I stumbled upon the following claim in a paper:

"We write the (Bochner) Laplacian in suffix notation: $\Delta_B = \nabla ^k \nabla_k$".

after this statement, the following is written: ($M$ is a riemannian manifold, $w:M\rightarrow \mathbb{R}$)

$\Delta_Bw=0 \stackrel{(*)}{\Rightarrow} \Delta_Bd_iw \stackrel{(1)}{=} \nabla^k\nabla_i d_kw \stackrel{(2)}{=} (Ric)_i^kd_kw$ where $Ric$ is the Ricci Curvature.

(note that the Ricci Curvature is (2,0)-tensor and the notation used here is of (1,1)-tensor, but this is not rellay important, since via the metric we can raise or lower indices as we wish)

My questions:

(1) What does $\nabla^k, \nabla_k,d_i$ mean? (I guess $d_iw$ is the derivative of $w$ with respect to the i-th coordinate). The meaning of these notations is not clarified in the paper.

I think there must be only one answer to that based on equality (2).

(2) Why the above equalities ((1),(2)) hold? In particular I do not understand the implication $\stackrel{(*)}{\Rightarrow}$.

(This index notation reminds me of taking trace, and I know that by definition $\Delta_B = \nabla ^* \nabla$ where $\nabla$ is a connection on the relevant bundle (here the trivial line bundle over $M$) and $\nabla^*$ is its formal adjoint etc..... but I am still confused about the indices notation)


The implication is a consequence of Ricci identity. The notations $\nabla_k$ is standard: Let, in general, that $A$ is a $(p, q)$-tensor, one can define a $(p, q+1)$ tensor $\nabla A$. In your specific example, there are two tensors:

  • The function, $\omega$, which is a $(0,0)$ tensor, and

  • It's exterior derivative $d\omega$, which is a one form (that is, a $(0,1)$-tensor). In your question you used $d_i\omega$ to denote the component of $d\omega$. That is, $(d\omega)_i := d_i\omega$. And you are right that $d_i\omega$ is the partial derivative of $\omega$ with respect to the $\partial _i$ direction. However, it is more convenient to consider your equation as an equation on tensor, cause this is how people write tensor equation, and this interpretaion make it easier to explain $\nabla_k$.

So now you have $\nabla d\omega$, which is a $(0,2)$-tensor. $\nabla_k d_i\omega$ is by definition the $(i, k)$ component of this tensor. Note that $\nabla_k$ is a quite standard notation, while $d_i\omega$ is not that standard: some might use $\nabla_k (d\omega)_i$ or $\nabla_k \omega_{,i}$.

Note that in this case, one more convenient notations should be used: As $\omega$ is a function, $d\omega$ as a $(0,1)$-tensor is the same as $\nabla\omega$. From now on, I will use $\nabla_i \omega$ instead of $d_i\omega$. Thus we have

$$\nabla_k (d\omega)_i = \nabla_k\nabla_i \omega.$$

$\nabla^k A$ is nothing but raising the $k$ index (to make it a tensor with one more upper index. We have $\nabla^k = g^{ik} \nabla_i$. Thus

$$\Delta_B = \nabla^k\nabla_k = g^{ik} \nabla_i\nabla_k.$$

Note that $\Delta_B$ acts on any $(p, q)$-tensor to return a $(p, q)$-tensor. Thus $\Delta_B d\omega_i$ should be interpreted as $(\Delta _B d\omega)_i$. So (1) and (2) are equality of $(0,1)$-tensors.

Now (1) and (2) are consequence of commuting the covariant differentiation, using the Ricci identity (Strange enough, I can't find one for general $(p,q)$-tensor online, but we need only this that for $(0,1)$-tensor). Now,

$$\begin{split} \Delta_B \nabla_i \omega &= \nabla^k \nabla_k \nabla_i \omega \\ &= g^{kj}\nabla_j\nabla_k\nabla_i \omega \\ &= g^{kj}\nabla_j\nabla_i\nabla_k \omega \end{split}$$

Note that we have $\nabla_k\nabla_i\omega = \nabla_i\nabla_k\omega$ as $\omega$ is a function. If we want to $\nabla_j$ and $\nabla_i$, we have to use Ricci identity: So

$$\begin{split} g^{kj}\nabla_j\nabla_i\nabla_k \omega &= g^{kj}\left(\nabla_i\nabla_j\nabla_k \omega - R^l_{\ kji}\nabla_l \omega\right) \\ &= \nabla_i \left(\Delta_B \omega\right) + g^{kj} R^l_{\ kij} \nabla_l \omega \\ &= \text{Ric}^l_i \nabla_l\omega. \end{split}$$


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