Find the number of independent components of an antisymmetric tensor of rank 2 indimensions

Find the number of independent components of an antisymmetric tensor of rank 2 in
dimensions

The Correct Answer and Explanation is:

To find the number of independent components of an antisymmetric tensor of rank 2 in n dimensions, we use the following principle:


Correct Answer:

The number of independent components of an antisymmetric tensor of rank 2 in nnn dimensions isn(n−1)2\frac{n(n-1)}{2}2n(n−1)​


Explanation:

An antisymmetric tensor of rank 2, often written as AμνA^{\mu\nu}Aμν, is a tensor for which the components satisfy the condition:Aμν=−AνμA^{\mu\nu} = -A^{\nu\mu}Aμν=−Aνμ

This means that the component flips sign when its two indices are exchanged.

Additionally, the diagonal components must satisfy:Aμμ=−Aμμ⇒2Aμμ=0⇒Aμμ=0A^{\mu\mu} = -A^{\mu\mu} \Rightarrow 2A^{\mu\mu} = 0 \Rightarrow A^{\mu\mu} = 0Aμμ=−Aμμ⇒2Aμμ=0⇒Aμμ=0

Hence, all diagonal elements are zero.

This antisymmetry significantly reduces the number of independent components compared to a general (non-symmetric) rank 2 tensor, which would have n2n^2n2 components in nnn dimensions.

To count how many independent components remain:

  • We only consider pairs of indices (μ,ν)(\mu, \nu)(μ,ν) such that μ<ν\mu < \nuμ<ν, because the components with μ>ν\mu > \nuμ>ν are determined by antisymmetry: Aμν=−AνμA^{\mu\nu} = -A^{\nu\mu}Aμν=−Aνμ.
  • The number of such unordered pairs (μ,ν)(\mu, \nu)(μ,ν) with μ<ν\mu < \nuμ<ν from nnn items is given by the binomial coefficient: (n2)=n(n−1)2\binom{n}{2} = \frac{n(n-1)}{2}(2n​)=2n(n−1)​

This formula gives the exact count of independent elements because each independent component corresponds to one unique pair of indices (μ,ν)(\mu, \nu)(μ,ν) with μ<ν\mu < \nuμ<ν.


Examples:

  • In 3 dimensions: 3(3−1)2=62=3independent components\frac{3(3-1)}{2} = \frac{6}{2} = 3 \quad \text{independent components}23(3−1)​=26​=3independent components
  • In 4 dimensions: 4(4−1)2=122=6\frac{4(4-1)}{2} = \frac{12}{2} = 624(4−1)​=212​=6
  • In 5 dimensions: 5(5−1)2=202=10\frac{5(5-1)}{2} = \frac{20}{2} = 1025(5−1)​=220​=10

Thus, for any nnn-dimensional space, the antisymmetric rank-2 tensor has n(n−1)2\frac{n(n-1)}{2}2n(n−1)​ independent components.

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