0708-1300/Class notes for Tuesday, November 13
| ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Typed Notes
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.
We begin with a review of last class. Since no one has typed up the notes for last class yet, I will do the review here.
Recall had an association [math]\displaystyle{ M\rightarrow\Omega^k(M) }[/math] which was the "k forms on M" which equaled [math]\displaystyle{ \{w:\ p\in M\rightarrow A^k(T_p M)\} }[/math]
where
[math]\displaystyle{ A^k(V):=\{w:\underbrace{V\times\ldots\times V}_{k\ times} \rightarrow\mathbb{R}\} }[/math] which is
1) Multilinear
2) Alternating
We had proved that :
1) [math]\displaystyle{ A^k(V) }[/math] is a vector space
2) there was a wedge product [math]\displaystyle{ \wedge:A^k(V)\times A^l(V)\rightarrow A^{k+l}(V) }[/math] via [math]\displaystyle{ \omega\wedge\lambda(v_1,\ldots,v_{k+l}) = }[/math][math]\displaystyle{ \frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}(-1)^{\sigma} \omega(v_{\sigma(1)},\ldots,v_{\sigma(k)})\lambda(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)}) }[/math] that is
a) bilinear
b) associative
c) supercommutative, i.e., [math]\displaystyle{ \omega\wedge\lambda = (-1)^{deg(w)deg(\lambda)}\lambda\wedge\omega }[/math]
From these definitions we can define for [math]\displaystyle{ \omega\in\Omega^k(M) }[/math] and [math]\displaystyle{ \lambda\in\Omega^l(M) }[/math] that [math]\displaystyle{ \omega\wedge\lambda\in\Omega^{k+l}(M) }[/math] with the same properties as above.
Claim
If [math]\displaystyle{ \omega_1,\ldots,\omega_n }[/math] is a basis of [math]\displaystyle{ A^1(V) = V^* }[/math] then [math]\displaystyle{ \{\omega_I = w_{i_1}\wedge\ldots\wedge\omega_{i_k}\ :\ I=(i_1,\ldots,i_k)\ with\ 1\leq i_1\lt \ldots\lt i_k\leq n\} }[/math] is a basis of [math]\displaystyle{ A^k(V) }[/math] and [math]\displaystyle{ dim A^k(V) = nCk }[/math]
If [math]\displaystyle{ \omega_1,\ldots,\omega_n\in\Omega^1(M) }[/math] and [math]\displaystyle{ w_1 |_p,\ldots,\omega_n |_p }[/math] a basis of [math]\displaystyle{ (T_p M)^*\ \forall p\in M }[/math] then any [math]\displaystyle{ \lambda }[/math] can be written as [math]\displaystyle{ \lambda = \sum_I a_I(p)\omega_I }[/math] where [math]\displaystyle{ a_I:M\rightarrow\mathbb{R} }[/math] are smooth.
The equivalence of these is left as an exercise.
Example
Let us investigate [math]\displaystyle{ \Omega^*(\mathbb{R}^3) }[/math] (the * just means "anything").
Now, [math]\displaystyle{ (T_p(\mathbb{R}^3))^* = \lt dx_1,dx_2,dx_3\gt }[/math] where [math]\displaystyle{ x_i:\mathbb{R}^3\rightarrow\mathbb{R} }[/math] and so [math]\displaystyle{ dx_i|_p:T_p\mathbb{R}^3\rightarrow T_{x_i(p)}\mathbb{R} =\mathbb{R} }[/math]
Hence, [math]\displaystyle{ dx_i|_p\in (T_p\mathbb{R}^3)^* }[/math]
Now, [math]\displaystyle{ dx_2(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}x_2 =\delta_{i2} }[/math] and hence we get a basis.
So, [math]\displaystyle{ \Omega^1(\mathbb{R}^3) = \{g_1 dx_1 + g_2 dx_2 + g_3 dx_3\}\approx\{g_1,g_2,g_3\}\approx }[/math] {vector fields on [math]\displaystyle{ \mathbb{R}^3 }[/math]}
where the [math]\displaystyle{ g_i:\mathbb{R}^3\rightarrow\mathbb{R} }[/math] are smooth.
[math]\displaystyle{ \Omega^0(\mathbb{R}^3) = \{f:\mathbb{R}^3\rightarrow\mathbb{R}\} }[/math]
This is because to each point p we associate something that takes zero copies of the tangent space into the real numbers. Thus to each p we associate a number.
[math]\displaystyle{ \Omega^3(\mathbb{R}^3) = \{kdx_1\wedge dx_2\wedge dx_3\} \approx }[/math] {functions} where again the k is just a smooth function from [math]\displaystyle{ \mathbb{R}^3 }[/math] to [math]\displaystyle{ \mathbb{R} }[/math].
[math]\displaystyle{ \Omega^2(\mathbb{R}^3) = \{h_1 dx_2\wedge dx_3 + h_2 dx_3\wedge dx_1 + h_3 dx_1\wedge dx_2\}\approx \{h_1,h_2,h_3\}\approx }[/math] {vector fields}
Aside
Recall our earlier discussion of how points and things like points (curves, equivalence classes of curves) pushfoward while things dual to points (functions) pullback and that things dual to functions (such as derivations) push forward. See earlier for the precise definitions.
Now differential forms pull back, i.e., for [math]\displaystyle{ \phi:M\rightarrow N }[/math] then [math]\displaystyle{ \phi^*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N) }[/math]
via
[math]\displaystyle{ \phi^*(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_* v_1,\ldots \phi_* v_k) }[/math]
The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via [math]\displaystyle{ \phi^*(\omega\wedge\lambda)=\phi^*(\omega)\wedge\phi^*(\lambda) }[/math]
Theorem-Defintion
Given M, [math]\displaystyle{ \exists }[/math] ! linear map [math]\displaystyle{ d:\Omega^*(M)\rightarrow\Omega^{k*+1}(M) }[/math] satisfies
1) If [math]\displaystyle{ f\in\Omega^0(M) }[/math] then [math]\displaystyle{ df(X) = X(f) }[/math] for [math]\displaystyle{ X\in TM }[/math]
2) [math]\displaystyle{ d^2 = 0 }[/math]. I.e. if [math]\displaystyle{ d_k:\Omega^k\rightarrow\Omega^{k+1} }[/math] and [math]\displaystyle{ d_{k+1}:\Omega^{k+1}\rightarrow\Omega^{k+2} }[/math] then [math]\displaystyle{ d_{k+1}\circ d_k = 0 }[/math].
3) [math]\displaystyle{ d(\omega\wedge\lambda) = d\omega\wedge\lambda + (-1)^{deg\omega}\omega\wedge d\lambda }[/math]
