Submanifolds

Most manifolds met in practice are cut out by equations: the sphere is $|x|^2 = 1$, the rotation group is $A^TA = I$, the positions of a mechanism are the solutions of its closure equations. When is such a solution set a smooth manifold, and of what dimension? The answer is the regular value theorem: if the equations are independent at every solution, in the sense that their derivatives are linearly independent, the solution set is a smooth submanifold whose dimension is the number of unknowns minus the number of equations. It is the implicit function theorem of 2B.9 The Inverse and Implicit Function Theorems in the language of manifolds.

The chapter's other theme is the difference between a map that is locally injective on tangent vectors (an immersion) and one that places a genuine copy of a manifold inside another (an embedding). A figure eight is an immersed circle but not an embedded one, and a line winding densely around a torus is an injective immersion that is not an embedding. The distinction matters whenever a curve or surface is drawn inside a larger space, as minimal surfaces and the necks of the Ricci flow are.

By the end of this chapter you will be able to:

• define the rank of a smooth map, and state the rank theorem;
• distinguish immersions, submersions and embeddings, with examples and counterexamples;
• define embedded submanifolds by slice charts;
• prove that regular level sets are submanifolds, find their dimension and tangent spaces;
• show that $O(n)$, $SO(3)$ and $SL(n)$ are submanifolds of the space of matrices, and compute their dimensions.

4.1The configuration space of a linkage

4.2Rank, immersions and submersions

The rank of a smooth map $F : M \to N$ at $p$ is the rank of $dF_p : T_pM \to T_{F(p)}N$ (8A.3 Tangent Vectors and Bundles). With $\dim M = m$ and $\dim N = n$:

• $F$ is an immersion if $dF_p$ is injective everywhere (rank $m$);
• $F$ is a submersion if $dF_p$ is surjective everywhere (rank $n$);
• $F$ is a local diffeomorphism if $dF_p$ is invertible everywhere (rank $m = n$).

The proof is the inverse function theorem (2B.9 The Inverse and Implicit Function Theorems) applied to a cleverly chosen map (Lee, chapter 4). So locally, immersions look like inclusions $\mathbb{R}^m \hookrightarrow \mathbb{R}^n$, and submersions like projections $\mathbb{R}^m \to \mathbb{R}^n$.

Immersions are not always embeddings. An embedding is an immersion that is also a homeomorphism onto its image (with the subspace topology). Two immersions of familiar spaces fail to be embeddings in instructive ways (Figure 4.2):

• the figure eight $t \mapsto (\sin2t, \sin t)$ immerses the circle $\mathbb{R}/2\pi\mathbb{Z}$ in the plane, but it is not injective: the point $t = 0$ and $t = \pi$ both go to the origin;
• the irrational line on the torus, $t \mapsto (e^{it}, e^{i\alpha t})$ with $\alpha$ irrational, is an injective immersion of $\mathbb{R}$ into $T^2$, but its image is dense, and points far apart in $\mathbb{R}$ come arbitrarily close in $T^2$, so it is not a homeomorphism onto its image (Exercise 4.5).

For a compact manifold, an injective immersion is automatically an embedding, by the closed map lemma (7A.2 Compactness and Compactification).

4.3Submanifolds and level sets

A subset $S \subseteq M$ is an embedded submanifold of dimension $k$ if each point of $S$ has a slice chart: a chart of $M$ in which $S$ is the coordinate plane $\{x^{k+1} = \dots = x^m = 0\}$. Then $S$ is itself a smooth manifold, the inclusion is an embedding, and $T_pS \subseteq T_pM$. The images of embeddings are exactly the embedded submanifolds.

For example, $S^n = f^{-1}(1)$ with $f(x) = |x|^2$, whose derivative $df_x(v) = 2x\cdot v$ is non-zero on the sphere; $T_xS^n = \{v : x\cdot v = 0\}$, as in 8A.3 Tangent Vectors and Bundles.

Matrix groups. The orthogonal group $O(n) = \{A : A^TA = I\}$ is the level set $F^{-1}(I)$ of $F(A) = A^TA$, a map from the $n^2$-dimensional space of matrices into the space $\operatorname{Sym}(n)$ of symmetric matrices, of dimension $\frac{n(n+1)}{2}$. Its derivative at $A$ is $dF_A(B) = B^TA + A^TB$, which is onto $\operatorname{Sym}(n)$: given symmetric $C$, $B = \frac12AC$ works when $A^TA = I$. So $I$ is a regular value, and

the skew-symmetric matrices (Exercise 4.3). The rotation group $SO(3) = \{A \in O(3) : \det A = 1\}$ is the open and closed piece of $O(3)$ containing $I$, of dimension $3$: a rotation has three degrees of freedom, an axis and an angle. As 7A.6 Covering Spaces showed, it is $\mathbb{R}P^3$. Similarly $SL(n) = \det^{-1}(1)$ has dimension $n^2 - 1$.

Transversality. The regular value theorem generalises to preimages of submanifolds: if $F$ is transverse to $Z \subseteq N$, then $F^{-1}(Z)$ is a submanifold of the same codimension as $Z$ (7A.7 Smooth Topology). Two submanifolds meeting transversally intersect in a submanifold; two surfaces in $\mathbb{R}^3$ meeting transversally meet in curves. Sard's theorem makes transversality generic.

4.4History

The implicit function theorem has a long history, from Lagrange and Cauchy to Ulisse Dini's 1877–78 lectures, which gave the modern form. Its manifold version, the constant rank theorem and the regular value theorem, became standard with Whitney's work in the 1930s. Franz Grashof stated his condition for four-bar linkages in his Theoretische Maschinenlehre of 1883.

4.5Exercises

Solution

(a) $F(A + tB) = A^TA + t(B^TA + A^TB) + O(t^2)$. For symmetric $C$ and $B = \frac12AC$: $B^TA + A^TB = \frac12CA^TA + \frac12A^TAC = C$. (b) $B \in \ker dF_A$ iff $A^TB$ is skew; put $X = A^TB$, so $B = AX$. (c) $\dim SO(n) = \frac{n(n-1)}{2}$: $6$ for $n = 4$, $3$ for $n = 3$.

Solution

$\gamma'(t) = (2\cos2t, \cos t)$ vanishes only if $\cos t = 0$ and $\cos2t = 0$, but $\cos t = 0$ gives $\cos2t = -1$; so $\gamma$ is an immersion. $\gamma(t) = \gamma(s)$ with $t \neq s$ only for $\{t, s\} = \{0, \pi\}$, both mapped to the origin. The image is not a submanifold: near the origin it is two crossing curves, and no neighbourhood of the origin in it is homeomorphic to an interval (removing the origin leaves four pieces, not two).

Solution

$f_\theta = 2(A - B)\cdot A'(\theta)$ and $f_\phi = -2(A - B)\cdot B'(\phi)$. $A'(\theta)$ is perpendicular to the crank and $B'(\phi)$ to the output link, so both vanish iff the coupler direction $A - B$ is parallel to both crank and output link: all collinear with the base. For the parallelogram, $\theta = \phi = 0$ is such a configuration. For the strict example, collinear configurations have $A = (\pm1, 0)$ and $B = (3 \pm 2.5, 0)$, so the coupler would need length $|A - B| \in \{0.5, 1.5, 4.5, 6.5\}$; none of these is $b = 3$, so there are no critical points on the configuration curve.

Solution

(a) $df_x(v) = 2(x_1v_1 + x_2v_2 + x_3v_3)$, non-zero on $C$ since $(x_1, x_2, x_3) \neq 0$; $C$ is the product of the sphere of radius $r$ in $\mathbb{R}^3$ with the $x_4$-line. (b) $T_xC = \{v : x_1v_1 + x_2v_2 + x_3v_3 = 0\}$, three-dimensional. (c) The Ricci tensor of a product is the sum of the factors' Ricci tensors; the line contributes $0$ and the sphere $\frac{1}{r^2}g_{S^2_r}$. With $g_{S^2_r} = r^2g_{S^2_1}$, $\partial_t(r^2) = -2$, so $r^2 = r_0^2 - 2t$, which reaches $0$ at $t = \frac{r_0^2}{2}$: the cross-sectional sphere shrinks to a point in finite time.