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A nonempty closed n-manifold cannot immerse in R-n for n at least one

Statement

Let $n \geq 1$ and let $M$ be a nonempty closed smooth $n$ -manifold. Then there is no immersion $M \to R^{n}$ ; consequently positive codimension is necessary in the equidimensional closed-source case, and every formal immersion of this nonempty $M$ into $R^{n}$ is non-holonomic. More generally, an equidimensional immersion $M^{n} \to N^{n}$ from a closed $M$ is a local diffeomorphism, hence an open map, and its image is open and closed in the target; since $M$ is compact and nonempty the image is nonempty compact and open, so it is a union of components of $N$ . For $N = R^{n}$ connected and noncompact this is impossible.

Facts & Assumptions

Given: $n \geq 1$ , a nonempty closed smooth $n$ -manifold $M$ , a smooth $n$ -manifold $N$ , and an immersion $f : M \to N$ .

[L1]

An immersion at $p$ has the local normal form $u \mapsto (u, 0)$ in adapted charts (Local normal form for immersions); in the equidimensional case $m = n$ there are no normal coordinates and the model is $u \mapsto u$ on an open subset of $R^{n}$ , so the restriction of $f$ is a diffeomorphism onto an open subset of $N$ (Diffeomorphisms and local diffeomorphisms of manifolds).

[L2]

The continuous image of a compact space is compact: pull an open cover of the image back to an open cover of the source, extract a finite subcover, and take its images (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it). A compact subset of a Hausdorff space is closed (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones); $M$ is compact and $N$ is Hausdorff (Smooth manifolds and their smooth charts, Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces).

[L3]

$R^{n}$ is connected for every $n \geq 1$ ( $R^{n}$ is polygonally connected, connected, locally path-connected and locally connected), and a nonempty subset that is both open and closed in a connected space is the whole space (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

Proof

technique · direct

1.1L1givenchoose

Fix $x \in M$ . Since $f$ is an immersion at $x$ and $\dim M = \dim N = n$ , [L1] gives charts near $x$ and $f (x)$ in which $f$ reads as the identity on an open subset of $R^{n}$ ; hence some open neighbourhood $U_{x}$ of $x$ is mapped diffeomorphically onto an open subset $f (U_{x})$ of $N$ . Therefore $f (M) = ⋃_{x \in M} f (U_{x})$ is open in $N$ , and $f$ is a local diffeomorphism.

2.1L2givenstep 1.1

Since $M$ is nonempty and compact, [L2] makes $f (M)$ nonempty and compact, and since $N$ is Hausdorff, [L2] makes $f (M)$ closed in $N$ .

3.1L3step 1.1step 2.1

An open and closed subset of a locally connected space is a union of components; in particular, if $N$ is connected then the nonempty clopen set $f (M)$ equals $N$ . Applying this with $N = R^{n}$ and [L3] gives $f (M) = R^{n}$ .

4.1step 3.1algebragiven∎

But $R^{n}$ is not compact: the open cover by the balls of radius $k$ , $k \geq 1$ , has no finite subcover, because a finite union of bounded sets is bounded while $R^{n}$ is unbounded. This contradicts step 2.1 with $f (M) = R^{n}$ . Hence no immersion $M \to R^{n}$ exists; since a smooth map $f$ is an immersion exactly when $(f, d f)$ is a formal immersion, no formal immersion of this nonempty $M$ into $R^{n}$ is holonomic, and equidimensional immersions into a general target $N$ have image a union of components of $N$ by step 3.1.

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