Riemann curvature of a metric defined with a coordinate basis

[Jianbing_Shao]

If given a non-trivial metric field $g_{ab}$, then we believe that we can get a non-zero Riemann curvature ${R^a}_{bcd}$ from such a metric. now if we start from a basis field:
$$e_a = {e_a}^I \frac{\partial}{\partial x^I}$$
Here $e_a$ is not necessarily to be a coordinate basis, So the commutation coefficients is:
$$[e_a, e_b] = {c_{ab}}^c e_c $$
$${c_{ab}}^c {e_c}^J = 2 {e_{[a}}^I {{e_{b]}}^J}_{,I}$$
We can define the metric field:
$$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$$
Then the corresponding Christoffel symbols is:
$$\Gamma_{abc} = \frac{1}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a} + c_{abc} + c_{acb} - c_{bca})$$
Substitute definitions of $g_{ab}$ and $c_{abc} = {c_{ab}}^d g_{dc}$. we can get:
$$\Gamma_{abc} = \frac{1}{2} (
{{e_a}^I}_{,c} e_{bI} + {{e_b}^I}_{,c} e_{aI}
+ {{e_a}^I}_{,b} e_{cI} + {{e_c}^I}_{,b} e_{aI}
- {{e_b}^I}_{,a} e_{cI} - {{e_c}^I}_{,a} e_{bI}
+ {{e_b}^I}_{,a} e_{cI} - {{e_a}^I}_{,b} e_{cI}
+ {{e_c}^I}_{,a} e_{bI} - {{e_a}^I}_{,c} e_{bI}
- {{e_c}^I}_{,b} e_{aI} + {{e_b}^I}_{,c} e_{aI} )$$
At last we get:
$${\Gamma^a}_{bc} = g^{ad} \Gamma_{dbc}= g^{ad} e_{dI} {{e_b}^I}_{,c} = {e^a}_I {{e_b}^I}_{,c}$$
Riemann curvature tensor:
$${R^a}_{bcd} = ({e^a}_I {{e_b}^I}_{,d})_{,c}
- ({e^a}_I {{e_b}^I}_{,c})_{,d}
+ {e^a}_I {{e_e}^I}_{,c} {e^e}_J {{e_b}^J}_{,d}
- {e^a}_I {{e_e}^I}_{,d} {e^e}_J {{e_b}^J}_{,c}
- 2 {e^a}_I {{e_b}^I}_{,e} {e^e}_I {{e_{[d}}^I}_{,c]}$$
Cancel out second derivatives:
$${R^a}_{bcd} =
{{e^a}_I}_{,c} {{e_b}^I}_{,d}
- {{e^a}_I}_{,d} {{e_b}^I}_{,c}
+ {e^a}_I {{e_e}^I}_{,c} {e^e}_J {{e_b}^J}_{,d}
- {e^a}_I {{e_e}^I}_{,d} {e^e}_J {{e_b}^J}_{,c}
- {e^a}_I {{e_b}^I}_{,e} {e^e}_J {{e_d}^J}_{,c}
+ {e^a}_I {{e_b}^I}_{,e} {e^e}_J {{e_c}^J}_{,d}$$
At last we get:
$${R^a}_{bcd} = 2 {e^a}_{I} {{e_b}^I}_{,e} {e^e}_J {{e_{[c}}^J}_{,d]} = -{\Gamma^a}_{be} {c_{cd}}^e$$
Now if the metric $g_{ab}$ is defined with a coordinate basis field $e_a$, the commutation coefficients ${c_{cd}}^e$ is zero, So the Riemann curvature tensor is also zero?

Then can we draw such a conclusion: metric with non-zero Riemann curvature only can be defined with non-coordinate basis ?

 

[Matterwave]

Then can we draw such a conclusion: metric with non-zero Riemann curvature only can be defined with non-coordinate basis ?

No, the Riemann curvature tensor is a proper geometric object. It is only 0 if it is 0 in all bases. If it is 0 in one basis, then it is 0 in every basis. Coordinate bases vs Non-coordinate bases don't factor into this.

 

[Jianbing_Shao]

No, the Riemann curvature tensor is a proper geometric object. It is only 0 if it is 0 in all bases. If it is 0 in one basis, then it is 0 in every basis. Coordinate bases vs Non-coordinate bases don't factor into this.

So why the Riemann curvature I get from metric $$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$$ ($e_a$ is coordinate basis) is zero?

 

[Matterwave]

So why the Riemann curvature I get from metric $$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$$ ($e_a$ is coordinate basis) is zero?

You can't arbitrarily assume a coordinate basis is a non coordinate basis and vice versa.

Think carefully about what your $I,J$ actually mean. Especially whether you were consistent in applying that meaning throughout your first post. If you go back and define your notation very clearly and carefully then you will make it easier to find your error.

 

[pervect]

I'm already a bit lost. If one has coordinates x^i, I would expect a coordinate basis e_i to be written as
$$e_i = \frac{\partial}{\partial x^i}$$

Thus if we have coordinates {$x^0, x^1, x^2, x^3$} the coordinate basis would be {$\partial / \partial x^0, \partial / \partial x^1, \partial / \partial x^2, \partial / \partial x^3$}

I don't really understand the notation in the original question.

Also, for a coordinate basis, the commutator should disappear, i.e. the coordinate basis is a holonomic basis where [e_i e_j] = 0.

The commutator vanishes because partial derivatives with respect to a smooth coordinate system commute, i.e. the partial derivative operators in a coordinate basis commute. So
$$\partial_x \partial_y = \partial_y \partial_x = \frac{\partial^2}{\partial x \partial y} = \frac{\partial^2}{\partial y \partial x}$$

So I suspect you might be actually asking about a non-coordinate basis?

 

[Jianbing_Shao]

You can't arbitrarily assume a coordinate basis is a non coordinate basis and vice versa.

Think carefully about what your $I,J$ actually mean. Especially whether you were consistent in applying that meaning throughout your first post. If you go back and define your notation very clearly and carefully then you will make it easier to find your error.

I just mean that if $x^I$ is a coordinate system, $e_a$ is an arbitrary basis.
then we can use $\frac{\partial}{\partial x^I}$ to describe the basis $e_a$.
$$e_a = {e_a}^I \frac{\partial}{\partial x^I}$$

 

[Jianbing_Shao]

I'm already a bit lost. If one has coordinates x^i, I would expect a coordinate basis e_i to be written as
$$e_i = \frac{\partial}{\partial x^i}$$

Thus if we have coordinates {$x^0, x^1, x^2, x^3$} the coordinate basis would be {$\partial / \partial x^0, \partial / \partial x^1, \partial / \partial x^2, \partial / \partial x^3$}

I don't really understand the notation in the original question.

Also, for a coordinate basis, the commutator should disappear, i.e. the coordinate basis is a holonomic basis where [e_i e_j] = 0.

The commutator vanishes because partial derivatives with respect to a smooth coordinate system commute.

So I suspect you might be actually asking about a non-coordinate basis?

I just mean $e_I$ is the basis to a Minkowski space, the corresponding Minkowski metric is $\eta_{IJ}$, then the basis $e_a$ is defined:
$$e_a = {e_a}^I \frac{\partial}{\partial x^I}$$
The basis is not necessarily to be a coordinate basis, It just means
$[e_a, e_b]$ can posssibly be non-zero. so it also can be a non-coordinate basis.
So I can use the basis to define a metric :
$$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$$
Then calculate the corresponding Riemann curvature from the metric.is there any problem here?

 

[Dale]

What is $e_a{}^I$? And is there any reason that $a$ is lower case and $I$ is upper case?

 

[Jianbing_Shao]

What is $e_a{}^I$? And is there any reason that $a$ is lower case and $I$ is upper case?

It is called tetrad, we use it to describe the relation between two different basis. if the two basis both are coordinate basis, then it is just the Jaccobi of cooedinate transformation. ${e_a}^I = \frac{\partial x^I}{\partial x^a}$.
Of course we can also define co-tetrad ${e^a}_I = \frac{\partial x^a}{\partial x^I}$

 

[PeterDonis]

It is called tetrad

More precisely, an orthonormal tetrad. You don't appear to have grasped the implications of that crucial property.

is there any problem here?

Only if you fail to realize that, in the formalism you are using, where $e_a^I$ defines an orthonormal tetrad, if you specify that $e_a^I$ is a coordinate basis, then you are also specifying that the spacetime must be flat. The calculations in the OP are basically just rediscovering this fact in a long-winded manner. From your first equation and the definition of a coordinate basis, we must have $e_a^I = \delta_a^I$. Plugging that into your equation for the metric gives $g_{ab} = \eta_{ab}$. That's why, when you plug a coordinate basis into your equations, you get a vanishing RIemann tensor--because a coordinate basis that is also an orthonormal tetrad can only exist in flat spacetime.

 

[Jianbing_Shao]

if you specify that $e_a^I$ is a coordinate basis,

A tetrad is not a vector, It just some coefficients . So you can not say a tetrad is a basis.

 

[Jianbing_Shao]

More precisely, an orthonormal tetrad. You don't appear to have grasped the implications of that crucial property.


From your first equation and the definition of a coordinate basis, we must have $e_a^I = \delta_a^I$. Plugging that into your equation for the metric gives $g_{ab} = \eta_{ab}$. That's why, when you plug a coordinate basis into your equations, you get a vanishing RIemann tensor--because a coordinate basis that is also an orthonormal tetrad can only exist in flat spacetime.

Don't you think that perhaps there exist some coordinate transformations in which $e_a^I$ is not $\delta_a^I$.

 

[PeterDonis]

A tetrad is not a vector, It just some coefficients .

Um, no. The very first equation in your OP says that each member of a tetrad is a vector, since $\partial / \partial x^I$ is a vector. Yes, there's a coefficient in front of it, but that doesn't mean it's not a vector.

So you can not say a tetrad is a basis.

You're the one that has been calling tetrads a basis. I think you have not thought through what you are saying.

In any case, yes, a tetrad field can be a basis--it just can't be a coordinate basis except in flat spacetime. It can be a non-coordinate basis in any spacetime.

Don't you think that perhaps there exist some coordinate transformations in which $e_a^I$ is not $\delta_a^I$.

Sure, but in such cases, $e_a^I$ cannot define a coordinate basis. Again, look at the very first equation in your OP. And then look up the definition of a coordinate basis.

Once more, I don't think you have thought through what you are saying.

 

[Jianbing_Shao]

Um, no. The very first equation in your OP says that each member of a tetrad is a vector, since $\partial / \partial x^I$ is a vector. Yes, there's a coefficient in front of it, but that doesn't mean it's not a vector.

Conbine the two formula: $$e_a = {e_a}^I \frac{\partial}{\partial x^I}$$
$${c_{ab}}^c e_c = [e_a, e_b]$$
I don't think it imply a tetrad is vector, If I say so it is not right.
$e_a$ is vector, $\frac{\partial}{\partial x^I}$ is vector, but ${e_a}^I$ is not.

 

[Dale]

A tetrad is not a vector, It just some coefficients

A tetrad is a set of 4 vectors. Usually with the further requirement that they be orthonormal. It is not just some coefficients.

 

[Jianbing_Shao]

A tetrad is a set of 4 vectors, not just some coefficients.

Vector $\frac{\partial}{\partial x^I}$ itsself is a derivative.
From vector $e_a$ we can define derivative $\partial_a={e_a}^I \frac{\partial}{\partial x^I}$,
Then from a tetrad ${e_a}^I$, How to define derivative corresponding to it?

 

[Dale]

How to define derivative corresponding to it?

The partial derivatives are vectors, but not all vectors are partial derivatives.

 

[Jianbing_Shao]

The partial derivatives are vectors, but not all vectors are partial derivatives.

Then a vector can be described with a basis $V=v^ie_i$, then how can you describe $e_a^i$ with basis $e_i$?

 

[Dale]

then how can you describe $e_a^i$ with basis $e_i$?

Isn't that what you wrote above: $$e_a = {e_a}^I \frac{\partial}{\partial x^I}$$

 

[Jianbing_Shao]

Isn't that what you wrote above: $$e_a = {e_a}^I \frac{\partial}{\partial x^I}$$

So $\frac{\partial}{\partial x^I}$ is vector, ${e_a}^I \frac{\partial}{\partial x^I}$ the sum of some vectors is also vector so $e_a$ is also vector, but $e_a^I$ is not.

 

[Dale]

So $\frac{\partial}{\partial x^I}$ is vector, ${e_a}^I \frac{\partial}{\partial x^I}$ the sum of some vectors is also vector so $e_a$ is also vector, but $e_a^I$ is not.

Correct. $e_a{}^I$ is a tetrad, which is a collection of 4 vectors, not a single vector.

 

[Jianbing_Shao]

Correct. $e_a{}^I$ is a tetrad, which is a collection of 4 vectors, not a single vector.

It is just the components of a vector, the components combine with the corresponding basis then give a vector, we can't call the components of vector vector.

 

[Matterwave]

There appears to be a lot of of confusion about notation and basic concepts of differential geometry. @Jianbing_Shao Would it be possible to tell us which reference(s) you use when studying differential geometry and general relativity? Or is it non-English? I think we can clear up a lot of issues if we clear up the notation and specifically what we mean by the notation.

I think one core confusion is this: a coordinate system defines, naturally, a coordinate basis (field). This coordinate basis (field) is a set of 4 vectors (fields). An orthonormal tetrad is also a set of 4 vectors (fields) but they don't necessarily arise from any given coordinate system. Tetrads are defined by "which vectors orthogonalize the metric?" (at each point).

The issue that I can't resolve is what do you mean with your notation.

So let me use Wald's notation because that is the book I have handy and it's clear. It is not the same as your notation (as far as I can tell) so don't attribute the same meaning to the same letters. In Wald's notation, an orthonormal tetrad is defined by the relation:

$$
(e_\mu)^a (e_\nu)_a=\eta_{\mu\nu}
$$

Where $\eta_{\mu\nu}=diag(-1,1,1,1)$ is a fixed diagonal matrix (to avoid confusion, one should not think of this as the Minkowski "metric" -- we don't use it to raise and lower the abstract indices here!). What do the letters mean though? Well, $\mu,\nu$ are simply labels for which tetrad vector (not components!) while the $a$ simply tells you that those objects are vectors (kind of like the little arrow we are used to in something like $\vec{x}$). Another note here: the upper and lower $a$ that you see in the above equation, that is simply a signal for contraction -- it is not Einstein summation convention. Usually we're good with that definition and we proceed to calculate the curvature via the (set of) connection 1-forms $\omega_{a\mu\nu}$. No where do we need to even try to bring a coordinate system into this picture.

If you demand that we bring in a coordinate system and a set of coordinate bases, then we can, and we will be able to linearly transform between our tetrad basis and the coordinate basis as such:

$$
(e_\mu)^a = \sum_\nu h_\mu^{\,\,\nu} (\frac{\partial}{\partial x^\nu})^a
$$

Notice though that my transformation does NOT use $(e_\mu)^a$ again! It's super confusing if you do! So in your first post, at the beginning, you used the symbol $e_a^{\, I}$ to mean something similar to my $h_\mu^{\,\,\nu}$ and then when you went to define the metric you reused that symbol without redefining it. Hence the confusion.

It's much more instructive if one figures this out for one's self which is why I tried not to give too much away in my first response (asking you instead to go back carefully and especially look at your $I,J$).

 

[Jianbing_Shao]

There appears to be a lot of of confusion about notation and basic concepts of differential geometry. @Jianbing_Shao Would it be possible to tell us which reference(s) you use when studying differential geometry and general relativity? Or is it non-English? I think we can clear up a lot of issues if we clear up the notation and specifically what we mean by the notation.

The notations I used just comes from the Sean Carroll's book 'SPACETIME AND GEOMETRY An introduction to General Relativity'.
In the appendix J(p483) in which noncoordinate bases is discussed.

I think one core confusion is this: a coordinate system defines, naturally, a coordinate basis (field). This coordinate basis (field) is a set of 4 vectors (fields). An orthonormal tetrad is also a set of 4 vectors (fields) but they don't necessarily arise from any given coordinate system. Tetrads are defined by "which vectors orthogonalize the metric?" (at each point).

The issue that I can't resolve is what do you mean with your notation.

In the book above, there he said:
The point of havvng a basis is that any vector can be expressed as a linear combination of basis vectors , Specifically, we can express our old basis vectors $e_\mu=\partial_\mu$ interms of the new ones:
$$e_\mu={e_\mu}^a e_a$$
The cornponents ${e_\mu}^a$ form a $n\times n$ invertible matrix (in accord with ourour usual practice of blurring the distinction between objects and their components , we will refer to the ${e_\mu}^a$ as tetrad or vielbein and often in the plural as vielbeins)

If you demand that we bring in a coordinate system and a set of coordinate bases, then we can, and we will be able to linearly transform between our tetrad basis and the coordinate basis as such:

$$
(e_\mu)^a = \sum_\nu h_\mu^{\,\,\nu} (\frac{\partial}{\partial x^\nu})^a
$$

Notice though that my transformation does NOT use $(e_\mu)^a$ again! It's super confusing if you do! So in your first post, at the beginning, you used the symbol $e_a^{\, I}$ to mean something similar to my $h_\mu^{\,\,\nu}$ and then when you went to define the metric you reused that symbol without redefining it. Hence the confusion.

It's much more instructive if one figures this out for one's self which is why I tried not to give too much away in my first response (asking you instead to go back carefully and especially look at your $I,J$).

So different basis $(e_\mu)^a$ and $(\frac{\partial}{\partial x^\nu})^a$ share the same index $\mu,\nu$?

 

[Matterwave]

The notations I used just comes from the Sean Carroll's book 'SPACETIME AND GEOMETRY An introduction to General Relativity'.
In the appendix J(p483) in which noncoordinate bases is discussed.

I see. That explains a lot.

In the book above, there he said:
The point of havvng a basis is that any vector can be expressed as a linear combination of basis vectors , Specifically, we can express our old basis vectors $e_\mu=\partial_\mu$ interms of the new ones:
$$e_\mu={e_\mu}^a e_a$$
The cornponents ${e_\mu}^a$ form a $n\times n$ invertible matrix

Ok got it. Carroll's ${e^\mu}_{a}$ is the $n\times n$ matrix that is my $h$ earlier. I won't reproduce the indices because they will only serve to confuse now. My convention followed Wald which is quite different than Carroll's.

But still be careful!! Carroll uses the spacing in indices and which index is Greek and which is Latin to track important distinctions.

(in accord with ourour usual practice of blurring the distinction between objects and their components , we will refer to the ${e_\mu}^a$ as tetrad or vielbein and often in the plural as vielbeins)

No offense to Carroll, but...

Ok so one problem (I'm not claiming this is the only problem):

$$e_a = {e_a}^I \frac{\partial}{\partial x^I}$$

I think here you are trying to reproduce eqn J.4 except you changed the index spacing (you should have written ${e^I}_a$ to stay consistent with Carroll) which confused you.

So when you got here (eqn J.6), you used the wrong matrices:

We can define the metric field:
$$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$$

Using your $I,J$ but Carroll's more consistent index spacing, to stay consistent with how you defined ${e_a}^I$ you would need to write:
$$g_{IJ}={e_I}^a {e_J}^b \eta_{ab}$$

Can you see why this is not the same equation as what you wrote? Can you see the difference between ${e^I}_a$ and ${e_I}^a$? The answer is in eqn J.3 in Carroll.

 

[Jianbing_Shao]

Using your $I,J$ but Carroll's more consistent index spacing, to stay consistent with how you defined ${e_a}^I$ you would need to write:
$$g_{IJ}={e_I}^a {e_J}^b \eta_{ab}$$

Can you see why this is not the same equation as what you wrote? Can you see the difference between ${e^I}_a$ and ${e_I}^a$? The answer is in eqn J.3 in Carroll.

From the eqn J.3 in Carroll, we can know that ${e_I}^a$ form an inverse with ${e^I}_a$,
So the two equations in eqn J.3 are equivalent with each other. isn't it?
Then what is the difference ?

 

[romsofia]

In a "natural" frame you get $$[\partial_a, \partial_b] = 0.$$

What you wrote is what relativists would say comes from a "moving" frame. I've seen it when we start from a "moving" coframe usually written something like $\theta^i = a^i_j dx^j, d \theta^i = -\frac{1}{2} C^i_{jk} \theta^j \wedge \theta^k$. These $C^i_{jk}$ are functions called structure coefficients, and it does take some work to get them into this form.

The duals to this + commutation we get $[e_i, e_j] = C^{k}_{ij} e_k$. Which is like something you start with.

In a "natural" frame, you can write the Riemann stuff as $$R_{\alpha\beta}{}^\lambda{}_\mu \equiv \frac{\partial}{\partial x^\alpha} \Gamma^\lambda_{\beta\mu} - \frac{\partial}{\partial x^\beta} \Gamma^\lambda_{\alpha\mu} + \Gamma^\lambda_{\alpha\rho} \Gamma^\rho_{\beta\mu} - \Gamma^\lambda_{\beta\rho} \Gamma^\rho_{\alpha\mu}$$

You can show, what I think you're trying to do, that in an arbitrary moving frame you can write the Riemann stuff as $$R_{\lambda\mu}{}^\alpha{}_\beta \equiv \partial_\lambda \omega^\alpha_{\mu\beta} - \partial_\mu \omega^\alpha_{\lambda\beta} + \omega^\alpha_{\lambda\rho} \omega^\rho_{\mu\beta} - \omega^\alpha_{\mu\rho} \omega^\rho_{\lambda\beta} - \omega^\alpha_{\rho\beta} C^\rho_{\lambda\mu}$$

I've never used Carroll, so not sure about his conventions and all that, but the above notation comes from Choquet Bruhat's intro GR book, and the arbitrary frame formula is exercise I.9.1 if you are able to get access to it.

 

[Dale]

Carroll, unfortunately, jumps back and forth between clear notation and unclear notation (as he admits). Since this is a point of confusion here, I would recommend using exclusively his clear notation. So (when he is being clear) he uses $\hat e_{(a)}$ to represent a tetrad (a set of four orthonormal vectors such that $g(\hat e_{(a)}, \hat e_{(b)})=\eta_{ab}$). The tetrad $\hat e_{(a)}$ forms a non-coordinate basis, but can be expressed in terms of the coordinate basis $\hat e_{(\mu)}$ as $$\hat e_{(\mu)}=e^a_\mu \hat e_{(a)}$$ where $e^a_\mu$ are the components of the tetrad in the coordinate basis.

Note carefully the distinction. $\hat e_{(a)}$ is the tetrad, a set of four vectors forming a non-coordinate basis, and $e^a_\mu$ are the components of those vectors in the coordinate basis.

Carroll admits that his notation is sloppy and uses it anyway. That is OK when one is not confused, but as soon as one becomes confused then it is a good idea to use the non-sloppy notation. So, can you go back to your original post and write it in the non-sloppy notation?

It is just the components of a vector, the components combine with the corresponding basis then give a vector, we can't call the components of vector vector.

Indeed. Which is why if we are not being sloppy then $e_\mu^a$ is not a tetrad, it is the components of a tetrad in a coordinate basis. Instead, $\hat e_{(a)}$ is a tetrad.

 

[Matterwave]

From the eqn J.3 in Carroll, we can know that ${e_I}^a$ form an inverse with ${e^I}_a$,
So the two equations in eqn J.3 are equivalent with each other. isn't it?
Then what is the difference ?

Reproducing the equations here:

$$
{e^\mu}_a {e_\nu}^a = \delta^\mu_\nu, \quad {e_\mu}^a{e^\mu}_b=\delta^a_b
$$

The two equations are logically equivalent but the inverse of a matrix is not the same as the matrix (unless the matrix is involute). I think that's where you got confused with the indices.

The two equations say, in matrix notation where $E={e^\mu}_a$ and $E^{-1}={e_\mu}^a$, $EE^{-1}=I$ and $E^{-1}E=I$ respectively. But you can't swap a matrix for its inverse, that's where you have to be careful.

So, can you go back to your original post and write it in the non-sloppy notation?

I would suggest you do as Dale says here and keep track of what each symbol actually means. We commit a lot of errors when we mistake one thing for another simply because the notation looks similar.

 

[pervect]

My $.02 and advice to the original poster. $e_a = e_a{}^I \frac{\partial}{\partial x^I}$ is not a coordinate basis, it's something more complicated.

I would recommend to the OP is that they first reviewing what a coordinate basis actually is. Rather than relying on PF posters get every little details absolutely correct, I'd suggest that the OP reviewing the subject using their textbook or reliable online source of choice. I believe I got it right, but the OP still seems to has doubts - which is fine, if they check their work and don't just blow off attempts at correction.

Hopefully a review will be sufficient, but it may turn out that there are more basic foundational elements that need clarification. For instance, I have some concern that the OP may be conflating vectors with ordered sets of numbers aka tuples. It is true that a tuple is one representation of a vector, but that doesn't make them the same thing.

After clearing the hurdle of what a coordinate basis is, I'd suggest re-writing the original question armed with this new knowledge to correct the errors. I would still suggest learning about the simpler coordinate basis first before attempting to tackle the more complex mathematical entity the OP originally asks about, which appears to be a more general basis than a coordinate basis.