Covariant derivative of determinant of the metric tensor

Question

Let $(M,g)$ be a Riemannian manifold and $g$ the Riemannian metric in coordinates $g=g_{\alpha \beta}dx^{\alpha} \otimes dx^{\beta}$, where $x^{i}$ are local coordinates on $M$. Denote by $g^{\alpha \beta}$ the inverse components of the inverse metric $g^{-1}$. Let $\nabla$ be the Levi-Civita connection of the metric $g$. Consider, locally, the function $\det((g_{\alpha \beta})_{\alpha \beta})$. It is known that $\nabla \det((g_{\alpha \beta})_{\alpha \beta}) = 0$ by using normal coordinates etc...

I would like to show this fact without using normal coordinates. Just by computation. Here is what I have so far:

$\nabla \det((g_{\alpha \beta})_{\alpha \beta}) = \left [ g^{\gamma \delta} \partial_{\delta} \det((g_{\alpha \beta})_{\alpha \beta}) \right ] \partial_{\gamma} = \left [ \det((g_{\alpha \beta})_{\alpha \beta}) g^{\gamma \delta} g^{\beta \alpha} \partial_{\delta} g_{\alpha \beta}\right ] \partial_{\gamma}.$`

Here: the first equality sign follows from the definition of the gradient of a function and the second equality sign is the derivative of the determinant.

Question: How do I continue from here without using normal coordinates? Or are there any mistakes? If yes, where and which?

Greetings, Phil

Answer 1

The determinant is a quantity associated to a linear operator not to a symmetric bilinear form.

On the other hand, given an inner product on a vector space $V$ we can identify a bilinear form (which is an element of $V^*\otimes V^*$) with a linear operator (which is an element of $V\otimes V^*$). Classically, this identification was called raising the indices.

If we use the metric $g$ to identify it with an endomorphism of $TM$ you obtain the identity endomorphism whose determinant is $1$. The derivative is clearly zero.

If instead you are thinking of the volume $dV_g$ form (assuming the manifold is oriented) then $\nabla dV_g=0$; see Proposition 4.1.44 in this book.

If the manifold is not oriented and you're thinking of the volume density $|dV_g|$ then again $\nabla|dV_g|=0$.

Answer 2

As you say, if $x^1, \dots, x^n$ are local coordinates, there is a symmetric matrix $[g_{\alpha\beta}]$ such that $g = g_{\alpha\beta}dx^\alpha dx^\beta$. This matrix depends on local coordinates and therefore so does the scalar function $\det [g_{\alpha\beta}]$. If you examine how this function changes when you change coordinates, you can see that, even if $\nabla (\det [g_{\alpha\beta}])$, whose definition does not use the Levi-Civita connection, vanishes in one set of coordinates, it will not in another.

On the other hand, the density $\Phi = |\det [g_{\alpha\beta}]|^{1/2}\,dx^1\wedge\cdots\wedge dx^n$ is invariant under all orientation preserving changes of coordinates. The definition of its covariant derivative $\nabla\Phi$ does use the Levi-Civita connection, and if you carry out the calculations, you find that the definition of the Levi-Civita connection in terms of $g$ (which is equivalent to the property $\nabla g = 0$) implies that $\nabla\Phi = 0$.

Answer 3

The determinant of a metric makes perfectly good sense, but it is not a function, rather a $2$-density. Formally, this means that it transforms as a section of the bundle associated with the frame bundle and a particular nontrivial character of the general linear group.

Let $\phi:V \to W$ be a linear map between $n$-dimensional vector spaces $V$ and $W$ (for simplicity, over a field of characteristic $0$). The induced map $\wedge^{n}\phi:\wedge^{n}V \to \wedge^{n}W$ can be identified with multiplication by a scalar when a generator is chosen for each of $\wedge^{n}V$ and $\wedge^{n}W$. When $V = W$, and the same generator is used on either side, the resulting scalar does not depend on the choice, and it is reasonable to call this scalar $\det \phi$ because it agrees with the usual determinant of an endomorphism.

A symmetric bilinear form $g$ on $V$ is identified with a linear map $V \to V^{\ast}$ to the dual vector space. The determinant $\det g$ is defined as in the preceding paragraph, as the induced map $\wedge^{n}V \to \wedge^{n}V^{\ast}$.

Consider the standard action of $GL(n) = GL(V)$ on $V$ and the induced actions on $V^{\ast}$ and tensor powers of $V$ and $V^{\ast}$. So the action of $\gamma \in GL(n)$ on $g$ is given by $(\gamma \cdot g)(u, v) = g(\gamma^{-1}\cdot u, \gamma^{-1}\cdot v)$ for $u,v \in V$. It follows straightforwardly that $\det (\gamma \cdot g) = (\det \gamma)^{-2}\det g$ for $\gamma \in GL(n)$.

Suppose the base field is $\mathbb{R}$. Thus $\det:S^{2}V^{\ast} \to \mathbb{R}$ is a $GL(n)$-equivariant map for the standard action of $GL(n)$ on $S^{2}V^{\ast}$ and the $1$-dimensional representation $\chi:GL(n) \to GL(1)$ given by $\chi(\gamma) = (\det \gamma)^{-2}$.

Now let $M$ be a smooth $n$-dimensional manifold with frame bundle $F\to M$. With each $GL(n)$ module $(\rho, W)$ there is associated a bundle of weighted tensors $F \times_{\rho}W$ whose fibers are linearly isomorphic to $W$. Applying this construction to the representations of the preceding paragraph one obtains a map $\det$ associating with a section of $S^{2}T^{\ast}M$ a section of the line bundle associated with the representation $\chi$, which can be interpreted as the tensor square of the top exterior power $\wedge^{n}T^{\ast}M$ (such a section is often called a $2$-density).

Let $g$ be a Riemannian metric on $M$ with Levi-Civita connection $\nabla$. Let $h$ be a section of $S^{2}T^{\ast}M$. Picking a local frame $\{E_{1}, \dots, E_{n}\}$ in $TM$ determines a local trivialization of $\wedge^{n}T^{\ast}M$ so also of all its tensor powers. With respect to this trivialization, $\det h$ equals the determinant of the matrix $h(E_{i}, E_{j})$. The connection determined by $g$ determines a convariant derivative on any associated bundle of the frame bundle and the covariant derivative $\nabla \det h$ is a section of the same line bundle as is $\det h$. If $h$ is everywhere full rank and $h^{-1}$ is the section of $S^{2}TM$ inverse to $h$, then $\nabla \det h = (h^{-1}\nabla h)\det h$ (the notation requires interpretation; in abstract index notation $h^{-1}\nabla h$ means $h^{pq}\nabla_{i}h_{pq}$). In particular, $\nabla \det g = 0$ because $\nabla g = 0$.