Coordinate and dual basis vectors and metric tensor

[Halaaku]

I have been reading an introductory book to General Relativity by H Hobson. I have been following it step by step and now I am stuck. It is stated in the book that:
"It is straightforward to show that the coordinate and dual basis vectors
themselves are related...
"e_a = g_abe^b ..."

I have been trying to prove it as follows:
e_a.e_b=g_ab=g_af[itex]\delta[/itex]^f_b
=g_af(e^f.e_b)
(e_a-g_afe^f).e_b=0
Because e_b≠0, the other side is and hence proved the given statement.
My question: is it a correct approach?

 

[Chestermiller]

Hi Halaaku. Welcome to Physics Forums.

Yes. This analysis looks good to me.

 

[FunWarrior]

Hi Halaaku,

What is important to understand here is that e_a and e^b are vectors, not numbers. Therefore, (e_a-g_afe^f).e_b = 0 only means that ea-gafef and eb are two orthogonal vectors (because the dot between them is a scalar product, not a multiplication). So the result can not be shownin this way.

A good way of doing it is the following. A vector w can be written in two ways: w = w^ae_a and w = w_be^b. Therefore, w^ae_a = w_be^b. From the previous equation in the book, you also know that w_b = g_bcw^c. It yields w^ae_a = g_bcw^ce^b. Now, the index c in the right-hand side is summed over. It means that it is a dummy index which can be renamed a. By doing it, it leads to w^ae_a = g_baw^ae^b. Since this relation holds for any arbitrary vector w of coordinates w^a, it is possible to "divide" by w^a on both sides. It follows that e_a = g_bae^b and finally e_a = g_abe^b since g_ab = g_ba.

I hope this helps.

Bye!

 

[stevendaryl]

I have been reading an introductory book to General Relativity by H Hobson. I have been following it step by step and now I am stuck. It is stated in the book that:
"It is straightforward to show that the coordinate and dual basis vectors
themselves are related...
"e_a = g_abe^b ..."

Hmm. Different books have different definitions, but the way I understand it, from Misner, Thorne and Wheeler's Gravitation, you would never write a vector and a co-vector as linear combinations of each other. They're just different types of objects.

A co-vector is a function that takes a vector and returns a scalar. A co-vector is a different type of mathematical object than a vector, so it doesn't make sense (in this way of defining things) to say that a co-vector is a equal to a certain combination of vectors.

 

[WannabeNewton]

That expression isn't saying that a covector is a combination of vectors. It's saying that covectors and vectors have a bijective correspondence under the musical isomorphism between the tangent and cotangent space, with the musical isomorphism being provided by the metric tensor. This is simply the formal way of saying that the metric tensor allows one to raise and lower indices.

It's analogous to the Reisz representation theorem.

 

[Chestermiller]

Hi Halaaku,

What is important to understand here is that e_a and e^b are vectors, not numbers. Therefore, (e_a-g_afe^f).e_b = 0 only means that ea-gafef and eb are two orthogonal vectors (because the dot between them is a scalar product, not a multiplication). So the result can not be shownin this way.

Since e_b can be arbitrary, subject only only to the condition that it cannot be expressed as a linear combination of the other coordinate basis vectors, it would seen that the dot product being equal to zero would imply that the vector in parenthesis is zero.

 

[Halaaku]

@FunWarrior: You are right in pointing out that if A.B=0, then A and B are orthogonal. Your proof totally makes sense. Thank you so much!

 

[Halaaku]

One more question again related to the proof:
Displacement between two nearby points P and Q=
ds= dx^a e_a=dx^'a e_'a
dx^a=([itex]\partial[/itex]x^a/[itex]\partial[/itex]x^'b)*dx^'b
I plug this in dx^ae_a
=([itex]\partial[/itex]x^a/[itex]\partial[/itex]x^'b)*dx^'be_a=dx^'ae_'a
I do not know what to do next. I am suppose to get:
e_'a=([itex]\partial[/itex]x^b/[itex]\partial[/itex]x^'a)e_b

 

[Halaaku]

Oh I just saw it... Relabelling of indices on the RHS to b. dx^'b would cancel with the dx^'b on the LHS and I get my answer.

 

[Halaaku]

For the dual basis transformation I take ds= dx_ae^a=dx_'ae^'a
I proceed as I had done for the vectors but I get the following:
e^'a=[itex]\partial[/itex]x_b/[itex]\partial[/itex]x_'a e^b

I should be getting:
e^'a=([itex]\partial[/itex]x_'a /[itex]\partial[/itex]x_b) e^b