# What does it mean that we can diagonalize the metric tensor

On a Riemannian manifold $M$, the matrix representation is diagonalisable, cause the tensor is symmetric. What is the physical meaning behind this? I mean, in Riemannian geometry, we always get a coordinate system defined by the charts(for simplicity I only use one) $\phi : M \rightarrow \mathbb{R}^k$ which we express here by $x_i := \pi_i \circ \phi,$ where $\pi_i$ is the projection on the $i-$th component in $\mathbb{R}^k.$ Now, by diagonalising the metric tensor, we can apparently make the transition to a basis of the tangent space $\partial_1,....,\partial_k$ at each point so that the metric tensor is the identity matrix. I am not sure how to interpret this?

Does this mean that the manifold looks locally euclidean? If so, does this mean that we can define locally a chart so that this chart would actually induce this locally euclidean coordinate system on the manifold? Probably, this chart would, if it exists, not be in general compatible with the atlas of the manifold $M$, is this correct?

If my attemt to interpret this is wrong, I would love to here an explanation of this fact.

• Does this mean that the manifold looks locally euclidean? Yes. Commented May 10, 2015 at 13:28
• @Demosthene what? There are local obstructions to isometry with euclidean space in general for a manifold with metric. To the OP, just because you have a family of vector fields that form an orthonormal basis at each point (this being the appropriate analogue of diagonalization) doesn't mean these can be realized as the coordinate vector fields of some chart. Indeed the commutator $[X_i, X_j]$ is an obvious obstruction.
– jxnh
Commented May 10, 2015 at 13:32
• @JHance: Demosthene is conditionally right: if one can diagonalize locally, then the manifold is locally isometric to the Euclidian space. Note the "if". Commented May 10, 2015 at 13:36
• @AlexM. when you say "diagonalize" do you mean that there is a basis of vector fields that diagonalize (which I think always happens) or that there is the above where they are the vector fields of some coordinate chart?
– jxnh
Commented May 10, 2015 at 13:43
• @JHance: I am thinking of local coordinates. Commented May 10, 2015 at 14:10

To amplify slightly on the two comments by Demosthene and JHance: it means that the manifold is very locally euclidean, i.e., pointwise. At each point, you can find a basis of tangent vectors in which things are Euclidean, but this basis cannot generally be extended even to a small neighborhood. A good approximation to extending it to a small neighborhood is given by "exponential coordinates"; on a surface, each ray emanating from your chosen point, in a kind of polar coordinates, is actually geodesic. The "circles" of the polar coordinate system, however, cannot be made as nice as the rays, alas.

Your initial assumption on which your whole question is constructed is mistaken: one cannot diagonalize the metric tensor locally in general (or else every Riemann manifold would be locally isometric to some Euclidean space, which is trivially not true); one can, though, diagonalize very special metrics locally and, also, one can diagonalize every metric pointwise (but not necessarily locally). Normal coordinates are one example of the latter situation. Young and deTurck also show that every Riemannian $$3$$-dimensional manifold admit local coordinates that make the metric tensor diagonal, yet it is clear that $$3$$-dimensional manifolds are not necessarily locally isometric.

The whole confusion, it seems, comes from two different uses of the words "diagonalization" / "rescaling":

• one can have the metric tensor in a non-diagonal form, and by a change of (local) coordinates make it diagonal (or even rescale it),

• or one can use linear algebra techniques to diagonalize (or even rescale) the metric tensor at every point of a coordinate chart, but this does not amount to a coordinate change.

If diagonalization and rescaling are performed like in the first case, the patch of coordinates is indeed locally Euclidean. The second case does not capture any geometric information about the underlying manifold.

Another way of looking at is is that what we're really interested in is not so much the metric itself as it is the norm on each tangent space we can derive from it. The norm is what actually gives structure to the manifold; representing it as a bilinear form is more of a technical convenience.

Viewed in this light, the important fact is not that "the matrix representation is diagonalizable", but the fact that there is a matrix representation at all. Once the norm has a matrix representation $A$ (such that $\|v\|^2 = v^TAv$), it also has a symmetric matrix representation, namely $\frac12(A+A^T)$.

However, the fact that we're only considering norms that can be represented as $\|v\|=\sqrt{f(v,v)}$ for some bilinear functional $f$ is a meaningful property of Riemannian manifolds in particular. Relaxing this condition leads to more general kinds of manifolds, which cannot be approximated as well with Euclidean space locally as Riemannian ones can.

• Nice point. OP, it's worth looking at Volume 3 of Spivak's Comprehensive Introduction to Differential Geometry, where there's a translation of Riemann's "inagural lecture", in which he discusses exactly this kind of generalization (albeit in terms that are hard for a modern reader to make sense of). Commented May 10, 2015 at 23:24
• It's also online: maths.tcd.ie/pub/HistMath/People/Riemann/Geom/WKCGeom.html Commented May 10, 2015 at 23:25