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The proof of the classical Weitzenböck formula

$$ \Delta (|f|^2)=|{\rm Hess}f|^2+\langle\nabla f, \nabla (\Delta f) +{\rm Ric} (\nabla f, \nabla f) \rangle $$

uses the local orthonormal frame field $X_i$ around any fixed point $p\in M$ satisfy $$ \langle X_i, X_j \rangle =\delta_{ij}, \ \ \nabla_{X_i}X_j(p)=0 $$ to simplify the calculation.

My question is: What if I start with arbitary orthonormal fram say $\{e_1, \cdots, e_n\}$. My calculation shows that for any fixed $\alpha=1,\cdots,n$, the following holds: $$ \begin{align} {\rm Hess}(|\nabla f|^2)(e_{\alpha}, e_{\alpha})= &2|\nabla f|^2 {\rm sec}(\nabla f, e_{\alpha}) + 2\nabla f \langle \nabla_{e_{\alpha}}\nabla f, e_{\alpha}\rangle +2 \langle \nabla _{e_{\alpha}}\nabla f, \nabla_{e_{\alpha}}\nabla f\rangle \\ &- 4\langle \nabla_{e_{\alpha}}\nabla f, \nabla_{\nabla f}e_{\alpha}\rangle \end{align} $$ Where the ${\rm sec}$ denotes the sectional curvature spaned by $\nabla f$ and $e_{\alpha}$, .

The only difference between the standard calculation using normal fram and mine is the term $$- 4\langle \nabla_{e_{\alpha}}\nabla f, \nabla_{\nabla f}e_{\alpha}\rangle $$ So it means after summing up over $1, \cdots , n$, we must get $0$. i.e. $$ \sum_{\alpha} - 4\langle \nabla_{e_{\alpha}}\nabla f, \nabla_{\nabla f}e_{\alpha}\rangle=0 $$ But this seems not obvious to me. Did I miss something?

The classical calculation can be found here: The Comparison Geometry of Ricci Curvature, by Shunhui Zhu, 221-262 http://library.msri.org/books/Book30/contents.html

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  • $\begingroup$ Can you check your calculuation/write-up? The second term on the RHS of your expression is a covector. Everything else are scalars. $\endgroup$ – Willie Wong Nov 15 '11 at 10:27
  • $\begingroup$ The term $\nabla f \langle \nabla_{e_{\alpha}}\nabla f, e_{\alpha}\rangle$ means the vector $\nabla f$ acts on the function $\langle \nabla_{e_{\alpha}}\nabla f, e_{\alpha} \rangle$ thing, so it's a scalar. $\endgroup$ – user17150 Nov 15 '11 at 13:47
  • $\begingroup$ sigh... you should really fix your notation. If you use $\nabla$ for the connection, you really should not also let it act on scalars as the metric gradient. $\endgroup$ – Willie Wong Nov 15 '11 at 14:09
  • $\begingroup$ Are you assuming that $e_\alpha$ is a ON field, or just a linear frame that happens to be ON at a point? $\endgroup$ – Willie Wong Nov 15 '11 at 14:25
  • $\begingroup$ It's ON field in a neighborhood of a fixed point say $p_0$. $\endgroup$ – user17150 Nov 15 '11 at 15:57
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First observe that the Hessian of a scalar function is a symmetric bilinear form.

Second observe that

$$ g(e_i, \nabla_X e_j) + g(\nabla_X e_i, e_j) = 0 $$

since $e_\alpha$ is ON. So when you take the sum of the expression you wrote down, it is the full contraction of a symmetric bilinear form with an antisymmetric bivector, hence must be zero.

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