3
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ The determinant of $g$ as a function is not intrinsic; that is, it depends on the coordinates you choose. If you choose normal coordinates, then it becomes clear that its derivative (in the usual sense) vanishes. Now, you can consider an intrinsic quantity which is the volume form $\mathrm{vol}(g)$ associated with $g$; it satisfies $\nabla \mathrm{vol}(g)=0$ as you may expect, using the same computation essentially. $\endgroup$ – Henri Apr 14 at 10:57
  • $\begingroup$ Is this statement also true locally? By this I mean consider only the locally defined function $\det(g_{\alpha \beta})$ and not the volume form (and continuing the above computation without using normal coordinates)? $\endgroup$ – Phillip Apr 14 at 11:13
  • $\begingroup$ I do not see why the component $g^{\gamma \delta}g^{\beta \alpha} \partial_{\delta} g_{\alpha \beta}$ should vanish ? $\endgroup$ – Phillip Apr 14 at 11:15
  • 2
    $\begingroup$ @Phillip You need to use $\nabla g = 0$ at some point. $\endgroup$ – Vít Tu?ek Apr 14 at 13:02
  • 2
    $\begingroup$ The point is that $\nabla\det g$ need not be zero. What miust be zero is the covariant derivative of the volume form. $\endgroup$ – Deane Yang Apr 14 at 13:22
7
$\begingroup$

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 $\nabla|dV_g|=0$.

$\endgroup$
4
$\begingroup$

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 $\nabla\Phi = 0$.

$\endgroup$
  • 4
    $\begingroup$ Don't you want a square root of the determinant? And even then I think it's only invariant under orientation-preserving changes of coordinates. $\endgroup$ – José Figueroa-O'Farrill Apr 14 at 20:09
  • $\begingroup$ @JoséFigueroa-O'Farrill, thanks. Yes. $\endgroup$ – Deane Yang Apr 14 at 20:24
2
$\begingroup$

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 $\nabla g = 0$.

$\endgroup$
  • $\begingroup$ Is this formula $\nabla \det (h) = h^{-1}\nabla h$ correct? Shouldn't it be $\nabla \det (h) = \det(h) \text{tr}(h^{-1}\nabla h)$ ? $\endgroup$ – Phillip Apr 17 at 8:41
  • $\begingroup$ Do you know any reference where the formula for $\nabla \det(h)$ is proven? $\endgroup$ – Phillip Apr 17 at 8:52
  • $\begingroup$ @Phillip: indeed, a $\det h$ was missing from the right-hand side. Corrected now. $\endgroup$ – Dan Fox Apr 17 at 14:16
  • $\begingroup$ Do you know any reference for this formula in the language of tensors? $\endgroup$ – Phillip Apr 17 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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