# Tensor multiplication 3 dimesnsions

#### dwsmith

##### Well-known member
\begin{alignat*}{3}
A_{ij}B_{ij} & = & (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})\\
& = & A_{(ij)}B_{(ij)} + A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} + A_{[ij]}B_{[ij]}
\end{alignat*}
$$A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} = \frac{1}{2}(A_{ji}B_{ij} - A_{ij}B_{ji})$$
Can I then say $A_{ji}B_{ij} = C_{jj} = C_{3\times 3}$ and $A_{ij}B_{ji} = C_{ii} = C_{3\times 3}$?
Therefore, $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} = 0$.

#### wmccunes

##### New member
Dustin, I think you basically solved it in line 1, since
$$A_{ij} = A_{(ij)} + A_{[ij]}$$
(decomposition of the tensor into symmetric $(A_{(ij)}$ and antisymmetric $(A_{[ij]})$ parts), so
$$A_{ij}B_{ij} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})$$

But how does
$$A_{(ij)}B_{(ij)} + A_{[ij]}B_{[ij]} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})$$
? Thanks

#### dwsmith

##### Well-known member
Dustin, I think you basically solved it in line 1, since
$$A_{ij} = A_{(ij)} + A_{[ij]}$$
(decomposition of the tensor into symmetric $(A_{(ij)}$ and antisymmetric $(A_{[ij]})$ parts), so
$$A_{ij}B_{ij} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})$$

But how does
$$A_{(ij)}B_{(ij)} + A_{[ij]}B_{[ij]} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})$$
? Thanks
If $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} = 0$ which I am not sure how to show.

#### wmccunes

##### New member
If $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} = 0$ which I am not sure how to show.
Yes nevermind I was looking at your solution backwards. Breaking up $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)}$ into elements it all cancelled out except
$$\frac{1}{2}(A_{ij}B_{ij} - A_{ji}B_{ji})$$
so if that equals zero we are good...

#### dwsmith

##### Well-known member
Yes nevermind I was looking at your solution backwards. Breaking up $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)}$ into elements it all cancelled out except
$$\frac{1}{2}(A_{ij}B_{ij} - A_{ji}B_{ji})$$
so if that equals zero we are good...
Yup