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We know that the Riemann tensor is antisymmetric with respect to the first two vectors (the vectors that we parallel transport the third vector around the parallelogram made by their integral curves).

But what is the geometric interpretation? Why do we have to expect that by changing the order of parallel transportation we will have the same vector with opposite orientation?

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up vote 13 down vote accepted

This is a re-flavouring of Alexander's answer but in a language I prefer.

Take two vectors $v,w \in T_p N$, and consider the `rectangle' $exp(xv+yw)$ where $0 \leq x \leq a$ and $0 \leq y \leq b$. The holonomy around the boundary of this rectangle is an orthogonal transformation of $T_p N$, and it looks approximately like

$$ Hol \simeq Id_{T_p N} + ab R(v,w) $$

where the approximation indicates the 2nd order taylor expansion of the holonomy with respect to the variables $a$ and $b$. Here $R(v,w)$ is the Riemann curvature tensor.

So from this point of view, the reason why it's anti-symmetric in the variables $v,w$ is that if you switch $v$ and $w$ you are essentially reversing the orientation of the rectangle you're computing the holonomy over, so the corresponding linear transformation (holonomy) is the inverse, which corresponds to negating the linear and quadratic parts of the Taylor expansion.

This is also the reason why the Riemann curvature tensor is skew-symmetric:

$$\langle R(v,w)z, y \rangle + \langle z, R(v,w)y \rangle = 0 $$

since the tangent space to the orthogonal group is consists of all the skew-symmetric linear transformations.

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@Ryan: Actually, the reason you give in your final paragraph is not related to the reason you give in the rest of your argument. One would have $R(v,w)=-R(w,v)$ for any connection whether it was Riemannian or not, and for the reason you give: Reversing the path will invert the holonomy transformation. However, only when you know that the holonomy preserves the metric will you know that it is an orthogonal transformation, so that the skew-symmetry of the tangent space to the orthogonal group comes into play, which is a different reason from the first, and what the OP is really asking about. – Robert Bryant Feb 10 '13 at 14:10
In case the OP doesn't know some of the sophisticated terminology and notation in the answer (I don't), here's what I think is the same explanation done with crayons. Say the vectors $v$ and $w$ looked like this placed tail to tail: |____. Changing $w$ to $-w$ gives this: ____|. Parallel transport takes you around the new rectangle in the opposite handedness. Since tensors are linear, flipping the sign on $w$ flips the sign of the result. Therefore reversing the direction of parallel transport flips the sign. Swapping $v$ and $w$ also reverses the direction, so it must also flip the sign. – Ben Crowell Feb 10 '13 at 15:22
@Robert: That wasn't the "this" I was referring to. :) I suppose I wrote my response too late in the evening. What I was getting at was that this interpretation forces us to think of the values of $R(v,w)$ as tangent vectors in the group of linear automorphisms of $T_p N$. – Ryan Budney Feb 10 '13 at 17:55
thanks Ryan . that is great .but I am really interested in reading more about what you said.where I can find these stuff? – DAVID Feb 11 '13 at 7:38
This argument appears around page 703 in Marcel Berger's "Panoramic View of Riemannian Geometry". My version of it is an elaboration on a fairly standard perspective on the curvature tensor, but most DG texts kind of skimp on it. Berger states it but does not go into as much detail as I describe above. I'll have some DG lecture notes up on my website in summertime where these kinds of details will be spelled-out in elaborate detail -- I'm teaching a DG course this semester so I'm thinking about these kinds of expositional issues at the moment. – Ryan Budney Feb 11 '13 at 8:48

More interpretations:

On (pseudo-) Riemannian manifolds: the numerator of sectional curvature $-\langle R(X,Y)X,Y\rangle$ is a symmetric bilinear form on the space of skew-symmetric bivectors. Skew symmetric bivectors describe measured 2-planes in the tangent space. Curvature in the form of $\langle R(X,Y)Z,W\rangle$ can be recomputed by polarization from this.

Another, more general point of view: On the (orthonormal) frame bundle, curvature is a 2-form with values in the Lie algebra of the structure group: $\Omega=d\omega+\omega\wedge\omega$ for matrix valued forms. This ties in well with the fact that curvature is the obstruction against integrability of the horizontal subbundle of $TTM$.

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Thanks , that is great and helpful. – DAVID Feb 11 '13 at 7:37

Probably the intuition should be that curvature is the commutator of infinitesimal parallel transports, so it is antisymmetric for the same reason that the Lie bracket is.

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