# Covariant derivative on associated vector bundle under change of section

Let $$(P,\pi,M;G)$$ be a principal bundle with connection form $$A\in\mathcal{C}(P)$$ and let $$\rho:G\rightarrow\mathrm{GL}(V)$$ be a representation of $$G$$ on some finite-dimensional vector space $$V$$. From these data we can construct an associated vector bundle $$E=P\times_\rho V$$ with typical fibre $$V$$. Using the parallel transport induced by $$A$$ on $$E$$ we can define a covariant derivative $$\nabla^A:\Gamma(E)\longrightarrow\Omega^1(M,E)$$ on $$E$$. There is also another (equivalent) way to introduce this covariant derivative: start with equation (1) (derived in the following) and show its invariance under a change of local section. $$\textbf{This is where I am stuck}$$. See below for details.

Let $$s:U\rightarrow P$$ be a local section of $$P$$ and $$\Phi:U\rightarrow E$$ a local section of $$E$$. Then we can find a smooth map $$\phi:U\rightarrow V$$ such that $$\Phi(x)=[s(x),\phi(x)]$$ on $$U$$. We can also pull back the connection $$A$$ to $$A_s=s^\ast A\in\Omega^1(U,\mathfrak{g})$$. The covariant derivative of $$\Phi$$ can then be written as

$$$$(\nabla^A_X\Phi)(x)=[s(x),\mathrm{d}\phi(X(x))+\rho_\ast(A_s(X(x))\phi(x)], \qquad\qquad\qquad\qquad (1)$$$$ where $$X\in\mathfrak{X}(U)$$ is a vector field.

There is another way to introduce the covariant derivative which is maybe more familiar to physicists: we start with the local formula (1) and show its covariance under a change of the section $$s$$. But this calculation is where I am stuck. Let $$s':U'\rightarrow P$$ be a nother section such that $$U\cap U'\neq\emptyset$$. Then there is a transition function $$g:U\cap U'\rightarrow G$$ of $$P$$ such that $$s=s'\cdot g$$. We also find another smooth map $$\phi':U'\rightarrow V$$ such that $$\Phi=[s',\phi']$$. By the definition of $$E$$ it then follows that $$[s,\phi]=[s',\phi']$$ if and only if $$\phi=\rho(g)^{-1}\phi'$$ on $$U\cap U'$$. We calculate

$$(\nabla^{A_s}_X\phi)(x)=\mathrm{d}(\rho(g(x))^{-1}\phi')(X(x)) +\rho_\ast\left(\mathrm{Ad}(g(x)^{-1})A_{s'}(X(x))+g^\ast\mu_G(X(x))\right)\rho(g(x))^{-1}\phi'(x)$$

I assume this has to be equal to $$\rho(g(x))^{-1}(\nabla^{A_{s'}}_X\phi')(x)$$. For the first differential I get

$$\mathrm{d}(\rho(g(x))^{-1}\phi')(X(x))=\rho(g(x))^{-1}\mathrm{d}\phi'(X(x))-\rho_\ast\left(g^\ast\mu_G(X(x))\right)\rho(g(x))^{-1}\phi'(x),$$ so that the second term cancels the last term in the equation before. The thing that bothers me is the adjoint representation in the remaining argument of $$\rho_\ast$$. How do I get rid of that? Or is my calculation of the differential wrong?

• We should have more questions on here at this level, so congrats for asking. But it's really hard to find your question. Would you consider editing it so that the question appears somewhere in the first paragraph? I gather you're deriving the same object two ways and trying to rectify the difference. Jun 13 at 12:09
• Is there any way to check on a simple example? Jun 13 at 12:10
• @MatthewLeingang Thanks for the suggestions! The examples I am familiar with and used for an example calculation were for abelian $G$. The adjoint representation doesn't play a role there... Jun 13 at 12:19
• What is $\mu_G$ in your formula? Jun 19 at 15:34
• @ahersh23 Its the Maurer-Cartan form of $G$. Jun 19 at 16:51

With the help of @ahersh23's answer I think I got it now:

It is $$\mathrm{d}\left(\rho(g(x))^{-1}\phi'\right)(X(x)) =\rho(g(x))^{-1}\mathrm{d}\phi'(X(x)) -\rho_\ast\left((g^\ast\mu_G)(X(x))\right)\rho(g(x))^{-1}\phi'(x),$$ so that we find $$\left(\nabla^{A_{s'}}_X\phi'\right)(x)=\rho(g(x))^{-1}\mathrm{d}\phi'(X(x))+\rho_\ast\left(\mathrm{Ad}\left(g(x)^{-1}\right)A_{s'}(X(x))\right)\rho(g(x))^{-1}\phi'(x).$$ Now the adjoint representation of $$G$$ is defined by $$\mathrm{Ad}:G\rightarrow\mathrm{GL}(\mathfrak{g}),\,g\mapsto\left(L_g\circ R_{g^{-1}}\right)_\ast.$$ With this we find for $$Y\in\mathfrak{g}$$ and $$h\in G$$ $$\rho_\ast\left(\mathrm{Ad}(h^{-1})Y\right) =\left(\rho\circ L_{h^{-1}}\circ R_h\right)_\ast(Y) =\left(\rho(h^{-1})\rho\circ R_h\right)_\ast(Y) =\rho(h^{-1})\left(\rho\circ R_h\right)_\ast(Y) =\rho(h^{-1})\rho_\ast(Y)\rho(h),$$ where the second equal sign follows from the fact that $$(\rho\circ L_{h^{-1}})(g)=\rho(h^{-1}g)=\rho(h^{-1})\rho(g)$$ and the last equal sign follows analogously.

I'm still new to this topic, and I'm having a little trouble figuring out everything you're doing in the second half of your question, but it may be helpful to look at how the connection coefficients transform under a change of section.

Given a trivialization, we can write our connections as follows:

$$\omega=g^{-1}dg+g^{-1}A_\mu dx^\mu g$$

where $$g^{-1}dg$$ is the maurer cartan form, and $$A_\mu dx^\mu$$ is a lie algebra valued one form on the base. We can think of $$A$$ as the pullback of $$\omega$$ with respect to a section corresponding to the trivialization:

$$A=s^*\omega$$

We can go from the trivialization corresponding to the section $$s$$ to a trivialization corresponding to a different section $$s'$$ by multiplying the section $$s$$ by some lie group valued function on the base $$h(x)$$:

$$s'=s(x)h(x)$$

The trivialization corresponding to $$s'$$ is then obtained by noting that:

$$(x,s'(x)g)=(x,s(x)h(x)g)$$

So the change of trivialization is given by $$g\rightarrow h(x)g$$. Under this transformation, $$\omega$$ becomes:

$$\omega^h=(h(x)g)^{-1}d(h(x)(g))+(h(x)g)^{-1}A_udx^u(h(x)g)$$

$$\omega^h=g^{-1}dg+g^{-1}h^{-1}(x)dh(x)g+g^{-1}h^{-1}(x)A_udx^uh(x)g$$

Thus we obtain:

$$A_udx^u\longmapsto h^{-1}dh(x)+h^{-1}A_udx^uh(x)$$

The Lie algebra valued one form $$A$$ are (as far as I have understood) the connection coefficients in the associated vector bundle, so under a change of section, which amounts to a change of trivialization, I believe that you would expect a term that looks like $$\text{Ad}_{h^{-1}}(A)$$. My apologies if this was unhelpful, or if I misunderstood what you are getting at.

Edit: You seem to be following Hamiltons Gauge Theory text, Im going to go look at the connections chapter and see if I can update this answer more.

Edit: I see you have already figured this out. For the sake of posterity here is my full solution to your question, sorry it took me so long to get around to this.

Fixing a connection $$A$$ on $$P$$ we have that:

$$\nabla^A_X\Phi=[s,d\phi(X_x)+\rho_*(A_s(X_x))\phi]$$

where $$\phi$$ is a map $$U\rightarrow V$$ and $$A_s=s^*A$$. For some map $$g:U\rightarrow G$$, let $$s'=s\cdot g$$ and $$\phi'=\rho(g^{-1})\phi$$, under this change of section we have:

$$A'=\text{Ad}_{g^{-1}}\circ A+g^*\mu_G$$

We have that:

$$d\phi'(X_x)=\rho(g^{-1})d\phi+\left(D_x\rho(g^{-1})(X_x)\right)\phi$$

and that:

$$\rho_*(A'(X_x))\phi'=\rho_*(\text{Ad}_{g^{-1}}A_s(X_x))\rho(g^{-1})\phi+\rho_*(\mu_G(D_xg(X_x)))\rho(g^{-1})\phi$$ Note that by chain rule:

$$D_x\rho(g^{-1})(X_x)=D_{g^{-1}}\rho\circ D_gi\circ D_x g(X_x)$$ where $$i$$ denotes the inversion map. Denote $$D_x g(X_x)$$ by $$\tilde{Z}\in T_gG$$, we then see that for any $$g\in G$$:

$$D_gi(\tilde{Z})= \lim_{t\rightarrow 0}\frac{d}{dt}g\exp(\mu_G(\tilde{Z})t)^{-1}$$ $$=\lim_{t\rightarrow 0}\frac{d}{dt}\exp(-\mu_G(\tilde{Z})t)g^{-1}$$ $$=-\mu_G(\tilde{Z})g^{-1}$$ Thus we obtain: $$D_x\rho(g^{-1})(X_x)=-D_g^{-1}\rho\left(\mu_G(\tilde{Z})g^{-1}\right)$$ $$=-D_{g^{-1}}\rho\circ D_e R_{g^{-1}}\left(\mu_G(\tilde{Z})\right)$$ $$=-D_{\rho(e)}R_{\rho(g^{-1})}\circ D_e\rho\left(\mu_G(\tilde{Z})\right)$$ $$=-\left(D_e\rho(\mu_G(\tilde{Z})\right)\rho(g^{-1})$$ $$=-\rho_*(\mu_G(D_xg(X_x)))\rho(g^{-1})$$

Therefore we have that:

$$d\phi'=\rho(g^{-1})d\phi-\rho_*(\mu_G(D_xg(X_x)))\rho(g^{-1})\phi$$

Now examine the following:

$$\rho_*(\text{Ad}_{g^{-1}}\circ A_s(X_x))=\rho_*(C_{g^{-1}*}\circ A_s(X_x))$$

$$=\rho_*\circ C_{g^{-1}*}\circ A_s(X_x)$$

$$=\left(\rho\circ C_{g^{-1}}\right)_*\circ A_s(X_x)$$ $$=\left(\rho\circ R_g\circ L_{g^{-1}}\right)_*A_s(X_x)$$ $$=\left(R_{\rho(g)}\circ L_{\rho(g^{-1})}\circ \rho\right)_*\circ A_s(X_x)$$ $$=\rho(g^{-1})\rho_*(A_s(X_x))\rho(g)$$

Thus we have that: $$\rho_*(A'_s(X_x))\phi'=\rho(g^{-1})\rho_*(A_s(X_x))\phi+\rho_*(\mu_G(D_xg(X_x)))\rho(g^{-1})\phi$$

Therefore:

$$d\phi'(X_x)+\rho_*(A'_s(X_x))\phi'=\rho(g^{-1})d\phi(X_x)+\rho(g^{-1})\rho_*(A_s(X_x))\phi$$ Putting it all together we obtain:

$$\nabla^A_X\Phi'=[s\cdot g,\rho(g^{-1})d\phi(X_x)+\rho(g^{-1})\rho_*(A_s(X_x))\phi$$ $$=[s,d\phi(X_x)+\rho_*(A_s(X_x))\phi]$$

as desired.

• Thank you very much for your answer and sorry for my late response! Isn't what you have shown 'just' the known transformation behaviour of the local (i.e pulled-back) connection forms? I actually used that in my calculation of $\nabla^{A_s}_X\phi)(x)$ already. It is probably that I am missing something here, but I do not really know how your calculation would me. And yes, I follow Hamiltons text. There he just writes that what I want to show "is a lengthy calculation" :D Jun 16 at 8:41
• @TwoStones yes it was, at the time I was mildly unfamiliar with Hamilton's notation and could not exactly parse the question. Jul 13 at 20:52
• @TwoStones I am unsure if you have done this problem in MGT, but it feels slightly similar to this one and I would welcome a fresh pair of eyes. math.stackexchange.com/questions/4492339/… Jul 13 at 21:21
• Thank your very much for your edit! I am not very familiar with general covariant derivatives on vector bundles but I will gladly have a look at your question (although your comment over there suggests that you already got it). Jul 14 at 15:32
• @TwoStones, I did, thank you for looking though. I will be updating with an answer shortly Jul 14 at 16:23

I've checked over your calculation for $$d(\rho(g(x)^{-1}\phi')$$and I think it is correct. The trick to pulling out the adjoint from inside $$\rho_*$$ is to remember that $$\rho$$ is a Lie Gp. homomorphism, hence $$\rho(Ad(h))(x) = Ad(\rho(h))\rho(x)$$. Let $$Y=A_s'(X)$$. Passing to differentials we have:

$$\rho_*(ad(g(x)^{-1})Y)(\rho(g(x)^{-1})\phi') = \rho_*(Ad_{g(x)^{-1}})_*(Y)(\rho(g(x)^{-1})\phi') = Ad_{\rho(g(x)^{-1})}(\rho_*(Y))(\rho(g(x)^{-1})\phi') = \rho(g(x)^{-1})\rho_*(Y)\phi' =\rho(g(x)^{-1})\rho_*(A_s')\phi'$$

which is what you want.

• Thank you very much for your answer, it looks very promising! Although, inside $\rho_\ast$ it is the adjoint representation of $G$, i.e. $\mathrm{Ad}$, that acts on $Y$ and not $\mathrm{ad}=\mathrm{Ad}_\ast$. I also don't see how you switch from $\mathrm{Ad}$ in your first calculation to $\mathrm{ad}$ in your second one. Jun 20 at 16:53
• Ignore the last sentence in my first comment, I was stupid! But another thing, how can you compose $\mathrm{Ad}$ and $\rho$? $\mathrm{Ad}$ maps into $\mathrm{GL}(\mathfrak{g})$ while $\rho$ starts from $G$. Jun 20 at 17:24
• I actually think the $Ad$ inside the $\rho_*$ should be $ad$ - on a matrix group, however, it amounts to the same thing! Jun 20 at 18:08
• I see your point, but $g$ is a curve in $G$, so only $\mathrm{Ad}$ can have $g(x)\in G$ as an argument. But nevertheless, your answer proved very helpful, thank your for that! Jun 20 at 18:15
• Sorry, what I said was more confusing than clarifying. To be explicit, if $Ad_h: x \to hxh^{-1}$, then what I meant was $ad_h = (Ad_{h})_* : T_x(G) \to T_{Ad_h(x)}(G)$. I realize this is confusing as $ad$ can also denote the Lie algebra rep $\mathfrak{g}\to gl(\mathfrak{g})$ Jun 20 at 21:51