Covariant derivative/connection of a local frame $\nabla \bar e = \nabla(e)g + edg$

Question

Suppose $e$ and $\bar e$ are two frames for a vector bundle $E \to M$ over $U \subset M$ such that $\bar e = eg$ for some $g: U \to \text{GL}(r,\Bbb R)$. Then $$\nabla \bar e = \nabla(e)g + edg.$$

I'm trying to understand this which is seemingly just the Leibniz rule, but I have some trouble with the definitions. A frame $e = (e_1,\dots,e_r)$ is an $r$-tuple of sections $e_i : U \to E$ such that $e(p)=(e_1(p),\dots,e_r(p))$ forms a basis for $E_{p}$.

There are some slight ambiguities here, first off is $e$ a map $e : U \to E$ also? I mean it has to be since otherwise $\nabla e$ would not make sense. Second, could someone help me understand how the above equation is the Leibniz rule? In general for a connection $\nabla$ on a vector bundle the Leibniz rule gives $$\nabla (fs)=df \otimes s + f\nabla s$$ where $s$ is a section $M \to E$, but $f :M \to \Bbb R$ is a smooth function not a map with codomain $\text{GL}(r,\Bbb R)$ so what gives?

Answer 1

So suppose we have a linear connection on a vector bundle $E$, then in any frame $(e_1,\dots, e_n)$, there exist one forms $\xi_i^j$ such that: $$\nabla e_i=\xi_i^j\otimes e_j$$ Now suppose that $\tilde{e}_i$ is another frame, related by a matrix of smooth function $g=g^i_j$, such that $\tilde{e}_i=e_jg^j_i$, then: \begin{align} \nabla(\tilde{e_i})=&\nabla(e_jg^j_i)\\ =&\nabla(e_j)g^j_i+e_jdg^j_i \end{align} This is really what your formula is saying, they're just removing the indices as the statement holds for all $i$, so you can apply it to the frame $e=(e_1,\dots, e_n)$.

Answer 2

What is a little confusing here is that the scalar multiplication is being written in the opposite of the standard order. In other words, given a real scalar $s \in \mathbb{R}$ and vector $v \in V$, the notation for scalar multiplication used here is $$ vs \text{ instead of the usual }sv. $$ The reason for this is the following: First, a standard convention for matrices is to use the superscript as a row index and the subscript as a column index. Second the standard convention is to write the basis of a vector space using subscripts for the index. This means that you can write a basis $(e_1, \dots, e_r)$ of a vector space formally as a row matrix, $$ e = \begin{bmatrix} e_1 & e_2 &\cdots & e_r \end{bmatrix}. $$ Then, any vector $v$ can be written as a linear combination of the basis vectors. The coefficients $c = (c^1, \dots, c^r)$ are written using superscripts. This allows the formula for $v$ to be written formally as matrix multiplication: $$ v = e_1c^1 + \cdots + e_rc^r = \begin{bmatrix} e_1 & \cdots & e_r \end{bmatrix}\begin{bmatrix} c^1 \\ \vdots \\ c^r \end{bmatrix} = ec. $$ If $\bar{e}$ is another basis, then since each $\bar{e}_j$ is a vector, there exist coefficients $g_j^i$ such that $$ \bar{e}_j = e_1g^1_j + \cdots + e_rg^r_j. $$ In other words, $g$ is an invertible matrix such that $$ \bar{e} = eg. $$

If you have a vector bundle and $e$ and $\bar{e}$ are frames, then the formulas above at each fiber $E_x$ gives you $$ \bar{e} = eg, $$ where $g$ is now a matrix of scalar functions. This is really a system of $r$ equations of the form $$ \bar{e}_j = e_1g^1_j + \cdots + e_rg^r_j. $$

If you now take the covariant derivative of each side of this equation, you get by the product rule, $$ \nabla\bar{e}_j = \nabla e_1 g^1_j + \cdots + \nabla e_r g^r_j + e_1dg^1_j + \cdots e_r dg^r_j. $$ This can be written in matrix form as $$ \nabla\bar{e} = (\nabla e)g + e(dg). $$

As you have already noted, there is no way to write $g$ directly in terms of $e$ and $\bar{e}$. It exists by the fact that $e$ and $\bar{e}$ are bases of a vector space (at each point in the manifold). As @psl2Z correctly observes, you can write $g$ in terms of $\bar{e}$ and the dual basis of $e$.

$\newcommand\R{\mathbb{R}}$A guick comment about how a frame is equivalent to a trivialization. If you have a frame over $O \subset M$ then there is a diffeomorphism \begin{align*} \Phi: O\times\R^r &\rightarrow E\\ (x,v) &\mapsto e_1(x)v^1+\cdots+e_r(x)v^r = ev \in E_x. \end{align*} The inverse map is a trivialization of $\pi^{-1}(O)$. Given another frame $\bar{e}$, you get another map \begin{align*} \overline\Phi: O\times\R^r &\rightarrow E\\ (x,v) &\mapsto \bar{e}_1(x)v^1+\cdots+\bar{e}_r(x)v^r = \bar{e}v \in E_x. \end{align*} Since $$ \bar{e}\bar{v} = eg\bar{v}, $$ The transition map between the two trivializations is \begin{align*} \Phi^{-1}\circ\overline{\Phi}: O\times\R^r &\rightarrow O\times\R^r\\ (x,\bar{v}) &\mapsto (x,g\bar{v}). \end{align*}

Answer 3

You should understand $e$ as $e = (e_1,...,e_r)$ and $\nabla e = (\nabla e_1, ..., \nabla e_r)$ and $dg = (dg^i_j)$ as matrix. Then, with Einstein summation convention, $\bar e = e g = (g^i_1 e_i, ..., g^i_r e_i)$ and $\nabla \bar e_j = \nabla (g^i_j e_i) = (dg^i_j) e_i + g^i_j \nabla e_i$ and therefore $\nabla \bar e = e dg + (\nabla e )g$.