# [Solutions] Ch 2 Manifolds in “Spacetime and geometry”

These are parts of the solutions to exercises from:

Carroll, S. M. (2005). Spacetime and geometry: An introduction to general relativity. Addison Wesley.

Solutions:

1. First consider mapping the infinite cylinder to a semi-infinite cylinder, then consider projecting points on it to a plane.

Let $L=1$,

$(acos\theta,asin\theta,z)\mapsto ^f (acos\theta,asin\theta,e^z)\mapsto ^g (acos\theta/e^z,asin\theta/e^z,1)$.

$g\circ f$ is the map desired whose image is an open set $R^2\backslash (0,0)$.

2. No. $\mathbb{R}^k$ must be k-dimension manifolds. Furthermore, the dimension of a manifold is unique.

Suppose that $U$ is a subset of a manifold $M$, if there are two charts with different dimension $m$ and $latex n$, then there exists $\Phi_1, \Phi_2\in C^{\infty}$, $\Phi_1:U\to \mathbb{R}^m$, $\Phi_2:U\to \mathbb{R}^n$. Then $\Phi_2\circ \Phi_1$ is a diffeomorphism from $\Phi_1(U)$ to $\Phi_2(U)$. Hence m must be equal to n.

Note: Here I use such an assertion( J. Milnor, Topology from Differentiable Viewpoint, P4).

Assertion. If $f$ is a diffeomorphism between open sets $U\subset R^k$ and $V\subset R^l$, then $k$ must equal $l$, and the linear mapping

$df_x:R^k\to R^l$

must be nonsingular.

Proof. The composition $f^{-1}\circ f$ is the identity map of U; hence $d(f^{-1})_v\circ df_x$ is the identity map of $R^k$. Similarly $df_x\circ d(f^{-1})_v$ is the identity map of $R^l$. Thus $df_x$ has a two-sided inverse, and it follows that $k=l$.

By the way, $k=l$ can also be seen from the $max\{rank(BA), rank(AB)\}=min\{rank(A),rank(B)\}$, $A,B$ denote the linear map $d(f^{-1})_v$ and $df_x$ respectively.

3. Trivial.

4. First two can be verified directly.

Composition formula:

$[X,Y]^{\mu}=X^{\mu}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu}$.

$[X,Y]^{\mu}=[X,Y](x^{\mu})=X(Y(x^{\mu}))-Y(X(x^{\mu}))=X(Y^{\mu})-Y(X^{\mu})\\ =X^{\mu}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu}$.

Transformation:

$[X,Y]^{\mu'}=X^{\mu'}\partial_{\lambda'}Y^{\mu'}-Y^{\lambda'}\partial_{\lambda'}X^{\mu'}\\ =X^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}}Y^{\mu})-Y^{\lambda}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}}X^{\mu})\\ =X^{\lambda}Y^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}})+\frac{\partial x^{\mu'}}{\partial x^{\mu}}x^{\lambda}\partial_{\lambda}Y^{\mu}-X^{\lambda}Y^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}})-\frac{\partial x^{\mu'}}{\partial x^{\mu}}Y^{\lambda}\partial_{\lambda}X^{\mu} \\ = \frac{\partial x^{\mu'}}{\partial x^{\mu}}(X^{\mu}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu})= \frac{\partial x^{\mu'}}{\partial x^{\mu}}([X,Y]^{\mu})$.

5. Set  $V_{1}=\partial _{x}$, $V_{2} = A(x, y)\partial _{x} + B(x, y)\partial _{y}$. The commutator will be given by

$[V_{1} , V_{2} ] = [\partial _{x} A(x, y)]\partial _{x} + [\partial _{x} B(x, y)]\partial _{y}$.

Set $V_{2} =x \partial _{x} + \partial _{y}$. Then $V_1$, $V_2$ are nowhere vanishing, and their commutator is nowhere vanishing as well.

6.

(a)

$r=\sqrt{x^2+y^2+z^2}=\sqrt{1+\lambda^2}$

$\theta=arccos \frac{z}{\sqrt{x^2+y^2+z^2}}=arccos\frac{\lambda}{\sqrt{1+\lambda^2}}$

$\phi=\lambda$

($\lambda\in [0,2\pi)$)

(b)

Cartesian: $(-sin\lambda, cos\lambda,1)$,

Polar:$(\frac{\lambda}{\sqrt{1+\lambda^2}},\frac{-1}{1+\lambda^2},1)$.

7.

(a) $\frac{\partial x^{\mu}}{\partial x^{\nu'}}=\frac{\partial(x,z)}{\partial(\chi,\theta)}=\left(\begin{array}{cc} {cosh\chi sin\theta}&{sinh\chi cos\theta}\\{sinh\chi cos\theta}&{-cosh\chi sin\theta} \end{array}\right)$
(b)$dx=cosh\chi sin\theta dx+sinh\chi cos\theta d\theta$,
$dz=sinh\chi cos\theta dx-cosh\chi sin\theta d\theta$,
$ds^2=(sin^2\theta+sinh^2\chi)(dx^2+d\theta^2)$.

8.
$d(\omega\wedge\eta)_{\mu_1\dots\mu_{p+q+1}}\\=\frac{(p+q)!(p+q+1)}{p!q!}\partial_{[\mu_1}\omega_{[\mu_2\dots\mu_{p+1}}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]}\\=\frac{(p+q)!(p+q+1)}{p!q!}[(\partial_{[\mu_1}\omega_{[\mu_2\dots\mu_{p+1}}).\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]}+\omega_{[\mu_{2}\dots\mu_{p+1}}.(\partial_{[\mu_1}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]})]\\=\frac{(p+q)!(p+q+1)}{p!q!}[(\partial_{[\mu_1}\omega_{[\mu_2\dots\mu_{p+1}}).\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]}+(-1)^p\omega_{[\mu_{1}\dots\mu_{p}}.(\partial_{[\mu_{p+1}}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]})]\\=(d\omega)\wedge\eta+(-1)^p\omega\wedge(d\eta)$

9.

(a)$d*F=3\partial_{[\mu}F_{\nu\rho ]}=1/2[\partial_{\mu}(\epsilon^{\nu_1\nu_2}_{\nu\rho}F_{\nu_1\nu_2})+\partial_{\nu}(\epsilon^{\nu_1\nu_2}_{\rho\mu}F_{\nu_1\nu_2})+\partial_{\rho}(\epsilon^{\nu_1\nu_2}_{\mu\nu}F_{\nu_1\nu_2})]=\epsilon^{\sigma}_{\mu\nu\rho}J_{\rho}=*J$
($\partial_{\mu}F^{\nu\mu}=J^{nu}$)
(b)$F=-(**F)$
$*F=\left( \begin{array}{cccc} 0&0 &0 &0 \\0 &0 &0 &0 \\0 &0 &0 &qsin\theta\\ 0&0 &-qsin\theta&0 \end{array}\right)$
$F=(- **F)=\left( \begin{array}{cccc} 0&-q/r^2 &0 &0 \\q/r^2 &0 &0 &0 \\0 &0 &0 &0\\ 0&0 &0&0 \end{array}\right)$
(c)$E_r=q/r^2$,$E_{\theta}=E_{\phi}=0$,
$B_{\mu}=0, \mu=r,\theta,\phi$
(d)$0$