# Why isn't this Young tableaux zero?

I am trying to understand Young tableaux in the context of $SU(2)$ irreducible representations (this is a physics course so we use them just as a tool without much care in the meaning).

As I understand a Young tableau is an arrangement of boxes where each box represents an index of a tensor. Columns denote the antisymmetrization of the tensor indices, while rows the symmetrization of the indices.

To calculate the dimension of a particular representation we count all the possible standard tableaux associated to a particular form.

"In a standard tableaux indices do not decrease from left to right in a row, and always increase from top to bottom in a column. The tensors corresponding to nonstandard tableaux either vanish or are not independent of the standard tableaux, i.e. can be expressed in terms of these."source

What I don't understand is that if $i=1, j=3, k=3$, in the following diagram, the tensor is zero applying the symmetry propertie. Yet this is considered a standard tableaux (well in seems that in mathematical context they are called semistandard Young tableux, but the naming convention shouldn't be relevant).

• The definition of a standard tableau includes the condition that all entries be distinct. Your definition is describing an increasing tableau. – Matt Samuel Jan 17 '17 at 2:49
• According to your notes, the tableau above corresponds to $$\psi_{133}+\psi_{313}-2\psi_{331}.$$ Please explain why you think it should be 0. – David Hill Jan 19 '17 at 18:41

Standard Tableaux and semistandard tableax are different but related things. And your confusion stems from that. If you regard a tableax as the instructions for a young symmetrizer (symmetrize with repect to the rows and then antisymmetrize with respect to the columns) then indeed your 3-box object would be zero as you are attempting to antymmetrize somthing (1 and 3) that has already been symmetrized. However Young symmetrizers are always made from standard tableaux where, in a tableau with $n$ boxes, we can only use the numbers 1 to $n$ with no repeats (thus your diagram is not allowed). The symmetrizer permutations act on the subscripts to the tensor indices --- so the permuation $(2,3)$ takes $$T_{i_1i_2i_3} \to T_{i_1i_3i_2}.$$ The permutation operators are acting on the tiny numbers 1,2,3 and it these necessarily distict numbers that appear in the boxes of a standard tableau.

For labeling the independent elements the tensor itself we are labeling the numbers $i_1,i_2,i_3$ themselves, so we can set $i_1=2, i_2=2, i_3=4$ to have the tensor component $T_{224}$. The numbers inserted into semistandard tableau (strictly increasing in columns, and not decreasing in the rows) are these $i_1$, $i_2$, $i_3$ indices (in your case 1, 3 and 3) The semistandard tableaux enumerate the independent tensor components. Other tensor components (corresponding to entries of numbers in non-standard tableax) can be computed from knowing only the standard tableau componets. For example, if $T_{ij}$ is symmetric, knowing the component $T_{34}$ (a semistandard tableau entry as the numbers are increasing) tells you what $T_{43}$ is.

• One more thing--- if we first symmetrize the rows and then anti-symmetrize the columns with the Young projector, then the resuting tensor is, naturally, antisymmetric in the column indices --- but it is no longer symmetric in the row indices. The antisymetrization has destroyed that property. – mike stone Jan 20 '17 at 12:52