Time derivative of time-dependent vector fields on a manifold

Question

Suppose $X = X(t,p)$ is a time dependent vector field on a Riemannian manifold $(M,g)$. How would one compute $\frac{d}{dt}g\big(X, X)?$ I know that the answer should be $2g\big(\frac{\partial}{\partial t}X,X\big)$ and uses the often used identity from basic calculus $$\frac{d}{dt} |f(t)|^2 = 2|f(t)| f'(t)$$ but I'm having some trouble filling in the details.

One way I thought to do this is to use local coordinates so that $$X = \sum_ia^i(t,p)\frac{\partial}{\partial x_i}$$ and so the time derivative amounts to taking the time derivative of the coefficient functions $a^i$.

Is there a coordinate free way of showing this? More generally, how do you take the (coordinate free) time derivative of a time dependent vector field?

Answer 1

Your identity from basic calculus is wrong. It should be $2\langle f(t),f’(t)\rangle$ assuming the norm comes from an inner product. The answer in your case is literally a direct application of this result.

Fix a point $p\in M$, and define $\psi(t)=X(t,p)$ and $f(t)=g_p(\psi(t),\psi(t))$. Note that $g_p$ is the inner product on the single vector space $T_pM$, and $\psi$ is a curve taking values in the single vector space $T_pM$, and $f$ is a composition of such maps. We’re working in a single vector space only so you can call $V=T_pM$ and forget about the manifold $M$ completely. Then, the product rule from elementary calculus (which generalizes immediately to the vector space level) now tells us \begin{align} f’(t)&=g_p(\psi’(t),\psi(t))+g_p(\psi(t),\psi’(t))=2g_p(\psi’(t),\psi(t)). \end{align} Hence, $\frac{\partial}{\partial t}\left(g(X,X)\right)=g\left(\frac{\partial X}{\partial t},X\right)$.

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Note, this situation shouldn’t be confused with a more difficult one, which is when you have a curve $\psi:I\to TM$ in the tangent bundle itself rather than a single vector space. In this case, if we set $f(t)=g(\psi(t),\psi(t))$, then we’d have to use the Levi-Civita connection (well technically, we’d have to induce it on the pullback bundle $\gamma^*(TM)$ where $\gamma=\pi\circ\psi:I\to M$ is the base curve) to get \begin{align} f’(t)&=2g(\nabla_t\psi,\psi(t)), \end{align} where $\nabla_t\psi\equiv D_t\psi\equiv\frac{D\psi}{dt}\equiv\frac{\nabla\psi}{dt}$ (these are all common notations for it) is the covariant derivative of $\psi$ along $\gamma$ ($\psi$ can be thought of as a section of the pullback bundle $\gamma^*(TM)$). See Lee or Spivak for more details about this technicality. But essentially, this comes from the Leibniz rule and metric compatibility: for any vector fields $X,Y,Z$ on $M$, we have \begin{align} \nabla_{X}\bigg(g(Y,Z)\bigg)&= g(\nabla_XY,Z)+g(Y,\nabla_XZ). \end{align}

Answer 2

On a Riemann manifold you have to go back to definitions, the metric on the tangent space. Start with the definition of the scalar product by the bilinear, symmetric map from tangent space to scalars $G: T(x) \times T(x) \to \mathbb R_+$ in a given coordinate map with tangent vectors in the canonical basis frame $e(x)_i$

$$g(x)_{i,k} = G(e(x)_i,e(x)_k)$$

and then differentiate with respect to translation in time along a path

$$ G\left( X(t+dt), X(t+dt)\right) - G\left(X(t) ,X(t)\right) = G\left(\sum_i e_i(X(t+dt)) \ X(t+dt)^i,\sum_k X(t+dt)^k e(X(t+dt))_k) \right)- G\left( \sum_i e_i(X(t)) \ X(t)^i,\sum_k X(t)^k e(X(t))_k \right) $$

Full linear and derivation machinery applied, generates the covariant total derivative as replacement for the time derivative with the affine Christoffel map $\Gamma$ as derivative of the moving basis frame

$$d X(t) = d\left(\sum_i X(t)^i e_i\left(X(t)\right)\right) = \sum_i d X(t)^i e_i + \sum_{i}\ X(t)^i d e_i\left(X(t)\right)$$ $$ = dt \ \left(\sum X'(t)^i e_i(X(t))\right) + \sum_{i}X(t)^i d e_i\left(X(z)\right)$$ $$= dt\ \left( \sum_i X'(t)^i e_i(X(t)) + \sum_{ik} X(t)^i \ \Gamma\left(X(t)\right)^k_i \ e_k\left( X(t)\right) \right)$$

Last step: insert derivatives, use product rule for a bilinear product, and use the fact, that G is covariantly a constant function (Levi Civita condition for a torsion free connection $\Gamma$).

So, the compact result of Euler-Langrange-Riemann-Einstein implies an unforseen additional assumption, because in principle, the connection map is an additional degree of freedom of rotations of the frame moving along curves, compared to classical transport prallel to itself.