How to recover the covariant derivative from the pull back from that on the principal bundle

Question

I am watching these lecture series by Fredric Schuller.

Covariant derivatives - Lec 25 - Frederic Schuller @minute 01:10:11

When we arrive at the covariant derivative from the principal bundle $P$ by pulling back to the base manifold $M$ we have:

$$\nabla _{T} S=dS(T)+\omega^{u, \phi} \triangleright S$$

where $S:u \to F$, $F$ any finite dimensional vector space on an open subset $u$ on $M$, $\phi: u \to P$, is the section on the principal G-bundle, $T \in T_{x}M$, is a tangent vector at point $x$ in the base manifold $M$, and $\omega^{u, \phi}$ is the Lie algebra valued one-form on the principal bundle .

Now my question is that how to recover, if possible step by step, the more familiar covariant derivative for e.g. a vector $V$ on the base manifold which is written as:

$$\nabla_{\mu}V^{\nu}=\partial _{\mu}V^{\nu}+\Gamma_{\mu \lambda}^{\nu}V^{\lambda}$$

I am a bit confused when I plug a vector $V$ instead of $S$ in the general equation above and how should I put the exterior derivative of this one-form vector-valued object and operate it on another vector $T$ and also how to operate a connection one form pulled back to the base manifold on the vector $V$?

Answer 1

Okay. I'll recap Schuller's notation and conventions.

$M$ a $d$-dimensional smooth manifold.

From lecture 20.

We regard the frame bundle $LM\to M$ as a $GL(d,\mathbb{R})$-principal bundle, and we construct the associated bundle $LM_{\mathbb{R}^{d}}\to M$ by $$LM_{\mathbb{R}^{d}} := (LM\times\mathbb{R}^{d})_{\displaystyle/\sim_{GL(d,\mathbb{R})}}$$ where $[e,f]\sim[e',f'] :\iff \exists_{g\in GL(d,\mathbb{R})}:(e'=e\triangleleft g)\land(f'=g^{-1}\triangleright f)$.

Then we know $LM_{\mathbb{R}^{d}}\to M$ is isomorphic as a bundle to $TM\to M$ via the map $$u:LM_{\mathbb{R}^{d}}\to TM \\ [e,f]\mapsto e_{i} f^{i}$$

We have a section $\sigma\in\Gamma(TM)$. We know for such a section there exists a G-equivariant function $\phi_\sigma:LM\to\mathbb{R}^{n}$

From lecture 25

We know that there is a bijective correspondence between sections $$\sigma:M\to LM_{\mathbb{R}^{d}}$$ and $GL(d,\mathbb{R})$-equivariant functions $$\phi:LM\to\mathbb{R}^{d}$$

On $LM$ we also have the action of the covariant exterior derivative $D$ whose action on $\mathbb{R}^{d}$-valued functions is given (for every $X\in T(LM)$) by $$D\phi(X) = \operatorname{d}\!\phi(X) + \omega(X)\triangleright\phi$$

Then

We want to have a covariant derivative $\nabla^{TM}$ which takes a $T\in T_pM$ and a $\Sigma\in\Gamma(TM)$ and gives us $\nabla_T^{TM}\Sigma \in T_pM$.
Using the isomorphism $u$ given above, we can equivalently construct a covariant derivative $\nabla$ which takes a $T\in T_pM$ and a $\eta\in\Gamma(LM_{\mathbb{R}^{d}})$ and gives us $\nabla_T \eta\in L_pM_{\mathbb{R}^{d}}$ (i.e., the ouptput is a point the fibre of $LM_{\mathbb{R}^{d}}$ over $p$), and then we can define the covariant derivative on $TM$ to be $$\nabla^{TM}_{T}\Sigma = u\left(\nabla_{T}\sigma\right)$$ where $\sigma:M\to LM_{\mathbb{R}^{d}}$ is the section given at every $p\in M$ by $\sigma(p) = u^{-1}(\Sigma(p))$

Okay, so what we do is:

1. We take our section $\sigma:M\to LM_{\mathbb{R}^{d}}$
2. We obtain the $GL(d,\mathbb{R})$-equivariant function $\phi_\sigma:LM\to\mathbb{R}^{d}$ that represents our section $\sigma$ in the frame bundle.
3. We apply to it the covariant exterior derivative and get the $\mathbb{R}^{d}$-valued 1-form $D\phi_\sigma:T(LM)\to \mathbb{R}^{d}$ on $LM$.
4. We pick an arbitrary local section $\varphi:U\to LM$, with $p\in U$ (hence $T\in T_pU$).
5. We pullback $D\phi_\sigma$ along $\varphi$, to get the $\mathbb{R}^{d}$-valued $1$-form $\varphi^{*}D\phi_\sigma:TU\to\mathbb{R}^{d}$ on $U$.
6. We apply $\varphi^{*}D\phi_\sigma$ to $T\in T_pU$ and get the $d$-tuple of real numbers $\left(\varphi^{*}D\phi_\sigma\right)\!(T)\in\mathbb{R}^{d}$
7. Finally, we define $\nabla_T\sigma$ to be exactly the equivalence class $\left[\varphi(p),\left(\varphi^{*}D\phi_\sigma\right)\!(T)\right]\in LM_{\mathbb{R}^{d}}$.

If we expand the covariant exterior derivative we get \begin{align*} \nabla_T\sigma &=\left[\varphi(p),\left(\varphi^{*}D\phi_\sigma\right)\!(T)\right]\\ &=\left[\varphi(p),\left(\varphi^{*}(\operatorname{d}\!\phi_\sigma + \omega\triangleright \phi_\sigma)\right)\!(T)\right]\\ &=\left[\varphi(p),\left(\varphi^{*}\operatorname{d}\!\phi_\sigma + \varphi^{*}(\omega\triangleright\phi_\sigma)\right)\!(T)\right]\\ &=\left[\varphi(p),\left(\operatorname{d}(\varphi^{*}\phi_\sigma) + (\varphi^{*}\omega)\triangleright(\varphi^{*}\phi_\sigma)\right)\!(T)\right] \end{align*} And finally, defining $\omega^{U,\varphi}:=\varphi^{*}\omega$ (the Yang-Mills field / local representation of the connection) and $S:=\varphi^{*}\phi_\sigma$ (an $\mathbb{R}^{d}$-valued function on $U$ which we interpret as the components of $\Sigma$ under the section $\varphi$), we obtain the formula

$$\nabla_T\sigma=\left[\varphi(p),\left(\operatorname{d}\!S + \omega^{U,\varphi}\triangleright S\right)\!(T)\right]\in L_pM_{\mathbb{R}^{d}}$$

Now, to get a tangent vector (an element of $TM$) we need to apply the map $u$ \begin{align*} \nabla_T\Sigma &= u\left(\nabla_T\sigma\right)\\ &=u\left(\left[\varphi(p),\left(\operatorname{d}\!S + \omega^{U,\varphi}\triangleright S\right)\!(T)\right]\right)\\ &=\varphi(p)_{i}\left(\operatorname{d}\!S + \omega^{U,\varphi}\triangleright S\right)\!(T)^{i}\\ &=\varphi(p)_{i}\left(\operatorname{d}\!S(T) + \omega^{U,\varphi}(T)\triangleright S\right)^{i} &\in T_pM \end{align*}

Sorry for the loooong introduction, I wanted to justify the formula I wrote in the comments. Now we can finally get to the answer.

How to get from here to the "normal" expression?

Easy. We choose a local chart $(U,x)$ on $M$ and we define our section $\varphi$ to be exactly the holonomic frame at every point w.r.t. $x$. I.e.: $$\varphi(p) := \left(\left(\frac{\partial}{\partial x^1}\right)_{p},\dots,\left(\frac{\partial}{\partial x^n}\right)_{p}\right) \in L_pM \subset LM$$

Then, for a section $V:M\to TM$ (a vector field) we obtain a section $v:M\to LM_{\mathbb{R}^{d}}$ (roughly, $v(p)$ gives us all possible decompositions of $V$ with respect to all possible bases at $p$) which we turn into a $GL(d,\mathbb{R})$-equivariant function $\phi_v:LM\to \mathbb{R}^{d}$ (given a frame $e\in LM$, $\phi_v(e)$ is the decomposition of $V(p)$ with respect to $e$), and finally we pull it back using $\varphi$ (this final object $\varphi^{*}\phi_v$, is an $\mathbb{R}^{d}$ valued function that when evaluated at $p\in M$, gives us the $d$ components of $V(p)$ with respect the basis $\left(\frac{\partial}{\partial x^{i}}\right)_p$. If we denote $\varphi^{*}\phi_v$ by $\bar{V}$ (so we can distinguish the section $V:M\to TM$ from its component functions $\bar{V}:M\to\mathbb{R}^{d}$), and denote $\omega^{U,\varphi}$ by $\Gamma$, we finally get the formula

\begin{align*} \nabla^{TM}_TV &=\varphi(p)_{i}\left(\operatorname{d}\!\bar{V}(T) + \Gamma(T)\triangleright \bar{V}\right)^{i}\\ &=\left(\frac{\partial}{\partial x^{i}}\right)_p\left(\operatorname{d}\!\bar{V}(T)^{i} + (\Gamma(T)\triangleright \bar{V})^{i}\right)\\ &=\left(\frac{\partial}{\partial x^{i}}\right)_p\left(\frac{\partial}{\partial x^{j}}\bar{V}^{i}\operatorname{d}\!x^{j}(T) + \left(\Gamma_{k}\operatorname{d}\!x^{k}(T)\triangleright \bar{V}\right)^{i}\right)\\ &=\left(\frac{\partial}{\partial x^{i}}\right)_p\left(\frac{\partial \bar{V}^{i}}{\partial x^{j}}T^{j} + \left(\Gamma_{k}T^{k}\triangleright \bar{V}\right)^{i}\right)\\ &=\left(\frac{\partial}{\partial x^{i}}\right)_p\left(\frac{\partial \bar{V}^{i}}{\partial x^{j}}T^{j} + {\Gamma^{i}}_{jk}T^{k} \bar{V}^{j}\right) &\in T_pM \end{align*}

Answer 2

Consider the associated bundle as $$LM_{T_{P}M} := (LM\times T_{p}M)/_{\sim}=TM$$

Now suppose that $$S=V: U \to T_{p}M$$ and $$\nabla_{T}V:T_{P}M \to T_{p}M$$

For $T \in T_{P}M$, $\sigma \in \Gamma(TM)$, $\phi_{\sigma}:LM \to T_{P}M$, and $\varphi:M \to LM$ we have on $M$:

$$\nabla_{T}V:=\varphi^{\star}D\phi_\sigma(T)=\operatorname{d}V(T)+\Gamma(T).V \in T_{p}M$$

Now choose a basis $\partial_{\mu}$ on $T_{p}M$

$$\nabla_{T}V=(\operatorname{d}V(T)+\Gamma(T).V)^{\mu}\partial_{\mu}$$

$$\nabla_{T}V=((\operatorname{d}V^{\mu}(T)+(\Gamma(T).V))^{\mu})\partial_{\mu}$$

$$\nabla_{T}V=(T(V^{\mu})+(\Gamma_{\nu}\operatorname{d}x^{\nu}(T^\alpha\partial_{\alpha}).V))^{\mu})\partial_{\mu}$$

$$\nabla_{T}V=(T^{\nu}\partial_{\nu}(V^{\mu})+(\Gamma_{\nu}T^{\nu}.V))^{\mu})\partial_{\mu}$$

$$\nabla_{T}V=T^{\nu}(\partial_{\nu}(V^{\mu})+(\Gamma_{\nu\lambda}^\mu V^{\lambda}))\partial_{\mu}$$

Now apply to $T=\partial_{\nu} \rightarrow T^{\nu}=1$

$$\nabla_{\partial_{\nu}}V=(\partial_{\nu}(V^{\mu})+(\Gamma_{\nu\lambda}^\mu V^{\lambda}))\partial_{\mu}$$

Then we have:

$$(\nabla_{\partial_{\nu}}V)^{\mu}=\partial_{\nu}V^{\mu}+\Gamma_{\nu\lambda}^\mu V^{\lambda}$$

$$\nabla_{\partial_{\nu}}V^{\mu}=\partial_{\nu}V^{\mu}+\Gamma_{\nu\lambda}^\mu V^{\lambda}$$

$$\nabla_{\nu}V^{\mu}:=\nabla_{\partial_{\nu}}V^{\mu}$$

$$\nabla_{\mu}V^{\nu}=\partial_{\mu}V^{\nu}+\Gamma_{\mu\lambda}^\nu V^{\lambda}$$