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A singular two-phase equation map with a four-dimensional solution set

Example

Take $C = 2$ components and $P = 2$ phases on the open coordinate domain $U = (0, \infty)^{2} \times (0, 1)^{2} \subset R^{4}$ . Set all four chemical-potential coordinate functions $μ_{i}^{α}$ identically equal to zero. The two coexistence equations then vanish identically, so the solution set is all of $U$ and has local dimension four. The regular formula would give $C - P + 2 = 2$ , but its nonsingular-minor hypothesis fails because the equation Jacobian is zero.

Facts & Assumptions

Given: The stated component and phase counts, open domain, and constant coordinate functions.

[L1]

Regular rank form of the finite-component phase dimension count: the dimension $C - P + 2$ is conditional on an invertible square derivative minor for the coexistence equation map.

Verification

technique · direct calculation

1.1givenalgebra

For $C = P = 2$ , there are $n = 2 + 2 (2 - 1) = 4$ coordinates and $q = 2 (2 - 1) = 2$ equations. Since each $μ_{i}^{α}$ is identically zero, both differences $μ_{i}^{2} - μ_{i}^{1}$ equal zero at every point of $U$ . Thus the zero set is exactly the open set $U$ , which is locally parameterized by its four original coordinates.

2.1L1step 1.1algebra∎

The derivative of each identically zero equation is the zero covector, so the equation Jacobian has rank zero and no $2 \times 2$ minor is invertible. Therefore [L1] does not apply; the example does not contradict its regular dimension count, and it shows why equation count alone does not determine local dimension.

Depends on

Regular rank form of the finite-component phase dimension count

Dependency tree · 2 levels

Sources

Faye Jackson, MATH 395 notes, Implicit Function Theorem, pp. 99–106 and 127–132