su(3) Root and Weight Diagrams

From su(2) to su(3): The Leap to Two Dimensions

Recall that the rank of a Lie algebra is the dimension of its Cartan subalgebra—the maximal number of commuting generators we can find. For su(2), the rank is one, with the Cartan subalgebra typically chosen as $\mathfrak{h}=\text{span}\{J_3\}$. All weights were single numbers (the eigenvalues of $J_3$), so the diagrams were one-dimensional.

The Lie algebra su(3) consists of all $3\times 3$ traceless anti-hermitian matrices. To find its rank, we look for the maximal number of commuting generators. A natural choice is the set of diagonal matrices. A general traceless diagonal $3\times 3$ matrix has the form $\text{diag}(a,b,-a-b)$, which depends on two independent parameters. Therefore, the Cartan subalgebra of su(3) is two-dimensional, and su(3) is a rank-2 algebra.

This means that every weight and root will be a pair of numbers—an ordered pair of eigenvalues corresponding to the two generators of the Cartan subalgebra. Consequently, our diagrams will be plotted on a 2D plane.

The following video introduces this key idea and shows how the two-parameter nature of the diagonal SU(3) matrices leads to two-dimensional weight diagrams.

The Weight Diagram of the Fundamental Representation

Let's begin by constructing the weight diagram for the simplest non-trivial representation of SU(3): the fundamental representation, where the matrices simply act on vectors in ${\mathbb{C}}^3$. This is often called the 3 representation.

The basis vectors are the standard basis of ${\mathbb{C}}^3$: $e_1=(1,0,0{)}^T$, $e_2=(0,1,0{)}^T$, and $e_3=(0,0,1{)}^T$. To find their weights, we act on them with a generic element of the maximal torus (the diagonal subgroup of SU(3)) and find the eigenvalues. As shown in the video, this matrix is:

The action on the basis vectors is straightforward:

• $D({\theta }_1,{\theta }_2)e_1=e^{i{\theta }_1}{e}_1=e^{i(1\cdot {\theta }_1+0\cdot {\theta }_2)}{e}_1$. The weight is $(k,l)=(1,0)$.
• $D({\theta }_1,{\theta }_2)e_2=e^{i{\theta }_2}{e}_2=e^{i(0\cdot {\theta }_1+1\cdot {\theta }_2)}{e}_2$. The weight is $(k,l)=(0,1)$.
• $D({\theta }_1,{\theta }_2)e_3=e^{-i({\theta }_1+{\theta }_2)}{e}_3=e^{i(-1\cdot {\theta }_1+-1\cdot {\theta }_2)}{e}_3$. The weight is $(k,l)=(-1,-1)$.

Plotting these three weights on the 120° grid introduced in the video yields an equilateral triangle, the characteristic shape of the fundamental representation.

The Root Diagram of su(3)

Now we turn to the central task of finding the structure of the su(3) algebra itself by constructing its root diagram. Recall that the roots of an algebra are simply the non-zero weights of its adjoint representation.

The vector space for the adjoint representation is the 8-dimensional Lie algebra su(3) itself. Our goal is to find the eigenvalues of the adjoint action of the Cartan subalgebra elements, ${\text{ad}}_H$, on the other basis vectors of the algebra.

• The vector space: The algebra sl(3, \mathbb{C}) (the complexification of su(3)).
• The Cartan elements $H$: Diagonal traceless matrices, like $H_{\theta }=\text{diag}({\theta }_1,{\theta }_2,{\theta }_3)$ with ${\theta }_1+{\theta }_2+{\theta }_3=0$.
• The other basis vectors: The six off-diagonal matrices $E_{ij}$ (with a 1 in the $i,j$-th position and 0s elsewhere, $i\ne j$). These will be our root vectors.
• The eigenvalue equation: ${\text{ad}}_{H_{\theta }}(E_{ij})=[H_{\theta },E_{ij}]=\alpha (H_{\theta })E_{ij}$.

A direct calculation of the commutator $H_{\theta }{E}_{ij}-E_{ij}{H}_{\theta }$ shows that:

The roots are therefore the linear functions ${\alpha }_{ij}(H_{\theta })={\theta }_i-{\theta }_j$. To get their coordinates, we use the constraint ${\theta }_3=-{\theta }_1-{\theta }_2$. This gives us six non-zero roots:

• ${\alpha }_{12}(H_{\theta })={\theta }_1-{\theta }_2$. Weight vector: $(1,-1)$.
• ${\alpha }_{21}(H_{\theta })={\theta }_2-{\theta }_1$. Weight vector: $(-1,1)$.
• ${\alpha }_{13}(H_{\theta })={\theta }_1-{\theta }_3=2{\theta }_1+{\theta }_2$. Weight vector: $(2,1)$.
• ${\alpha }_{31}(H_{\theta })={\theta }_3-{\theta }_1=-2{\theta }_1-{\theta }_2$. Weight vector: $(-2,-1)$.
• ${\alpha }_{23}(H_{\theta })={\theta }_2-{\theta }_3={\theta }_1+2{\theta }_2$. Weight vector: $(1,2)$.
• ${\alpha }_{32}(H_{\theta })={\theta }_3-{\theta }_2=-{\theta }_1-2{\theta }_2$. Weight vector: $(-1,-2)$.

What about the weight of the Cartan generators themselves? Since they commute with any other Cartan generator $H$, their associated weight is $(0,0)$. As the Cartan subalgebra is two-dimensional (e.g., spanned by $H_1=\text{diag}(1,0,-1)$ and $H_2=\text{diag}(0,1,-1)$), the zero-weight space has a dimension of 2.

Plotting these six non-zero roots and the two-dimensional zero weight at the origin gives the root diagram of su(3).

The resulting hexagonal pattern is the defining structure of the su(3) algebra.

Connecting to Physics: Isospin and Hypercharge

In the context of particle physics, the two commuting generators are chosen to be the third component of isospin, $T_3$, and hypercharge, $Y$.

The root vectors are combinations of the other Gell-Mann matrices. The roots themselves become vectors in the $(T_3,Y)$ plane. For instance, the root corresponding to the $u_{+}$ raising operator is found by calculating its commutator with a general Cartan element $H=a{T}_3+bY$.

If you plot the root vectors $\alpha (H)$ derived in this reading—such as $(1,0)$ for $t_{+}$, $(1/2,1)$ for $v_{+}$, and $(-1/2,1)$ for $u_{+}$, along with their negatives—you get the exact same hexagonal structure, just plotted on conventional orthogonal axes. This is the diagram you've likely seen for the meson or baryon octets.

Another Example: The 6 Representation

To complete our exploration, let's construct the weight diagram for another representation. A useful example is the symmetric tensor product of two fundamental representations, ${\text{Sym}}^2({\mathbb{C}}^3)$. This is a 6-dimensional representation, often called the 6.

Its basis vectors can be thought of as quadratic polynomials in the fundamental basis vectors: $\{e_1^2,e_2^2,e_3^2,e_1{e}_2,e_1{e}_3,e_2{e}_3\}$. The weight of a product state is the sum of the individual weights. Using the weights $(1,0),(0,1),(-1,-1)$ for $e_1,e_2,e_3$, we find the weights of the 6:

• $e_1^2\to (1,0)+(1,0)=(2,0)$
• $e_2^2\to (0,1)+(0,1)=(0,2)$
• $e_3^2\to (-1,-1)+(-1,-1)=(-2,-2)$
• $e_1{e}_2\to (1,0)+(0,1)=(1,1)$
• $e_1{e}_3\to (1,0)+(-1,-1)=(0,-1)$
• $e_2{e}_3\to (0,1)+(-1,-1)=(-1,0)$

Plotting these six weights on our 120° grid reveals another hexagonal diagram.

Conclusion

In this lesson, we have successfully extended the geometric framework of roots and weights from su(2) to su(3). By moving from a rank-1 to a rank-2 algebra, we have seen the emergence of beautiful and highly symmetric two-dimensional patterns.

Key Takeaways:

• su(3) is a rank-2 Lie algebra, meaning its Cartan subalgebra is two-dimensional. Its roots and weights are therefore represented by vectors in a 2D plane.
• The root diagram of su(3) consists of six non-zero root vectors forming a hexagon, and a two-dimensional zero-weight space at the origin. This diagram is the geometric signature of the algebra's structure.
• The weight diagrams of su(3) representations, such as the fundamental 3 and the symmetric 6, also form symmetric geometric patterns (triangles, hexagons) in this 2D plane.
• The choice of axes for plotting is a convention. Using axes at 120° reveals the symmetries most clearly, while the orthogonal $(T_3,Y)$ axes used in particle physics are physically motivated but represent the same underlying structure.

These diagrams are far more than just pictures. They are a complete encoding of the representation's structure. In our next lesson, we will learn how to classify all irreducible representations of su(3) using the "theorem of the highest weight," and explore the properties of the most important low-dimensional representations, including the 3, its conjugate 3-bar, and the adjoint 8, which form the basis of the Eightfold Way.