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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 $V$?

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  • $\begingroup$ Did you solve this question? I have the derivation in a notebook. Hint. Following Schuller's notation, the actual formula is $$\nabla _{T} S=\left[\phi,\operatorname{d}\!S(T)+[\omega^{u, \phi}(T)] \triangleright S\right]$$. I guess prof. Schuller was in a hurry to end the lecture. $\endgroup$ – Jackozee Hakkiuz Jun 28 '18 at 6:45
  • $\begingroup$ No I didn't. Yes he was in a hurry. A source of confusion for me is the action on $S$ from the left. If you follow it must be the identity element acting in the principal bundle. For a free action acting from the left this should yield the same zero form on the principal bundle. And then pulled back on the base manifold. Also don't quite get your hint. Are you suggesting that the co-variant lives as an equivalent class on the associated bundle. Will appreciate it if you share from your notebook as the answer and help me linking it to the equation for $V^{\nu}$. $\endgroup$ – user169903 Jun 28 '18 at 7:00
  • $\begingroup$ Actually, in my equation, the left hand side should be $\nabla_T\sigma$. Remember that the covariant derivative acts on sections of $TM$, regarded as an associated bundle to $LM$, so yes, the output of the covariant derivative is an equivalence class which can be mapped to $TM$ via the isomorphism introduced at the minute 22:22 of lecture 20. $\endgroup$ – Jackozee Hakkiuz Jun 28 '18 at 7:09
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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*}

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  • $\begingroup$ Thank you very much for your time. But are you sure this part in your item 3. is correct $\Gamma(T(LM))$? Shouldn't be $T(LM)$ $\endgroup$ – user169903 Jun 28 '18 at 14:47
  • $\begingroup$ Right. $D\phi_\sigma$ can be seen as a map $\Gamma(T(LM))\to C^{\infty}(LM,\mathbb{R}^{d})$, but in this case it is easier (and more appropiate) to see it as a map $T(LM)\to \mathbb{R}^{d}$ $\endgroup$ – Jackozee Hakkiuz Jun 28 '18 at 16:37
  • $\begingroup$ Something similar can be said about the pullback. I already edited the answer. Thanks :) $\endgroup$ – Jackozee Hakkiuz Jun 28 '18 at 16:39
  • $\begingroup$ By the way, is it necessary to define the map $u$ when we can directly construct the associated bundle simply as $LM_{TM}$ to begin with? $\endgroup$ – user169903 Jun 29 '18 at 5:50
  • $\begingroup$ Well... the definition would be $$LM_{TM} := (LM\times TM)_{\displaystyle/\sim_{GL(d,\mathbb{R})}'}$$ for some equivalence relation $\sim_{GL(d,\mathbb{R})}'$, so this is not immediately isomorphic to $TM$. I'm not even sure if it's possible, because you'd need to find a $\sim_{GL(d,\mathbb{R})}'$ on $LM\times TM$ that makes it isomorphic to $TM$ itself (i.e., kind of "factoring" the whole $LM$) but in such a way that it is still tied to $LM$. Also you need a $GL(d,\mathbb{R})$ left action on $TM$. You might want to rewatch lecture 20, I think the motivation is pretty clear. $\endgroup$ – Jackozee Hakkiuz Jun 29 '18 at 7:24
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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:

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

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

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

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

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  • $\begingroup$ I think you need to provide the left $GL(d,\mathbb{R})$ action $\triangleright$ that you put on $T_pM$ and then show that $$\forall_{e\in LM}:\forall_{v\in T_pM}:[e,v]\sim[e',v'] :\iff \exists_{g\in GL(d,\mathbb{R})}:(e'=e\triangleleft g)\land(v'=g^{-1}\triangleright v)$$ defines an equivalence relation such that your claim $$(LM\times T_{p}M)_{\displaystyle/\sim}\cong TM$$ holds. $\endgroup$ – Jackozee Hakkiuz Jun 30 '18 at 8:00
  • $\begingroup$ Yes. Of course. Thanks again $\endgroup$ – user169903 Jun 30 '18 at 9:55

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