Taylor expansion of the square of the distance function

[shooride]

Does it make a sense to define the Taylor expansion of the square of the distance function? If so, how can one compute its coefficients? I simply thought that the square of the distance function is a scalar function, so I think that one can write
$$
d^2(x,x_0)=d^2(x'+(x-x'),x_0)=d^2(x',x_0) + (x-x')^\mu\partial_\mu d^2(x',x_0)\\+1/2(x-x')^\mu(x-x')^\nu\nabla_\mu\partial_\nu d^2(x',x_0)+\dots
$$
Is this anywhere near correct?

 

[Dr. Courtney]

I think the Taylor expansion of any polynomial is just the original polynomial.

 

[shooride]

I think the Taylor expansion of any polynomial is just the original polynomial.

I couldn't get your point. Could you please explain it more?! Where am I doing anything wrong?! I think that I miss some points about the distance and displacement..

 

[mfb]

What is the definition of your distance function? The Euclidean distance is just a polynomial so the taylor expansion is trivial.

 

[shooride]

What is the definition of your distance function? The Euclidean distance is just a polynomial so the taylor expansion is trivial.

I'd like to consider the taylor expansion of the square of the distance function (geodesic distance) $d^2:M\times M\to R$ on a smooth Riemannian manifold $(M,g)$. BTW, AFAIK, one can only define the Euclidean distance on a Euclidean space, right?!

 

[zinq]

Before we can have a Taylor expansion on a manifold, it's necessary to specify a system of local coordinates in which to express this expansion. For instance, say we want to express the squared distance D(p,q) between two points on the unit sphere S² in ℝ³. And suppose we are considering the coordinate patch that has a positive z=coordinate, and is defined by projection (X,Y,Z) → (X,Y).

Then for any pair of points (x,y) and (u,v) in local coordinates, we can calculate their squared distance on the sphere:

D((x,y),(u,v)) = (arccos(xu + yv + √((1-x²-y²) (1-u²-v²)))²

Like any smooth function of finitely many variables, this has a Taylor series. But calculating it explicitly looks complicated.

 

[shooride]

Before we can have a Taylor expansion on a manifold, it's necessary to specify a system of local coordinates in which to express this expansion.

You are right! For this, I used normal coordinates, then wrote
$$
d^2(x,x_0)=d^2(x'+(x-x'),x_0)=d^2(x',x_0) + (x-x')^\mu\partial_\mu d^2(x',x_0)\\+1/2(x-x')^\mu(x-x')^\nu\nabla_\mu\partial_\nu d^2(x',x_0)+\dots
$$

Like any smooth function of finitely many variables, this has a Taylor series.

I see your points.. but the thing is that I'm not sure the squared distance is a smooth function..

 

[zinq]

Good idea to use a smart choice of coordinates. On a manifold M, I am sure that the squared distance function

dsq_q: M → ℝ

via dsq_q(p) := dist(p,q)² from a given point q is (unlike the unsquared distance) a smooth function in some neighborhood of q.

Which is just the condition we need for dsq_q to have a Taylor series at q.

I will try to come up with a proof later.

But, the squared distance function is never differentiable at every point p of any compact manifold. (E.g., consider a circle of circumference = 1 and graph its squared distance function. When p is at the maximum distance = 1/2 from q, the function dsq_q is clearly not differentiable there.)

(It is a difficult theorem that, if there exists some point q such that a manifold's squared distance function dsq_q is differentiable at all points of the manifold, then the manifold is diffeomorphic to Euclidean space ℝⁿ.)

 

[zinq]

As promised, here is a proof that the squared distance function dsq_q (notation as above) on a manifold M is smooth in some neighborhood of the point q:

The exponential map in differential geometry:

exp_q: T_qM → M​

taking any vector v in some neighborhood U of the origin in the tangent space T_qM at the point q, to the manifold M, via

exp_q(v) := γ_v(1),​

where γ_v denotes the unique distance-minimizing constant-speed geodesic curve with γ_v(0) = q, whose velocity γ'_v(0) at time equals v.* (For an arbitrarily large neighborhood, there is usually no unique distance minimizing curve.)

It is a standard fact in differential geometry that this exponential map is a diffeomorphism from the open neighborhood of the origin U of T_qM, onto its image γ_v(U), which we call W. W will be an open set of M. For simpler notation, let's call this diffeomorphism h:

h: U → W​

Every diffeomorphism has an inverse diffeomorphism and h is no exception:

h^inv: W → U.​

Also, the map

lensq: U → ℝ
​

via

lensq(v) := ||v||²
​

is easily shown to be smooth. Finally, the composition

lensq ⚬ h^inv: W → ℝ​

gives the distance of any point in the neighborhood W (of q in M) to the point q, and since it is the composition of two smooth maps, must be smooth itself.
__________________
* If all geodesics can be extended indefinitely in M, then M is called a complete Riemannian manifold. In this case, the exponential map is then defined for all real multiples of v via

exp_q(tv) := γ_v(t)​

for any t in ℝ, where γ_v is the same uniquely defined geodesic curve as above. In this case, the exponential map exp_q will be onto the manifold M, but in most cases will not be a diffeomorphism T_qM → M.

 

[Ssnow]

Assuming that $M$ is a Riemannian manifold, fixing a local coordinate system (a normal coordinate system, a preferred coordinate system...) and using the exponential map it is possible to find the expansion of the distance that is a polynomial with coefficients that probably depends by Riemannian invariants of the manifold $M$ as the Riemann scalar curvature $R$ the Ricci tensor $R_{ij}$ or the Riemann tensor $R_{ijkl}$...