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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2/6/b/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 6 / b / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

First separate the symmetric tensor and antisymmetric tensor parts:

P_{ij} = \frac{1}{2} (P_{ij} + P_{ji}) + \frac{1}{2} (P_{ij} - P_{ji}) .

(1)

The trace of the antisymmetric part vanishes. In three dimensions, remove one third of the trace from the symmetric part, and identify the remaining antisymmetric part with an axial vector. The scalar, symmetric-traceless and axial tensor decomposition is

P = \frac{1}{3} P_{ii}, S_{ij} = \frac{1}{2} (P_{ij} + P_{ji}) - \frac{1}{3} P_{kk} δ_{ij}, A_{k} = \frac{1}{2} ϵ_{kij} P_{ij} .

(2)

Here

S_{ji} = S_{ij}

and

S_{ii} = P_{ii} - P_{kk} = 0

. The contraction of two Levi-Civita symbols verifies the reconstruction:

ϵ_{ijk} A_{k} = \frac{1}{2} ϵ_{ijk} ϵ_{k ℓ m} P_{ℓ m} = \frac{1}{2} (δ_{i ℓ} δ_{jm} - δ_{im} δ_{j ℓ}) P_{ℓ m} = \frac{1}{2} (P_{ij} - P_{ji}) .

(3)

Therefore

P_{ij} = P δ_{ij} + S_{ij} + ϵ_{ijk} A_{k} .

(4)

The pieces contain

1

5

and

3

independent components respectively, accounting for all nine entries. Taking the trace, then the traceless symmetric part, then the antisymmetric part shows that the decomposition is unique.

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