Whether $(\Delta E) v=\Delta(E(v))$?

Question

Consider a Riemannian manifold $(M,g)$, and $$ E=E_i^j dx^i\otimes \partial_j ~~~~v=v^i\partial_i $$ namely, $E\in \Gamma(TM^*\otimes TM), v\in \Gamma (TM)$, I want to know whether $(\Delta E) v=\Delta(E(v))$. Therefore, assuming $u=u_i dx^i$, then $$ [(\Delta E) v]u = (\Delta E)(v,u) = g^{ij} \nabla^2E (\partial_i, \partial_j, u,v) = \text{complex} \\ [\Delta(E(v))] u = [\Delta(E_i^jv^i\partial_j)] u =\text{then, I don't know how to do} $$ I get stuck here. In fact, I think there should be an easy way. I feel use the abstract index maybe easy, but I fail.

Answer 1

Using the abstract index notation you can make it look easy, but you still need to understand what is going on under the hood.

First of all, what is $E(v)$? Well, in abstract indices we write it as $E^a{}_b v^b$ and the latter notation really means the same thing as the former, by the virtue of the natural isomorphism $V^* \otimes W \cong Hom(V,W)$.

In fact, we use a lot of identifications without mentioning to make our language concise. Another example of such an identification, which is relevant to the context of this question, is using the Riemannian metric as a chosen isomorphism between $V$ and $V^*$. We use it to lower and raise indices (the "musical" isomorphisms).

Notice, that I deliberately abuse the language and notation and mix vector spaces and vector bundles (whatever $V$ and $W$ are!), because in finite dimensions I can do so (with care).

Secondly, let's do the calculation. $$ \Delta(T \otimes S) := g^{a b} \nabla_a \nabla_b (T \otimes S) = \\ = g^{a b} \nabla_a \big( (\nabla_b T) \otimes S + T \otimes \nabla_b S \big) = \\ = g^{a b} \big( (\nabla_a \nabla_b T) \otimes S + (\nabla_b T) \otimes \nabla_a S + (\nabla_a T) \otimes \nabla_b S + T \otimes \nabla_a \nabla_b S \big) = \\ = (\Delta T) \otimes S + 2 (\nabla^a T) \otimes \nabla_a S + T \otimes \Delta S $$

Please pay attention how we use parentheses to restrict the action of differential operators such as $\nabla$ and $\Delta$.

Now you can forget about abstract indices, if you want, and use this knowledge to your liking.

In particular, your equality does not hold. Instead, we have: $$ \Delta (E(v)) = (\Delta E)(v) + 2 g^{i j} (\nabla_i E)(\nabla_j v) + E(\Delta v) $$

where the indices $i, j$ are thought now as concrete (e.g. coordinate or frame) indices, with the Einstein convention applied. Do you see what each pair of parentheses means in the above display?