Formal Immersions and the Smale Hirsch Theorem — Examples

In the standard trivializations $T{\mathbb{R}}^2\cong {\mathbb{R}}^2\times {\mathbb{R}}^2$ the map $F$ reads $(x,\xi )\mapsto (x,\mathrm{diag}(1,0)\xi )=(x,({\xi }_1,0))$, a smooth map covering the identity whose restriction to each fibre is linear; by [F2] it is a smooth bundle map over $f={\mathrm{id}}_{{\mathbb{R}}^2}$.

At every $x\in {\mathbb{R}}^2$ the fibre map is $F_x=\mathrm{diag}(1,0):{\mathbb{R}}^2\to {\mathbb{R}}^2$, whose kernel contains the nonzero vector $(0,1)$; hence $F_x$ is not injective at any point.

By [F1] the pair $(\mathrm{id},F)$ therefore fails the defining fibrewise-injectivity condition at every point and is not a formal immersion, although the base map is even a diffeomorphism. Injectivity on a dense open set or off a proper closed subset gives no conclusion at the remaining points: if any such fibre is noninjective, [F1] excludes a formal immersion; if all fibres are injective, the condition is satisfied. Neither surjectivity of the base map nor linearity of $F$ replaces that condition.

For $x\in S^2$ the tangent space $T_x{S}^2$ is the orthogonal complement of the radial line $\mathbb{R}x$ in ${\mathbb{R}}^3$, and $d{j}_x$ is the inclusion $T_x{S}^2\hookrightarrow {\mathbb{R}}^3$. Hence the normal line of $dj$ at $x$ is spanned by $x$, and the outward unit normal $N(x):=x$ is a global smooth nonvanishing section of ${\nu }_{dj}$.

A line bundle with a global nonvanishing section is trivial, so ${\nu }_{dj}\cong {\epsilon }^1$; the tangent-normal identity of [L1] then gives $T{S}^2\oplus {\epsilon }^1\cong j^{\ast }T{\mathbb{R}}^3={\epsilon }^3$, and the isomorphism $(v,a)\mapsto v+ax$ pulls back the standard basis to the three smooth sections $(e_i-⟨e_i,x⟩x,⟨e_i,x⟩)$, which are a global frame because their images are a basis in every fibre.

Choose a global frame of $TM$ by [F1] and the standard frame of $T{\mathbb{R}}^m$; for any smooth $f:M\to {\mathbb{R}}^m$, define $F$ to be the bundle map over $f$ that carries the frame of $TM$ to the standard frame of $T{\mathbb{R}}^m$ fibrewise. Then $F$ is a fibrewise linear isomorphism, hence fibrewise injective, and $(f,F)$ is a formal immersion by [L2].

Since $M$ has no compact component, [L1] applies in the equidimensional case $m=n$: the formal immersion $(f,F)$ is homotopic through formal immersions to a genuine immersion $g:M\to {\mathbb{R}}^m$, so $M$ immerses in ${\mathbb{R}}^m$ and every formal immersion is homotopic through formal immersions to a genuine one.

For $m\ge 1$, a nonempty closed $m$-manifold is excluded: by the equidimensional obstruction A nonempty closed n-manifold cannot immerse in R-n for n at least one no nonempty closed $m$-manifold immerses in ${\mathbb{R}}^m$, so openness of the source is essential; the examples $M={\mathbb{R}}^m$ minus a point, the open annulus $S^1\times (0,1)\subset {\mathbb{R}}^2$, and open subsets of ${\mathbb{R}}^m$ have trivial tangent bundles and no compact component, so the theorem applies to them. For $m=0$, the no-compact-component condition forces $M=\mathrm{\varnothing }$, since each point is a compact component; its unique map to ${\mathbb{R}}^0$ is an immersion. The countable-choice assumption of [L1] is inherited.

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 $T{T}^n\cong T^n\times {\mathbb{R}}^n$ by [F1]; pairing this frame with the standard frame of $T{\mathbb{R}}^n$ 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 $T^n\to {\mathbb{R}}^n$ for every $n\ge 1$, and the rank data $T{T}^n\oplus \nu \cong {\epsilon }^n$ with $\nu$ of rank zero are formally realized.

Yet no nonempty closed $n$-manifold, in particular not $T^n$, admits an immersion into ${\mathbb{R}}^n$, 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.

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.

Since $\tau (x)$ is a basis of $T_x{S}^1$, $F_x$ is determined by the nonzero vector $v(x)=F_x\tau (x)$, not just its unoriented image line. The smooth positive length $\mathrm{\ell }(x)=\mathrm{\mid }v(x)\mathrm{\mid }$ contracts to $1$ through positive functions, while the base map contracts to the zero map in ${\mathbb{R}}^2$ without changing $v$ in the target's standard trivialization. Thus a formal pair deforms to $(0,u)$, where $u=v\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$ is a smooth unit-vector map.

For a map $u:S^1\to S^1$, normalize its value at $1$ by $\alpha (x)=u(x)\overline{u(1)}$, a based loop. Choose one angular path from $u(1)$ to $1$; multiplying $u$ by this path gives a free homotopy to $\alpha$. A free homotopy $u_s$ normalizes to the based homotopy $u_s(x)\overline{u_s(1)}$, so its degree is invariant. Conversely equal degrees give a based homotopy of the normalized maps by [L2], and the angular paths undo the normalizations. Hence free homotopy classes are exactly the integer degrees. This classification applies to smooth maps and smooth homotopies: lifting a smooth normalized map to a real angle on $[0,1]$, its angle is $kt+h(t)$ with $h$ smooth periodic; interpolation of periodic $h$ to zero gives a smooth homotopy to $e^{2\pi ikt}$. Smoothness of the lift follows locally from the exponential's smooth inverse on an arc.

For an immersion $f$, the derivative direction is $u(x)=d{f}_x(\tau (x))\mathrm{/}\mathrm{\mid }d{f}_x(\tau (x))\mathrm{\mid }$. For the standard inclusion it is $ix$; after normalization this is $x=e^{2\pi it}$, of degree $+1$ by [L2]. The contractions in step 1.1 and the degree classification in step 2.1 identify formal homotopy classes with $\mathbb{Z}$; [L1] transfers this classification to regular homotopy classes of the parametrized immersions.