Generalizing Fermi-Walker Derivative/Transport to General Vector Bundles

Question

The usual definition of the Fermi derivative (eg as given in Hawking and Ellis) is to consider a Lorentzian manifold $(M,g)$ and a unit timelike curve $\gamma$, a smooth vector field $X$ along $\gamma$, and define the Fermi-Walker derivative of $X$ along $\gamma$ to be the quantity \begin{align} \frac{D_{\text{Fermi}}X}{ds}&=\frac{DX}{ds}-g\left(X,\frac{D\dot{\gamma}}{ds}\right)\dot{\gamma}+ g\left(X,\dot{\gamma}\right)\frac{D\dot{\gamma}}{ds}, \end{align} where $\frac{D}{ds}$ denotes the covariant derivative along the path $\gamma$. Following this definition, Hawking and Ellis state that we can extend this to all tensor fields along $\gamma$ simply by requiring $\frac{D_{\text{Fermi}}}{ds}$ to commute with contractions, obey the Leibniz rule with respect to tensor products, and such that $\frac{D_{\text{Fermi}}f}{ds}=\frac{df}{ds}$ for smooth functions $f$.

My question is whether one can define an analogue of this in arbitrary vector bundles $(E,\pi,M)$, say equipped with a bundle metric $g$ (and say a corresponding metric-compatible connection on $E$)? I ask because for a general vector bundle, it is not possible to start with a curve $\gamma$ in the base manifold and lift it naturally to a curve in $E$. Next, I considered the following as a candidate definition:

Let $(E,\pi,M,g)$ be a vector bundle with a metric $g$ and a compatible connection. Let $\gamma:I\to M$ be a smooth curve and fix a lifting $\gamma_1:I\to E$, which is normalized (so $|g(\gamma_1,\gamma_1)|=1$ identically). For any smooth lifting $X:I\to E$ of $\gamma$ we define the Fermi derivative of $X$ with respect to $\gamma_1$ to be \begin{align} \frac{D_{\text{Fermi}}X}{ds}&=\frac{DX}{ds}-g\left(X,\frac{D\gamma_1}{ds}\right)\gamma_1+ g(X,\gamma_1)\frac{D\gamma_1}{ds}\tag{$*$} \end{align}

So, the curve $\gamma_1$ which I am fixing is playing the role of the velocity $\dot{\gamma}$. However, the problem I'm facing with this definition is that if I consider two vector bundles $(E_i,\pi_i,M,g_i)$ for $i=1,2$ and consider a single base curve $\gamma:I\to M$ and two fixed liftings $\gamma_i:I\to E_i$ and $X_i:I\to E_i$, then the Fermi derivative of $X_1\otimes X_2$ (a lifting of $\gamma$ to $E_1\otimes E_2$) with respect to $\gamma_1\otimes \gamma_2$ doesn't satisfy the Leibniz rule with respect to $\otimes$ anymore.

So, I'm wondering if there's a way to 'fix' the definition so that I have a general notion of Fermi-derivative on an arbitrary vector bundle from which I can prove that when specialized to the tangent bundle $TM$, and the corresponding tensor bundles, the Leibniz rule is obeyed. (I know that I can start with $(*)$ as a definition and then just force by hand for the Leibniz rule to hold, but I'm not a fan of such algebraic extensions, hence the question).

Answer 1

Rather than viewing the Fermi-Walker derivative as a specific differential operator (which is what is causing the issues with generalization and the dependence on the curves $\gamma_1,\gamma_2$), one can instead view it as the covariant derivative operator with respect to a modified linear connection, called the Fermi-Walker connection (see Sachs and Wu for this view-point); this is a linear connection on the pullback bundle $\gamma^*E\to I$ not on the original bundle $(E,\pi, M)$. Having defined this connection, we can thus extend it (by the usual theory of linear connections) to arbitrary tensor bundles/ Hom bundles/ direct/Whitney-sum bundles etc.

We can be slightly more general as follows. Let $E$ be a vector bundle over $M$, with bundle metric $g$ and a compatible connection $\nabla$. Suppose now that we have a smooth map $f:N\to M$. We can then consider the pullback bundle $(f^*E,\pi, N)$, and we can endow $f^*E$ with the induced metric from $g$ and the induced connection from $\nabla$ (which by abuse of notation we shall still denote as $g$ and $\nabla$). Now, suppose that for each $z\in N$ we have an orthogonal direct sum decomposition of the fiber $E_{f(z)}=V_{f(z)}\oplus W_{f(z)}$; let $P_z,Q_z$ be the projections onto $V_{f(z)}$ and $W_{f(z)}$ respectively, and assume that the maps $z\mapsto P_z$ and $z\mapsto Q_z$ are smooth (from $N$ into $\text{Hom}(E,E)$; and we can also shrink the target space to $\text{Hom}(f^*E,f^*E)$, so that $P,Q$ are sections of the endomorphism bundle $\text{End}(f^*E)$ over $N$). Then, we define a connection $\mathcal{D}$ on $f^*E$ by \begin{align} \mathcal{D}_{\xi_z}X&:= P_z\left(\nabla_{\xi_z}(PX)\right) + Q_z\left(\nabla_{\xi_z}(QX)\right), \end{align} for all sections $X$ of $f^*E$ and all $z\in N,\xi_z\in T_zN$. Strictly speaking, on the right we should write $f^*\nabla$. Now that we have a connection $\mathcal{D}$ on $f^*E$, we can induce, in a natural way, connections on the various tensor bundles.

Here are some basic properties of the connection $\mathcal{D}$. Suppose $\{v_1(z),\dots, v_k(z)\}$ is a smoothly-varying pointwise orthonormal basis for $V_{f(z)}$; let us write $\epsilon_i=g(v_i,v_i)\in \{-1,1\}$. Then, some tedious algebra shows that \begin{align} \mathcal{D}_{\xi}X= \nabla_{\xi}X &+\sum_{i=1}^k\epsilon_i\left[g(X,\nabla_{\xi}v_i)v_i-g(X,v_i)\nabla_{\xi}v_i\right] \\ &+2 \sum_{i<j}\epsilon_i\epsilon_jg(\nabla_{\xi}v_i,v_j)\left[g(X,v_i)v_j-g(X,v_j)v_i\right] \end{align} So, $\mathcal{D}$ is like $\nabla$ plus some "rotation terms", the first one being rotations in the plane $\nabla_{\xi}v_i\,\wedge v_i$ (think the plane spanned by velocity and their orthogonal accelerations), and the second term consists of the planes spanned by $v_i\wedge v_j$. So we can think of being parallel with respect to $\mathcal{D}$ as saying we "$\nabla$ should only rotate vectors as much as is permitted by the freedom available in the subspaces $V_{f(z)}$ as we vary $z$".

Some properties of the connection (mostly following by some straight-forward manipulations of the above formula) include:

• If $\nabla v_i=0$ for all $i$ then the connections $\mathcal{D}$ and $\nabla$ on $f^*E$ are identical.
• $\mathcal{D}$ is metric compatible.
• If $X$ is a section of $f^*E$ for which the projection $QX=0$ identically (and hence $PX=X$), then from the definition, $\mathcal{D}X= P(\nabla X)$.

In the case where $N=I$ is an interval, and we consider the tangent bundle $E=TM$, and a timelike curve $\gamma:I\to M$, we can take the subspaces $V_{\gamma(s)}=\text{span}\{\dot{\gamma}(s)\}$ and $W_{\gamma(s)}$ the orthogonal complement. Since $V_{\gamma(s)}$ is 1-dimensional, we have $k=1$ in the above formula, and $\epsilon_1=-1$ due to timelike-ness. Now, many of the terms will end up vanishing and we're left with \begin{align} \mathcal{D}_{\xi}X&=\nabla_{\xi}X-g\left(X,\nabla_{\xi}\dot{\gamma}\right)\dot{\gamma}+ g(X,\dot{\gamma})\nabla_{\xi}\dot{\gamma}, \end{align} and hence by taking $\xi=\frac{d}{ds}$ the standard vector field on the interval, we recover the Fermi-Walker derivative. In this sense, we can view the Fermi-Walker derivative as being precisely the covariant derivative along $\gamma$ with respect to a modified connection. Note also that in this specific 1-dimensional example, we have that

• $\dot{\gamma}$ is parallel with respect to the connection $\mathcal{D}$ (i.e $\mathcal{D}\dot{\gamma}=0$ identically, or equivalently, $\frac{\text{D}_{\text{Fermi}}\dot{\gamma}}{ds}=0$). This is because in the general formula above, we have $k=1$ so the second summation isn't there, and we have only a single term $g(\nabla_{\xi}\dot{\gamma},\dot{\gamma})$ which vanishes since $g(\dot{\gamma},\dot{\gamma})=-1$ is constant.

Now, the notion of Fermi-Walker transport is the same thing as parallel-transport with respect to the above defined Fermi-Walker connection $\mathcal{D}$ and so on. So, by the general theory of parallel-transport, we have existence and uniqueness of Fermi-Walker transport and so on.

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So, tldr: think of a modified connection, which plays nicely with respect to a parametrized family of orthogonal subspaces of $E$.