False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-01

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Intersecting submanifolds need not be transverse

Statement

False claim: any two embedded submanifolds with nonempty intersection are transverse.

Facts & Assumptions

Given: In $R^{2}$ , the embedded submanifolds $S = R \times {0}$ and $T = R \times {0}$ .

[F1]

Transversality means that the tangent spaces span the ambient tangent space at each intersection point (Transverse embedded submanifolds).

Refutation

technique · direct

1.1

The intersection $S \cap T$ is all of $R \times {0}$ , so it is nonempty.

given

1.2

At every point $p \in S \cap T$ , one has $T_{p} S = T_{p} T = R \times {0}$ . Their sum is still the $x$ -axis, not $T_{p} R^{2} = R^{2}$ . Thus [F1] fails.

F1givenalgebra

2.1

Therefore nonempty intersection does not force transversality.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

6 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

Marco Gualtieri, Topology I: Smooth Manifolds, cumulative notes (standard reference, not scraped)