De Rham Cohomology- I

by ayushkhaitan3437

This will be a rambling progression to the De Rham cohomology. First we have a map from an open set in the manifold to an open subset of \Bbb{C}^n. A collection of such charts over a cover of the manifold is known as an atlas. There can be many charts and hence many atlases. Do these charts have to agree on the intersection of open sets? No. But \phi_2\circ\phi_1^{-1} has to be diffeomorphic. Why do we have this condition? As discussed in an earlier post, we cannot have consistently map any manifold to \Bbb{C}. The flat “mattress” of \Bbb{C} is a fairly restrictive shape. Hence, we need to weaken this condition.

What is the union of two atlases? You’re mapping the same point on the manifold to different open sets of \Bbb{C}. The more the number of atlases, the more the number of different points that the same point on the manifold may be mapped to. However, as even a neighbourhood of that point is mapped to those very different open sets, differentiability is preserved. In a manner of speaking, things are still very much under control. We can still do calculus. The maximal atlas I suppose is the union of all these atlases. Now it’s possible that the union of two smooth atlases may not be smooth. Hence, we form equivalence classes based on the “compatibility” of atlases- two atlases are called compatible if their union is smooth. Hence union of all atlases in the same equivalence class is a maximal atlas.

Let us now talk about manifolds with boundary. A manifold has a boundary of dimension n if each point has a neighbourhood that can be mapped homeomorphically to an open set of \Bbb{H}^n. A smooth map f:M\to N between manifolds is such that for any U\subset M, there exists V\subset N such that f(U)\subset V, and there exist charts (U,\phi) and (V,\psi):\Bbb{R}^m\to\Bbb{R}^n such that \psi\circ f\circ \phi^{-1} is differentiable. Here the manifold M is assumed to be m-dimensional and N is supposed to be n-dimensional. If f is a smooth homeomorphism and so it f^{-1}, then M and N are called diffeomorphic manifolds. Turns out smooth manifolds form a category called Man. Morphisms are smooth maps, and it is easy to see by the chain rule why morphisms compose to give smooth morphisms.

Now we shall discuss tangent spaces. These, as intuition would suggest, are determined by differentiation. However, differentiation cannot be done on the manifold itself. It has to be done on the image of a chart on each point (to be precise, a neighbourhood around that point). Hence, we take a path containing that point, and then map that path to \Bbb{C}^n. We then find the derivative. This derivative of course is path dependant, but independent of the chart. Two paths are called equivalent if their derivatives at the image of p are the same. The tangent space is the vector space generated by all such equivalence classes of path. Why is the dimension of the tangent space the same as the dimension of the manifold? Just think about paths in all m dimensions of the manifold. Think about why they would be linearly independent.

Let f:M\to N be a smooth map. Then the pushforward or f_* is a map from T_pM \to T_{f(p)}N. It maps [\alpha] to [f\circ\alpha]. This map is well-defined. If \alpha_1'(0)=\alpha_2'(0), then (f\circ\alpha_1)'(0)=(f\circ\alpha_2)'(0). Now we defined a basis for T_pM. The basis is defined as \frac{\partial}{\partial x_i}|_p=(\phi^{-1})_*(e_i) for the chart \phi:M\to \Bbb{C}. What does this mean? We just want to determine the equivalence class of paths that maps to e_i. So the equivalence class of paths that “looks like” that standard ith basis vector.

The tangent bundle is a way of considering the union of all the tangent spaces. Say for p\in M, the fiber \phi^{-1}(p) is T_pM. Then for U\subset M, we have \phi^{-1}(U) as homeomorphic to U\times \Bbb{C}^n. Also, if \pi is the projection operator from U\times \Bbb{C}^n to U, and f is the homeomorhism from \phi^{-1}(U) to U\times \Bbb{C}^n, then \pi\circ f\circ \phi^{-1}(p)=p. What does this condition imply? That (p)\times\Bbb{C}^n is also a sort of tangent space on the manifold M, and T_pM maps to p\times\Bbb{C}^n. The vector bundle is smooth if E and M are both smooth manifolds, and \phi^{-1}:M\to E is also smooth. Let TM be the tangent bundle of the smooth manifold M. Why is TM a smooth manifold? What we need to do is the following: take any chart on M, and extend it to TM (in the obvious way). Then just prove that the transition functions are smooth. They will be smooth because we’re just extending the Jacobian by increasing its rank by, say, n, and just appending I_n to it.

What is a smooth bundle map between two smooth bundles? Remember that the bundle is now a smooth manifold. Hence, we can think of such a map as just a normal smooth map. In fact, we can now talk about the tangent space on this bundle. The maps between tangent spaces should be linear. The bundle map and the map between manifolds should be smooth. And the map between bundles and the map between the manifolds should commute in the obvious way.

A cotangent space is the dual vector space on T_pM on any p\in M, where M is a manifold. The basis vectors dx^i|p of this cotangent space are developed in the usual way, and are called differentials. Pushforwards (f_*) become pullbacks (f^*) the same way that the direction of the map between vectors spaces gets reversed when mapping duals to duals. f^* takes a functional on a certain vector in T_{f(p)}N, and maps it to f_* composed with that functional on the pre-image of that vector. A cotangent bundle is constructed in a way that is similar to the way that a tangent bundle is constructed, and it is also a smooth manifold for the very same reasons. You have a homeomorphism from \phi^{-1}(U) to U\times \Bbb{C}^n.

A differential 1-form is a smooth map f:M\to T^*M such that \pi\circ f=id_M. It is important to note that this is a smooth map. For a map f:M\to N between manifolds, the pullback between differential forms is defined in the obvious way.