Root and Weight of Lie Group

This article is one of Lie Group & Representation contents.

Weights

To categorize state maximally, we have to find the largest set of commuting hermitian generators to diagonalize as much as possible.

In particular irreducible representation, $D$, there will be a number of hermitian generators, $H_i$ for $i=1$ to $m$, corresponding to the elements of the Cartan Subalgebra called the Cartan generators satisfying $$\begin{align}H_i=H^\dagger_i,\ \mbox{and } [H_i,H_j]=0.\end{align}$$

The Cartan generators form a linear space. Thus we can choose a basis in which they satisfy $$\begin{align}\mbox{Tr}(H_iH_j)=k_D\delta_{ij}\ \mbox{for } i,j=1\mbox{ to }m\end{align}$$ where $k$ is some constant that depends on the representation and on the normalization of the generators. The integer $m$, the number of independent Cartan generators, is called the rank of the algebra.

Of course, the point is that the Cartan generators can be simultaneously diagoanlized. After diagonalization of the Cartan generators, the states of the representation $D$ can be written as $\left| \mu, x, D\right\rangle$ where $$\begin{align}H_i \left| \mu, x, D \right\rangle = \mu_i \left| \mu, x, D\right\rangle\end{align}$$ and $x$ is any other label that is necessary to specify the state.

The eigenvalues $\mu_i$ are called weights. They are real, because they are eigenvalues of hermitian operators. The $m$-component vector with components $\mu_i$ is the weight vector. We will often use a vector notation in which $$\begin{align}\alpha \cdot \mu \equiv \alpha_i \mu_i\ \mbox{and } \alpha^2\equiv \alpha_i \alpha_i\end{align}$$

Adjoint Representation

The state in the adjoint representation corresponding to an arbitrary generator $X_a$ as $$\begin{align}\left| X_a \right\rangle.\end{align}$$

Linear combinations of these states correspond to linear combinations of the generactors - $$\begin{align} \alpha \left| X_a\right\rangle + \beta \left| X_b \right\rangle = \left| \alpha X_a + \beta X_b \right\rangle. \end{align}$$

A convenient scalar product on this space is the following: $$\begin{align}\left\langle X_a | X_b \right\rangle = \lambda^{-1} \mbox{Tr} (X_a^\dagger X_b),\end{align}$$ ($\lambda$ is what we called $k_D$ for the adjoint representation). Now using (6), we can compute the action of a generator on a state, as follows: $$\begin{align} X_a \left| X_b \right\rangle = \left| X_c \right\rangle \left\langle X_c \left| X_a \right| X_b \right\rangle = \left| X_c \right\rangle [T_a]_{cb}=-if_{acb} \left| X_c\right\rangle if_{abc} \left| X_c \right\rangle= \left| i f_{abc}X_c \right\rangle = \left| [X_a,X_b]\right\rangle.\end{align}$$

Roots

The roots are the weights of the adjoint representation. Because $[H_i,H_j]=0$, the states corresponding to the Cartan generators have zero weight vectors $$\begin{align}H_i \left| H_j\right\rangle= = \left| [H_i, H_j]\right\rangle = 0\end{align}$$

Furthermore, all states in the adjoint representation with zero weight vectors correspond to Cartan generators. Because of (2), the Cartan states are orthonormal, $$\begin{align} \left\langle H_i | H_j \right\rangle = \lambda^{-1} \mbox{Tr} (H_iH_j)=\delta_{ij}\end{align}$$

The other states of the adjoint represenation, those not corresponding to the Cartan generators, have non-zero weight vectors, $\alpha$, with components $\alpha_i$, $$\begin{align} H_i \left| E_\alpha \right\rangle = \alpha_i \left| E_\alpha \right\rangle\end{align}$$ which means that the corresponding generators satisfy $$\begin{align} [H_i,E_\alpha]=\alpha_i E_\alpha \end{align}$$

It will turn out (and we will prove it below) that for the adjoint representation, the non-zero weights uniquely specify the corresponding states, so there is no need for another paorameter (like $x$ in (6) in the arbitrary representation $D$).

Like the $SU(2)$ raising and lowering operators, the $E_\alpha$ are not hermitian. They cannot be hermitian because we can take the adjoint of (12) and get $$\begin{align} [H_i,E_\alpha^\dagger]=-\alpha_i E_\alpha^\dagger\end{align} $$ thus we can take $$\begin{align} E_\alpha^\dagger = E_{-\alpha}.\end{align}$$

This should remind you of the raising and lowering operators $J^+$ and $J^-$ in $SU(2)$.

States corresponding to different weights must be orthogonal, because they have different eigenvalues of at least one of the Cartan generators. Thus we can choose the normalization of the states in the adjoint representation (that is , the generators) so that $$\begin{align} \left\langle E_\alpha | E_\beta \right\rangle = \lambda^{-1} \mbox{Tr}(E_\alpha^\dagger E_\beta)=\delta_{\alpha\beta} (=\prod_i \delta_{\alpha_i\beta_i}).\end{align}$$

The weights $\alpha_i$ are called roots, and the special weight vector $\alpha$ with components $\alpha_i$ is a root vector.

Raising and Lowering

The $E_{\pm \alpha}$ are raising and lowering operators for the weights, because the state $E_{\pm\alpha}\left| \mu,D\right\rangle$ has weight $\mu\pm \alpha$ - $$\begin{align} H_i E_{\pm\alpha} \left| \mu, D\right\rangle = [H_i,E_{\pm \alpha}]\left| \mu, D\right\rangle + E_{\pm \alpha} H_i \left| \mu, D\right\rangle = (\mu \pm \alpha)_i E_{\pm \alpha} \left| \mu, D\right\rangle.\end{align}$$

At this point, we have no notion of positivity, so it doesn't make sense to ask which is rasing and which is lowering. But we will introduce this later.

Equation (16) is true for any representation, but it is particularly important for the adjoint representation. To see why, consider the state $E_\alpha \left| E_{-\alpha} \right\rangle$.

This has weight $\alpha-\alpha=0$, thus it is a linear combination of states corresponding of Cartan generators: $$\begin{align} E_\alpha \left| E_{-\alpha}\right\rangle = \beta_i \left| H_i\right\rangle = \left| \beta_i H_i\right\rangle = \left| \beta \cdot H\right\rangle = \left| [E_\alpha,E_{-\alpha}]\right\rangle.\end{align}$$

But we can actually compute $\beta$- $$\begin{align} \beta_i=\left\langle H_i \left| E_\alpha \right| E_{-\alpha} \right\rangle = \lambda^{-1} \mbox{Tr} (H_i[E_\alpha,E_{-\alpha}])=\lambda^{-1} \mbox{Tr} (E_{-\alpha}[H_i,E_\alpha])=\lambda^{-1}\alpha_i \mbox{Tr} (E_{-\alpha}E_\alpha)=\alpha_i\end{align}$$ (use (8). property of Tr, (12), (15))

Thus $$\begin{align} [E_\alpha,E_{-\alpha}]=\alpha\cdot H.\end{align}$$

This should remind you of the $SU(2)$ commutation relation $[J^+,J^-]=J_3$. It is this analogy that we will exploit to learn about compact Lie groups and their representations.

Lots of $SU(2)$s

For eacg non-zero pair of root vectors, $\pm \alpha$, there is an $SU(2)$ subalgebra of the group, with generators $$\begin{align}E^\pm \equiv |\alpha |^{-1} E_{\pm \alpha},\ E_3\equiv |\alpha|^{-2}\alpha\cdot H.\end{align}$$

To see this, note that $$\begin{align} [E_3, E^\pm] = |\alpha|^{-3} [\alpha\cdot H,E_{\pm\alpha}]=|\alpha|^{-3}\alpha\cdot(\pm\alpha)E_{\pm\alpha}=\pm |\alpha|^{-1}E_{\pm \alpha}=\pm E^\pm \end{align}$$ and from (19) $$\begin{align} [E^+,E^-]=|\alpha|^{-2}[E_\alpha,E_{-\alpha}]=|\alpha|^{-2}\alpha\cdot H=E_3.\end{align}$$

We know on general gounds that the states of each irreducible representation of the full algebra can be deomposed into irreducible rerpesentations of any one of these $SU(2)$ subalgebras. and we already know everything about the irreducible rerpesentations of $SU(2)$. This puts very strong constraints on the nature of the roots. For example, we can now easily prove that the root vectors correspond to unique generators. Suppose the contrary, so that there are two generators, $E_\alpha$ and $E'_\alpha$, We can choose linear combinations of these two so that they are orthogonal in the adjoint representation(I will use the same names for the two generators, assuming that I chose then to be orthogonal from the beginning, just to avoid useless notation) - thus we can write $$\begin{align} \left\langle E_\alpha | E'_\alpha \right\rangle = \lambda^{-1} \mbox{Tr} (E^\dagger_\alpha E'_\alpha)=\lambda^{-1} \mbox{Tr} (E_{-\alpha}E'_\alpha)=0.\end{align}$$

Consider the behavior of the state $\left| E'_\alpha \right\rangle$ under the action of the $SU(2)$ subalgebra (2). $E^-\left| E'_\alpha \right\rangle$ has zero weight vector, and thus it is a linear combination of Carten states. But $$\begin{align}\left\langle H_i \left| E^- \right| E'_\alpha \right\rangle = \lambda^{-1} \mbox{Tr}(H_i[E^-,E'_\alpha])=-\lambda^{-1} Tr(E^- [H_i,E'_\alpha])=-\alpha_i \lambda^{-1} \mbox{Tr} (E'_\alpha E^-)=0\end{align}$$ for all $i$, and thus the coefficient of every Cartan state in $E^-\left| E'_\alpha \right\rangle$ vanishes, and therefore $$\begin{align} E^- \left| E'_\alpha \right\rangle = 0\end{align}$$

But we also have $$\begin{align}E_3\left| E'_\alpha \right\rangle = | \alpha|^{-2} \alpha\cdot H \left| E'_\alpha \right\rangle = \left| E'_\alpha \right\rangle .\end{align}$$

Equations (25) and (26) are inconsistent, because (25) implies that $\left| E'_\alpha\right\rangle$ is the lowest $J_3$ state in an $SU(2)$ representation, and (26) implies that it has $J_3=1$. But the lowest $J_3$ state of an $SU(2)$ representation cannot have positive $J_3$ - the $J_3$ value is always - $j$ for a non-negatice half integer $j$. Thus $E'_\alpha$ cannot exist, and we have shown, as promised, that $\left| E_\alpha \right\rangle$ is uniquely specified by $\alpha$ - no other labels are required.

In fact, if $\alpha$ is a root, then no non-zero multiple of $\alpha$ (except $-\alpha$) is a root. To see this, note that the three states $\left| E_3\right\rangle$ and $\left| E^\pm \right\rangle$ form a spin $1$ representation of the algebra (20), because they form the adjoint representation. Now suppose $k\alpha$ a root for $k\ne \pm 1$. Clearly, $k$ must be a half-integer, because the $E_3$ value of the corresponding state must be a half integer. But if $k$ is an integer not equal to $\pm 1$, the state is part of a representation that contains another state with root $\alpha$, which is impossible, by the argument we just gave. And if $k$ is half an odd integer, then there is a state with root $\alpha/2$, and we can repeat the argument using the $SU(2)$ associated with that generator and get a contradiction in the same way.

Angle between roots

More generally, for any weight $\mu$ of a representation $D$, the $E_3$ value is $$\begin{align} E_3 \left| \mu,x,D\right\rangle = \frac{\alpha\cdot\mu}{\alpha^2}\left| \mu, x, D \right\rangle.\end{align}$$

Because the $E_3$ values must be integers or half integers, $$\begin{align} \frac{2\alpha\cdot \mu}{\alpha^2}\mbox{ is an integer.}\end{align}$$

The general state $\left| \mu, x, D\right\rangle$ can always be written as a linear combination of states transforming according to definite representations of the $SU(2)$ defined by (20). Supoose that the highest spint state that appears in the linear combination is $j$. Then there is some non-negative integer $p$ such that $$\begin{align} (E^+)^p \left| \mu, x, D\right\rangle \ne 0\end{align}$$ with weight $\mu+p\alpha$ is the highest $E_3$ state of the $SU(2)$ spint $j$ representation, so that $$\begin{align} (E^+)^{p+1}\left| \mu,x,D\right\rangle=0.\end{align}$$

The $E_3$ value of the state (29) is $$\begin{align}\frac{\alpha\cdot (\mu+p\alpha)}{\alpha^2}=\frac{\alpha\cdot \mu}{\alpha^2}+p=j.\end{align}$$

Likewise, there is some non-negative integer $q$ such that $$\begin{align} (E^-)^q \left| \mu, x, D \right\rangle \ne 0\end{align}$$ with weight $\mu-q\alpha$ is the lowest $E_3$ state of the $SU(2)$ spint $j$ representation, so that $$\begin{align} (E^-)^{q+1}\left| \mu,x,D\right\rangle = 0\end{align}$$ and the $E_3$ value of the state (32) is $$\begin{align} \frac{\alpha \cdot (\mu-q\alpha)}{\alpha^2}=\frac{\alpha\cdot \mu}{\alpha^2}-q=-j.\end{align}$$

Adding (31) and (34) gives $$\begin{align} \frac{2\alpha\cdot \mu}{\alpha^2}+p-q=0\end{align}$$ or $$\begin{align}\ frac{\alpha\cdot \mu}{\alpha^2}=-\frac{1}{2}(p-q).\end{align}$$

We will refer to (36) as the 'master formula'. The relations (31), (34) and (36) are the basic relations that lead to a geometrical classification of all the compact Lie groups. They don't look like much, but when we augment them with some geometrical intuition, we can exploit them to great effect, as you will see.

Here is a simple first step. Applying (36) to the roots gives a particularly strong constraint, because we can apply it twice for any pair of distinct roots, $\alpha$ and $\beta$. Defining the $SU(2)$ algebra with $E_\alpha$ gives $$\begin{align} \frac{\alpha\cdot \beta}{\alpha^2}=-\frac{1}{2}(p-q).\end{align}$$

Defining the $SU(2)$ algebra with $E_\beta$ gives $$\begin{align} \frac{\beta\cdot\alpha}{\beta^2}=-\frac{1}{2}(p'-q').\end{align}$$

Multiplying these gives a remarkable formula for the angle $\theta_{\alpha\beta}$ between the roots $\alpha$ and $\beta$: $$\begin{align} \cos^2 \theta_{\alpha\beta}=\frac{(\alpha\cdot\beta)^2}{\alpha^2\beta^2}=\frac{(p-q)(p'-q')}{4}.\end{align}$$

What is remarkable about this is that $(p-q)(p'-q')$ must be an integer, so (because it must be non-negative) there are only four interesting possibilities (up to complements) for angles between roots!

$$\begin{align} (p-q)(p'-q') & & \theta_{\alpha\beta}\\ 0 & & 90^\circ \\ 1 & & 60^\circ \mbox{ or } 120^\circ\\ 2 & & 45^\circ \mbox{ or } 135^\circ\\ 3 & & 30^\circ \mbox{ or } 150^\circ\end{align}$$

Simple Roots

When first non-zero component is positive, the weight called positive.

Simple roots are positive roots that cannot be written as a sum of other positive roots.

Simple root can construct whole Lie algebra.

The Cartan Matrix

There is a useful way of keeping track of the integers $p^i$ and $q^i$ associated with the action of a simple root $\alpha^i$ on an state $\left| \phi\right\rangle$ for a positive root $\phi$ that eliminates the need for tedious geometrical calculations. The idea is to label the roots directly by their $q^i-p^i$ values. The $q^i-p^i$ of any weight, $\mu$, is simply twice its $E_3$ value, where $E_3$ is the Cartan generator of the $SU(2)$ associated with the simple root $\alpha^i$, because $$2E_3\left| \mu \right\rangle = \frac{2H\cdot \alpha^i}{\alpha^{i2}}\left| \mu \right\rangle = \frac{2\mu \cdot \alpha^i}{\alpha^{i2}}\left| \mu \right\rangle = (q^i-p^i)\left| \mu \right\rangle.$$

Because the $\alpha^i$ are complete and linearly independent, the $q^i-p^i$ values for the weights contain the same information as the values of the components of the weight vector, so we can use them to label the weights. The advantage of doing under the $SU(2)$s generated by the simple roots.

Since a positive root, $\phi$, can be written as $\phi=\sum_j k_ja^j$, the master formula can be written as $$q^i-p^i=\frac{2\phi\cdot \alpha^i}{\alpha^{i2}}=\sum_j k_j \frac{2\alpha^j\cdot \alpha^i}{\alpha^{i2}}=\sum_j k_j A_{ji}$$ where $A$ is the Cartan matrix $$A_{ji} \equiv \frac{2\alpha^j \cdot \alpha^i}{\alpha^{i2}}$$

The matrix element $A_{ji}$ is the $q-p$ value for the simple root $\alpha^i$ acting on the state $\left| \alpha^j\right\rangle$, twice the $E_3$ value, thus all the entries of $A$ are integers. THe diagonal entries are all equal to 2. This is obious from the definition, and also from the fact that the simple roots have $E_3=1$ because the $SU(2)$ generators themselves transform like the adjoint, spint $1$ representation. The off-diagonal entries, all 0, -1, -2, or -3, record the angles between simple roots and their relative lengths - the same information as the Dynkin diagram, and they tell us how the simple roots fit into representations of the $SU(2)$s associated with the other simple roots. It is easy to see that the Cartan matrix is invertivle because the $\alpha^j$ are complete and linearly independent. Note that the $j$th row of the Cartan matrix consists of the $q_i - p_i$ values of the simple root $\alpha^j$.

Fundamental Weights

Suppose that the simple roots of some simple Lie algebra are $\alpha^j$ for $j=1$ to $m$. The highest weight, $\mu$, of an arbitrary irreducible representation, $D$ has the property that $\mu+\phi$ is not a weight for any positive root, $\phi$. From the preceding discussion, it is clearly sufficient that $\mu+\alpha^j$ not be a weight in the representation for any $j$, beacuse then $$\begin{align}E_{\alpha^j}\left| \mu \right\rangle = 0\ \forall j\end{align}$$ which implies that all positive roots annihilate the state, because any positive root can be expressed as a multiple commutator of simple roots. We will see soon that this is an if and only if statement, because an entire irreducible representation can be constructed by applying lowering operators to any state with this property. So if (45) is true, then $\mu$ is the highest weight of an irreducible representation. This means that for every $E_{\alpha^j}$ acting on $\left| \mu \right\rangle$, $p=0$, and thus $$\begin{align}\frac{2\alpha^j \cdot \mu}{\alpha^{j2}}=l^j\end{align}$$ where the $l^j$ are non-negative integers. Beacuse the $\alpha^j$s are linearly independent, the integers $l^j$ completly determine $\mu$. Every set of $l^j$ gives a $\mu$ which is the highest weight of some irreducible representation.

Thus the irreducible representations of a rank $m$ simple Lie algebra can be labeled by a set of $m$ non-negative integers, $l^j$. These integers are sometimes called the Dynkin coefficients.

It is useful to consider the weight vectors, $\mu^j$, satisfying $$\frac{2\alpha^j\cdot \mu^k}{\alpha^{j2}}=\delta_{jk}$$

Every highest weight, $\mu$, can be uniquely written as $$\begin{align}\mu = \sum^m_{j=1} l^j \mu^j\end{align}$$

We can therefore build the representation with highest weight $\mu$ by constructing a tensor product of $l^1$ representation of highest weight $\mu^1$, $l^2$ with highest weight $\mu^2$, and so on. This representation will generally be reducible, but we can always then pick out the representation $\mu$ by applying lowering operators to $\left| \mu \right\rangle$.

The vectors $\mu^j$ are called the fundamental weights and the $m$ irreducible representation that have these as highest weights are called the fundamental representations. We will somtimes denote them by $D^j$. Remember that the superscripts are just labels. The vectors also have vector indices. It's potentially confusing because both run from $1$ to $m$.

There is more to say about the Dynkin coefficients. Since the fundamental weights form a complete set, we can expand any weight of any representation in terms of them, as in (47). Then we can run the argument backwards and get (46) which implies that for a general $\mu$, $$l^j=q^j-p^j$$ that is $l^j$ is the $q^j-p^j$ value for the simple root $\alpha^j$. Thus the matrix elements of the vectors we were manipulating in constructing the positive roots of various algebras were just the Dynkin coefficients of the roots (though of course, for a general weight or root, the Dynkin coefficients will not necessarily be positive). In particular, the highest box in the construction is just the highest weight in the adjoint representation. The rows of the Cartan matrix are the Dynkin coefficients of the simple roots.

Constructing the states

Suppose we have an irreducible representation with highest weight $\mu$. Then there will be one or more states, $\left|\mu\right\rangle$, and all the states in the representation can clearly be written in the form $$\begin{align} E_{\phi_1}E_{\phi_2} \cdots E_{\phi_n} \left| \mu \right\rangle\end{align}$$ where the $\phi_i$ are any roots. But any such with a positive $\phi$ can be dropped, because if a positive $\phi$ appears, we can move it to the right using the commutation relations until it acts on $\left| \mu \right\rangle$, which gives zero. So we can take all the $\phi_i$ in (48) to be negative.

But since every negative root is a sum over the simple roots with non-positive integer coefficients, we can take only states of the form $$\begin{align} E_{-\alpha^{\beta_1}} E_{-\alpha^{\beta_2}}\cdots E_{-\alpha^{\beta_n}}\left| \mu \right\rangle\end{align}$$ where $\alpha^{\beta_i}$ are simple roots without losing any of the states.

Now it is clear if it wasn't before that the highest weight state is unique. If there were two states with the highest weight, the representation would break up into two independent sets of the form (49) above, and it would not be irreducible.

In addition, any state that can be obtained in only one way by the action of simpel lowering operators on $\left| \mu \right\rangle$ is unique. This shows that all the states in the $(0,1)$ representation are unique.

This technique of starting with the highest weight state and building the other states by the action of lowering operators actually allows us to build the representation explicitly. We can compute the norms of the states and the action of the raising and lowering operators on the states is built in by construction. We will not actually ues this to build the representation matrices, because there are ususally easier ways, but we could do it if we wanted to. The point is that all we need to do to consturct the representation explicitly is to undertand the structure of the space sppaned by the kets in (49).

The first thing to notice is that two states produced by a different set of $E_{-\alpha^j}$s in (49) are orthogonal. This is clear because the linear independence of the $\alpha^j$s implies that the two states have different weights. But then they are orthogonal because they are eigenstates of the hermitian Carten generators with different eigenvalues.

The norm of any state can be easily computed using the commutatiuon relation of the simple root generators. The norm has the form $$\begin{align} \left\langle \mu \left| E_{\alpha^{\beta_n}}\cdots E_{\alpha^{\beta_2}}E_{\alpha^{\beta_1}}E_{-\alpha^{\beta_1}}E_{-\alpha^{\beta_2}}\cdots E_{-\alpha^{\beta_n}}\right| \mu \right\rangle\end{align}$$

The rasing operators, starting with $E_{\alpha^{\beta_1}}$ can be successively moved to the right until they annihilate $\left| \mu \right\rangle$. On the way, they commute with the lowering operators, unless the roots are a positive and negative pair, in which case the commutator is a linear combination of Cartan generators, which can be evaluated in terms of $\mu$ and the $\alpha^{\beta_j}$s, leaving a matrix element with one less raising and lowering operator. This is tedious, but completely systematic procedure - you can easily write a recursive program to do it explicitly.

If two states in (49) have the same set of $\beta_j$s but in different order, then they have tha same weight. In this case, we must compute thier norms ans matrix elements to understand the structure of the space. The matrix elements can be calculated in the same way as the norm in (50) because they have a similar structure: $$\begin{align} \left\langle \mu \left| E_{\alpha^{\gamma_n}}\cdots E_{\alpha^{\gamma_2}}E_{\alpha^{\gamma_1}}E_{-\alpha^{\beta_1}}E_{-\alpha^{\beta_2}}\cdots E_{-\alpha^{\beta_n}}\right| \mu \right\rangle\end{align}$$ where $\gamma_j$ and $\beta_j$ are two lists of the same set of integers but in a different order. Again, simply moving the raising operators to the right gives a completely systematic way of computing the matrix element.

One you have the norms and the matrix elements, you can construct an orthonormal absis in the Hilbert space using the Gram-Schmidt procedure.

Reference

Matthew Foster - Lie group lecture note 3,4,5

Brian C. Hall - Lie Groups, Lie Algebras, and Representations

Georgi - Lie Algebras in Particle Physics Chapter 6, 8