Tensor representations of $SU(N)$

Section 5.5 Tensor representations of $SU(N)$

Objectives

You should be able to:

Recognize tensors as objects that transform according to representations of $SU(N)\text{.}$
Determine the dimension of the irreducible representations of $SU(N)$ by looking at the corresponding tensors.
Show that the defining representation of $SU(2)$ is pseudo-real.
List the irreducible representations of $SU(3)$ as tensor representations.

In Section 5.1 we studied representations of

$SO(3)$ by looking at how tensors transform under rotations. In this section we use a similar approach to study tensor representations of

$SU(N)\text{.}$

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Subsection 5.5.1 Tensor representations of $SO(N)$

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Let us start with a brief recap of the construction of tensor representations of

$SO(3)$ in Section 5.1, which applies just as well to tensor representations of

$SO(N)\text{.}$ The idea was to construct representations of

$SO(N)$ by constructing objects that transform according to these representations. These objects were called “tensors”, and denoted by a letter with many indices, for instance:

$T_{ijk}\text{.}$ The number of indices (called the “rank” of the tensor) tells us how these objects transform under a rotation

$R \in SO(3)\text{:}$

$\begin{equation} T'_{abc} = \sum_{i,j,k=1}^3 R_{ai} R_{bj} R_{ck} T_{ijk}.\label{equation-trans-tensors-son}\tag{5.5.1} \end{equation}$

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If we were to put all independent components of

$T_{ijk}$ in a column vector, the transformation rule above would define a matrix representation of

$SO(3)\text{,}$ which in this case would be the

$27$ -dimensional tensor product representation

$\mathbf{27} = \mathbf{3} \otimes \mathbf{3} \otimes \mathbf{3}\text{.}$

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We also noticed that symmetry properties of the objects

$T_{ijk}$ under permutations of indices are preserved by the transformation rules (for instance, symmetric tensors are mapped to symmetric tensors). Thus restricting to tensors with specific symmetry properties defines proper invariant subspaces, or subrepresentations. In this way we can build all kinds of tensor representations of

$SO(3)\text{.}$ We claimed in Section 5.1 that for

$SO(3)\text{,}$ all irreducible representations correspond to symmetric traceless tensors, which are indexed by a non-negative integer

$j$ (the rank of the tensor, or the number of indices), and have dimension

$2j + 1\text{.}$

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The case of

$SO(N)$ is very similar. Tensors are still defined by how they transform under

$n$ -dimensional rotations, as in (5.5.1), but with

$R \in SO(N)\text{.}$ For instance, a tensor

$T_{ijk}\text{,}$ as in (5.5.1), transforms in the

$N^3$ -dimensional tensor product representation

$\mathbf{N}^3 = \mathbf{N} \otimes \mathbf{N} \otimes \mathbf{N}\text{.}$ As for

$SO(3)\text{,}$ tensors with particular symmetry properties define subrepresentations of these tensor product representations. However, for

$SO(N)$ irreducible representations are not simply given by symmetric traceless tensors anymore. The study of irreducible representations of

$SO(N)$ in terms of symmetry properties of tensors uses Young tableaux, which we will not discuss in this class (but feel free to research them if you are interested!). The idea is to look at the general tensor product representation

$\mathbf{N} \otimes \cdots \otimes \mathbf{N}\text{,}$ which is generally reducible, and look at how it decomposes as a direct sum of irreducible representations. This involves the study of Clebsch-Gordan coefficients, which is another interesting topic that we will not study in this class.

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Let us also remark that we can identify the special representation known as the adjoint representation (which is the exponentiated version of the adjoint representation for the Lie algebra) in terms of tensors for

$SO(N)$ in general. Rank two anti-symmetric tensors

$A_{ij} = - A_{ji}$ transform under the adjoint representation. Indeed, a rank two anti-symmetric tensor has

$\frac{1}{2} N (N-1)$ independent components, which is the dimension of the adjoint representation of

$SO(N)$ (also the dimension of the Lie group).

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Subsection 5.5.2 Tensor representations of $SU(N)$

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Let us now move on to the study of tensor representations of

$SU(N)\text{.}$ We now want to contruct tensors that transform in certain ways under special unitary transformations

$U \in SU(N)\text{.}$ The key difference with

$SO(N)$ is that the matrices

$U$ are unitary, and, in particular, complex-valued. So the transformations

$U$ and

$U^\dagger$ are different transformations. So we should really construct objects that transform according to products of

$U$ s and

$U^\dagger$ s. More precisely, what we are saying here is that the defining

$\mathbf{N}$ representation of

$SU(N)$ is complex-valued, so we can construct tensor products of

$\mathbf{N}$ with itself, but also with its complex conjugate representation

$\overline{\mathbf{N}}\text{.}$

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There is a neat way of keeping track of objects that transform according to

$U$ and

$U^\dagger\text{.}$ We will use upper and lower indices. An object with upper indices will transform according to products of

$U$ s, while an object with lower indices will transform according to products of

$U^\dagger$ s. We can of course also have objects with both lower and upper indices, which will transform accordingly. Concretely, what we are doing is identifying the operation of complex conjugation as raising or lowering an index:

$\begin{equation*} \psi_i := (\psi^*)^i. \end{equation*}$

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This is of course just a notational convention, but it is very useful to keep track of the transformation properties of tensors.

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We then denote the components of a unitary matrix

$U$ by

${U^i}_a\text{.}$ Its transpose matrix

$U^\dagger$ then have components

${(U^\dagger)^a}_i\text{.}$ With this convention the upshot is that to construct objects with appropriate transformation properties under unitary transformations, we should always sum only over one index that is upper and one that is lower. This follows because of the unitary property that

$U^\dagger U = I\text{,}$ or, in index notation,

$\begin{equation*} \sum_{j=1}^N {(U^\dagger)^b}_j {U^j}_a = \delta^b_a. \end{equation*}$

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As an example, a tensor

$T^{ij}_k$ transforms as

$\begin{equation*} (T')^{ij}_k = \sum_{a, b, c =1}^N {U^i}_a {U^j}_b {(U^\dagger)^c}_k T^{a b}_c. \end{equation*}$

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It thus defines the tensor product representation

$\mathbf{N} \otimes \mathbf{N} \otimes \overline{\mathbf{N}}\text{.}$

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The task now is to identify proper invariant subspaces, or subrepresentations, of these tensor representations. Using the exact same argument as for

$SO(N)\text{,}$ it is straightforward to show that symmetry properties of tensor according to permutations of upper indices and lower indices separately are preserved by the transformation rule. For instance, a tensor that is fully symmetric under permutations of its upper indices is mapped to a tensor that is also fully symmetric under permutations of its upper indices. However, we cannot exchange upper and lower indices; symmetries under such permutations would not be preserved by the transformation rule.

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Is there an analog of the trace that was preserved by special orthogonal transformations? Yes, but it must now involve summing over an upper and lower index. Indeed, one sees that, for instance,

$\begin{align*} \sum_{i=1}^N (T')^{ji}_i =\amp \sum_{i=1}^N \sum_{a,b,c=1}^N {U^j}_a {U^i}_b {(U^\dagger)^c}_i T^{a b}_c\\ =\amp \sum_{a,b,c=1}^N {U^j}_a \left( \sum_{i=1}^N {U^i}_b {(U^\dagger)^c}_i \right) T^{a b}_c\\ =\amp \sum_{a,b,c=1}^N {U^j}_a \left( \delta_b^c\right) T^{a b}_c\\ =\amp \sum_{a=1}^N {U^j}_a \left(\sum_{b=1}^N T^{a b}_b \right). \end{align*}$

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Therefore, the object

$\sum_{i=1}^N T^{j i}_i$ transforms like a tensor with one single upper index (i.e. in the

$\mathbf{N}$ defining representation of

$SU(N)$ ). So summing over one upper and one lower index for any tensor creates a subrepresentation. We call this operation taking the trace of a tensor.

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Since taking the trace always defines a subrepresentation, to understand irreducible representations of

$SU(N)$ we know that we should look for traceless tensors that have appropriate symmetry properties upstairs and downstairs. In general, the study of irreducible representations from this point of view involves Young tableaux, as already mentioned for

$SO(N)\text{.}$ The idea is similar, where we now look at general tensor products

$\mathbf{N} \otimes \cdots \mathbf{N} \otimes \overline{\mathbf{N}} \otimes \cdots \otimes \overline{\mathbf{N}},$ and study their decompositions as direct sums of irreducible representations using Clebsch-Gordan coefficients.

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We will not discuss Young tableaux in this course, but let us at least enumerate the first few non-trivial tensor representations of

$SU(N)\text{:}$

Tensor	Symmetry property	Dimension of the representation
$T^i$	-	$N$
$T^{ij} = - T^{ji}$	Anti-symmetric	$\frac{1}{2}N (N-1)$
$T^{ij} = T^{ji}$	Symmetric	$\frac{1}{2}N(N+1)$
$T^i_j$	Traceless ( $\sum_{i=1}^N T^i_i = 0$ )	$N^2-1$

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Table 5.5.1. The first few tensor representations of

$SU(N)$

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We note that the

$N$ -dimensional,

$\frac{1}{2}N (N-1)$ -dimensional and

$\frac{1}{2}N (N+1)$ -dimensional representations also have complex conjugate representations, corresponding to similar tensors but with lower indices. We also note that the representation given by the traceless tensor

$T^i_j$ is the adjoint representation of

$SU(N)\text{,}$ which is equivalent to its complex conjugate.

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Subsection 5.5.3 Tensor representations of $SU(2)$

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We already know the irreducible representations of

$SU(2)\text{,}$ since they are in one-to-one correspondence, through exponentiation, with the irreducible representations of its Lie algebra

$\mathfrak{su}(2)\text{.}$ We already constructed those using the highest weight construction, and obtained an infinite family indexed by a non-negative half-integer

$j\text{,}$ with dimensions

$2j+1\text{.}$ Let us see how those arise from the point of view of tensor representations.

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Just as for

$SO(3)\text{,}$ it turns that

$SU(2)$ is very special. In the case of

$SO(3)\text{,}$ what was particular about it is that irreducible representations were all given by symmetric traceless tensors. In the case of

$SU(2)\text{,}$ it turns out that all irreducible representations are constructed from symmetric tensors with only upper indices. In other words, for

$SU(2)\text{,}$ we do not need tensors with lower indices. We will see in a second why.

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So all irreducible representations of

$SU(2)$ correspond to fully symmetric tensors

$T^{i_1 \cdots i_n}\text{.}$ So we obtain an infinite family of irreducible representations, indexed by a non-negative integer

$n\text{.}$ What is the dimension of such a representation? Recall that the indices

$i_1$ to

$i_n$ can only take values

$1$ or

$2\text{,}$ since we are considering

$SU(2)\text{,}$ and its defining representation is two-dimensional. Thus the independent components of a fully symmetric tensor

$T^{i_1 \cdots i_n}$ are:

$\begin{equation*} T^{1 1 \cdots 1}, \qquad T^{2 1 \cdots 1}, \qquad T^{2 2 1 \cdots 1}, \qquad \ldots \qquad T^{2 2 \cdots 2}. \end{equation*}$

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There are exactly

$n+1$ of them. So we obtain an infinite family of irreducible representations indexed by a non-negative integer

$n$ with dimensions

$n+1\text{.}$ This is exactly what we obtained before with the highest weight construction, if we let

$n= 2 j$ with

$j$ a half-integer. Great!

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But why is it that we only have to consider tensors with upper indices for

$SU(2)\text{?}$ Concretely, this means that we are only considering tensor products of the defining representation

$\mathbf{2}$ with itself: we do not consider tensor products with its complex conjugate

$\overline{\mathbf{2}}\text{.}$ Why?

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Well, let us look at the defining representation

$\mathbf{2}$ more closely. Recall that it is given by the Pauli matrices (4.4.1):

$\begin{equation*} T_1 = \frac{1}{2} \begin{pmatrix} 0 \amp 1 \\ 1 \amp 0 \end{pmatrix},\qquad T_2 = \frac{1}{2} \begin{pmatrix} 0 \amp -i \\ i \amp 0 \end{pmatrix},\qquad T_3 = \frac{1}{2} \begin{pmatrix} 1 \amp 0 \\ 0 \amp -1 \end{pmatrix}. \end{equation*}$

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The question is: is the group representation obtained by exponentiating the Pauli matrices complex, real, or pseudo-real (see Section 2.12)? If it was real or pseudo-real, this would mean that it is equivalent to its complex conjugate representation, which would justify only looking at tensor products of

$\mathbf{2}$ with itself (since

$\overline{\mathbf{2}}$ would be equivalent to

$\mathbf{2}$ ).

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Lemma 5.5.2. The defining representation of $SU(2)$ is pseudo-real.

The defining, two-dimensional, representation of $SU(2)$ is pseudo-real.

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Proof.

Recall that a representation is real or pseudo-real if it is equivalent to its complex conjugate representation. That is, there exists an invertible matrix $S$ such that

\begin{equation*} T(g)^* = S T(g) S^{-1} \end{equation*}

for all $g \in SU(2)\text{.}$ It is real if $S$ is symmetric, and pseudo-real if $S$ is anti-symmetric.

The defining representation of $SU(2)$ is obtained by exponentiating a general real linear combination of Pauli matrices:

\begin{equation*} T(g) = \exp(i \sum_{k=1}^3 \theta_k T_k ). \end{equation*}

Then its complex conjugate is:

\begin{equation*} T(g)^* = \exp(- i \sum_{k=1}^3 \theta_k T_k^* ). \end{equation*}

Thus,

\begin{equation*} T(g)^* = S T(g) S^{-1} \end{equation*}

for some $S$ if and only if

\begin{equation*} T_k^* = - S T_k S^{-1} \end{equation*}

for the Pauli matrices $T_k\text{,}$ $k=1,2,3\text{.}$

One can check using matrix multiplication that Pauli matrices satisfy the property that

\begin{equation*} T_i T_j = - T_j T_i \qquad \text{for } i \neq j, \qquad T_i^2 = I. \end{equation*}

Since $T_1$ and $T_3$ are real-valued, it thus follows that

\begin{equation*} T_2 T_1^* T_2 = T_2 T_1 T_2 = - T_1, \qquad T_2 T_3^* T_2 = T_2 T_3 T_2 = - T_3. \end{equation*}

As for $T_2\text{,}$ we see that $T_2^* = - T_2\text{,}$ and thus

\begin{equation*} T_2 T_2^* T_2 = - T_2 T_2 T_2 = - T_2. \end{equation*}

Thus, if we set $S=T_2\text{,}$ and hence $S^{-1} = T_2\text{,}$ we get that $T_k^* = - S T_k S^{-1},$ as required. Finally, since $T_2$ is anti-symmetric, we conclude that the defining representation of $SU(2)$ is pseudo-real.

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This is the key property that makes

$SU(2)$ special. Since its defining representation is pseudo-real, that is,

$\mathbf{2} \cong \overline{\mathbf{2}},$ we do not need to consider tensor products involving both

$\mathbf{2}$ and

$\overline{\mathbf{2}}\text{,}$ since

$\begin{equation*} \mathbf{2} \otimes \cdots \otimes \mathbf{2} \otimes \overline{\mathbf{2}} \otimes \cdots \otimes \overline{\mathbf{2}} \cong \mathbf{2} \otimes \cdots \otimes \mathbf{2}. \end{equation*}$

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Concretely, what this means is that we only need to consider tensors with upper indices for

$SU(2)\text{.}$

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Note that this is certainly not true for

$SU(N)$ in general though. The defining representation is pseudo-real only for

$SU(2)\text{.}$ For

$N>2\text{,}$ the defining representation is complex.

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Subsection 5.5.4 Tensor representations of $SU(3)$

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Let us now look at

$SU(3)\text{.}$ In this case, the defining representation

$\mathbf{3}$ is complex, and we need to consider general tensor products

$\mathbf{3} \otimes \cdots \mathbf{3} \otimes \overline{\mathbf{3}} \otimes \cdots \otimes \overline{\mathbf{3}}.$

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For

$SU(3)\text{,}$ the statement is that all irreducible representations are given by traceless tensors that are fully symmetric upstairs an downstairs individually. That is, tensors of the form

$T^{i_1 \cdots i_p}_{k_1 \cdots k_q}$ that are traceless and fully symmetric under permutations of the

$i_a$ and the

$k_b$ separately.

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Thus we conclude that irreducible representations of

$SU(3)$ are indexed by two non-negative integers

$(p,q)\text{,}$ corresponding to the number of upper and lower indices respectively. What is the dimension of such a representation?

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Lemma 5.5.3. Dimension of irreducible representations of $SU(3)$ .

A traceless tensor $T^{i_1 \cdots i_p}_{k_1 \cdots k_q}$ that is fully symmetric under permutations of the $i_a$ and the $k_b$ separately has

$\begin{equation*} \frac{1}{2}(p+1)(q+1)(p+q+2) \end{equation*}$

independent components. Thus, the dimension of the $(p,q)$ irreducible representation of $SU(3)$ is $\frac{1}{2}(p+1)(q+1)(p+q+2)\text{.}$

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Proof.

A tensor $T^{i_1 \cdots i_p}_{k_1 \cdots k_q}$ that is fully symmetric under permutations of the $i_a$ and the $k_b$ separately has

\begin{equation} \left( \frac{1}{2}(p+1)(p+2) \right) \left( \frac{1}{2}(q+1)(q+2) \right)\label{equation-number-fully}\tag{5.5.2} \end{equation}

independent components. We then need to impose the tracelessness condition, which will impose a number of constraints. Let us count the number of constraints, which are independent, and then substract this number from the number of above to get the number of independent components of a traceless tensor that is fully symmetric upstairs and downstairs.

To impose that a tensor is traceless, we need to sum over one upper and one lower index. We can choose any of those, since the tensor is fully symmetric upstairs and downstairs. Thus we need to impose the conditions

\begin{equation*} \sum_{a=1}^3 T^{a i_2 \cdots i_p}_{a k_2 \cdots k_p} = 0. \end{equation*}

How many constraints does this give? Well, the indices $i_2, \ldots, i_p$ and $k_2,\ldots, k_p$ can take any values here. So the number of constraints is the number of independent components of a tensor with $p-1$ upper indices and $q-1$ lower indices that is fully symmetric upstairs and downstairs. By (5.5.2), this is

\begin{equation} \left( \frac{1}{2}(p)(p+1) \right) \left( \frac{1}{2}(q)(q+1) \right).\label{equation-number-constraints}\tag{5.5.3} \end{equation}

Therefore, the number of independent components of a traceless tensor $T^{i_1 \cdots i_p}_{k_1 \cdots k_q}$ that is fully symmetric under permutations of the $i_a$ and the $k_b$ separately is (5.5.2) minus (5.5.3), that is,

\begin{equation*} \frac{1}{4} (p+1)(p+2)(q+1)(q+2) - \frac{1}{4} p (p+1)q(q+1) = \frac{1}{2}(p+1)(q+1)(p+q+2). \end{equation*}

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To end this section, let us list the first few irreducible representations of

$SU(3)$ in terms of the integers

$(p,q)\text{.}$ We write the representation in terms of dimension, and give its name when appropriate.

$(p,q)$	Representation (name)
$(0,0)$	$\mathbf{1}$ (trivial)
$(1,0)$	$\mathbf{3}$ (fundamental)
$(0,1)$	$\overline{\mathbf{3}}$ (anti-fundamental)
$(2,0)$	$\mathbf{6}$
$(0,2)$	$\overline{\mathbf{6}}$
$(1,1)$	$\mathbf{8}$ (adjoint)
$(3,0)$	$\mathbf{10}$
$(0,3)$	$\overline{\mathbf{10}}$

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Table 5.5.4. The first few irreducible representations of

$SU(3)$

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In fact, there is an interesting story here. In the early 1960s, nine short-lived baryonic particles had been observed. They all shared similar properties and masses. So it was believes that those should live some irreducible representation of the symmetry group of the theory. Using the fact that 8 spin 0 mesons and 8 spin 1/2 baryons, Gell-Mann proposed that the symmetry group of the strong force should be

$SU(3)\text{,}$ and that those came in the adjoint representation of

$SU(3)\text{.}$ Then, since there is no

$9$ -dimensional irreducible representation for

$SU(3)\text{,}$ he made the striking prediction that there should be a 10th short-lived baryonic particle with similar properties and masses as the nine other ones, so that they transformed in the

$10$ -dimensional irreducible representations of

$SU(3)\text{.}$ The missing particle was soon found. Gell-Mann's prediction is a remarkable achivement of representation theory in particle physics!

Section 5.5 Tensor representations of SU(N)SU(N)

Objectives

Subsection 5.5.1 Tensor representations of SO(N)SO(N)

Subsection 5.5.2 Tensor representations of SU(N)SU(N)

Subsection 5.5.3 Tensor representations of SU(2)SU(2)

Lemma 5.5.2. The defining representation of SU(2)SU(2) is pseudo-real.

Proof.

Subsection 5.5.4 Tensor representations of SU(3)SU(3)

Lemma 5.5.3. Dimension of irreducible representations of SU(3)SU(3).

Proof.

Section 5.5 Tensor representations of $SU(N)$

Subsection 5.5.1 Tensor representations of $SO(N)$

Subsection 5.5.2 Tensor representations of $SU(N)$

Subsection 5.5.3 Tensor representations of $SU(2)$

Lemma 5.5.2. The defining representation of $SU(2)$ is pseudo-real.

Subsection 5.5.4 Tensor representations of $SU(3)$

Lemma 5.5.3. Dimension of irreducible representations of $SU(3)$ .