[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.



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

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,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}.


[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.




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


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


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



(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,



(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
*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)
B_{\mu}=0, \mu=r,\theta,\phi




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