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.
Best Answer
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: