Differential Geometry. Honours 1996

[malawi_glenn]

http://www.maths.adelaide.edu.au/michael.murray/dg_hons/

Contents
Co-ordinate independent calculus.
Introduction
Smooth functions
Derivatives as linear operators.
The chain rule.
Diffeomorphisms and the inverse function theorem.
Differentiable manifolds
Co-ordinate charts
Linear manifolds.
Topology of a manifold
Smooth functions on a manifold.
The tangent space.
The derivative of a function.
Co-ordinate tangent vectors and one-forms.
How to calculate.
Submanifolds
Tangent space to a submanifold
Smooth functions between manifolds
The tangent to a smooth map.
Submanifolds again.
Vector fields.
The Lie bracket.
Differential forms.
The exterior algebra of a vector space.
Differential forms and the exterior derivative.
Pulling back differential forms
Integration of differential forms
Orientation.
Integration again
Stokes theorem.
Manifolds with boundary.
Stokes theorem.
Partitions of unity.
Vector fields and the tangent bundle.
Vector fields and derivations.
Tensor products
About this document ...

 

[robphy]

There is a .pdf of the 2008 version on http://www.maths.adelaide.edu.au/michael.murray/publications.html .

 

[RedX]

On page 2 of the pdf, section 1.3, what does that funny arrow pointing t to f(x+tv) mean?

I just ignored the t and the arrow and only had f(x+tv) to be differentiated and it makes sense, but I still can't figure out what the t and the arrow mean.

The linear map is v to the derivative so if the arrow means some sort of mapping I don't get it as t is not mapped, v is.

 

[Fredrik]

The arrow, together with the symbols on the left and on the right, mean "the function that takes the thing on the left to the thing on the right". That part of his notation is totally standard, so I'm surprised you don't recognize it. For example, $x\mapsto f(x)$ is the function that takes x to f(x). That function is of course usually written as $f$. This guy is writing [tex]t\mapsto f(x+tv)[/itex] to emphasize that that's the function he's taking the derivative of. Note that f(x+tv) isn't a function. It's a member of the range of f.

In this case, it's completely pointless to emphasize that, since writing the derivative operator as d/dt is enough to show the reader what function he's taking the derivative of. It would have made more sense if he had written $(t\mapsto f(x+tv))'(0)$.

 

[RedX]

Got it, thanks. I think the bar at the end of the arrow threw me off. The notes have

f: R --->R and not f: R |--->R

but when you insert something more specific it's:

t |---> f(x+tv)