What is the definition of trace for n-indexed tensor in group theory?

[dontknow]

TL;DR

Reducing Representations of tensors

I was reading zee's group theory in a nutshell.
I understand that we can decompose a 2 index tensor for rotation group into an antisymmetric vector(3), symmetric traceless tensor(5) and a scalar(trace of the tensor). Because "trace is invariant" it put a condition on the transformation of symmetric tensor elements, reduces it further from dimension from 6 to 5. But when we go for 3-index tensor and try to take a trace with 2 dimension Kronecker delta, we don't get an invariant quantity (like trace(scalar) for 2 index tensor). Can we really make a "symmetric traceless" 3 index tensor by subtracting a quantity which kind of transform like a vector?
What's the definition of trace for n indexed tensor (if possible specify reasons)?
Sorry for not posting any mathematical equation.
Thanks in advance

 

[Paul Colby]

For rotations in three dimensions one can use addition of angular momentum. Let a vector be one unit of angular momentum, $\mathbf{1}$. The two index tensors looks like.

$\mathbf{1}\otimes \mathbf{1} = \mathbf{0} \oplus \mathbf{1} \oplus \mathbf{2}$

where $\mathbf{1}$ transforms like a vector and $\mathbf{2}$ is your trace free symmetric tensor. This all follows from the Clebach-Gordan coefficients. So, just add an additional vector to the above to get the three index tensors. Clearly the only way to get a scalar in the three index case is to add a vector to the $\mathbf{1}$ representation above. There may be more than one way to do this since the order in which one couples the indices may be important.