# Invariant properties of metric tensor

• I
Gold Member

## Main Question or Discussion Point

Which properties of metric tensor are invariant of basevectors transforms? I know that metric tensor depends of basevectors, but are there properties of metric tensor, that are basevector invariant and describe space itself?

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fresh_42
Mentor
Which properties of metric tensor are invariant of basevectors transforms? I know that metric tensor depends of basevectors, but are there properties of metric tensor, that are basevector invariant and describe space itself?
A metric tensor ##g## above an affine point space ##A## with a real translation space ##V## is a map form ##A## into the space of scalar products on ##V ##, i.e. ##g(P)\, : \,V \times V \longrightarrow \mathbb{R}## is a symmetric, positive definite bilinear form on ##V## for every ##P\in A##.

No basis vectors anywhere around.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Or more generally, nothing about any tensor is basis dependent except its components given a particular basis.

Gold Member
##g_{i;j}=e_i\otimes e_j##.

in base 1:
##g_{i;j}=
\begin{bmatrix}
\vec{e_0}\cdot\vec{e_0}&\vec{e_0}\cdot\vec{e_1} \\
\vec{e_1}\cdot\vec{e_0}&\vec{e_1}\cdot\vec{e_1}
\end{bmatrix}=
\begin{bmatrix}
1&0 \\
0&1
\end{bmatrix}##

in base 2:
##\vec{e´_0}=2*\vec{e_0}##
##\vec{e´_1}=\vec{e_1}##

##g_{i;j}=
\begin{bmatrix}
\vec{e´_0}\cdot\vec{e´_0}&\vec{e´_0}\cdot\vec{e´_1} \\
\vec{e´_1}\cdot\vec{e´_0}&\vec{e´_1}\cdot\vec{e´_1}
\end{bmatrix}=
\begin{bmatrix}
(2*\vec{e_0})\cdot(2*\vec{e_0})&\vec{e_0}\cdot\vec{e_1} \\
\vec{e_1}\cdot\vec{e_0}&\vec{e_1}\cdot\vec{e_1}
\end{bmatrix}=
\begin{bmatrix}
4&0\\
0&1
\end{bmatrix}##

fresh_42
Mentor
This is the old difficulty to distinguish vectors and their coordinates. It is meaningless to ask about a description of a vector (matrix, tensor) once you described them by coordinates. Coordinates are the tool, not the object. It is just difficult to describe the object without coordinates, but the definition in post #2 does it, namely as a map.

DEvens
Orodruin
Staff Emeritus
Homework Helper
Gold Member
their coordinates
Or their components

But I agree.

Gold Member
for example minkowsky metric tensor is often given only by components
##
\begin{bmatrix}
-1 & 0 &0 &0 \\
0& 1 & 0 &0 \\
0& 0 & 1 &0 \\
0& 0 & 0 &1
\end{bmatrix}
##
without specifiyng base vectors. Do they assume some specific base vectors? Which ones?

Is there something invariant in the components?
How can spaces with different elemens be compared by their metric tensors if they have different elements and therefore we cant choose same basevectors there?

Last edited:
Orodruin
Staff Emeritus
Homework Helper
Gold Member
The standard assumption on Minkowski space is that you are using a set of standard affine Minkowski coordinates.

Gold Member
The standard assumption on Minkowski space is that you are using a set of standard affine Minkowski coordinates.
What are these?

There should be something invariant in components of metric tensor because it is probably impossible to choose base where minkowsky metric has components
##\begin{bmatrix}
1 & 0 &0 &0 \\
0& 1 & 0 &0 \\
0& 0 & 1 &0 \\
0& 0 & 0 &1
\end{bmatrix}##

Gold Member
it is probably impossible to choose base where minkowsky metric has components
##\begin{bmatrix}
1 & 0 &0 &0 \\
0& 1 & 0 &0 \\
0& 0 & 1 &0 \\
0& 0 & 0 &1
\end{bmatrix}##
I was wrng it is possible if ##\vec{e_0'}=\sqrt{-1}*\vec{e_0}##

Gold Member
Is there any relation between metric tensor and transformation matrix? Can I derive lorentz tranformation matrix from minkowsky metric tensor?

haushofer
Is there any relation between metric tensor and transformation matrix? Can I derive lorentz tranformation matrix from minkowsky metric tensor?
The metric is a rank 2 tensor under general coordinate transformations, and hence transforms as such (with "two transformation matrices"). The Lorentz transformations are those transformations which keep the Minkowski metric form invariant. These special transformations, which are a subset of the general coordinate transformations, connect al those observers who would use the very same components for the Minkowski metric and we call them inertial observers. So yes, you can derive the Lorentz transformations from this property ("which transformations keep the form of the metric the same? "). It's covered in any basic book about GR, I guess.

Technically, one says that "the isometries of a metric break the general coordinate transformations down to a subgroup of them." This means that the metric transforms as a tensor under general coordinate transformations, but is kept invariant under a subgroup of this group. And those form the isometries (=symmetries) of the spacetime this metric describes.

Hope this helps ;)

Orodruin
Staff Emeritus