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Wald - General Relativity

  1. Chapter 2 - Manifolds and tensor fields
    1. Problem 2.1
    2. Problem 2.2
    3. Problem 2.3
    4. Problem 2.4

Chapter 2 - Manifolds and tensor fields

Problem 2.1 : a) Show that the overlap functions $f^{\pm}_i \circ (f^\pm_j)^{-1}$ are $C^\infty$, thus completing the demonstration given in section 2.1 that $S^2$ is a manifold.

b) Show by explicit construction that two coordinate systems (as opposed to the six used in the text) suffice to cover $S^2$. (it is impossible to cover $S^2$ with a single chart, as follow from the fact that $S^2$ is compact, but every open subset of $\mathbb{R}^2$ is noncompact; see appendix A).

Solution : For a) : As a reminder, here is Wald's partial demonstration that $S^2$ is a manifold.

The sphere $S^2$ is defined as a subset of $\mathbb{R}^3$ in the usual way :

\begin{equation} S^2 = \left\{ (x_1, x_2, x_3) \in \mathbb{R}^3 | x_1^2 + x_2^2 + x_3^2 = 1 \right\} \end{equation}

The six open sets covering the spheres are the six possible hemispheres cut along the three coordinate planes of $\mathbb{R}^3$ :

\begin{equation} O_i^\pm = \left\{ (x_1, x_2, x_3) \in S^2 | \pm x^i > 0 \right\} \end{equation}

and are mapped onto the open disk $D^2 \subset \mathbb{R}^2$ via the projection of those coordinates onto the planes cutting the hemispheres :

\begin{eqnarray} f^\pm_1 (x_1, x_2, x_3) &=& (x_2, x_3)\\ f^\pm_2 (x_1, x_2, x_3) &=& (x_1, x_3)\\ f^\pm_3 (x_1, x_2, x_3) &=& (x_1, x_2)\\ \end{eqnarray}

The inverse map is done by finding the appropriate third coordinates from the first two in the projected plane, which will simply be obtained by using the sphere's equation :

\begin{eqnarray} (f^\pm_1)^{-1} (x_2, x_3) &=& (\pm \sqrt{1 - x_2^2 - x_3^2}, x_2, x_3)\\ (f^\pm_2)^{-1} (x_1, x_3) &=& (x_1, \pm \sqrt{1 - x_1^2 - x_3^2}, x_3)\\ (f^\pm_3)^{-1} (x_1, x_2) &=& (x_1, x_2, \pm \sqrt{1 - x_1^2 - x_3^2})\\ \end{eqnarray}

Trivially $f^\pm_i \circ (f^\pm_i)^{-1} = \mathrm{Id}_{D^2}$ since this projects back to the appropriate coordinates, and conversely, $(f^\pm_i)^{-1} \circ f^\pm_i = \mathrm{Id}_{O^\pm_i}$, since the coordinates obtained obey the appropriate equation for the sphere's appropriate hemispheres.

What is now the status of its transition maps? Every open set $O^\pm_i$ overlaps with the others, except for the one in the other half of the same plane. To keep things general let's consider some coordinates of an abstract plane $(x,y)$. The transition map $f^{\pm_k}_i \circ (f^{\pm_l}_j)^{-1}$ is given by

\begin{eqnarray} f^{\pm_k}_i \circ (f^{\pm_l}_j)^{-1} (x, y) &=& f^{\pm_k}_i (x_j = \pm_l \sqrt{1 - x^2 - y^2}, x_{\neq j} = x, y)\\ &=& \begin{cases} (x_j = \pm_l \sqrt{1 - x^2 - y^2}, \text{$x$ or $y$}), & \text{if $i = j$}\\ (x, y), & \text{otherwise} \end{cases} \end{eqnarray}

All of this may be up to some switching of the coordinates depending on the case. All of these are smooth functions of $x$ and $y$ and therefore they form an appropriate chart of $S^2$.

For b) : Wald probably refers here to the famous stereographic maps from $S^2$ to $\mathbb{R}^2$. These are obtained by considering two planes tangent to $S^2$'s poles, let's say along the $x_3$ axis :

\begin{equation} P^\pm = \left\{ (x_1, x_2, x_3) \in \mathbb{R}^3 | x_3 = \pm 1 \right\} \end{equation}

Let's now consider the two sets of every point on $S^2$ except for a pole each :

\begin{equation} S^\pm = \left\{ (x_1, x_2, x_3) \in \mathbb{R}^3 | x_1^2 + x_2^2 + x_3^2 = 1, x_3 \neq \pm 1 \right\} \end{equation}

The stereographic map is composed by a line going from the excluded pole to the point considered, and then its intersection with the plane $P^\pm$. The line between the two points is given by the vector of origin $(0,0,\mp 1)$ and the point on the sphere $(x_1, x_2, x_3)$, with direction vector $(x_1, x_2, x_3 \mp 1)$.

The intersection of this line and the plane $P^\pm$ is given by the usual formula of intersection of line and plane, where our plane has the origin point $(0,0,\pm 1)$ and normal vector $(0,0,\mp 1)$. The intersection of the two is then given by the equation

\begin{equation} ((x_1, x_2, x_3 \mp 1) \cdot (0, 0, \mp 1)) d + ((0,0,\mp 1) - (0,0,\pm 1)) \cdot (0, 0, \mp 1) = 0 \end{equation} \begin{equation} (\mp x_3 + 1) d + 2 = 0 \end{equation}

This forms a triangle composed of the axis $x_3 \in [-1, 1]$, the line between the excluded pole and point, and the included pole and projection onto the plane. The degenerate case of the line between the two poles maps the included pole to $(0,0)$ on the appropriate plane. The point $(x,y)$ on the projected plane is of length $\sqrt{x^2 + y^2}$, so that our triangle is of sides $(2, \sqrt{x^2 + y^2}, \sqrt{4 + \sqrt{x^2 + y^2}})$.

...

Problem 2.2 : Prove that any smooth function $F : \mathbb{R}^n \to \mathbb{R}$ can be written in the form equation (2.2.2). (Hint : for $n = 1$, use the identity \begin{equation} F(x) - F(a) = (x - a) \int_0^1 F'[t(x-a) + a] dt \end{equation} then prove it for general $n$ by induction).

Solution: First a reminder of equation (2.2.2) : for a function $F : \mathbb{R}^n \to \mathbb{R}$, for each $a \in \mathbb{R}^n$, there exists smooth functions $H_\mu$ such that for $x \in \mathbb{R}^n$,

\begin{equation} F(x) - F(a) = \sum_{\mu = 1}^n (x^\mu - a^\mu) H_\mu(x) \end{equation}

with

\begin{equation} H_\mu(a) = \frac{\partial F}{\partial x^\mu}|_{x = a} \end{equation}

For $n = 1$, the given formula

\begin{equation} F(x) - F(a) = (x - a) \int_0^1 F'[t(x-a) + a] dt \end{equation}

Simply derives from the fundamental theorem of calculus. We simply have that, after the change of variable $u = t(x-a) + a$, with $du = x dt$, we obtain the integral

...

This is then true for $n = 1$. Now if the formula is true for $n$, let's consider the case $n + 1$. In this case, the function $F_{n+1}$ can be interpreted as

...

Problem 2.3 :

  1. Verify that the commutator, defined by equation (2.2.14), satisfies the linearity and Leibniz properies, and hence defines a vector field.
  2. Let $X, Y, Z$ be smooth vector fields on a manifold $M$. Verify that their commutator satisfies the Jacobi identity : \begin{equation} [[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] = 0 \end{equation}
  3. Let $Y_1, \ldots, Y_n$ be smooth vector fields on an $n$-dimensional manifold $M$ such that at each $p \in M$ they form a basis of the tangent space $V_p$. Then, at each point, we may expand each commutator $[Y_\alpha, Y_\beta]$ in this basis, thereby defining the functions ${C^\gamma}_{\alpha\beta} = - {C^\gamma}_{\beta\alpha}$ by \begin{equation} [[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] = 0 \end{equation} Use the Jacobi identity to derive an equation satisfied by ${C^\gamma}_{\alpha\beta}$. (This equation is a useful algebraic relation if the ${C^\gamma}_{\alpha\beta}$ are constants, as will be the case if $Y_1, \ldots, Y_n$ are left [or right] invariant vector fields on a Lie group).

Solution : For a) : Equation (2.2.14) for the Lie bracket is

\begin{equation} [v, w](f) = v[w(f)] - w[v(f)] \end{equation}

The linearity and Leibniz rule for vector fields are defined by

\begin{eqnarray} v(af + bg) &=& av(f) + bv(g)\\ v(fg) &=& f v(g) + g v(f) \end{eqnarray}

If $v$ and $w$ are vector fields, then the linearity of the bracket is given by

\begin{eqnarray} [v, w](a f + b g) &=& v[w(a f + b g)] - w[v(a f + b g)]\\ &=& v[aw(f) + bw(g)] - w[av(f) + bv(g)]\\ &=& av[w(f)] + b v[w(g)] - aw[v(f)] - bw[v(g)] \\ &=& a(v[w(f)] - w[v(f)]) + b (v[w(g)] - w[v(g)])\\ &=& a [v,w](f) + b [v,w](g) \end{eqnarray}

Therefore the Lie bracket is linear. For the Leibniz property :

\begin{eqnarray} [v, w](fg) &=& v[w(fg)] - w[v(fg)]\\ &=& v[f w(g) + g w(f)] - w[f v(g) + g v(f)]\\ &=& v[f w(g)] + v[g w(f)] - w[f v(g)] - w[g v(f)]\\ &=& f v[w(g)] + v(f) w(g) + g v[w(f)] + v(g) w(f) - f w[v(g)] - w(f) v(g) - g w[v(f)] - w(g) v(f)\\ &=& f v[w(g)] + g v[w(f)] - f w[v(g)] - g w[v(f)] \\ &=& f [v, w](g) + g [v, w](f) \end{eqnarray}

The Lie bracket then also obeys the Leibniz property, and the Lie bracket of two vector fields is itself a vector field.

For b) : Let's take the Jacobi identity of those three vector fields :

\begin{eqnarray} ([[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y])(f) &=& [[X, Y], Z](f) + [[Y, Z], X](f) + [[Z, X], Y](f)\\ &=& [X, Y][Z(f)] - Z[[X,Y](f)] + [Y,Z][X(f)] - X[[Y,Z](f)] + [Z,X][Y(f)] - Y[[Z,X](f)]\\ &=& X[Y[Z(f)]] - Y[X[Z(f)]] - Z[X[Y(f)]] + Z[Y[X(f)]] + Y[Z[X(f)]] - Z[Y[X(f)]]\\ && - X[Y[Z(f)]] + X[Z[Y(f)]] + Z[X[Y(f)]] - X[Z[Y(f)]] - Y[Z[X(f)]] + Y[X[Z(f)]]\\ &=& 0 \end{eqnarray}

The Lie bracket then does obey the Jacobi identity.

For c) : Given the basis $\{ Y_\mu \}$, and the coefficients $[Y_\alpha, Y_\beta] = {C^\gamma}_{\alpha\beta} Y_\gamma$, the Jacobi identity of the basis is given by

\begin{eqnarray} [[Y_\alpha, Y_\beta], Y_\gamma] + [[Y_\beta, Y_\gamma], Y_\alpha] + [[Y_\gamma, Y_\alpha], Y_\beta] &=& {C^\delta}_{\alpha\beta} [Y_\delta, Y_\gamma] + {C^\delta}_{\beta\gamma} [Y_\delta, Y_\alpha] + {C^\delta}_{\gamma\alpha} [Y_\delta, Y_\beta] \\ &=& {C^\delta}_{\alpha\beta} {C^\sigma}_{\delta\gamma} + {C^\delta}_{\beta\gamma} {C^\sigma}_{\delta\alpha} + {C^\delta}_{\gamma\alpha} {C^\sigma}_{\delta\beta} \\ &=& 0 \end{eqnarray}

This is the Jacobi identity for the structure constants of the basis.

Problem 2.4 :

  1. Show that in any coordinate basis, the components of the commutator of two vector fields are given by \begin{equation} [v,w]^\mu = \sum_\nu (v^\nu \frac{\partial w^\mu}{\partial x^\nu} - w^\nu \frac{\partial v^\mu}{\partial x^\nu}) \end{equation}
  2. Let $Y_1, \ldots, Y_n$ be as in problem 3(c). Let $Y^{1*}, \ldots, Y^{n*}$ be the dual basis. Show that the components $(Y^{\gamma*})_\mu$ of $Y^{\gamma*}$ in any coordinate basis satisfy \begin{equation} \frac{\partial (Y^{\gamma*})_\mu}{\partial x^\nu} - \frac{\partial (Y^{\gamma*})_\nu}{\partial x^\mu} = \sum_{\alpha, \beta} {C^\gamma}_{\alpha\beta} (Y^{\alpha*})_\mu (Y^{\beta*})_\nu \end{equation} (Hint : Contract both sides with $(Y_\sigma)^\mu (Y_\rho)^\nu$)

Solution : For a) : In some coordinate patch $\{ x^\mu \}$, with coordinate basis $\partial_\mu$, the vectors can be decomposed as

\begin{eqnarray} [v,w] &=& [v^\mu \partial_\mu, w^\nu \partial_\nu ] \end{eqnarray}

As the components of the vector fields are scalar functions, by linearity and the Leibniz property, we get

\begin{eqnarray} [v,w] &=& [v^\mu \partial_\mu, w^\nu \partial_\nu ] \end{eqnarray}

In that given basis, the action of a vector as a derivative can be expressed as

\begin{eqnarray} v(f) = \sum_{\mu} v^\mu \partial_\mu(f) = \sum_{\mu} v^\mu \frac{\partial f}{\partial x^\mu} \end{eqnarray}

Applying our Lie bracket to a function, this gives us

\begin{eqnarray} [v,w](f) &=& v^\mu w^\nu [ \partial_\mu, \partial_\nu ] (f)\\ &=& v^\mu v^\nu ( \partial_\mu [\partial_\nu(f)] - \partial_\nu [\partial_\mu(f)])\\ &=& v^\mu v^\nu ( \frac{\partial}{\partial x^\mu} \frac{\partial f}{\partial x^\nu} - \frac{\partial}{\partial x^\nu} \frac{\partial f}{\partial x^\mu}])\\ \end{eqnarray}