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 Incoming links: Levi-Civita formulas for a three-dimensional determinant and inverse

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 6 / Solution 2026-10-18

In Cartesian coordinates a rank-

r

tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:

T_{i_{1} \dots i_{r}}^{'} = Q_{i_{1} j_{1}} \dots Q_{i_{r} j_{r}} T_{j_{1} \dots j_{r}} .

An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeys

Q_{ia} Q_{jb} Q_{k c} ϵ_{ab c} = (det Q) ϵ_{ijk} = ϵ_{ijk}

for

Q \in SO (3)

, proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant

- 1

its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.

Contract the determinant identity with

ϵ_{ijk}

. Since

ϵ_{ijk} ϵ_{ijk} = 3! = 6

, one obtains

det A = \frac{1}{6} ϵ_{ijk} ϵ_{l mn} A_{i l} A_{jm} A_{kn} .

Every index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently,

A^{'} = Q A Q^{T}

gives

det A^{'} = (det Q)^{2} det A = det A

.

For the matrix inverse, define

C_{ab} = \frac{1}{2} ϵ_{ajk} ϵ_{bmn} A_{jm} A_{kn} .

This is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with

A_{ib}

and use the determinant identity again:

A_{ib} C_{ab} = \frac{1}{2} ϵ_{ajk} (det A) ϵ_{ijk} = (det A) δ_{ia} .

Here the contraction of two Levi-Civita symbols gives

ϵ_{ajk} ϵ_{ijk} = 2 δ_{ai}

. Therefore

A C^{T} = (det A) I

and, when

det A \neq = 0

,

(A^{- 1})_{ba} = \frac{C _{ab}}{det A} = \frac{1}{2 det A} ϵ_{ajk} ϵ_{bmn} A_{jm} A_{kn} .

These are the Levi-Civita formulas for a three-dimensional determinant and inverse.

Now differentiate the cubic determinant formula with respect to the individual entry

A_{ab}

. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. Thus

\frac{\partial d e t A}{\partial A _{ab}} = \frac{3}{6} ϵ_{ajk} ϵ_{bmn} A_{jm} A_{kn} = C_{ab} .

Substituting the inverse expression proves the entrywise derivative of the determinant:

\frac{\partial det A}{\partial A _{ab}} = (det A) (A^{- 1})_{ba} = (det A) (A^{- 1})_{ab}^{T} .

The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.

For the time-dependent Jacobian matrix, the stated smooth expansion implies

A_{ij} (t, x) = δ_{ij} + t \partial_{i} u_{j} (x) + O (t^{2}), A (0, x) = I .

Applying the chain rule and the entrywise determinant derivative gives Jacobi's formula near

t = 0

:

\frac{d}{d t} det A = \sum_{ab} C_{ab} \dot{A}_{ab} = (det A) tr (A^{- 1} \dot{A}) .

At zero,

C = I

, so

\partial_{t} det A ∣_{0} = \sum_{a} \partial_{a} u_{a} = \nabla \cdot u

. A Taylor expansion therefore yields

det A (t, x) = 1 + t \nabla \cdot u (x) + O (t^{2}) = 1 + O (t^{2}) if \nabla \cdot u = 0.

The Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.

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