\documentclass{article} \def\ZZ{\hbox{\bf Z}} \def\RR{\hbox{\bf R}} \def\CC{\hbox{\bf C}} \def\QQ{\hbox{\bf Q}} \def\Ga{\Gamma} \def\ga{\gamma} \def\al{\alpha} \def\De{\Delta} \def\de{\delta} \def\La{\Lambda} \def\la{\lambda} \def\eps{\varepsilon} \def\ka{\kappa} \def\oM{\overline{M}} \def\oX{\overline{X}} \def\onabla{\overline{\nabla}} \def\om{\omega} \def\Om{\Omega} \def\sup{\rm sup} \def\grad{\rm grad} \def\pa{\partial} \def\bpa{\bar{\partial}} \def\spec{\hbox{spec}} \def\Dom{\hbox{Dom}} \def\bb{\bar{b}} \def\bB{\bar{B}} \def\bM{\overline{M}} \def\bx{\bar{x}} \def\bw{\bar{w}} \def\bxi{\bar{\xi}} \def\bz{\bar{z}} \def\dist{{\rm dist}} \def\bpar{\bar{\partial}} \def\dimg{\dim_{\Gamma}} \def\codimg{\hbox{codim}_{\Gamma}} \def\cO{{\cal O}} \def\Re{\hbox{Re}\,} \def\Im{\hbox{Im}\,} \def\Ker{\hbox{Ker}\,} \begin{document} \centerline {\bf Riemannian Geometry and General Relativity} \centerline{\bf (Differential Geometry-2, MTH 3412)} \centerline{\bf Home Assignment 2. Solutions of selected problems.} \centerline{\bf Professor M.Shubin} \centerline{\bf Winter 1997} \medskip\noindent {\bf 1.} Derive the transformation rule for the Christoffel symbols $\Ga^\nu_{\mu\alpha}$ under a change of curvilinear coordinates. \medskip{\bf Solution.} By the definition of the Christoffel symbols we have $$\nabla_{\pa\over\pa x^\mu}{\pa\over\pa x^\nu}=\Ga^\sigma_{\mu\nu}{\pa\over\pa x^\sigma}\;. \leqno (1)$$ Let $x^{\mu^\prime}$ be other local coordinates. Then by the chain rule $${\pa\over\pa x^\nu}=x^{\nu^\prime}_{\ ,\nu}{\pa\over\pa x^{\nu^\prime}}\;.$$ Therefore (1) can be rewritten in the form $$x^{\nu^\prime}_{\ ,\nu\mu}{\pa\over\pa x^{\nu^\prime}}+{x^{\mu^\prime}_{\ ,\mu} x^{\nu^\prime}_{\ ,\nu}\nabla_{\pa\over\pa x^{\mu^\prime}}} {\pa\over\pa x^{\nu^\prime}}= \Ga^\sigma_{\mu\nu}{x^{\sigma^\prime}_{\ ,\sigma}{\pa\over\pa x^{\sigma^\prime}}}\;,$$ or $$\bigl(x^{\sigma^\prime}_{\ ,\nu\mu}+{x^{\mu^\prime}_{\ ,\mu} x^{\nu^\prime}_{\ ,\nu}\Ga^{\sigma^\prime}_{\mu^\prime\nu^\prime}\bigr)} {\pa\over\pa x^{\sigma^\prime}}= \Ga^\sigma_{\mu\nu}{x^{\sigma^\prime}_{\ ,\sigma}{\pa\over\pa x^{\sigma^\prime}}}\;,$$ hence $$x^{\sigma^\prime}_{\ ,\nu\mu}+{x^{\mu^\prime}_{\ ,\mu} x^{\nu^\prime}_{\ ,\nu}\Ga^{\sigma^\prime}_{\mu^\prime\nu^\prime}}= \Ga^\sigma_{\mu\nu}{x^{\sigma^\prime}_{\ ,\sigma}}\;.$$ Multiplying both sides by $x^\al_{\ ,\sigma^\prime}$ and using the identity $$x^{\sigma^\prime}_{\ ,\sigma}x^\al_{\ ,\sigma^\prime}=\de^\al_\sigma\,,$$ we see that $$\Ga^\al_{\mu\nu}= x^{\mu^\prime}_{\ ,\mu}x^{\nu^\prime}_{\ ,\nu}x^\al_{\ ,\sigma^\prime} \Ga^{\sigma^\prime}_{\mu^\prime\nu^\prime}+ x^{\sigma^\prime}_{\ ,\nu\mu}x^\al_{\ ,\sigma^\prime} $$ or $$\Ga^\sigma_{\mu\nu}= x^{\mu^\prime}_{\ ,\mu}x^{\nu^\prime}_{\ ,\nu}x^\sigma_{\ ,\sigma^\prime} \Ga^{\sigma^\prime}_{\mu^\prime\nu^\prime}+ x^{\sigma^\prime}_{\ ,\mu\nu}x^\sigma_{\ ,\sigma^\prime}\;,\leqno (2) $$ which is the desired transformation law. % $\ecarre$ \bigskip\noindent {\bf 2.} Show that $\Ga^\sigma_{\mu\nu}$ is not a tensor. \medskip {\bf First solution.} Using (2) above we see that $\Ga^\sigma_{\mu\nu}$ has a tensor transformation law only in case when $x^{\sigma^\prime}_{\ ,\mu\nu}=0$ which means that the elements of the Jacobi matrix $x^{\sigma^\prime}_{\ ,\mu}$ are constants which implies that the transformation of the coordinates is linear. For all non-linear transformations the tensor law of transformation does not hold. % $\ecarre$ \medskip {\bf Second solution.} Consider the flat Euclidean space in the standard coordinates $x^\mu$, so that the components of the metric tensor are constants. It follows that $\Ga^\sigma_{\mu\nu}=0$ at all points. Therefore the parallel transport is the usual parallel transport which does not change the coordinates of the vectors (and components of any tensors as well). But in other curvilinear coordinates $x^{\mu^\prime}$ the same parallel vectors can obviously have variable coordinates, therefore $\Ga^{\sigma^\prime}_{\mu'\nu'}$ can not be identically zero. Hence $\Ga^\sigma_{\mu\nu}$ is not a tensor. % $\ecarre$ \bigskip\noindent {\bf 3.} On $\RR^n$ show that $\Ga^\sigma_{\mu\nu}$ can be arbitrary $C^\infty$ functions i.e. for any choice of such $n^3$ functions there exists a unique affine connection with the Christoffel symbols $\Ga^\sigma_{\mu\nu}$ in the canonical coordinates on $\RR^n$. \medskip {\bf Plan of solution.} It follows from (1) and the properties of the covariant derivative that we can calculate $\nabla_XY$ for any vector fields $X,Y$ as soon as we know $\Ga^\sigma_{\mu\nu}$ . This means that the desired connection is unique. In this way we can easily write formulas for $\nabla_XY$ is terms of the components of the fields $X,Y$ and the Christoffel symbols $\Ga^\sigma_{\mu\nu}$ . But then it is easy to check that all the axioms defining connection are satisfied for the covariant derivative given by these formulas. \bigskip\noindent {\bf 5.} Prove that for any connection with the Christoffel symbols $\Ga^\sigma_{\mu\nu}$ the quantities \noindent $\tilde\Ga^\sigma_{\mu\nu}={1\over 2}(\Ga^\sigma_{\mu\nu}+\Ga^\sigma_{\nu\mu})$ are the Christoffel symbols of a symmetric connection. \medskip {\bf First solution.} Interchanging $\mu$ and $\nu$ in the transformation law (2) we see that $\Ga^\sigma_{\nu\mu}$ has the same transformation law: $$\Ga^\sigma_{\nu\mu}= x^{\mu^\prime}_{\ ,\mu}x^{\nu^\prime}_{\ ,\nu}x^\sigma_{\ ,\sigma^\prime} \Ga^{\sigma^\prime}_{\nu^\prime\mu^\prime}+ x^{\sigma^\prime}_{\ ,\mu\nu}x^\sigma_{\ ,\sigma^\prime}\;,\leqno (2') $$ because $x^{\sigma^\prime}_{\ ,\mu\nu}=x^{\sigma^\prime}_{\ ,\nu\mu}$. Therefore adding (2) and (2') and dividing by 2, we obtain the same transformation law for $\tilde\Ga^\sigma_{\mu\nu}$. This means that $\tilde\Ga^\sigma_{\mu\nu}$ are the Christoffel symbols of a connection (which is obviously symmetric because $\tilde\Ga^\sigma_{\mu\nu}=\tilde\Ga^\sigma_{\nu\mu}$). % $\ecarre$ \medskip {\bf Second solution.} Assume that $\nabla$ is a connection with the Christoffel symbols $\Ga^\sigma_{\mu\nu}$ and define $$\nabla'_XY=\nabla_YX+[X,Y]\;.$$ It is easy to check that this is again a connection with the Christoffel symbols $\Ga^{\prime\sigma}_{\mu\nu}=\Ga^\sigma_{\nu\mu}$. Now we can use the following \medskip\noindent {\bf Lemma.} {\it If $\nabla$ and $\nabla'$ are connections then $\nabla_t=t\nabla+(1-t)\nabla'$ is a connection for any $t\in\RR$. Its Christoffel symbols are $\Ga^\sigma_{\nu\mu}(t)=t\Ga^\sigma_{\nu\mu}+(1-t)\Ga^{\prime\sigma}_{\nu\mu }$ where $\Ga^\sigma_{\nu\mu}$ and $\Ga^{\prime\sigma}_{\nu\mu}$ are the christoffel symbols of $\nabla$ and $\nabla'$ respectively.} \medskip The proof is straightforward. % $\ecarre$ \medskip {\bf Remark.} Lemma above means that connections form an affine space where the corresponding vector space is the space of tensors of type (1,2). \medskip\noindent {\bf 7.} Describe geometrically the parallel transport along a parallel $c$ of latitude $\theta$ on the standard unit sphere $S^2\subset \RR^3$. \medskip\noindent {\it Hint.} Consider the cone $C$ tangent to $S^2$ along $c$ and show that the parallel transport of any tangent vector along $c$ is the same whether taken relative to $S^2$ or to $C$. \medskip\noindent {\it Notational convention.} In the solution below the latitude $\theta$ is counted from the equator (as in geography), i.e. $\theta=0$ on the equator and $\theta=\pi/2$ at the North Pole. \medskip {\bf Solution.} The Levi-Civita connection on any surface $M$ imbedded into $\RR^3$ with respect to the induced metric is obtained by the usual infinitesimal parallel displacement (in $\RR^3$) followed by the orthogonal projection to the tangent space to $M$. Therefore if two such surfaces have a common curve $c$ and common tangent spaces at all points of this curve, then the parallel transport along $c$ will be the same on these surfaces. In particular this is true for a common parallel $c$ of $S^2$ and $C$ provided $S^2$ and $C$ are tangent along $c$. Denote the vertex of $C$ by $V$. The parallel transport along $c$ on the cone $C$ is easy to understand if we notice that $C$ is locally isometric to $\RR^2$ with the standard metric. Let us take a generating line $L$ of $C$ (the straight line on $C$ passing through $V$). Denote by $C_\theta$ the part of $C$ between $V$ and the parallel $c$. Then cutting $C_\theta$ along $L$ we obtain a piece of surface which can be isometrically straightened out to a piece of $\RR^2$ which is a sector $S$ of a disc. An elementary geometric consideration shows that the radius of this disc (which is equal to the distance from $V$ to $c$) is $R_\theta=\cot\theta$, and the length of the circular arc $l$ of the sector ( the length of the parallel $c$) is $l_\theta=2\pi\cos\theta$. In $\RR^2$ the radius of the disc connecting $V$ with the arc $l$ rotates by the angle $l_\theta/R_\theta=2\pi\sin\theta$ as we traverse the whole arc. At the same time the parallel transported vector does not change. This means that the parallel transported vector rotates by the same angle with respect to the tangent vector to the parallel because this tangent vector and the radius remain perpendicular. This means that the angular velocity of the rotation of the parallel transported vector is $\omega_\theta=(2\pi\sin\theta)/(2\pi\cos\theta)=\tan\theta$. (It is clear that the angular velocity is constant due to the symmetry: it should be the same at all points of the arc $l$ because any such point can be moved to any other by an isometry preserving $c$.) % $\ecarre$ \vfill\eject \medskip\noindent {\bf 8.} Calculate the Christoffel symbols on the Lobachevsky plane $\RR^2_+=\{(x,y)\in\RR^2|\,y>0\}$ (with the metric $ds^2=y^{-2}(dx^2+dy^2)$). \medskip {\bf Solution.} We will use notations $x^1=x, x^2=y$. Then $$g_{11}=g_{22}={1\over y^2},\quad g_{12}=g_{21}=0;\qquad g^{11}=g^{22}=y^2, \quad g^{12}=g^{21}=0\;.$$ Using the formula $$\Ga_{\sigma\mu\nu}={1\over 2}(g_{\sigma\mu,\nu}+g_{\sigma\nu,\mu}-g_{\mu\nu.\sigma})\;,$$ we obtain by straightforward calculation that $$\Ga_{111}=\Ga_{122}=\Ga_{212}=\Ga_{221}=0, \qquad \Ga_{112}=\Ga_{121}=\Ga_{222}=-{1\over y^3}\;, \qquad \Ga_{211}={1\over y^3}\;.$$ Now $g^{12}=0$ implies that $\Ga^{\sigma}_{\mu\nu}=g^{\sigma\sigma}\Ga_{\sigma\mu\nu}\ (\hbox{no summation!})= y^2\Ga_{\sigma\mu\nu}$, so we find $$\Ga^1_{11}=\Ga^1_{22}=\Ga^2_{12}=\Ga^2_{21}=0, \qquad \Ga^1_{12}=\Ga^1_{21}=\Ga^2_{22}=-{1\over y}, \qquad \Ga^2_{11}={1\over y}\;.$$ \medskip\noindent {\bf 9.} On the Lobachevsky plane (see Problem 8 above) describe the parallel transport along the ``curve" $x=t, y=1$. \medskip {\bf Solution.} We will use the notations $x^1=x, x^2=y$. The parallel transport along a curve $\ga=\ga(t)$ is defined by the differential equations $${dv^\sigma(t))\over dt}+\Ga^{\sigma}_{\mu\nu}(\ga(t)){d\ga^\mu(t)\over dt}v^\nu(t)=0\;,$$ where $v^\sigma(t)$ are the components of the transported vector at the point $\ga(t)$. In our case we have $\ga^1(t)=t, \ga^2(t)=1$, hence ${d\ga^1(t)\over dt}=1,\ {d\ga^2(t)\over dt}=0$. Also $\Ga^\sigma_{\mu\nu}$ are constants along our curve. Therefore the equations can be rewritten as $${dv^\sigma(t))\over dt}+\Ga^{\sigma}_{1\nu}v^\nu(t)=0\;.$$ Substituting the values of $\Ga^{\sigma}_{1\nu}$ given in the solution of Problem 8, we obtain $$\cases{{dv^1\over dt}+\Ga^1_{12}v^2=0 \cr {dv^2\over dt}+\Ga^2_{11}v^1=0\cr}$$ and $$\cases{{dv^1\over dt}-v^2=0 \cr {dv^2\over dt}+v^1=0\cr}$$ The solution of this system has the form $$\cases {v^1(t)=v^1(0)\cos t +v^2(0)\sin t \cr v^2(t)=-v_1(0)\sin t +v^2(0)\cos t\cr}$$ i.e. the parallel transport is the uniform rotation with the angular velocity 1. \vfill\eject \medskip\noindent {\bf 10.} Show that for any Riemannian or pseudo-Riemannian metric $g=(g_{\mu\nu})$ and arbitrary tensor $T=(T^\alpha_{\mu\nu})$ with $T^\alpha_{\mu\nu}=-T^\alpha_{\nu\mu}$ there exists unique affine connection which is compatible with $g$ and has the given torsion tensor $T$ (see Problem 4). \medskip {\bf Solution.} The simplest way is to repeat the argument which was used for the construction of the Levi-Civita connection. Let $X,Y\in\hbox{Vect}(M)$ i.e. $X,Y$ are vector fields on $M$. Instead of the relation $$\nabla_XY-\nabla_YX=[X,Y]$$ (which is equivalent to the symmetry of the connection) we should use the relation $$\nabla_XY-\nabla_YX=[X,Y]+T(X,Y)\;,$$ where $T(X,Y)$ is the torsion tensor which is a bilinear antisymmetric tensor map on vector fields $\hbox{Vect}(M)\times \hbox{Vect}(M)\to \hbox{Vect}(M)$, such that $T({\pa\over\pa x^\mu},{\pa\over\pa x^\nu})=T^\al_{\mu\nu}{\pa\over\pa x^\al}$. Compatibility with $g$ means: $$X\langle Y,Z\rangle=\langle \nabla_XY,Z\rangle+\langle Y,\nabla_XZ\rangle\;.\leqno (a)$$ Making cyclic permutations of $X,Y,Z$ we obtain $$Y\langle Z,X\rangle=\langle \nabla_YZ,X\rangle+\langle Z,\nabla_YX\rangle\;,\leqno (b)$$ $$Z\langle X,Y\rangle=\langle \nabla_ZX,Y\rangle+\langle X,\nabla_ZY\rangle\;.\leqno (c)$$ Adding (a) and (b) and subtracting (c), we obtain $$X\langle Y,Z\rangle+Y\langle Z,X\rangle-Z\langle X,Y\rangle=$$ $(\langle \nabla_XY,Z\rangle+\langle\nabla_YX, Z\rangle)+(\langle \nabla_XZ,Y\rangle- \langle \nabla_ZX,Y\rangle)+(\langle \nabla_YZ,X\rangle-\langle \nabla_ZY,X\rangle)=$ \noindent $\langle 2\nabla_X Y-[X,Y]-T(X,Y),Z\rangle+\langle[X,Z]+T(X,Z) ,Y\rangle +\langle[Y,Z]+T(Y,Z) ,X\rangle\,,$ \medskip\noindent therefore $$2\langle\nabla_XY,Z\rangle= X\langle Y,Z\rangle+Y\langle Z,X\rangle-Z\langle X,Y\rangle$$ \noindent $+\langle [X,Y],Z\rangle-\langle [Y,Z],X\rangle+\langle[Z,X],Y\rangle+ \langle T(X,Y),Z\rangle+\langle T(Z,X) ,Y\rangle -\langle T(Y,Z) ,X\rangle\,.$ \medskip\noindent A routine check shows that the right-hand side is a tensor with respect to $X,Z$ and satisfies the Leibniz rule with respect to $Y$. Therefore defining $\nabla_XY$ by this relation, we get a connection. Another straightforward calculation shows that its torsion equals $T$. % $\ecarre$ \end{document}