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.