Second Bianchi identity

Question

This is q. 7 of ch. 4 from Do Carmo's book on Riemannian Geometry . Prove that: $$ \nabla R(X,Y,Z,W,T) + \nabla R(X,Y,W,T,Z) + \nabla R(X,Y,T,Z,W)=0.$$

Let $\{e_i\}$ a geodesic frame on $p$ , it is enough to prove that the identity for :

$$ \nabla R(e_i,e_j,e_k,e_l,e_h) + \nabla R(e_i,e_j,e_l,e_h,e_k) + \nabla R(e_i,e_j,e_h,e_k,e_l)=0.$$

The author says that the left side expression is :

$$ R(e_l,e_h,\nabla_{e_k} e_i,e_j) + R(e_h,e_k,\nabla_{e_l} e_i,e_j) + R(e_k,e_l,\nabla_{e_h} e_i,e_j).$$

I'd like to know why this is indeed the case.(you can check that: $$\nabla R(e_i,e_j,e_k,e_l,e_h)=\langle \nabla_{e_h} \nabla_{e_l} \nabla_{e_k}e_i - \nabla_{e_h} \nabla_{e_k} \nabla_{e_l}e_i +\nabla_{e_h} \nabla_{[e_k,e_l]}e_i , e_j \rangle).$$

Answer 1

This is indeed a rather cumbersome calculation. Here it goes:

$$\nabla R(e_i,e_j,e_k,e_l,e_h)(p)+\nabla R(e_i,e_j,e_h,e_k,e_l)(p)+\nabla R(e_i,e_j,e_l,e_h,e_k)(p)=$$

$$\langle \nabla_{e_h} \nabla_{e_l} \nabla_{e_k}e_i - \nabla_{e_h} \nabla_{e_k} \nabla_{e_l}e_i +\nabla_{e_h} \nabla_{[e_k,e_l]}e_i$$

$$+\nabla_{e_l} \nabla_{e_k} \nabla_{e_h}e_i - \nabla_{e_l} \nabla_{e_h} \nabla_{e_k}e_i +\nabla_{e_l} \nabla_{[e_h,e_k]}e_i$$

$$+\nabla_{e_k} \nabla_{e_h} \nabla_{e_l}e_i - \nabla_{e_k} \nabla_{e_l} \nabla_{e_h}e_i +\nabla_{e_k} \nabla_{[e_l,e_h]}e_i, e_j \rangle (p) =$$

$$ \langle R(e_l,e_h)\nabla_{e_k}e_i-\nabla_{[e_l,e_h]}\nabla_{e_k}e_i+R(e_k,e_l) \nabla_{e_h}e_i -\nabla_{[e_k,e_l]}\nabla_{e_h}e_i+R(e_k,e_h)\nabla_{e_l}e_i-\nabla_{[e_k,e_h]}\nabla_{e_l}e_i+\nabla_{e_h} \nabla_{[e_k,e_l]}e_i+\nabla_{e_l} \nabla_{[e_h,e_k]}e_i+\nabla_{e_k} \nabla_{[e_l,e_h]}e_i,e_j\rangle (p)$$

$$= \langle R([e_l,e_h],e_k)e_i-\nabla_{[[e_l.e_h],e_k]}e_i+R([e_k,e_l],e_h)e_i-\nabla_{[[e_k.e_l],e_h]}e_i+R([e_k,e_h],e_l)e_i-\nabla_{[[e_k.e_h],e_l]}e_i,e_j\rangle(p) =$$

$$\langle R(\nabla_{e_l}e_h-\nabla_{e_h}e_l,e_k)e_i + R(\nabla_{e_k}e_l-\nabla_{e_l}e_k,e_h)e_i+R(\nabla_{e_k}e_h-\nabla_{e_h}e_k,e_l)e_i- \nabla_{[[e_l.e_h],e_k]+[[e_k.e_l],e_h]+[[e_k.e_h],e_l]}e_i,e_j\rangle(p) = 0$$

In this last line the Jacoby Identity and the symmetry of the connection is used. Observe that every time that elements of the form $\nabla_{e_l}e_h$ are dropped, what is being used is that of course $\nabla_{e_l}e_h(p)=0$, and the tensorial property of the Riemannian tensor (observe also the necessity that everything must be evaluated on p). Since p is arbitrary, and because of the linearity of everything involved, the identity is proven.