# Does a connection satisfying $(\nabla_X g)(Y,Z)-(\nabla_Y g)(X,Z)=0$ have a special name or satisfy a special property?

Let $$\nabla$$ be a flat torsion-free connection on a smooth manifold. Let $$g$$ be a metric on $$M$$ ($$\nabla$$ may not be the Levi-Civita connection of $$g$$). Suppose that

$$(\nabla_X g)(Y,Z)-(\nabla_Y g)(X,Z)=0.$$ Does $$\nabla$$ receive a special name? Or does it imply $$\nabla$$ is related with the Levi-Civita connection or any other thing stronger than the above relation itself? Maybe the existence of some coordinates, I don't know.

I ask this because in Dombrowski's paper On the geometry of the Tangent Bundle, it is proved that the presymplectic form defined as

$$\omega(X,Y)=g_S(X,JY),$$

where $$J$$ is the canonical complex form on $$TM$$ defined using the split induceb by $$\nabla$$ and $$g_S$$ denotes the Sasaki metric defined in terms of the same split and the metric $$g$$, is closed whenever $$\nabla$$ is the Levi-Civita connection.

However, the computation for the general case shows that the $$d\omega$$ is proportional to the above factors when evaluated on some vector fields. But I'm sure the $$\omega$$ is closed. In fact, I would say, under certain conditions, it is the pullback of the canonical 2-form on $$T^*M$$ along $$g$$. But I don't know exactly what conditions are necessary and if they are related to my expression above.

I typed the question on the phone. Let me describe in detail Dombrowski's construction now I have a computer avaible.

It is well known every linear connection on $$M$$ defines an split on $$TTM$$ into vertical and horizontal subbundles. Each of them is isomorphic to $$TM$$ and for every vector field $$Z\in\Gamma(TTM)$$ there are vector fields $$X,Y\in\Gamma(TM)$$ such that $$Z=X^h+Y^v$$, where the superscripts v and h denote the verticla and the horizontal litf, respectively. Hence, every tensor on $$TM$$ is completely characterised once we know how it acts on horizontal and vertical vector fields.

Now, Dombroski defines a complex tructure $$J:TTM\rightarrow TTM$$ as

$$JX^h, X^v, \qquad JX^v= - X^h.$$

He proves $$J$$ is integrable if and only if $$\nabla$$ is flat and torsion-free. But in the third appendix we goes beyon. The aim is to prove $$TM$$ is a Kähler. For that he considers a Riemannian metric $$g$$ on $$M$$ (again the flat and torsion-free connection $$\nabla$$ may not be the Levi-civita connection for $$g$$) and define the Sasaki metric

$$g_S(X^h,Y^h)=g_S(X^v,Y^v)= \pi_{TM}\circ g(X,Y), \qquad g_S(X^h,Y^v)=g_S(X^v,Y^h)=0$$

(Notice I compose on the right and I denote the differential of a map $$f$$ as $$Tf$$). It is not difficult to see that $$J$$ leaves $$g_S$$ invariant. Hence $$\omega(X,Y)=g_S(X,JY)$$ is 2-form.

Finally, he particularise for the case when $$\nabla$$ is the Levi-Civita connection for $$g$$. In this case he proves $$\omega$$ is locally and non-locally an exact 2-form. But the computation can be done for a non-metric connection. In that case one finds the relation I have written above.

I claim under certain conditions, $$\omega$$ is the pullback along $$g$$ of the canonical 2-form defined on $$T^*M$$. For that, let me write firstly the pullback of the canonical 1-form:

$$g^*\theta(Z)|_{p,X}= g_p(X,(T\pi)_{(p,X)}Z), \qquad Z\in T_{(p,X)}TM.$$

In coordinates $$g^*\theta$$ reads simply as $$y_ig_{ij}d\tilde{x}_j$$. If I compute the diferential I get

$$d(g^*\theta)= -g_{ij}d\tilde{x}_i\wedge dy_j -y_i\frac{\partial g_{ij}}{\partial x_k} d\tilde{x}_i\wedge d\tilde{x}_k,$$ while $$\omega$$ reads in coordinates as $$-g_{ij}d\tilde{x}_i\wedge dy_j$$ (or the opposite, I'm not 100% sure about the sign). Since the connection has not been used, it is logic to expect $$dg^*\theta$$ cannot match always with the above $$\omega$$. However, there are certain conditions under it should. What are they? Is it related to the relations I am asking for?

• Here there is a usefull paper named Almost Hermitian structures on tangent bundles. Its main theorem proves that $\omega$ is closed iff the dual connection of $\nabla$ is torsion-free, and $(TM,J,g_S,\omega)$ is Kähler iff both $\nabla$ and its adjoint are flat and torsion-free, which the author says it is eqivalent to have a Hessian metric on $M$. I will try to write an answer later, but meanwhile feel free to add your own answers or post bibliography about all this stuff (Hermitian manifolds, adjoint of a connection...). – Dog_69 May 3 '19 at 21:12
• I had forgotten to say the mpst important thing: My expression, the one I'm asking for, turns out to be the torsion tensor (with the covariabt index lowered with the metric $g$ on $M$) of the adjoint connection of $\nabla$. Namaley, if we denote this adjoint by $\nabla^*$, then: $$g(T^{\nabla^*}(X,Y),Z)=(\nabla_Xg)(Y,Z)-(\nabla_Yg)(X,Z) +g(T^\nabla(X,Y),Z)$$ (formula (2.3) of the paper I have cited above). I should also have mentioned that when the torsion of $\nabla^*$ vanishes, not only $\omega$ is integrable but also it is the pullback of the canonical 2-form on $T^*M$ by $g$ (Remark 1.2). – Dog_69 May 3 '19 at 22:26

I'm not familiar with the specific construction you're studying, I don't have a direct answer, but this will be too long for a comment. If $$E \to M$$ is a vector bundle with connection $$\nabla$$, you can define a exterior derivative $${\rm d}^\nabla\colon \Omega^k(M;E) \to \Omega^{k+1}(M;E)$$ acting on $$E$$-valued differential forms, with the aid of $$\nabla$$. Namely, in the usual formula for the exterior derivative, everytime a field $$X_i$$ is acting on a smooth function, you write $$\nabla_{X_i}$$ instead. So, e.g., $${\rm d}^\nabla\alpha(X,Y) = \nabla_X\alpha(Y) - \nabla_Y\alpha(X) - \alpha([X,Y]),$$and so on. Note that we do not have $${\rm d}^\nabla \circ {\rm d}^\nabla = 0$$ is $$\nabla$$ is not flat. Any $$B \in \Gamma(T^*M\otimes E^*)$$ (that is, is $$B\colon \mathfrak{X}(M)\times \Gamma(E) \to \mathscr{C}^\infty(M)$$ is $$\mathscr{C}^\infty(M)$$-bilinear) can be seen as a $$E^*$$-valued $$1$$-form, by $$X \mapsto B(X,\cdot),$$and so it makes sense to talk about the exterior derivative of that. We have $$({\rm d}^\nabla B)(X,Y)\psi = (\nabla_XB)(Y,\psi) - (\nabla_YB)(X,\psi),$$whenever we use any auxiliary torsion-free connection in $$TM$$ to form covariant derivatives of $$B$$. When $$E = TM$$, we say that $$B$$ is a Codazzi tensor if $${\rm d}^\nabla B = 0$$. So in your case, you're looking for the connections which make the metric itself a Codazzi tensor.