# Coordinate vector fields associated to normal coordinates

Let $$(M,g)$$ be a pseudo-Riemann manifold, $$p\in M$$ and $$\mathcal{B}_p=\{X_i^p \,:\,i=1,\ldots,n\}\in T_pM$$ an orthonormal basis of the tangent space of the point $$p \in M$$.

Attached to this basis $$\mathcal{B}_p$$ and using the exponential map (induced by the Levi-Civita connection), we have the so-called normal coordinate system around a neighbourhood $$U \subset M$$ of $$p$$

$$\phi:U \subset M \rightarrow \phi(U) \subset \mathbb{R}^n$$.

My question is about the local coordinate vector fields attached to this coordinate system:

$$\left\{ \left.\frac{\partial}{\partial \phi_1}\right|, \ldots, \left.\frac{\partial}{\partial \phi_n}\right| \right\} \subset \mathfrak{X}(U)$$ .

It is not difficult to see that these vector fields evaluated at $$p$$ coincide with the original tangent vectors used to define the normal coordinate system, i.e.

$$\left.\frac{\partial}{\partial \phi_i}\right|_p = X_i^p \;\;\;\;$$ for all $$i=1,\ldots, n$$ ,

and hence they form an orthonormal basis of $$T_pM$$.

I would like to know whether the coordinates vector fields evaluated at a different point $$U \ni q \neq p$$

$$\left\{ \left.\frac{\partial}{\partial \phi_1}\right|_q, \ldots, \left.\frac{\partial}{\partial \phi_n}\right|_q \right\} \subset T_qM$$

form an orthonormal basis? If not, in which cases this is true?

I'm thinking that, perhaps, the above situation is related with the curvature of the spacetime. Perhaps, something like "coordinate vector fields attached to normal coordinates are orthogonal on $$U$$ if and only if the spacetime is locally flat on $$U$$ (the Riemann tensor vanishes on $$U$$)" may hold?

If the coordinate basis is orthonormal on an open set $$U$$, then on that set we have

$$g_{ij} = \left\langle \frac{\partial}{\partial \phi^i}, \frac{\partial}{\partial \phi^j} \right\rangle = \ \eta_{ij},$$

which is only possible if the metric is flat on $$U$$.

Also, with the usual meaning (at least in physics) of "locally flat", every manifold is locally flat, because the metric can always be set equal to the Euclidean/Minkowski metric plus second order terms. You're using a slightly different meaning, so be careful with that.

• Are you sure that physicists would say that any manifold is locally flat? I think that everyone, physicists and mathematicians alike, agree that "flatness" means vanishing of the curvature tensor. This cannot be made to happen, even locally, in most cases. – Danu May 7 '20 at 21:45
• The definition of locally flat I know from physics books is what I said: at a given point, one can use coordinates where the metric is Euclidean and the Christoffel symbols are zero. I can find this usage in Schutz, for example. – Javier May 7 '20 at 22:13
• That's right: A manifold is flat at a point if you can find a neighborhood of the point and coordinates on that neighborhood so that the Christoffel symbols vanish on the neighborhood. But the Christoffel symbols cannot be made to vanish locally around a point unless the curvature tensor vanishes identically at that point. They can be made to vanish exactly at the one point, but not in a coordinate patch around it! I think your confusion arises because of the distinction (which is significant in this context) of vanishing at a point and vanishing on a neighborhood around it. – Danu May 9 '20 at 7:17
• I understand what you're saying - I'm not confused :) I understand what OP means by locally flat and the difference with what I said. I'm just saying that in my experience, what physicists (probably the less mathematically minded physicists) tend to call "locally flat" is something different. It's all over Zee's book, for example. – Javier May 9 '20 at 15:20
• OK! All clear, just wanted to make sure. – Danu May 9 '20 at 17:29

See exponential mapping which uniquely maps $$T(M)_{p}$$ to a neighborhood along a geodesic - also known normal coordinates.

The value of the vector field in the neighborhood depends on the geodesic curve - assuming a symmetric connection.

And it would be in the neighborhood of $$T(M)_{p}$$ - it wouldn't be $$T_q(M)$$ unless $$M$$ was flat.

In a sense, you can think of it as extending the basis at $$T(M)_p$$ into the neighborhood to first order.