3

My question is simply wrote on the title. (I'm using Einstein's contraction rule.)

In the case of three dimensions, I can construct the Levi-Civita-like tensor as follows. \begin{align} e_{ijk} = \begin{cases} 1 \quad for \ i=j=k \\ 1 \quad for \ (i,j,k) \in even \ or \ odd \ permutations \ of \ (1,2,3) \\ 0 \quad otherwise \end{cases}. \end{align} This satisfies the equation \begin{align} e_{ijk} e_{lmk} = \delta_{il} \delta_{jm}. \end{align} Is it possible to construct such tensors in higher dimensions? And if it is, how can I get some of these practically?

Thank you in advance for any suggestions.

kohta
  • 61
  • It is $e_{ijk} e_{lmk} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$ . This is true only in three dimensions. In general, for n dimensions, one can write the product of two Levi-Civita symbols as a determinant, see e.g. https://math.stackexchange.com/questions/1946493/product-of-levi-civita-symbol-is-determinant or https://en.wikipedia.org/wiki/Levi-Civita_symbol – Quillo May 05 '20 at 12:37

1 Answers1

1

That equation is incorrect; I'm not used to this notation but I think it's actually

$$e_{ijk} e_{lmk} = \delta_{il} \delta_{jm} + \delta_{im} \delta_{jl}.$$

In fact a tensor with this property does not exist in any dimension at least $2$. The reason is that it would imply the existence of a pair of linear maps $f : V \otimes V \to V$ and $g : V \to V \otimes V$ (where $V$ is a vector space of the given dimension) such that the composite

$$g \circ f : V \otimes V \to V \otimes V$$

is the identity. But this is impossible when $\dim V \ge 2$ because $\dim V \otimes V = (\dim V)^2 > \dim V$, and so neither $f$ nor $g$ can have full rank.

Qiaochu Yuan
  • 419,620
  • @Muphrid: well, yes, but this question isn't about the Levi-Civita tensor. – Qiaochu Yuan Apr 02 '15 at 04:58
  • Thank you for your suggestion. I might get your discussion. But the equation $e_{ijk} e_{lmk} = \delta_{il} \delta_{jm}$ was derived from the other discussion I want to solve. So a equation I want to satisfy is the above. It might mean the tensor product of identity maps on each vector spaces, not the identity on the tensor product of vector spaces. Does my discussion or equation make no sense? – kohta Apr 03 '15 at 09:31
  • @kohta: the tensor product of identity maps is the identity map on the tensor product. The equation makes sense, it just doesn't admit any solutions in dimension at least $2$. – Qiaochu Yuan Apr 03 '15 at 09:32
  • Well, yes, I've got it. My example on the top has been incorrect. There's no tensors satisfying my condition in dimensions higher than 2. Thank you very much. – kohta Apr 03 '15 at 10:07