Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I encountered the following identities while reading this article on global calculus (p. 10):

$$ d(\|df\|^2)=2\mathop{\iota_{\mathop{\mathrm{grad}} f}} \mathop{\mathrm{Hess}} f, $$

$$ \mathop{\mathrm{grad}}(\|df\|^2)=2\mathop{\nabla_{\mathop{\mathrm{grad}} f}}\mathop{\mathrm{grad}} f $$

Here $\mathop{\mathrm{Hess}} f = \nabla d f$ is the covariant derivative of the 1-form $df$, and the norm $\|\cdot\|$ is given by Riemannian metric: $\|\upsilon\|^2=g(\upsilon,\upsilon)$.

I wonder how does one usually derive these identities (without using coordinates)?

share|cite|improve this question
up vote 7 down vote accepted

By simply using the definition. The first thing to remember is that the covariant derivative commutes with metric and inverse metric. The second thing to remember is that the covariant and exterior derivatives agree when acting on a scalar function. So,

$$ d(\|df\|^2) = \nabla(\|df\|^2) = \nabla g^{-1}(df,df) = g^{-1}(\nabla df, df) + g^{-1}(df, \nabla df) $$

Next you use the symmetry of $g^{-1}$ as a bilinear form, and you use the definition for raising indices that for any one form $\omega$,

$$ \iota_{\mathrm{grad}~f} = \omega(\mathrm{grad}~f) = g^{-1}(\omega, df) $$

to arrive at

$$ d(\|df\|)^2 = 2 \iota_{\mathrm{grad}~f} \mathrm{Hess}~f $$

The computation for the second identity is similar. (Note that for the notations to make sense, you also need to use the fact that for a scalar function $f$, the second derivatives commute. In other words, the Hessian is a symmetric (0,2) tensor.)

share|cite|improve this answer
As a nice exercise: the Levi-Civita connection $\nabla$ is defined to be the unique derivation/affine connection on the tangent bundle that is torsion free and parallel w.r.t. the metric $g$. In particular, it is only originally defined to act on vector fields. Think how you can, in a coordinate free manner, generalize the Levi-Civita connection to act on arbitrary (a) one forms (b) contravariant tensors (c) $p$-forms (d) covariant tensors. – Willie Wong May 7 '11 at 2:44
Perhaps someone should ask your exercise as a question, it is indeed very good for someone learning differential geometry to do this :). – Glen Wheeler May 8 '11 at 19:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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