The Reference Frame: Classification of simple compact Lie groups

The goal of this text is to concisely explain why the simple compact Lie groups are divided to the groups \[

A_\ell, B_\ell,C_\ell, D_\ell, E_{6,7,8}, F_4, G_2.

\] The focus will be on the representation of every step, not on the detailed explanation of each step. In other words, I will assume that the reader is busy and smart and can rediscovery many things for herself.

It's a complementary texts to some previous ones such as Exceptional Lie groups.

Some definitions

First, what do the words in the title mean? A group is a set \(G\) of elements that include \(1\) with an operation "product" (if the group is Abelian, i.e. the ordering in the "product" doesn't matter, we often talk about a "sum") which is associative\[

(gh)k = g(hk)

\] and which contains an identity element \(1\) as well as the inverse element \(g^{-1}\) for every element. Now, a Lie group (named after Sophus Lie) is "continuous" which means that you should imagine this whole set as a manifold that can be labeled by continuous coordinates. Groups of continuous rotations such as \(SO(N)\) are examples.

A compact Lie group is a Lie group that is "compact" i.e. that has a finite volume. If you imagine it as a manifold and define a "volume density form" on this manifold that is nonzero and invariant under the group action, the total volume is finite. Effective, compactness means that when written in terms of matrices, all the matrix entries are bounded. So \(SO(3)\) is compact but \(SO(3,1)\) is not. The former group manifold, \(SO(3)\), is actually geometrically equivalent to a sphere, \(S^3\), well, modulo a \(\ZZ_2\).

For a compact Lie group, I may simplify the definition of a "simple group" by saying that it cannot be written as (a group isomorphic to) \(G\times H\) which is a direct product. For example, \(SO(3)\times SO(4)\), the direct product of two groups whose product is defined\[

(g,h)(g',h') = (gg', hh'),\\ \quad \{g,g'\}\subseteq SO(3), \quad \{h,h'\}\subseteq SO(4),

\] is not simple while \(SO(3)\) is. Well, \(SO(4)\) actually isn't simple either because it is isomorphic to \(SU(2)\times SU(2)\), well, quotiented by a discrete group and all these subtleties will be ignored in this text.

Let's roll.

Lie algebras

For a given Lie group, we define the corresponding Lie algebra. Take all elements on the group manifold that are very close to the identity \(1\), for example all rotations by small angles (and their compositions).

So the group element (or the matrix representing it) may be written as \(1+\epsilon\) in an approximation. Well, we actually write it as \(\exp(iM)\) where \(i\) is a subtlety chosen by physicists' conventions ? useful for making \(M\) Hermitian rather than anti-Hermitian ? while the exponential (which reduces to \(1+iM\) in a Taylor expansion) is great because it's exact even for finite, not just infinitesimal, \(M\). It's because of a formula for the exponential\[

\exp(iM) = \lim_{N\to\infty} \zav{ 1+ \frac{iM}{N} }^N.

\] You see that it is a product of infinitely many infinitesimal generators. The space of all possible values of \(M\) is the Lie algebra. What's important is the "commutator" \([M,N]\) on this Lie algebra. It may be calculated from the Lie group by considering expressions of the sort\[

\zav{ \exp\frac AN \exp\frac BN \exp\frac{-A}N\exp\frac{-B}{N} }^{N^2} = \dots

\] Ignore the \(N^2\) exponent for a while. You see a product of four exponentials, four elements of the Lie group. They would cancel and the result would be \(1\) if \(AB=BA\) because each action is undone. However, they're undone in the opposite order so if the exponentials don't commute with one another. Now, expand each of the exponentials up to the order \(1/N^2\) which is enough if we want the finite piece of the \(N^2\)-th power in the large \(N\) limit. What you get is\[

\dots =\left[1+\frac{AB-BA}{N^2} + o(1/N^2)\right]^{N^2} \to \exp ([A,B]).

\] The commutator just comes from all the pairs in which you had to exchange the order of \(A,B\). So we see that by combining infinitesimal transformations, \(\exp([A,B])\) is actually also obtainable which means that \([A,B]\) is an element of the Lie algebra. Well, in physics conventions, we would insert factors of \(i\) and say that \(\pm i[A,B]\) is an element of the Lie algebra for all \(A,B\).

Great. We have a Lie algebra. As long as \([A,B]\) is defined as \(AB-BA\) for some particular matrices (we can multiply), it's easy to verify that\[

[A,[B,C]]+[B,[C,A]] + [C,[A,B]] = 0.

\] The left hand side contains \(3\times 2\times 2 = 12\) terms, various permutations of \(ABC\), and they cancel in pairs. So it's a tautology (we call it "identity") if the commutator "means" \(AB-BA\). However, it's useful to imagine that the commutator is a "more abstract" operation not defined by any particular matrices in which case the identity above, the Jacobi identity, has to be required as a defining condition for the commutator to generate a Lie algebra (aside from the bilinearity which is obvious).

For \(SO(N)\), the Lie algebra called \({\mathfrak so}(N)\) ? yes, these are Gothic fonts ? consists of all antisymmetric \(N\times N\) matrices which are real in mathematicians' conventions and pure imaginary in physicists' conventions. That's because a small rotation around the \(z\) axis just adds a small multiple of \(y\) to \(x\) and vice versa with a minus sign.

Killing form

On the Lie algebra, we may define a bilinear "Killing form"\[

b(A,B) = {\rm Tr}(AB^{T}).

\] Whenever the Lie algebra is given by matrices, this inner product on the Lie algebra may be written as the trace above, up to an undetermined overall normalization. Whenever it's abstract, we want the Killing form to be invariant under the action of the group etc. ? to obey the same properties that the trace above does. The Killing form in combination with the commutator obey a fun identity\[

b([X,A]B) + b(A,[X,B]) = 0.

\] The Killing form is "antisymmetric" with respect to the commutator. This is also required even if the Killing form isn't given by explicit matrices. Now, the operation \[

A\mapsto \exp(-X) A \exp(X)

\] is an isometry (symmetry preserving the length) relatively to the metric encoded by the Killing form.

I will leave the discussion of general representation theory for another day. Some next-to-rudimentary material has been covered, e.g. in Why are there spinors?.

Maximum tori, Cartan subalgebras

To classify simple compact Lie groups, we need to define the maximal tori. It's subgroups isomorphic to \(U(1)^\ell=SO(2)^\ell\) which is, as you can see, a commutative (Abelian) group. The maximum value of \(\ell\) is known as the rank of the Lie group or the Lie algebra.

For example, the group \(U(\ell)\) has the maximum commuting subgroup \(U(1)^\ell\). It's the maximum torus of it. The Lie algebra of the maximum torus is the Cartan subalgebra, \[

{\mathfrak u}(1)\oplus \dots \oplus {\mathfrak u}(1)

\] which has \(\ell\) terms. Similarly, the groups \(SO(2\ell)\) and \(SO(2\ell+1)\) have the same (isomorphic) maximal torus and Cartan subalgebra consisting of all rotations of the coordinates 1-2 into one another composed with a rotation of axes 3-4 into one another, and so on. For \(SO({\rm odd})\), one coordinate is left and cannot be used to extend the maximal torus anymore.

Why did I call the subgroups tori rather than "subspaces"? It's because they're periodic ? largely because \(exp(2\pi i k)=1\). This is particularly clear in terms of the so-called Stiefel diagrams drawn in the Cartan subalgebra ? chosen to respect the angles inherited from the Killing form. For a few random easy groups, they look like this:

Click to zoom in. "Jin? pohled" means "A different perspective".

The empty or filled disks or squares correspond to elements of the Lie algebra (its Cartan subalgebra) whose exponential with the correct \(i\) if needed maps to the identity of the Lie group indicated in the legend. We also indicate the straight lines which are co-dimension 1 places for which all the roots take integral values under the most natural map. I won't really need the lines.

Root systems

Why is it clever to consider the maximum torus or its Lie algebra, the Cartan subalgebra of the original Lie algebra? It's because they define what the quantum physicists call "a maximum set of commuting observables". In particular, we can simultaneously diagonalize all these \(\ell\) (rank) generators.

Note that all these generators ? a basis of the Cartan subalgebra ? are operators that have well-defined actions on any "representation" of the Lie group (or Lie algebra, it's the same representation). A particularly important example of a representation of a Lie group or its Lie algebra is the same Lie algebra itself. The action of a Lie algebra element \(M\) on an element of the representation \(V\), which is just another element of the same Lie algebra, is simply \([M,V]\), the commutator. We call this action of the Lie algebra (or group) on itself "the adjoint representation".

Just like you may find states of the Hydrogen atom that are simultaneous eigenstates of \((n,l,m,s_z)\), a maximum set of commuting observables, you may look for collections of eigenvalues \((m_1,m_2,\dots , m_\ell)\) and the corresponding "shared eigenstates" in the Lie algebra. Here, \(m_i\) are some basis of the Cartan subalgebra. In the Cartan subalgebra space, the collection of these numbers ? eigenvalues \(m_i\) of \(M_i\) corresponding to an eigenstate ? looks like a vector. We call this vector a "weight of a given representation". Weights of the adjoint representation are known as "roots".

Note that \(\ell\) roots are zero (vanishing vectors in the Cartan subalgebra). They correspond to the "vectors of the representation" which are elements of the Cartan subalgebra itself, and because this subalgebra is commuting, \([M,V]=0\) for all pairs. We usually don't consider these "vanishing roots" to be "roots". So a "root" is really a weight for the adjoint representation that is nonzero. Their number is therefore \(d-\ell\) where \(\ell\) is the rank and \(d\) is the dimension of the Lie group or the Lie algebra, i.e. the number of its independent generators.

What do these roots (or their set for a given algebra, the so-called root system) look like? They're "stars" distributed in some "lattice" around the origin of the Cartan subalgebra. One may show that the introduction to the roots above implies that a root system

* doesn't contain the vanishing vector (by convention)
* for every \(\alpha\in\Sigma\), \(c\alpha\in\Sigma\) iff \(c=\pm 1\)
* the set \(\Sigma\) is symmetric with respect to mirroring defined by any plane normal to any root \(\alpha\in\Sigma\)
* the number\[

\{\alpha,\beta\} = \frac{2b(\alpha,\beta)}{b(\beta,\beta)} \in\ZZ

\] is integer. The second condition among the four heuristically says that the generator associated with a root commutes with itself so you can't get any roots that are longer but going in the same direction. It's not quite a proof but believe me it's the case for the root systems we may construct from actual groups ? and we may demand this condition to be valid for "abstract root systems", too.

The third condition reflects the existence of an automorphism associated with a given root. The last condition about the integrality may be proven but I chose to omit this point. However, it's the most powerful axiom because it implies that any pair of roots \(\alpha\neq \pm \beta\) ? note that these are just two vectors in some Euclidean space ? must have one of the following relative geometric positions:

(0) they're orthogonal to one another
(1) they're equally long and the angle between them is 60? or 120?
(2) one of them is \(\sqrt{2}\) times longer than the other one and the angle between them is 45? or 135?
(3) one of them is \(\sqrt{3}\) times longer than the other one and the angle between them is 30? or 150?

I will prove this assertion, however. Four times the cosine of the angle \(\omega\) between them may be written as\[

4\cos^2\omega = \frac{ 2b(\alpha,\beta) 2b(\beta,\alpha) }{ b(\alpha,\alpha)b(\beta,\beta) }

\] from the usual cosine interpretation of an inner product. Because of the left hand side, it is a number between 0 and 4. The right hand side shows that it is a product of two integers, due to an assumption a few paragraphs above, and it can't be \(4\) because we assumed that the roots weren't equal (not even up to a sign) and they couldn't be parallel if \(\alpha\neq \pm\beta\), due to another assumption. So the angle \(\omega\) equal to zero or a multiple of \(\pi\) is excluded and both sides must actually be equal to 0, 1, 2, or 3.

For 0, the roots are orthogonal to one another because the inner product has to be zero. What remains is to write numbers 1,2,3 in all possible ways as products of two integers and make the corresponding interpretation of the values \(b(\alpha,\beta)/b(\alpha,\alpha)\) and so on. We obtain the possible length ratios and angles listed above.

Finally, constructing root systems from Dynkin diagrams

Using the warp speed approach, we quickly got to the fun part of the game with many pictures and possibilities. We want to classify all root systems. Root systems are finite collections of (nonzero) roots in the \(\ell\)-dimensional space. The relative geometric relationships between the roots are given by the propositions (0), (1), (2), (3). One of them.

First, we will choose a subset of \(\ell\) roots which may act as a basis of the Cartan subalgebra, \(\RR^\ell\). A convenient choice is the choice of "positive roots". In some Cartesian coordinate system, they're roots whose "first" nonzero coordinate is positive. Note that we have made many steps. From a given Lie group, we constructed a Lie algebra. We found its Cartan subalgebra. We found the roots. From the root system, we chose the positive roots.

Now, we will draw the Dynkin diagram for the root system.

The Dynkin diagram is a collection of \(\ell\) (rank) nodes (small circles) and each two nodes are connected by 0, 1, 2, 3 lines if the relative geometric position (and length ratio) of the two roots is given by the propositions (0), (1), (2), (3) above, respectively. Moreover, if the link between two roots is of the (2) or (3) type, we draw an arrow on the link between the two nodes. By convention, it's directed towards the shorter root, so that it can be interpreted as \(\lt\) or \(\gt\) for the lengths of the roots.

(Someone uses the opposite convention but we can't do anything about the human freedom.)

Our final task is to draw all possible Dynkin diagrams. That's a cool task. The result will be

The possible Dynkin diagrams are of the \(A,B,C,D,E,F,G\) type. \(A\) is special unitary, \(B\) is odd orthogonal, \(C\) is symplectic, \(D\) is even orthogonal, and \(E,F,G\) refers to the five exceptional groups. How do we understand that these Dynkin diagrams are possible and the only possible ones?

We will go through a cute algorithm by showing that the geometrically possible Dynkin diagrams can't have loops; triple links combined with any other links; double links with extra links on both sides, and that the diagrams with simple links only (the so-called simply laced diagrams) can't be more convoluted than the diagrams \(A,D,E\) above.

Let us begin. But first, let's dedicate one paragraph to a refinement of our strategy.

Our strategy will be to look at a "candidate" Dynkin diagram and to construct a vector ? an integer linear combination of the positive roots (represented by the nodes of the diagram) ? whose squared length (computed via the inner product) is non-positive. Because any nontrivial combination of these bases vector is a nonzero vector in the Euclidean space, its length has to be positive. So if we find out it's zero or negative, the Dynkin diagram just can't be realized by any roots in an actual Euclidean space!

It just happens that all the linear combinations we will really care about will produce vectors of vanishing length. We will write the coefficients in front of each positive root as labels next to each node and the fact that exactly one combination of the positive roots is a "zero norm" vector actually means that all these diagrams will be "extended Dynkin diagrams" appropriate for "affine Lie groups". These are infinite-dimensional groups in which the original Lie group element is chosen for each point on a closed string.

I wanted to draw these extended Dynkin diagrams with the coefficients ? called "marks" in the diagram below ? but finally I saved one hour after I found the complete product somewhere:

Click to zoom in.

In all cases, the Dynkin numbering is just a particular way to label the nodes by numbers \(0,1,2,\dots ,\ell-1\): there are \(\ell\) (rank) of them. It's the marks we care about.

The text will be completed and proofread later...

Source: http://motls.blogspot.com/2012/10/classification-of-simple-compact-lie.html

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