In particular, reall that we have provisionally defined a subset $S\subset \mathbb{R}^n$ to be a smooth subset of dimension $k$ if every $s\in S$ has an open neighborhood $U\subset \mathbb{R}^n$ on which there exists a smooth function $f:U\to f(U)\subset\mathbb{R}^n$ with smooth inverse $g:f(U)\to U$ such that $f(S\cap U) = \mathbb{R}^k\cap f(U)$. (I.e., $S$ 'looks like' $\mathbb{R}^k\subset\mathbb{R}^n$ locally, up to invertible smooth maps.)
On the way to defining the notion of a smooth manifold (more in the next lecture), I defined the idea of an $n$-atlas on a set $M$: A $C^k$ $n$-atlas (where $0\le k\le \infty$) on a set $M$ is a collection $\mathcal{A}=\{(f_\alpha,U_\alpha)\ |\ \alpha\in A\}$ where, for each $\alpha,\beta$ in the index set $A$, we have
Discussion of some famous theorems, such as "For $n\not=4$,any two smooth structures that induce the usual topology on $\mathbb{R}^n$ are diffeomorphic", but "There are infinitely many mutually non-diffeomorphic smooth $4$-manifolds that are homeomorphic to $\mathbb{R}^4$ [Donaldson]", and "There are 28 mutually non-diffeomorphic manifolds that are homeomorphic to $S^7$ [Milnor]."
Unfortunately, I got the hypothesis wrong in this problem. It's not enough that the group action be discrete to ensure that $\Gamma\backslash M$ is Hausdorff; it has to be proper, which in this case, means that, for every compact set $K\subset M$, the set of $g\in\Gamma$ for which $g{\cdot}K\cap K$ is nonempty, is finite. I'm sorry about this mistake. I should have been more careful. Here's an example to show that you need this: Let $M$ be $\mathbb{R}^2$ minus the origin $O = (0,0)$, and let $\Gamma= \mathbb{Z}$, with the action $n\cdot(x,y)= (2^nx,2^{-n}y)$. Then, in the quotient topology, $\Gamma\backslash M$ is not Hausdorff, since $(1,0)$ and $(0,1)$ will not have disjoint open neighborhoods in the quotient. Do you see why?
On the other hand, I didn't actually ask you to show that $\Gamma\M$ is Hausdorff, so the problem is technically not wrong, but I certainly wanted to use this method to induce a smooth manifold structure on the quotient, so not warning you about this was actually a mistake of mine. Meanwhile, in the specific examples that I gave, the actio is proper, so the quotients are smooth manifolds.
For example, $\Gamma = \mathbb{Z}^n\subset\mathbb{R}^n$ acts freely, discretely and smoothly on $\mathbb{R}^n$ by vector addition. The resulting quotient manifold $\mathbb{Z}^n\backslash\mathbb{R}^n$ is a smooth $n$-manifold called the $n$-torus, $\mathbb{T}^n$. There are many other examples. For example, $\mathbb{Z}_2$ acts freely, discretely, and smoothly on $S^n$ by $(\pm 1)\cdot x = \pm x$. The quotient manifold $\mathbb{Z}_2\backslash S^n$ is diffeomorphic to $\mathbb{RP}^n$. Do you see why? (N.B.: Because $\mathbb{Z}^n$ and $\mathbb{Z}_2$ are abelian groups, there is not really any difference between left actions and right actions for these groups, so most books will write $\mathbb{T}^n = \mathbb{R}^n/\mathbb{Z}^n$ and $\mathbb{RP}^n = S^n/\mathbb{Z}_2$.)
Note: I also, as a point of interest, proved that $TS^2$ is not just $S^2\times\mathbb{R}^2$. That won't be important right away, and we'll see a different proof later in the course.
Smooth vector fields on $M$ as smooth maps $Z:M\to TM$ that satisfy $\pi\circ Z = \mathrm{id}_M$. For a coordinate chart $f = (x^i):U\to\mathbb{R}^n$, the smooth vector fields $X_i = {{\partial\ }\over{\partial x^i}}$ on $U$, defined so that $D_{X_i(p)}:C^\infty(M)\to\mathbb{R}$ satisfies $D_{X_i(p)}(x^j) = \delta^i_j$ for all $i$ and $j$ and all $p\in U$.
The differential $df$ of a function $f\in C^\infty(M)$ as a function $df:TM\to\mathbb{R}$ that is linear on each $T_pM$ by $df(v) = D_vf$. For a coordinate chart, the identity $dx^i( X_j) = \delta^i_j$ for all $i$ and $j$. The definition of a (smooth) $1$-form $\alpha$ as a (smooth) function $\alpha:TM\to\mathbb{R}$ that is linear on each vector space $T_pM$. Regarding $\alpha$ as a section of the cotangent bundle $\pi: T^*M\to M$, whose fiber $\pi^{-1}(p) = T^*_pM$ is the vector space dual to $T_pM$.
The equivalence between vector fields on $M$ (i.e., sections of $\pi:TM\to M$) and smooth derivations of the ring of smooth functions on $M$, i.e., $C^\infty(M)$.
The commutator $[\delta_1,\delta_2] = \delta_1\delta_2-\delta_2\delta_1$ of smooth derivations $\delta_1$ and $\delta_2$ on $M$ is a smooth derivation on $M$. Formal identities for the commutator. The Lie bracket of vector fields $[X,Y]$ as a vector field, and its explicit expression in a coordinate chart.
Review of the basic existence and uniqueness theorem for smooth ODE.
Integral curves of a vector field $X$ and the manifold version of the ODE existence theorem.
The statement of the FlowBox Theorem and the Simultaneous FlowBox Theorem. (Proofs next time.)
Let $D\subset\mathbb{R}^n$ be an open set and let $f:D\to\mathbb{R}^n$ be a smooth map. The Existence and Uniqueness Theorem for Ordinary Differential Equations states that there is an open neighborhood $\mathscr{D}\subset\mathbb{R}\times D$ of $\{0\}\times D$ with the property that, for each $p\in D$, the set $\mathscr{D}_p = \{ t\in\mathbb{R} \ |\ (t,p)\in \mathscr{D}\}$ is an open interval containing $0\in\mathbb{R}$ and a smooth mapping $F:\mathscr{D}\to D$ with the property that, for $p\in D$, the set $\mathscr{D}_p$ is the largest open interval in $\mathbb{R}$ on which the ODE $\dot x = f(x)$ with initial value $x(0)=p$ has a solution in $D$ and that solution, which is unique, is $x(t) = F(t,p)$ for $t\in \mathscr{D}_p$.
Show that if $s\in \mathscr{D}_p$, then $\mathscr{D}_{F(s,p)} = \{ u- s\ |\ u\in \mathscr{D}_p\}$ and $F(t,F(s,p)) = F(t+s,p)$ for $t\in\mathscr{D}_{F(s,p)}$. (Hint: Use the uniqueness of solutions of ODE with given intial values and the Chain Rule.)
For the ODE $\dot x = x^2$ on $M=\mathbb{R}$, determine the domain of the function $F:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ such that, for all $p\in\mathbb{R}$, the function $x(t) = F(t,p)$ is the maximally extended solution of the initial value problem $\dot x = x^2$ with $x(0)=p$.
(Optional) Let $M$ be a smooth manifold. If $c:\mathbb{R}\to M$ is a smooth curve and $h:\mathbb{R}\to\mathbb{R}$ is a smooth mapping, show, using the definition of $c':\mathbb{R}\to TM$, that $(c\circ h)'(t) = h'(t)\,c'(h(t))$. Conclude that if $h(0)=0$ and $h(1)=1$, then for any smooth $1$-form $\alpha:TM\to M$, we have $\int_0^1 \alpha\bigl(c'(t)\bigr)\,dt = \int_0^1 \alpha\bigl((c\circ h)'(u)\bigr)\, du$.
(Optional) If $M=\mathbb{R}^2$ with standard coordinates $(x,y)$, let $(x,y,u,v)$ be coordinates on $TM\simeq\mathbb{R}^2\times\mathbb{R}^2$ such that, if $c(t) = \bigl(x(t),y(t)\bigr)$ is a curve in $M$, then $c'(t) = \bigl(x(t),y(t),x'(t),y'(t)\bigr)$. Write down the expressions for $\mathrm{d}x$ and $\mathrm{d}y$ (as functions on $TM$ in terms of the coordinates $(x,y,u,v)$. Show that, for $f = f(x,y)$ a smooth function on $M$, we have $\mathrm{d}f = f_x\,\mathrm{d}x + f_y\,\mathrm{d}y$. Conclude that the $1$-form $\alpha = x\,\mathrm{d}y - y\,\mathrm{d}x$ is not of the form $\mathrm{d}f$ for any smooth function $f$ on $M$.
Manifold smooth ODE existence and uniqeness: Let $X:M\to TM$ be a smooth vector field on a manifold $M$. There exists an open neighborhood $\mathscr{D}\subset\mathbb{R}\times M$ of $\{0\}\times M$ and a smooth map $\Phi:\mathscr{D}\to M$ with the property that, for each $p\in M$, the set $\mathscr{D}_p = \{t\in\mathbb{R} |(t,p)\in\mathscr{D}\}$ is an open interval containing $0\in\mathbb{R}$ and the curve $x(t)=\Phi(t,p)$ for $t\in\mathscr{D}_p$ is the maximally defined solution to the initial value problem $$ x(0) = p,\qquad x'(t) = X\bigl(x(t)\bigr). $$ If $s\in \mathscr{D}_p$, then $\mathscr{D}_{\Phi(s,p)} = \{ u-s\ |\ u\in \mathscr{D}_p\}$. Moreover, $\Phi\bigl(t,\Phi(s,p)\bigr) = \Phi(t{+}s,p)$ for $t\in \mathscr{D}_{\Phi(s,p)}$. (The map $\Phi$ is called the local flow of $X$.)
The Simultaneous Flowbox Theorem: . If $X_1,\ldots,X_m$ are smooth vector fields on $M$ satisfying $[X_i,X_j] = 0$ for $1\le i,j\le m$ and $p\in M$ is such that the vectors $X_1(p),\ldots,X_m(p)$ are linearly independent in $T_pM$, then there is a $p$-centered coordinate chart $(x,U)$ where $U\subset M$ is an open neighborhood of $p$, such that, on $U$, we have $$ X_i = {{\partial\ }\over{\partial x^i}}\quad \mathrm{for}\ 1\le i\le m. $$
If $a = (a^i_j)\in M_n(\mathbb{R})$ is an $n$-by-$n$ matrix, define a vector field on $\mathbb{R}^n$ by (summation convention assumed) $$ X_a = a^i_j\,x^j\,{{\partial\ }\over{\partial x^i}}. $$ Show that, for $a,b\in M_n(\mathbb{R})$, we have $[X_a,X_b] = -X_{[a,b]}$ where $[a,b] = ab-ba$. (This 'unexpected' minus sign is one reason that some references define the Lie bracket to be the negative of our Lie bracket, which is the most common convention.)
(Optional) Consider a pair of first order partial differential equations for a function $u(x,y)$: $$ u_x(x,y) = f\bigl(x,y,u(x,y)\bigr)\,\qquad u_y(x,y) = g\bigl(x,y,u(x,y)\bigr) $$ where $f$ and $g$ are defined in an open domain in $xyz$-space.
These equations do not always have a solution. For example, if $f \equiv y$ and $g\equiv-x$, show that there is no solution by expanding the identity $(u_x)_y - (u_y)_x = 0$ in this case.
Show that a function $u(x,y)$ on some domain in the $xy$-plane satisfies the above equations if and only if the surface $z = u(x,y)$ is tangent to the vector fields $$ X = {{\partial\ }\over{\partial x}} + f(x,y,z)\,{{\partial\ }\over{\partial z}} \quad\text{and}\quad Y = {{\partial\ }\over{\partial y}} + g(x,y,z)\,{{\partial\ }\over{\partial z}}. $$ Show that $$ [X,Y] = \bigl(g_x - f_y + f g_z - g f_z\bigr)\,{{\partial\ }\over{\partial z}}. $$ Apply the Simultaneous Flowbox Theorem (assuming that $f$ and $g$ satisfy $g_x - f_y + f g_z - g f_z=0$) to show that, in a neighborhood of any point $p=(x,y,z)$, there are coordinates $(v^1,v^2,v^3)$ in which $$ X = {{\partial\ }\over{\partial v^1}}\quad\text{and}\quad Y = {{\partial\ }\over{\partial v^2}}\,. $$ Explain why the level sets of $v^3$ (in the domain of the $v^i$) are locally graphs of solutions to the original system of equations.
Notation: The space ${\mathfrak{X}}(M)$ of smooth vector fields on $M$ and $\Omega^1(M)$, the space of smooth $1$-forms on $M$. The action of $C^\infty(M)$, the ring of smooth functions on $M$, on ${\mathfrak{X}}(M)$ and $\Omega^1(M)$, making these two spaces into modules over $C^\infty(M)$. The natural pairing ${\mathfrak{X}}(M)\times\Omega^1(M)\to C^\infty(M)$.
The definition of the classical covariant bilinear $\mathrm{d}\alpha$ of a $1$-form $\alpha\in \Omega^1(M)$ as a $C^\infty(M)$-bilinear pairing $\mathrm{d}\alpha: {\mathfrak{X}}(M)\times {\mathfrak{X}}(M)\to C^\infty(M)$: $$ \mathrm{d}\alpha(X,Y) = X\bigl(\alpha(Y)\bigr)-Y\bigl(\alpha(X)\bigr) - \alpha\bigl([X,Y]\bigr). $$ The verification that, if $\alpha = \mathrm{d}f$, then $\mathrm{d}\alpha = 0$ and the computation of an example to show that $\mathrm{d}\alpha$ need not be zero. Thus, showing that there is no straightforward `flowbox theorem' for $1$-forms.
The general definition of a $2$-form on $M$ as a $C^\infty(M)$-bilinear pairing $\beta: {\mathfrak{X}}(M)\times {\mathfrak{X}}(M)\to C^\infty(M)$ that is skew-symmetric, i.e., $\beta(X,Y) = -\beta(Y,X)$. The space of such $2$-forms is denoted $\Omega^2(M)$, and it is naturally a $C^\infty(M)$-module. The definition of the exterior product (aka wedge product) of two $1$-forms $\alpha_1,\alpha_2\in\Omega^1(M)$ as $$ (\alpha_1\wedge\alpha_2)(X,Y) = \alpha_1(X)\,\alpha_2(Y)-\alpha_1(Y)\,\alpha_2(X). $$ This defines a pairing $\wedge:\Omega^1(M)\times\Omega^1(M)\to \Omega^2(M)$, and, noting that $\alpha_1\wedge\alpha_2 = - \alpha_2\wedge\alpha_1$, we see that this pairing is skew-symmetric. Also, the covariant bilinear operation defines a mapping $\mathrm{d}:\Omega^1(M)\to\Omega^2(M)$. Note that this pairing is not $C^\infty(M)$-linear, since $\mathrm{d}(f\alpha) = \mathrm{d}f\wedge\alpha + f\,\mathrm{d}\alpha$ for any $f\in C^\infty(M)$ and $\alpha\in\Omega^1(M)$.,
The spaces $\Omega^1(\mathbb{R}^3)$ and $\Omega^2(\mathbb{R}^3)$ and expressions in coordinates for the exterior derivative (AKA 'differential') of functions $f$ (henceforth to be called $0$-forms from time to time), and $1$-forms. Comparison with the operations of $\mathrm{grad}$ and $\mathrm{curl}$ in $3$-D vector calculus.
An analogous expression for differentiating $2$-forms: If $\beta$ is a $2$-form and $X_1,X_2,X_3$ are vector fields on $M$, then we define $$ \begin{aligned} \mathrm{d}\beta(X_1,X_2,X_3) &= X_1\bigl(\beta(X_2,X_3)\bigr) +X_2\bigl(\beta(X_3,X_1)\bigr) +X_3\bigl(\beta(X_1,X_2)\bigr) \\ &\quad -c\bigl(\beta(X_1,[X_2,X_3])+\beta(X_2,[X_3,X_1])+\beta(X_3,[X_1,X_2])\bigr)\\ (&= X_1\bigl(\beta(X_2,X_3)\bigr) +X_2\bigl(\beta(X_3,X_1)\bigr) +X_3\bigl(\beta(X_1,X_2)\bigr) \\ &\quad +c\bigl(\beta([X_2,X_3],X_1)+\beta([X_3,X_1],X_2)+\beta([X_1,X_2],X_3)\bigr)\quad), \end{aligned} $$ with the 'universal' constant $c$ chosen so that $\mathrm{d}\beta(X_1,X_2,fX_3) = f\,\mathrm{d}\beta(X_1,X_2,X_3)$. (Exercise: Determine $c$!). Note that $\mathrm{d}\beta(X_1,X_2,X_3)$ is fully skew-symmetric in its entries, i.e., $$ \mathrm{d}\beta(X_1,X_2,X_3) = \mathrm{d}\beta(X_2,X_3,X_1) =\mathrm{d}\beta(X_3,X_1,X_2) = -\mathrm{d}\beta(X_1,X_3,X_2) = -\mathrm{d}\beta(X_2,X_1,X_3) = -\mathrm{d}\beta(X_3,X_2,X_1). $$ Let $\Omega^3(M)$ denote the space of $C^\infty(M)$-trilinear, skew-symmetric maps $$ \gamma:{\mathfrak{X}}(M)\times {\mathfrak{X}}(M)\times {\mathfrak{X}}(M)\to C^\infty(M), $$ so that $\mathrm{d}:\Omega^2(M)\to\Omega^3(M)$. The computation of this map when $M=\mathbb{R}^3$ and comparison with the operator $\mathrm{div}$ from $3$-D vector calculus.
Show, using the definitions, that, for $f\in C^\infty(M)$ (also denoted $\Omega^0(M)$ sometimes) and $\alpha\in \Omega^1(M)$, we have $\mathrm{d}(f\alpha) = \mathrm{d}f\wedge\alpha + f\,\mathrm{d}\alpha$. (Hint: To verify this formula, you can assume that the vector fields $X_1$ and $X_2$ that you use to check the equality of the two sides satisfy $[X_1,X_2]=0$. Indeed, you can assume that these vector fields are coordinate vector fields. Why?)
(Optional) Define a 'wedge product' pairing $\wedge:\Omega^1(M)\times\Omega^2(M)\to\Omega^3(M)$ by the formula $$ (\alpha \wedge \beta)(X_1,X_2,X_3) = \alpha(X_1)\beta(X_2,X_3) + \alpha(X_2)\beta(X_3,X_1) + \alpha(X_3)\beta(X_1,X_2), $$ for $\alpha\in\Omega^1(M)$ and $\beta\in\Omega^2(M)$, and define $\beta\wedge\alpha$ to be $\alpha\wedge\beta$ for $\alpha\in\Omega^1(M)$ and $\beta\in\Omega^2(M)$. Verify that $\alpha_1\wedge(\alpha_2\wedge\alpha_3) = \alpha_2\wedge(\alpha_3\wedge\alpha_1)$ and that $(\alpha_1\wedge\alpha_2)\wedge\alpha_3 = \alpha_1\wedge(\alpha_2\wedge\alpha_3)$.
(Optional) Show that for $f\in C^\infty(M)$ and $\beta\in\Omega^2(M)$, we have $\mathrm{d}(f\beta) = \mathrm{d}f\wedge\beta + f\,\mathrm{d}\beta$ and that, for $\alpha_1,\alpha_2\in\Omega^1(M)$, we have $\mathrm{d}(\alpha_1\wedge\alpha_2) = \mathrm{d}\alpha_1\wedge\alpha_2 -\alpha_1\wedge\mathrm{d}\alpha_2$. (Hint: To verify these formulae, you can assume that the vector fields $X_i$ that you use to check the identities satisfy $[X_i,X_j]=0$ for all $i$ and $j$.)
On a smooth manifold $M$, a smooth exterior differential $p$-form $\beta$ on $M$ is a $C^\infty(M)$-multilinear, skew-symmetric mapping $$ \beta:\underbrace{{\mathfrak{X}}(M)\times \ldots \times {\mathfrak{X}}(M)}_{p\ \rm times}\to C^\infty(M). $$ The set of such $p$-forms is denoted $\Omega^p(M)$, and it is a $C^\infty(M)$-module. The value of the function $\beta(X_1,\ldots,X_p)$ at $m\in M$ depends only on the values of the vector fields $X_i$ at $m$, so there is a well-defined skew-symmetric multilinear map $$ \beta_m:\underbrace{T_mM\times \ldots \times T_mM}_{p\ \rm times}\to \mathbb{R}. $$ (The proof uses bump functions.). We have $\Omega^p(M) = (0)$ unless $0\le p\le n = \dim M$. By convention, $\Omega^0(M)$ is defined to be $C^\infty(M)$.
The definition of the wedge product $\wedge:\Omega^p(M)\times\Omega^q(M) \to \Omega^{p+q}(M)$, which makes the direct sum $\Omega^*(M) = \bigoplus_{p=0}^n\Omega^p(M)$ into a graded ring that is associative and commutative in the graded sense: $\alpha\wedge\beta = (-1)^{pq}\beta\wedge\alpha$ if $\alpha\in\Omega^p(M)$ and $\beta\in\Omega^q(M)$.
For any smooth mapping $\phi:M\to N$, there is an induced pullback mapping $\phi^*:\Omega^p(N) \to \Omega^p(M)$ defined so that, for $v_1,\ldots,v_p\in T_mM$, we have $$ (\phi^*\alpha)_m(v_1,\ldots,v_p) = \alpha_{\phi(m)}\bigl(D\phi(v_1),\ldots,D\phi(v_p)\bigr). $$ If $\phi:M\to N$ and $\psi:N\to P$ are smooth maps, then $(\psi\circ\phi)^* = \phi^*\circ\psi^*$ (which follows immediately from the Chain Rule: $D(\psi\circ\phi) = D\psi\circ D\phi$) and $\phi^*:\Omega^*(N)\to\Omega^*(M)$ is a homomorphism of graded algebras.
The definition of the exterior derivative $\mathrm{d}:\Omega^p(M)\to\Omega^{p+1}(M)$ and a discussion of its defining properties:
Show that, if $\alpha^1,\ldots,\alpha^p\in\Omega^1(M)$ and $X_1,\ldots,X_p\in{\mathfrak{X}}(M)$, then $\alpha^1\wedge\cdots\wedge\alpha^p(X_1,\ldots,X_p) = \det\bigl( \alpha^i(X_j)\bigr)$. (Hint: Use induction on $p$ and an expansion property of determinants.)
Show that, if the $1$-forms $\alpha^1,\ldots\alpha^p$ are linearly dependent in the sense that there exist smooth functions $\lambda_i$, not all simultaneously vanishing, such that $\lambda_i\,\alpha^i = 0$, then $\alpha^1\wedge\cdots\wedge\alpha^p = 0$. Conversely, if $\alpha^1_m,\ldots\alpha^p_m$ are linearly independent at $m\in M$, show that $(\alpha^1\wedge\cdots\wedge\alpha^p)_m\not=0$.
Suppose that $\alpha^1,\ldots,\alpha^p$ are everywhere linearly independent $1$-forms on $M$. For a strictly increasing sequence $I = (i_1,\ldots,i_k)$ where $1 \le i_1 < i_2 < \cdots < i_k \le p$, set $|I| = k$ and $\alpha^I = \alpha^{i_1}\wedge\cdots\wedge\alpha^{i_k}$. Show that the set of $k$-forms $\{\ \alpha^I\ |\ |I| = k\ \}$ is linearly independent everywhere on $M$. (Hint: Find vector fields $X_1,\ldots X_k$ on $M$ (locally near every point if necessary) such that $\alpha^i(X_j) = \delta^i_j$ and use those.)
Prove Cartan's Lemma: If $\alpha^1,\ldots\alpha^p\in\Omega^1(M)$ are linearly independent at every point of $M$ and $\beta_1,\ldots,\beta_p\in\Omega^1(M)$ satisfy $\beta_i\wedge\alpha^i = 0$, then there exist smooth functions $s_{ij} = s_{ji}$ on $M$ such that $\beta_i = s_{ij}\,\alpha^j$. (Hint: First, wedge the equation $\beta_i\wedge\alpha^i = 0$ with $p{-}1$ of the $\alpha$s to conclude that $\beta_i\wedge\alpha^1\wedge\cdots\wedge\alpha^p = 0$. What does that tell you?)
Computation of simple examples: $M = \mathbb{R}$, $S^1$, or $\mathbb{R}^2$.
The Homotopy Lemma: For any manifold $M$, there is a linear map $h:\Omega^*(M\times\mathbb{R}) \to \Omega^*(M)$ of degree $-1$ (i.e., if $\phi$ has degree $p$ then $h(\phi)$ has degree $p{-}1$) that satisfies $$ h(\mathrm{d}\phi) +\mathrm{d}\bigl(h(\phi)\bigr) = \iota_1^*\phi - \iota_0^*\phi, $$ where $\iota_a:M\to M\times\mathbb{R}$ is given by $\iota_a(m) = (m,a)\in M\times\mathbb{R}$.
The Poincaré Lemma: If $U\subset\mathbb{R}^n$ is star-shaped with respect to some point $p\in U$, then $H^*_{dR}(U) = H^0_{dR}(U) \simeq\mathbb{R}$. (In other words, every closed smooth form of positive degree on $U$ is exact.)
Show, by direct calculation, that $H^n_{dR}(\mathbb{R}^n) = (0)$. (We did the case $n=1$ in class. Hint: If $\alpha = f(x)\,\mathrm{d}x^1\wedge\mathrm{d}x^2\wedge\cdots\wedge\mathrm{d}x^n$, look for a simple $\beta$ satisfying $\mathrm{d}\beta = \alpha$ of the form $F(x)\,\mathrm{d}x^2\wedge\cdots\wedge\mathrm{d}x^n$.)
Assuming the Poincaré Lemma, show that $H^1_{dR}(S^n)=0$ for $n>1$. (Hint: Let $U_+\subset S^n$ (respectively, $U_-\subset S^n$) be the open set obtained by removing the north pole (respectively, the south pole). Look back at Day 3, Exercise 1, and argue that $H^1_{dR}(U_\pm)=0$, so that, for $\alpha\in Z^1(S^n)$, there exist a smooth function $a_+$ on $U_+$ and a smooth function $a_-$ on $U_-$ such that $\alpha = \mathrm{d}a_\pm$ on $U_\pm$. How much ambiguity is there in the choice of such functions, and what do you know about $a_+-a_-$ on $U_+\cap U_-$?
Two smooth maps $f_0,f_1:M\to N$ are said to be smoothly homotopic if there is a smooth map $F:M\times \mathbb{R}\to N$ such that $f_0(m) = F(m,0)$ and $f_1(m) = F(m,1)$. Use the Homotopy Lemma to show that, in such a situation, the induced maps $f_0^*$ and $f_1^*$ of $H^*_{dR}(N)$ to $H^*_{dR}(M)$ are the same.
A submanifold $S\subset M$ is said to be a a deformation retract of $M$ if there is a smooth mapping $F:M\times\mathbb{R}\to M$ such that $F(m,0) = m$ for all $m\in M$ while $F(m,1)\in S$ for all $m\in M$ and $F(s,1) = s$ for all $s\in S$. Show that, in this case, the inclusion mapping $\iota:S\to M$ induces an isomorphism $\iota^*:H^*_{dR}(M)\to H^*_{dR}(S)$.
The local Frobenius Theorem: If $E\subset TM$ is a smooth involutive $k$-plane field on $M$, then each $p\in M$ has an open neighborhood $U\subset M$ on which there exists a smooth coordinate chart $x = (x^i):U\to\mathbb{R}^n$ such that, in $U$, the $k$-plane field $E$ is generated by $\{\partial/\partial x^1,\ ,\partial/\partial x^2,\ ,\ldots,\ \partial/\partial x^k\}$.
The notion of an $E$-atlas for an involutive $k$-plane field $E\subset TM$.
A curve $\gamma:(a,b)\to M$ is an $E$-curve if $\gamma'(t)\in E_{\gamma(t)}$ for all $t\in(a,b)$.
The global Frobenius Theorem: Let $E\subset TM$ be a smooth, involutive $k$-plane field on $M$. For each $p\in M$, let $L\subset M$ denote the set of points of the form $\gamma(1)$ where $\gamma:[0,1]\to M$ is an $E$-curve with $\gamma(0)=p$. Then $L$ can be given the structure of a connected, smooth $k$-manifold for which the inclusion mapping $\iota:L\to M$ is a smooth mapping and the differential $D\iota:TL\to TM$ is injective. ($L$ is called the leaf of $E$ through $p$. The collection of all such leaves of $E$ is called the foliation of $M$ associated to $E$.)
The dual formulation of the Frobenius Theorem in terms of the subbundle $E^\perp\subset T^*M$ associated to a smooth $k$-plane field $E\subset TM$: $E$ is involutive if and only if the algebraic ideal in the exterior forms generated by the $1$-forms that annihilate $E$ is closed under exterior derivative.
For any Lie subalgebra $\mathfrak{g}\subset M_n(\mathbb{R}) = \mathrm{End}(\mathbb{R}^n$, we showed the existence (and uniqueness) of a (path-)connected smooth subgroup $G\subset \mathrm{GL}(n,\mathbb{R})$ for which $T_{I_n}G = \mathfrak{g}$. Explain why our construction using the Frobenius Theorem implies that $\exp(\mathfrak{g})\subseteq G$ and that $\exp:\mathfrak{g}\to G$ is a diffeomorphism from a neighborhood of $0_n\in\mathfrak{g}$ to a neighborhood of $I_n\in G$ (endowed with the topology that makes it a smooth manifold of dimension equal to the dimension of $\mathfrak{g}$).
Show that if $a,b\in M_n(\mathbb{R})$ satisfy $[a,b]=0$, then $\exp(ta+sb) = \exp(ta)\exp(sb) = \exp(sb)\exp(ta)$ for all $t,s\in\mathbb{R}$. Also show that $$ \exp\begin{pmatrix} t & s\\0 & 0\end{pmatrix} = \begin{pmatrix} e^t & \bigl((e^t{-}1)/t\bigr)s\\0 & 1\end{pmatrix}. $$
Define two subsets $G_\pm\subset \mathrm{SO}(4)\subset \mathrm{GL}(4,\mathbb{R})$ as the set of elements $q_+(x)\in G_+$ and $q_-(x)\in G_-$ where $$ q_+(x) = \begin{pmatrix}x_0&-x_1&-x_2&-x_3\\x_1&x_0& x_3 & -x_2\\ x_2 & -x_3 & x_0 & x_1\\x_3&x_2&-x_1&x_0\end{pmatrix} \quad\text{and}\quad q_-(x) = \begin{pmatrix}x_0&-x_1&-x_2&-x_3\\x_1&x_0&-x_3 & x_2\\ x_2 & x_3 & x_0 &-x_1\\x_3&-x_2&x_1&x_0\end{pmatrix}. $$ and $x_0^2+x_1^2+x_2^2+x_3^2 = 1$. Observe that $G_\pm$ is diffeomorphic to $S^3$. Show that each of $G_\pm$ is a subgroup of $\mathrm{SO}(4)$. (Hint: Identify $q_+(x)$ with the $2$-by-$2$ complex matrix $$ \begin{pmatrix}x_0+i\,x_1&-x_2+i\,x^3\\x_2+i\,x_3&x_0-i\,x_1\end{pmatrix} $$ and observe that this identification preserves multiplication. Then, show that conjugation by $C = \begin{pmatrix}-1 & 0\\0 & I_3\end{pmatrix} = C^{-1}$ exchanges $G_+$ and $G_-$. How do these facts help?). Describe $\mathfrak{g}_pm = T_{I_4}G_\pm$, show that $\mathfrak{so}(4) = \mathfrak{g}_+\oplus\mathfrak{g}_-$, and show that $[a_+,a_-]=0$ for $a\pm\in G_\pm$. Assume that $\mathrm{SO}(4)$ is connected (if you don't know this already), and deduce L. Euler's observation that matrix multiplication $\mu:G_+\times G_-\to \mathrm{SO}(4)$ defines a surjective group homomorphism. What is the kernel of this homomorphism?
Look back at the exercises for Day 11, and, in each case, write down a nonvanishing $1$-form $\alpha$ on $\mathbb{R}^3$ that evaluates to zero on $X$ and $Y$. Show that, for Exercise $1$, $\mathrm{d}\alpha$ is a multiple of $\alpha$ while, for Exercise 2, $\mathrm{d}\alpha$ is not a multiple of $\alpha$.
Definition of Lie group , a smooth manifold $G$ together with a smooth map $\mu:G\times G\to G$ such that $(G,\mu)$ satisfies the axioms of a group. Left multiplication and right multiplication by $a\in G$ define diffeomorphisms $L_a,R_a:G\to G$. The proof that the inversion mapping $\iota:G\to G$ defined by $\iota(a) = a^{-1}$ is smooth. The proof that, if $G$ is connected, then the union of the successive 'powers' of any open neighborhood $U$ of $e\in G$ is equal to $G$. The definition of left- and right- invariant vector fields on $G$ and the proof that these are smooth vector fields on $G$ and that the Lie bracket of two left-invariant vector fields is again left-invariant, implying that $\mathfrak{g}=T_eG$ inherits a natural Lie algebra structure. Left-invariant and right-invariant vector fields are complete.
The 'canonical' left-invariant $1$-form $\omega_G:TG\to {\mathfrak g}$, defined by $\omega_G(v) = (D L_{a^{-1}})(v)$ for $v\in T_aG$, which satisfies $(L_a)^*(\omega_G) = \omega_G$ for all $a\in G$ and $\omega_G(v) = v$ for $v\in {\mathfrak g} = T_eG$. When $G\subset \mathrm{GL}(n,\mathbb{R})$ is a matrix Lie group, and $g:G\to M_n(\mathbb{R})$ is the inclusion mapping, regarded as a $M_n(\mathbb{R})$-valued function on $G$, we have $\omega_G = g^{-1}\,\mathrm{d}g$ and $\omega_G$ satisfies $\mathrm{d}\omega_G = -\omega_G \wedge \omega_G$. For general Lie groups, $\mathrm{d}\omega_G = -\tfrac12 [\omega_G,\omega_G]$, (where, for $\mathfrak g$-valued $1$-forms $\alpha$ and $\beta$, we define $[\alpha,\beta](v,w) = [\alpha(v),\beta(w)]-[\alpha(w),\beta(v)]$). When $G$ is understood, we just write $\omega$ instead of $\omega_G$.
If $x_1,\ldots,x_n$ is a basis of $\mathfrak g$, then there are constants $c^k_{ij} = -c^k_{ji}$ such that $[x_i,x_j] = c^k_{ij}\,x_k$. The Jacobi identity is then a set of quadratic equations for the $c^k_{ij}$. If we write $\omega = x_i\,\omega^i$ where $\omega^i$ is the 'dual' basis for the left-invariant $1$-forms, then we have $\mathrm{d}\omega^i = -\tfrac12 c^i_{jk}\,\omega^j\wedge\omega^k$. (These are known as the structure equations of Maurer and Cartan.)
The kernel of $\pi$ is a discrete normal subgroup of $\tilde G$. Show that this kernel lies in the center of $\tilde G$. In fact, show that, for any connected Lie group $G$, any discrete normal subgroup $H\subset G$ lies in the center of $G$. (Hint: For any $z\in H$, the connected set $\{aza^{-1}\ |\ a\in G\}$ must also lie in $H$.)
Show that the center of the simply connected Lie group $$ G = \left\{\pmatrix{a&b\cr0&1\cr} \biggm| a\in\mathbb{R}^+,\ b\in\mathbb{R} \right\} $$ is trivial, so, by the previous exercise, any connected Lie group with the same Lie algebra is actually isomorphic to $G$.
Up to isomorphism, there are 2 Lie algebras of dimension $2$. The corresponding simply-connected Lie groups.
Up to isomorphism, there are $8$ 'isolated' Lie algebras of dimension $3$ plus two '$1$-parameter' connected families of $3$-dimensional Lie algebras.
$\mathrm{SL}(2,\mathbb{R})$ is diffeomorphic to $\mathbb{R}^2\times S^1$.
Applications: Any connected, simply-connected Lie group of dimension $2$ is diffeomorphic to $\mathbb{R}^2$. Any connected, simply-connected Lie group of dimension $3$ is diffeomorphic either to $S^3$ or $\mathbb{R}^3$.
Smooth quotients of Lie groups: If $G$ is a Lie group and $H\subset G$ is a closed (Lie) subgroup, then the left coset space $G/H$ carries the structure of a smooth manifold of dimension $\dim G - \dim H$, uniquely defined by the properties
Examples of such coset spaces are the Grassmannian $\mathrm{Gr}(k,n)$ of $k$-dimensional subspaces of $\mathbb{R}^n$, which is $\mathrm{GL}(n,\mathbb{R})/P_k$ where $P_k\subset \mathrm{GL}(n,\mathbb{R})$ is the (closed) subgroup of linear transformations that preserve $\mathbb{R}^k\subset\mathbb{R}^n$, the set ${\mathcal J}(\mathbb{R}^2n)= \{ J\in \mathrm{GL}(2n,\mathbb{R})\ |\ J^2 = -I_{2n}\ \}$, which is $\mathrm{GL}(2n,\mathbb{R})/\mathrm{GL}(n,\mathbb{C})$, etc.
The definition of smooth left actions $\lambda:G\times M\to M$ (and right actions $\rho:M\times G\to M$). Examples. Orbits of group actions. Effective actions, free actions. The orbit theorem that every orbit $G\cdot m\subset M$ of a group action is a smooth submanifold of $M$ diffeomorphic to the coset space $G/G_m$ where $G_m\subset G$ is the closed subgroup of $G$ that stabilizes $m$, i.e., $G_m = \{ g\in G\ |\ g\cdot m = m\ \}$.
A left action $\lambda:G\times M\to M$ induces a linear 'anti-homomorphism $\lambda_*:{\mathfrak g}\to {\mathcal{X}}(M)$ (the smooth vector fields on $M$), i.e., $\lambda_*([v,w]) = -\left[\lambda_*(v),\lambda_*(w)\right]$. The example of the left action of $\mathrm{SL}(2,\mathbb{R})$ acting on $\mathbb{RP}^1$ given by $$ \begin{pmatrix} a& b\\ c & d\end{pmatrix}\cdot \left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}ax+by\\cx+dy\end{matrix}\right], $$ and the induced mapping of Lie algebras, leading to the Lie subalgebra of $\mathcal{X}(\mathbb{R})$ that is spanned by the three vector fields $\partial/\partial x$, $x\,\partial/\partial x$, and $x^2\,\partial/\partial x$.
The definition of a local left action $\lambda:U\to M$ where $U\subset G\times M$ is an open neighborhood of $\{e\}\times M$. The proof (using the Frobenius Theorem) that, if $\mathfrak{g}\subset\mathcal{X}(M)$ is isomorphic to the Lie algebra of a (finite-dimensional) Lie group $G$, then it comes from a local left action $\lambda:U\to M$ for some open neighborhood $U\subset G\times M$ of $\{e\}\times M$.
The definition of the ordinary differential equations called Lie equations that are associated to a local Lie group action. The classical Riccati equation in ODE theory corresponds to a Lie subalgebra $\mathfrak{g}\subset\mathcal{X}(M)$ where $\mathfrak{g}$ is isomorphic to ${\mathfrak{sl}}(2,\mathbb{R})$.
Review of the Fundamental Theorem of Calculus, Green's Theorem. The Jacobian change of variables formula. Orientation-preserving and orientation-reversing diffeomorphisms between open sets $U$ and $V$ in $\mathbb{R}^n$.
The definition of an oriented atlas $\mathcal{A}$ on a smooth $n$-manifold $M$, i.e., the transisition diffeomorphisms $\tau:x(U\cap V)\to y(U\cap V)$ for coordinate charts $(U,x)$ and $(V,y)$ in $\mathcal{A}$ are orientation preserving. An orientation of $M$ is a choice of an oriented atlas. Two oriented atlases $\mathcal{A}$ and $\mathcal{B}$ are equivalent if $\mathcal{A}\cup\mathcal{B}$ is an oriented atlas. An orientation of $M$ is a choice of a maximal oriented atlas. $M$ is orientable if it has an orientation, in which case, it has exactly two orientations. $M^n$ is orientable if and only if it has a non-vanishing smooth $n$-form. (We proved 'if' but not 'only if; more on that later.)
The examples of $\mathbb{R}^n$ and $S^n$ as oriented manifolds. The proof that $\mathbb{RP}^2$ is not orientable. (More generally, $\mathbb{RP}^{2n}$ is not orientable while $\mathbb{RP}^{2n+1}$ is orientable.)
The space $\Omega^n_0(M)\subset\Omega^n(M)$ of compactly supported $n$-forms on an $n$-manifold $M$. The goal of proving that, if $M$ is oriented, there is a well-defined linear map $\int_M:\Omega^n_0(M)\to \mathbb{R}$ (called 'integration over $M$'). Two pieces of the proof: First, the special case of $n$-forms with support in a given coordinate chart $(U,x)\in\mathcal{A}$ (where $\mathcal{A}$ is an oriented atlas), proving that this does not depend on the choice of coordinate chart containing the support of the $n$-form (uses the Jacobian Change of Variable Theorem). Second, using partitions of unity to cover the case when the support of the form is not contained in a coordinate chart. (I introduced a Lemma about partitions of unity that I did not have time to prove in class. To be continued.)
The Partition of Unity Lemma.
Domains $D\subset M^n$ with smooth boundary $\partial D$: $D\subset M$ is the closure of the set of its interior points $\mathrm{int}(D)$ and it boundary $\partial D = D\setminus \mathrm{int}(D)$ is a closed, embedded smooth submanifold of $M$ of dimension $n{-}1$.
Orientation of a $0$-dimensional manifold $M$ as a function $\sigma:M\to \{\pm 1\}$. The orientation induced on the $(n{-}1)$-manifold $\partial D$, when $M^n$ is oriented.
The statement of Stoke's Theorem: Let $M^n$ be oriented and let $\Omega^{n-1}_0(M)$ denote the vector space of smooth, compactly supported $(n{-}1)$ forms on $M$. If $D \subset M$ is a domain with smooth boundary with inclusion $\iota:\partial D\to M$ is given its induced orientation, then the pullback $\iota^*:\Omega^{n-1}_0(M)\to \Omega^{n-1}(\partial D)$ has image in $\Omega^{n-1}_0(\partial D)$ and, moreover, $$ \int_D \mathrm{d}\phi = \int_{\partial D} \iota^*(\phi) $$ for all $\phi\in \Omega^{n-1}_0(M)$.
Given a manifold $M$ and a Lie group $G$, a principal right $G$-bundle over $M$ is a triple $(B,\pi,\rho)$ where $B$ is a smooth manifold, $\pi:B\to M$ is a surjective smooth submersion, and $\rho:B\times G\to B$ is a smooth right action satisfying the following local triviality condition: Each $m\in M$ has an open neighborhood $U$ for which there exists a smooth diffeomorphism $\tau:\pi^{-1}(U)\to U \times G$ (called a local trivialization) such that $\tau(b) = \bigl(\pi(b),f(b)\bigr)$ with $f(b\cdot g) = f(b)g$ for all $b\in \pi^{-1}(U)$ and $g\in G$. (Note that we are using the usual convention that $b\cdot g$ means $\rho(b,g)$.). A principal right $G$-bundle $(B,\pi,\rho)$ over $M$ is said to be trivial if there is a global trivialization $\tau:B\to M\times G$.
Example 1. If $H\subset G$ is a closed Lie subgroup, then the left coset projectiion $\pi:G\to G/H = M$ is a principal right $H$-bundle over the manifold $M=G/H$. (These examples are often called homogeneous bundles.)
Example 2. (The coframe bundle of a smooth manifold) If $M$ is an $n$-manifold, then the set $\mathcal{F}(M)$ whose elements are the isomorphisms $u:T_xM\to \mathbb{R}^n$ for $x\in M$ has a natural projection (the basepoint mapping) $\pi:\mathcal{F}(M)\to M$ that sends $u:T_xM\to\mathbb{R}^n$ to $x = \pi(u)$. The Lie group $G = \mathrm{GL}(n,\mathbb{R})$ (which consists of the linear isomorphims of $\mathbb{R}^n$ with itself) has a natural right action on $\mathcal{F}(M)$ defined by $u\cdot g = g^{-1} u$. It only remains to define a smooth structure on the set $\mathcal{F}(M)$ so that the map $\pi:\mathcal{F}(M)\to M$ is a smooth submersion and the right action of $\mathrm{GL}(n,\mathbb{R})$ on $\mathcal{F}(M)$ is smooth and the local triviality condition is satisfied. We do this by means of a smooth atlas $\mathcal{A}$ on $M$. For any smooth chart $(U,\phi)$ in $\mathcal{A}$, we define a bijection $\tau_{U,\phi}:\pi^{-1}(U)\to U\times \mathrm{GL}(n,\mathbb{R})$ by $$ \tau_{U,\phi}(u) = \bigl(\pi(u), D\phi_{\pi(u)}\circ u^{-1}\bigr). $$ (Note that $D\phi_{\pi(u)}:T_{\pi(u)}\to\mathbb{R}^n$ and $u:T_{\pi(u)}\to\mathbb{R}^n$ are both isomorphisms, so $D\phi_{\pi(u)}\circ u^{-1}:\mathbb{R}^n\to\mathbb{R}^n$ is an isomorphism, i.e., an element of $\mathrm{GL}(n,\mathbb{R})$.) We give $\mathcal{F}(M)$ the smooth structure for which $\tau_{U,\phi}$ is a diffeomorphism for every $(U,\phi)\in\mathcal{A}$.
(Associated Bundles) If $(B,\pi,\rho)$ is a smooth principal right $G$-bundle over $M$ and $F$ is a smooth manifold on which $G$ acts on the left, $\lambda:G\times F\to F$, then we can consider the quotient manifold $B\times_\lambda F$, which is the set of $G$-orbits of the right action $(b,f)\cdot g = (b\cdot g,\ g^{-1}\cdot f)$, and we let $[b,f]\in B\times_\lambda F$ denote the $G$-orbit of $(b,f)$. Then the quotient map $q:B\times F\to B\times_\lambda F$ is a principal right $G$-bundle over the manifold $B\times_\lambda F$, and this manifold has a natural map $\mu:B\times_\lambda F\to M$ defined by $\mu\bigl([b,f]\bigr)= \pi(b)$. This $\mu$ is a submersion, and each of the fibers $\mu^{-1}(x)\subset B\times_\lambda F$ is a closed submanifold diffeomorphic to $F$. In fact, $\mu:B\times_\lambda F\to M$ is locally trivial in the sense that each $x\in $M has an open neighorhood $U\subset M$ on which there exists a local trivialization, i.e., a diffeomorphism $\sigma:\mu^{-1}(U)\to U\times F$. The bundle $\mu:B\times_\lambda F\to M$ is said to be associated to the principal bundle $\pi:B\to M$.
Example: If $\mathcal{F}(M)\to M$ is the coframe bundle of $M$ and $\lambda:\mathrm{GL}(n,\mathbb{R})\times \mathbb{R}^n \to\mathbb{R}^n$ is the standard right action $\lambda(g,v) = gv$ for $g\in \mathrm{GL}(n,\mathbb{R})$ and $v\in \mathbb{R}^n$, then there is a natural identification of $\mathcal{F}(M)\times_\lambda \mathbb{R}^n$ with $TM$ defined by sending $[u,v]$ to $u^{-1}(v)\in T_{\pi(u)} M$. (Check that this is well-defined!) This natural identification is an isomorphism, so the tangent bundle of $M$ is naturally constructed as an associated bundle of $\mathcal{F}(M)\to M$. In general, if $h:\mathrm{GL}(n,\mathbb{R})\to \mathrm{GL}(m,\mathbb{R})$ is any homomorphism, we can define a left action $\lambda_h:\mathrm{GL}(n,\mathbb{R})\times\mathbb{R}^m\to \mathbb{R}^m$, and the bundle $\mathcal{F}(M)\times_{\lambda_h}\mathbb{R}^m$ is a vector bundle over $M$ defined in a natural way by the representation $h$. For example, if $h(g) = (g^T)^{-1}$, then the corresponding bundle is identifiable with $T^*M$, and if $h(g) = \det(g)^{-1}$, then the corresponding bundle is the bundle whose sections are the $n$-forms on $M$.
In the definition of the smooth structure on $\mathcal{F}(M)$, if $(U,\phi)$ and $(U,\psi)$ are both smooth charts in $\mathcal{A}$, show that $\tau_{U,\phi}$ and $\tau_{U,\psi}$ induce the same smooth structure on $\pi^{-1}(U)\subset\mathcal{F}(M)$, i.e., $\tau_{U,\phi}\circ \left(\tau_{U,\psi}\right)^{-1}:U\to\mathrm{GL}(n,\mathbb{R}) \times U\to\mathrm{GL}(n,\mathbb{R})$ is a diffeomorphism.
Explain how, starting with a local trivialization $\tau:\pi^{-1}(U)\to U\times G$ of a principal right $G$-bundle $\pi:B\to M$, we can construct a local trivialization $\sigma:\mu^{-1}(U)\to U\times F$ of the bundle $B\times_\lambda F$ associated to $B$ via the left action $\lambda:G\times F\to F$.
Define a map $\eta:T\mathcal{F}(M)\to\mathbb{R}^n$ by the formula $ \eta_u(v) = u\bigl(D\pi_u(v)\bigr) $ for $u\in \mathcal{F}(M)$ and $v\in T_u\mathcal{F}(M)$. Show that $\eta$ is a smooth $\mathbb{R}^n$-valued $1$-form on $\mathcal{F}(M)$.
A smooth vector bundle of rank $r$ over a base manifold $M$ is a smooth manifold $B$ together with a surjective submersion $\pi:B\to M$ together with a choice, for each $m\in M$, of the structure of a vector space of dimension $r$ on the fiber $B_m = \pi^{-1}(m)$ satisfying the local triviality property that, each $m\in M$ has an open neighborhood $U\subset M$, there exists a diffeomorphism $\tau = (\pi,\phi):\pi^{-1}(U)\to U\times\mathbb{R}^r$ such that, for each $q\in U$, the map $\phi:\pi^{-1}(q)\to \mathbb{R}^r$ is an isomorphism of vector spaces.
Examples: $TM$, $T^*M$, the normal bundle of a submanifold $M^n\subset\mathbb{R}^{n+r}$, $S^2(T^*M)$ (the bundle of quadratic forms on the tangent spaces of $M$).
A representation of a Lie group $G$ on a vector space $V$ is a Lie group homomorphism $\rho:G\to\mathrm{GL}(V)$. Two representations $\rho_i:G\to \mathrm{GL}(V_i)$ are isomorphic (aka equivalent) if there is an isomorphism of vector spaces $\iota:V_1\to V_2$ such that $\rho_2(g) = \iota\circ\rho_1(g)\circ\iota^{-1}$. Given a representation $\rho:G\to\mathrm{GL}(V)$, it has a dual (aka contragredient) representation $\rho^\dagger: G\to\mathrm{GL}(V^*)$ defined by $\bigl(\rho^\dagger(g)(\xi)\bigr)(v) = \xi\bigl(\rho(g)^{-1}v)\bigr)$ for all $v\in V$ and $\xi\in V^*$. Given a pair of representations $\rho_i:G\to\mathrm{GL}(V_i)$ for $i=1,2$, there is the direct sum representation, $\rho_1\oplus\rho_2:G\to\mathrm{GL}(V_1\oplus V_2)$ and the tensor product representation $\rho_1\otimes\rho_2:G\to\mathrm{GL}(V_1\otimes V_2)$, where $V_1\otimes V_2$ is the vector space (unique up to isomorphism) with the universal mapping property that there is a bilinear mapping $\otimes:V_1\times V_2\to V_1\otimes V_2$ such that, for any bilinear mapping $\beta:V_1\times V_2\to W$ (where $W$ is a vector space), there is a unique linear mapping $\hat\beta:V_1\otimes V_2\to W$ such that $\beta = \hat\beta\circ\otimes$. (Note: We usually write $v_1\otimes v_2$ instead of $\otimes(v_1,v_2)$ for $v_i\in V_i$.). We have the usual equivalences (isomorphisms) $$ V_1\oplus V_2 \simeq V_2\oplus V_1\,,\quad (V_1\oplus V_2)\oplus V_3 \simeq V_1\oplus (V_2\oplus V_3),\quad V_1\otimes V_2 \simeq V_2\otimes V_1\,,\quad (V_1\otimes V_2)\otimes V_3 \simeq V_1\otimes (V_2\otimes V_3). $$ The set of equivalent classes of $G$-representations forms a semi-ring, as tensor product distributes over direct sum: $V_1\otimes(V_2\oplus V_3) \simeq V_1{\otimes}V_2 \oplus V_1{\otimes}V_3$.
Examples: There is a natural isomorphism $\mathrm{Hom}(V_1,V_2)\simeq V_1^*\otimes V_2$. There is a natural decomposition $V\otimes V = S_2(V)\oplus\Lambda_2(V)$, where $S_2(V)$ is the subspace spanned by elements of the form $v\otimes w + w\otimes v$ while $\Lambda_2(V)$ is the subspace spanned by elements of the form $v\otimes w - w\otimes v$.
If $\pi:B\to M$ is a principal right $G$-bundle and $\rho:G\to\mathrm{GL}(V)$ is a representation of $G$ then $B\times_\rho V$, the set of equivalence classes $[b,v]$, where $(b,v)\sim (b{\cdot}g,\ \rho(g)^{-1}v)$, is naturally a smooth vector bundle over $M$ with projection $\pi\bigl([b,v]\bigr) = \pi(b)$. In the case $\rho:\mathrm{GL}(n,\mathbb{R})\to \mathrm{GL}(V)$ is a representation, and $\pi:\mathcal{F}(M)\to M$ is the coframe bundle, then $\mathcal{F}(M)\times_\rho V$ is a smooth vector bundle over $M$ called the tensor bundle of type $\rho$ on $M$ and a section of this vector bundle over $M$ is called a tensor (field) of type $\rho$ on $M$.
Any vector bundle $\pi:E\to M$ has a canonical section, the zero section, $z:M\to E$ defined by $z(m) = 0_m\in E_m$, where $0_m$ is the zero element in the vector space $E_m$. Show that the zero section is smooth, and show that, for any $m\in M$ and any element $v\in E_m$, there is a smooth section $s:M\to E$ such that $s(m) = v$. (Hint: Consider an open $m$-neighborhood $U\subset M$ on which there exists a smooth 'bump function' $\lambda$ that is nonzero at $m$, but whose support is a compact subset of $U$. How can you use that?)
Given a smooth $n$-manifold $M$, a Riemannian metric is a smooth function $g:TM\to\mathbb{R}$ such that, for all $m\in M$, the mapping $g_m:T_mM\to\mathbb{R}$ is a positive definite quadratic form. The pair $(M,g)$ is said to define a Riemannian manifold.
For example, if $f = (f^i):M\to\mathbb{R}^{n+r}$ is a smooth immersion, then $g_f = (\mathrm{d}f^1)^2 + \cdots + (\mathrm{d}f^{n+r})^2$ is a Riemannian metric on $M$. More generally, if $g$ is a Riemannian metric on $N$ and $f:M\to N$ is a smooth immersion, then seting $g_f(v) = g\bigl(Df(v)\bigr)$ for $v\in TM$ defines a Riemannian metric $g_f$ on $M$, called the pullback metric.
If $x = (x^i):U\to \mathbb{R}^n$ is a smooth coordinate chart on $U\subset M$, then there exist unique functions $g_{ij}=g_{ji}\in C^\infty\bigl(x(U)\bigr)$ for which $$ g_{|TU} = g_{ij}(x)\,\mathrm{d}x^i\mathrm{d}x^j. $$ The matrix $G(p) = \bigl(g_{ij}(p)\bigr)$ for $p\in x(U)$ is positive definite. A Riemannian metric $g$ is said to be (locally) flat if, for each $m\in M$, there exists a coordinate chart $x:U\to\mathbb{R}^n$ with $m\in U$ for which $g_{ij} = \delta_{ij}$, i.e., $g_{|TU} = (\mathrm{d}x^1)^2 + \cdots + (\mathrm{d}x^{n})^2$.
A Riemannian metric $g$ on $M$ can be used to define the $g$-length of a (piecewise) smooth curve $\gamma:[0,1]\to M$, by the formula $$ \mathcal{L}_g(\gamma) = \int_0^1 \sqrt{g\bigl(\gamma'(t)\bigr)}\,\mathrm{d}t. $$ By the Fundamental Theorem of Calculus, $\mathcal{L}_g(\gamma) = \mathcal{L}_g(\gamma\circ h)$ where $h:[0,1]\to[0,1]$ is any monotone (piecewise) smooth function. When $M$ is connected, one can define a function $\delta_g:M\times M\to [0,\infty)$ by $$ \delta_g(p,q) = \inf\left\{ {\mathcal{L}}_g(\gamma)\ |\ \gamma:[0,1]\to M,\ \gamma(0)=p,\ \gamma(1)=q\ \right\}. $$ (See Exercise 1 below for the verification that $\delta_g$ is a metric on $M$ in the usual topological sense.)
Given a Riemannian metric $g$ on $M^n$, a smooth local coframing $\eta = (\eta_i):TU\to\mathbb{R}^n$ on an open set $U\subset M$ is said to be $g$-orthonormal if $g_{|TU} = \eta^{\mathsf{T}}\eta = {\eta_1}^2 + \cdots + {\eta_n}^2$. Each point of $M$ lies in an open set $U$ on which a $g$-orthonormal coframing exists, and, if $\eta$ and $\bar\eta$ are $g$-orthonormal coframings on $U\subset M$, then there exists a unique smooth mapping $h:U\to\mathrm{O}(n)$ such that $\bar\eta = h^{-1}\,\eta$.
You proved, in Exam 2, that for any smooth coframing $\eta:TU\to\mathbb{R}^n$, there exists a unique $1$-form $\theta = (\theta_{ij})$ on $U$ taking values in ${\mathfrak{so}}(n)$ (i.e., such that $\theta$ satisfies $\theta^{\mathsf{T}} = -\theta$) and satisfying $\mathrm{d}\eta = -\theta\wedge\eta$. (This is a version of the so-called Fundamental Lemma of Riemannian Geometry.) When $\bar \eta = h^{-1}\,\eta$ where $h:U\to\mathrm{O}(n)$, then $\bar\theta = h^{-1}\,\mathrm{d}h + h^{-1}\theta h$ satisfies $\mathrm{d}\bar\eta = -\bar\theta\wedge\bar\eta$ and $\bar\theta^{\mathsf{T}} = -\bar\theta$. Moreover, direct computation shows that $\mathrm{d}\bar\theta + \bar\theta\wedge\bar\theta = h^{-1}\bigl(\mathrm{d}\theta + \theta\wedge\theta\bigr)h$.
Criterion for local flatness: A Riemannian metric $g$ on $M^n$ is locally flat if and only if, when $\eta$ is any local $g$-orthonormal coframing on a simply-connected $U\subset M$ and $\theta$ is the unique ${\mathfrak{so}}(n)$-valued $1$-form on $U$ satisfying $\mathrm{d}\eta = -\theta\wedge\eta$, then $\mathrm{d}\theta + \theta\wedge\theta = 0$. (The proof of sufficiency is via the Frobenius Theorem.)
The Riemann curvature tensor: If $g$ is a Riemannian metric on $M^n$, there is a unique $C^\infty(M)$-quadrilinear mapping $R_g:\mathcal{X}(M)\times\mathcal{X}(M)\times\mathcal{X}(M)\times\mathcal{X}(M)\to C^\infty(M)$ with the property that, for any $g$-orthonormal coframing $\eta$ with domain $U$, we have $$ R_g(X,Y,Z,W)_{|U} = \eta(X)^{\mathsf{T}}\bigl(\mathrm{d}\theta + \theta\wedge\theta\bigr)(Z,W)\,\eta(Y). $$ (The above formulae for a change of $g$-orthonormal coframing show that the expression on the right-hand side does not depend on the choice of $(\eta,\theta)$.). In particular, $g$ is locally flat if and only if $R_g\equiv0$. Relative to a $g$-orthonormal coframing $\eta = (\eta_{i})$ with $\theta=(\theta_{ij}) = (-\theta_{ji})$, we have $$ \mathrm{d}\eta_i = -\theta_{ij}\wedge\eta_j\quad\text{and}\quad \mathrm{d}\theta_{ij} = -\theta_{ik}\wedge\theta_{kj} + \tfrac12 R_{ijkl}\,\eta_k\wedge\eta_l\,, $$ where $R_{ijkl} = -R_{jikl} = - R_{ijlk}$ and $R_{ijkl}+R_{iklj}+R_{iljk} = 0$, with this last following from the above equations and the identity $\mathrm{d}^2=0$, i.e., $$ 0 = \mathrm{d}(\mathrm{d}\eta_i) = -\mathrm{d}\theta_{ij}\wedge\eta_j +\theta_{ij}\wedge(-\theta_{jk}\wedge\eta_k) = -\tfrac12 R_{ijkl}\,\eta_j\wedge\eta_k\wedge\eta_l = -\tfrac16(R_{ijkl}+R_{iklj}+R_{iljk})\,\eta_j\wedge\eta_k\wedge\eta_l\,. $$ Note that these identities also imply that $R_{ijkl} = R_{klij}$. This is because $$ \begin{align} 2 R_{ijkl} &= R_{ijkl}-R_{jikl} = -R_{iklj}-R_{iljk}+R_{jkli}+R_{jlik}\\ 2 R_{klij} &= R_{klij}-R_{lkij} = -R_{kijl}-R_{kjli}+R_{lijk}+R_{ljki} \end{align} $$ and the symmetries $R_{ijkl} = -R_{jikl} = - R_{ijlk}$ show that the terms on the right hand sides of these two lines agree.
For each vector field $X$ on $M$, there is a linear map $\nabla_X:\mathcal{X}(M)\to\mathcal{X}(M)$ (called covariant differentiation by $X$) that satisfies $$ \eta\bigl(\nabla_X Y\bigr) = \mathrm{d}(\eta(Y))(X) + \psi(X)\,\eta(Y) $$ on an open set $U\subset M$ and $\eta:TU\to\mathbb{R}^n$ is a $g$-orthonormal coframing and $\psi:TU\to{\mathfrak{so}}(n)$ is the unique $1$-form on $U$ that satisfies $\mathrm{d}\eta = -\psi\wedge\eta$. (This does not depend on the choice of $\eta$.). This operator has the following properties for $X,X_i,Y,Y_i\in \mathcal{X}(M)$, $f\in C^\infty(M)$ and constants $c_i$:
If $\gamma:[a,b]\to M$ is a differentiable curve, and $Z:[a,b]\to TM$ is a vector field in $M$ along $\gamma$ (i.e., $Z(t)\in T_{\gamma(t)}M$ for all $t\in[a,b]$, then we can also define $\frac{DZ}{dt}:[a,b]\to TM$ where $\frac{DZ}{dt}(t)$ lies in $T_{\gamma(t)}M$ for all $t\in[a,b]$ by $$ \eta\left(\frac{DZ}{dt}(t)\right) = \frac{\mathrm{d}\bigl(\eta(Z)\bigr)}{\mathrm{d} t} + \psi(\gamma'(t))\,\eta(Z)\,. $$ This gives us a way of measuring how a vector field changes as one moves along a curve. We say that $Z$ is parallel along $\gamma$ if $DZ/dt \equiv 0$.
Let $\pi:FM\to M$ be the $g$-orthonormal coframe bundle, thus, an element $u\in \pi^{-1}(m)$ is an isometry of vector spaces $u:T_mM\to\mathbb{R}^n$. The group $\mathrm{O}(n)$ acts on $FM$ on the right by $R_a(u) = u\cdot a = a^{-1}\circ u$ for $u\in FM$ and $a\in \mathrm{O}(n)$, making $FM$ into a principal right $\mathrm{O}(n)$-bundle over $M$. For each $u\in F$, there is a smooth embedding $\iota_u: \mathrm{O}(n)\to FM$ defined by the right action via $\iota_u(a) = u\cdot a = a^{-1}\circ u$. There is a (smooth) tautological $\mathbb{R}^n$-valued $1$-form $\omega:TFM\to \mathbb{R}^n$ defined by $\omega_u(v) = u\left( D\pi_u(v)\right)$. It satisfies $R_a^*(\omega) = a^{-1}\omega$. There is also a unique ${\mathfrak{so}}(n)$-valued $1$-form $\theta:TFM\to {\mathfrak{so}}(n)$ such that $\mathrm{d}\omega = -\theta\wedge\omega$. This $1$-form satisfies $R_a^*(\theta) = a^{-1}\theta a$. It also satisfies that $\iota_u^*(\theta)$ is the canonical left-invariant $1$-form on $\mathrm{O}(n)$.
As a particular example, if, in a local coordinate chart $(U,x)$, we have $g = g_{ij}(x)\,\mathrm{d}x^i\mathrm{d}x^j$, then taking $\alpha = \mathrm{d}x$ and $s = (g_{ij})$, we find that there exists a matrix $\gamma = (\gamma^i_j)$ of $1$-forms satisfying $-\gamma\wedge\mathrm{d}x = \mathrm{d}(\mathrm{d}x) = 0$ and $\mathrm{d}g_{ij} = g_{ik}\gamma^k_j + g_{kj}\gamma^k_i$. Then $\gamma^i_j = \Gamma^i_{jk}\,\mathrm{d}x^k$, where the $\Gamma^i_{jk}=\Gamma^i_{kj}$ are the famous Christoffel symbols of the metric $g$ in the local coordinate system $x = (x^i)$.
Let $G$ be a Lie group with Lie algebra ${\mathfrak{g}}= T_eG$ and canonical left-invariant $1$-form $\omega_G: TG\to {\mathfrak{g}}$ (which satisfies $\mathrm{d}\omega_G +\tfrac12[\omega_G,\omega_G]=0$).
Let $M^n$ be a smooth $n$-manifold and let $\pi:B\to M$ be a principal right $G$-bundle. Thus, $\pi:B\to M$ is a surjective submersion and the fibers of $\pi$ are the orbits of a smooth free right action $\rho:B\times G \to B$ of $G$ on $B$. Let $R_a:B\to B$ denote the right action by $a\in G$: $R_a(b) = b\cdot a = \rho(b,a)$, and, for $b\in B$, let $\iota_b:G\to B$ be the $b$-orbit embedding: $\iota_b(g) = b\cdot g$. Note that $\iota_{b\cdot a} = R_a\circ \iota_b\circ C_a$, where $C_a:G\to G$ is $C_a(g) = aga^{-1}$ (since $\iota_{b\cdot a}(g) = (b\cdot a)\cdot g = b\cdot(ag) = (b\cdot aga^{-1})\cdot a = R_a(\iota_b(C_a(g)))$.
Since the right action is free, $(D\iota_b)_e:{\mathfrak{g}}\to T_b(b{\cdot}G)$ is an isomorphism, and, hence, the inverse mapping $\tau(b):T_b(b{\cdot}G)\to{\mathfrak{g}}$ is well-defined. In particular, the $\pi$-vertical plane field $V_b = T_b(b{\cdot}G)$ is canonically trivial, with $\tau:V\to{\mathfrak{g}}$ a smooth mapping. The equivariance $\tau\circ (DR_a) = \mathrm{Ad}(a^{-1})\circ\tau$ follows from the identity $\iota_{b\cdot a} = R_a\circ \iota_b\circ C_a$.
A connection on $B$ is a smooth $n$-plane field $H\subset TB$ with the properties that (1) $V_b \cap H_b = \{0_b\}\subset T_bB$ and that (2) $DR_a(H_b) = H_{b\cdot a}\subset T_{b\cdot a}B$ for all $b\in B$ and $a\in G$ Thus, $H\subset TB$ is invariant under $R_a:B\to B$ for all $a\in G$.
Since $T_bB = V_b\oplus H_b$ for any connection $H\subset TB$, specifying $H$ defines a unique extension of $\tau:V\to {\mathfrak{g}}$ to a ${\mathfrak{g}}$-valued $1$-form $\theta:TB\to {\mathfrak{g}}$ with the property that $\theta(H) = \{0\}$. This $1$-form satisfies $R_a^*(\theta) = \mathrm{Ad}(a^{-1})\bigl(\theta\bigr)$ for all $a\in G$.
Conversely, given a $1$-form $\theta:TB\to {\mathfrak{g}}$ whose restriction to $V$ is $\tau$ and that satisfies $R_a^*(\theta) = \mathrm{Ad}(a^{-1})\bigl(\theta\bigr)$, then $H = \ker(\theta)\subset TB$ is a connection on $B$. Thus, the two notions are equivalent: The right-invariant horizontal plane field $H$ and the $\mathrm{Ad}$-equivariant $1$-form $\theta$.
Example: The $1$-form $\theta$ constructed on the $g$-orthonormal coframe bundle $\pi:FM\to M$ defines a connection on $FM$ as a principal right $\mathrm{O}(n)$-bundle.
Let $H\subset TB$ be a connection on $B$ with defining $1$-form $\theta:TB\to{\mathfrak{g}}$. For $s:U\to B$ a local section of $\pi:B\to M$ defined over the open set $U\subset M$, define a trivialization $S:U\times G\to\pi^{-1}(U)\subset B$ by $S(p,g) = s(p)\cdot g$. If $\psi = s^*(\theta)$, then one has the local formula $$ S^*(\theta) = \omega_G + \mathrm{Ad}(g^{-1})(\psi). $$ (When $G\subset\mathrm{GL}(m,\mathbb{R})$, this takes the familiar form $S^*(\theta) = g^{-1}\,\mathrm{d}g + g^{-1}\psi g$.) In particular, $$ S^*(\mathrm{d}\theta+\tfrac12[\theta,\theta]) = \mathrm{Ad}(g^{-1})(\mathrm{d}\psi+\tfrac12[\psi,\psi]), $$ so the curvature form $\Theta = \mathrm{d}\theta+\tfrac12[\theta,\theta]$ is `semi-basic', i.e., $\Theta(X,Y)=0$ if $X$ is a vertical vector field on $B$, i.e., $X(b)\in V_b$ for all $b\in B$.
Because $D\pi_b:H_b\to T_{\pi(b)}M$ is an isomorphism for all $b\in B$, it follows that any smooth vector field $X$ on $M$ has a unique `lift' to a horizontal vector field $\hat X$ on $B$, i.e., $\hat X (b)\in H_b$ for all $b\in B$, and $D(R_a)(\hat X) = \hat X$, i.e., $\hat X$ is $R_a$-invariant for all $a\in G$. Since $\hat X$ is $\pi$-related to $X$, it follows that $[\hat X,\hat Y]$ is $\pi$-related to $[X,Y]$ for vector fields $X$ and $Y$ on $M$. Consequently, $$ [\hat X,\hat Y] - \widehat{[X,Y]} $$ is a vertical vector field on $B$, and $ \Theta(\hat X, \hat Y) = -\theta\bigl([\hat X,\hat Y]\bigr)$. Consequently, $H\subset TB$ satisfies the Frobenius condition if and only if $\Theta \equiv 0$. The connection $H$ is said to be (locally) flat in this case. When the connection is locally flat, each point of $M$ has an open neighborhood $U\subset M$ on which there exists a flat local section $s:U\to B$, i.e., $s$ satisfies $s^*(\theta) = 0$.
Given a representation $\rho:G\to \mathrm{GL}(r,\mathbb{R})$ with induced Lie algebra homomorphism $\rho_*:{\mathfrak{g}}\to {\mathfrak{gl}}(r,\mathbb{R})$, one has an associated vector bundle $E = B\times_\rho\mathbb{R}^r$ over $M$. A section $\sigma:M\to E$ corresponds to a function $s:B\to\mathbb{R}^r$ that satisfies $s(b\cdot g) = \rho(g^{-1})s(b)$. For any vector field $X$ on $M$, we can define a `derivative' $D_Xs:B\to\mathbb{R}^r$ by $D_Xs = \mathrm{d}s(\hat X)$, which satisfies $D_Xs(b\cdot g) = \rho(g^{-1})D_xs(b)$. Thus, there is a section $\nabla_X\sigma:M\to E$ corresponding to $D_Xs$, and this operation $\nabla_X:C^\infty(E)\to C^\infty(E)$ is a derivation of the sections of $E$ over the ring of smooth functions on $M$, known as the covariant derivative associated to the connection $H$.