How Is the Second Covariant Derivative of a Scalar Computed?

For a scalar field $f$, the first covariant derivative is just its partial derivative. But after one derivative, the result has a lower index: it is a covector. Taking a second covariant derivative therefore requires the covector derivative rule, which adds a Christoffel-symbol term.

With the convention that the first index on the derivative is written first, the components are

$$\nabla_\mu\nabla_\nu f =\partial_\mu\partial_\nu f -\Gamma^\lambda_{\mu\nu}\,\partial_\lambda f.$$

This is the covariant Hessian of $f$. It is a rank-two covariant tensor, not generally just the array of second partial derivatives.

Why a connection term appears

A scalar has no indices, so its first covariant derivative is

$$\nabla_\nu f=\partial_\nu f.$$

Now set $A_\nu=\nabla_\nu f$. The covariant derivative of a covector is

$$\nabla_\mu A_\nu =\partial_\mu A_\nu-\Gamma^\lambda_{\mu\nu}A_\lambda.$$

Substituting $A_\nu=\partial_\nu f$ gives the Hessian formula. The minus sign is the one used for a lower (covariant) index. In general relativity, the usual choice is the Levi-Civita connection determined by the metric.

Worked example: $f=r^2$ in polar coordinates

Consider the Euclidean plane in polar coordinates $(r,\theta)$, with metric

$$ds^2=dr^2+r^2d\theta^2.$$

The nonzero Christoffel symbols needed here are

$$\Gamma^r_{\theta\theta}=-r, \qquad \Gamma^\theta_{r\theta}=\Gamma^\theta_{\theta r}=\frac{1}{r}.$$

Take $f(r,\theta)=r^2$. Its first derivatives are $\partial_r f=2r$ and $\partial_\theta f=0$. The coordinate second partial derivatives are $\partial_r^2f=2$, $\partial_r\partial_\theta f=0$, and $\partial_\theta^2f=0$.

For the $rr$ component, the relevant connection symbols vanish:

$$\nabla_r\nabla_r f =\partial_r^2f-\Gamma^\lambda_{rr}\partial_\lambda f =2.$$

For the mixed component,

$$\nabla_r\nabla_\theta f =\partial_r\partial_\theta f-\Gamma^\lambda_{r\theta}\partial_\lambda f =0,$$

because the only possibly relevant symbol is $\Gamma^\theta_{r\theta}=1/r$, and $\partial_\theta f=0$. The $\theta r$ component is also zero.

For the angular component, the connection term does not vanish:

$$\nabla_\theta\nabla_\theta f =0-\Gamma^r_{\theta\theta}\partial_r f =-(-r)(2r)=2r^2.$$

Thus, in the coordinate basis $(dr,d\theta)$, the Hessian components are

$$(\nabla_a\nabla_b f) =\begin{pmatrix}2&0\\0&2r^2\end{pmatrix}.$$

The nonzero angular component is not a contradiction: tensor components depend on the coordinate basis. In Cartesian coordinates the same Hessian for $f=x^2+y^2$ has components $2,2$ on the diagonal. The polar-coordinate connection term accounts for the changing coordinate basis.

Symmetry and the scalar Laplacian

For a torsion-free connection, such as the Levi-Civita connection, the Hessian of a scalar is symmetric:

$$\nabla_\mu\nabla_\nu f=\nabla_\nu\nabla_\mu f.$$

This follows because ordinary partial derivatives commute and the Christoffel symbols are symmetric in their lower indices. If a connection has torsion, this symmetry need not hold.

Contracting the Hessian with the inverse metric gives the scalar Laplacian, also called the Laplace–Beltrami operator:

$$\Box f=g^{\mu\nu}\nabla_\mu\nabla_\nu f.$$

In a Lorentzian spacetime this is often written $\Box f$; in a Riemannian setting it is commonly written $\Delta f$. The sign convention for the metric can affect the displayed operator’s sign, so check the convention used in your text.

Common mistakes

• Using only second partial derivatives. Those are not tensor components under general coordinate changes. Include the connection term unless it vanishes in the coordinates and at the point being considered.
• Using the wrong sign. The gradient’s index is lower, so its covariant derivative has a minus Christoffel term.
• Treating the first derivative as a scalar. $\partial_\nu f$ is a covector component, which is why the second differentiation uses the covector rule.
• Confusing the Hessian with its trace. $\nabla_\mu\nabla_\nu f$ has two indices; $g^{\mu\nu}\nabla_\mu\nabla_\nu f$ is the scalar Laplacian.

If you evaluate the Hessian along an affinely parametrized geodesic with tangent $u^\mu$, its contraction $u^\mu u^\nu\nabla_\mu\nabla_\nu f$ equals the second derivative of $f$ along that geodesic. This is one reason the covariant Hessian is useful in geometry and relativity.