# Relation between Levi-Civita connection and any another metric connection.

On a Riemannian manifold $$(M,g),$$ we have $$D^g,$$ Levi-Civita connection, the only connection that is metric (i.e. $$D^g g=0$$) and without torsion (i.e. $$T^{D^g}=0$$).

My question is that if I have say $$\nabla$$ another connection on $$(M,g)$$ that is metric (i.e. $$\nabla g =0$$) and it has torsion (otherwise $$\nabla=D^g$$) is there any relation between $$\nabla$$ and $$D^g$$ ?

The difference between two connections on $$TM$$ is a tensor which eats two vector fields and spits out one vector field. On the other hand, the sum of a connection and such a tensor is another connection. Hence, fixing a connection $$\nabla^0$$, the space of connections is identified with the space of tensors of type $$(1,2)$$ through the map $$\nabla\mapsto S(\nabla):=\nabla-\nabla^0.$$

Now, if the connection we fix, $$\nabla^0$$, is compatible with a Riemannian metric $$g$$, then a connection $$\nabla$$ is also compatible with the metric if and only if the tensor $$S=S(\nabla)$$ satisfies the condition $$g(S(X,Y),Y)=0$$for any two vector fields $$X$$ and $$Y$$. Indeed, if $$\nabla g=0$$, then\begin{align}g(S(X,Y),Y)&=g(\nabla_XY,Y)-g\left(\nabla^0_XY,Y\right)\\ &=\frac{1}{2}X(g(Y,Y))-\frac{1}{2}X(g(Y,Y))\\&=0.\end{align} On the other hand, suppose $$\nabla$$ satisfies the above condition and let $$X$$ and $$Y$$ be vector fields such that $$\nabla_XY=0$$ (maybe at a point). Then \begin{align}X(g(Y,Y))&=2g\left(\nabla^0_XY,Y\right)\\&=2g\left(\nabla_XY,Y\right)\\&=0,\end{align} and this means that parallel translation with respect to $$\nabla$$ preserves $$g$$.

• I was expecting a formula involving $D^g$ and $\nabla$ and maybe its torsion but thanks anyway for the answer! Apr 4, 2019 at 21:29
• @HurjuiIonut Well, as you can see beyond any doubt, the condition is on the difference between the two connections. Apr 4, 2019 at 21:33

Let $$\nabla$$ be any linear connection on the manifold $$M$$ with torsion $$T(X, Y) = T^{\nabla}(X, Y) = \nabla_X Y - \nabla_Y X - [X,Y] \tag{1}$$

Notice that $$T(Y,X) = - T(X,Y)$$, which in the abstract index notation can be stated as $$T_{(a b)}{}^{c} = 0$$. Here, for a tensor $$t_{ab}$$, I denote by $$t_{(ab)} = \tfrac{1}{2}(t_{a b} + t_{b a})$$ its symmetric part, and by $$t_{[ab]} = \tfrac{1}{2}(t_{a b} - t_{b a})$$ its anti-symmetric part. Thus, any tensor $$t_{ab}$$ can be represented as $$t_{ab} = t_{(a b)} + t_{[a b]}$$. When $$t_{(ab)} = 0$$, tensor $$t_{a b}$$ is anti-symmetric. Similarly, we can talk about symmetries over more than two indices.

Fact. The connection $$\nabla'$$ given by $$\nabla'_X Y = \nabla_X Y - \tfrac{1}{2}T(X,Y) \tag{2}$$ is torsion-free.

Proof. Let us compute the torsion $$T' = T^{\nabla'}$$: \begin{align} T'(X, Y) & = \nabla'_X Y - \nabla'_Y X - [X,Y] \\ & = \nabla_X Y - \tfrac{1}{2}T(X,Y) - \nabla_Y X + \tfrac{1}{2}T(Y,X) - [X,Y] \\ & = \nabla_X Y - \nabla_Y X - [X,Y] - 2\cdot\tfrac{1}{2}T(X,Y) = 0 \end{align}

The following fact, proposed in the answer by Amitai Yuval, we can to reformulate in an equivalent form (the equivalence is easily established using polarization).

Fact. The difference tensor $$S(X,Y) = \nabla_X Y - \nabla'_X Y$$ of two metric connections $$\nabla$$ and $$\nabla'$$ is anti-symmetric in the two last indices: $$S_{a(bc)} = 0$$

Proof. Subtracting the identities expressing the metricity of connections $$\nabla'$$ and $$\nabla$$ \begin{align} X g(Y,Z) - g(\nabla'_X Y, Z) - g(Y, \nabla'_X Z) & = 0 \\ X g(Y,Z) - g(\nabla_X Y, Z) - g(Y, \nabla_X Z) & = 0 \end{align} we get $$g(\nabla_X Y - \nabla'_X Y, Z) + g(Y, \nabla_X Z - \nabla'_X Z) = 0$$ or $$g(S(X,Y), Z) + g(S(X,Z), Y) = 0 \tag{*}$$

Rewriting the identity (*) in the abstract index notation, we obtain $$g_{c d} X^a Y^b S_{a b}{}^c Z^d + g_{c d} X^a Z^b S_{a b}{}^c Y^d = 0 \tag{**}$$

Introducing the difference tensor in the covariant form as $$S_{a b c} = S_{a b}{}^d g_{c d}$$ we can rewrite (**) as $$S_{a b c} X^a Y^b Z^c + S_{a b c} X^a Z^b Y^c = 0$$ or, renaming the indices, as $$S_{a b c} X^a Y^b Z^c + S_{a c b} X^a Y^b Z^c = 0 \\$$

Factoring out $$X^a Y^b Z^c$$, $$(S_{a b c} + S_{a c b}) X^a Y^b Z^c = 0$$ and taking into account that $$X$$, $$Y$$, and $$Z$$ are arbitrary, we obtain the claim.

Combining these two facts together, we can formulate the following relation sought in the question:

Theorem. Let $$g$$ be a Riemannian metric on a smooth manifold $$M$$, and $$\nabla$$ be a metric connection ($$\nabla g = 0$$) with torsion $$T$$ (see equation (1)). Then the connection $$\nabla'$$ as in equation (2) is the Levi-Civita connection for metric $$g$$ if and only if the torsion tensor $$T$$ is totally anti-symmetric: $$T_{(ab)c} = 0 \\ T_{a(bc)} = 0 \\ T_{(a|b|c)} = 0$$

The curious can find some similar observation in [1]. I have no association with this article, though.

It would be nice to hear in the comments, if there are other, deeper, relations known.

References.

[1]. D. Lindstrom, H. Eckardt, M. W. Evans. On Connections of the Anti-Symmetric and Totally Anti-Symmetric Torsion Tensor, August 5, 2016, available here