ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-01

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The intersection of coordinate spheres as a transverse level set

Example

In $R^{4}$ , let

F (x_{1}, x_{2}, x_{3}, x_{4}) = (x_{1}^{2} + x_{2}^{2}, x_{3}^{2} + x_{4}^{2}) .

Then

F^{- 1} (1, 1) = S^{1} \times S^{1}

is a transverse codimension- $2$ level set.

Facts & Assumptions

Given: The smooth map $F : R^{4} \to R^{2}$ above and the point $(1, 1) \in R^{2}$ .

[L1]

The transverse preimage theorem identifies transverse fibres as embedded submanifolds (The transverse preimage theorem).

Verification

technique · direct

1.1

If $F (x) = (1, 1)$ , then $(x_{1}, x_{2}) \neq (0, 0)$ and $(x_{3}, x_{4}) \neq (0, 0)$ . The Jacobian matrix $[\begin{array}{cccc} 2 x_{1} & 2 x_{2} & 0 & 0 \\ 0 & 0 & 2 x_{3} & 2 x_{4} \end{array}]$ therefore has rank $2$ , so $(1, 1)$ is a regular value.

givenalgebra

2.1

The fibre equation is exactly $x_{1}^{2} + x_{2}^{2} = 1, x_{3}^{2} + x_{4}^{2} = 1,$ which is $S^{1} \times S^{1}$ . By [L1], this fibre is an embedded codimension- $2$ submanifold of $R^{4}$ .

L1step 1.1

3.1

Thus the product of the two coordinate circles appears as a transverse level set.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)