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A closed manifold with formally plausible rank data needs positive codimension

Statement refuted

The $n$ -torus $T^{n} = S^{1} \times \dots \times S^{1}$ admits formal immersions into $R^{n}$ for every $n \geq 1$ : it is a product of circles, the standard angular fields give a global frame of $T T^{n} ≅ T^{n} \times R^{n}$ , and the identity bundle map is fibrewise injective. Yet no nonempty closed $n$ -manifold — in particular not $T^{n}$ — admits an immersion into $R^{n}$ . Thus the rank data $T M \oplus ν ≅ ε^{n}$ with $ν$ of rank zero is formally plausible but geometrically impossible, and the equidimensional closed-source case genuinely needs the positive-codimension hypothesis; the Smale–Hirsch theorem is not contradicted, because its closed-source form requires $m < n$ .

Facts & Assumptions

Given: The $n$ -torus $T^{n} = S^{1} \times \dots \times S^{1}$ for $n \geq 1$ .

[F1]

Use the finite product atlas obtained from the standard two-arc atlas of each circle. Its tangent charts have smooth derivative transitions and, together with the angular frame, explicitly identify $T T^{n}$ with the smooth product $T^{n} \times R^{n}$ ; the finite atlas and a fixed rational-ball basis establish the tangent total-space structure without choice. Thus $T T^{n}$ is trivial: the product of the standard angular fields of the circle factors is a global frame, and a vector bundle with a global frame is trivial (Local and global frames of a vector bundle, A vector bundle is trivial if and only if it has a global frame, Smooth vector bundles, rank, fibres, and trivial bundles).

[L1]

A vector bundle map over a base map that is a fibrewise linear isomorphism of trivialized bundles is fibrewise injective, hence determines a formal immersion (Formal immersion between smooth manifolds).

[L2]

No nonempty closed $n$ -manifold admits an immersion into $R^{n}$ (A nonempty closed n-manifold cannot immerse in R-n for n at least one); the closed-source form of the Smale–Hirsch theorem requires positive codimension $m < n$ (The Smale–Hirsch immersion theorem).

Counterexample

technique · direct

1.1F1L1givenconstruct

For the explicitly constructed tangent bundles of [F1], the standard angular fields of the circle factors give a global frame of $T T^{n}$ , so $T T^{n} ≅ T^{n} \times R^{n}$ by [F1]; pairing this frame with the standard frame of $T R^{n}$ defines the identity bundle map over any chosen smooth base map, in particular over a constant map, and this map is a fibrewise linear isomorphism, hence fibrewise injective. Thus $(f, F)$ is a formal immersion $T^{n} \to R^{n}$ for every $n \geq 1$ , and the rank data $T T^{n} \oplus ν ≅ ε^{n}$ with $ν$ of rank zero are formally realized.

2.1L2step 1.1

Yet no nonempty closed $n$ -manifold, in particular not $T^{n}$ , admits an immersion into $R^{n}$ , by [L2]; the equidimensional obstruction applies verbatim to $T^{n}$ , which is closed and nonempty. Hence the formal datum of step 1.1 is not holonomic, and the equidimensional closed-source case genuinely needs the positive-codimension hypothesis.

3.1L2step 2.1∎

The Smale–Hirsch theorem is not contradicted: its closed-source form requires $m < n$ , so it makes no assertion about this example; the example shows the rank data alone do not force the existence of an immersion when the source is closed and the codimension is zero.

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