Lie Algebras & Relativity: Usefulness & Purpose

[kent davidge]

What's the usefulness of a Lie Algebra? As I see on Wikipedia a Lie Algebra is a vector space with an operation on it called Lie Bracket.. This seems to be the formal definition.

In Relativity, we have the Lie Algebra of the Lorentz group, the Lie Algebra of the Poincaré Group, which are those familiar commutation relations... I also have read about the Lie Algebra of the metric.. and so on..

But in pratice, in what a Lie Algebra help us? What's the purpose of knowing the Lie Algebra of a symmetry group?

 

[fresh_42]

What's the usefulness of a Lie Algebra? As I see on Wikipedia a Lie Algebra is a vector space with an operation on it called Lie Bracket.. This seems to be the formal definition.

In Relativity, we have the Lie Algebra of the Lorentz group, the Lie Algebra of the Poincaré Group, which are those familiar commutation relations... I also have read about the Lie Algebra of the metric.. and so on..

But in pratice, in what a Lie Algebra help us? What's the purpose of knowing the Lie Algebra of a symmetry group?

Yes, you basically included all key words. Very roughly it is:

physical invariant → Noether's theorem → symmetries → group generated by symmetries → Riemannian manifold → tangent space → Lie algebra

The symmetry groups in physics are (all?) Lie groups and their tangent spaces are the Lie algebras. It is a very small collection of as well possible groups as possible Lie algebras. At the end of this chain above, the plan is to relate the behavior of Lie algebras to the invariants we started with. E.g. in string theory, this chain is simply prolonged with $\mathbb{Z}_2$ graded Lie algebras. Other ideas of unification theories simply enlarged the collection above by some more, but still classical, semisimple Lie algebras.

 

[kent davidge]

and their tangent spaces are the Lie algebras

In what could it be useful to bring the symmetries down to a vector space? I mean, when we talk about a Lie Algebra of a Lie Group we are regarding the group generators as vectors on a vector space. Why would we want this?

 

[fresh_42]

In what could it be useful to bring the symmetries down to a vector space? I mean, when we talk about a Lie Algebra of a Lie Group we are regarding the group generators as vectors on a vector space. Why would we want this?

Mainly because their representations, i.e. the way they act on other vector spaces, resp. what they do to vectors we are interested in, are closely related via the exponential function. In addition, we do what is always done in physics: study complex behavior by (local) linear approximations. Lie algebras are easier to handle than groups, and many properties translate directly: center, simplicity, representations, normal subgroups and ideals etc. A finding in the theory of Lie algebras has good chances to have a mirroring property for groups. I don't know whether this is true, because I'm no physicist, but today I've read, that the coupling constants in S-matrices are the structure constants of Lie algebras. If physicists don't use the same word for different objects (which they sometimes do), then the structure constants of Lie algebras are simply the factors in their multiplication tables. It is a bit as if you asked: Why should I regard $t \mapsto v(t)=\dot{x}(t)$ if I'm only interested in position $x(t)$? Lie groups have the property, that their tangent space at the identity element can be transported to any other point (in its connection component of course) and still is the same vector space. So whatever you find out about the vector space, applies locally to all points of the group. Furthermore is Noether's theorem a statement about differentials, so looking at tangent spaces is a natural thing to do.

 

[PeterDonis]

the structure constants of Lie algebras are simply the factors in their multiplication tables

Not quite. They are the constants that appear in the commutation relations of the generators of the Lie algebra.

 

[fresh_42]

As you've mentioned it, I'd like to post a question in this realm, which by itself might be a bit too speculative. However, I think it has to be allowed. The standard model is based on $SU(n)$, which are simple groups. I once asked the question here, why physicists are mainly interested in (semi-)simple groups and their Lie algebras. The best answer I received was, because the Killing form is non-degenerated and can be used as a measure for angles and lengths, which is widely used in the classification theory of semisimple Lie algebras.

Now you also mentioned the Poincaré group, which is not semisimple, and I would like to add the Heisenberg group(s). I'm not sure about the Lorentz group, as it is rather close to one: $SO(3,1)$ instead of $SO(4)$, however not exactly. So the question which naturally arises is: Is it possible, that our difficulties to find a unification theory has to do with the fact, that those groups are not semisimple?

 

[fresh_42]

Not quite. They are the constants that appear in the commutation relations of the generators of the Lie algebra.

Can you explain this to me? If we consider e.g.
$$\mathfrak{sl}(2,\mathbb{C}) = \langle X_1,X_2,X_3\,\vert \,[X_1,X_2]=2X_2\; , \;[X_1,X_3]=-2X_3\; , \;[X_2,X_3]=X_1 \rangle $$
then $\pm 2 \; , \; \pm 1\; , \;0$ are the structure constants as I've learned it, i.e. the factors in the multiplication table.

 

[PeterDonis]

If we consider e.g.

$$
\mathfrak{sl}(2,\mathbb{C}) = \langle X_1,X_2,X_3\,\vert \,[X_1,X_2]=2X_2\; , \;[X_1,X_3]=-2X_3\; , \;[X_2,X_3]=X_1 \rangle
$$

then $\pm 2 \; , \; \pm 1\; , \;0$ are the structure constants as I've learned it, i.e. the factors in the multiplication table.

The things you wrote here, like $[X_1,X_2]$, are commutators (also called "Lie brackets"). $X_1$, $X_2$, and $X_3$ are the generators of the Lie algebra. So the structure constants are the commutators (or Lie brackets) of the generators.

Some sources might use the term "multiplication" to refer to the Lie bracket of the Lie algebra, but I think this is misleading; a commutator $[X_1, X_2] = X_1 X_2 - X_2 X_1$ is a more specific operation than simple multiplication, and requires more structure (since you have to be able to subtract two products). Also, since the full Lie algebra consists of all possible linear combinations of the generators (which means you also have to be able to add elements of the algebra as well as multiply them), the structure constants obviously do not equal the commutators (let alone the simple products) of all possible elements of the algebra.

 

[samalkhaiat]

As you've mentioned it, I'd like to post a question in this realm, which by itself might be a bit too speculative. However, I think it has to be allowed. The standard model is based on $SU(n)$, which are simple groups. I once asked the question here, why physicists are mainly interested in (semi-)simple groups and their Lie algebras. The best answer I received was, because the Killing form is non-degenerated and can be used as a measure for angles and lengths, which is widely used in the classification theory of semisimple Lie algebras.

Now you also mentioned the Poincaré group, which is not semisimple, and I would like to add the Heisenberg group(s). I'm not sure about the Lorentz group, as it is rather close to one: $SO(3,1)$ instead of $SO(4)$, however not exactly. So the question which naturally arises is: Is it possible, that our difficulties to find a unification theory has to do with the fact, that those groups are not semisimple?

Yes, the invariant metric (inner product) on the Lie algebra in question needs to be non-degenerate so that the action integral contains kinetic terms for each component of the gauge field (the force carrier): $$\mathcal{L} \sim g_{ab}F^{a}_{\mu\nu}F^{b \mu\nu}$$

 

[Orodruin]

The things you wrote here, like $[X_1,X_2]$, are commutators (also called "Lie brackets"). $X_1$, $X_2$, and $X_3$ are the generators of the Lie algebra. So the structure constants are the commutators (or Lie brackets) of the generators.

Some sources might use the term "multiplication" to refer to the Lie bracket of the Lie algebra, but I think this is misleading; a commutator $[X_1, X_2] = X_1 X_2 - X_2 X_1$ is a more specific operation than simple multiplication, and requires more structure (since you have to be able to subtract two products). Also, since the full Lie algebra consists of all possible linear combinations of the generators (which means you also have to be able to add elements of the algebra as well as multiply them), the structure constants obviously do not equal the commutators (let alone the simple products) of all possible elements of the algebra.

I don't agree with this. A priori, the Lie bracket is the only multiplication available in the Lie algebra. The "simple multiplication" is not an operation in the Lie algebra since typically the Lie algebra will not be closed under it. It is also something that will typically not be of interest for an abstract Lie algebra, but only becomes something to worry about once you start looking at matrix representations (and in those cases you should most often conclude that the operation is not an operation that is of any particular interest). I think that the characterisation of the structure constants as the multiplication table of the Lie bracket or, more precisely, the multiplication table of the generators of the Lie group, is not a bad one. The full multiplication table follows directly from the linear properties of the Lie bracket.

 

[PeterDonis]

The "simple multiplication" is not an operation in the Lie algebra since typically the Lie algebra will not be closed under it.

Ah, I see. Yes, this is a fair point.