For a vector bundle $\xi:E\to M$, why is Trace a parallel section of the bundle $End(\xi)^*$? For any $L\in$ $End(\xi)$, $Tr(L)$ has a well defined value independent of the choice of basis, so we have a globally defined function $Tr:End(\xi)\to\mathbb{R}$. This function is tensorial, and thus represents a section of $End(\xi)^*$.
My book (Walschap) claims that this section is parallel, but I don't understand his argument. Suppose we have a basis $\beta$ of parallel sections of $\xi$ and a parallel section $X$ of $End(\xi)$, all taken along a curve $c$ in $M$. Then according to my book $\beta^{-1}(t)\circ X(t)\circ \beta(t)$ is constant in the space of $n\times n$ matrices. Why? 
First of all I assume $\beta(t)$ here is supposed to be a matrix whose columns are the same as the chosen basis, in a particular coordinate chart. Then I see that $X(t)\circ\beta(t)$ is parallel. I'm not sure where to go from here.
 A: Let me try and explain what the author means by $\beta^{-1}(t) \circ X(t) \circ \beta(t)$ (which doesn't even compile at first glance). First, given a linear map $T \colon V \rightarrow V$ on an $n$-dimensional real vector space and some basic $\beta$ for $V$, I'll denote by $[T]_{\beta} \in M_n(\mathbb{R})$ the matrix which represents $T$ with respect to the basis $\beta$.
Fix a curve $c \colon I \rightarrow M$ and let $\beta(t) = (e_1(t), \dots, e_n(t))$ be a parallel frame (basis) along $c$ and let $X(t)$ be a parallel section of $\operatorname{End}(\xi)$ along $c$. Note that $\beta(t)$ is just a list of $n$ vectors in $\xi_{c(t)}$ and we cannot identify it with a matrix without making some extra choice. However, for each $t \in I$ we have a linear map $X(t) \colon \xi_{c(t)} \rightarrow \xi_{c(t)}$ and a basis $\beta(t)$ of $\xi_{c(t)}$. What the author says is that $[X(t)]_{\beta(t)}$ (the matrix representation of the parallel map with respect to a parallel frame) is constant. This makes sense without any extra choices.
How can we relate it to the author's notation? The pullback bundle $c^{*}(\xi)$, being a vector bundle over an interval, is isomorphic to a trivial bundle. Choose arbitrary some trivialization $c^{*}(\xi) \cong I \times \mathbb{R}^n$ and use it to represent sections of $\xi$ along $c$ as (curves of) column vectors in $\mathbb{R}^n$. Similarly, use it to denote sections of $\operatorname{End}(\xi)$ along $c$ as (curves of) $n \times n$ matrices. Then, we can think of our parallel frame $\beta(t)$ as a list of $n$ columns vectors $\hat{e}_1(t), \dots, \hat{e}_n(t) \in \mathbb{R}^n$. Now we can take the vectors and put them as columns in a $n \times n$ matrix which we will denote by $\hat{\beta}(t)$. Now, the expression $\hat{\beta}^{-1}(t) \circ \hat{X}(t) \circ \hat{\beta}(t)$ will give us precisely the matrix representation of the linear operator $X(t)$ with respect to the basis $\beta(t)$ and it seems that your book doesn't use a notation which makes this identification explicit.
The resulting vectors $\hat{e}_i(t)$ and matrices $\hat{X}(t)$ depend very much on our choice of initial identification. For example, we could have used the trivialization of $c^{*}(\xi)$ coming from the parallel frame $e_1(t), \dots, e_n(t)$ and then $\hat{\beta}(t)$ would just be the identify matrix and $\hat{X}(t)$ would be $[X(t)]_{\beta(t)}$.

Finally, since I really don't like to make unnecessary identifications, let me offer a much more "choice-free" proof of the fact that $\operatorname{Tr}$ is parallel. Fix a curve $c \colon I \rightarrow M$ and let $e_1(t), \dots, e_n(t)$ be a parallel frame along $c$. Denote by $e^1(t), \dots, e^n(t)$ the dual frame of $E^{*}$ along $c$. By the definition of the induced connection/parallel transport on $E^{*}$, the dual frame is also parallel. Finally, Note that $\operatorname{Tr}$ can be written as $e_i(t) \otimes e^i(t)$ (where we identify $\operatorname{Hom}(E)$ with $E^{*} \otimes E$ and $\operatorname{Hom}(E)^{*}$ with $\left( E^{*} \otimes E \right)^{*} \cong \left( E^{*} \right)^{*} \otimes E^{*} \cong E \otimes E^{*}$). But then
$$ \frac{D \operatorname{Tr}}{dt}(t) = \frac{D (e_i \otimes e^i)}{dt}(t) = \frac{De_i}{dt}(t) \otimes e^i(t) + e_i(t) \otimes \frac{De^i}{dt}(t) = 0.$$ 
