# Connection forms

I find the following paragraph in the book of Manifolds and differential geometry by Jeffrey M.Lee (section 12.2 Connection forms page 506):

Let $$\pi : E \longrightarrow M$$ be a rank r vector bundle with a connection $$\nabla$$. Recall that a choice of a local frame field over an open set $$U \subset M$$ is equivalent to a trivialisation of the restriction $$\pi_U : {E|}_U \longrightarrow U .$$ Namely, if $$\phi = (\pi, \Phi)$$ is such a trivialisation over U, then defining $$e_i(x)= \phi ^{-1}(x,e_i)$$, where $$(e_i)$$ is the standard basis of $$\mathbb{F}^n$$, we have a frame field $$(e_1,...,e_k).$$ We now examine the expression for a given covariant derivative from the viewpoint of such local field. It is not hard to see that for every such frame field there must be a matrix of 1-forms $$\omega = {(\omega_j^i)}_{1≤i,j≤r}$$ such that for $$X \in \Gamma(U,E)$$ we may write $$\nabla_Xe_j =\sum\limits_{i=1}^k {\omega_j^i}(X)e_i$$

My question is how can prove the existence of such a matrix of 1-forms ?

• Hint: act on both sides of the equation with an element of the dual coframe $e^l$. – Kajelad Oct 16 '20 at 1:31

We know that $$\nabla_Xe_i$$ is a smooth section of $$E$$, and working locally over a frame $$(U,e_1,\ldots, e_k)$$ as you are we have $$\nabla_Xe_i=\sum_{j=1}^n \omega^j_i(X)e_j$$ where $$\omega^j_i:\mathfrak{X}(U)\to C^\infty(U).$$ It is easy to see using $$\nabla_{aX+bY}=a\nabla_X+b\nabla_Y$$ that $$\omega^j_i$$ is linear, but more is true. $$\omega^j_i$$ is tensorial because $$\nabla_{fX}=f\nabla_X$$ so that $$\omega^j_i(fX)=f\omega^j_i(X)$$. Hence, $$\omega^j_i$$ is a smooth $$(0,1)-$$tensor field over $$U$$. That is, the $$(\omega_j^i)$$ form a $$k\times k$$ array of smooth $$1-$$forms over $$U$$.
You want to express the matrix $$(\omega_j^i)_{i,j}$$ defined locally as a section of a bundle of endomorphisms. It is possible to define vector bundle valued differential forms. Actually, this is not so hard. In a local trivialization of $$E$$ given by $$(U,e_1,\ldots, e_r)$$ an $$E-$$valued differential $$k-$$form is an expression of the form $$\eta=\eta^1\otimes e_1+\cdots +\eta^r\otimes e_r$$ for each $$\eta^i$$ a $$k-$$form. That is, we have a $$k-$$form in front of each frame element, so that given $$X_1,\ldots, X_k$$ vector fields on the same open neighborhood, $$\eta(X_1,\ldots, X_k)=\sum_{i=1}^r \eta^i(X_1,\ldots,X_k)e_i$$ defines a section of $$E$$ over $$U$$. With all this being said, if $$E$$ is locally trivial with $$e_1,\ldots, e_r$$ a local frame, then $$\operatorname{End}(E)$$ is also trivialized, with frame given by the standard matrices $$E_{i,j}$$ (all zeroes except a $$1$$ in the $$(i,j)$$ position). That is, $$E_{i,j}$$ represents the transformation sending $$e_j\mapsto e_i$$ and all other $$e_k\mapsto0$$. Then a $$\operatorname{End}(E)-$$valued $$k-$$form is exactly a sum of the form $$\sum_{i,j}\underbrace{\eta^{i,j}}_{\in \Omega^k(U)}E_{i,j}= \begin{bmatrix} \eta^{1,1}&\cdots& \eta^{1,r}\\ \vdots&\ddots&\vdots\\ \eta^{r,1}&\cdots& \eta^{r,r} \end{bmatrix}.$$ In the case where $$k=1$$, this is called an $$\operatorname{End}(E)-$$valued $$1-$$form. Note that this is a special case of the general construction for vector bundles, since $$\operatorname{End}(E)$$ is a vector bundle. This is what the author is referring to as $$\mathcal{A}^1(-,\operatorname{End}(E))$$.
• Thanks a lot @Alekos Robotics for your answer ! But I still have a question if you don't mind ! We know that the 1- form $\omega _i^j \in \mathcal{A}(U)$ . How can one express the space where it belongs in terms of an endomorphisme bundle ? I have this question because when I read the part of the book where the author calculate the difference between two covariant derivatives say $\nabla_1 , \nabla_2$ , he get an operator $D = \nabla_1-\nabla_2$ which is $C^\infty -$linear and then he concludes that $D \in \mathcal{A} ^1(M, End(E)$? – Maria Oct 16 '20 at 23:03