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Let $(M, g)$ be a Riemannian manifold. Is it possible to construct two different affine (or metric) connections, say $\nabla$ and $\nabla'$, which induce the SAME curvature tensor, i.e. $R(X, Y)Z=R'(X, Y)Z$ for any $X, Y, Z\in\Gamma(TM)$?
  I am not interested about answers related to metrics conformal to $g$. Are there any references related to this question? Thank you very much.

Let $(M, g)$ be a Riemannian manifold. Is it possible to construct two different affine (or metric) connections, say $\nabla$ and $\nabla'$, which induce the SAME curvature tensor, i.e. $R(X, Y)Z=R'(X, Y)Z$ for any $X, Y, Z\in\Gamma(TM)$?
  Are there any references related to this question? Thank you very much.

Let $(M, g)$ be a Riemannian manifold. Is it possible to construct two different affine (or metric) connections, say $\nabla$ and $\nabla'$, which induce the SAME curvature tensor, i.e. $R(X, Y)Z=R'(X, Y)Z$ for any $X, Y, Z\in\Gamma(TM)$? I am not interested about answers related to metrics conformal to $g$. Are there any references related to this question? Thank you very much.

Source Link
314159.
314159.
  • 1.2k
  • 1
  • 15
  • 24

connections and curvature

Let $(M, g)$ be a Riemannian manifold. Is it possible to construct two different affine (or metric) connections, say $\nabla$ and $\nabla'$, which induce the SAME curvature tensor, i.e. $R(X, Y)Z=R'(X, Y)Z$ for any $X, Y, Z\in\Gamma(TM)$?
Are there any references related to this question? Thank you very much.