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

  • $\begingroup$ Can you check your calculuation/write-up? The second term on the RHS of your expression is a covector. Everything else are scalars. $\endgroup$ Nov 15, 2011 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, 2011 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. $\endgroup$ Nov 15, 2011 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? $\endgroup$ Nov 15, 2011 at 14:25
  • $\begingroup$ It's ON field in a neighborhood of a fixed point say $p_0$. $\endgroup$
    – user17150
    Nov 15, 2011 at 15:57

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.