connection laplacian on general vector bundles As the title says, my question is about how to define the connection laplacian on general vector bundles.

I think I understand how to define the connection laplacian on the tensorbundles:
Let $M$ be a Riemannian manifold and $\mathcal{T}^k_l(M)$ be the space of smooth section of the vector bundle of $(k,l)$-tensors on $M$. Call elements in $\mathcal{T}^k_l(M)$ smooth $(k,l)$-tensor fields.
We think of a smooth $(k,l)$-tensor field as a $C^{\infty}(M)$-multilinear map
$F\colon \Omega^1(M)\times\ldots\times\Omega^1(M)\times\mathfrak{X}(M)\times\ldots\times\mathfrak{X}(M)\rightarrow C^\infty(M)$,
where $\Omega^1(M)$ is the space of $1$-forms on $M$, $\mathfrak{X}(M)$ is the space of vector fields on $M$, $\Omega^1(M)$ is taken $l$-times and $\mathfrak{X}(M)$ is taken $k$-times.
For each $k,l$ he Levi-Civita-Connection on $M$ induces a connection $\nabla$ on the bundle of $(k,l)$-tensors, so we have maps
$\nabla\colon \mathcal{T}^k_l(M)\rightarrow \mathcal{T}^{k+1}_l(M)$ given by
$\nabla F(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k,X):=(\nabla_XF)(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k)$
It turns out that $(\nabla^2F)(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k,Y,X):=(\nabla\nabla F)(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k,Y,X)=(\nabla_X\nabla_YF-\nabla_{\nabla_XY}F)(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k)$
Finally we define the connection laplacian
$\Delta\colon \mathcal{T}^k_l(M)\rightarrow \mathcal{T}^k_l(M)$ by $(\Delta F)(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k):=tr_g((Y,X)\mapsto\nabla^2F(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k,Y,X))$
where $tr_g$ is to be understood as follows: if $G$ is a $(2,0)$-tensor, we transform it into a $(1,1)$-tensor by via the metric (i.e. by applying the #-operator). A  $(1,1)$ tensor can be understood as an endomorphism of $T_pM$ of which the trace can be taken.
If $(e_i)$ is a local orthonormal frame, we have $(\Delta F)(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k)=\sum_i \nabla^2F(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k, e_i,e_i)$.

Now, let $E$ be a vector bundle over $M$ with a connection $\nabla$. For any smooth section $\varphi$ of $E$, we define
$\nabla^2\varphi (X,Y):=\nabla_X\nabla_Y\varphi - \nabla_{\nabla_XY}\varphi$
where $X$ and $Y$ are vector fields. For a local orthonormal frame $(e_i)$ we set
$\Delta \varphi :=\sum_i\nabla^2\varphi (e_i,e_i)$.
However, this definition is unsatisfying for me:

Question 1: Is it possible to define a "trace" in the setting of general vector bundles $E$ so that $\Delta \varphi$ turns out to be trace($\nabla^2\varphi$) just as in the case of the tensor bundles? Edit: I found a reference that defines the connection laplace via trace (Lawson, Spin Geometry, p. 154). Could someone explain to me how the trace is to be understood in that context?
Question 2: Is there more behind the definition of $\nabla^2\varphi (X,Y)$ (as in the case of the tensor bundles, where $\nabla^2 F$ is $\nabla\nabla F$)? That is, do $\nabla$ and the Levi-Civita-Connection induce a connection $\nabla$ on $T^*M\otimes E$ in a way that $\nabla\nabla\varphi=\nabla^2\varphi$?

I also would appreciate any kind of reference where this is explained.
 A: The answer to both of your questions is yes, and I think that it works completely analogously  to what describe in the first part of your question. More specifically,


*

*The idea of the trace is the same one that you discuss in the
first part of your question. The "$E$-Hessian" $\nabla^2 \varphi$ is a section of $E
    \otimes T^\ast M \otimes T^\ast M$. Use the metric (the musical
isomorphism $\sharp$) to identify $T^\ast M$ with $TM$ and obtain a section
of $E \otimes T^\ast M \otimes TM \cong E \otimes \text{End}(TM)$.
Take the trace of the endomorphism piece to obtain a section of $E$.
Note that $\nabla^2 \phi$ may not be symmetric in its entries, so we have a choice
as to which factor of $T^\ast M$ we apply $\sharp$ to, but this choice is irrelevant once 
we take the trace.
In terms of a local frame $\{ e_i \}$ for $TM$, we have
$$ \Delta \phi = \text{tr}_g \nabla^2 \varphi = g^{ij} [\nabla^2 \varphi] (e_i, e_j).$$
Of course, if the frame is orthonormal, this reduces to $[\nabla^2 \varphi] (e_i, e_i)$,
as you stated.

*We have a connection $\nabla^E$ on $E$ and the Levi-Civita connection $\nabla^{LC}$ on $T^\ast M$. These induce a connection $\tilde{\nabla}^E$ on
$E \otimes T^\ast M$, which is defined by the "product rule":
$$\tilde{\nabla}^E (\varphi \otimes \omega) = (\nabla^E \varphi) \otimes \omega + \varphi \otimes (\nabla^{LC} \omega)$$
where $\varphi$ is a section of $E$ and $\omega $ is a one-form.
This connection gives a map $\tilde{\nabla}^E: \Gamma(E \otimes T^\ast M) \to \Gamma(E \otimes T^\ast M \otimes T^\ast M)$, and as you guessed, $\nabla^2$ as you defined it is precisely the composition of the connections
$$\tilde{\nabla}^E \circ \nabla^E: \Gamma(E) \to \Gamma(E \otimes T^\ast M) \to \Gamma(E \otimes T^\ast M \otimes T^\ast M).$$
It's a good exercise to prove this!

A note, as much for my own understanding as anything: $\tilde{\nabla}^E$ as defined above is an extension of $\nabla^E$ which maps $\Gamma(E \otimes T^\ast M) \to \Gamma(E \otimes T^\ast M \otimes T^\ast M)$. There is another interesting extension of $\nabla^E$ to $E \otimes T^\ast M$, which I will call $d^E$. $d^E$, in contrast to $\tilde{\nabla}^E$, is a map
$$d^E: \Gamma(E \otimes T^\ast M) \to \Gamma(E \otimes \Lambda^2 T^\ast M),$$
i.e., $d^E$ gives a two-form with $E$-coefficients. More generally, one can define $d^E$ as a map on $E$-valued forms of any degree:
$$d^E: \Gamma(E \otimes \Lambda^k T^\ast M) \to \Gamma(E \otimes \Lambda^{k+1} T^\ast M),$$
defined by 
$$d^E(\varphi \otimes \omega) = (\nabla^E \varphi) \wedge \omega + \varphi \otimes (d\omega),$$
where $\wedge$ means "wedge the one-form part of $\nabla^E \varphi$ with $\omega$".
I suppose one should think of $d^E$ as a generalization of the de Rham exterior derivative $d$ on forms. (If $E$ is the trivial bundle $M \times \mathbb{R}$ with trivial connection $\nabla^E = d$, we recover $d$.) Note that $d^E$ does not require a connection on $TM$.
The curvature $R^E$ of the connection $\nabla^E$ is the $\text{End}(E)$-valued two-form defined by the composition
$$R^E:=(d^E)^2 \text{ (or }d^E \circ \nabla^E): \Gamma(E) \to \Gamma(E \otimes T^\ast M) \to \Gamma(E \otimes \Lambda^2 T^\ast M) .$$
It's a standard computation to show that $R^E$ really lives in $\text{End}(E)$, i.e., that it's $C^\infty(M)$-linear, and that
$$ R^E(X,Y) = \nabla^E_X \nabla^E_Y - \nabla^E_Y \nabla^E_Y - \nabla^E_{[X,Y]} $$
As a final remark to relate this back to the Hessian, notice that the antisymmetric part of the Hessian is precisely the curvature, i.e.,
$$[\nabla^2 \varphi](X, Y) - [\nabla^2 \varphi](Y, X) = R^E(X,Y) \varphi.$$
The Hessian of a smooth function is symmetric, which is equivalent to the the fact that the de Rham "curvature" $d^2$ is zero.
References: Here are a couple of books I found useful in reminding myself how some of this works:


*

*Jost, Riemannian Geometry and Geometric Analysis. See chapter 4.

*Taylor, Partial Differential Equations I. See Appendix C (on his website) and chapter 2.

