I Confused about the inner product

[Rick16]

well this seems to be what you are asking me to do, if I agree to say that
A*.( ) equals A instead of A**.

Yes, sorry, I did not realize that #44 was connected to #40, because there was another post between them, and I did not understand #44 correctly. Actually, I would not even have thought that this could be an issue. If V* is the dual vector space of V, I would automatically assume that V is the dual vector space of V*. The word "dual" seems to imply that there are two of them, and that there is reciprocity between them. But as usual, things are apparently not as easy as an amateur believes them to be.

 

[Rick16]

So, fuzzy as this sounds, Fleisch seems to be saying he wants to identify V and its inner product with R^n and the standard inner product. But if he does this, then a vector A has to be identified with the coordinate vector (a^1,...,a^n) when thought of as a vector, but with the coordinate vector (a1,...,an) when thought of as a covector........???? yoicks. (As the unpicky rat in Ratatouille said when the gourmet rat asked what he was eating, I don't really know.)

Do you really mean Fleisch now, or are you still talking about Bernacchi?

 

[mathwonk]

I am sure you can guess my answer to that: i.e. I meant what I said.
I do not have a copy of Bernacchi and have not read past page 11 online, since I had to wait 20 seconds for every page to load. So I have not reached Bernacchi's discussion of what happens with an arbitrary inner product on a space. But Bernacchi freely discusses the space V* in pages 1-11, as different from the space V, which Fleisch does not.

In post#50, right before the words: "So, fuzzy as this sounds", insert the words: "Now what does this tell us about the discussion in Fleisch, where V* and its natural dual basis {fj} do not appear, but the (inner product dependent) 'dual basis' {u^j} for V does appear?"

 

[Rick16]

I want to recapitulate what I have understood so far. At the heart of the matter are expressions of the type $V_i\vec e^i$. Here is one more time Fleisch's equation from page 133: $\vec A=A^i\vec e_i=A_i\vec e^i$. This equation means that $A^i\vec e_i$ represents vector A in terms of contravariant components and basis vectors, and $A_i\vec e^i$ represents vector A in terms of covariant components and dual basis vectors.

And here is equation 1.5 from Bernacchi (page 7), showing the expansion of a covector: $\tilde P=\tilde e^\alpha P_\alpha$. The expressions $A_i\vec e^i$ and $\tilde e^\alpha P_\alpha$ use slightly different notational conventions and different variables, but they are still the same expression. From this fact--that these expressions are the same--I deduced that $\vec V=\tilde V$. But $\vec V\neq \tilde V$. Therefore $A_i\vec e^i$ and $\tilde e^\alpha P_\alpha$ can not mean the same thing, although they look the same.

Where is the difference between these two expressions? $\tilde e^\alpha P_\alpha$ represents a covector, and therefore the dual basis vector $\tilde e^\alpha$ should logically belong to the dual vector space V*. $A_i\vec e^i$, on the other hand, represents a vector and therefore the dual basis vector $\vec e^i$ should logically belong to the vector space V.

You already spent a lot of time explaining this back in post #12, and here is an important quote from this:

In particular the elements e^j of the "dual basis", are not covectors, nor are they a basis of the dual space. Rather given a basis e1,...,en for the space V, with "dual basis" e^1,...,e^n, the operators e^1.( ), ..., e^n( ), i.e. the corresponding covectors, give the corresponding good basis of the dual space V*.

Do I understand correctly that the elements e^j from the above quote correspond to Fleisch's dual basis vectors $\vec e^i$ and that the operators e^j.( ) correspond to Bernacchi's dual basis vectors $\tilde e^\alpha$?

The difficulty for me in understanding this lies in the fact that you treat a covector as a function and you treat a vector as not a function. This is apparently how it is generally done, as other posts on this thread use the same approach. But I am not familiar with this approach. My two major sources are Fleisch and Bernacchi. Fleisch does not use the functional approach at all and only mentions it in a note at the end. Bernacchi treats everything as functions, covectors and vectors alike. I must find a text that explains the approach with covectors as functions in order to see clearer in all this. Can you for now tell me if my conclusion from above goes in the right direction, i.e. that your elements e^j correspond to Fleisch's dual basis vectors $\vec e^i$ and that the operators e^j.( ) correspond to Bernacchi's dual basis vectors $\tilde e^\alpha$?

 

[PeroK]

The difficulty for me in understanding this lies in the fact that you treat a covector as a function and you treat a vector as not a function. This is apparently how it is generally done, as other posts on this thread use the same approach.

That seems a natural approach to me. We start with a vector space with an inner product and we define the dual space, which is also an inner product space. We can then treat the dual space as our vector space and define the dual space of that. In the cases of interest here, the dual of the dual is isomorphic to the original vector space. And you have complete symmetry.

At that point, you really need to move on with the physics. This has become a serious mental block for you, where you're spending a lot of your limited time exhausting a subject that is tangential to the actual physics.

You need to find a way to move on from this.

 

[Rick16]

You need to find a way to move on from this.

Yes, I will try to move on. I have actually ordered Hartle's book based on your recommendation, but while waiting for it, I thought I could still spend some time pondering this question.

 

[jbergman]

I want to recapitulate what I have understood so far. At the heart of the matter are expressions of the type $V_i\vec e^i$. Here is one more time Fleisch's equation from page 133: $\vec A=A^i\vec e_i=A_i\vec e^i$. This equation means that $A^i\vec e_i$ represents vector A in terms of contravariant components and basis vectors, and $A_i\vec e^i$ represents vector A in terms of covariant components and dual basis vectors.

That equation is wrong in my opinion. It should read $\vec A=A^i\vec e_i \cong \tilde A =A_i\vec e^i$, where $\cong$ means isomorphic not equal and the isomorphism is given by $g(\vec A) = \tilde A$.

As far as the functional approach versus the component approach they are essentially the same if you know enough linear algebra to make the connections. This is due to the fact that a linear function is completely determined by its values on a basis of a vector space. As a others have said, you should probably move on as to fully understand this you probably need a text that compares all the perspectives or you could read @fresh_42 insight articles on tensors.

 

[mathwonk]

Everything depends on definitions. In Fleisch's book, the symbol e^j is defined as a vector in V, such that ui.u^j = 1 iff i=j and 0, otherwise. To me this makes his equation A = Aje^j correct, i.e. consistent with his (possibly non standard) definitions.
In other books apparently the same symbol e^j denotes instead a covector in V* whose value at ei equals 1 iff i=j and 0 otherwise. With this other definition, indeed ATilda = Aje^j, but with Fleisch's definition, I agree that A = Aje^j (in V), and ATilda = A.( ) = Aj(e^j.( )) (in V*), [although AFAIK Fleisch does not mention V* and hence does not write this last equation].

I also agree as to the wisdom of moving on, and will try to.

 

[jbergman]

Everything depends on definitions. In Fleisch's book, the symbol e^j is defined as a vector in V, such that ui.u^j = 1 iff i=j and 0, otherwise. To me this makes his equation A = Aje^j correct, i.e. consistent with his (possibly non standard) definitions.
In other books apparently the same symbol e^j denotes instead a covector in V* whose value at ei equals 1 iff i=j and 0 otherwise. With this other definition, indeed ATilda = Aje^j, but with Fleisch's definition, I agree that A = Aje^j (in V), and ATilda = A.( ) = Aj(e^j.( )) (in V*), [although AFAIK Fleisch does not mention V* and hence does not write this last equation].

I also agree as to the wisdom of moving on, and will try to.

I looked briefly at his chapter 4 on covariant and contravariant tensors and you appear to be correct. I would say that his treatment is far from the standard and would recommend something by Schutz to for a more standard presentation or maybe even MTW if you are adventurous.

There are definitely some nuggets in there and it is an interesting geometric explanation of dual vectors, but his conception of dual vectors is not what we typically describe as a covector. You can get from one to the other using the isomorphisms from the inner product but he doesn't mention them or at least the brief part I read, didn't.