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I'm trying to solve the following problem in preparation for an exam :

(i) Let $g_0$ be a Riemannian metric and define $g=cg_0$ where $c$ is a positive constant. Prove that $g_0$ and $g$ have the same Levi-Civita connection.

(ii) Use this to find as many diffeomorphisms as you can that preserve geodesics of the metric $$g=y(dx^2+dy^2)$$ on the set $M=\{(x,y)\in\mathbb{R}^2: y>0\}$

My trouble is I'm not exactly sure how to calculate the LC connection for these metrics as the notes available to me don't cover it in much detail and so I'm not sure where to begin. If anyone has an example of how to perform such a calculation it would be greatly appreciated. I think seeing how a calculation like this is performed will be helpful for part (ii) but any other hints would be very helpful. Thanks.

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  • $\begingroup$ 3 years old on Math.SE, 29 questions and no accepted answer? Precisely the type of user to avoid. $\endgroup$ – Alex M. Mar 29 '17 at 21:10
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For (i), you can think about Christoffel symbols $\Gamma_{ij}^k$, which is defined as: $$\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial x_i}=\sum_{k=1}^n\Gamma_{ij}^k\frac{\partial}{\partial x_k}.$$ In terms of the metric, Christoffel symbol $\Gamma_{ij}^k$ can be computed as: $$\Gamma_{ij}^k=\sum_{l=1}^n\frac{g^{kl}}{2}\left(\frac{\partial g_{il}}{\partial x_j} +\frac{\partial g_{jl}}{\partial x_i}-\frac{\partial g_{ij}}{\partial x_l}\right).\tag{1}$$ (You can find these facts from any standard textbook in Riemannian Geometry. Do Carmo's Riemannian Geometry is my favorite.)

It is easy to see that Christoffel symbols determine the Levi-Civita connection: if Christoffel symbols of two metrics are the same, then their Levi-Civita connections are the same (Try to prove it yourself if you haven't seen it before). Therefore, if $g=cg_0$, you can use $(1)$ to check that their Christoffel symbols are the same.

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For (i), the correct argument using coordinates has already been provided by @Paul, so I shall provide one written in index-free notation. Any metric is linked to its Levi-Civita connection by Koszul's formula

$$g(\nabla_X Y, Z) = \frac 1 2 \{X g(Y, Z) + Y g(X, Z) - Z g(X, Y) + g( [X,Y], Z) - g( [Y,Z], X) - g( [X,Z], Y) \} .$$

If $\nabla_0$ and $\nabla$ are the Levi-Civita connections of $g_0$ and, respectively, $g$, then

$$g(\nabla_X Y, Z) = \frac 1 2 \{X g(Y, Z) + Y g(X, Z) - Z g(X, Y) + g( [X,Y], Z) - g( [Y,Z], X) - g( [X,Z], Y) \} = \\ \frac 1 2 \{X (cg_0)(Y, Z) + Y (cg_0)(X, Z) - Z (cg_0)(X, Y) + (cg_0)( [X,Y], Z) - (cg_0)( [Y,Z], X) - (cg_0)( [X,Z], Y) \} = \\ c \frac 1 2 \{X g_0(Y, Z) + Y g_0(X, Z) - Z g_0(X, Y) + g_0( [X,Y], Z) - g_0( [Y,Z], X) - g_0( [X,Z], Y) \} = \\ c g_0((\nabla_0)_X Y, Z) = g((\nabla_0)_X Y, Z) ,$$

so $g(\nabla_X Y, Z) = g((\nabla_0)_X Y, Z)$, for all tangent fields $X, Y, Z$, i.e. $g(\nabla_X Y - (\nabla_0)_X Y, Z) = 0$ for all $Z$, so $\nabla_X Y - (\nabla_0)_X Y = 0$ for all $X,Y$, so $\nabla = \nabla_0$.


For (ii), let me notice that the formulation "find as many diffeomorphisms as you can" is very subjective. In particular, you are not required to find all those diffeomorphisms. It is not clear how "as many as you can" will be considered "sufficiently many" by an examinator.

Remember that the coordinate-free definition of a geodesic $\gamma$ is to satisfy the identity $\nabla _{\dot \gamma} \dot \gamma = 0$. Let $f : (M, g) \to (M, g_0)$ be an isometry, i.e. $f$ is a diffeomorphism and $f^* g_0 = g = cg_0$. In general, an isometry takes geodesics to geodesics, but in this very special situation we can say even more: $f$ will take geodesics of $g$ into geodesics of $g_0$ and since these two metrics have the same Levi-Civita connection, and the definition of geodesics only requires the connection (not the metric), it follows that $g_0$ and $g$ have the same geodesics, i.e. $f$ preserves the geodesics of $M$.

To conclude, every isometry $f : (M,g) \to (M, g_0)$ with $c>0$ arbitrary will be a geodesic-preserving diffeomorphism of $M$. Notice that this is true for general $M$, not just for the one given in the question.


It might be interesting to get a more concrete feeling about how these maps look like for the manifold given in the problem. Since $M$ admits global coordinates, let $\big( u(x,y), v(x,y) \big) = f(x,y)$. The first condition to impose is $v > 0$. Next, $f^* [y (\Bbb d x^2 + \Bbb d y^2)] = c y (\Bbb d x^2 + \Bbb d y^2)$ is rewritten as $v (\Bbb d u^2 + \Bbb d v^2) = c y (\Bbb d x^2 + \Bbb d y^2)$. Since $\Bbb d u = \frac {\partial u} {\partial x} \Bbb d x + \frac {\partial u} {\partial y} \Bbb d y$ and $\Bbb d v = \frac {\partial v} {\partial x} \Bbb d x + \frac {\partial v} {\partial y} \Bbb d y$, we get

$$v \left[ \left( \frac {\partial u} {\partial x} \Bbb d x + \frac {\partial u} {\partial y} \Bbb d y \right)^2 + \left( \frac {\partial v} {\partial x} \Bbb d x + \frac {\partial v} {\partial y} \Bbb d y \right)^2 \right] = c y (\Bbb d x^2 + \Bbb d y^2)$$

which can be rewritten as

$$\left\{ \begin{align} & v \left[ \left( \frac {\partial u} {\partial x} \right)^2 + \left( \frac {\partial v} {\partial x} \right)^2 \right] &=& &cy \\ & v \left[ \left( \frac {\partial u} {\partial y} \right)^2 + \left( \frac {\partial v} {\partial y} \right)^2 \right] &=& &cy \\ & \frac {\partial u} {\partial x} \frac {\partial u} {\partial y} + \frac {\partial v} {\partial x} \frac {\partial v} {\partial y} &=& &0 \end{align} \right.$$

(in the last equation a silent division by $2v$ has been performed since $v>0$). Like most non-linear systems, this too gets pretty ugly once you attempt to solve it explicitly.


Since the general form of $f$ seems difficult to obtain, let us settle for something less: we shall look only for those $f$ given by matrix multiplication, i.e. $\big( u(x,y), v(x,y) \big) = (x, y) \begin{pmatrix} A & C \\ B & D \end{pmatrix}$.

The condition $v>0$ becomes $xC + yD > 0$ for all $(x,y) \in M$. Choosing $x=0$ gives us $yD>0$, so $D>0$ (because $y>0$ by the definition of $M$). If $C>0$ then fixing $y$ and letting $x \to -\infty$ produces $-\infty \ge 0$, which is impossible. Similarly, if $C<0$ then fixing $y$ and letting $x \to \infty$ again produces $-\infty \ge 0$, which is impossible. It remains that $C=0$.

The above system then gets simpler:

$$\left\{ \begin{align} & D A^2 &=& &c \\ & D (B^2 + D^2) &=& &c \\ & AB &=& &0 . \end{align} \right.$$

From the last equation, if $A=0$ then the matrix becomes $\begin{pmatrix} 0 & 0 \\ B & D \end{pmatrix}$, so $f$ is not bijective, which is impossible. It remains that $B=0$ and the system becomes

$$\left\{ \begin{align} & D A^2 &=& &c \\ & D^3 &=& &c \\ \end{align} \right.$$

whence $D = \sqrt[3] c$ and $A = \pm \sqrt[3] c$, so that $f$ becomes $f(x,y) = \sqrt[3] c (\pm x, y)$.

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  • $\begingroup$ Thank you for your answer, it was very helpful in understanding what my lecturer meant in his notes! $\endgroup$ – Crunch Apr 1 '17 at 19:56

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