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I'm trying to fit the different notions of covariant derivatives I know in a unified framework. I already know how to relate connections between general, principal and vector bundles but now I'm only looking at covariant derivatives.

Let $F \hookrightarrow E \overset{\pi}{\to} M$ be a fiber bundle and $VE := \ker \pi_*$ the vertical bundle. Let $H \subset TE$ be a connection, i.e. $e \mapsto H_e$ smooth and $T_eE = V_eE \oplus H_e$, and let $v: TE \to VE$ be the projection (1-form) associated with $H$.

We can define a covariant derivative on $E$ as the vertical part of the tangent mapping: $$D: \Gamma(E) \to \Gamma(VE \otimes T^*M) \quad \sigma \mapsto v \circ \sigma_*$$ We then have: $$D_X\sigma : \Gamma(E) \to \Gamma(VE)\quad \sigma \mapsto v(\sigma_*X)$$ The first time I saw this definition it blew my mind. It is simple and has a strong geometric flavour.

Q1: How to explicitly relate the general covariant derivative $D$ to the exterior covariant derivative on principal bundles $D^\omega \alpha = d\alpha^H$?

For example in vector bundles we have $VE \simeq E$ and so we get a map $\nabla: \Gamma(E) \to \Gamma(E\otimes T^*M)$ that corresponds to the usual notion of covariant derivative.

Proceding as for vector bundles, I could use $VP \simeq P \times \mathfrak{g}$ and write: $$D: \Gamma(P) \to \Gamma((P \times \mathfrak{g}) \otimes T^*M)$$ but seems to me this is not the right thing to do. In principal bundle we differentiate vector valued forms (functions for $k=0$) and not sections of $P$...

I am aware that I can indirectly pass trough an associated vector bundle of $P$ and then recover the exterior covariant derivative.

I'm sorry if this question is a bit messy, I'm trying to bring order in my head.

Now the original question from where I started this:

Q2: Is there a way to define a general notion of covariant derivative in general fiber bundles that comprehend, as a special cases, the covariant derivatives on vector and principal bundles?

Thanks

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I cannot comment, so I give you this video as a possile answer to your question.

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