**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.




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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.



How can I prove this Weitzenbock formula
$\newcommand{\pa}{\partial}$Edit: The answer is now LaTeXified.


I happen to have written notes on this. :)


\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.


