The affine connection is not in general defined uniquely by the smooth structure and the Riemannian metric. Can you give some demonstration with some examples?

  • $\begingroup$ A smooth manifold with an affine connection need not be endowed with a Riemannian metric. If it is, the Levi-Chivita connection is one possibility. See more details see en.wikipedia.org/wiki/Affine_connection. $\endgroup$ Sep 5, 2014 at 13:40
  • $\begingroup$ what is the difference between affine connection and Levi-civita connection? $\endgroup$
    – phy_math
    Sep 5, 2014 at 13:56
  • $\begingroup$ I learned that Levi-civita connection is a torsion-free connection, is it relevant to distinguish them? $\endgroup$
    – phy_math
    Sep 5, 2014 at 14:00
  • $\begingroup$ The Levi-Civita connection preserves a given Riemannian matric. It is unique with this respect. An affine connection can exist without any Riemmannian metric. $\endgroup$ Sep 5, 2014 at 14:01
  • $\begingroup$ Levi-Civita connection is also required to be torsion-free: without this assumption, the Riemannian connection is not unique. $\endgroup$ Sep 5, 2014 at 16:24

1 Answer 1


Take $M = \mathbb{R}^2$ with its standard metric. With respect to the standard coordinates $(x,y)$ each affine connection on $M$ is written as

$$ \nabla = \mathrm{d} + A $$

where $A$ is a 2 by 2 matrix of $1$-forms on $M$. Remarks:

  • $\nabla$ is compatible with the metric if and only if $A$ is skew-symmetric, i.e. $A \in \mathfrak{o}(2)$

  • $\nabla$ is the Levi-Civita connection when $A = 0$

Let $\omega$ be a $1$-form on $M$ and take

$$ A = \begin{pmatrix}0 & \omega \\ -\omega & 0 \end{pmatrix} $$

My computations lead to

$$ \nabla_X Y - \nabla_Y X = [ X,Y ] + T_\omega(X, Y) $$

where, if $X = (X^1, X^2)$ and $Y = (Y^1,Y^2)$,

$$ T_\omega(X,Y) = (\omega(X)\,Y^2 - \omega(Y)\, X^2, -\omega(X)Y^1 + \omega(Y)X^1) $$

For a suitable $\omega$ (e.g. $\omega = \mathrm{d}x$), there are vector fields $X$ and $Y$ such that $T_\omega(X,Y) \neq 0$, thus the corresponding $\nabla$ is a metric connection which is not torsionfree.

  • $\begingroup$ Thanks @Yvoz, your explanation helps me a lot $\endgroup$
    – phy_math
    Sep 6, 2014 at 0:51
  • $\begingroup$ @phy_math Notice that the condition $T_\omega (X,Y) = 0 \,\,\forall X,Y$ (i.e. $\nabla$ torsionfree) is equivalent to $\omega = 0$. Thus the example above provides a proof of the uniqueness, on $\mathbb{R}^2$, of the metric torsionfree connection. $\endgroup$
    – Ivo
    Sep 6, 2014 at 7:57

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