What is the meaning of craniocaudal dimensions?

Algebra: What does "any fully antisymmetric third order system in three dimensions is proportional to the corresponding e-symbol"  mean?

• I'm with some difficulty to understand clearly this statement:  "Any fully antisymmetric third order system in three dimensions is proportional to the corresponding e-symbol" ( Dalarsson, page 35) Someone can explain what is the meaning of this ? The book's example was not too clear in my opinion. Thanks;

• Answer:

  The "e-symbol" is the Levi-Civita symbol, defined as follows: ϵ123=ϵ231=ϵ312=+1ϵ123=ϵ231=ϵ312=+1\epsilon_{123} = \epsilon_{231} = \epsilon_{312} = +1 ϵ321=ϵ213=ϵ132=−1ϵ321=ϵ213=ϵ132=−1\epsilon_{321} = \epsilon_{213} = \epsilon_{132} = -1 All other entries are zero. The Levi-Civita symbol is totally antisymmetric, meaning that if you swap two of its indices, the value's sign changes. For example, ϵ123=−ϵ213ϵ123=−ϵ213\epsilon_{123} = -\epsilon_{213}. It's not hard to see that any totally antisymmetric third-rank tensor in three dimensions has to be a multiple of the Levi-Civita symbol. First of all, if any index is repeated, the value has to be zero, just like in the Levi-Civita symbol. For example, T112=0T112=0T_{112} = 0. This is because you can just transpose the two equal indices and this is supposed to change the sign: T112=−T112T112=−T112T_{112} = -T_{112}. Obviously, this implies that it's zero. Now, you can choose T123T123T_{123} to be whatever value you want, but once you've done that, you can just permute the indices by repeated transpositions. For example, suppose a totally antisymmetric third-rank tensor has T123=−5T123=−5T_{123} = -5. Then, by antisymmetry, we see that T213=−T123=+5T213=−T123=+5T_{213} = -T_{123} = +5 by transposing the first two indices T132=−T123=+5T132=−T123=+5T_{132} = -T_{123} = +5 by transposing the last two indices T321=−T123=+5T321=−T123=+5T_{321} = -T_{123} = +5 by transposing the first and last index Furthermore, T231=−T213=−5T231=−T213=−5T_{231} = -T_{213} = -5 by transposing the last two indices T312=−T213=−5T312=−T213=−5T_{312} = -T_{213} = -5 by transposing the first and last index So, in fact, we find that T=−5ϵT=−5ϵT = -5\epsilon; our totally antisymmetric tensor is a multiple of the Levi-Civita symbol. Once we've chosen just one component of the tensor, our hands are tied; the rest can be deduced by antisymmetry alone. The result is that you always get a multiple of the Levi-Civita symbol. This result holds in any number of dimensions: in N dimensions, any totally antisymmetric rank N tensor is proportional to the N-dimensional Levi-Civita symbol.