How to prove $T_{(a_1 \dots [a_i \dots a_j] \dots a_n)} = 0$?

Question

How to prove this property of tensor:

$$T_{(a_1 \dots [a_i \dots a_j] \dots a_n)} = 0,$$

where $()$ is symmetrization and $[]$ is antisymmetrization ?

I do not know where to start.

Thank you for any help.

Answer 1

Provided $j > i$, then what you state is true. To see this, note that by symmetry, we have

$$T_{a_1\dots [a_{i+1} a_i \dots a_j] \dots a_n} = T_{a_1\dots [a_i a_{i+1} \dots a_j] \dots a_n}.$$

However, by skew-symmetry, we have

$$T_{a_1\dots [a_{i+1} a_i \dots a_j] \dots a_n} = -T_{a_1\dots [a_i a_{i+1} \dots a_j] \dots a_n}.$$

So

$$T_{a_1\dots [a_i a_{i+1} \dots a_j] \dots a_n} = -T_{a_1\dots [a_i a_{i+1} \dots a_j] \dots a_n}$$

and hence

$$T_{a_1\dots [a_i a_{i+1} \dots a_j] \dots a_n} = 0.$$