# Invariant tensors in a general representation and their physical meaning

I'm trying to use tensor methods to find invariant elements of representations. Specifically I'm looking at representations of $SU(5)$.

I can show that the invariant element in $5\otimes\bar{5}$ (or equivalently the $1$ in the $5\otimes\bar{5} =$1$\oplus 24$ representation?) is $\delta_i^j$: this is straightforward because $X \in SU(5)$ acts by $[X\delta]_j^i = X_\lambda^i\delta_j^\lambda - X^\lambda_j\delta^i_\lambda = 0$.

1. I'm wondering how we find the $1$ more generally. E.g. how do we find the invariant tensor in a decomposition $5\otimes10\otimes10$ etc. is there a general method for this?

2. Secondly I'm wondering what is the physical content of a $1$ representation generally?

3. Thirdly I'm trying to find the branching of such tensors under various subgroups of $SU(5)$.

• Related $SU(3)$ posts: physics.stackexchange.com/q/167680/2451 , physics.stackexchange.com/q/219710/2451 and links therein; especially the answer physics.stackexchange.com/a/14586/2451 and links therein. – Qmechanic Jan 6 '16 at 12:45
• Okay thank you for the links. So we also have the invariant $\epsilon^{ijk}$ in $SU(3)$ - is there an equivalent in $SU(5)$? – Wooster Jan 6 '16 at 13:01
• That would be $\epsilon^{ijk\ell m}$. – Qmechanic Jan 6 '16 at 13:04
• Okay thanks. So $\delta$ gives an invariant tensor in $5\otimes5$ and $\epsilon$ gives an invariant tensor in $5\otimes5\otimes5$ (or conjugates if we use lower indices) - In my example then, I'm not really sure how we then get an invariant tensor in $5\otimes10\otimes10$? – Wooster Jan 6 '16 at 13:08
• The is no guarantee that there is any trivial subrepresentation inside such a tensor product. One general way to decompose $\mathrm{SU}(N)$ representations is by Young diagrams, cf. this note. – ACuriousMind Jan 6 '16 at 13:12

1. Apply the Young calculus (per ACuriousMind's suggestion in the comments). For finding the multiplicity of the trivial representation in a tensor product of representations of $SU(n)$, note that each irreducible representation $D$ of $SU(n)$ has a unique conjugate irreducible representation $\bar D$ such that the Young calculus allows $D\otimes \bar D$ to include a rectangular Young diagram of full height $n$ (which is invariant under $SU(n)$). As suggested by Wooster in the comments, in order for $D\otimes\bar D$ to accommodate such a Young diagram, the joint symmetry types of $D$ and $\bar D$ must be compatible with (i.e. have nonzero overlap with) some tensor power/outer product of the fully antisymmetric tensor $\epsilon_{i_1,\dots,i_n}$. In $SU(n)$, Young diagrams of this type correspond to the invariant or trivial representation.
2. If you view the $n$ dimensional representation of $SU(n)$ as a sort of single-particle Hilbert space, then the invariants formed from tensor products of this representation can be thought of as '$SU(n)$-neutral' many-particle states. More abstractly you could interpret representations $SU(n)$ very differently as a sort of gauge theory where 'particle number' is gauged.
3. The branching problem has been solved in a number of special cases. For example, there is an explicit formula for the branching of representations for $SU(n)\rightarrow SU(n-1)$. For low-rank representations, the Young calculus is a powerful general-purpose tool for determining branching. One strategy is to decompose the fundamental representation of $SU(5)$ into representations of $H\subset SU(5)$, and then iteratively compare how tensor products decompose. As an example, consider the problem of decomposing rank 2 representations of $SU(5)$ into representations of $SU(2)\subset SU(5)$. The fundamental (vector) representation of $SU(5)$ breaks up as $5_5 \rightarrow 2_2\oplus (1_2 \oplus 1_2 \oplus 1_2)=2_2\oplus 3\times 1_2$. Next, we have $5_5\otimes 5_5=10_5^A\oplus 15_5^S \rightarrow (2_2\oplus 3\times 1_2)\otimes (2_2\oplus 3\times 1_2)=(2_2\otimes 2_2)\oplus 3 \times (2_2\otimes 1_2) \oplus 3\times (1_2\otimes 2_2)\oplus 9\times 1_2=3_2^S\oplus 1_2^A\oplus 3\times 2_2^S\oplus 3\times 2_2^A\oplus 3\times 1_2^A\oplus 6\times 1_2^S$. Grouping terms according to symmetry, we see that $10_5^A\rightarrow 4\times 1_2^A\oplus 3\times 2_2^A$, and $15_5^S\rightarrow 3_2^S\oplus 3\times 2_2^S\oplus 6\times 1_2^S$.

## Background on the Young calculus

In physics, irreducible representations are often labeled by their dimension. This notation is compact, but obscures the underlying algebraic structure. Young diagrams provide a more transparent notation based on a deep result, Schur-Weyl duality, that relates irreducible representations of $GL(n)$ to those of the permutation group $S_r$ on $r$ symbols (here $r$ is the rank of a tensor representation). Ultimately, Schur-Weyl duality comes from the fact that finite dimensional representations of $GL(n)$ can all be constructed out of tensor products of a single fundamental representation (this is the analogue of the $\frac{1}{2}$ representation of $SU(2)$ from elementary quantum mechanics). For now, all you need to know is that there is a 1-1 correspondence between representations of $GL(n)$ and the set of all Young diagrams with maximum height $n$. Young diagrams greatly simplify the task of decomposing tensor products of representations of $GL(n)$, as well as many subgroups of $GL(n)$ with 'similar' structure (e.g. $U(n)$, $SL(n)$, $SU(n)$, etc.). They also make it easier to notice certain partial solutions to the branching problem, such as determining how representations of $GL(n)$ decompose into representations of $GL(n-1)$.

Let $r$ be a positive integer. Young diagrams are associated with partitions of $r$: sequences of integers $\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_k\geq 0$ such that $\sum_j \lambda_j = r$. Given a partition $(\lambda_1,\dots,\lambda_k)$, draw a Young diagram as follows: (i) draw a horizontal row of $\lambda_1$ boxes, (ii) draw a horizontal row of $\lambda_{j+1}$ boxes starting from the left below the $j$th row, $1\leq j<k$. For example, the partition $(2,1,1)$ of $r=4$ would correspond to the diagram As mentioned above, each diagram with at most $n$ rows corresponds to an irreducible representation of $GL(n)$. Again, this fact is useful because $GL(n)$ is closely related to many other groups of interest in physics. A Young diagram can be thought of as an efficient way to keep track of the symmetrization of tensor indices: after placing tensor indices $i_1$ through $i_r$ in the squares of a Young diagram, the corresponding irreducible tensors are symmetric (even) under permutations that preserve rows, and antisymmetric (odd) under permutations that preserve columns. There is a general formula for the dimension of a $GL(n)$ representation labelled by a Young diagram, but in practice the dimension can be computed more efficiently for low rank using the decomposition rules for tensor products, to be explained now.

The tensor-product decomposition rules for $GL(n)$ follow from a special sort of 'inverse branching' problem for the permutation group $S_r$. In the end, one obtains the following rules:

Let $L=(\lambda_1,\dots,\lambda_k)$ and $M=(\mu_1,\dots,\mu_\ell)$ be two irreducible representations of $GL(n)$, given by their Young diagrams.

1. Draw the diagrams corresponding to $L$ and $M$. In the diagram for $M$, choose a distinct symbol for each row (e.g. $a$ for the first row, $b$ for the second, $c$ for the third, etc.), and write the symbol in each box of that symbol's row.
2. Find all ways in which $\mu_1$ $a$'s can be added to the Young diagram of $L$ so that no two $a$'s appear in the same column, and the resulting graph is another Young diagram (i.e. the length of rows is nonincreasing).
3. For each larger Young diagram obtained above, find all the ways in which $\mu_2$ $b$'s can be placed with no two in the same column, together with an additional constraint: when reading the added symbols from right to left, top to bottom, the number of $a$'s that have been read must match or exceed the number of $b$'s that appear at any step.
4. Repeat for the $\mu_3$ $c$'s, then $\mu_4$ $d$'s, etc., except that now when imposing the last constraint mentioned in step 3, the recorded number of $c$'s cannot exceed the number of $b$'s (etc.).
5. The tensor product $L\otimes M$ decomposes into a direct sum of all Young diagrams obtained this way.

As an example, consider the following tensor product: To decompose this, first we label the second diagram with $A$'s and $B$'s: Next we find all the ways of adding $A$ blocks, and then $B$ blocks, to the Young diagram of $L$ according to the above rules:

Note that diagrams like the ones below are not permitted: The first two diagrams contain two $A$'s in the same column, while the last one isn't allowed because when reading the added symbols right-left top-bottom, we obtain $ABBA$, which has more $B$'s than $A$'s after the third letter (this is from the rule stated in step 3).

Now it turns out that all irreducible representations of $GL(n)$ remain irreducible when restricted to $SU(n)$. However, some representations of $GL(n)$ that were previously distinct become isomorphic. This comes from the fact that it is possible for two irreducible representations of $GL(n)$ to differ from each other only by powers of the determinant homomorphism: $\det(gh)=\det(g)\det(h)$. Once the determinant is set to unity in $SU(n)$ (or $SL(n)$ for that matter) this distinction vanishes, and representations that only differed in their power of $\det(g)$ are isomorphic. Fortunately there is a simple way to account for this redundancy: under $SU(n)$, the representations $(\lambda_1,\dots,\lambda_n)$ and $(\lambda_1+s,\dots,\lambda_n+s)$ are equivalent. To account for the redundancy, we simply choose $s=-\lambda_n$ and label representations of $SU(n)$ with only $n-1$ non-increasing integers instead of $n$. A consequence of this is that if $\lambda_1=\lambda_2=\dots=\lambda_n$, then $[\lambda_1,\dots,\lambda_n]\cong[0,0,\dots,0]$: tensors corresponding to rectangular Young diagrams are invariant under $SU(n)$. For finding the multiplicity of the trivial representation in tensor products, you can check from the decomposition rules that each irreducible representation $V$ of $SU(n)$ has a unique conjugate $\bar V$ such that $V\otimes\bar V$ includes the trivial representation.