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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 $D^g$ ?

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2 Answers 2

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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 $g$.

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

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