# Orthonormal basis => vanishing Riemann curvature tensor

#### przemek

Hey!

If a (pseudo) Riemannian manifold has an orthonormal basis, does it mean that Riemann curvature tensor vanishes? Orthonormal basis means that the metric tensor is of the form

$$(g_{\alpha\beta}) = \text{diag}(-1,+1,+1,+1)$$

what causes Christoffel symbols to vanish and puts Riemann curvature tensor equal to 0.

I know that I am doing something wrong here, but I don't know where.

Thanks!

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#### George Jones

Staff Emeritus
Gold Member
Hey!

If a (pseudo) Riemannian manifold has an orthonormal basis, does it mean that Riemann curvature tensor vanishes? Orthonormal basis means that the metric tensor is of the form

$$(g_{\alpha\beta}) = \text{diag}(-1,+1,+1,+1)$$

what causes Christoffel symbols to vanish and puts Riemann curvature tensor equal to 0.

I know that I am doing something wrong here, but I don't know where.

Thanks!
This is true if there exists (in a region) an orthonormal coordinate basis, but an orthonormal basis does not have to be a coordinate basis.

#### quasar987

Homework Helper
Gold Member
The christoffel symbols are defined with respect to some coordinate chart on the manifold.
They are defined by the equation
$$\nabla_{\partial_i}\partial_j = \sum_k\Gamma_{ij}^k\partial_k$$
If there is a coordinate chart whose induced coordinate vector fields $\partial_i$ are orthonormal, then yes, the Christoffel symbols are 0 and the curvature is 0.

But from the existence of an orthonormal basis, you can't conclude anything. (By Gram-Schmidt, such a basis exists near every point!)

#### przemek

Thanks for help :) Let me get this straight - the general affine connection coefficients are defined by

$$\nabla_{\mathbf{e}_j}\mathbf{e}_k=\mathbf{e}_i \omega ^{i}_{jk}$$

where $$\mathbf{e}_i$$ are the vectors of any basis. Orthonormality of the basis does not imply here that connection coefficients vanish. If the basis is a coordinate basis

$$\mathbf{e}_i=\partial_i$$

and orthonormal

$$g_{ij}=\left\langle\mathbf{e}_i,\mathbf{e}_j\right\rangle=\delta_{ij}$$

then the connection coefficients are given by the Christoffel symbols

$$\omega^{i}_{jk}=\Gamma^{i}_{jk}$$

which are equal to 0. So an orthonormal coordinate frame exists only in flat manifolds?

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#### quasar987

Homework Helper
Gold Member
So an orthonormal coordinate frame exists only in flat manifolds.
Well, flatness is a local property. So a more correct statement would be that an orthonormal coordinate frame exists only in a neighborhood where the manifold is flat.

#### lavinia

Gold Member
Well, flatness is a local property. So a more correct statement would be that an orthonormal coordinate frame exists only in a neighborhood where the manifold is flat.
The opposite direction seems difficult. That is, suppose the curvature tensor vanishes. Does that imply an orthonormal coordinate frame around any point?

For instance, for a surface zero curvature means that the connection 1-form is closed. How does this imply the existence of an orthonormal coordinate frame?

It does of course, since every flat manifold is covered by flat Euclidean space via the action of a group of isometries

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#### przemek

I have found a theorem in Frenkel's book attributed to Riemann which says:

In a Riemannian manifold, one can introduce local coordinates $y$ such that the metric assumes the euclidean or flat form

$$\mathrm{d}s^2=(\mathrm{d}y^1)^2 + ... + (\mathrm{d}y^n)^2$$

if and only if curvature vanishes, $\theta=0$.

#### lavinia

Gold Member
I have found a theorem in Frenkel's book attributed to Riemann which says:

In a Riemannian manifold, one can introduce local coordinates $y$ such that the metric assumes the euclidean or flat form

$$\mathrm{d}s^2=(\mathrm{d}y^1)^2 + ... + (\mathrm{d}y^n)^2$$

if and only if curvature vanishes, $\theta=0$.
interesting. How does the theorem work?

#### quasar987

Homework Helper
Gold Member
The opposite direction seems difficult. That is, suppose the curvature tensor vanishes. Does that imply an orthonormal coordinate frame around any point?
Yes! Pick a point p and a local coordinate chart (xi) around p such that the coordinate vectors $\partial_i|_p$ at p are orthonormal (ex: normal coordinates!). Define near p n vector fields Y_i like so: Set Y_i(p):=$\partial_i|_p$ and then define Y_i along the x1 coordinate line by parallel transport of Y_i(p). Then for each point q on that line, consider the x2 coordinate line passing through there and extend the Y_i's along those lines by parallel transport again, etc. Since parallel transport is an isometry, the vector fields Y_i's so constructed are orthonormal.

To show that they are the coordinate vector fields of some coordinate system (yi) around p, it suffices to show that [Y_i,Y_j]=0 for each i,j (because in this case, their flow $\varphi^{Y_i}_{t_i}$ commutes and it is easy to see that $(-\epsilon,\epsilon)^n\rightarrow M:(t_1,\ldots,t_n)\mapsto \varphi^{Y_n}_{t_n}\circ\ldots\circ\varphi^{Y_1}_{t_1}(p)$ defines a coordinate chart around p).

But this we get for free since the Levi-Civita connection is torsion free: $[Y_i,Y_j]=\nabla_{Y_i}Y_j - \nabla_{Y_j}Y_i = 0$!

#### lavinia

Gold Member
I am unsure of your argument. - parallel translation may not be independent of path even when curvature is zero. Don't you need to show that it is in a coordinate chart ? -

I don't see why the vector fields are parallel with respect to each other. You have only shown that they are parallel along the coordinate curves. Yes?

#### quasar987

Homework Helper
Gold Member
I am unsure of your argument. - parallel translation may not be independent of path even when curvature is zero. Don't you need to show that it is in a coordinate chart ? -
Show that what is in a coordinate chart? ((by construction, I start with a coordinate chart and use it to build the n vector fields Y_i in this chart)

I don't see why the vector fields are parallel with respect to each other. You have only shown that they are parallel along the coordinate curves. Yes?
They Y_i are the coordinate vector fields of some system of coordintes (y^i). And they are orthonormal. So the metric is the identity matrix wrt that coordinate chart. So the Chritoffel symbols of the Levi-Civita connection wrt to (y^i) are all 0 since this is expressed as derivatives of the metric. So the Y_i are parallel vector fields for the Levi-Civita connection.

But this wasn't the point, was it? You asked for a coordinate system of whose coordinate vector field form an orthonormal basis of T_pM at each point.

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#### lavinia

Gold Member
I thought your argument was in 2 parts. - first construct n indepedent mutually orthogonal unit vector fields by parallel translating a frame at a point along the coordinate curves. - then appeal to the symmetry of the connection to conclude that the vector fields form an
Involutive distributrion.

- the parallel translation construction fails if it is not independent of the path taken along the coordinate curves - no?
- using the symmetry of the connection works if the covariant derivatives of the vector fields wrsp to each other are zero. Why is this?

#### quasar987

Homework Helper
Gold Member
I thought your argument was in 2 parts. - first construct n indepedent mutually orthogonal unit vector fields by parallel translating a frame at a point along the coordinate curves. - then appeal to the symmetry of the connection to conclude that the vector fields form an
Involutive distributrion.

- the parallel translation construction fails if it is not independent of the path taken along the coordinate curves - no?
<
In what way does it fail? (Granted, if you parallel translate along x² and then along x¹ it may yield a different vector field as if you translate along x¹ and then along x², but so what? Pick an order in which to translate and stick with it.)

using the symmetry of the connection works if the covariant derivatives of the vector fields wrsp to each other are zero. Why is this?
Oh, right, because I use the parallelism of the Y_i in the proof that they are in involution, and in my previous post, I used the fact that the Y_i are orthonormal to prove that they are parallel. This is nonsence: I need to prove paralelism independantly of orthonormality.

I will do this in the the 2-dimensional case to convey the idea. The general case can be done by induction using the idea.

Fix i=1 or 2. By construction, $\nabla_{\partial_1}Y_i =0$ on the coordinate submanifold {x²=0} and $\nabla_{\partial_2}Y_i =0$ everywhere. So we want to show that $\nabla_{\partial_1}Y_j =0$ everywhere. For this, it suffices to show that $\nabla_{\partial_2}\nabla_{\partial_1}Y_i=0$. But since $[\partial_1,\partial 2]=0$, the vanishing of the riemmanian curvature tensor yields the equation $\nabla_{\partial_2}\nabla_{\partial_1}Y_i=\nabla_{\partial_1}\nabla_{\partial_2}Y_i$, and this later field is 0.

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By the way, you said "parallel translation may not be independent of path even when curvature is zero". In any vector bundle with connection D, dropping the orthonormality stuff, the above argument show that D is flat iff around each point, there is a paralllel local frame. This implies that parallel translation along a sufficiently small loop is the identity. And this implies that in a small nbdh of a point, parallel translation is independant of the path.