# Reimannian metric tensor g

1. Feb 20, 2008

### ejays

How can I calculate reimannian metric tensor for the vector.

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

but dont know what it will be for vector w

2. Feb 20, 2008

### mrandersdk

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.

3. Feb 20, 2008

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 .

4. Feb 20, 2008

### ejays

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

5. Feb 21, 2008

### mrandersdk

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 thats 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 thats 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.

6. Feb 21, 2008