How to compute the Laplacian of this function involving a $2$-tensor?

Question

Let $(M^n,g)$ be a Riemannian manifold of dimension $n \geq 2$, $T$ a symmetric covariant $2$-tensor on $M$ and $u \in C^\infty(M)$. Could you help me calculate the Laplacian (= $\operatorname{div} \nabla$) of the function $f(x) = T(\nabla u(x), \nabla u(x))$, please? I have no idea on how to begin.

Answer 1

To avoid getting confused, let's call $V = \nabla u$. It's just a vector field.

Now work in coordinates:

$$ f(x) = T(V(x), V(x)) = \sum_{ij} T_{ij}(x) V_i(x) V_j(x) $$ and now you can apply the coordinate expression of the Laplacian:

$$ \Delta f = \sum_{ijkl} \frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x_k} \bigg( \sqrt{|g|} g^{kl}\frac{\partial}{\partial x_l} \big(T_{ij}V_iV_j\big) \bigg) $$

Then just happily simplify!

Answer 2

Just do it like a physicist. Let $V = \nabla u$ and write $$\begin{align}\triangle f &= g^{ij}\nabla_i\nabla_jf \\ &= g^{ij}\nabla_i\nabla_j(T_{k\ell} V^kV^\ell) \\ &= g^{ij}\nabla_i( (\nabla_jT_{k\ell})V^kV^\ell + T_{k\ell} (\nabla_jV^k)V^\ell + T_{k\ell}V^k(\nabla_jV^\ell) ) \\ &= g^{ij}\nabla_i((\nabla_jT_{k\ell})V^kV^\ell + 2T_{k\ell}V^k (\nabla_jV^\ell)) \\ &= g^{ij}\nabla_i((\nabla_jT_{k\ell})V^kV^\ell)+2g^{ij}\nabla_i(T_{k\ell}V^k(\nabla_jV^\ell)). \end{align}$$Let's continue to compute each term separately.

For the first one: $$\begin{align}g^{ij}\nabla_i((\nabla_jT_{k\ell})V^kV^\ell) &= g^{ij}(\nabla_i\nabla_jT_{k\ell})V^kV^\ell + g^{ij}(\nabla_jT_{k\ell})(\nabla_iV^k) V^\ell + g^{ij}(\nabla_jT_{k\ell})V^k\nabla_iV^\ell \\ &= (\triangle T)_{k\ell}V^kV^\ell + 2 g^{ij}(\nabla_jT_{k\ell})(\nabla_iV^k)V^\ell \\ &= (\triangle T)_{k\ell}V^kV^\ell + 2 g^{ij}(\nabla_jT_{k\ell})(\nabla_iV^k)V^\ell \end{align}$$And for the second one: $$\begin{align}2g^{ij}\nabla_i(T_{k\ell}V^k(\nabla_jV^\ell)) &= 2g^{ij}(\nabla_iT_{k\ell})V^k(\nabla_jV^\ell) \\ &\qquad+2g^{ij}T_{k\ell}(\nabla_iV^k)(\nabla_jV^\ell) + 2g^{ij}T_{k\ell}V^k\nabla_i\nabla_jV^\ell \\ &= 2g^{ij}(\nabla_jT_{k\ell})(\nabla_iV^k)V^\ell +2g^{ij}T_{k\ell}(\nabla_iV^k)(\nabla_jV^\ell) + 2T_{k\ell}V^k(\triangle V)^\ell.\end{align}$$Putting it all together: $$\triangle f = (\triangle T)_{k\ell}V^k V^\ell+4g^{ij} (\nabla_jT_{k\ell})(\nabla_iV^k)V^\ell+2g^{ij}T_{k\ell}(\nabla_iV^k)(\nabla_jV^\ell) +2T_{k\ell}V^k(\triangle V)^\ell.$$In a coordinate-free way, we have $$\begin{align} \triangle f &= (\triangle T)(\nabla u,\nabla u) + 4\,{\rm tr}_g\big((X,Y)\mapsto (\nabla_XT)(\nabla_Y(\nabla u), \nabla u) \big) \\ & \qquad+2\left\langle T, {\rm tr}_g\big((X,Y)\mapsto (\nabla_X(\nabla u))\otimes (\nabla_Y(\nabla u))\big)\right\rangle + 2T(\nabla u, \triangle(\nabla u)). \end{align}$$I'd guess that you may be able to find equivalent expressions for $\triangle (\nabla u)$ with suitable Ricci-Weitzenbock identities (since it's the Laplacian of a gradient), but you'll know better.

I will admit, however, that my computation here doesn't really seem to have much "content": I did it blindly and just translated it back to "mathematics" at the end. Extra information about $T$ or $u$ (e.g., $\nabla\nabla u = 0$) would certainly make this look more palatable.

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Edited to add: recall that for two smooth functions $f_1$ and $f_2$ on a Riemannian manifold $(M,g)$, the "product rule" $$\triangle (f_1f_2) = (\triangle f_1)f_2 + f_1(\triangle f_2) + 2 g(\nabla f_1,\nabla f_2)$$holds. So, taking the Laplacian of $f = T(\nabla u, \nabla u)$, one should certainly expect $(\triangle T)(\nabla u, \nabla u)$ and $2T(\nabla u,\triangle (\nabla u))$ (as $T$ is symmetric and both of its arguments are equal). The remaining terms in the expression I obtained should account for the "product between the gradients" and the fact that here we have a "triple" product ($T$ times $\nabla u$ times $\nabla u$). You can use the product rule above twice to compute $\triangle (f_1f_2f_3)$ for more heuristics.