Calculating Riemannian Metric Tensor for a Vector

• ejays
In summary: I will try to find a way to do it myself .In summary, the riemannian tensor is a map from the tangent space (bundle) x tangents space (bundle) to the real numbers. You take two vectors and plug them into the riemann tensor and then get a number.
ejays
How can I calculate reimannian metric tensor for the vector.

I know about matrix it is equivalent to W(tranpose)W

but don't know what it will be for vector w

not sure you have understood what the metric tensor is.

It is a map from the tangent space (bundle) x tangents space (bundle) to the real numbers.

So you take two vectors and plug them into the riemann tensor and then get a number.

If you have understood it, please reformulate your question, because then it is unclear.

Hi , ejays
I don't know if the Reimannian metric tensor is different from the metric tensor that I know , so I'll tell you what I know and you see if it helps ( the following method is for the metric tensor of a surface and it can be generalized for spaces )
Suppose M is a surface determined by $$$\vec X\left( {u,v} \right) \subset E^3$$$ and $$$\vec c\left( t \right)$$$ is a curve on M , $$$t \in \left[ {a,b} \right]$$$. Then we can write $$$\vec c\left( t \right) = \vec X\left( {u\left( t \right),v\left( t \right)} \right)$$$ .Now by taking the derivative of the previous equation we obtain :
$$$\vec c'\left( t \right) = \frac{{\partial \vec X}}{{\partial u}}\frac{{du}}{{dt}} + \frac{{\partial \vec X}}{{\partial v}}\frac{{dv}}{{dt}} = u'\vec X_1 + v'\vec X_2$$$
and If $$$s\left( t \right)$$$ represents the arc length along the curve Then :
$$$s\left( t \right) = \int\limits_a^b {\left| {\left| {\vec c'\left( t \right)} \right|} \right|} dt$$$
and :
$$$\frac{{ds}}{{dt}} = \left| {\left| {\vec c'\left( t \right)} \right|} \right|$$$
so :
$$$\begin{array}{l} \left( {\frac{{ds}}{{dt}}} \right)^2 = \left| {\left| {\vec c'\left( t \right)} \right|} \right|^2 = \vec c'.\vec c' = \left( {u'\vec X_1 + v'\vec X_2 } \right).\left( {u'\vec X_1 + v'\vec X_2 } \right) \\ = u'^2 \left( {\vec X_1 .\vec X_1 } \right) + 2u'v'\left( {\vec X_1 .\vec X_2 } \right) + v'^2 \left( {\vec X_2 .\vec X_2 } \right) \\ \end{array}$$$
and by putting :
$$$\begin{array}{l} \vec X_1 .\vec X_1 = g_{11} \\ \vec X_1 .\vec X_2 = g_{12} \\ \vec X_2 .\vec X_2 = g_{22} \\ \end{array}$$$
then we have :
$$$\left( {\frac{{ds}}{{dt}}} \right)^2 = g_{11} \left( {\frac{{du}}{{dt}}} \right)^2 + 2g_{12} \left( {\frac{{du}}{{dt}}\frac{{dv}}{{dt}}} \right) + g_{22} \left( {\frac{{dv}}{{dt}}} \right)^2$$$
or :
$$$ds^2 = g_{11} du^2 + 2g_{12} dudv + g_{22} dv^2 = \sum\limits_{i,j} {g_{ij} } dx^i dx^j$$$
and the metric tensor ( as a matrix ) is :
$$$g_{ij} = \left( {\begin{array}{*{20}c} {g_{11} } & {g_{12} } \\ {g_{21} } & {g_{22} } \\ \end{array}} \right)$$$

(Note that $$$g_{12} = g_{21}$$$ )
I hope this will help .

Thanks, this is indeed a help, i would require some further elaboration but let me first test it with my Natural gradient problem, the one based upon reimannian tensor.

Thanks again

the shadow, as far as I see you are correct. You are working in what is called local coordinates (what most physicist do), I personally like to not work in coordinates as long as possible, and only do the last calculations in some coordinates, but that's because I work mostly with theoretical physics, and I would say that to do that and fully understand the underlying math, you would need a course in manifold theory and riemannian geometry. But you can learn the skill to work in coordinates all the time, if you only need to do calculations, but maybe then the whole thing can seems like a mystery, but guess that's ok, if you don't care why it works, but wan't to use the fact that it do, to make calculations on concrete problems.

Hi , mrandersdk
thank you for your comment , but I need to tell you that I'm new at the field of differential geometry , so I don't know all the methods of it .
What I know that in a problem like this I use the method that I wrote , but in theoratical physics I use the riemannian tensor to calculate the metric ( like the metrics in GR ) , but I don't know how to use the riemannian tensor method in a problem like this .
Could you tell me how , please ??
thanks again

1. What is a Reimannian metric tensor g?

The Reimannian metric tensor g is a mathematical object used in differential geometry to define the distance and angle between points in a curved space. It is a symmetric, positive definite matrix that assigns a length to each tangent vector at a point in the space.

2. How is the Reimannian metric tensor g used?

The Reimannian metric tensor g is used to calculate the length of curves, the angle between curves, and the curvature of a space. It is also used in the study of general relativity, where it describes the curvature of spacetime.

3. What are the properties of the Reimannian metric tensor g?

The Reimannian metric tensor g is symmetric, positive definite, and covariantly constant. This means that it is unchanged under coordinate transformations, it only takes positive values, and its components do not depend on the choice of coordinates.

4. How is the Reimannian metric tensor g related to the Levi-Civita connection?

The Levi-Civita connection is a way of defining parallel transport of vectors in a curved space. It is uniquely determined by the Reimannian metric tensor g, and vice versa. This means that the metric tensor and the Levi-Civita connection are closely related and together define the geometry of a space.

5. What is the difference between the Reimannian metric tensor g and the Minkowski metric tensor?

The Reimannian metric tensor g is used in curved spaces, while the Minkowski metric tensor is used in flat spacetime. The Minkowski metric has a signature of (-1,1,1,1) and is used in special relativity, while the Reimannian metric has a positive definite signature and is used in general relativity and differential geometry.

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