Is My Understanding of Covariant Derivative Right?

Question

Let $E\overset{\pi}{\twoheadrightarrow}M$ be a fiber bundle, and $\sigma$ be a smooth local section, i.e. $\pi\circ\sigma=\mathrm{id}$. For $\forall X\in T_{x}M$, if $E$ is trivial, then one can talk about the directional derivative $d\sigma(x)(X)$, where $d\sigma(x)$ is a tangent map $T_{x}M\rightarrow T_{\sigma(x)}E$. This is because there's an obvious way to parallelly transport each element in $E$.

For example, let $\gamma(t)$ is a curve in $M$, such that $\gamma(0)=x$, and $\gamma^{\prime}(0)=X$, then, since $E\overset{\pi}{\twoheadrightarrow}M$ is trivial, one has $$d\sigma(x)(X)=\frac{d}{dt}\Bigg|_{t=0}\sigma(\gamma(t))=\lim_{t\rightarrow 0}\frac{\sigma(\gamma(t))-\sigma(\gamma(0))}{t},$$

where $\sigma(\gamma(t))\in E_{\gamma(t)}$. It makes sense because $E_{\gamma(t)}$ is isomorphic to $E_{\gamma(0)}$ in a canonical way.

In general, if $E$ is none trivial, then one has to use the covariant differential $\nabla\sigma(x)$ to define the tangent map. The covariant derivative is usually defined by introducing the parallel transport:

For each curve $\gamma(t)$ in manifold $M$, the collections of diffeomorphisms $$\Gamma(\gamma)_{s}^{t}: E_{\gamma(s)}\rightarrow E_{\gamma(t)}$$ such that \begin{align} &1.\,\,\,\Gamma(\gamma)_{s}^{s}=\mathrm{Id}_{E_{\gamma(s)}} \\ &2.\,\,\,\Gamma(\gamma)_{\epsilon}^{t}\circ\Gamma(\gamma)_{s}^{\epsilon}=\Gamma(\gamma)_{s}^{t} \\ &3.\,\,\,\Gamma(\gamma)_{s}^{t}\,\,\,\mathrm{depends\,\,on\,\,\gamma,\,\,s,\,\,\mathrm{and}\,\,t\,\,\mathrm{smoothly}.} \end{align}

Then, for a given curve $\gamma(t)$ in M, such that $\gamma(t)=x$, and $\gamma^{\prime}(t)=X$, one defines the covariant derivative $$\nabla\sigma(x)(X)=\nabla\sigma(x)(\gamma^{\prime}(t))\equiv\nabla_{X}\sigma(x)\equiv\frac{d}{d\epsilon}\Bigg|_{\epsilon=0}\Gamma(\gamma)_{t+\epsilon}^{t}\circ\sigma(\gamma(t+\epsilon)).$$

What confused me a lot recently is that under the diffeomorphism $\Gamma(\gamma)_{t+\epsilon}^{t}: E_{\gamma(t+\epsilon)}\rightarrow E_{\gamma(t)}$, the section $\sigma(\gamma(t+\epsilon))\in E_{\gamma(t+\epsilon)}$ is mapped to another section $\varsigma_{\epsilon}(\gamma(t))\in E_{\gamma(t)}$. For convenience, I denote $$\frac{d}{d\epsilon}\Bigg|_{\epsilon=0}\varsigma_{\epsilon}(\gamma(t))=\xi(t)\in T_{\varsigma_{\epsilon}(\gamma(t))}E.$$

Then, from the canonical projection $E\overset{\pi}{\twoheadrightarrow}M$, which locally gives $$\pi\circ\sigma=\pi\circ\varsigma_{\epsilon}=\mathrm{id},$$

one has $$d\pi(\xi(t))=\frac{d}{d\epsilon}\Bigg|_{\epsilon=0}\pi(\varsigma_{\epsilon}(\gamma(t)))=\frac{d}{d\epsilon}\Bigg|_{\epsilon=0}\gamma(t)=0.$$

In other words, it seems to me that the push forward $d\pi(\nabla\sigma(x)(X))=0$.

Or, if one views the covariant differential $\nabla\sigma(x)$ as a tangent map $T_{x}M\rightarrow T_{\sigma(x)}E$, then the above calculation really showed that the covariant derivative actually maps the tangent vector $X\in T_{x}M$ to the vertical subspace $V_{\sigma(x)}E$. i.e. $$\nabla\sigma(x): T_{x}M\rightarrow V_{\sigma(x)}E.$$ Is that correct?

Answer 1

I found the same claim from several sources. For example, in this lecture notes, Chris Wendl also claimed that the covariant derivative is the vertical part of the tangent map. So I believe that my understanding of the covariant derivative is correct.

In the following, I will prove the result for a vector bundle from a different perspective. From this Wikipedia page, given a connection $\nabla$ on a vector bundle $E$ over $M$, and $\gamma(t)$ a smooth curve in $M$, a section $\sigma$ of $E$ is called parallel if $$\nabla_{\dot{\gamma}(t)}\sigma=0. \tag{0}$$

This is supposed to be equivalent to the definition of $\Gamma(\gamma)_{s}^{t}$ of parallel transport in a vector bundle. I will use this equivalence to prove that for a generic section, its covariant derivative is indeed vertical.

Starting from equation (0), locally, in an neighborhood $U\subset M$ with local coordinates $\left\{q^{\mu}\right\}$, the parallel section $\sigma$ can be expressed as $\sigma|_{U}=\sigma^{i}e_{i}$, where $\left\{e_{i}\right\}_{i=1,\cdots,m}$ is a local frame. For convenience, I will denote $\sigma^{i}(\gamma(t))\equiv\sigma(t)^{i}$, and $q^{\mu}(\gamma(t))=q^{\mu}(t)$. Denoting the connection $1$-form of the vector bundle $E$ by $A_{i}^{j}=A_{i\mu}^{j}dq^{\mu}$. Then, one has $$\nabla e_{i}=A_{i}^{j}e_{j}.$$

Using the Leibniz rule, one has $$\nabla_{\dot{\gamma}(t)}\sigma=\nabla_{\dot{\gamma}(t)}(\sigma^{i}e_{i})=\left(\frac{d\sigma^{i}}{dt}+\sigma^{j}\frac{dq^{\mu}}{dt}A_{j\mu}^{i}\right)e_{i}. \tag{1}$$

So the condition that the section $\sigma$ is parallel implies that the differential equation $$\frac{d\sigma^{i}}{dt}+\sigma^{j}\frac{dq^{\mu}}{dt}A_{j\mu}^{i}=0 \tag{2}$$

has a solution, which is certainly true according to the theory of ordinary differential equations.

Lemma: Let $\nabla$ be a connection on the vector bundle $E\overset{\pi}{\twoheadrightarrow}M$, with the vector space $\mathbb{V}$ its typical fiber. Let $x\in M$, and $\sigma\in\pi^{-1}(x)$. Then, the horizontal subspace $H_{\sigma}E$ is isomorphic to $T_{x}M$ under the push forward $d\pi_{\sigma}$.
proof: Let $\left\{q^{\mu}\right\}$ be the local coordinates of $x\in U\subset M$. The vector bundle $E|_{U}$ has local trivialization $\varphi: U\times\mathbb{V}\rightarrow\pi^{-1}(U)$. Denote $\left\{f_{i}\right\}_{i=1,\cdots,m}$ as the basis of the vector space $\mathbb{V}$, and $e_{i}\equiv\varphi(x,f_{i})$. Accordingly, $\left\{\frac{\partial}{\partial q^{\mu}}\right\}$ is a local frame on $TM|_{U}$, and $\left\{e_{i}\right\}$ is a local frame of the vector bundle $E|_{U}$. Then, any local section $\sigma|_{U}$ can be written as $\sigma=\sigma^{i}e_{i}$. For any point $\sigma\in\pi^{-1}(x)$, there's a local coordinate $\left\{\tilde{q}^{\mu},p^{i}\right\}$ of $\sigma\in\pi^{-1}(U)$ such that \begin{equation} \begin{cases} \tilde{q}^{\mu}(\sigma):=q^{\mu}(x) \\ p^{i}(\sigma):=\sigma^{i}(x). \end{cases} \end{equation} Denote the frame of $TE|_{U}$ by $\left\{X_{\mu},Y_{i}\right\}$, i.e. $$X_{\mu}=\frac{\partial}{\partial\tilde{q}^{\mu}},\quad Y_{i}=\frac{\partial}{\partial p^{i}}.$$ Let $\gamma(t)$ be an arbitrary smooth curve in $M$, such that $\gamma(0)=x$. Then, one can consider its horizontal lift $\tilde{\gamma}(t)$ in $E$ (i.e. it is parallel along $\gamma(t)$), such that $\tilde{\gamma}(0)=\sigma$. Using the above coordinate, one can express the parallel curve $\tilde{\gamma}(t)$ as $$\left\{\tilde{q}^{\mu}(t),p^{i}(t)\right\},$$ where $\tilde{q}^{\mu}(t)\equiv q^{\mu}(\gamma(t))=q^{\mu}(t)$, and $p^{i}(t)\equiv\sigma^{i}(\gamma(t))=\sigma^{i}(t)$. Then, by definition, it has to satisfy the equation $$\frac{dp^{i}}{dt}+p^{j}\frac{d\tilde{q}^{\mu}}{dt}A_{j\mu}^{i}=0.$$ Then, the tangent vector $\tilde{\gamma}^{\prime}(0)\in T_{\sigma}E$ is \begin{align} \tilde{\gamma}^{\prime}(0)&=\frac{d\tilde{q}^{\mu}}{dt}\frac{\partial}{\partial\tilde{q}^{\mu}}\Bigg|_{\sigma}+\frac{dp^{i}}{dt}\frac{\partial}{\partial p^{i}}\Bigg|_{\sigma} \\ &=\frac{d\tilde{q}^{\mu}}{dt}\left(X_{\mu}(\sigma)-p^{j}(\sigma)A^{i}_{j\mu}(x)Y_{i}(\sigma)\right). \tag{3} \end{align} Comparing the above vector with $$\gamma^{\prime}(0)=\frac{dq^{\mu}}{dt}\frac{\partial}{\partial q^{\mu}}\Bigg|_{x}\in T_{x}M,$$ one finds that the horizontal lift of the tangent vector $\gamma^{\prime}(0)$ is given by (3). Under the map $$\frac{\partial}{\partial q^{\mu}}\Bigg|_{x}\longleftrightarrow Z_{\mu}(\sigma)\equiv X_{\mu}(\sigma)-p^{j}(\sigma)A^{i}_{j\mu}(x)Y_{i}(\sigma),$$ there is an isomorphism between $T_{x}M$ and $H_{\sigma}E$. Under the projection $\pi_{\ast}|_{\sigma}$, $Z_{\mu}(\sigma)$ is mapped to $\frac{\partial}{\partial q^{\mu}}\Bigg|_{x}$.

The above statement implies that the tangent space $T_{\sigma}E$ has a decomposition

1. $T_{\sigma}E=H_{\sigma}E\oplus V_{\sigma}E,$
2. $d\pi_{\sigma}:H_{\sigma}\rightarrow T_{\pi(\sigma)}M$ is an isomorphism.

Then, finally, one has the following theorem:

Theorem: Let $E\overset{\pi}{\twoheadrightarrow}M$ be a vector bundle, with connection $\nabla$ and the connection $1$-form $A^{i}_{j}$. Let $\gamma:(-\epsilon,\epsilon)\rightarrow M$ be a smooth curve, and $\sigma\in\Gamma(E)$ be any section of $E$. Then, $$\nabla_{\gamma^{\prime}(0)}\sigma=\Pi_{\mathrm{v}}(\sigma^{\prime}(0)),$$ where $\Pi_{\mathrm{v}}$ is the projection on the vertical subspace under the decomposition $T_{\sigma}E=H_{\sigma}E\oplus V_{\sigma}E$.
proof: The section $\sigma\in\Gamma(E)$ along the curve $\gamma(t)\subset M$ is also a curve in the total space $E$. Denote this curve $\sigma|_{\gamma(t)}$ by $\tilde{\gamma}(t)$, then using local coordinates, one has $$\tilde{\gamma}^{\prime}(t)=\frac{d\tilde{q}^{\mu}}{dt}\frac{\partial}{\partial\tilde{q}^{\mu}}\Bigg|_{\sigma(t)}+\frac{dp^{i}}{dt}\frac{\partial}{\partial p^{i}}\Bigg|_{\sigma(t)},$$ where $\sigma(t)\equiv\sigma(\gamma(t))$. Then, using the decomposition $T_{\sigma}E=H_{\sigma}E\oplus V_{\sigma}E$, one has $$\tilde{\gamma}^{\prime}(t)=\frac{d\tilde{q}^{\mu}}{dt}\left(\frac{\partial}{\partial\tilde{q}^{\mu}}\Bigg|_{\sigma(t)}-p^{j}(t)A^{i}_{j\mu}(\gamma(t))\frac{\partial}{\partial p^{i}}\Bigg|_{\sigma(t)}\right)+\left(\frac{dp^{i}}{dt}+p^{j}(t)\frac{d\tilde{q}^{\mu}}{dt}A^{i}_{j\mu}(\gamma(t))\right)\frac{\partial}{\partial p^{i}}\Bigg|_{\sigma(t)}.$$ Then, one has $$\Pi_{\mathrm{v}}(\tilde{\gamma}^{\prime}(t))=\left(\frac{dp^{i}}{dt}+p^{j}(t)\frac{d\tilde{q}^{\mu}}{dt}A^{i}_{j\mu}(\gamma(t))\right)\frac{\partial}{\partial p^{i}}\Bigg|_{\sigma(t)}.$$ Then, using equation (1), one has $$\Pi_{\mathrm{v}}(\tilde{\gamma}^{\prime}(t))=\nabla_{\gamma^{\prime}(t)}\sigma(\gamma(t)), \tag{4}$$ under the natural identification $$\frac{\partial}{\partial p^{i}}\longleftrightarrow e_{i},\quad\mathrm{or}\quad T_{\sigma(t)}(\pi^{-1}(\gamma(t)))\simeq V_{\sigma(t)}E.$$

The above theorem implies that the covariant differential as a tangent map is indeed vertical on a vector bundle. It acts on a section by taking ordinary differential and then eating up the horizontal component. Unsurprisingly, if the section is parallel (i.e. horizontal), its covariant derivative vanishes.