Symmetric compatibility between a bundle map and a connection Let $E$ be a vector bundle over a smooth manifold $M$, equipped with a connection $\nabla$. 
Let $A:TM \to E$ be a bundle homomorphism. 
We say $A$ is symmetric-compatible with $\nabla$ if for every $X,Y \in \Gamma(TM)$
$$
A([X,Y])=\nabla_X^E \big(A(Y)\big) - \nabla_Y^E \big(A(X)\big). \tag{1}
$$
Here is the basic example:
Let $N$ be a manifold, and let $\nabla^{TN}$ be any symmetric connection on $TN$. Suppose $f:M \to N$ is a smooth map.
Take $E=f^*{TN}$, $A=df$. Then, indeed
$$ df([X,Y])= \nabla^{f^*TN}_X df(Y) -  \nabla^{f^*TN}_Y df(X). $$ 

Question:


*

*Fix a bundle $E$. Does every connection $\nabla$ on $E$ admit a non-zero symmetric-compatible operator? 

*What are the possible dimensions of the vector space $$\{ A \in \operatorname{Hom(TM,E)} \,|\, \text{ is symmetric-compatible with $\nabla$} \}?$$
Can it be zero? Can it be infinite? I tried to write condition $(1)$ locally in coordinates. 


*Is this notion has been studied? Are there any other "natural" examples besides the one I described above?


I am mostly interested in the case where the rank of $E$ equals $\dim M$.

Edit:
The first reasonable thing to do is to understand the local version of the problem:
Suppose $\operatorname{rank}(E)=\dim M=d$:
Let $U\subset M$ be a neighborhood on which $M$ and $E$ are trivial. Let $x_1,\ldots,x_d$ be coordinates on $U$, and let $e_1,\ldots,e_d$ be a frame of $E$ on $U$. Let $\Gamma_{ij}^k$ represent the connection $\nabla$ on $U$. That is, $$\nabla_ie_j=\Gamma_{ij}^ke_k.$$ For the bundle morphism $A$, write $$A\left(\frac{\partial}{\partial x^i}\right)=A_i^ke_k.$$ So, $$\nabla_iA\left(\frac{\partial}{\partial x^j}\right)=\nabla_iA_j^ke_k=\frac{\partial A_j^k}{\partial x^i}e_k+A_j^k\Gamma_{ik}^le_l=\left(\frac{\partial A_j^l}{\partial x^i}+A_j^k\Gamma_{ik}^l\right)e_l.$$ Finally, $A$ is symmetric compatible with $\nabla$ if and only if the equality $$\frac{\partial A_j^l}{\partial x^i}+A_j^k\Gamma_{ik}^l=\frac{\partial A_i^l}{\partial x^j}+A_i^k\Gamma_{jk}^l$$ holds for every $1\leq i<j\leq d$ and $1\leq l\leq d$.
(Since it is enough to verify condition $(1)$ on different pairs of coordinate fields, as observed by Amitai Yuval).
In general this is a system of $\frac{d^2(d-1)}{2}$ scalar equations.
Let us consider the first non-trivial case: $d=2$:
In that case the only surviving pair is $(i,j)=(1,2)$:
$$l=1: \, \,\frac{\partial A_2^1}{\partial x^1}+A_2^1\Gamma_{11}^1+A_2^2\Gamma_{12}^1=\frac{\partial A_1^1}{\partial x^2}+A_1^1\Gamma_{21}^1+A_1^2\Gamma_{22}^1,$$
and
$$l=2: \frac{\partial A_2^2}{\partial x^1}+A_2^1\Gamma_{11}^2+A_2^2\Gamma_{12}^2=\frac{\partial A_1^2}{\partial x^2}+A_1^1\Gamma_{21}^2+A_1^2\Gamma_{22}^2$$
I still need to think how many degrees of freedom do we have for this system?

Note that even if we can solve this locally, it is not clear how to extend a "symmetric morphism" defined locally to a global symmetric morphism (bumb functions won't work I think).
 A: The following may be the beginning of a solution to 1. and/or 2.
Let $U\subset M$ be a neighborhood on which $M$ and $E$ are trivial. Let $x_1,\ldots,x_n$ be coordinates on $U$, and let $e_1,\ldots,e_m$ be a frame of $E$ on $U$. Let $\Gamma_{ij}^k$ represent the connection $\nabla$ on $U$. That is, $$\nabla_ie_j=\Gamma_{ij}^ke_k.$$ For the bundle morphism $A$, write $$A\left(\frac{\partial}{\partial x^i}\right)=A_i^ke_k.$$ So, $$\nabla_iA\left(\frac{\partial}{\partial x^j}\right)=\nabla_iA_j^ke_k=\frac{\partial A_j^k}{\partial x^i}e_k+A_j^k\Gamma_{ik}^le_l=\left(\frac{\partial A_j^l}{\partial x^i}+A_j^k\Gamma_{ik}^l\right)e_l.$$ Finally, $A$ is symmetric compatible with $\nabla$ if and only if the equality $$\frac{\partial A_j^l}{\partial x^i}+A_j^k\Gamma_{ik}^l=\frac{\partial A_i^l}{\partial x^j}+A_i^k\Gamma_{jk}^l$$ holds for every $1\leq i,j\leq n$ and $1\leq l\leq m$. Note that this is a first order linear equation. 
For the time being, focus on the case where $n=2$ and $m=1$. In this case we are looking for two functions, $A_1,A_2:U\to\mathbb{R}$, satisfying $$\frac{\partial A_2}{\partial x^1}=\frac{\partial A_1}{\partial x^2}+\Gamma_2A_1-\Gamma_1A_2.$$ We can choose $A_1$ as we like. Then, $A_2$ needs to satisfy a first order linear condition involving only $\frac{\partial}{\partial x^1}$. This means that we can choose $A_2$ as we like on the axis $\{x_1=0\}$ and solve the equation separately on every line $\{x_2=c\}$. All together, there are many solutions.
I seems that it shouldn't be too hard to extend a symmetric morphism defined only on a part of $U_\alpha$. This may help to solve the global problem. You are encouraged to verify that. You are also encouraged to check what changes when increasing $n$ and $m$.
A: Interesting question. I'm not sure what was your original motivation and whether I'm reverse engineering something you already noticed, but a bundle morphism $A$ is symmetric-compatible with $\nabla$ iff $d_{\nabla}(A) = 0$ where $d_{\nabla} \colon \Omega^1(M;E) \rightarrow \Omega^2(M;E)$ is the exterior covariant derivative on $E$-valued one forms and we think of $A$ as an element of $\Omega^1(M;E$). The reason is that by definition
$$ d_{\nabla}(A)(X,Y) = \nabla_{X}(A(Y)) - \nabla_{Y}(A(X)) - A([X,Y]). $$
Before analyzing this operator, it is instructive to see what we can say about the operator $d_{\nabla} \colon \Omega^0(M;E) \rightarrow \Omega^1(M;E)$ acting on $E$-valued zero forms (section of $E$). Set $n = \dim M \geq 1$ and $k = \operatorname{rank}(E) > 0$.


*

*In local coordinates, the equation $d_{\nabla}(s) = 0$ is a system of $nk$ linear partial differential equations for the $k$ components of $s$ on $\mathbb{R}^n$. If $n > 1$ this is an overdetermined system so we don't expect any non-trivial solutions in the generic case.

*When $n = 1$, this becomes a linear ODE system and so it has locally a $k$-dimensional solution space. When the topology is not trivial (i.e $M = S^1$), the local solutions don't necessarily extend to global solutions. One obstruction comes from the fact that because this is a linear system, a global solution which is non-zero at a single point must be non-zero everywhere so we can't have global solutions if $E$ doesn't have a global non-zero section (as it is the case for the Mobius line bundle). Even if the bundle $E$ is trivial, local solutions might not glue to global solutions because of monodromy. We can start with a non-zero vector, obtain a local solution, extend it until we reach the starting point and find out that the solution doesn't agree with the initial vector. In other words, even if the line bundle is trivial and the connection is flat we can have non-trivial parallel transport.

*When $n > 1$, there are no solutions in general. A full obstruction for the existence of $k$ linearly-independent local solutions is given by $R_{\nabla} = 0$. Even if $R_{\nabla} \neq 0$ there might be some solutions. For example, if $E$ splits as $E = E_1 \oplus E_2$ and the connection splits as $\nabla = \nabla_1 \oplus \nabla_2$ with $R_{\nabla_1} = 0$ and $R_{\nabla_2} \neq 0$ then we will find some local solutions. Whether the curvature vanishes or not, local solutions (on a connected neighborhood) are completely determined by their values at a single point so we always have a finite dimensional space of solutions of $\dim \leq k$.


Let us move now to $d_{\nabla} \colon \Omega^1(M;E) \rightarrow \Omega^2(M;E)$.


*

*When $n = 1$, this is the zero operator because there aren't any antisymmetric two-forms on a one-dimensional space. Hence, there are an infinite dimensional space of solutions in this case.

*When $n \geq 2$, the equation $d_{\nabla}(A) = 0$ is a system of ${ n \choose 2 }k$ linear partial differential equations for the $nk$ components of $A$ on $\mathbb{R}^n$. When $n = 2$, this is an underdetermined system so we expect infinitely many solutions while if $n \geq 4$ this is an over-determined system so we expect no local solutions in the generic case.

*A non-generic case occurs when $\nabla$ is flat. Then $(\Omega^{*}(M;E),d_{\nabla})$ is a complex and we have an infinite dimensional space of "trivial" solutions. Namely, given any section $s \in \Gamma(E) = \Omega^{0}(M;E)$ we can set $A = d_{\nabla}(s)$ and then $d_{\nabla}(A) = d_{\nabla}^2(s) = 0$. Those are the trivial solutions which correspond to $d_{\nabla}$-exact one-forms. Depending on the vector bundle $E$, the topology of $M$ and the connection $\nabla$ there might be additional non-trivial (families) of solutions which correspond to non-zero elements in $H^1(M;E,\nabla)$. 

*If $\nabla$ is not flat, we don't get a complex but we can still extract a necessary condition (but definitely not sufficient condition) for the existence of a one form $A$ with $d_{\nabla}(A) = 0$ because we must have
$$ 0 = d_{\nabla}(0) = d_{\nabla}(d_{\nabla}(A)) = d^2_{\nabla}(A) = R_{\nabla} \wedge A$$
where we think of $R_{\nabla}$ as an element of $\Omega^{2}(M;\operatorname{End}(E))$. More explicitly, we must have
$$ (R_{\nabla} \wedge A)(X,Y,Z) = R(X,Y)(A(Z)) + R(Y,Z)(A(X)) - R(X,Z)(A(Y)) = 0. $$
This imposes ${n \choose 3}k$ algebraic equations on coefficients of $A$ which I suspect are satisfied generically only by $A = 0$ when $n \geq 4$.

*By the functoriality of all the constructions involved, if for some reason we have managed to find a solution $A$ to the equation, we can pull it back using any smooth map to obtain solutions over a different base. That is, if $d_{\nabla^{E}}(A) = 0$ and $f \colon N \rightarrow M$ then $f^{*}(A) \in \Omega^1(N;f^{*}(E))$ satisfies $d_{\nabla^{f^{*}(E)}}(f^{*}(A)) = 0$. Your basic example comes from the fact that if $E = TM$ and $A = \operatorname{id}$ is the tautological one-form then
$$ d_{\nabla}(A)(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y] = T_{\nabla}(X,Y) $$
is the torsion of $\nabla$. Hence, if $\nabla$ is torsion free we get a solution and then for any smooth $f \colon N \rightarrow M$ a solution $f^{*}(\operatorname{id}) = df$ on $(N,f^{*}(E),f^{*}(\nabla))$. Note that in this case, the equation $R_{\nabla} \wedge \operatorname{id} = 0$ is just the Binachi identity for a torsion free connection.


That's all I can say in full generality. Let me try and analyze the case of a trivial line bundle and extract a necessary and sufficient condition for the existence of local solutions. Assume $E = M \times \mathbb{R}$. A connection $\nabla$ on $E$ can be written uniquely as $\nabla = d + \omega$ for some one-form $\omega$ and the curvature of $\nabla$ can be identified with $d\omega$. An endomorphism $TM \rightarrow M \times \mathbb{R}$ can also be identified with a one-form $A$. The equation $d_{\nabla}(A) = 0$ then becomes
$$ dA + \omega \wedge A = 0. $$
A non-zero (local) solution of the equation defines a distribution $\ker(A)$ and the equation above implies that if $A(X) = A(Y) = 0$ then
$$ (dA + \omega \wedge A)(X,Y) = XA(Y) - Y(A(X)) - A([X,Y]) + \omega(X)A(Y) - \omega(Y)A(X) = A([X,Y]) = 0 $$
so the distribution is integrable. Hence, we can choose coordinates $(x^1,\dots,x^n)$ so that $A$ has the form $A = gdx^n$ with $g \neq 0$. Write $\omega = \omega_i dx^i$. Then
$$ 0 = d(dA + \omega \wedge A) = d\omega \wedge A = \sum_{i < j \leq n-1} g\left( \frac{\partial \omega_j}{\partial x^i} - \frac{\partial \omega_i}{\partial x^j} \right) dx^i \wedge dx^j \wedge dx^n $$
which implies that
$$ \frac{\partial \omega_j}{\partial x^i}  = \frac{\partial \omega_i}{\partial x^j} $$
for all $1 \leq i < j \leq n - 1$. This means if we restrict $\omega$ to the leaves $x^n = c$ of the foliation defined by $A$ we get closed one-forms! 

If the equation $dA + \omega \wedge A = 0$ has a solution $A$ on a neighborhood of a point $p$ with $A|_{p} \neq 0$ then there exists a coordinate system $(x^1,\dots,x^n)$ on a possibly smaller neighborhood around $p$ such that the restriction of $\omega$ to the submanifolds $x^n = c$ is closed.

In other words, the existence of a non-zero solution implies that a neighborhood of $p$ is foliated by codimension one submanifolds on which 
that restriction of $E$ to each submanifold (with the induced connection) is flat. Let us call this condition flat in codimension one. Clearly if $(E,\nabla)$ is flat then $d\omega = 0$ and the restriction of $\omega$ to any submanifold is closed so this appears (and we'll see that it is) a weaker condition. 
Conversely, if $E$ is flat in codimension one let us show that we can construct infinitely many solutions to our equation. We will try to construct solutions of the form $A = gdx^n$. We have
$$ dA + \omega \wedge A = \sum_{i=1}^{n-1} \left( \frac{\partial g}{\partial x^i} + g \omega_i \right) dx^i \wedge dx^n = 0 $$
which gives us the $n - 1$ equations
$$ \frac{\partial g}{\partial x^i} + g \omega_i = 0, \,\,\, 1 \leq i \leq n - 1$$
for $g$. Since $\omega$ satisfies the compatibility condition, this system can be solved by
$$ g(x^1,\dots,x^{n-1},c) = \exp \left(-\int_{(0,\dots,0,c)}^{(x^1,\dots,x^{n-1},c)} \omega|_{x^n = c} \right) h(c) $$
where $h$ is an arbitrary function. Geometrically, $h(c) = g(0,\dots,0,c)$ determines the initial value of a section of $E$ on the line $x^1 = \dots = x^{n-1} = 0$ and then we obtain $g(x^1,\dots,x^{n-1},c)$ by parallel transporting the section along any path between $(0,\dots,0,c)$ and $(x^1,\dots,x^{n-1},c)$ which lies on the hyperplane $x^{n} = c$.
Finally, let me give an explicit example for which there are no solutions. While the criterion above gives us a sufficient and necessary condition on $\omega$ for the existence of a local non-zero solution, it is not clear how to check it in practice so let us derive a necessary condition that is easier to check. We have seen that if there exists a non-zero solution $A$ then we can find coordinates $x^1,\dots,x^n$ such that if we write $\omega$ as $\omega = \omega_i dx^i$ we have $\frac{\partial \omega_j}{\partial x^i} = \frac{\partial \omega_i}{\partial x^j}$ for $1 \leq i < j < n$. But then
$$ d\omega = \sum_{i < j} \left( \frac{\partial \omega_j}{\partial x^i} -\frac{\partial \omega_i}{\partial x^j} \right) dx^i \wedge dx^j = \sum_{i=1}^{n-1} \left( \frac{\partial \omega_n}{\partial x^i} -\frac{\partial \omega_i}{\partial x^n} \right) dx^i \wedge dx^n $$
which implies that
$$ d(\omega \wedge d\omega) = d\omega \wedge d\omega = 0$$
because $dx^n$ appears in all the summands of $d\omega$. This condition is trivially satisfied if $n \leq 3$ but if $n \geq 4$, this restricts $\omega$. For example, if $\omega = -ydx + xdy -wdz + zdw$ then $d\omega = 2(dx \wedge dy + dz \wedge dw)$ and $d\omega \wedge d\omega = 8 dx \wedge dy \wedge dz \wedge dw$. For such a connection, there aren't even local non-zero solutions.
