## Wednesday, April 25, 2012 ... /////

### Exceptional Lie groups

Twenty years ago, when I started to read various articles on particle physics, I often encountered exceptional Lie groups in various articles on grand unified theories and related topics. And I couldn't understand why they exist and what they really are. Even when I got to the college, it took half a year to find the right sources that made it clear.

There might be some people who are facing the same puzzles so let me offer an exposition that would have clarified my confusions when I was a kid. This blog entry will try to clarify the composition of all the simple compact Lie groups.

Groups, compact Lie groups

Pretty much everyone may understand the group $SO(N)$. It is the group of symmetries of the $N$-dimensional unit ball$x_1^2 + x_2^2 + \dots x_N^2 \leq 1.$ The coordinates $x_1,\dots , x_N$ may be rotated into each other by orthogonal transformations.

You may represent the rotations by linear transformations$\pmatrix{x_1\\x_2\\ \vdots\\ x_N} \to M \cdot \pmatrix{x_1\\x_2\\ \vdots\\ x_N}$ where the dot represents the matrix multiplication and $M$ is an $N\times N$ matrix. If the matrix is orthogonal i.e. if it obeys$M M^T = M^T M = {\bf 1},$ then it is easy to see that the Pythagorean norm of a vector $\vec x$ is preserved when $\vec x$ is replaced by $M\vec x$:$ds^2 = \vec x^T\cdot \vec x = (M \vec x)^T \cdot (M \vec x) = \vec x^T\cdot M^T\cdot M \cdot \vec x$ because the matrix multiplication $(\cdot)$ is associative and because the product $M^TM$ may be replaced by the identity matrix that may be erased because ${\bf 1}\cdot \vec x = \vec x$. Fine. One may consider transformations that are very modest, very close to the identity. There are $N(N-1)/2$ of independent generators that produce such transformations because each pair of the $(x_i,x_j)$ coordinates may be rotated into each other.

Because $M^T M = {\bf 1}$ and because$\det(M^T M ) = \det M^T \det M = (\det M)^2 = +1,$ we see that $\det M=\pm 1$. That means that the group manifold $O(N)$ – the set of all possible matrices that satisfy the orthogonality condition – is composed of two independent continuous pieces. The determinants of the matrix in these two pieces are $\pm 1$, respectively. If we only consider the matrices that obey$M^T M = {\bf 1},\qquad \det M = {\bf 1},$ then we define the so-called group $SO(N)$ which is also $N(N-1)/2$-dimensional and only has one component (every element of the group is continuously connected to every other element of the group).

Unitary groups

There exists a straightforward generalization of the orthogonal groups $O(N)$ and $SO(N)$ for complex linear spaces. Just replace $x_i$ by complex variables $z_i$ and require that the following norm$|z_1|^2+|z_2|^2+\dots+|z_N|^2 = 1$ is preserved by the linear transformations acting on the complex $N$-dimensional space. The complex matrix $M$ that acts on $\vec z$ has to obey$M^\dagger M = {\bf 1}, \qquad M^\dagger \equiv (M^T)^*$ for the bilinear expression constructed from the absolute values of $z$ to stay constant. The condition above defines the so-called unitary group, $U(N)$. One may see that in this case, the determinant satisfies$\abs{\det M}^2 = 1$ which means that $\det M$ is any number whose absolute value is equal to one. In other words, the determinant has to belong to the one-dimensional manifold of the complex numbers of the type $\exp(i\delta)$ for $\delta\in\RR$. If one demands $\det M = +1$ again, we obtain the so-called special unitary group, $SU(N)$. Note that the real dimension of the $U(N)$ group is $N^2$ while one more dimension is removed for $SU(N)$; its real dimension is therefore $N^2-1$.

Symplectic groups

Those were simple enough things but there actually exists one more infinite class of similar groups. The quaternions are associative so one may construct nice associatively-multiplicative matrices with quaternionic entries. The group $USp(2N)$ is defined as nothing else than $U(N,{\mathbb H})$. Recall that quaternions are numbers of the type$q = a+bi+cj+dk$ where the most important defining property is the multiplication table for the unit imaginary quaternions,$\eq{ i^2&=j^2=k^2=-1,\\ ij=-ji=k, \quad jk&=-kj=i,\quad ki=-ik=j }$ If you consider vectors with quaternionic entries $q_1,q_2,\dots, q_N$, the group of quaternionic-valued matrices $M$ obeying $M^\dagger M={\bf 1}$ will be defined as the symplectic group $USp(2N)$. Note that the dagger $\dagger$ involves both transposition and the complex conjugation which changes the sign of the three coefficients in front of all the $i,j,k$ imaginary units.

One may offer an equivalent, quaternion-free definition of the group. Write the quaternions as $q=z_1+j z_2$ where $z_1,z_2$ are complex numbers. That's possible because $z_1$ takes care of the real multiples of $1$ and $i$ while $z_2$ takes care of the $j$ and $k=ij$ terms. Now, replace each entry $q_i$ of the quaternionic vector as the column with two entries $z_1,z_2$ above each other. The quaternionic multiplication may be mimicked by a procedure acting on the complex $2\times 2$ matrices and the quaternionic matrix $M$ may be emulated by a larger complex matrix $M_\CC$ of the size $2N\times 2N$ which obeys$M_\CC^\dagger M_\CC = {\bf 1}, \qquad M_\CC^T \cdot a \cdot M_\CC = a$ where $a$ is a specific antisymmetric matrix, namely$a = {\rm diag} \zav{ \pmatrix{0&-1\\+1&0}, \,\, \pmatrix{0&-1\\+1&0}, \cdots \pmatrix{0&-1\\+1&0} }$ where the $2\times 2$ basic block is repeated $N$ times on the diagonal. This is another way to define the symplectic group. Note that the real dimension of $USp(2N)$ is the same as the real dimension of the $SO(2N+1)$ group, namely $(2N+1)N$.

The orthogonal, unitary, and symplectic groups are all the possible examples of "infinite families" of simple compact Lie groups. The adjective "simple" means that the group can't be written as $G_1\times G_2$ where both $G_1,G_2$ are nontrivial groups. The adjective "Lie" named after Sophus Lie (picture at the top) means that the groups are continuous, i.e. their elements are parameterized by continuous parameters. The word "compact" means that if you imagine the set of all the elements, it is a manifold, and with some natural "measure" on that manifold that is invariant under the group operation itself, the volume of the group manifold is finite. There are no directions in which you may "escape to infinity" away from the identity.

Off-topic: David Gross gave a talk at Chapman University a month ago (95 minutes) on the state of physics.

Isomorphisms

Now, there are lots of important isomorphisms between the elements of the three infinite classes of the compact simple Lie groups, especially for small values of $N$. I will use the symbol $=$ for the isomorphism which will tolerate the fact that one of the groups that are "identified" may be factored to be really isomorphic to the other one. Equivalently, the isomorphisms below hold at the level of the Lie algebras. We have$\eq{ U(N) &= SU(N)\times U(1)\\ U(1) &= SO(2) \\ SO(3) &= SU(2) = USp(2)\\ SO(4) &= SU(2) \times SU(2) = USp(2)\times USp(2)\\ SO(5) &= USp(4)\\ SO(6) &= SU(4) }$ where the most complicated isomorphisms require you to learn about the spinor representations. I don't want to expand this blog entry into a huge book so I must be very brief. The orthogonal groups $SO(N)$, more precisely $Spin(N)$ which are "refined" versions of $SO(N)$ that, unlike $SO(N)$, distinguish the rotation by 0 degrees (identity) from the rotation by 360 degrees (while the rotation by 720 degrees is the identity again), i.e. groups obeying$SO(N) \approx Spin(N)/\ZZ_2$ have a nice new type of a representation whose dimension is something like $2^{[N/2]}$, the so-called spinors. The existence of such representations is linked to our ability to define the Dirac gamma matrices whose width as well as height is the same power of two. These gamma matrices obey$\gamma_i \gamma_i + \gamma_j \gamma_i = 2\cdot \delta_{ij}\cdot{\bf 1}, \qquad i,j=1,2,\dots, N.$ Such matrices may be defined as tensor products of the Pauli matrices (and the identity matrix). And if the key identities above hold, we may also see that $\gamma_{ij}/2=\gamma_i\gamma_j/2$ for $i\neq j$ obey commutation relations that coincide with the commutation relations for the usual generators of $SO(N)$. The commutator of $G_{ij}$ and $G_{jk}$ is equal to $i=\sqrt{-1}$ times $G_{ik}$ for $i\neq j\neq k\neq i$. This isomorphism between the commutation relations guarantees that one may construct matrices of the same size as the gamma matrices whose products are related by the same dictionary as the products of the usual $SO(N)$ matrices we started with.

The spinors – which change the sign if you rotate the coordinates by 360 degrees – are the new and unexpected kind of representations of the orthogonal groups you have to master when you really want to understand the compact Lie groups. In some sense, the spinor index which takes one of the $2^{[N/2]}$ possible values may be interpreted as "one-half" of the vector index of $SO(N)$. Otherwise, you may build the most general representations as tensors with many vector indices – which may be constrained by various symmetric or antisymmetric rules as well – but you must also allow the spinor indices.

Going exceptional

So far, the text was unexceptional. We haven't gotten anywhere yet. ;-) I want you to understand the exceptional groups. They're five more simple compact Lie groups that don't belong to the orthogonal, unitary, and symplectic infinite families listed above. It turns out that these groups are $G_2,F_4,E_6,E_7,E_8$ where the subscript indicates the rank, i.e. the maximum number of independent $U(1)$ generators of these groups that commute with each other (any pair commutes).

As you can see, the largest group seems to be $E_8$, so we will start with it. It's the "master example" of an exceptional group; all other exceptional groups may be identified with some subgroups of $E_8$. We will construct $E_8$ as a particular "extension" of $SO(16)$, one of the ordinary orthogonal groups.

Start with the generators $J_{IJ}$ where $I,J=1,2,\dots 16$. They have the commutators of the type$-i[J_{IJ},J_{KL}] = [\delta_{IL} J_{JK} - (K\leftrightarrow L)] - (I\leftrightarrow J)$ where I added the factor of $(-i)$ to agree with some conventions and to allow all the generators to be Hermitian (rather than antihermitian). The "exchange arrow" terms guarantee the two antisymmetry conditions. These matrices $J_{IJ}$ may be obtained as $16\times 16$ matrices whose only nonzero entries are at positions $IJ$ and $JI$ and they are equal to $+1$ and $-1$, respectively. Alternatively, as I suggested, you may also define $J_{IJ}$ as $\gamma_{IJ}/2$, a product of two gamma matrices (the matrix is defined to vanish for $I=J$ i.e. it is antisymmetrized in $IJ$.

We have $16\times 15/ 2\times 1 = 120$ generators like that.

Now, we add $128$ generators $Q_\alpha$ that transform as a real chiral spinor of $Spin(16)$ i.e. "essentially" $SO(16)$. According to the general rule, the Dirac spinor for $Spin(16)$ has $2^{16/2}=2^8=256$ components. One may show that because $16$ is a multiple of eight, it is a real spinor: all the rotation matrices may be expressed as real matrices in an appropriate basis.

However, because $16$ is even, the rotation matrices decompose into a block-diagonal form in a nicely chosen basis. So the $256\times 256$ matrices for the rotations in the "Dirac spinor representation" may be decomposed into two $128\times 128$ matrices acting on the "chiral", i.e. Weyl spinor; one of the spinors is left-handed and one of them is right-handed. Let's take only one of the spinors which has, as I said, $128$ real components.

What does it mean for $128$ bonus generators $Q_\alpha$ to transform as spinors? It means that$[J_{IJ},Q_\alpha] = \frac {i}2 (\gamma_{IJ})_{\alpha\beta} Q_\beta$ because the commutator is how the symmetry generators $J_{IJ}$ act on the objects $Q_\alpha$ if these objects are meant to be operators themselves and not just "vectors". We obtain some linear combinations of the same $Q_\beta$ generators and the coefficients are the matrix entries of the $Spin(16)$ generators in the spinor matrix.

To prove that we have defined a Lie algebra, one must verify the Jacobi identities$[A,[B,C]]+ [B,[C,A]] + [C,[A,B]] = 0$ where the three terms differ by cyclic permutations. This identity – which is tautologically satisfied whenever $[A,B]$ may be expressed as $AB-BA$: just expand the damn commutators – has to be obeyed for every triplet of generators $A,B,C$ if we actually want some representation(s) of the commutator in terms of the product differences to exist. Now, each of the generators $A,B,C$ may be either of the $J_{IJ}$ type, i.e. the generator of $Spin(16)$, or the $Q_\alpha$ type, the bonus spinor generator. The $JJJ$ Jacobi identity holds because it holds in $SO(16)$, a nice group. The $JJQ$ Jacobi identity also holds because it just says that the commutator of two $J$'s acts on a $Q$ in the same way as if you compute the commutator of the two actions.

One must also verify the $JQQ$ and $QQQ$ Jacobi identity. Only the latter calculation turns out to be hard enough; one has to use various identities for the spinors. Both $JQQ$ and $QQQ$ identities require you to know the commutator of two $Q_\alpha$ generators:$[Q_\alpha,Q_\beta] = J_{IJ}\cdot (\gamma_{IJ})_{\alpha\beta}.$ The right hand side is actually the only candidate you may construct – a linear combination of generators that has the $\alpha,\beta$ indices ($I,J$ are contracted) which preserves the $SO(16)$ symmetry. We are obliged to use nice $SO(16)$-preserving defining relations because we want to keep this orthogonal group as our symmetry (a subgroup of the emerging $E_8$ group). The only freedom is the overall normalization. The coefficient has to be real. If it were rescaled by a positive number, the change might be absorbed to the normalization of $Q_\alpha$.

The only freedom that is left is the sign. If you change the sign of the commutator above, you switch between $E_8$ and its noncompact version $E_{8(8)}$ which also contains the $SO(16)$ subgroup.

Now, if we have the Lie algebra, the Lie group may be defined as the set of the exponentials of the Lie algebra matrices$E_8^{\rm group} = \left\{ \exp\left(i \sum_a G_a \omega_a\right),\quad \omega_{a}\in \RR \right \}$ where $G_a$ with $a=1\dots 248$ stands for $J_{IJ}$ as well as $Q_\alpha$ – or, equivalently, as the automorphism group of the Lie algebra. We have added a 128-dimensional real spinor to the 120-dimensional real Lie algebra so we have created a 248-dimensional Lie algebra, the $E_8$. The decomposition of the adjoint representation of $E_8$ under the $SO(16)$ subgroup is${\bf 248} = {\bf 120}_{\rm adj} \oplus {\bf 128}_{\rm spinor}.$ The strategy to construct a new Lie algebra by adding a spinor to an adjoint of an orthogonal group works in 3 cases; only in 3 cases, all the Jacobi identities, especially the hard $QQQ$ identity, may be satisfied. Aside from the $E_8$ case above, one may also construct the 52-dimensional $F_4$ exceptional Lie algebra by adding the 16-dimensional real spinor of $SO(9)$ to the 36-dimensional $SO(9)$ adjoint. Note that there is no chirality in 9 dimensions which is an odd number.

The final, third example is all about the addition of the 8-dimensional real chiral spinor to the 28-dimensional $SO(8)$ adjoint.

The triality of $SO(8)$ may be seen as the permutation symmetry between the three "branches" of the Dynkin diagram of this group.

In this way, we obtain a 36-dimensional group. What is it? Well, it's an $SO(9)$. It may remind you of a simple way to construct $SO(9)$ out of $SO(8)$: add an 8-vector of new generators $J_{i9}$ to the $SO(8)$ generators $J_{ij}$. The addition of an 8-dimensional left-handed spinor is related to the addition of the 8-dimensional vector by an operation called "triality": the third sibling is the addition of the right-handed spinor (which is just a left-right reflection of the addition of the left-handed one). The triality symmetry is a special symmetry of the $SO(8)$ group and no other compact Lie group. It permutes its representations, too. In particular, the two spinor representations and the vector representation (each of them has 8 dimensions) are permuted.

One may say that we have defined $E_8$ and $F_4$ by the "adjoint plus spinor" strategy above. We may define all the remaining exceptional groups by constructions that only depend on ordinary Lie groups discussed at the beginning.

First, what is a $G_2$? It is the centralizer of $F_4$ embedded into $E_8$. Of course, there are other ways to define it but this is one of them. We constructed $F_4$ out of $SO(9)$. But $SO(9)\times SO(7)$ may be embedded in the obvious "additive" way to $SO(16)$ which is inside $E_8$. So a centralizer or $F_4$ is a centralizer of an "extension of $SO(9)$" which is inevitably a "subgroup of $SO(7)$". Note that a similar statement with the opposite terms "extension" and "subgroup" omitted is also true: the centralizer of $SO(9)$ in $SO(16)$ and even in $E_8$ is nothing else than $SO(7)$.

The decomposition of the fundamental representation of $E_8$ (which is the same as the adjoint for this group and only this group) under the $F_4\times G_2$ maximal subgroup ("maximal" means that one can't embed any other embedding that works on both sides in between) is${\bf 248} = ({\bf 52},{\bf 1}) \oplus ({\bf 1}, {\bf 14}) \oplus ({\bf 26},{\bf 7})$ which is the direct sum of the two adjoint representations and the tensor product of the fundamental representations.

However, one may also find the groups $E_6,E_7$ inside $E_8$. They're centralizers of $SU(3)$ and $SU(2)$ inside $E_8$, respectively. These special unitary groups have to be embedded into $SU(4)=SO(6)$ where $SO(6)$ is a part of $SO(6)\times SO(10)$ embedded into $SO(16)$ and $E_8$ in the obvious way. The decompositions of the representations may be rather easily derived to be$\eq{ {\bf 248} &= ({\bf 78},{\bf 1}) \oplus ({\bf 1}, {\bf 8}) \oplus ({\bf 27},{\bf 3})\oplus {\rm c.c.}\\ {\bf 248} &= ({\bf 133},{\bf 1}) \oplus ({\bf 1}, {\bf 3}) \oplus ({\bf 56},{\bf 2}) }$ under $E_6\times SU(3)$ and $E_7\times SU(3)$, respectively. The abbreviation ${\rm c.c.}$ stands for the complex conjugate representation to the previous one, i.e. $(\bar{\bf 27},\bar{\bf 3})$.

There are tons of other basic things to be said about the five exceptional groups and the compact simple Lie groups in general but I suspect that some of the readers may already be tired at this point so let me stop.

Any questions? ;-)