# How can I prove this Weitzenbock formula

Update: It is almost sure that the expression of $$\kappa$$ in coordinates given in the book, namely

$$\kappa(u_t)=h_{\alpha\beta}\frac{\partial u_t^\alpha}{\partial x^i}\frac{\partial u_t^\beta}{\partial x^j}$$

is a typo. I have edited the question and given my version of this below.

In the book Variational Problems in Geometry by Seiki Nishikawa, two Weitzenbock formulae are given (Proposition 4.2, page 124):

Let $$u\in C^\infty(M\times(-\delta,\delta),N)$$, where $$M,N$$ are Riemannian manifolds, be a solution to the equation $$\frac{\partial u}{\partial t}=\tau(u_t)$$, where $$\tau(u_t)$$ is the tension field of $$u_t(-):=u(-,t):M\to N$$. Then we have

1. $$\frac{\partial e(u_t)}{\partial t}=\Delta e(u_t)-|\nabla\nabla u_t|^2-\sum_{i=1}^m\left\langle du_t\left(\sum_{j=1}^mRic^M(e_i,e_j)e_j\right),du_t(e_i)\right\rangle\\ +\sum_{i,j=1}^mR^N(du_t(e_i),du_t(e_j),du_t(e_j),du_t(e_i)$$

2. $$\frac{\partial\kappa(u_t)}{\partial t}=\Delta\kappa(u_t)-\left|\nabla\frac{\partial u_t}{\partial t}\right|^2+\sum_{i=1}^mR^N\left(du_t(e_i),\frac{\partial u_t}{\partial t},\frac{\partial u_t}{\partial t},du_t(e_i)\right)$$

where $$Ric^M$$ is the Ricci tensor on $$M$$, $$R^N$$ the curvature tensor on $$N$$, $$\Delta$$ the Laplace operator on $$M$$, $$e_i$$ an orthonormal basis of tangent spaces of $$M$$, and $$\left\langle,\right\rangle$$ is the inner product on the tangent spaces of $$N$$.

The quatities $$e(u_t)$$, $$\kappa(u_t)$$ are defined on page 123 of the book

$$e(u_t)=\frac{1}{2}|du_t|^2=\frac{1}{2}g^{ij}h_{\alpha\beta}\frac{\partial u_t^\alpha}{\partial x^i}\frac{\partial u_t^\beta}{\partial x^j}$$

$$\kappa(u_t)=\frac{1}{2}\left|\frac{\partial u_t}{\partial t}\right|^2=\frac{1}{2}g^{ij}u_{t;ij}^\alpha g^{rs}u_{t;rs}^\beta h_{\alpha\beta}$$

where $$g_{ij}$$, $$h_{\alpha\beta}$$ are the components of the metric tensor on $$M$$, $$N$$ respectively, and a semicolon together with subscripts denotes covariant derivatives. Note that the coordinate representation of $$\kappa(u_t)$$ is not given as a definition, but a result of calculation under the assumption that $$\frac{\partial u_t}{\partial t}=\tau(u_t)=g^{ij}u_{t;ij}^\alpha\frac{\partial}{\partial y\alpha}$$ holds.

Question:

The formula about $$e(u_t)$$ has a complete proof in the book. The book says the second one can be proven in a similar manner to the first one. However, the original coordinate representation of $$\kappa(u_t)$$ given in the book must be wrong (see the top of this post), since the RHS has indexes $$i,j$$ while the LHS does not. I have given my version of $$\kappa(u_t)$$ in coordinates in the question, which is nothing similar to the expression of $$e(u_t)$$ in coordinates. And I now really don't know how to prove the second Weitzenbock formula "in a similar manner to the first one". Can you provide a sketch of proof of the second formula? A reference is also okay.

• I have not looked at this carefully, but if you believe the coordinate-free definition of $\kappa$, then isn't the correct version in coordinates $$\frac{1}{2}h_{\alpha\beta}\partial_tu^\alpha\partial_tu^\beta?$$ That's in fact consistent with the second Weitzenbock formula. Feb 13, 2020 at 14:42
• @DeaneYang I think so too, the book may have made a typo while writing down the coordinate representation of $\kappa$. But this is not the main problem. I am struggling to prove the second Weitzenbock formula, even after fixing the proof. What do you mean by "consistent with", are you saying there is an easy way to see why the second formula holds? Feb 13, 2020 at 15:30

$$\newcommand{\pa}{\partial}$$Edit: The answer is now LaTeXified.
\begin{aligned} \frac{\pa\kappa(u_t)}{\pa t}&=h_{\alpha\beta}(u_t)\frac{\pa^2 u_t^\alpha}{\pa t^2}\frac{\pa u_t^\beta}{\pa t}\\&=\Big\langle\frac{\pa^2u_t}{\pa t^2},\frac{\pa u_t}{\pa t}\Big\rangle. \end{aligned} By Ricci's identity, $$\nabla_i\nabla_t\nabla_ju_t^\alpha-\nabla_t\nabla_i\nabla_ju_t^\alpha=R^{\alpha}_{\beta\gamma\delta}(u_t)\frac{\pa u_t^\beta}{\pa x^i}\frac{\pa u_t^\gamma}{\pa t}\frac{\pa u_t^\delta}{\pa x^j}.$$ Thus \begin{aligned} \Delta_g\kappa(u_t)&=g^{k\ell}\nabla_k\nabla_\ell\kappa(u_t)\\&=g^{k\ell}\nabla_k\Big(h_{\alpha\beta}(u_t)\nabla_\ell\nabla_tu_t^\alpha\nabla_t u_t^\beta\Big)\\&=g^{k\ell}h_{\alpha\beta}(u_t)\bigg(\nabla_k\nabla_\ell\nabla_tu_t^\alpha\frac{\pa u_t^\beta}{\pa t}+\nabla_\ell\frac{\pa u_t^\alpha}{\pa t}\nabla_k\frac{\pa u_t^\beta}{\pa t}\bigg)\\&=g^{k\ell}h_{\alpha\beta}(u_t)\nabla_k\nabla_t\nabla_\ell u_t^\alpha\frac{\pa u_t^\beta}{\pa t}+\Big|\nabla\frac{\pa u_t}{\pa t}\Big|^2\\&=g^{k\ell}h_{\alpha\beta}(u_t)\nabla_t\nabla_k\nabla_\ell u_t^\alpha\frac{\pa u_t^\beta}{\pa t}\\&\phantom{{}={}}+g^{k\ell}h_{\alpha\epsilon}(u_t)R^{\alpha}_{\beta\gamma\delta}(u_t)\frac{\pa u_t^\beta}{\pa x^k}\frac{\pa u_t^\gamma}{\pa t}\frac{\pa u_t^\delta}{\pa x^\ell}\frac{\pa u_t^\epsilon}{\pa t}+\Big|\nabla\frac{\pa u_t}{\pa t}\Big|^2\\&=h_{\alpha\beta}(u_t)\nabla_t\big(g^{k\ell}\nabla_k\nabla_\ell u_t^\alpha\big)\frac{\pa u_t^\beta}{\pa t}\\&\phantom{{}={}}-g^{k\ell}R_{\beta\gamma\epsilon\delta}(u_t)\frac{\pa u_t^\beta}{\pa x^k}\frac{\pa u_t^\gamma}{\pa t}\frac{\pa u_t^\epsilon}{\pa t}\frac{\pa u_t^\delta}{\pa x^\ell}+\Big|\nabla\frac{\pa u_t}{\pa t}\Big|^2. \end{aligned} This gives the second formula.