# Can an arbitrary curve locally be made into a geodesic through a conformal change of the metric?

Given a nowhere null curve $\gamma: I \rightarrow M$ in a pseudo-Riemannian manifold $(M,g)$. (I is some open interval) Let $t_0 \in I$ arbitrary. Does a conformally equivalent metric $\hat{g}=e^{2\sigma} g$ and some $\epsilon >0$ exist, such that $\gamma \mid _{(t_0-\epsilon, t_0 + \epsilon)}$ is a geodesic in $(M,\hat{g})$?

$\gamma$ being a geodesic with respect to $(M, \hat{g})$ means

$$0=\hat{\nabla}_{\gamma'}\gamma'=\nabla_{\gamma'}\gamma'+ 2 \gamma' \cdot \gamma'(\sigma) - g(\gamma', \gamma') \cdot \operatorname{grad} \sigma,$$

where we used the transformation law for the Levi Civita connection under a conformal change of metrics. Now it feels there should be some $\sigma$ satisfying this equation locally, because we have very big freedom. However I am having difficulties carrying out the construction of $\sigma$ and I don't know of general fact that would give existence of a solution.

• What is $\sigma$? Is it a real-valued function on $M$? When you write $\sigma(\gamma')$, it makes me think not. – John Hughes Dec 15 '16 at 12:29
• @JohnHughes: I think the dot is supposed to represent the directional derivative of $\sigma$ in the direction $\gamma'$. It would be clearer to write $0=\hat{\nabla}_{\gamma'}\gamma'=\nabla_{\gamma'}\gamma'+ 2 (\gamma' \cdot \sigma)\gamma' - g(\gamma', \gamma') \cdot \operatorname{grad} \sigma$. – Jack Lee Dec 15 '16 at 16:19
• Thanks -- that makes some sense. – John Hughes Dec 15 '16 at 18:13
• Sorry, there was indeed a typo in the formular, but Jack Lee has got it right. $\sigma$ is a real-valued function on $M$. $\gamma'(\sigma)$ denotes the directional derivative of $\sigma$ in direction $\gamma'$. $\sigma(\gamma')$ was a typo and has no meaning. Thank you for clarifying this! – 761838257287 Dec 15 '16 at 21:38

Interesting question. The answer seems to be yes. (The problem with @JohnHughes's counterexample is that his curve $\gamma$ does not satisfy $\nabla_{\gamma'}\gamma'=0$ with respect to his original metric $g$. If my back-of-the envelope computation is correct, it should be $\nabla_{\gamma'}\gamma'=\frac12 \partial/\partial y$.)

Here's a sketch of a proof that it's always possible to conformally change the metric in a neighborhood of a point to make $\gamma$ a geodesic there. The proof comes in several steps. (Note that all of these constructions are merely local.)

Step 1. Choose local coordinates $(x^1,\dots,x^n)$ in which $\gamma(t) = (t,0,\dots,0)$. Because we're assuming $\gamma'(t)$ is never zero, this is possible by the rank theorem.

Step 2. Make a preliminary conformal change to $g$ to make $\gamma$ a unit-speed curve. In the coordinates above, $|\gamma'(t)|^2_g = g_{11}(t,0,\dots,0)$, so this can be accomplished by defining $\overline g = f g$, where $f(x^1,\dots,x^n) = 1/ g_{11}(x^1,0,\dots,0)$. Let's replace the original metric by $\overline {g}$.

Step 3. Now change $g$ again to make $\hat \nabla_{\gamma'}\gamma'=0$. Write the coordinate representation of $\nabla_{\gamma'(t)}\gamma'(t)$ as $\sum_j a^j(t)\partial_j|_{\gamma(t)}$. Differentiating the equation $|\gamma'(t)|^2 \equiv 1$ shows that $\nabla_{\gamma'}\gamma'$ is orthogonal to $\gamma'$ along $\gamma$, which means that $$\sum_j g_{1j}(t,0,\dots,0) a^j(t) \equiv 0.\tag{1}$$ Define a function $\sigma$ by $$\sigma(x^1,\dots,x^n) = 1 + \sum_{i,j=1}^n g_{ij}(x^1,0,\dots,0) x^i a^j(x^1) = 1 + \sum_{i=2}^n\sum_{j=1}^n g_{ij}(x^1,0,\dots,0)x^i a^j(x^1) ,$$ where the second equality follows from $(1)$, and set $\hat g = e^{2\sigma}g$. Note that along the image of $\gamma$, where $x^2=\dots =x^n=0$, we have $\sigma\equiv 1$, so $\gamma$ is still unit-speed with respect to $\hat\gamma$. Also along the image of $\gamma$, we have $\partial_i \sigma = \sum_j g_{ij}a^j$: for $i=1$, this follows by noting that $\partial_1\sigma=0$ since $\sigma=1$ there, and applying $(1)$; while for $i>1$, it follows by differentiating the formula for $\sigma$ and setting $x^2=\dots= x^n=0$. Therefore, $$\operatorname{grad} \sigma = \sum_{i,k} g^{ik} \partial_i\sigma \partial_k = \sum_{i,j,k} g^{ik} g_{ij} a^j \partial_k =\sum_j a^j \partial_j = \nabla_{\gamma'}{\gamma'}.$$ Moreover, we also have $\gamma'(\sigma) = \partial\sigma/\partial x^1 = 0$ and $g(\gamma',\gamma') = 1$ along $\gamma$, so it follows from the OP's formula that $\hat \nabla_{\gamma'}\gamma' \equiv 0$. $\square$

There might be a more efficient way to do this, but this was the best I could come up with on the spur of the moment.

• Thank you very much, your proof is easy to follow and helps me a lot. For your initial back-of-the-envelope calculation: Using the Koszul formula, it seems that $\nabla _{\gamma'} \gamma' = \frac{1}{2t} \cdot \frac{\partial}{\partial y}$. But setting aside the exact value, we can agree that the suggested curve is not parallel with respect to $\nabla$. This also follows from the fact, that it is not a unit speed curve, which is easier to check. On an unrelated note: I'm a big fan of your "Smooth Manifolds" book, it sure has helped a lot of students. – 761838257287 Jan 9 '17 at 10:38
• @Daniel -- Thanks! I'm glad it's been helpful. – Jack Lee Jan 9 '17 at 14:53

Following Jack Lee's interpretation, I think that the answer is "no".

Let's look at the case where $M$ is the upper half-plane, and the metric at location $(x, y)$ is given, in the standard basis, by

\begin{align} g(x, y) = \begin{bmatrix} 1 & 0 \\ 0 & y \end{bmatrix} \end{align}

And let's also look at the path $$\gamma(t) = (t, t)$$

Suppose that there were a scaling function $\sigma$ on the plane that made $\gamma$ into a geodesic.

Since $\gamma'$ is constant, the first term in $$0=\nabla_{\gamma'}\gamma'+ 2 (\gamma') \cdot \gamma'(\sigma) - g(\gamma', \gamma') \cdot \operatorname{grad} \sigma$$ is zero [NB: Per the comments, this is in doubt]. The second term is just twice the directional derivative of $\sigma$ in the direction $[1, 1]^t$, times the vector $[1, 1]^t$. The $g$ part of the third term is $1 + t^2$, and the gradient is ... well, the gradient.

But this gives us something of the form $$c \begin{bmatrix}1\\1 \end{bmatrix} = (1+t^2) \operatorname{grad} \sigma$$ which tells us that the two components of the gradient must be equal. So $\sigma$ is constant on the lines where $x + y = c$, hence we can write $$\sigma(x, y) = f(x + y)$$ for some function $f$. Now $$\operatorname{grad} \sigma (x, y) = \begin{bmatrix} f'(x+ y) \\ f'(x+ y)\end{bmatrix}$$ and so our equation becomes \begin{align} 2 (\gamma' \cdot \sigma) \begin{bmatrix}1\\1 \end{bmatrix} &= (1+t^2) \operatorname{grad} \sigma \\ 2 (2f'(t+ t)) \begin{bmatrix}1\\1 \end{bmatrix} &= (1+t^2) \begin{bmatrix} f'(t+ t) \\ f'(t+ t)\end{bmatrix} \end{align}

That seems to say that $4 = 1 + t^2$ (or that $f$ is constant). The first is false, so $f$ must be constant ... but a constant $f$ does not make this path a geodesic.

I've probably screwed up somewhere obvious, but ... that's what I've got.

As mentioned, the claim about the directional derivative of $\gamma'$ is in doubt...but the rest of the analysis shows that this directional derivative, if it's not zero, must be in the $[1, 1]$ direction.
• Is there an easy way to see that $\nabla _{\gamma'} \gamma'=0$ in your example? For the Levi-Civita connection of the standard flat metric that would be obvious, but for your altered metric I don't see it immediately. I will think more about your example during the next days, thank you for your answer already! – 761838257287 Dec 15 '16 at 21:36
• To be honest, I thought it denoted the ordinary gradient in the $\gamma'$ direction. (As I said, I surely made some stupid mistake...) I'm more familiar with explicit everything-in-coordinates diff'l geometry. :( – John Hughes Dec 15 '16 at 21:38