construct a connection such that a given tensor is parallel wrt it Let $\omega$ be a symplectic form on a smooth manifold $M$. How does one construct a connection on $TM$ such that $\omega$ is parallel to it? 
It's easy to construct a connection on a dual bundle that makes a given section of the bundle parallel (if the section is non-vanishing), but I am somewhat confused as to what can one do when one wants to make a tensor parallel.
More generally, given a vector bundle $B$ and tensor $b \in H^0(M,B^{\otimes n})$ when can one construct a connection such that $\nabla(b)=0$?
update: here's my thinking, given an arbitrary connection $\nabla_0$, compute it on $b$ and get a section $\nabla_0 b \in H^0(M, B \otimes \Omega^1)$. Connections being a torsor under $End(B) \otimes \Omega^1$ we would like to add a (form-valued) endomorphism $A$ such that $(\nabla_0 + A) b = 0$. So it seems that we only need to solve the equation
$$
A b = -\nabla_0 b
$$
for $A$ (here the action of $A$ is the diagonal action on $B^{\otimes n}$).
Now I am not quite sure what condition to impose on $b$ to make sure this can be solved. In the most stupid case $n=1$, $b \in B$ one clearly wants $b$ to be non-vanishing when $\nabla_0 b$ vanishes. What would one require for tensors?
 A: Let $(M,\omega)$ be our symplectic manifold. We want a (symplectic) connection $\nabla$ such that


*

*it is torsionless, $T(X,Y)=\nabla_X Y-\nabla_Y X -[X,Y]=0,$ and

*it is parallel, $(\nabla_X \omega)(Y,Z)=X(\omega(Y,Z))-\omega(\nabla_X Y, Z)-\omega(Y,\nabla_X Z)=0.$


To prove that it exists, we can mimic a Levi-Civita connection and define $$\nabla_Y X = \tilde\nabla_Y X + \frac{1}{2}[S(X,Y)+S(Y,X)]$$
where $S$ is some $(1,2)$-tensor, and we have symmetrized it to make everything similar to the Levi-Civita case. We pick $\tilde\nabla$ to satisfy $$\tilde\nabla_X \omega(Y,Z)=a\,\omega(S(X,Y),Z)$$
with $a\in\mathbb R$ a constant. 
Now we can go back to check the parallel condition, switch to this "Levi-Civita" connection and compute: 
$$(\nabla_X \omega)(Y,Z)=X(\omega(Y,Z))-\omega(\nabla_X Y, Z)-\omega(Y,\nabla_X Z)\\=\Big(X(\omega(Y,Z))-\omega(\tilde\nabla_X Y, Z)-\omega(Y,\tilde\nabla_X Z)\Big)-\frac{1}{2}\omega(S(X,Y),Z)-\frac{1}{2}\omega(S(Y,X),Z)-\frac{1}{2}\omega(Y,S(X,Z))-\frac{1}{2}\omega(Y,S(Z,X))
\\=(\tilde\nabla_X\omega)(Y,Z)-\frac{1}{2}\omega(S(X,Y),Z)-\frac{1}{2}\omega(S(Y,X),Z)-\frac{1}{2}\omega(Y,S(X,Z))-\frac{1}{2}\omega(Y,S(Z,X))$$
This can be rearranged if we note that due to skew-symmetry $\omega(S(X,Y),Z)=\frac{1}{a}\tilde\nabla_X\omega(Y,Z)=-\frac{1}{a}\tilde\nabla_X\omega(Z,Y)=-\omega(S(X,Z),Y)$. The closedness of $\omega$, $\mathrm d\omega=0$ implies the Bianchi identity $$\sum_{cycl.}\tilde\nabla_X\omega(Y,Z)=\sum_{cycl.}\omega(N(X,Y),Z)=0.$$
So, using both:
$$a\,\omega(S(X,Y),Z)-\frac{1}{2}\omega(S(X,Y),Z)-\frac{1}{2}\omega(S(Y,X),Z)-\frac{1}{2}\omega(Y,S(X,Z))-\frac{1}{2}\omega(Y,S(Z,X))\\=(a-\frac{1}{2}-\frac{1}{2})\omega(S(X,Y),Z)+\frac{1}{2}\omega(S(X,Z),Y)=0,$$
if we pick $a=\frac{3}{2}$ and transform the last term again.
However, if $S$ is symmetric and $\nabla$ is symplectic, such that for some $\bar\nabla$ we have (mimicking the previous calulation) $$\bar\nabla_X\omega(Y,Z)=\nabla_X\omega(Y,Z)-\omega(S(X,Y),Z)-\omega(Y,S(X,Z))=-\omega(S(X,Y),Z)+\omega(S(X,Z),Y)$$
This vanishes if $\omega(S(\cdot,\cdot),\cdot)$ is symmetric in the last two arguments, as well as the first two, meaning that it is a fully symmetric tensor. So, high nonuniqueness.
Note that we depended on $\omega$ being a closed 2-form. For an n-form $b$, we have
$$\nabla_{X_0}b(X_1,\dots,X_n)=X_0( b(X_1,\dots,X_n))-\sum_{j=1}^n b(X_1,\dots,\nabla_{X_0}X_j,\dots,X_n)$$ which implies using 
$$\nabla_{X_0}X_j=\nabla_{X_0}X_j+\text{Sym } S(X_0,X_j)$$ (where I abbreviated just here using new notation):
$$\nabla_{X_0}b(X_1,\dots,X_n)=\tilde\nabla_{X_0}b(X_1,\dots,X_n)-\frac{1}{2}\sum_{j=1}^n b(X_1,\dots,S(X_0,X_j)+S(X_j,X_0),\dots,X_n)
//=a\,b(S(X_0,X_1),X_2,\dots,X_n)-\frac{1}{2}\sum_{j=1}^n b(X_1,\dots,S(X_0,X_j)+S(X_j,X_0),\dots,X_n).$$
I don't have time to pursue this calculation further now. We have once again $$b(\dots,S(X_0,X_j),X_{j+1},\dots)=-b(\dots,S(X_0,X_{j+1}),X_j,\dots)$$
etc. I might attempt it another day, or the reader could. Hope this helped!
A: upon some thinking here's what I have arrived at (continuing from the update).
In order for an endomorphism $A$ satisfying $A b = - \nabla_0 b$  to exist, $b(x)$ ought to belong to the the same orbit of the action of $GL(V)$ on $B_x^{\otimes n}$ for all $x$ (here we identify various $B_x$ using parallel transport induced by $\nabla_0$). In particular, the stabilizer group of $b(x)$ ought to be the same (up to conjugation) for all $x$.
For a symplectic form $\omega$ this group is isomorphic to $Sp(n) \subset GL_n$ at every point $x \in M$, so the condition is satisfied.
