I Understanding dual basis

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1. Mar 13, 2016

Vanille

Hello all!
I've just started to study general relativity and I'm a bit confused about dual basis vectors.
If we have a vector space $\textbf{V}$ and a basis $\{\textbf{e}_i\}$, I can define a dual basis $\{\omega^i\}$ in $\textbf{V}^*$ such that: $$\omega^i(\textbf{e}_j) = \delta^i_j$$But in some pdf and documents I found this relationship: $$\omega^i\cdot\textbf{e}_j = \delta^i_j$$ So I don't understand why these two relationships are equals.
In fact I know that there's an isomophism $\Phi: \textbf{V}\rightarrow\textbf{V}^*$ induced by the inner product in such a way that: $$\tilde v(\textbf{u}) = \textbf{v}\cdot\textbf{u}\qquad\forall\textbf{u}\in\textbf{V}$$Where $\tilde v$ is the covector associated to the vector $\textbf{v}$ by the isomorphism $\Phi$.
So I expect that the basis associated to the dual basis is exactly the reciprocal basis: $$\omega^i(\textbf{e}_j) = \textbf{e}^i\cdot\textbf{e}_j =\delta^i_j$$.So the dual basis $\{\omega^i\}$ seems to be equal to the reciprocal basis $\{\textbf{e}^i\}$.
I think I'm doing a very bad mistake.
Can anyone help me, please? Thank you!

2. Mar 13, 2016

Orodruin

Staff Emeritus
In order to understand one-forms, it really helps to consider what they are independently from the existence of a metric, i.e., elements of the dual of the tangent vector space. Without the metric, things such as $\vec v \cdot \vec w$, where $\vec v$ and $\vec w$ are tangent vectors do not really make any sense whatsoever. However, the dual vector space still exists and it is sometimes common to simply write $\omega(X)$, where $\omega$ is a one-form and $X$ a tangent vector as $\omega \cdot X$. In particular, when a metric does exists it largely erases the need to separate one-forms from tangent vectors. The only difference between the expressions is one of notation.

3. Mar 13, 2016

Vanille

right,so i think that my mistake is an abuse of notation... in fact in the first case the operator between $\omega^i$ and $\textbf{e}_j$ is a bilinear operator on $\textbf{V}^*\times \textbf{V}$ whose properties resemble the properties of an inner product. In the second case, i use the inner product, defined as a bilinear operator on $\textbf{V}\times \textbf{V}$. So if i denote the first operator with the symbol $<\cdot,\cdot>$, i can write the first case in this way: $$<\omega^i,\textbf{e}_j> = \delta^i_j$$

But i'm not sure of this argument. For example, if i want calculate the covariant derivative of a covector, say $\tilde v = v_i \omega^i$, i can do: $$\frac{\partial \tilde v }{\partial x^k} = \frac{\partial }{\partial x^k}(v_i \omega^i) = v_i\frac{\partial \omega^i}{\partial x^k}$$ Now, since: $$<\omega^i,\textbf{e}_j> = \delta^i_j$$ and we've defined the Christoffel symbols as: $$\frac{\partial \textbf{e}_i}{\partial x^k} = \Gamma^j_{ik} \textbf{e}_j$$ thus: $$\frac{\partial }{\partial x^k}<\omega^i,\textbf{e}_j> = <\frac{\partial \omega^i}{\partial x^k},\textbf{e}_j> + <\omega^i,\frac{\partial \textbf{e}_j}{\partial x^k}> = <\frac{\partial \omega^i}{\partial x^k},\textbf{e}_j> + \Gamma^i_{jk} = 0$$ Now, we can expand $\frac{\partial \omega^i}{\partial x^k}$ in terms of the dual basis as follow: $$\frac{\partial \omega^i}{\partial x^k} = \Pi^i_{nk} \omega^n$$ So finally: $$\Pi^i_{jk} = - \Gamma^i_{jk}$$ I've seen only one book do this operation in this way, all the others use the identity $\omega^i\cdot\textbf{e}_j = \delta^i_j$.
I can't believe that many people use this abuse of notation, it's more probably i'm doing something wrong...

EDIT: i don't know why i have "math processing error"

Last edited by a moderator: Mar 14, 2016
4. Mar 13, 2016

DrGreg

I think you need to write a two-letter suffix as "_{jk}" instead of "_j_k" e.g. $\Gamma^i{}_{jk}$ and not $\Gamma^i_j_k}$

If you're reading this soon after I wrote it, you should be able to go back and edit your previous message.

5. Mar 14, 2016

Orodruin

Staff Emeritus
This was indeed the problem. I have used my supernatural powers to fix this.

You are missing the partial derivative of the vector component here. Also, you really should not be using the partial derivative notation here, but the notation of the affine connection.

But it is the same identity, just a different notation! In my opinion, using the partial derivative to denote the affine connection is a greater problem in general manifolds although I understand where you come from - I introduce the covariant derivative in a general coordinate system in a Euclidean space in exactly this fashion in my lecture notes and I find it most intuitive to think of the Christoffel symbols as derivatives of the coordinate basis.

6. Mar 14, 2016

Ibix

Off topic, but I would find that a rather disturbing statement coming from someone with your name, were I in Vanille's place...

7. Mar 14, 2016

Orodruin

Staff Emeritus
Agh burzum-ishi krimpatul.

You are assuming everyone is familiar with this?

8. Mar 14, 2016

Ibix

It translates to "And in the Google find them", if my Black Speech is up to the task. But no, I guess not everyone will recognise a Sindarin name for Mount Doom.

9. Mar 14, 2016

Vanille

Yes sorry, my bad! i was focused on dual basis and i missed the partial derivative of the vector component! Also, you're right again; it's better using the notation of the affine connection $\nabla$.

So am i right? i think there's an abuse of notation. For the first expression: $\omega^i\cdot\textbf{e}_j = \delta^i_j$ with the operator $\cdot$ i mean the bilinear operator between a vector and a covector or one-form. For the second expression: $\textbf{e}^i\cdot\textbf{e}_j = \delta^i_j$ with the operator $\cdot$ i mean the inner product in the vector space $\textbf{V}$.

10. Mar 14, 2016

Orodruin

Staff Emeritus
I would say the abuse of notation is more severe when you write $\textbf{e}^i\cdot\textbf{e}_j = \delta^i_j$. This is not defined without referring to a metric tensor and it is not even clear what you intend by $\textbf{e}^i$. What is well defined without referring to a particular metric is $\omega^i(\vec e_j) = \delta^i_j$. From this point, you only need to introduce the affine connection in order to find the relation between the dual basis and how it changes with respect to the connection coefficients.

11. Mar 14, 2016

Vanille

$\textbf{e}^i$ are the reciprocal basis. Ok, now i've understood my mistake; the key is to see the scalar product between a vector and a covector as a tensor of rank (1,1), that's the application of a covector to a vector or vice versa; while the scalar product between two vectors as a tensor of rank (0,2), that's the metric tensor; and finally the scalar product between two covectors as a tensor of rank (2,0), that's the inverse metric tensor. So, you're right again; i must define a metric tensor as above and then my relationships aren't in conflict. Thank you! now it's all clearer :)

12. Mar 14, 2016

stevendaryl

Staff Emeritus
I probably shouldn't bring this up, since it's probably bad pedagogy, but I have actually seen textbooks that make use of a double set of basis vectors:

$e^i$ and $e_j$

with the relationship $e^i \cdot e_j = \delta^i_j$. This was in a context where there was a metric (it was talking about a non-orthogonal basis for $R^n$).

For example, http://physastro-msci.tripod.com/webonmediacontents/notes1.pdf (see Equation 17)

13. Mar 14, 2016

Orodruin

Staff Emeritus
I do this as well, but only when I discuss general coordinate transformations in $\mathbb R^n$. I find it usually helps students to produce a mental image and more intuition. Then when I get to more general manifolds I discuss what parts of these definitions we can keep and which we should throw out of the window, essentially letting go of seeing $\vec e_i$ and $\vec e^i$ as spanning the same vector space and instead seeing the former as spanning the dual of the latter.

Last edited: Mar 14, 2016
14. Mar 14, 2016

pervect

Staff Emeritus
I'd agree with other posters that this is the clearest and most fundamental notation.

I'd interpret this as follows. "i" is something I'll call a selector (there may be a better name). It select one of four different one-forms. The notation is confusing because these selectors are mixed in with tensor notation where the tensor indices have a different interpretation. Similarly, j selects one of four different vectors from a set of vectors.

Thus $\{ \omega^0, \omega^1, \omega^2, \omega^3\}$ are all one-forms, rank-one tensors. Assuming a 4-d space-time, each of these tensors has four components. Thus, considering the tensor $\omega^0$ we might write it's 4 components as ${(\omega^0)}_a$ (in abstract index notation) or $(\omega^0)_\mu$ (in component notation).

So we interpret $\omega^i$ as some member of the set $\{ \omega^0, \omega^1, \omega^2, \omega^3\}$ , each member of the set which is a tensor in its own right. Each tensor in the set has 4 components, but when we write $\omega^i$ we've omitted writting down the components - this is called index free notation. The selector i in $\omega^i$ is confusingly because it's not the usual tensor index, but something I'm calling a selector (there may be a better name for it). It selects which one form in the set we are talking about.

As you've correctly noted, you don't really need a metric to compose a one-form with a vector to get a scalar, because a one-form is a map from a vector to the scalar. The index-free notation obscures this a bit. Hopefully this post will aid in interpreting those books and papers which insist on using a more confusing notation.

Last edited: Mar 14, 2016
15. Mar 14, 2016

Ibix

If I'm following this correctly (and I may not be), Carroll's lecture notes use brackets to identify what you're calling a selector. So the $\mu$th basis vector is $\hat{e}_{(\mu)}$. It's not a clash with symmetrisation notation because the brackets can only contain one "index".