I am trying to find the dimension (n° of components) of the antisymmetric traceless tensor $A^{ij}_k$ which is a representation in $SU(N)$. As it is antisymmetric in $ij$, we know that $A^{ij}_k = - A^{ji}_k$. My attempt is as follows:
- $i$ can take $N$ elements
- $j$ can take $N-1$ elements (because as it is antisymmetric, all elements (N in total) with $i=j$ are $0$)
- In order to avoid linearly dependent terms, we divide by 2 and get $$\frac{1}{2}N(N-1)$$
- $k$ can take $N$ elements, so we have in total $$\frac{1}{2}N^2(N-1)$$
However, the solution says we in fact get $$\frac{1}{2}N^2(N-1) - N$$ Where was my mistake? It also says that $T^i_j$ antisymmetric in $i$ has $N^2-1$ dimensions. Why?
I appreciate your answers.
Thanks to Nicholas Todoroff. My problem was with the definition of traceless. For tensors in $SU(N)$ of the form $A^{ij}_k$ it means that $A^{ij}_i = 0$. For a constant $j$, $1$ component of the $N$ $A^{ij}_i$'s is linearly dependent of the other $N-1$ components. That is $1$ component less. However, as $j$ runs to $N$ we in total have $N$ components less.