Are connection forms tensorial objets?

Question

I think the standard answer is they are not, for example, from wiki. Under a coordinate transformation of $e' =eA$ where $e$ is a set of frames and $A$ the transformation matrix, connection form $\omega$ transform as $$\omega'=A^{-1}\omega A + A^{-1}\mathrm{d}A$$ Apperatently apart from the usual tensoril transformation term $A^{-1}\omega A$, there is an additional $A^{-1}\mathrm{d}A$ term.

However, due to the $\mathcal{F}$-linear w.r.t. $W$ as in $\nabla_W V$, one can consider $\nabla_{(\cdot)}V$ with a fixed vector field $V$ a $(1, 1)$ tensor field (reference: pg. 204 of [1]).

Aren't connection forms the same construction as $\nabla_{(\cdot)}V$? i.e. $$\nabla_Ve_a = \omega^b_a(V)e_b$$ This is just picking a set of frame fields $e_a$ in place of $V$ in $\nabla_{(\cdot)} V$, and we decompose the result on to the same set of frames?

How do I reconcile these two understandings?

[1]: Schutz, Bernard - Geometrical methods of mathematical physics. Cambridge university press 1980

Answer 1

Say that $U$ is the domain of the frame $e$. In parts:

• Each $\omega^a_{~b}$ is a $1$-form on $U$ precisely due to $\mathcal{F}$-linearity of the connection $\nabla$ in the lower argument, yes.
• The whole matrix $(\omega^a_{~b})_{a,b=1}^n$ is a $\mathfrak{gl}_n(\Bbb R)$-valued $1$-form on $U$.
• However, $(U,\{\omega^a_{~b} \mid a,b=1,\ldots,n\})$ are not the local components of any tensor on $M$, because whenever $(\widetilde{U} ,\{\widetilde{\omega}^a_{~b}\mid a,b=1,\ldots, n\})$ corresponds to a second frame $\widetilde{e}$ on $\widetilde{U}$ (with $\widetilde{e} = eA$ on $U\cap\widetilde{U}\neq\varnothing$), they do not satisfy the tensor transformation law (due to $A^{-1}{\rm d}A$).