# Coordinate-free differentiation techniques in Riemannian geometry

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

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

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)?

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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.)

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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