# Tensor operator

I have come across the following expression:

H:E

where, H = e(levi-cita symbol)*a constant which means a 3rd order tensor with 27 components

E = 2nd order tensor,

now, what does H:E mean? I know that the result has to be a vector (3 components).

• – user147263 Nov 25 '14 at 21:19
• – janmarqz Nov 26 '14 at 2:42
• @janmarqz There are two double dot products $A:B$ and $A..B$. In a Cartesian basis, they are $A_{ij}B_{ij}$ and $A_{ij}B_{ji}$, but some authors use the opposite convention. They give different results, in general, unless one of the tensors is symmetric. – user_of_math Nov 26 '14 at 11:27

Assuming that $E=E_{ij}\textbf{e}^i\textbf{e}^j$ is a symmetric tensor such that $E_{ij}=E_{ji}$, So you have: $$H^{ijk}=\alpha\cdot\epsilon^{ijk}$$ $$E=E_{ij}$$ In order to produce a first order tensor (vector) with three components, we arrange the tensors in such a way to produce a vector, with components $V^i$: $$H:E=H^{ijk}E_{jk}=V^i$$ So therefore: $$V^i=\alpha\cdot\epsilon^{ijk}E_{jk}=\alpha\cdot\epsilon^{ijk}E_{kj}$$