# Metric,Propery of manifold?

#### ShayanJ

Gold Member
From the things I've read on manifold geometry,metric is a property of the manifold.Maybe you can call it intrinsic.
But consider e.g. a euclidean manifold and two coordinate systems on it.say,cartesian and spherical.As you know,the metric for this two coordinate systems is different.
So what?Is metric a property of manifold or coordinate system?

thanks

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#### chiro

From the things I've read on manifold geometry,metric is a property of the manifold.Maybe you can call it intrinsic.
But consider e.g. a euclidean manifold and two coordinate systems on it.say,cartesian and spherical.As you know,the metric for this two coordinate systems is different.
So what?Is metric a property of manifold or coordinate system?

thanks
Hey Shyan.

The manifold defines the coordinate system used and the metric defines how to calculate distance between points either in a differential way or in an explicit way like you get in a cartesian system.

Don't think about a manifold like say a sphere (surface of a sphere with 2 dimensions: latitude and longitude): think about the sphere as a two-dimensional geometry with an appropriate metric that defines how close to points on the manifold itself.

So to answer your question the metric is a result of the actual geometry you are using and the coordinate system is often seen as a way to parametrize the system using the minimum number of dimensions needed.

Remember that a geometry in n dimensions can have different parameterizations: for example think about polar and cartesian systems in three dimensions: Although (x,y,z) and (r,theta,gamma) refer to different ways of representing a point, they are still valid. This is what we use tensors for: we go from one system to another.

When we deal with manifolds, we could introduce some kind of parameterization going from the (lat,long) model to some other model in the same way that we did for the cartesian to polar as long as we preserve the properties of the geometry which are encoded in the metric if it's well defined.

Don't think about a manifold being something embedded inside a cartesian geometry: the manifold is the geometry.

#### ShayanJ

Gold Member
Thanks but let's just consider a euclidean 3-manifold.
We can use different coordinate systems on it.But the metric should be $\delta_{ij}$ ,right?
But in spherical coordinates,we get a different metric.
If metric is the property of manifold,we reach to this conclusion that there is sth special about cartesian coordinates.
Is this correct?

#### chiro

Thanks but let's just consider a euclidean 3-manifold.
We can use different coordinate systems on it.But the metric should be $\delta_{ij}$ ,right?
But in spherical coordinates,we get a different metric.
If metric is the property of manifold,we reach to this conclusion that there is sth special about cartesian coordinates.
Is this correct?
The Euclidean system has very nice attributes and we understand how to represent things very well from unidimensional to multidimensional representations as well as things like derivatives and integrals in this space.

But apart from this, the Euclidean system is not really any more special than the others. The fact is that the Euclidean system is intuitive for us because we can relate to these kinds of systems due to our experience in perceiving and trying to make sense of the world.

The metric for euclidean space is well understood by the Pythagorean theorem. The theorem itself is intuitive from high school math and it's not too difficult to prove. The 3D version is easy to visualize and the next dimension can be proved inductively in the kind of way that the 3D version is derived from 2D.

Again you have to realize that the Cartesian representation is well understood because its intuitive and because all of the results have been formed in the Euclidean frame.

I imagine that maybe if our sensory perception was not like this, then we might of have possibly developed mathematics in hyperbolic or even spherical geometries instead of Euclidean, but because Euclidean has familiar properties of linearity (which motivates things like vector spaces), again it makes it easy to relate to and digest.

Think about how we have gone from simple numbers to vectors and think about what might have happened if we saw the world in a way that mirrored say spherical geometry instead of Euclidean: in other words imagine if Euclid wrote a treatise on geometry that described spherical geometry instead of Euclidean geometry and what that would have done on mathematical development and especially if this was the intuitive geometry to us.

#### ShayanJ

Gold Member
thank you.
I'm telling when we switch from cartesian coordinates to spherical coordinates when we're doing calculations in euclidean space,why the metric changes while the manifold is not changed?

#### Rasalhague

The metric tensor field on a Riemannian manifold is a function which specifies for each point of the manifold a non-degenerate, positive-definite symmetric bilinear form. The value of the metric tensor field at each point is a metric tensor which is is function that takes as its inputs a pair of tangent vectors from the tangent space at that point. It is defined intrinsically in this way, without reference to coordinates. But coordinates are useful for doing calculations. For every coordinate system (Cartesian, spherical, etc.), there is, at each point of the manifold, a special basis for the tangent space at that point called a coordinate basis. If a coordinate function is labeled $x^i$, the coordinate basis vector field corresponding to this coordinate function is typically written

$$\frac{\partial }{\partial x^i}.$$

The corresponding dual basis vector field, for this coordinate basis, is $\mathrm{d}x^i$; these functions specify a basis for the cotangent space at each point of the manifold. The metric tensor field, as a type (0,2) tensor field, can be expressed as a linear combination of tensor products of pairs of dual basis vector fields.

For example, at every point of 3d Euclidean space, the Euclidean metric tensor field has the following coefficients (=components) with respect to the coordinate basis for Cartesian coordinates: $( \delta_{ij} )$

$$=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}$$

That is to say, the metric tensor field is

$$\mathbf{g}=\mathrm{d}x \otimes \mathrm{d}x + \mathrm{d}y \otimes \mathrm{d}y + \mathrm{d}z \otimes \mathrm{d}z.$$

But with respect to a coordinate basis for spherical coordinates on 3d Euclidean space, $( r, \zeta, \alpha )$ (where $r$ is the radial coordinate function, $\zeta$ the zenith angle coordinate function, and $\alpha$ the azimuth angle coordinate function), this same Euclidean metric tensor field has a different set of coefficient functions, namely:

$$\begin{pmatrix} 1 & 0 & 0\\ 0 & r^2 & 0\\ 0 & 0 & r^2 \sin(\zeta)^2 \end{pmatrix}$$

where $r$ and $\zeta$ are functions from the manifold, Euclidean space, to the real numbers. (The coefficient functions of the metric tensor field are not constant with respect to this coordinate basis, as they were with respect to a Cartesian coordinate basis.)

That is to say, the metric tensor field can also be expressed

$$\mathbf{g}=\mathrm{d}r \otimes \mathrm{d}r + r^2 \mathrm{d}\zeta \otimes \mathrm{d}\zeta + r^2 \sin(\zeta)^2 \mathrm{d}\alpha \otimes \mathrm{d}\alpha.$$

Same tensor field, just expressed with respect to different bases. A tensor is a kind of vector, so this is exactly the same idea as expressing any vector as a linear combination of basis vectors. When you change basis, you have to change the scalar coefficients in the sum if you want to describe the same vector.

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#### Rasalhague

Since, no matter what coordinate system you use, these are dual bases, we have:

$$\mathrm{d}x^i\left ( \frac{\partial }{\partial x^j} \right ) = \delta^i_j,$$

which means that

$$\mathbf{g}(\mathbf{V},\mathbf{W})$$

$$\sum_{ij} g_{ij}\mathrm{d}x^i \otimes \mathrm{d}x^j \left ( \sum_m V^m \frac{\partial }{\partial x^m}, \sum_n W^n \frac{\partial }{\partial x^n} \right )$$

$$=\sum_{ij} g_{ij}V^iW^j,$$

where superscripts are indices rather than exponents (traditionally written this way so that the summation signs can be omitted), and the sums are over, say, {1,2,3} in the case of a 3-dimensional manifold, or {1,2,3,4} in the case of a 4-dimensional manifold.

When the coordinates are changed, if we make the appropriate change to switch to the new coordinate basis for each tangent space, and also make the appropriate change to switch to the new coordinate basis for each space of type (0,2) tensors, then we can do calculations with the coefficients using matrix algebra, without needing to worry about the various basis vectors and tensors. The basis vectors for the tangent spaces and their dual bases for the cotangent spaces (and tensor products of the latter) "cancel out".

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#### ShayanJ

Gold Member
Thanks
I understand now
Another question.
Is the spherical basis as below:
$\frac{\partial}{\partial r} \\ \frac{1}{r^2} \frac{\partial}{\partial \zeta}\\ \frac{1}{r^2 \sin^2{\zeta}}\frac{\partial}{\partial \alpha}\\$

thanks

#### Rasalhague

No. For a coordinate system $(x^1,x^2,...,x^n)$, the ith coordinate basis vector for each tangent space is

$$\frac{\partial }{\partial x^i}.$$

So for spherical coordinates, as I labeled them, that's

$$\frac{\partial }{\partial r},\frac{\partial }{\partial \zeta},\frac{\partial }{\partial \alpha}.$$

The coefficient of each coordinate basis vector with respect to that same basis is 1; think how you'd express it as a linear combination:

$$\frac{\partial }{\partial r}=1\frac{\partial }{\partial r}+0\frac{\partial }{\partial \zeta}+0\frac{\partial }{\partial \alpha}.$$

But with respect to another basis, such as a Cartesian coordinate basis:

$$\frac{\partial }{\partial r^k}=\sum_i \mathrm{d}x^i\left ( \frac{\partial }{\partial r^k} \right )\frac{\partial }{\partial x^i}=\sum_i \frac{\partial x^i}{\partial r^k}\frac{\partial }{\partial x^i}.$$

For example,

$$\frac{\partial }{\partial r}=\frac{\partial x}{\partial r}\frac{\partial }{\partial x}+\frac{\partial y}{\partial r}\frac{\partial }{\partial y}+\frac{\partial z}{\partial r}\frac{\partial }{\partial z}$$

$$=\sin(\zeta)\cos{\alpha}\frac{\partial }{\partial x}+\sin(\zeta)\sin(\alpha)\frac{\partial }{\partial y}+\cos(\zeta)\frac{\partial }{\partial z}.$$

In other words, the coefficients are taken from the first column of the Jacobian matrix of the coordinate transformation from polar to Cartesian. The coefficients for the other polar basis vectors, with respect to a Cartesian coordinate basis, are taken from the second and third columns of the Jacobian matrix respectively.

#### WWGD

Gold Member
Let me just make sure I got it right:

The coefficient gij of the metric has the (i,j)th coefficient of the

change-of-basis matrix from cartesian to some other system, so that, in this

case, you have the partial ∂fi/∂yj , where

f=(f1(y1,,,.,yn),...,fi(y1,...yn,...,fk(y1,...,yn) ?

#### ShayanJ

Gold Member
Thanks Rasalhague

That's wrong WWGD!
imagine we want to calculate the distance between point P and a point very near to it.
we have:
$ds^2 = dx^2 + dy^2$
but we know the transformation equations between cartesian and polar coordinates are:
$x=r \cos{\alpha} \\ y=r \sin{\alpha}$
Now we calculate the differentials:
$dx=\cos{\alpha}\ dr - r \sin{\alpha}\ d \alpha \\ dy=\sin{\alpha}\ dr + r \cos{\alpha}\ d \alpha$
Now we substitute differentials above,in the distance formula:
$ds^2=(\cos{\alpha}\ dr - r \sin{\alpha}\ d \alpha)^2+(\sin{\alpha}\ dr + r \cos{\alpha}\ d \alpha)^2$
If you do the calculations,the coefficients of $dr^2$ will be the 1-1 element of metric tensor and the coefficients of $d \alpha ^2$ the 2-2.Other elements will be zero because the polar coordinate system is orthogonal.

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#### Rasalhague

You're welcome, Shyan.

WWGD: In general, if we know the $n \times n$ scalar fields which are the coefficient functions of a metric tensor field with respect to the coordinate bases of one coordinate system, say $x=(x^1,x^2,...,x^n)$, then we can find the coefficient functions with respect to the coordinate bases of another coordinate system, say $y=(y^1,y^2,...,y^n)$, on the intersection of the domains of $x$ and $y$, using the formula

$$g_{ab}(y)= \sum_r\sum_s \frac{\partial x^r}{\partial y^a}\frac{\partial x^s}{\partial y^b}g_{rs}(x)$$

where

$$\frac{\partial x^p}{\partial y^q}$$

is the $(p,q)$th entry of the Jacobian matrix, $J$, a.k.a. change-of-basis matrix, of the coordinate transformation from $y$ to $x$. In other words,

$$g(y)=J^Tg(x)J.$$

(This is because a metric tensor is a tensor of type (0,2); it has two "covariant indices"; it takes two tangent vectors as its arguments. If we we were dealing with a (2,0) tensor - that is, a tensor with two contravariant argument slots, such as, for example, an inverse metric tensor - then we'd use the Jacobian of the coordinate transformation from $x$ to $y$.)

For Euclidean space, the metric tensor field has a particularly simple expression in Cartesian coordinates where its scalar coefficient functions are constant over the whole manifold, with values $\delta_{rs}$ (Kronecker's delta, i.e. the entries of the $n \times n$ identity matrix) at each point. Knowing this, we can find the coefficients for another coordinate system using the above formula by letting $x$ be our Cartesian coordinates, and $y$ the other system. In that case, the formula simplifies to

$$g_{ab}(y)= \sum_r \frac{\partial x^r}{\partial y^a}\frac{\partial x^r}{\partial y^b}$$

which in matrix terms, letting $J$ be the Jacobian matrix (=change-of-basis matrix) of the coordinate transformation from the other system to Cartesian, is

$$g(y)=J^TJ.$$

#### WWGD

Gold Member
I see, thanks, both.

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