Note:
These were taken during a reading course on smooth manifolds.
Smooth Manifolds
D. Zack Garza
University of Georgiadzackgarza@gmail.com
Last updated: 2020-10-25
Note:
These were taken during a reading course on smooth manifolds.
Last updated: 2020-10-25
Interesting things to know:
A subset \(A\subseteq X\) is saturated with respect to \(p:X\to Y\) if whenever \(p^{-1}(\left\{{y}\right\}) \cap A \neq \emptyset\), then \(p^{-1}(\left\{{y}\right\}) \subseteq A\).
Equivalently, \(A = p^{-1}(B)\) for some \(B\subseteq Y\), i.e. it is a complete inverse image of some subset of \(Y\), i.e. \(A\) is a union of fibers \(p^{-1}(b)\).
A continuous surjective map \(p: X\twoheadrightarrow Y\) is a quotient map if \(U\subseteq Y\) is open iff \(p^{-1}(U) \subset X\) is open.
Note that \(\implies\) comes from the definition of continuity of \(p\), but \(\impliedby\) is a stronger condition.
Equivalently, \(p\) maps saturated subsets of \(X\) to open subsets of \(Y\).
For \(\pi :X\to Y\) a quotient map, if \(g:X\to Z\) is a map that is constant on each \(p^{-1}(\left\{{y}\right\})\), then there is a unique map \(f\) making the following diagram commute:
Fact: an injective quotient map is a homeomorphism.
Fact: a product of quotient maps need not be a quotient map.
If \(X\) and \(\iota_S: S\hookrightarrow Y\) is a subspace, then every continuous map \(f: X\to S\) lifts to a continuous map \(\tilde f: X\to Y\) where \(\tilde f \mathrel{\vcenter{:}}=\iota_S \circ f\):
Note that we can view \(\iota_S \mathrel{\vcenter{:}}={\left.{\text{id}_Y}\right|_{S}}\). The subspace topology is the unique topology for which this property holds.
Some properties of subspace:
Why these differ: in \({\mathbb{R}}^\infty\), the set \(S = \prod (-1, 1)\) is open in the box topology but not the product topology, since \(\left\{{0}\right\}^\infty\) is not contained in any basic open neighborhood contained in \(S\).
Some properties of products:
A function \(f:(a, b) \to {\mathbb{R}}\) is differentiable at \(x\) iff there is a number \(y \in {\mathbb{R}}\) such that \[\begin{align*} \qty{ {f(x+h) - f(x) \over h} - y } \overset{h\to 0}\to 0 \end{align*}\] where \(h\in {\mathbb{R}}\).
The number \(f'(x) \mathrel{\vcenter{:}}= y\) is the derivative of \(f\) at \(x\).
Note that this equivalently says \[\begin{align*} f(x+h) - f(x) = f'(x)h + r(h) \text{ where } {r(h) \over h}\overset{h\to 0}\to 0 .\end{align*}\]
For \(\mathbf{f}: (a,b) \to {\mathbb{R}}^n\), \(\mathbf{\mathbf}{f}'(x)\) is the vector \(\mathbf{y} \in {\mathbb{R}}^n\) such that \[\begin{align*} \qty{ {\mathbf{f}(x+h) - \mathbf{f}(x) \over h} - \mathbf{y} } \overset{h\to 0}\to 0 \iff {{\left\lvert { \mathbf{f}(x+h) - \mathbf{f}(x) - h \mathbf{\mathbf}{y}} \right\rvert} \over {\left\lvert {h} \right\rvert}} \overset{h\to 0}\to 0 \end{align*}\] where \(h\in {\mathbb{R}}\).
The vector \(\nabla f \mathrel{\vcenter{:}}=\mathbf{y}\) is the derivative (or gradient) of \(f\) at \(\mathbf{x}\).
Note that this equivalently says \[\begin{align*} \mathbf{f}(x + h) - \mathbf{f}(x) = h\nabla \mathbf{f} + \mathbf{r}(h) \quad\text{ where } {\mathbf{r}(h) \over h}\overset{h\to 0}\to \mathbf{0} .\end{align*}\]
A function \(\mathbf{f}: {\mathbb{R}}^n \to {\mathbb{R}}^m\) is differentiable iff there exists a linear transformation \(\mathbf{Y}\) such that \[\begin{align*} {{\left\lVert { \mathbf{f}(\mathbf{x}+ \mathbf{h}) - \mathbf{f}(\mathbf{x}) - \mathbf{Y} \mathbf{h}} \right\rVert}_{{\mathbb{R}}^m} \over {\left\lVert {\mathbf{h}} \right\rVert}_{{\mathbb{R}}^n} } \overset{\mathbf{h} \to \mathbf{0}}\to 0 .\end{align*}\]
The matrix \(D_f(\mathbf{x}) \mathrel{\vcenter{:}}=\mathbf{Y}\) is the total derivative of \(f\) at \(\mathbf{x}\).
Note that this equivalently says \[\begin{align*} \mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}( \mathbf{x}) = D_f \mathbf{h} + \mathbf{r}(\mathbf{h}) \quad\text{ where } { {\left\lVert {\mathbf{r}(\mathbf{h})} \right\rVert} \over {\left\lVert {\mathbf{h}} \right\rVert} }\overset{\mathbf{h}\to \mathbf{0}}\to \mathbf{0} .\end{align*}\]
Note that we can write \((\nabla f)(\mathbf{x}) = \sum_{i=1}^n {\frac{\partial f}{\partial x_i}\,} \mathbf{e}_i\).
Warning: \(f\) continuous and existence of all \({\frac{\partial f_i}{\partial x_j}\,}\) does not imply differentiability. If \(f\) is differentiable, however, then \(D_f\) is the Jacobian matrix.
Suppose \(f\in C^1({\mathbb{R}}^n, {\mathbb{R}}^n)\) and \(D_f(\mathbf{a}) \in \operatorname{GL}(n, {\mathbb{R}})\) for some \(\mathbf{a}\) and \(\mathbf{b} = f(\mathbf{a})\).
Then there exist \(U\ni \mathbf{a}\) and \(V\ni \mathbf{b}\) such that \(f(U) = V\) and \({\left.{f}\right|_{U}}\) is bijective with inverse \(g\in C^1(V)\).
Let \(A: {\mathbb{R}}^n\times{\mathbb{R}}^m \to {\mathbb{R}}^n\) and suppose \(A_x: {\mathbb{R}}^n\to {\mathbb{R}}^n\) is invertible.
Then for every \(\mathbf{k}\in {\mathbb{R}}^m\) there exists a unique \(\mathbf{h}\in {\mathbb{R}}^n\) such that \[\begin{align*} A(\mathbf{h}, \mathbf{k}) = \mathbf{0} {\quad \text{and} \quad} \mathbf{h} = -A_x^{-1}A_y \mathbf{k} .\end{align*}\]
Pages 1- 29.
A topological space \(M\) that satisfies
The last property says \(p\in M \implies \exists U\) with \(p\in U \subseteq M\), \(\widehat{U}\subseteq {\mathbb{R}}^n\), and a homeomorphism \(\phi: U \to \widehat{U}\).
Note that second countability is primarily needed for existence of partitions of unity.
Thus any open subset of a topological manifold with the subspace topology is again a topological manifold.
\(U\) is the coordinate domain and \(\phi\) is the coordinate map.
Note that we can write \(\phi\) in components as \(\phi(p) = {\left[ {x^1(p), \cdots, x^n(p)} \right]}\) where each \(x^i\) is a map \(x^i: U \to {\mathbb{R}}\). The component functions \(x^i\) are the local coordinates on \(U\).
Shorthand notation: \({\left[ {x^i} \right]} \mathrel{\vcenter{:}}={\left[ {x^1, \cdots, x^n} \right]}\).
Define \[\begin{align*} \Gamma(f) = \left\{{(x, y) \in {\mathbb{R}}^{n} \times{\mathbb{R}}^{k} {~\mathrel{\Big|}~}x\in U,~ y = f(x) \in \widehat{U} }\right\} .\end{align*}\]
This is a topological manifold since we can take \(\phi: \Gamma(f) \to U\) by restricting \(\pi_1: {\mathbb{R}}^{n}\times{\mathbb{R}}^k \to {\mathbb{R}}^n\) to the subspace \(\Gamma(f)\). Projections are continuous, restrictions of continuous functions are continuous.
This is a homeomorphism because the map \(g: x \mapsto (x, f(x))\) is continuous and \(g\circ \pi_1 = \text{id}_{{\mathbb{R}}^n}\) is continuous with \(\pi_1 \circ g = \text{id}_{\Gamma(f)}\). Note that \(U \cong \Gamma(f)\), and thus \((U, \phi) = (\Gamma(f), \phi)\) is a single global coordinate chart, called the graph coordinates of \(f\).
Note that this works in greater generality:: “The same observation applies to any subset of \({\mathbb{R}}^{n+k}\) by setting any \(k\) of the coordinates equal to some continuous function of the other \(n\).”
\(S^n\) is a subspace of \({\mathbb{R}}^{n+1}\) and is thus Hausdorff and second-countable by exercise .
To see that it’s locally Euclidean, take \[\begin{align*} U_i^+ &\coloneqq\left\{{{\left[ {x^1, \cdots, x^n} \right]} \in {\mathbb{R}}^{n+1} {~\mathrel{\Big|}~}x^i > 0}\right\} {\quad \text{for} \quad} 1 \leq i \leq n+1 \\ U_i^- &\coloneqq\left\{{{\left[ {x^1, \cdots, x^n} \right]} \in {\mathbb{R}}^{n+1} {~\mathrel{\Big|}~}x^i < 0}\right\} {\quad \text{for} \quad} 1 \leq i \leq n+1 .\end{align*}\]
Define \[\begin{align*} f: {\mathbb{R}}^{n} &\to {\mathbb{R}}^{\geq 0} \\ \mathbf{x} &\mapsto \sqrt{1 - {\left\lVert {\mathbf{x}} \right\rVert}^2} .\end{align*}\]
Note that we immediately need to restrict the domain to \({\mathbb{D}}^n \subset {\mathbb{R}}^n\), where \({\left\lVert {x} \right\rVert}^2 \leq 1\implies 1 - {\left\lVert {x} \right\rVert}^2 \geq 0\), to have a well-defined real function \(f: {\mathbb{D}}^n \to {\mathbb{R}}^{\geq 0}\).
Then (claim) \[\begin{align*} U_i^+ \cap S^n {\quad \text{is the graph of} \quad} & x^i = f(x^1, \cdots, \widehat{x^i}, \cdots, x^{n+1}) \\ U_i^- \cap S^n {\quad \text{is the graph of} \quad} &x^i = -f(x^1, \cdots, \widehat{x^i}, \cdots, x^{n+1}) .\end{align*}\]
This is because \[\begin{align*} \Gamma(x^i) &\coloneqq\left\{{(\mathbf{x}, f(\mathbf{x})) \subseteq {\mathbb{R}}^n \times{\mathbb{R}}}\right\} \\ &= \left\{{ {\left[ {x_1, \cdots, \widehat{x^i}, \cdots, x^{n+1}} \right]}, f\qty{{\left[ {x_1, \cdots, \widehat{x^i}, \cdots, x^{n+1} } \right]}}\subseteq {\mathbb{R}}^n \times{\mathbb{R}}}\right\} \\ &= \left\{{ {\left[ {x_1, \cdots, \widehat{x^i}, \cdots, x^{n+1} } \right]}, \qty{1 - \sum_{\substack{j=1 \\ j\neq i}}^{n+1} (x^j)^2}^{1\over 2} \subseteq {\mathbb{R}}^n \times{\mathbb{R}}}\right\} \\ \end{align*}\]
and any vector in this set has norm satisfying \[\begin{align*} {\left\lVert {(\mathbf{x}, y)} \right\rVert}^2 = \sum_{\substack{j=1 \\ j\neq i}}^{n+1} (x^j)^2 + \qty{1 - \sum_{\substack{j=1 \\ j\neq i}}^{n+1} (x^j)^2} = 1 \end{align*}\] and is thus in \(S^n\).
To see that any such point also has positive \(i\) coordinate and is thus in \(U_i^+\), we can rearrange (?) coordinates to put the value of \(f\) in the \(i\)th coordinate to obtain \[\begin{align*} \Gamma(x_i) = \left\{{{\left[ {x^1, \cdots, f(x^1, \cdots, \widehat{x^i}, \cdots, x^n), \cdots, x^n } \right]}}\right\} \end{align*}\] and note that the square root only takes on positive values.
Thus each \(U_i^{\pm} \cap S^n\) is the graph of a continuous function and thus locally Euclidean, and we can define chart maps \[\begin{align*} \phi_i^{\pm}: U_i^{\pm} \cap S^n &\to {\mathbb{D}}^n \\ {\left[ {x^1, \cdots, x^n} \right]} &\mapsto [x^1, \cdots, \widehat{x^i}, \cdots, x^{n+1}] \end{align*}\] yield \(2(n+1)\) charts that are graph coordinates for \(S^n\).
Define \({\mathbb{RP}}^n\) as the space of 1-dimensional subspaces of \({\mathbb{R}}^{n+1}\) with the quotient topology determined by the map \[\begin{align*} \pi: {\mathbb{R}}^{n+1}\setminus\left\{{0}\right\}&\to {\mathbb{RP}}^n\\ \mathbf{x} &\mapsto {\operatorname{span}}_{\mathbb{R}}\left\{{\mathbf{x}}\right\} .\end{align*}\]
Notation: for \(\mathbf{x} \in {\mathbb{R}}^{n+1}\setminus\left\{{0}\right\}\) write \([\mathbf{x}] \mathrel{\vcenter{:}}=\pi(\mathbf{x})\), the line spanned by \(\mathbf{x}\).
Define charts: \[\begin{align*} \tilde U_i \coloneqq\left\{{\mathbf{x} \in {\mathbb{R}}^{n+1}\setminus\left\{{0}\right\}{~\mathrel{\Big|}~}x^i \neq 0}\right\}, \quad U_i = \pi(\tilde U_i) \subseteq {\mathbb{RP}}^n \\ .\end{align*}\]
and chart maps \[\begin{align*} \tilde \phi_i: \tilde U_i &\to {\mathbb{R}}^n \\ {\left[ {x^1, \cdots, x^{n+1}} \right]} &\mapsto {\left[ {{x^1 \over x^i}, \cdots \widehat{x^i}, \cdots {x^{n+1} \over x^i} } \right]} .\end{align*}\]
Then (claim) this descends to a continuous map \(\phi_i: U_i \to {\mathbb{R}}^n\) by the universal property of the quotient:
The restriction \(\pi_U: \tilde U_i \to U_i\) of \(\pi\) is still a quotient map because \(\tilde U_i = \pi_U^{-1}(U_i)\) where \(U_i\subseteq {\mathbb{RP}}^n\) is open in the quotient topology and thus \(\tilde U_i\) is saturated.
Thus \(\pi_U\) sends saturated sets to open sets and is thus a quotient map.
\(\tilde \phi_i\) is constant on preimages under \(\pi_U\): fix \(y\in U_i\), then \(\pi_U^{-1}(\left\{{y}\right\}) = \left\{{\lambda \mathbf{y} {~\mathrel{\Big|}~}\lambda \in {\mathbb{R}}\setminus\left\{{0}\right\}}\right\}\), i.e. the point \(y \in {\mathbb{RP}}^n\) pulls back to every nonzero point on the line spanned by \(\mathbf{y}\in {\mathbb{R}}^n\).
But \[\begin{align*} \tilde \phi_i(\lambda \mathbf{y}) &= \phi_i \qty{ {\left[ {\lambda y^1, \cdots, \lambda y^i, \cdots, \lambda y^n} \right]} } \\ &= {\left[ {{\lambda y^1 \over \lambda y^i}, \cdots, \widehat{\lambda y^i}, \cdots, {\lambda y^{n+1} \over \lambda y^i}} \right]} \\ &= {\left[ {{y^1 \over y^i}, \cdots, \widehat{y^i}, \cdots, {y^{n+1} \over y^i}} \right]} \\ &= \tilde \phi_i(\mathbf{y}) .\end{align*}\]
So this yields a continuous map \[\begin{align*} \phi_i: U_i \to {\mathbb{R}}^n .\end{align*}\]
We can now verify that \(\phi\) is a homeomorphism since it has a continuous inverse given by
\[\begin{align*} \phi_i^{-1}: {\mathbb{R}}^n &\to U_i \subseteq {\mathbb{RP}}^n \\ \mathbf{u} \coloneqq{\left[ {u^1, \cdots, u^n } \right]} &\mapsto {\left[ {u^1, \cdots, u^{i-1}, {\color{red}1}, u^{i+1}, \cdots, u^n} \right]} .\end{align*}\]
It remains to check:
Let \(M \mathrel{\vcenter{:}}= M_1 \times \cdots \times M_k\) be a product of manifolds of dimensions \(n_1, \cdots, n_k\) respectively. A product of Hausdorff/second-countable spaces is still Hausdorff/second-countable, so just need to check that it’s locally Euclidean.
Let \(\mathbf{p} \in \prod_{i=1}^N M_i\), so \(p_i \in M_i\)
Choose a chart \((U_i, \phi_i)\) with \(p_i\in U_i\) and assymble a product map: \[\begin{align*} \Phi \coloneqq\prod \phi_i: \prod U_i \to \prod R^{n_i} \cong {\mathbb{R}}^{\Sigma n_i} \coloneqq{\mathbb{R}}^N .\end{align*}\]
Claim: \(\Phi\) is a homeomorphism onto its image in \(R^N\).
Let \(M\) be a topological manifold.
Two charts are smoothly compatible iff \(U\cap V = \emptyset\) or \(\psi \circ \phi^{-1}\) is a diffeomorphism.
Attempting to define a function \(f: M\to {\mathbb{R}}\) to be smooth iff \(f\circ \phi^{-1}: {\mathbb{R}}^n \to {\mathbb{R}}\) is smooth for each \(\phi\) may not work because many atlases give the “same” smooth structure in the sense that they all determine the same collection of smooth functions on \(M\).
For example, take the following two atlases on \({\mathbb{R}}^n\): \[\begin{align*} \begin{array}{l} {{\mathcal{A}}}_{1}=\left\{\left(\mathbb{R}^{n}, \operatorname{Id}_{\mathbb{R}^{n}}\right)\right\} \\ {{\mathcal{A}}}_{2}=\left\{\left({\mathbb{D}}_{1}(\mathbf{x}), \text{id}_{{\mathbb{D}}_{1}(\mathbf{x})}\right){~\mathrel{\Big|}~}\mathbf{x} \in \mathbb{R}^{n}\right\} \end{array} .\end{align*}\]
Claim: a function \(f:{\mathbb{R}}^n \to {\mathbb{R}}\) is smooth wrt either atlas iff it is smooth in the usual sense.
Let \(M\) be a topological manifold.
A set \(B\subseteq M\) is a regular coordinate ball if there is a smooth coordinate ball \(B'\) such that \(\operatorname{cl}_M(B) \subseteq B'\), and a smooth coordinate map \(\phi: B'\to {\mathbb{R}}^n\) such that for some positive numbers \(r < r'\),
This says \(B\) “sits nicely” insane a larger coordinate ball.
Define \(\psi(x) = x^3\); then \({\mathcal{A}}_1 \mathrel{\vcenter{:}}=\left\{{({\mathbb{R}}^n, \phi)}\right\}\) defines a smooth structure.
Then \({\mathcal{A}}_1 \neq {\mathcal{A}}_0\), which follows because \(\qty{\text{id}_{{\mathbb{R}}^n} \circ \phi^{-1}}(x) = x^{1\over 3}\), which is not smooth at \(\mathbf{0}\).
Note: helpful theorem, two smooth structures induced by two smooth atlases \({\mathcal{A}}_1, {\mathcal{A}}_2\) are equivalent iff \({\mathcal{A}}_1 \cup{\mathcal{A}}_2\) is again a smooth atlas. So it suffices to check pairwise compatibility of charts.
Show that if \(M^n\neq \emptyset\) is a topological manifold of dimension \(n\geq 1\) and \(M\) has a smooth structure, then it has uncountably many distinct ones.
Hint: show that for any \(s> 0\) that \(F_s(x) \mathrel{\vcenter{:}}={\left\lvert {x} \right\rvert}^{s-1}x\) defines a homeomorphism \(F_x: {\mathbb{D}}^n \to {\mathbb{D}}^n\) which is a diffeomorphism iff \(s=1\).
Solution:
Define \[\begin{align*} F_s: {\mathbb{R}}^n &\to {\mathbb{R}}^n \\ \mathbf{x} &\mapsto {\left\lVert {\mathbf{x}} \right\rVert}^{s-1} \mathbf{x} .\end{align*}\]
Claim: \(F_s\) restricted to \({\mathbb{D}}^n\) is a continuous map \({\mathbb{D}}^n \to {\mathbb{D}}^n\).
Note that if \({\left\lVert {\mathbf{x}} \right\rVert}\leq \varepsilon< 1\) then \[\begin{align*}{\left\lVert {F_s(\mathbf{x})} \right\rVert} = {\left\lVert { {\left\lVert {\mathbf{x}} \right\rVert}^s \widehat{\mathbf{x}} } \right\rVert} = {\left\lVert {\mathbf{x}} \right\rVert}^s \leq {\left\lVert {\mathbf{x}} \right\rVert} \leq \varepsilon< 1,\end{align*}\] so \(F_s({\mathbb{D}}^n) \subseteq {\mathbb{D}}^n\) and moreover \(F_s({\mathbb{D}}_\varepsilon^n) \subseteq {\mathbb{D}}_\varepsilon^n\).
This is a continuous function on the punctured disc \({\mathbb{D}}_0^n \mathrel{\vcenter{:}}={\mathbb{D}}^n\setminus\left\{{\mathbf{0}}\right\}\), since it can be written as a composition of smooth functions:
For any \(s\geq 0\), continuity at zero follows from the fact that \({\left\lVert {F_s(\mathbf{x})} \right\rVert} \leq {\left\lVert {\mathbf{x}} \right\rVert} \to 0\), so \(\lim_{\mathbf{x} \to \mathbf{0}}F_s(\mathbf{x}) = \mathbf{0}\) and the sequential definition of continuity applies. So \(F_s\) is continuous on \({\mathbb{D}}^n\) for every \(s\).
Here we are taking for granted the fact that taking norms, exponentiating, and multiplying are all smooth functions away from zero.
Claim: \(F_s\) is a bijection \({\mathbb{D}}^n\setminus{\mathbf{0}}{\circlearrowleft}\) that extends to a bijection \({\mathbb{D}}^n{\circlearrowleft}\).
We can note that \[\begin{align*} F_s(\mathbf{x}) = \begin{dcases} {\left\lVert {\mathbf{x}} \right\rVert}^s{\mathbf{x} \over {\left\lVert {\mathbf{x}} \right\rVert}} \coloneqq{\left\lVert {\mathbf{x}} \right\rVert}^s \widehat{\mathbf{x}} & \text{if } {\left\lVert {\mathbf{x}} \right\rVert} \neq 0 \\ \mathbf{0} & \text{if } {\left\lVert {\mathbf{x} } \right\rVert} = 0 \end{dcases} \end{align*}\]
This follows because we can construct a two-sided inverse that composes to the identity, namely \(F_{1\over s}\), for \(\mathbf{x}\neq \mathbf{0}\), and note that \(F_s(\mathbf{0}) = \mathbf{0}\). Using the fact that \({\left\lVert {t \mathbf{x}} \right\rVert} = t{\left\lVert {\mathbf{x}} \right\rVert}\) for any scalar \(t\), we can check that \[\begin{align*} \qty{F_s \circ F_{1\over s}}(\mathbf{x}) &= F_s({\left\lVert {\mathbf{x}} \right\rVert}^{1\over s} \widehat{\mathbf{x}}) \\ &= {\left\lVert { {\left\lVert {\mathbf{x}} \right\rVert}^{1\over s} \widehat{\mathbf{x}} } \right\rVert}^{s} \cdot \widehat{{\left\lVert {\mathbf{x}} \right\rVert}^{1\over s} \widehat{\mathbf{x}}} \\ &= \qty{ {\left\lVert {\mathbf{x}} \right\rVert}^{1\over s}}^s \cdot {\left\lVert { \widehat{\mathbf{x}} } \right\rVert}^{s} \cdot { {\left\lVert {\mathbf{x}} \right\rVert}^{1\over s} \widehat{\mathbf{x}} \over {\left\lVert { {\left\lVert {\mathbf{x}} \right\rVert}^{1\over s} \widehat{\mathbf{x}} } \right\rVert} } \\ &= {\left\lVert {\mathbf{x}} \right\rVert} \cdot 1^s \cdot \qty{{\left\lVert {\mathbf{x}} \right\rVert}^{1\over s} \over {\left\lVert {\mathbf{x}} \right\rVert}^{1\over s}} \cdot {\widehat{\mathbf{x}} \over {\left\lVert {\widehat{\mathbf{x}}} \right\rVert} } \\ &= {\left\lVert {\mathbf{x}} \right\rVert} \widehat{\mathbf{x}} \\ &= \mathbf{x} .\end{align*}\]
and similarly \[\begin{align*} \qty{F_{1\over s} \circ F_s}(\mathbf{x}) &= F_{1\over s} \qty{ {\left\lVert {\mathbf{x}} \right\rVert}^s \widehat{\mathbf{x}} } \\ &= {\left\lVert {{\left\lVert {\mathbf{x}} \right\rVert}^s \widehat{\mathbf{x}} } \right\rVert}^{1\over s} \cdot \widehat{ {\left\lVert {\mathbf{x}} \right\rVert}^s \widehat{\mathbf{x}} } \\ &= \qty{{\left\lVert {\mathbf{x}} \right\rVert}^s}^{1\over s} {\left\lVert {\widehat{\mathbf{x}} } \right\rVert}^{1\over s} \cdot {{\left\lVert {\mathbf{x}} \right\rVert}^s \widehat{\mathbf{x}} \over {\left\lVert { {\left\lVert {\mathbf{x}} \right\rVert}^s \widehat{\mathbf{x}} } \right\rVert} } \\ &= {\left\lVert {\mathbf{x}} \right\rVert} \cdot 1^{1-s} \cdot \qty{{\left\lVert {\mathbf{x}} \right\rVert}^s \over {\left\lVert {\mathbf{x}} \right\rVert}^s } \cdot {\widehat{\mathbf{x}} \over {\left\lVert {\widehat{\mathbf{x}}} \right\rVert}} \\ &= {\left\lVert {\mathbf{x}} \right\rVert} \widehat{\mathbf{x}} \\ &= \mathbf{x} .\end{align*}\]
Claim: \(F_s\) is a homeomorphism for all \(s\).
This follows from the fact that the domain \({\mathbb{D}}^n\) is compact and the codomain \({\mathbb{D}}^n\) is Hausdorff, and a continuous bijection between such spaces is a homeomorphism.
Claim: \(F_s\) is a diffeomorphism iff \(s=1\).
If \(s=1\), \(F_s = \text{id}_{{\mathbb{D}}^n}\) which is clearly a diffeomorphism.
Otherwise, we claim that \(F_s\) is not a diffeomorphism because either \(F_s\) or \(F_s^{-1}\) will fail to be smooth at \(\mathbf{x} = \mathbf{0}\).
We now show that we can produce infinitely many distinct maximal atlases on \(M\). Let \({\mathcal{A}}\) by any smooth atlas on \(M\) and fix \(p_0\in M\).
Claim: We can modify \({\mathcal{A}}\) to obtain an atlas \({\mathcal{A}}'\) where \(p_0\) is in exactly one chart \((V, \psi)\) with \(\psi(p_0) = \mathbf{0} \in {\mathbb{R}}^n\).
Claim: We can define a new atlas \({\mathcal{A}}_s\) from \({\mathcal{A}}^2\) by only replacing the single chart \((V, \psi)\) with \((V, F_s \circ \psi)\).
Claim: If \(s\neq t\) then \({\mathcal{A}}_s\) and \({\mathcal{A}}_t\) are not smoothly compatible, and thus generate distinct maximal smooth atlases.
\(\hfill\blacksquare\)
Let \(N\mathrel{\vcenter{:}}={\left[ {0, \cdots, 1} \right]} \in S^n\) and \(S\mathrel{\vcenter{:}}={\left[ {0, \cdots, -1} \right]}\) and define the stereographic projection \[\begin{align*} \sigma: S^n\setminus N &\to {\mathbb{R}}^n \\ {\left[ {x^1, \cdots, x^{n+1}} \right]} &\mapsto {1 \over 1-x^{n+1}} {\left[ {x^1, \cdots, x^n} \right]} \end{align*}\] and set \(\tilde\sigma(x) = -\sigma(-x)\) for \(x\in S^n\setminus S\) (projection from the South pole)
For any \(x\in S^n\setminus N\) show that \(\sigma(x) = \mathbf{u}\) where \((\mathbf{u}, 0)\) is the point where the line through \(N\) and \(x\) intersects the linear subspace \(H_{n+1} \mathrel{\vcenter{:}}=\left\{{x^{n+1} = 0}\right\}\).
Similarly show that \(\tilde \sigma(x)\) is the point where the line through \(S\) and \(x\) intersects \(H_{n+1}\).
Show that \(\sigma\) is bijective and \[\begin{align*} \sigma^{-1}(\mathbf{u}) = \sigma^{-1}\qty{{\left[ {u^1, \cdots, u^n } \right]}} = {1\over {\left\lVert {\mathbf{u}} \right\rVert}^2 + 1} {\left[ {2u^1, \cdots, 2u^n, {\left\lVert {\mathbf{u}} \right\rVert}^2 - 1} \right]} .\end{align*}\]
Compute the transition map \(\tilde \sigma \circ \sigma^{-1}\) and verify that the atlas \[\begin{align*} {\mathcal{A}}\coloneqq\left\{{(S^n\setminus N, \sigma), (S^n\setminus S, \tilde \sigma) }\right\} \end{align*}\] define a smooth structure on \(S^n\).
Show that this smooth structure is equivalent to the standard smooth structure: Put graph coordinates on \(S^n\) as outlined in to obtain \(\left\{{(U_i^\pm, \phi_i^{\pm})}\right\}\).
For indices \(i<j\), show that \[\begin{align*} \phi_i^\pm \circ (\phi_j^\pm)^{-1}{\left[ {u^1, \cdots, u^n} \right]} = {\left[ {u^1, \cdots, \widehat{u^i}, \cdots, \pm\sqrt{1 - {\left\lVert {\mathbf{u}} \right\rVert}^2} ,\cdots u^n} \right]} \end{align*}\] where the square root appears in the \(j\)th position. Find a similar formula for \(i>j\). Show that if \(i=j\), then \[\begin{align*} \phi_i^\pm \circ (\phi_j^\pm)^{-1}= \phi_i^- \circ (\phi_i^+)^{-1}= \text{id}_{{\mathbb{D}}^n} .\end{align*}\]
Show that these yield a smooth atlas.
Solution (1):
Parameterize the line through \(\mathbf{x}\in S^n\) and \(\mathbf{N}\): \[\begin{align*} \ell_{N, \mathbf{x}}(t) &= t\mathbf{x} + (1-t) \mathbf{N} \\ &= t{\left[ {x^1, \cdots, x^n, x^{n+1}} \right]} + (1-t){\left[ {0, \cdots, 1} \right]} \\ &= {\left[ {tx^1, \cdots, x^n, tx^{n+1} + (1-t)} \right]} \\ &= {\left[ {tx^1, \cdots, x^n, 1 - t\qty{1-x^{n+1}}} \right]} \\ .\end{align*}\]
Evaluate at \(t = {1 \over 1 - x^{n+1}}\) to obtain \({1\over x^{n+1}}{\left[ {x^1, \cdots, x^n, 0} \right]} = {\left[ { \sigma(\mathbf{x}), 0} \right]}\).
For \(\tilde \sigma(\mathbf{x})\): Todo .
Solution (2):
Solution (3):
Computing the transition maps: \[\begin{align*} (\tilde \sigma \circ \sigma^{-1})(\mathbf{u}) &= -\sigma\qty{\qty{-1 \over {\left\lVert {\mathbf{u}} \right\rVert}^2 + 1} {\left[ {2u^1, \cdots, 2u^n, {\left\lVert {\mathbf{u}} \right\rVert}^2 - 1} \right]} } \\ &= -1 \cdot {\left[ { {{ -2 u^1 \over {\left\lVert {\mathbf{u}} \right\rVert}^2 + 1} \over 1 - {1 - {\left\lVert {\mathbf{u}} \right\rVert}^2 \over 1 + {\left\lVert {\mathbf{u}} \right\rVert}^2 }} , \cdots_n } \right]} \\ &= {\left[ { {2u^1 \over {\left\lVert {\mathbf{u}} \right\rVert}^2 + 1} \cdot {1 + {\left\lVert {\mathbf{u}} \right\rVert}^2 \over 1 + {\left\lVert {\mathbf{u}} \right\rVert}^2 - (1 - {\left\lVert {\mathbf{u}} \right\rVert}^2)}, \cdots_n} \right]} \\ &= {\left[ { {2u^1 \over 2{\left\lVert {\mathbf{u}} \right\rVert}^2}, \cdots_n } \right]} \\ &= {\mathbf{u} \over {\left\lVert {\mathbf{u}} \right\rVert}^2} \\ &\coloneqq\widehat{\mathbf{u}} ,\end{align*}\] which is a smooth function on \({\mathbb{R}}^n\setminus\left\{{\mathbf{0}}\right\}\).
Todo: computing \((\sigma \circ \tilde\sigma^{-1})(\mathbf{u}) = \widehat{\mathbf{u}}\)
Todo: argue that it suffices that these are smooth on \({\mathbb{R}}^n\setminus\left\{{\mathbf{0}}\right\}\)
Solution (4):
We want to argue that these define the same maximal smooth atlas, for which it suffices to the charts from each are pairwise smoothly compatible.
Define an angle function on \(U\subset S^1\) as any continuous function \(\theta:U\to{\mathbb{R}}\) such that \(e^{i\theta(z)} = z\) for all \(z\in U\).
Show that \(U\) admits an angle function iff \(U\neq S^1\), and for any such function \(\theta\), \((U, \theta)\) is a smooth coordinate chart for \(S^1\) with its standard smooth structure.
Note that \(f: {\mathbb{R}}\to S^1\) given by \(f(x) = e^{ix}\) is a covering map (in fact the universal cover).
\(\implies\):
\(\impliedby\):
Alternatively:
\((U, \theta)\) is a smooth coordinate chart:
Fact: \(C^\infty(M) \mathrel{\vcenter{:}}=\left\{{f:M\to {\mathbb{R}}}\right\}\) is a vector space
Fact: \(f\) is smooth \(\iff\) \(f\) is smooth (in the above sense) in some smooth chart about each point.
Fact: taking \(N = V = {\mathbb{R}}^k\) and \(\psi=\text{id}\) recovers the previous definition.
If \(F:M\to N\), then
Proving a map between manifolds is smooth:
Fact: projection maps from products are smooth
Notation: \(\mathbf{v}_a\) denotes the tangent vector at \(\mathbf{v}\), i.e. the pair \((\mathbf{a}, \mathbf{v})\). Think of this as a vector with its base at the point \(\mathbf{a}\).
Picking the standard basis for \({\mathbb{R}}^n_a = \left\{{\mathbf{e}_{i, a}}\right\}_{i=1}^n\) and expanding \(\mathbf{v} = \sum_{i=1}^n v^i \mathbf{e}_{i, a}\), we can explicitly write \[\begin{align*} D_v\Big|_a f = \sum_{i=1}^n v^i {\frac{\partial f}{\partial x^i}\,}(a) .\end{align*}\]
Denote the space of all derivations of \(C^\infty({\mathbb{R}}^n)\) at \(a\) as
\[\begin{align*} T_a {\mathbb{R}}^n \coloneqq\left\{{w \in \hom_{{\mathbb{R}}{\hbox{-}}\text{mod}}(C^\infty({\mathbb{R}}^n), {\mathbb{R}}) {~\mathrel{\Big|}~}w(fg) = f(a)w(g) + g(a)w(f)}\right\} ,\end{align*}\] i.e. a derivation \(w\) is an \({\mathbb{R}}{\hbox{-}}\)linear map satisfying the Leibniz Rule (LR).
Facts:
Claim: if \(f\in C^\infty({\mathbb{R}}^n)\) is constant, say \(f(\mathbf{p}) = 1\) for all \(\mathbf{p}\in {\mathbb{R}}^n\), then \(w(f) = 0\) for any derivation \(w\).
Proof: WLOG suppose \(f(\mathbf{p}) = 1\in {\mathbb{R}}\). Note that \(f(\mathbf{p}) = f(\mathbf{p}) \cdot f(\mathbf{p})\), so \[\begin{align*} w(f) = w(f\cdot f) \stackrel{\mathclap{\scriptscriptstyle{LR}}}{=} f(\mathbf{p})w(f) + w(f)f(\mathbf{p}) = 2f(\mathbf{p})w(f) = 2w(f) \quad\text{since } f(\mathbf{p}) = 1 ,\end{align*}\] and thus \(w(f) = 2w(f) \in {\mathbb{R}}\) forcing \(w(f) = 0\).
More precisely,
Note that \(f\in C^\infty(N)\) implies that \(f\circ F \in C^\infty(M)\), and since \(v\in T_p M\) is a functional in \(C^\infty(M)^\vee\), \(v\) can act on such objects.. Moreover, \(dF_p(v)\) is in fact a derivation at \(F(p)\), since \[\begin{align*} dF_p(v)(fg) &= v((fg) \circ F) \\ &= v((f\circ F) \cdot (g\circ F) ) \hspace{8em}{Why?} \\ &= (f \circ F)(p) \cdot v(g\circ F) + v(f\circ F) \cdot (g \circ F)(p) \quad\text{since $v$ is a derivation}\\ &\coloneqq(f\circ F)(p) dF_p(v)(g) + (g\circ F)(p) dF_p(v)(f) \\ &\coloneqq f(F(p)) dF_p(v)(g) + g(F(p)) dF_p(v)(f) ,\end{align*}\] which puts it in the form \({{\partial}}(fg) = f(q){{\partial}}(g) + {{\partial}}(f) g(q)\) where \(q = F(p)\).
Facts:
Warning: the action of a derivation depends only on the values of a function in arbitrarily small neighborhoods, and in particular, can only be applied to functions defined in a neighborhood of \(p\) (not necessarily on all of \(M\)).
In words: the tangent space of any submanifold is isomorphic to the tangent space of the ambient manifold.
For a vector space \(V\), there is a natural smooth structure (Example 1.24) and for any \(\mathbf{a}, \mathbf{v}\in V\) we can similarly define a map \[\begin{align*} D_{\mathbf{v}}\Big|_{\mathbf{a}}: C^\infty(V) &\to {\mathbb{R}}\\ f &\mapsto D_{\mathbf{v}}f(\mathbf{a})\coloneqq{\frac{\partial }{\partial t}\,}\Big|_{t = 0} f(\mathbf{a} + t\mathbf{v}) .\end{align*}\]
Let \(p\in M\) and \(C_p^\infty(M, {\mathbb{R}})\) be the \({\mathbb{R}}{\hbox{-}}\)algebra of germs of functions at \(p\). Let \(D_p M\) denote the vector space of derivations of \(C_p^\infty(M, {\mathbb{R}})\). Show that the map
\[\begin{align*} \Phi: D_p M &\to T_p M \\ \qty{\Phi_v} f &= v([f]_p) \end{align*}\] is an isomorphism.
First, clarify that this is the map \[\begin{align*} \Phi: D_p M &\to T_p M \\ v &\mapsto \qty{ f \mapsto v([(f, U)]_p) } ,\end{align*}\] where \(\Phi_v\) is the image of \(v\) and \([(f, U)]\) is a germ, i.e. an equivalence class of ordered pairs.
We note that \(v: C_p^\infty(M) \to {\mathbb{R}}\). For \(w\in T_p M\), we have \(w: C^{\infty}(M) \to {\mathbb{R}}\), so define an inverse map \[\begin{align*} \Phi^{-1}: T_p M &\to D_p M \\ w &\mapsto \qty{ [(f, U)] \mapsto w(\tilde f) } ,\end{align*}\] where \(\tilde f\) is to be defined.
Note that \(w\) can’t act directly on \(f\), since \(f\) is only defined on a subset \(U\subseteq M\) whereas \(w\) needs to act on functions defined on all of \(M\). So take \(\tilde f: M\to {\mathbb{R}}\) to be \(f\) extended by smooth bump functions to all of \(M\).
Things to check:
Three categories of maps:
If \(F:M \to N\) is a smooth map of manifolds and \(p\in M\), then the rank of \(F\) at \(p\) is the rank of the linear map \(dF_p: T_p M \to T_{F(p)} N\)
This is equivalently the rank of the Jacobian of \(F\) in any chart, or the dimension of \(\operatorname{im}({d})F_p \subseteq T_{F(p)}N\). The rank may vary from point to point.
The positive integer \(\operatorname{rank}(F)\) is bounded above by \(\min \left\{{\dim M, \dim N}\right\}\); if it achieves this maximum we say \(F\) has full rank.
A smooth map \(F:M\to N\) is a submersion iff \(dF_p\) is surjective for every \(p\in M\), or equivalently \(F\) has constant rank \(\operatorname{rank}(F) = \dim N\).
Analogy: surjective linear maps.
A smooth map \(F:M\to N\) is an immersion iff \(dF_p\) is injective for every \(p\in M\), or equivalently \(F\) is of constant rank \(\operatorname{rank}(F) = \dim M\).
Analogy: injective linear maps.
Note that this can fail if \({{\partial}}M \neq \emptyset\), but will hold when \(F(M)\subseteq N^\circ\). This always happens at points \(p\) where \(dF_p\) is invertible.
Conversely, if \(dF_p\) is an isomorphism at each point, the inverse function theorem supplies neighborhoods on which \(F\) is a diffeomorphism.
If \(F:M\to N\) with \(\dim(M) = m,~\dim(N) = n\) and about each \(p\in M\) there exist charts for which \(F\) has a coordinate representation
\[\begin{align*} \widehat{F}(x^1, \cdots, x^r, x^{r+1}, \cdots, x^m) &= (x^1, \cdots, x^r, 0, \cdots, 0) \\ \widehat{F}(x^1, \cdots, x^n, x^{n+1}, \cdots, x^m) &= (x^1, \cdots, x^n) \quad\text{if $F$ is a submersion} \\ \widehat{F}(x^1, \cdots, x^n, x^{n+1}, \cdots, x^m) &= (x^1, \cdots, x^m, 0, \cdots, 0) \quad\text{if $F$ is an immersion}\\ .\end{align*}\]
I.e., submersions are projections onto the first \(n = \dim N\) coordinates, and immersions are inclusions of the first \(m=\dim M\) coordinates.
Suppose \(F:M\to N\) and \(M\) is connected, then TFAE:
I.e. constant rank maps locally behave like their differentials.
Let \(F:M \to N\) be smooth of constant rank, then
One additional more general case: manifolds with boundary, where the domain includes a boundary point. In this case, if \(F:M\to N\) is an immersion with \(p\in {{\partial}}M\) then there exist charts such that \(F\) has a coordinate representation of inclusion of the first \(m\) coordinates. The situation is more complicated when the codomain includes boundary points, since the image may intersect \({{\partial}}N\) in unpredictable ways.
Note that this is not just a smooth topological embedding, it additionally must be an immersion.
Examples:
Counterexamples:
Failing immersion: The curve \(\gamma: {\mathbb{R}}\to {\mathbb{R}}^2\) where \(t\mapsto {\left[ {t^3, 0} \right]}\) is a smooth topological embedding but not a smooth embedding since \(\gamma'(0)= 0\) (so it fails to be an immersion).
Failing topological embedding: the figure-eight curve \(\beta: (-\pi ,\pi) \to {\mathbb{R}}^2\) where \(t\mapsto {\left[ {\sin(2t), \sin(t)} \right]}\) is an injective smooth immersion since \(\beta'(t) \neq 0\) for any \(t\), but not an embedding because its image is compact and its domain is not.
Failing topological embedding: the curve \(\gamma: {\mathbb{R}}\to T^2\) given by \(t \mapsto {\left[ {e^{2\pi i t}, e^{2\pi i \alpha t}} \right]}\) where \(\alpha\) is irrational is a smooth immersion since \(\gamma'(t) \neq 0\) and injective since \(\gamma(t) = \gamma(s) \implies t-s, \alpha(t-s) \in {\mathbb{Z}}\) which can not happen.
But one can use Dirichlet’s approximation theorem to show that \(\gamma(0)\) is a limit point of \(\gamma({\mathbb{Z}})\), whereas \({\mathbb{Z}}\) has no limit points in \({\mathbb{R}}\), so \(\gamma\) can not be a homeomorphism onto its image.
If any of the following hold for \(F:M\to N\), then \(F\) is a smooth embedding:
Application: \(\iota: S^n \hookrightarrow{\mathbb{R}}^{n+1}\) is smooth and \(d\iota_p\) is injective at every \(p\in S^n\). Since \(S^n\) is compact, \(\iota\) is a smooth embedding.
This leads to formulating the following definition:
I.e., smooth submersions admit enough smooth local sections.
Note: a surjective smooth submersion is a topological quotient map.
If \(\pi:M\to N\) is a surjective submersion, a map \(F:N\to P\) is smooth \(\iff F\circ \pi\) is smooth, as in the following diagram
If \(F:M\to P\) is constant on the fibers of \(\pi:M\to N\) then it descends to a map \(\tilde F: N\to P\):
Need the sheets to be mapped diffeomorphically, as opposed to just smooth homeomorphisms.
Note that if \(E\) is simply connected, then \(E\) is the universal cover of \(M\).
An injective smooth covering map is a diffeomorphism
A topological covering map is a smooth covering map \(\iff\) it is a local diffeomorphism.
Note that smooth covering maps are surjective smooth submersions, so all previous theorems work. E.g. theorems about descending to submersions can be used to define maps out of the base of a covering space.
Exercises:
If \(M\) is a connected smooth manifold, then there exists a simply-connected smooth manifold \(\tilde M\), its universal cover, and a smooth covering map \(\pi \tilde M\to M\).
It is unique in the sense that if \(\tilde M'\) is any other such cover, there exists a diffeomorphism \(\Phi: \tilde M \to \tilde M'\) such that \(\pi' \circ \Phi = \pi\).
Even if a map is known to be a surjective local diffeomorphism, it is difficult to tell if it is a smooth covering map.
Proof uses the fact that local diffeomorphisms are open and proper maps are closed.
Todo
The most important type of manifolds: embedded submanifolds. Most often described as the level set of a smooth map, but needs extra conditions. The level sets of constant rank maps are always embedded submanifolds.
More general: immersed submanifolds. Locally embedded, but may have global topology different than the subspace topology.
For \(S\subseteq M\) in the subspace topology, with a smooth structure such that the inclusion \(S\hookrightarrow M\) is smooth. If \(S\hookrightarrow M\) is a proper map, then \(S\) is properly embedded.
An embedded submanifold of codimension 1.
A subset \(S\subseteq M\) of codimension zero is an embedded submanifold iff \(S\) is an open submanifold.
A way to produce submanifolds: :::{.proposition} If \(F:N\to M\), then \(F(N)\) is a submanifold of \(M\) with the subspace topology and a unique smooth structure making \(F\) a diffeomorphism onto its image and \(F(N)\hookrightarrow M\) and embedding. :::
Thus every embedded submanifold is the image of an embedding, namely its inclusion.
Embedded submanifolds are exactly the images of smooth embeddings: :::{.proposition} The slices \(M\times\left\{{p}\right\}\) for \(p\in N\) are embedded submanifolds of \(M\times N\) diffeomorphic to \(M\). :::
For \(f:U\to N\) with \(U\subseteq M\), \[\begin{align*} \Gamma(f) \mathrel{\vcenter{:}}=\left\{{(x, f(x)) \in M\times N {~\mathrel{\Big|}~}x\in U}\right\} \hookrightarrow M\times N \end{align*}\] is an embedded submanifold.
Note: any manifold that is locally the graph of a smooth function is an embedded submanifold.
\(S\hookrightarrow M\) is a properly embedded submanifolds \(\iff\) \(S\) is a closed subset of \(M\). Thus every compact embedded submanifold is properly embedded.
Embedded submanifolds are locally modeled on the standard embedding \({\mathbb{R}}^k \hookrightarrow{\mathbb{R}}^n\) where \(\mathbf{x} \mapsto {\left[ {\mathbf{x}, \mathbf{0}} \right]}\).
\(S\subseteq M\) satisfies the local \(k{\hbox{-}}\)slice condition iff each \(s\in S\) is in the domain of a smooth chart \((U, \phi)\) such that \(S\cap U\) is a single \(k{\hbox{-}}\)slice in \(U\).
\(S\hookrightarrow M\) is an embedded \(k{\hbox{-}}\)dimensional submanifold \(\iff\) \(S\) satisfies the local \(k{\hbox{-}}\)slice condition. Moreover, there is a unique smooth structure on \(S\) for which this holds.
For manifolds with boundary, \({{\partial}}M \hookrightarrow M\) is a proper embedding. Every such manifold can be embedded in a larger manifold \(\tilde M\) without boundary.
For \(\phi:M\to N\) and \(c\in N\), \(\phi^{-1}(c)\) is a level set of \(\phi\).
Examples:
Every closed \(S\subset M\) is the zero set of some smooth function \(M\to {\mathbb{R}}\).
For \(\phi: M\to N\) with constant rank \(r\), each level set of \(\phi\) is a properly embedded codimension \(r\) submanifold.
If \(\phi: M\to N\) is a smooth submersion, then the level sets are properly embedded of codimension \(\dim N\).
Every smooth submersion has constant rank equal to the dimension of the codomain.
Analogy: for \(L:{\mathbb{R}}^m\to {\mathbb{R}}^r\) a surjective linear map, \(\ker L \leq {\mathbb{R}}^m\) has codimension \(r\) by rank-nullity. Surjective linear maps are analogous to smooth submersions.
If \(\phi: M\to N\) is smooth, \(p\in M\) is a regular point if \(d\phi\) is surjective and a critical point otherwise. A point \(c\in N\) is a regular value if every point in \(\phi^{-1}(c)\) is a regular point, and a critical value otherwise. A set \(\phi^{-1}(c)\) is a regular level set iff \(c\) is a regular value.
Note that if every point of \(M\) is critical then \(\dim M < \dim N\), and every point is regular \(\iff\) \(F\) is a submersion. The set of regular points is always open.
Every regular level set of a smooth map \(\phi: M\to N\) is a properly embedded submanifold of codimension \(\dim N\).
If \(S\hookrightarrow M\) is an embedded submanifold, a defining map for \(S\) is the smooth map \(\phi: M\to N\) such that \(S\) is a regular level set of \(\phi\), if such a map exists.
Example: \(f(\mathbf{x}) = {\left\lVert {\mathbf{x}} \right\rVert}^2\) is the defining map for \(S^n\).
Not every embedded submanifold is the level set of a smooth submersion globally, but this does hold locally. I.e., every embedded submanifold admits a local defining map: :::{.proposition} \(S_k\hookrightarrow M_m\) is an embedded \(k{\hbox{-}}\)dimensional submanifold \(\iff\) every \(s\in S\) admits a neighborhood \(U\) such that \(U\cap S\) is the level set of a smooth submersion \(U\to {\mathbb{R}}^{m-k}\). :::
Immersed submanifolds: more general than embedded submanifolds. Encountered when studying Lie subgroups, where subsets will be the images of injective immersions but not necessarily embeddings (example: figure eight curve).
A subset \(S\subseteq M\) equipped with some topology for which the inclusion \(S\hookrightarrow M\) is a smooth immersion is said to be an immersed submanifold.
Convention: smooth submanifolds always denote immersions, whereas embeddings are a special case.