Bott and Tu - Differential Forms in Algebraic Topology



Springer GTM 82.

"The guiding principle in this book is to use differential forms as an aid in exploring some of the less digestible aspects of algebraic topology. Accordingly, we move primarily in the realm of smooth manifolds and use the de Rham theory as a prototype of all of cohomology."

=Chapter I: de Rham Theory= Notes for Sections 1-6 (includes a deep treatment of vector bundles for section 5): https://themodularperspective.com/2019/03/25/graduate-text-notes-current-as-of-3-25-2019/

Section I.1: The de Rham Complex on $$\mathbb{R}^n$$
$$ \displaystyle\frac{\Omega^1_c(\mathbb{R}^n)}{\ker \int_{\mathbb{R}}} \cong \mathbb{R}. $$ To see this, assume that f dx and g dx are compactly-supported one-forms on R, and that $$\int_{\mathbb{R}} f \, dx = \int_{\mathbb{R}} g \, dx.$$ Then it follows that (f-g) dx is a compactly supported one-form (its support is contained in the union of supp f and supp g) with integral zero, and therefore $$(f - g) \, dx \in \ker \int_{\mathbb{R}} = {\rm im}\, d.$$ Therefore the equivalence class of a compactly-supported one-form in $$H_{c,DR}^1(\mathbb{R})$$ is completely determined by its integral.
 * Exercise 1.6 It wasn't immediately clear to me that

Section I.2: The Mayer-Vietoris Sequence
http://math.stackexchange.com/questions/94691/why-is-the-pullback-completely-determined-by-d-f-ast-f-ast-d-in-de-rham-co/94702#94702
 * The commutativity with d defines $$f^*$$ uniquely : We need the assumption that $$f^*$$ will be an $$\mathbb{R}$$-algebra homomorphism here -- i.e., it should be $$\mathbb{R}$$-linear and preserve multiplication.

=Chapter 2: The Čech-de Rham Complex=

=Chapter 3: Spectral Sequences and Applications=

=Chapter 4: Characteristic Classes=