots\ldots\ldots g\in R\, , \,h\in M##. It is this situation which I discussed and answered to, not the concept of tangents to integral curves.

Physicists often don't distinguish between the group and its tangent space, call all operations representations, which they are, but different ones, throw in some generators at best, which means they changed to the tangent space, and mix all this wildly.
 
  • #7
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OK. I recognize the fact there is some 'backwards and forwards' between the 2 disciplines. I am a retired EE so am somewhat impartial. So the bottom line is that In the first case, forget about rings and modules since g, h ∈ G is not compatible with g ∈ G and h ∈ M. This makes perfect sense now. In the second case, the tangent space carries the structure of the Lie algebra. It is a vector space but one that differs from the abstract definition, Is this a good place to stop?
 
  • #8
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OK. I recognize the fact there is some 'backwards and forwards' between the 2 disciplines. I am a retired EE so am somewhat impartial. So the bottom line is that In the first case, forget about rings and modules since g, h ∈ G is not compatible with g ∈ G and h ∈ M. This makes perfect sense now. In the second case, the tangent space carries the structure of the Lie algebra. It is a vector space but one that differs from the abstract definition, Is this a good place to stop?
I don't know how it should differ from the abstract definition. If we don't specify the group and only take the requirements (smooth structure), then the resulting tangent space, which is a Lie algebra, will also be quite abstract, namely left-invariant vector fields. I've calculated an example here:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/
and some calculations for ##\operatorname{SU}(2)## here:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/

I'm not sure whether this is interesting for you, but it fits into the context.

To close the circle on rings and modules, this usually refers to commutative and associative rings, which Lie algebras are not. And a ring doesn't necessarily have a scalar multiplication. Modules are basically vector spaces, where the scalars are taken from a ring like ##\mathbb{Z}## or ##\mathbb{M}_n(\mathbb{R})## instead of a field like ##\mathbb{R}## or ##\mathbb{C}##. The branch of mathematics which deals with them is called commutative algebra or ring theory. A matrix ring and the vector space those matrices apply to is indeed an example of ring and module; although not a Lie group.
 
  • #9
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Maybe abstract is the wrong term to use. The tangent space has the structure of the Lie algebra which seems more elaborate than just vector addition and scalar multiplication. That is the source of my confusion. I understand the relevance of the LIVF but I can't connect the dots between this and the definition if vector spaces that one reads about in basic texts. Am I making any sense?
 
  • #10
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It is basically this way:

We have something, a Lie group ##G##, which has a tangent space ##\mathfrak{g}##. The former is at first just a differentiable manifold and as such has a tangent space. The latter is at first only a vector space. But a Lie group is more than an object with integral curves that can be differentiated to get tangents. It also has a multiplication and an inversion which can be differentiated. These additional structures on what originally was only a differentiable manifold give rise to an additional structure of the tangent space, namely a multiplication of tangent vectors. These live in ##\mathfrak{g}##, so it is more than just a vector space: it has a multiplication ##(X,Y) \mapsto [X,Y]## with the properties of anti-commutativity and Jacobi identity. All together it is called a Lie algebra.

One could as well start with a Lie algebra and simply define: A Lie algebra ##\mathfrak{g}## is a vector space with a multiplication which is anti-commutative and satisfies the Jacobi identity. In this approach no Lie group is necessary. But if we have a Lie group, the tangent space gets a Lie algebra. We cannot multiply vectors in general, but in case of a tangent space to a Lie group we can. The connection between Lie group and Lie algebra is even deeper, since all matrix Lie algebras are also a tangent space to some Lie group, so the correspondence is quite close, esp. as Ado says that all complex, finite dimensional Lie algebras are indeed matrix Lie algebras.

The reason to consider Lie algebras instead of Lie groups is the same as always with differentials: they are easier to calculate and locally a good approximation to the object which they are tangents of, and the correspondence transports information from one to the other. The details are of course a bit more complicated, but in general this is the principle behind. In the second link I gave, there is an example of an easy Lie group which isn't one of the standard matrix groups.

Another example for the other way around is here in problem #9:
https://www.physicsforums.com/threads/intermediate-math-challenge-may-2018.946386/
There is a Lie algebra ##\mathfrak{A(g)}## defined which is a matrix Lie algebra and thus is the tangent space of some Lie group, but I have no idea how this Lie group looks like in general. This shows that different approaches are possible, coming from the differentiable manifold with a group structure, or from the Lie algebra side without any group near.
 
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  • #11
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Thank you. I think this helps a lot. I realize now that modules and rings were somewhat of a red herring. However, I learned something that I didn't know before. Again, thanks for your patience and time.
 

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