# "Pedantic" derivation of geodesic equation using pullback bundles

I'm trying to get more comfortable with manipulations involving connections and vector fields so I've tried to derive the geodesic equations without having to resort to any familiarities using standard calculus, everything computed "properly" from the definitions.

For a Reimannian manifold $$(M,g)$$ I have a curve $$\gamma : \mathbb{R} \rightarrow M$$ which in local coordinates can be written as $$\gamma(t) = \left(x^1(t), \ldots, x^n(t)\right)$$. If I wish $$\gamma(t)$$ to be a geodesic then I want its tangent vector to be auto-parallel.

The tangent vector is given by the pushforward of coordinate vector field on $$\mathbb{R}$$ $$\hspace{01mm}\dot{\gamma}(t)\hspace{-0.5mm}" := \gamma_*\left(\frac{\partial}{\partial t}\right) = \dot{x}^i(t) \frac{\partial}{\partial x^i}$$

I want $$\nabla_\dot\gamma \dot{\gamma} = 0$$, but this expression is misleading since the vector field $$\dot{\gamma}(t)$$ only exists along the image of $$\gamma(t)$$, but we can consider the pullback of $$M$$ by $$\gamma$$ and take the connection and vector bundle with us. If $$\nabla$$ is the Levi-Civita connection on $$(M,g)$$ denote $$\widetilde{\nabla}$$ as its pullback connection by $$\gamma$$

Then $$\gamma$$ is geodesic if $$\widetilde{\nabla}_\frac{\partial}{\partial t}\gamma_*\left(\frac{\partial}{\partial t}\right) = 0$$. At this point I start to get stuck, I have the following definition from a worked exam question that inspired me to do this exercise:

I'm not quite sure if I'm in the lucky situation where $$\gamma_*\left(\frac{\partial}{\partial t}\right)$$ is already of the form $$v \circ u$$, and I'm not so sure what this even means, if my bundle is the tangent bundle of $$M$$, then my sections $$e_i = \frac{\partial}{\partial x^i}$$ are vector fields, how does one compose a vector field with a map?

I think this has something to do with where we are evaluating $$\gamma_*\left(\frac{\partial}{\partial t}\right)f = \left.\dot{x}^i(t) \frac{\partial f}{\partial x^i}\right|_{\gamma(t)}$$, that is, $$e_i = \left.\frac{\partial}{\partial x^i}\right|_p$$ for $$p \in M$$ whereas $$e_i \circ \gamma = \left.\frac{\partial}{\partial x^i}\right|_{\gamma(t)}$$. I'm not sure how to properly justify this but it certainly feels more correct that $$\gamma_*\left(\frac{\partial}{\partial t}\right)$$ should be "evaluating" on $$\gamma(t)$$ rather than any old $$p$$ since the whole point of this pullback stuff was to differentiate along the curve.

If we accept the above handwaving then my calculation is as follows: $$\widetilde{\nabla}_\frac{\partial}{\partial t}\gamma_*\left(\frac{\partial}{\partial t}\right) := \nabla_{\gamma_*\left(\frac{\partial}{\partial t}\right)}\gamma_*\left(\frac{\partial}{\partial t}\right) = \nabla_{\dot{x}^i(t) \frac{\partial}{\partial x^i}}\dot{x}^j(t) \frac{\partial}{\partial x^j}$$

Using $$C^\infty(M)$$ linearity of a connection in the lower argument and the Liebnitz rule gives

$$= \dot{x}^i(t) \nabla_{\frac{\partial}{\partial x^i}}\dot{x}^j(t) \frac{\partial}{\partial x^j} = \dot{x}^i(t)\frac{\partial}{\partial x^i}\left(\dot{x}^j\right)\frac{\partial}{\partial x^j} + \dot{x}^i(t)\dot{x}^j(t)\nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}$$

The second term is $$\dot{x}^i(t)\dot{x}^j(t)\Gamma_{ij}^k\frac{\partial}{\partial x^k}$$ which starts to look on the right tracks, but I have no idea what to do to the first term to get a second time derivative, and if my approach is even correct.

Apologies for the wall of equations, but I wanted to get down all my thoughts and where my confusions lie, I am looking for how to finish the derivation and an explanation of all this stuff with pullback bundles and correct any misunderstandings I have. Thanks in advance.

• This is a bit old, but you already have the geodesic equation as follows. Note that by the chain rule $$\sum_i \frac{\partial \dot{x}^j}{\partial x^i} \cdot \dot{x}^i = \sum_i \frac{\partial \dot{x}^j}{\partial x^i} \cdot \frac{\partial x^i}{\partial t} = \frac{d^2 x^j}{dt^2}$$
– JMK
Commented Jan 12, 2022 at 8:33

This is a good question. Here's how to do it: given coordinates $$(x^1,\ldots, x^n)$$ around (some particular point) $$\gamma(t)$$, where $$\gamma\colon I \to M$$, we have that $$\dot{\gamma}(t) = \sum_{i=1}^n \dot{x}^i(t)\frac{\partial}{\partial x^i}\bigg|_{\gamma(t)}$$for all $$t \in I$$. Then indeed $$\nabla_{\dot{\gamma}(t)}\dot{\gamma}$$ does not immediately makes sense, but we have the pull-back bundle $$\gamma^*(TM) \to I$$, and a connection $$\gamma^*\nabla$$. Then $$\partial/\partial t$$ is a vector field on the "base manifold" $$I$$, and $$\gamma_\ast(\partial/\partial t) = \dot{\gamma}$$, and this is what allows us to use the defining property of $$\gamma^*\nabla$$: \begin{align}\frac{D\gamma'}{{\rm d}t}(t) &= (\gamma^*\nabla)_{(\partial/\partial t)|_t}(\dot{\gamma}) = (\gamma^*\nabla)_{(\partial/\partial t)|_t}\left(\sum_{j=1}^n \dot{x}^j \left(\frac{\partial}{\partial x^j}\circ \gamma\right)\right) \\ &= \sum_{j=1}^n \ddot{x}^j(t) \frac{\partial}{\partial x^j}\bigg|_{\gamma(t)}+ \sum_{j=1}^n \dot{x}^j(t) (\gamma^*\nabla)_{(\partial/\partial t)|_t}\left(\frac{\partial}{\partial x^j}\circ \gamma\right) \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} + \sum_{j=1}^n \dot{x}^j(t) \nabla_{\dot{\gamma}(t)}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} + \sum_{j=1}^n \dot{x}^j(t) \nabla_{\sum_{i=1}^n \dot{x}^i(t) (\partial/\partial x^i)|_{\gamma(t)}}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} + \sum_{i,j=1}^n \dot{x}^i(t)\dot{x}^j(t) \nabla_{(\partial/\partial x^i)|_{\gamma(t)}}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} + \sum_{i,j,k=1}^n \Gamma_{ij}^k(\gamma(t))\dot{x}^i(t)\dot{x}^j(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} \\ &= \sum_{k=1}^n \left(\ddot{x}^k(t) + \sum_{i,j=1}^n \Gamma_{ij}^k(\gamma(t))\dot{x}^i(t)\dot{x}^j(t)\right)\frac{\partial}{\partial x^k}\bigg|_{\gamma(t)}.\end{align}
• I see, so on the second line you are differentiating $\dot{x}^i(t)$, which is a function on the manifold $I$ with the connection $\gamma^* \nabla$ directly rather than doing things on the $M$ side? Commented Jul 13, 2020 at 21:49
• Yes! The Leibniz rule for $\gamma^*\nabla$ as opposed to the one for $\nabla$ is what gives $$\frac{\partial}{\partial t}\bigg|_t \dot{x}^j = \ddot{x}^j(t).$$ Commented Jul 13, 2020 at 21:51
• A nice exercise is to show that if $(M_1,g_1)$ and $(M_2,g_2)$ are Riemannian manifolds with Levi-Civita connections $\nabla^1$ and $\nabla^2$, the Levi-Civita connection of $(M_1\times M_2, g_1\oplus g_2)$ is $\pi_1^*\nabla^1 \oplus \pi^2\nabla^2$, where $\pi_i\colon M_1\times M_2 \to M_i$ are the projections. Note that the pull-backs are $\pi_i^*(TM_i) \to M_1\times M_2$, and so we have the Whitney sum $\pi_1^*(TM_1)\oplus \pi_2^*(TM_1) \to M_1\times M_2$. And $$\pi_1^*(TM_1)\oplus \pi_2^*(TM_2) \cong TM_1\times TM_2,$$and $TM_1\times TM_2$ is not the same thing as $TM_1\oplus TM_2$. Commented Jul 13, 2020 at 21:57
• If $\psi \in \Gamma(E)$, then $\psi \circ u \in \Gamma(u^*(E))$ is defined by taking $x \in M$ to $\psi(u(x)) \in E_{u(x)} = (u^*E)_x$, there's nothing to it. It's an actual set-theoretic composition of functions. Commented Jul 13, 2020 at 22:10