Let \(f(x)\) be a smooth map from an open set in \(\mathbb{R}^{n}\) to \(\mathbb{R}^{m}\). \(\forall h \in \mathbb{R}^{n}\), the directional derivative of \(f\) (in direction of h) at \(X\) is defined as \[df_{x}(h) = \lim\limits_{t \to 0} \frac{f(x + th) - f(x)}{t}\] Thus, for each \(x\) in the domain of \(f\), we can define the mapping \(df_{x}(h): \mathbb{R}^{n} \mapsto \mathbb{R}^{m}\) by \(df_{x}(h) \in \mathbb{R}^{m}\). Thus, \(df_{x}\) gives the derivative of \(f\) at \(x\) applied to all \(h \in \mathbb{R}^{n}\).

So if we have \(f(y) = (f_{1}(y), ..., f_{m}(y))\) then \(df_{x}\) may be represented as

So if we have \(f(y) = (f_{1}(y), ..., f_{m}(y))\) then \(df_{x}\) may be represented as

thus, \(df_{x}\) is a linear map.

Chain Rule

- Suppose \(U \subset \mathbb{R}^{n}\), \(V \subset \mathbb{R}^{m}\) are open sets

- and \(f: U \mapsto V\), \(g: V \mapsto \mathbb{R}^{l}\) are smooth maps

Then for each \(x \in U\), \[d(g \circ f)_{x} = dg_{f(x)} \circ df_{x}\] If we have

- and \(f: U \mapsto V\), \(g: V \mapsto \mathbb{R}^{l}\) are smooth maps

Then for each \(x \in U\), \[d(g \circ f)_{x} = dg_{f(x)} \circ df_{x}\] If we have

with derivative maps

- Suppose \(X \subset \mathbb{R}^{N}\), \(\phi: U \mapsto X\) is a local parametrization around \(x\), and \(U\) is an open set in \(\mathbb{R}^{k}\)

- assume that \(\phi(0) = x\)

- then the best linear approximation to \(\phi: U \mapsto X\) at 0 is \(U \mapsto \phi(0) + d\phi_{0}(u) = x + d\phi_{0}(u)\)

- assume that \(\phi(0) = x\)

- then the best linear approximation to \(\phi: U \mapsto X\) at 0 is \(U \mapsto \phi(0) + d\phi_{0}(u) = x + d\phi_{0}(u)\)

tangent space (of X at x): | the image of the map \(d\phi_{0}: \mathbb{R}^{k} \mapsto \mathbb{R}^{N}\) denoted \(T_{x}(X)\) |

for a linear transformation \(T: V \mapsto W\)...

image: | the image im\((T)\) is the set \(\{T(\vec{v}): \vec{v} \in V\}\) that is, all vectors in \(W\) which \(= T(\vec{v})\) for some \(\vec{v} \in V\) |

kernel: | the kernel ker\((T) = \{\vec{v} \in V : T(\vec{v}) = \vec{0}\}\) |

Thus the tangent space \(T_{x}(X)\) given by the image of \(d\phi_{0}: \mathbb{R}^{k} \mapsto \mathbb{R}^{N}\) is a vector subspace of \(\mathbb{R}^{N}\) and \(x + T_{x}(X)\) is the closest flat approximation to \(X\) through \(x\).

tangent vector: | (to \(X\subset \mathbb{R}^{N}\) at \(x \in X\)): a point \(v\) of \(\mathbb{R}^{N}\) that lies in the vector subspace \(T_{x}(X)\) of \(\mathbb{R}^{N}\) |

\(\cdot\) The dimension of the vector space \(T_{x}(X)\) is the dimension \(k\) of \(X\).

"We can now cosntruct the best linear approximation of a smooth map of arbitrary manifolds \(f: X \mapsto Y\) at a point \(x\)."

\(\cdot\) if \(f(x) = y\), the derivative should be a linear transformation of tnagent spaces: \(df_{x}: T_{x}(X) \mapsto T_{y}(Y)\)

Suppose

\(\cdot\) \(\phi:U \mapsto X\) parameterizes \(X\) about \(x\)

\(\cdot\) \(\psi:V \mapsto Y\) parameterizes \(Y\) about \(y\)

\(\cdot\) \(U \subset \mathbb{R}^{k}\) and \(V \subset \mathbb{R}^{l}\)

\(\cdot\) \(\phi(0) = x\) and \(\psi(0) = y\)

for small \(U\), we have:

"We can now cosntruct the best linear approximation of a smooth map of arbitrary manifolds \(f: X \mapsto Y\) at a point \(x\)."

\(\cdot\) if \(f(x) = y\), the derivative should be a linear transformation of tnagent spaces: \(df_{x}: T_{x}(X) \mapsto T_{y}(Y)\)

Suppose

\(\cdot\) \(\phi:U \mapsto X\) parameterizes \(X\) about \(x\)

\(\cdot\) \(\psi:V \mapsto Y\) parameterizes \(Y\) about \(y\)

\(\cdot\) \(U \subset \mathbb{R}^{k}\) and \(V \subset \mathbb{R}^{l}\)

\(\cdot\) \(\phi(0) = x\) and \(\psi(0) = y\)

for small \(U\), we have:

Taking derivatives and applying the **chain rule:**

Since \(d\phi_{0}\) is an isomorphism, \(df_{x}\) must be \[df_{x} = d\psi_{0} \circ dh_{0} \circ d\phi_{0}^{-1}\]

isomorphism:

- a mapping of an open set \(U\)... \[f: U \subset \mathbb{R}^{n} \mapsto \mathbb{R}^{m}\]

is **smooth **if it has continuous partial derivatives

- a mapping of an arbitrary set \(X\)... \[f: X \subset \mathbb{R}^{n} \mapsto \mathbb{R}^{m}\]

\(\cdot\) smooth if for each \(x \in X\), \(\exists\) and open set \(U \subset \mathbb{R}^{n}\) and a smooth map \(F: U \mapsto \mathbb{R}^{m}\)

\(\cdot\) results in \(F = f\) on \(X \cap U\)

\(\cdot\) "smoothness is a local property"

\(\cdot\) smooth if for each \(x \in X\), \(\exists\) and open set \(U \subset \mathbb{R}^{n}\) and a smooth map \(F: U \mapsto \mathbb{R}^{m}\)

\(\cdot\) results in \(F = f\) on \(X \cap U\)

\(\cdot\) "smoothness is a local property"

diffeomorphism:

given \(X \subset \mathbb{R}^{n}\) and \(Y \subset \mathbb{R}^{m}\) and a smooth map \(f: X \mapsto Y\), \(f\) is a diffeomorphism iff it is one-to-one, onto, and \(f^{-1}: Y \mapsto X\) is smooth

if there exists a diffeomorphism between some \(X \subset \mathbb{R}^{n}\) and \(Y \subset \mathbb{R}^{m}\), then \(X\) and \(Y\) are diffeomorphic

Suppose \(X \subset \mathbb{R}^{N}\) (ambient Euclidean space) and for each \(x \in X\), \(\exists\) a neighborhood \(V\) in \(X\) diffeomorphic to an open set \(U\) of some space \(\mathbb{R}^{k}\) via a diffeomorphic \(\phi : U \mapsto V\)

\(\cdot \phi =\) a parametrization of \(V\)

\(\cdot \phi^{-1}: V \mapsto U =\) a coordinate system on \(V\)

\(\cdot \phi =\) a parametrization of \(V\)

\(\cdot \phi^{-1}: V \mapsto U =\) a coordinate system on \(V\)

Example

consider the circle:

\[S^{1} = \{(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} = 1\}\]

\(\cdot\) for upper semicircle with \(y > 0\), \(\phi_{1}: W \mapsto\) upper \(S^{1}\) with \(\phi_{1}(x) = (x, \sqrt{1 - x^{2}})\) and \(W = (-1, 1)\) is a parametrization

\(\cdot\) for the lower semicircle we instead have \(\phi_{2}(x) = (x, -\sqrt{1 - x^{2}}\)

\(\cdot\) for \((1, 0)\) we have \(\phi_{3}(y) = (\sqrt{1 - y^{2}}, y)\)

\(\cdot\) for \((-1, 0)\) we have \(\phi_{4}(y) = (-\sqrt{1 - y^{2}}, y)\)

Since each parametrization is a map of a point in \(\mathbb{R}^{1} \mapsto \mathbb{R}^{2}\), it follows that \(S^{1}\) is a 1 - dimensional manifold.

\[S^{1} = \{(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} = 1\}\]

\(\cdot\) for upper semicircle with \(y > 0\), \(\phi_{1}: W \mapsto\) upper \(S^{1}\) with \(\phi_{1}(x) = (x, \sqrt{1 - x^{2}})\) and \(W = (-1, 1)\) is a parametrization

\(\cdot\) for the lower semicircle we instead have \(\phi_{2}(x) = (x, -\sqrt{1 - x^{2}}\)

\(\cdot\) for \((1, 0)\) we have \(\phi_{3}(y) = (\sqrt{1 - y^{2}}, y)\)

\(\cdot\) for \((-1, 0)\) we have \(\phi_{4}(y) = (-\sqrt{1 - y^{2}}, y)\)

Since each parametrization is a map of a point in \(\mathbb{R}^{1} \mapsto \mathbb{R}^{2}\), it follows that \(S^{1}\) is a 1 - dimensional manifold.

Suppose \(X\) is a manifold in \(\mathbb{R}^{N}\) and \(Y\) a manifold in \(\mathbb{R}^{M}\). That is, \(X \subset \mathbb{R}^{M}\), \(Y \subset \mathbb{R}^{M}\) and \(X \times Y \subset \mathbb{R}^{N} \times \mathbb{R}^{M} = \mathbb{R}^{N + M}\). Let dim\(X = k\) and dim\(Y = l\). Let \(W \subset \mathbb{R}^{k}\) be an open set st. \(\phi: W \mapsto X\) is a local parametrizatio around \(x \in X\) and let \(U \subset \mathbb{R}^{l}\) be an open set st. \(\psi: U \mapsto Y\) is a local parametrization around y. Define \(\phi \times \psi: W\times U \mapsto X\times Y\) by \[\phi \times \psi(w, u) = (\phi(x), \psi(u))\] where \(W\times U\) is an open set in \(\mathbb{R}^{k + l}\) and \(\phi \times \psi\) is a local parametrization around \((x, y) \in X \times Y\).

Thus...

If \(X\) and \(Y\) are manifolds, so is \(X \times Y\) and dim\(X \times Y =\) dim\(X +\) dim\(Y\).

if \(X\) and \(Y\) are manifolds in \(\mathbb{R}^{N}\), and \(Y \subset X\), then \(Y\) is a submanifold of \(X\). \(X\) is also a submanifold of \(\mathbb{R}^{N}\).