# Covariant derivative in associated bundle

Let $\pi:P→M$ be an arbitrary principal $G$-bundle, $F$ a vector space and $\pi_F : P × _G F → M$ the associated fiber bundle.

If $U$ is an open subset of $M$, then a smooth map $ϕ : \pi^{−1} (U)→ F$ is said to be equivariant if

$ϕ(p · g) = g^{-1} · ϕ(p)$

Given a section $s : V → \pi_F^{-1}(U)$ we can show that there is a one-to-one and onto correspondence between $ϕ(p)$ and $s$.

Assuming now that $\pi:P→M$ has defined on it a connection form $ω$, we define the covariant exterior derivative $d^ω ϕ$ of $ϕ$ by

$(d^ω ϕ)_p(v)= (dϕ)_p(v^H)$

Where $dϕ$ is the exterior derivative of $ϕ$ , $v ∈ T_pP$ and $v^H$ is the horizontal part of it.
It also can be show that $d^ω ϕ=dϕ+ω\cdotϕ$ where $ω\cdotϕ$ means $ω$ acting on $ϕ$

Now given a section on the principal bundle $\sigma : V → \pi^{-1}(U)$ we can show also that

$(\sigma^*d^ωϕ)(T)=d(\sigma^*ϕ)(T)+(\sigma^*ω)(T)\cdot\sigma^*ω$

Identifying $(\sigma^*d^ωϕ)(T)$ as $\nabla_T S$ where $S=\sigma^*ϕ$ we obtain

$\nabla_T S=dS(T)+(\sigma^*ω)(T)\cdot S$

Now my question is if $ϕ$ is a function a $0$ form then $S$ the pullback of a $0$ form is a function but by definition of covariant derivative we should have

$\nabla_T S=dS(T)$

Only if if we identify $S$ as a section that we would have the extra piece. Then how can we say that $\nabla_T$ is a covariant derivative?

• I don't understand the question. Can you please rephrase it? But in general, $S$ will be a fixed vector-space valued function. The reason you get the extra terms is because $S$ highly depends on the local section $\sigma$. Basically, you need to think of $S$ as the local component representation of a geometric object. Which basis is used to take these components is determined by $\sigma$. If the components were independent of the basis, then $\phi$ would not just be equivariant, but completely "constant" with respect to the fibrewise transitive group action... Oct 14, 2017 at 19:47
• ... and then the extra term would not appear. But I really don't understand your problem so please rephrase. Oct 14, 2017 at 19:48
• My main problem is that if $S$ is a function covariant derivative should act as $\nabla_T S=dS(T)$ by definition of covaraint derivative Oct 14, 2017 at 20:21
• $S$ is the component representation of a section of the associated vector bundle, so while technically it is a function, the covariant derivative should be interpreted to act on the underlying abstract section, not on its representative. But see my answer. Oct 14, 2017 at 20:23
• I have locked this post since it is only 3 votes short of a migration to Mathematics. This post is discussed in the hbar chat room here. Oct 30, 2017 at 12:36

Okay, I still don't completely understand you, but I think I pinpointed the problem.

Using your notation, $S$ is not really a section of the associated bundle, but rather the component representation of such. So the notation $\nabla_T S$ is either incorrect, or just requires a lot of interpretation.

Let $F$ be a real, $k$ dimensional vector space, to see an example. Let us assume, that the latin indices $a$ range from $1$ to $k$.

We actually have some separate things here. We have a principal fiber bundle $(P,\pi,M,G)$ and an associated vector bundle $(E,\pi,M)$ with $E=P\times_\rho F$, where $\rho:G\rightarrow GL(F)$ the representation that defines the associated vector bundle.

The thing is, if $\psi:M\rightarrow E$ is a (smooth) section of $E$, then there exists a unique corresponding equivariant map $\phi:P\rightarrow F$, but for all intents and purposes, they are not the same.

You should instead think about them like this: If ${e_{a}}$ is a local frame of $E$, then we have locally $\psi=\psi^ae_a$, where the components $\psi^a$ depend on two things - they depend on the manifold points $x\in M$, but also on the local frame $e_a(x)$ (at $x$) chosen to represent it, so $$\psi^a=\psi^a(x,\{e_a(x)\}).$$

But if $E$ is an associated vector bundle to $P$, then, essentially, $P$ is an associated principal bundle to $E$, so $P$ can be identified as the frame bundle of $E$ (because a local frame $e_a$ provides a local trivialization of both $E$ and $P$, so local sections of $P$ can be understood as frames of $E$). Well, not the total frame bundle, but some restriction of the frame bundle to a $G$-subbundle.

So the point is, that because the components $\psi^a$ depend on points of $P$, rather than $M$, it is an $F$-valued (essentially $\mathbb{R}^k$-valued) function on $P$. But it needs to be equivariant. Why? Because if $\Lambda^a_{\ b}(x)$ is a local frame transformation on $E$, then the components transform by $\Lambda^{-1}$. But a local frame transformation (acting on the model fiber $F$) is equivalent to the fibrewise transitive right action of $G$ on $P$, so $$\psi^a(x,e(x)\Lambda(x))=(\Lambda^{-1})^a_{\ b}(x)\psi^b(x,e(x)).$$

Essentially, invariant fields are sections of $E$, component representations of it are realized basis independently as equivariant functions on $P$. Equivariance guarantees that the components transform "properly" under frame transformations (frame transformation = group action on $P$).

With this discussed, let us concatenate notation:

• $\psi$ is the invariant section of $E$.

• $\psi^a$ are components of it taken with respect to a local trivialization of either $E$ or $P$ (they agree).

• $\phi$ is the equivariant function on $P$ satisfying $$\phi(x,e(x))=(\psi^1(x,e(x)),...,\psi^k(x,e(x))),$$ where $(x,e(x))$ is a point of $P$.

What actually is, is that $$\nabla_T\psi=(d(\sigma^*\psi^a)(T)+(\sigma^*\omega)^a_{\ b}(T)(\sigma^*\psi^b))e_a,$$ where the frame $e_a$ is the frame that generates the local trivialization/section $\sigma$.

Since this got quite complicated, and the otherwise simple essence gets lost, what your "$\nabla_TS$" is, is an analogue of $\nabla_T X^\mu$ for tangent vectors, when, in fact, what interests you is not $\nabla_T X^\mu$, but $\nabla_TX=\nabla_T(X^\mu\partial_\mu)$.

You got the component representation of the covariant derivative. Which is fine, just the notation confuses you, because $\nabla_T$ should not act on the components, but the invariant section.