A first simple question: Do you mean AGA or do you mean AtGA? Only to understand the meaning of your symbols.

Thank you

Johann Hartl

Johann, thank you for the interest!

I'm not sure right now, I think A^{T}GA (I'll check it in the book on Monday). The idea is that after introducing the new coordinates x=x(z) and having a curve z=z(t) (and therefore x=x(z(t)) ) the tangent vector to the curve in the point t_0 (that is a vector z'(t_0)) in the old system coordinates and in the new system of coordinates would be the same in particular the same length. The length is defined using the metric g_ij as the integral from t_0 to t_1 from quadratic form with matrix G(t) from the tangent vector.

In general currently I have rather vague understanding of the formal definition of the method. As far as I understand we have point space with coordinates (the idea of the coordinate system is only the different coordinates for different points). In each point we consider the tangent vector from the vector space (again as far as I understand). By definition while transforming the coordinates we use the Jacobian of the transformation connected with the particular point. That is the vector space connected with the point z_0 is transformed with the linear transformation with the matrix A(the Jacobian of the transformation in the point z_0). Further we transform the metric to keep the tangent vector lengths.

The first question that is of interest for me is this formal definition of the spaces is enough or my understanding of the point space is not enough for obtaining the above mentioned result and we need other restrictions?

With respect,

Maksim

The tangent vector of a curve is defined in a way such that it is the same in different coordinate systems. But the coordinates of the tangent vector will in general be different in different coordinate systems. The length of the tangent vector in a point of a curve is defined using the metric g_ij. The integral from t_0 to t_1 over the square root of the quadratic form with matrix G(t) is the length of the curve from z(t_0) to z(t_1).

If you are asking for the definition of a Riemannian manifold, you have - at least implicitly - all restrictions mentioned. What you did not explicitly mention: g_ij = g_ji and G should be positive definit.

> If you are asking for the definition of a Riemannian manifold, you have - at least >implicitly - all restrictions mentioned. What you did not explicitly mention: g_ij = >g_ji and G should be positive definite.

You are right with the two latter restrictions, I checked the book.

The definition of the Euclidean space in the book is the above definition with additional restriction that there are the coordinates x_1,...x_n, x=x(z), with det(A)!=0 that the metric in this coordinates is G=A^{T}A, where A is the Jacobian of the transformation of the coordinates, as far as I understand in each point of the point space.

And initial question: Is every movement(or motion, I'm not sure regarding the term, see the definition above)of the Euclidean space affine? Could you tell if this proposition is true?

There is something else I would like to ask. Is there some definition in the manifold that when we consider the point of the point space z_0 and vector v in the vector space attached to this point we have some correspondence (z_0,v) into the other point of the point space z_1. And after the transformation of the coordinates x=x(z) we have (x(z_0), Av) goes to x(z_1)? That is after the transformation the vector from the image goes to the image of the target point?

I am not sure that we use the same terms. But I will tell you something, and you will see, if it is conformal with your book.

A Euclidean space with Cartesian coordinates x_1, x_2, ..., x_n is a point space, where every pair of points X, Y has a connecting vector v such that X + v = Y, in coordinates x_1 + v_1 = y_1, ..., x_n + v_n = y_n. The vector v is a vector from the tangent space in the point X.

The metric G is given in such a coordinate system by the identity matrix I.

You get new Cartesian coordinates z_1, z_2, ..., z_n with a Matrix D in the form X = DZ + r, where D is orthogonal, that means D^(T)D = I, the identity matrix.

But if you introduce for example in a twodimensional Euclidean space polar coordinates r, phi in the form x_1 = r cos phi, x_2 = r sin phi, then r, phi are not Cartesian coordinates, and you cannot add vector coordinates (v_1,v_2) and (r_0,phi_0) to geht an endpoint (r_1,phi_1). In this new coordinate system the metric is given by G=A^{T}A, where A is the Jacobian of the transformation of the coordinates.

Perhaps this will help you.

With best wishes

Johann Hartl

I think the things you tell have sense. As you mentioned, we may look at the point space of the x coordinates as vector space. That is the vector from each point goes to some other point: (X, v) -> Y. Furthermore we may easily obtain the transformation with the fixed origin by considering the transformation x'(z)=x(z)-x(0). That is if we fix the origin we may consider the point space in the coordinate system x as the vector space , that is (0,v)->v that have coordinates x and we can tell that vector v is actually vector x. Additionally, by definition of the tangent vector space transformation we have v=Au or x=Au, where A is Jacobian in the origin, v is the image vector and u is the vector of the tagent vector space of the origin in the coordinate system z. That is we have x=Au. But here I'm not sure regarding the connection of u and z. u is some vector from the tangent vector space that have a orthonormal basis. But z is some coordinates (actually obtained from the "good" coordinates x by some continuously differentiable transformation, not necessarily linear) and I'm not sure that we can tell that point space with coordinates system z is a vector space with a basis. I guess we need to use the fact that the transformation is a movement here but currently I'm out of ideas regarding the way to use it.

> You get new Cartesian coordinates z_1, z_2, ..., z_n with a Matrix D in the form X > = DZ + r, where D is orthogonal, that means D^(T)D = I, the identity matrix.

This is true that the affine transformation with orthogonal matrix is a movement of a euclidean space as you shown above. The problem is to prove that an arbitrary transformation that is a movement is affine.

> But if you introduce for example in a twodimensional Euclidean space polar >coordinates r, phi in the form x_1 = r cos phi, x_2 = r sin phi, then r, phi are not >Cartesian coordinates, and you cannot add vector coordinates (v_1,v_2) and

>(r_0,phi_0) to geht an endpoint (r_1,phi_1). In this new coordinate system the >metric is given by G=A^{T}A, where A is the Jacobian of the transformation of the >coordinates.

This has sense, but this transformation is not a movement since the metric matrix for the point r_0, phi_0 that corresponds to the point x_1(r_0, phi_0), x_2(r_0, phi_0) is G=[1 0; 0 r_0^2].

A movement does not change distances.

In a Euclidean space the distance between two points x, y is measured as the square root of

(x_1-y_1)^2 + ... + (x_n-y_n)^2.

Therefore the set of points which have the same distance from two fixed points a, b is a hyperplane with an equation of the form

2(a_1-b_1)x_1 + ... + 2(a_n-b_n)x_n + b_1^2 + ... b_n^2 - a_1^2 - ... - a_n^2 = 0.

Every hyperplane can b described as the set of points which have the same distance from two fixed points.

As a movement does not change distances, the image of every hyperplane is a hyperplane, and the preimage of a hyperplane is a hyperplane as well.

Therefore the hyperplane with the equation x_1 = 0 has as an image (or preimage, as you take it) a hyperplane with an equation of the form

d_11 z_1 + d_12 z_2 + ... + d_1n z_n + r_1 = 0,

and so on. So X = DZ + r.

Thank you Johann!

I'll think on this and get back with the comments.

Johann, first question. Look at the transformation of the hyperplane (r, Phi) -> (1/r, Phi) except (0, Phi)-> (0,Phi). It is not linear but the image is a hyperplane, preimage is the hyperplane.

Let me present the other possibility of dealing with the argument.

First building blocks.

We consider point space with coordinates x. In each point we have a tangent vector space with vectors v. We have a transformation of the coordinates x=x(z). z is the other coordinates system again in each point in this coordinate system we have a tangent space with vectors u. By definition we set v=Au, where A is the Jacobian in the point of interest. We may define vectors v_1=x_1=(1,0,...,0),v_2=x_2=(0,1,...,0),...,v_n=x_n=(0,0,...,1) and look at our spaces as the affine space. That is we have a correspondence (x,v)->y, where x,y are points of the point space and v is the vector from the space of tangent vectors in the point x. By definition we may assume that x + v = y in natural addition system. After the transformation we also must have the affine space and we may set by definition that (z_1, u)-> z_2, where (x(z_1), v)->x(z_2) and of course v=Au, where A is the Jacobian in the point z_1.

Assume we have a transformation with a fixed origin, above I explained the way to obtain this transformation from the transformation x(z).

Further, we extend the vector space structure to the point space, namely:

(0, v_1 + v_2) -> x_3, (0, v_1)->x_1, (0, v_2)->x_2 then x_1+x_2=x_3.

(0,Lambda v_1)->x_3, (0, v_1)->x_1 then Lambda x_1 = x_3.

The same with the image point space

(0, u_1 + u_2) -> z_3, (0, u_1)->z_1, (0, u_2)->z_2 then z_1+z_2=z_3.

(0,Lambda u_1)->z_3, (0, u_1)->z_1 then Lambda z_1 = z_3.

Note that we have

z_1 + z_2 = z_3 iff x(z_1)+x(z_2)=x(z_3) and

Lambda z_1 = z_3 iff Lambda x(z_1)= x(z_3)

Here we made all necessary building arrangements and now we proceed with the argument.

Look at the basis v_1, v_2,..., v_n of the tangent vector space in the origin and simulteneously the basis of the point space in the coordinates x. Look at the images of these vectors u_1, u_2,..., u_n under the linear transformation A^{-1}v in the image tangent space(this linear transformation is orthogonal because the transformation x(z) is a movement and keeps the Euclidean metric). Look at the vectors of the points vector space (0, u_1)=z_1, (0, u_2)=z_2,...,(0,u_n)=z_n. Look at z_1, z_2,..., z_n as the basis of the built above points vector space (it is indeed basis as one may easily check). Now consider vectors z_1^0=(1,0,...,0), z_2^0=(0,1,...,0), z_n^0=(0,0,...,1). Determine the coordinates of the vector z_i^0 in the basis z_1, z_2,...,z_n. We have Betta1*z_1+Betta2*z_2+...+Betta_n*z_n=z_i^0 or coming back to the original space Betta1*v_1+Betta2*v_2+...+Betta_n*v_n=x(z_i^0) or x(z_i^0) is the vector of coordinates of the vector z_i^0 in the basis z_1,z_2,...,z_n.

Now we have one point space z and on the basis of this point set z we have two vector spaces: the one defined above with the basis z_1, z_2,..., z_n, and the other with the basis z_1^0, z_2^0,..., z_n^0 with ordinary addition and multiplication by the scalar. The question is to determine the coordinates of the point in the latter space given we know the coordinates of the point in the former space. Johann, is it possible to determine this?

In a Euclidean plane with polar coordinates r, phi the set of all points with constant r is not a hyperplane (or in this case a straight line) but a circle. The transformation (r, Phi) -> (1/r, Phi) is an inversion with respect to the unit circle.

If in your argument

z_1 + z_2 = z_3 iff x(z_1)+x(z_2)=x(z_3) and

Lambda z_1 = z_3 iff Lambda x(z_1)= x(z_3),

then x = x(z) should be of the form x = Az with a constant matrix A, where A is the Jacobian, and A is the same Jacobian in all points of the space.

But x = Az can be interpreted in at least two ways. You can have two different spaces, where x = (x1, ..., xn)^t consists of the coordinates of a point X in a coordinate system in one space and z = (z1, ..., zn)^t consists of the coordinates of another point Z in another coordinate system in the other space. You can also have one space where x are the coordinates of a point P in one coordinate system and z are the coordinates of the same point P in another coordinate system of the same space.

I guess you have one space and a coordinate transformation x = Az in this space. Is this correct?

Then x are the coordinates of a point P in one coordinate system (O; x_1, ..., x_n), and z are the coordinates of the same point P in another coordinate system (O; z_1, ..., z_n) in the same space. (O is the same origin of both coordinate systems.) Now you have a third coordinate system (O; z_1^0, ..., z_n^0), where

z_i^0 = Beta1i*z_1 + ... + Betani*z_n (i = 1, 2, ..., n) or

z_k = Gamma1k*z_1^0 + ... + Gammank*z_n (k = 1, 2, ..., n),

where the matrix (Gammajk)_0< j, k < n+1 is the inverse of the matrix (Betajk)_0< j, k < n+1.

What are the coordinates z^0 = (z1^0, ..., zn^0) in the coordinate system (O; z_1^0, ..., z_n^0) of the point P with coordinates z = (z1, ..., zn) in the coordinate system (O; z_1, ..., z_n)?

z1*z_1 + ... + zn*z_n =

z1*(Gamma11*z_1^0 + ... + Gamman1*z_n^0) + ... +

zn*(Gamma1n*z_1^0 + ... + Gammann*z_n^0) =

(z1*Gamma11 + .. zn* Gamma1n) * z_1^0 + .. +

(z1*Gamman1 + .. zn* Gammann) * z_n^0.

Therefore the coordinates z1^0, ..., zn^0 in the coordinate system (O; z_1^0, ..., z_n^0) are

z1^0 = Gamma11*z1 + .. + Gamma1n*zn, ...,

zn^0 = Gamman1*z1 + .. + Gammann*zn or

zk^0 = Gammak1*z1 + ... + Gammakn*zn (k = 1, ..., n).

In the transformation of the basis vectors

z_k = Gamma1k*z_1^0 + ... + Gammank*z_n (k = 1, 2, ..., n)

the sum runs over the first index of the Gammas, and in the transformation of the coordinates

zk^0 = Gammak1*z1 + ... + Gammakn*zn (k = 1, ..., n)

the sum runs over the second index of the Gammas.

If I understood your question correctly, perhaps this was a possible answer to your question.

Thank you Johann!

ok, in the orthogonal euclidean coordinates system (x,y,z), the transformation of the hyperplane is (0,0,z)->(0,0,z), (x,y,z)->(1/sqrt(x^2+y^2), 1/sqrt(x^2+y^2), z ). it is not linear, but it preserves the plane z=z_0.

>What are the coordinates z^0 = (z1^0, ..., zn^0)

>in the coordinate system (O; >z_1^0, ..., z_n^0)

>of the point P with coordinates z = (z1, ..., zn) in the coordinate >system (O; z_1, .>.., z_n)?

>z1*z_1 + ... + zn*z_n =

>z1*(Gamma11*z_1^0 + ... + Gamman1*z_n^0) + ... +

>zn*(Gamma1n*z_1^0 + ... + Gammann*z_n^0) =

>(z1*Gamma11 + .. zn* Gamma1n) * z_1^0 + .. +

>(z1*Gamman1 + .. zn* Gammann) * z_n^0.

Here I see the difficulty. We have the same point space, that is the basis set z, but we may potentially have two different vector spaces in that sense that multiplication by a constant and sum of vectors may be different. That is you have indeed that z1*z_1+...+zn*z_n=z1*(Gamma11*z_1^0+...Gamman1*z_n^0)+...+...

But multiplication z1*r, where r = Gamma11*z_1^0+...Gamman1*z_n^0 is a differently defined operation then the multiplication Gamma11*z_1^0 for example. In other words, is it true that z1*r (where z1 is a constant, and r is a vector) in one vector space that is with one definition of multiplication will have coordinates (z1*Gamma11, ..., z1*Gamman1) in vector space with basis (z_i^0) with differently defined multiplication?

Dear Maksim Sergeevich,

it is difficult to write and to read unformatted texts, so I wrote some lines in latex. I add a pdf file. If I did not answer your last question, you may ask again.

Thank you Johann for the efforts and for the answer!

I need to think for a while on your note.

At every point of E you have a tangent space which is not only isomorphic to V

but which can be identi ed with V . If you change the coordinates in E from one

Cartesian coordinate system to another Cartesian coordinate system by a coordinate

transformation of the form X = AZ +b then A is the Jacobian, and the coordinates

in V and in every tangent space will be transformed by v = Aw.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

That is clear. The question is that the transformation x=x(z) may be potentially not affine. The only thing we know that the transformation is a movement of the space, in other words it keeps metrics at every point.

In your terms we deal with the Euclidean affine space.

Is there ideas regarding the answer in this settings?

Finally I've done it! The answer is the coordinate in one vector space depend on the coordinates in other vector space linearly. The idea is that both spaces have one scalar product. If someone interested let me know I'll write the details.

Thank you Johann for helping the process!

I hope that my questions do not disturb your time.