Dot product between basis vectors and dual basis vectors

[Rick16]

TL;DR Summary

Why does the dot product between a dual basis vector and the original basis vector with the same index equal one?

In "A Student's Guide to Vectors and Tensors" Daniel Fleisch presents basis vectors and dual basis vectors like this:

Then he writes: "The second defining characteristic for dual basis vectors is that the dot product between each dual basis vector and the original basis vector with the same index must equal one (so $\vec e^1 \cdot \vec e_1=1$ and $\vec e^2 \cdot \vec e_2=1$). The figures show that basis vectors and dual basis vectors with different indices are perpendicular to each other, so their dot product equals zero. So far, so good. Then, in order for the dot product between basis vectors and dual basis vectors with the same index to equal one, they should be in the same direction. But the figures show that they are not in the same direction. How can their dot product then equal one?→

 

[martinbn]

The defining property for $\vec e^1$ is not only $\vec e^1 \cdot \vec e_1=1$, but also $\vec e^1 \cdot \vec e_2=0$, so it has to be orthogonal to $\vec e_2$.

 

[Rick16]

Yes, $\vec e^1$ has to be orthogonal to $\vec e_2$. This is obvious from the figures and this is not my problem. My question is, why is $\vec e^1 \cdot \vec e_1=1$? According to the figures, this should not be the case.

 

[martinbn]

Yes, $\vec e^1$ has to be orthogonal to $\vec e_2$. This is obvious from the figures and this is not my problem. My question is, why is $\vec e^1 \cdot \vec e_1=1$? According to the figures, this should not be the case.

You cannot tell that from the figure. It is chosen so that holds. The fact that is orthogonal to $e_2$ fixes the line it lies on. Then it cannot be orthogonal to $e_1$, therefore $e^1\cdot e_1 \not = 0$. If this is not equal to $1$, then change the length of $e^1$ to make this product equal to $1$.

 

[Rick16]

I see. So this is a matter of definition. $\vec e^1\cdot \vec e_1=1$ by definition. Thank you very much.

 

[martinbn]

I see. So this is a matter of definition. $\vec e^1\cdot \vec e_1=1$ by definition. Thank you very much.

Yes, you said that the author wrote:

The second defining characteristic for dual basis vectors is that the dot product between each dual basis vector and the original basis vector with the same index must equal one.

 

[Rick16]

Yes, he wrote that. But I thought that this characteristic could be inferred from first principles. Apparently this is not the case.

 

[Orodruin]

The dual basis is not necessarily parallel to the original basis. It is the case if and only if the original basis is orthogonal. The relation $\vec e^a \cdot \vec e_b= \delta^a_b$ is what defines the dual basis.

There is no requirement that either the dual basis nor the original basis consists of normalized vectors. In fact, in many situations you may not even have an inner product but the dual space and original vector space are still well defined.

 

[WWGD]

I see. So this is a matter of definition. $\vec e^1\cdot \vec e_1=1$ by definition. Thank you very much.

Yes, these are orthonormal bases; it won't hold otherwise.

 

[Orodruin]

Yes, these are orthonormal bases; it won't hold otherwise.

No. An orthonormal basis has normalized basis vectors that are all orthogonal to each other. This is not (necessarily) the case here. See above.

 

[WWGD]

No. An orthonormal basis has normalized basis vectors that are all orthogonal to each other. This is not (necessarily) the case here. See above.

It seems your post agrees with what I wrote. If $<e_i, e_j>=\delta_i^j$, isn't that the definition of an orthonormal basis?

 

[Orodruin]

It seems your post agrees with what I wrote. If $<e_i, e_j>=\delta_i^j$, isn't that the definition of an orthonormal basis?

It is not $\vec e_i \cdot e_j$, it is $\vec e^i \cdot e_j$.

$\vec e^i$ is not $\vec e_i$

 

[Orodruin]

More specifically, in Euclidean space in arbitrary coordinates (including oblique and curvilinear) $y^i$, $\vec e_a = \partial \vec x/\partial y^a$ but $\vec e^a = \nabla y^a$.

 

[WWGD]

It is not $\vec e_i \cdot e_j$, it is $\vec e^i \cdot e_j$.

$\vec e^i$ is not $\vec e_i$

Ok, I thought it was a basis with itself. But I'll just bail out.

 

[Rick16]

I have a follow-up question about the notation. Shouldn't the dot product be reserved for the homogeneous inner product?

Since the homogeneous inner product is defined as $\vec A \cdot \vec B=\mathbf g(\vec A , \vec B)=\langle \vec A, \tilde B\rangle$, we cannot write the heterogeneous inner product as a dot product, i.e. as $\langle \vec A, \tilde B\rangle=\vec A \cdot \tilde B$, because this would imply that $\vec A \cdot \vec B = \vec A \cdot \tilde B$, which is clearly wrong.

Why is it then legitimate to write $\vec e^i \cdot \vec e_j$, instead of $\langle \vec e^i, \vec e_j\rangle$ or $\vec e^i(\vec e_j)$?

 

[mathwonk]

the reason this was so confusing to me is I had not read the book referred to, and it uses different conventions from what I assume as a mathematician. (see link below for the book) in particular he is not working with orthonormal bases in spite of using the letters ej that usually imply that. as always, one has to know his definitions in order to understand his statements. (See post #8, Orodruin, second sentence, for a succinct statement of the correct definition. I wrote the following lengthy explanation before noticing post#8.)

recall that. if e1, e2 is an orthonormal basis, and we write any vector v = a1e1 + a2e2 in terms of this basis, then we can recover the coefficients a1, a2, just by dotting v with e1 and e2. i.e. a1 = v.e1 and a2 = v.e2.

In the author's system this fact means the basis is self dual. I.e. for him the dual basis B* of a basis B is the basis that gives the coefficients of B-expansions by dotting.

This is actually the usual mathematical definition of dual basis. I.e. if v,w is any basis B for the space V, then the dual basis B* is a basis of linear functions, i.e., a basis f,g for the dual space V* of linear functions on V, such that f(v) = g(w) = 1, and f(w) = g(v) =0. Then if u = av+bw, we get a as f(u) and we get b as g(u).

Now when we have a dot product on V, we can ignore V*, since then every linear function on V has form u.( ) for some u in V. Thus the dual basis for the basis B = {v,w}, will be the unique basis {u,z} also for B, such that u.v = z.w = 1, and u.w = z.v = 0.

Thus in the author's system, given a basis e1,e2, NOT assumed orthonormal, the dual basis is the unique basis e^1, e^2, such that e1.e^1 = e2.e^2 = 1, and e1.e^2 = e2.e^1 = 0. (This agrees with Orodruin's second sentence In post #8.) This dual basis will also not be orthonormal if the original basis was not.

Also, if the original basis was not orthogonal, then since e^1 is perpendicular to e2, but e1 is not, then e^1 will not be parallel to e1. etc...

see pages 113-115:
https://web.unica.it/static/resources/cms/documents/FleischVectorsAndTensors.pdf

 

[Rick16]

Thank you for your answer. However, I am still confused about something. My confusion seems to boil down to the role of the metric. By definition, $\vec A \cdot \vec B = \mathbf g(\vec A, \vec B)=\langle \vec A, \tilde B\rangle \neq \vec A \cdot \tilde B$. But perhaps $\langle \vec e^a , \vec e_b\rangle = \vec e^a \cdot \vec e_b$?

Perhaps I should clarify the notation that I use: $\vec A=A^a \vec e_a$ and $\tilde A=A_a \vec e^a$.

When I write out the action of the metric on a vector in detail, I get this:$g_{ab} \vec e^a \vec e^b(A^c \vec e_c)=g_{ab} A^c \vec e^a \vec e^b(\vec e_c)=g_{ab} A^c \vec e^a \delta ^b_c=g_{ab}A^b \vec e^a =A_a \vec e^a$. This seems to be okay.

But when I try the same with a basis vector, I get stuck: $g_{ab} \vec e^a \vec e^b(\vec e_c)=g_{ab}\vec e^a \delta ^b_c=?$

What exactly happens here? Is it true that $\langle \vec e^a , \vec e_b\rangle = \vec e^a \cdot \vec e_b$ although $\langle \vec A, \tilde B\rangle \neq \vec A \cdot \tilde B$? If I understand correctly, $\vec A \cdot \tilde B$ is not even defined.

 

[Rick16]

Perhaps I should be more explicit. This whole thing may seem trivial to experts, but I am still confused.

Here is the thing: In order to take the scalar product of two vectors, we cannot simply write $A^a \vec e_a \cdot B^b \vec e_b$ because this expression does not result in a scalar.

Instead, we have to operate with the metric tensor on one of the vectors to turn it into a covector, and then we can take the scalar product: $\langle A^a \vec e_a, B_b \vec e^b\rangle=A^aB_b\langle \vec e_a, \vec e^b\rangle =A^aB_b \delta^b_a=A^aB_a=scalar$. An alternative notation would be $A^a \vec e_a (B_b \vec e^b)$. Here is an example from a popular text: Schutz, A First Course in General Relativity, third edition, page 68: $\tilde V(\vec e_a)=\vec V \cdot \vec e_a$. Two things happen here: the covector/one-form turns into a vector, and the notation changes from bracket notation to dot notation. Is this use of dot notation and bracket notation mandatory? If it is, I wonder why the dot notation is used for basis vectors and basis covectors, like $\vec e^a \cdot \vec e_b$ instead of using the bracket notation like it is done for vectors and covectors.

And here is a really confusing point for me: If we used the dot notation for the scalar product between a vector and a covector, we would have $\langle A^a \vec e_a, B_b \vec e^b\rangle=A^a \vec e_a \cdot B_b \vec e^b$, which would imply that $A^a \vec e_a \cdot B^b \vec e_b=A^a \vec e_a \cdot B_b \vec e^b$, which in turn would imply that $B^b \vec e_b=B_b \vec e^b$. This last equation is actually correct, because both expressions represent the same object. If we have two expressions representing the same object, do we actually have to distinguish between these two expressions at all?