calculus review 03
manifolds
New stuff (to me)
definition
manifold: a subset \(M\subset \mathbb{R}^n\) is a smooth \(k\)-dimensional manifold
if locally it is the graph of a \(C^1\) mapping \(f\) expressing \(n-k\) variables as functions of the other \(k\) variables
The definition seems highly related to the implicit function theorem
Therefore, we can quickly catch the spirit below
let \(\mathbf{F}:U\rightarrow \mathbb{R}^{n-k}\) be a \(C^1\) mapping, where \(U\subset\mathbb{R}^n\).
Some subset \(M\) of the domain \(U\) is a \(k\)-dimensional manifold if \(\mathbf{F}(z)=0\) and \([D\mathbf{F}(z)]\) is onto for every \(z\in M\).
(converse) if \(M\) is a smooth \(k\)-dimensional manifold, then for every \(z\in M\), there exists \(\mathbf{F}\) s.t., \([D\mathbf{F}(z)]\) is onto and \(\mathbf{F}(y)=0\) with a neighborhood of \(z\) as the domain
\(\star\) it says we can virtually claim a manifold by \(\mathbf{F}(z)=0\)
intuitively, the definition of manifold should not depend on the coordinates
\(k\)-dimensional manifold \(M\subset\mathbb{R}^m\), \(f\) is a mapping with some properties 1
then \(f^{-1}(M)\) is a submanifold of \(\mathbb{R}^n\) of dimension \(k+n-m\)
1 \(f:U\rightarrow\mathbb{R}^m\) where \(U\) is an open subset of \(\mathbb{R}^n\) and \([Df(x)]\) is surjective at \(\forall x\in f^{-1}(M)\)
as a corollary, manifolds are independent of coordinates if \(f\) is an affine transformation
parametrization of manifolds
parametrization is useful for analysing manifolds since taking a manifolds as domain directly is rather cumbersome
a parametrization of a \(k\)-dimensional manifold \(M\subset\mathbb{R}^n\) is a mapping \(\gamma:U\subset\mathbb{R}^k\rightarrow M\), s.t.,
\(U\) is open
\(\gamma\) is \(C^1\), one to one, and onto \(M\)
\([D\gamma(u)]\) is one to one for every \(u\in U\)
as a comparison, linear algebra and differential calculus have mucn in common
row reduction \(\leftrightarrow\) newton’s method
kernels \(\leftrightarrow\) defining manifolds by equations ( e.g., \(f(x)=0\) )
images \(\leftrightarrow\) defining manifolds by a parametrization
tangent space
tangent space is similar to the linear approximation of a function \(f\), but it is used on a manifold.
- (derivative) replace a function \(f\) locally by a linear map
- (tangent space) replace a manifold locally by a linear space
for a manifold \(M=\left\{\left(\begin{matrix} x\\y \end{matrix}\right)\in\mathbb{R}^n\mid f(x)=y\right\}\)
fix \(z_0\in M\).
the change of \(z\) should be approximated by a linear mapping
\[ y-y_0=[Df(x_0)](x-x_0) \]
in short \(\Delta y=[Df(x_0)]\Delta x\)
the graph of \([Df(x_0)]\) is the tangent space, which is denoted as \(T_{z_0}M\)
the linear approximation to the graph is the graph of the linear approximation
how to find tangent space
if you have an equation form \(F(x)=0\) of the manifold, then you are lucky.
the kernel space of the derivative of \(F\) is the tangent space.
\[ T_{z_0}M=\ker [DF(z_0)] \]
again, this will certainly remind you the implicit function theorem, where we also claim the derivative of the implicit function
here is another approach to get the tangent space
Let \(U\subset\mathbb{R}^k\) be open, and let \(\gamma:U\rightarrow\mathbb{R}^n\) be a parametrization of manifold \(M\), then
\[ T_{\gamma(u)}M=\textrm{img}[D\gamma(u)]. \]
it might look strange at first — why taking the image instead of the kernel like above.
quick calculation:
\[ [D(F\circ\gamma)(u)]=[DF(\gamma(u))]\circ[D\gamma(u)]=[0] \]
derivative is not quite like what we think as the origin function
here \(\textrm{img}[D\gamma(u)]\) and \(\textrm{img}[DF(\gamma(u))]\) are orthogonal complementary subspaces
maps defined on manifolds
TBD :)