# About the term $-\nabla_{[u,v]}w$ in the definition of Riemann curvature tensor

As we know, in the definition of Riemann curvature tensor, we require

$$R(u,v)w=\nabla_u\nabla_v w-\nabla_v\nabla_u w-\nabla_{[u,v]}w$$

Could somebody tell me why we need $-\nabla_{[u,v]}w$ appearing in this definition. Is there any geometric meaning in it? Because the geometric meaning of $\nabla_u\nabla_v w-\nabla_v\nabla_u w$ is pretty clear, but for $-\nabla_{[u,v]}w$ I feel not so direct.

In my opinion, because $[u,v]=\cal L_u v$, then $-\nabla_{[u,v]}w=-\nabla_{\cal L_u v}w$. So it looks like that this term is kind of correction of $\nabla_u\nabla_v w-\nabla_v\nabla_u w$, I mean, to neutralize the effects of the vector fields. This is a very naive hunch. Could you give me a more clear answer?

If $u,v,w$ are the vector fields $\partial_i,\partial_j,\partial_k$ then the term $\nabla_{[\partial_i,\partial_j]}\partial_k$ vanishes. This is because $[\partial_i,\partial_j]=0.$

However, this doesn't hold for arbitrary vectors. Note that in the Riemann curvature tensor there are involved second derivatives of a vector. So:

$$\nabla_u (\nabla_v w)=(\nabla_u \nabla_v)w+\nabla_{\nabla_uv}w,$$ or equivalently

$$(\nabla_u \nabla_v)w=\nabla_u (\nabla_v w)-\nabla_{\nabla_uv}w.$$

That is,

$$(\nabla_u \nabla_v)w-(\nabla_v \nabla_u)w=\nabla_u (\nabla_v w)-\nabla_{\nabla_uv}w-\nabla_v (\nabla_u w)+\nabla_{\nabla_vu}w \\ =\nabla_u (\nabla_v w)-\nabla_v (\nabla_u w)-\nabla_{[u,v]}w.$$

• Cheers, it looks good for me, but I don't know why $\nabla_u(\nabla_v w)=(\nabla_u\nabla_v) w+ \nabla_{\nabla_u v} w$. Oct 24, 2014 at 6:00

It is necessary, because $\nabla_u\nabla_v w-\nabla_v\nabla_u w$ is not tensor.

• It's a good point. Thanks for your advice. Oct 25, 2014 at 8:54