# I Invariant properties of metric tensor

#### olgerm

Gold Member
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

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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

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Or more generally, nothing about any tensor is basis dependent except its components given a particular basis.

#### olgerm

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

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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.

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#### olgerm

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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

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The standard assumption on Minkowski space is that you are using a set of standard affine Minkowski coordinates.

#### olgerm

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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
$\begin{bmatrix} 1 & 0 &0 &0 \\ 0& 1 & 0 &0 \\ 0& 0 & 1 &0 \\ 0& 0 & 0 &1 \end{bmatrix}$

#### olgerm

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 $\vec{e_0'}=\sqrt{-1}*\vec{e_0}$

"Invariant properties of metric tensor"

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