How to calculate basis vectors from metric tensor (as a matrix)?

[Tachytaenius]

TL;DR

AIUI, the metric tensor is what describes the shape of a manifold. Most things tend to be derived from it, so how can I derive basis vectors? Or have I misunderstood?

As I understand from my self-study (and I'm well aware I may be wrong, so please correct me and/or point me in the right direction etc if so), the metric tensor is more or less what describes the shape of a manifold. It's the fundamental object from which most things tend to be derived from, e.g. distances, angles, Christoffels, etc. And IME so far is most generally useful for more things when represented as a matrix, not as a line element, thought it can be harder to get a picture of the space from a matrix.

So, how can I derive basis vectors for each coordinate from it, in a general way for any metric tensor? Is the idea complicated and case-specific, or has my research just not shown the general way to me yet? (E.g. with (second kind) Christoffels I thought it was going to be super hard but it turns out to be derivable from the metric tensor with simple for loops in software). I'm somewhat aware it's easier to do with diagonal metric tensor matrices because that means you have an orthonormal basis. But have I misunderstood the metric tensor's role?

Also, quick check: basis vectors for particular coords at a particular point on a manifold have length 1 according to the line element, right? And, for orthonormal bases, only point in one coordinate's direction?

Thank you!

And, again, please remember that I self-study for my own purposes so I'm always having slightly alternative interpretations of things at first :)

 

[Dale]

I don’t understand the question. Basis vectors are not things that are derived. They are things that are chosen. What do you mean by “derive” in this context?

 

[Tachytaenius]

I don’t understand the question. Basis vectors are not things that are derived. They are things that are chosen. What do you mean by “derive” in this context?

By derive the basis vectors from the metric tensor I mean: acquire a formula to calculate the values of the basis vectors, at any point on the manifold, using the metric tensor.

In my current mental model, the "starting point" is to choose coordinates (and define anywhere they wrap around) and then define a metric tensor as a formula for a matrix that depends on those coordinates.

For example:
The time-constant equatorial cross section of the Morris-Thorne wormhole can be represented in r and theta coordinates, where theta wraps around at every full turn.
Its metric tensor is then:
$$ \begin{bmatrix}
1 & 0 \\
0 & b^2+r^2
\end{bmatrix} $$
And from that we can derive its line element, which is:
$$ ds^2=dr^2+(b^2+r^2)d\theta^2 $$
And I found a way to get basis vectors from that for each coordinate:
Set $ds^2$ to 1, all d variables to 0 except for the coordinate in question, then solve for the change in the coordinate in question. Which should get a formula in terms of the manifold's constants (b) and the position coordinates (r and theta), which is the value for the relevant component in the basis vector (all others are zero).

But I'm 99% sure this isn't gonna work for non-orthonormal basis vectors. I think I've gotten metric->bases the wrong way 'round. Can I even reverse basis->metric every time, or is it only for orthonormal bases?

 

[Orodruin]

You’ve gotten things wrong. The basis is a choice. The typical coordinate basis depends only on the coordinates chosen - the manifold does not even need to have a metric.

 

[martinbn]

May be you are trying to ask how to find an orthonormal basis?

 

[Dale]

Its metric tensor is then:
$$ \begin{bmatrix}
1 & 0 \\
0 & b^2+r^2
\end{bmatrix} $$

If you can write the matrix above then you have already chosen a basis. The obvious basis is $$ \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} $$ $$ \begin{bmatrix}
0& 1\\
0 & 0
\end{bmatrix} $$ $$ \begin{bmatrix}
0& 0 \\
1& 0
\end{bmatrix} $$ $$ \begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix} $$

And from that we can derive its line element, which is:
$$ ds^2=dr^2+(b^2+r^2)d\theta^2 $$

So to go from the previous statement to this one you have now explicitly named the basis vectors: $$ \begin{bmatrix}
1& 0 \\
0 & 0
\end{bmatrix} =dr^2$$ $$ \begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix} = d\theta^2$$ and presumably the other two basis vectors would be written $d\theta \ dr$ and $dr \ d\theta$.

And I found a way to get basis vectors from that for each coordinate:
Set $ds^2$ to 1, all d variables to 0 except for the coordinate in question, then solve for the change in the coordinate in question. Which should get a formula in terms of the manifold's constants (b) and the position coordinates (r and theta), which is the value for the relevant component in the basis vector (all others are zero).

This sounds to me not like you are trying to find a basis but you are trying to normalize your basis. In other words, your new basis vectors will have the same directions as the previous basis vectors, just normalized so that their length is each 1.

 

[pervect]

A recap of what others said. If one defines coordinates on a manifold, (to be more formal, one might talk about a coordinat chart), one automatically define a coordinate basis. But, that's not the only possible basis one might use. Frequently, one is interested in some other basis, such as an orthonormal basis, rather than a coordinate basis. It eventually becomes important to know how to work with non-coordinate bases, especially to get any sort of intuitive understanding. For a while, one might get away with using only coordinate base, but one will ultimately need to learn how to use other bases.

Texts such as Sean Caroll go into more detail when they discuss manifolds, see for instance https://preposterousuniverse.com/wp-content/uploads/grnotes-two.pdf. There's actually quite a bit of stuff that Caroll talks about with regard to manifolds before they get to the part about vectors at all. The key insight regarding vectors in GR is that they live in (or define) a tangent space at every point on the manifold, and that the vectors in this tangent space can be identified with partial derivative operators. Some sources (such as Caroll) use directional derivative operators rather than partial derivative operators.

Metrics come into play when one wants to define things like the lengths and angles of the vectors. The vectors or vector spaces exist without a metric, the metric defines the lengths of a vector, or the angle between two vectors. I talked a bit about orthonormal bases - they are very useful, but to talk about an orthonormal basis, one does need the concept of the length of a vector, and the angle between two vectors, hence one needs a metric to talk about them.