# Writing a coordinate formula for $|df|^2$

I need to write a coordinate formula for $$|df|^2$$, where $$f$$ is a smooth function. This is in the context of Riemannian Geometry.

I know that $$|X|^2$$ can be written as $$\langle X, X\rangle$$, which is $$g_{ij}X^iX^j$$. However, what about $$|df|^2$$? Is this equal to $$\langle \nabla f,\nabla f\rangle=g_{ij}(\nabla f)^i(\nabla f)^j$$?

## 2 Answers

The Riemannian metric induces an isomorphism $$g$$ between each tangent and cotangent space, so you get an inner product on the cotangent space by pulling back the inner product on the tangent space: $$\langle \theta_1, \theta_2\rangle_{T^*M} = \langle g(\theta), g(\theta)\rangle_{TM}$$

As in coordinates $$g(\theta)^{i} = g^{ij}\theta_j$$ and $$\langle X, Y\rangle_{TM} = g_{ij}X^iY^j$$ we have that $$\langle \theta_1,\theta_2\rangle_{T^*M} = g_{ij}g^{ik}\theta_kg^{jl}\theta_l = g^{kl}\theta_k\theta_l$$ That is, multiply by the inverse of the local coordinate expression of the Riemannian metric.

I leave it to you to compute $$|df|^2$$.

• In the last line, do you mean $g^{kl}\theta_k\theta_l$? – Anju George Apr 4 at 2:54
• Yes, thank you :) – Neal Apr 4 at 2:55

I think, that Neal's answer is all correct and valid, but I would like to add my perspective on this issue, for the sake of completeness, and for further references.

By definition, for any connection $$\nabla$$ on a (smooth) manifold and any smooth function $$f: M \to \mathbb{R}$$, $$\nabla f := \textrm{d}f$$

Using the abstract index notation, we can write $$\nabla_i f$$ for $$\nabla f$$ to indicate that $$\nabla f$$ is a $$1$$-form on $$M$$. For a vector (field) $$X$$ we would write $$X^i$$.

A metric tensor $$g$$ is denoted as $$g_{i j}$$, whereas its inverse is written as $$g^{i j}$$. For instance, $$\langle X, Y \rangle = g_{i j} X^i Y^j$$

We can extend the inner product $$\langle \cdot, \cdot \rangle$$ from vector fields to any tensor field by using the metric $$g_{i j}$$ to contract vector (contravariant) indices, and the inverse metric $$g^{i j}$$ to contract co-vector (covariant) indices.

For instance, $$|df|^2 = \langle \nabla f, \nabla f \rangle = g^{ij} ( \nabla_i f) \nabla_j f \tag{*}$$ so that you were almost there.

Notice, however, that (*) trivially holds for any (not only the Riemannian, aka Levi-Civita!) connection $$\nabla$$.