Why is the curvature tensor defined to be anti-symmetric?

Question

I'm reading Chern's Lectures on Differential Geometry. On page 117 of section 4-2, he points out that \begin{eqnarray} d\omega_i^j - \omega_i^h \wedge\omega_h^j &=& \frac{\partial\Gamma^j_{ik}}{\partial u^l}du^l\wedge du^k - \Gamma^h_{il}\Gamma^j_{hk} du^l \wedge du^k \\ &=& \frac{1}{2}\left(\frac{\partial\Gamma^j_{il}}{\partial u^k} - \frac{\partial\Gamma^j_{ik}}{\partial u^l}+\Gamma^h_{il}\Gamma^j_{hk} - \Gamma^h_{ik}\Gamma^j_{hl}\right)du^k\wedge du^l. \end{eqnarray}

Recall the expression on the left are the components of the connection matrix, which uniquely represents the connection itself.

He then defines the curvature tensor $R^j_{ikl}$ to be the expression in parentheses.

Clearly, $R^j_{ikl}$ is anti-symmetric in $k$ and $l$. For any tensor $Q^j_{ikl}$ that is symmetric in $k$ and $l$, we also have $$ d\omega_i^j - \omega_i^h \wedge\omega_h^j = \frac{1}{2}\left(R^j_{ikl} + Q^j_{ikl}\right)du^k\wedge du^l. $$

This begs the question: why is the curvature tensor defined to be $R^j_{ikl}$ and not $R^j_{ikl} + Q^j_{ikl}$ for arbitrary symmetric $Q^j_{ikl}$? If we took $R^j_{ikl} + Q^j_{ikl}$ as the definition, there would of course be curvature tensors, rather than a uniquely defined curvature tensor.

Is there anything, other than convenience, that suggests defining the curvature tensor to be anti-symmetric? Does its anti-symmetry lead to great simplifications or important theorems that would otherwise be false if we were not to constrain it to be anti-symmetric?

Answer 1

Actually, what Ted Shifrin wrote results in an anti-symmetry in the indices $i$ and $j$ in $R^j_{ikl}$ (for metric compatible connections), but Will Nelson asked about the other two. And this is quite easy to explain.

The reason for those to be anti-symmetric is simply that we are dealing with differential forms here. Note that for every $i$ and $j$ $\Omega^j_i = d\omega_i^j - \omega_i^h\wedge\omega_h^j$ is a differential 2-form, hence an anti-symmetric tensor with components $(\Omega^j_i)_{kl} = R^j_{ikl}$ (with respect to the tensor basis $du^k \otimes du^l$). Writing this with basis forms gives an additional factor of two and we end up with the formula $$ \Omega^j_i = \frac 1 2 R^j_{ikl} du^k \wedge du^l. $$ Note that there is no ambiguity in here! This is simply an expansion in a basis which is of course unique. I think you got confused because of the Einstein notation which is used here. Of course it looks like you can "add" a symmetric tensor in this notation, but the expansion actually stays they same.

To see the point here consider this example in two dimensions: For every anti-symmetric 2-tensor $A$ and symmetric 2-tensor $S$ we have $$ \frac 1 2(A_{kl}+S_{kl}) du^k \wedge du^l = \frac 1 2(A_{12} + S_{12}) du^1 \wedge du^2 + \frac 1 2(A_{21} + S_{21}) du^2 \wedge du^1 = A_{12} du^1\wedge du^2. $$ So the component of $\frac 1 2(A_{kl}+S_{kl}) du^k \wedge du^l$ in this basis is $A_{12}$. Loosely speaking, the LHS is not a "proper" linear combination since basis forms are appearing twice.

Answer 2

The structure group for a Riemannian structure is $O(n)$. The curvature 2-form matrix naturally takes values in the Lie algebra of the structure group. Here $\mathfrak o(n)$ consists of all skew-symmetric matrices. I guess I should add the comment that there are plenty of connections (not compatible with the Riemannian structure in the case of the tangent bundle, or on a more general vector bundle) whose curvature is not skew-symmetric.