I Confused about the inner product

[Rick16]

TL;DR Summary

confused about the inner product

I posted a follow-up question on an earlier question, but did not get a response. So I will start a new thread.

A vector $\vec V$ and a covector/one-form $\tilde V$ represent the same object, right? $\vec V=V^i \vec e_i = V_i \vec e^i = \tilde V$.

We also have $\mathbf g(\vec V)=\tilde V$, where $\mathbf g$ is the metric tensor.

Putting these two together, I get $\mathbf g(\vec V)=\vec V$. That's odd. How can I make sense of this?

And when I look at equations like $\tilde V(\vec A):= \mathbf g(\vec V, \vec A)=\vec V \cdot \vec A$ (Schutz, A First Course in General Relativity, 3rd edtion, page 68), I wonder if it is really necessary to do it like this. Couldn't I simply write $\tilde V(\vec A)=\vec V(\vec A)$, or $\tilde V \cdot \vec A=\vec V \cdot \vec A$, since $\vec V$ and $\tilde V$ are the same object?

Is the distinction between bracket notation and dot notation purely conventional, is it always done like this, is it necessary to do it like this?

 

[anuttarasammyak]

TL;DR Summary: confused about the inner product

A vector V→ and a covector/one-form V~ represent the same object, right? V→=Vie→i=Vie→i=V~.

It seems that these two vectors which do not share same direction in general cannot be same when each multiplied with number.

 

[fresh_42]

Question: Are the number two and the function $x\longmapsto 2x$ the same object?

 

[Rick16]

Why do they not share the same direction? Here is equation 4.25 from Fleisch (Daniel Fleisch, A Student's Guide to Vectors and Tensors, page 116): $\vec A=A^x \vec e_1 +A^y \vec e_2=A_x \vec e^1 +A_y \vec e^2$.

 

[Rick16]

And here is a citation from Fleisch (page 156): "... the traditional approach tends to treat contravariant and covariant components as representations of the same object, whereas in the modern approach objects are classified either as "vectors" or as "one-forms" (also called "covectors")."

So I learned from Fleisch that the two expressions $V^i \vec e_i$ and $V_i \vec e^i$ represent one and the same object. If the modern approach treats them differently, how can that change the underlying fact that they are the same object?

 

[martinbn]

And here is a citation from Fleisch (page 156): "... the traditional approach tends to treat contravariant and covariant components as representations of the same object, whereas in the modern approach objects are classified either as "vectors" or as "one-forms" (also called "covectors")."

So I learned from Fleisch that the two expressions $V^i \vec e_i$ and $V_i \vec e^i$ represent one and the same object. If the modern approach treats them differently, how can that change the underlying fact that they are the same object?

There is no such underling fact! They are not the same object. He doesn't say that they are. If you have a metric you can identify covectors and vectors and treat them as if they were the same. But they are not. And if you have a different metric they will be identified in a different way.

 

[Ibix]

Vector and covector aren't the same object. They live in different vector spaces, for a start. However, if you have a metric then there is a one-to-one map between vectors and covectors, so you can always get uniquely and reversibly from one to the other and in that case you might not care too much about the distinction.

In physics, as far as I'm aware, you can always assume the presence of a metric. So you can always turn a vector into its dual and vice versa, so it's possible not to care whether you work with a vector or covector except to check that the indices match up when you do maths. I suspect that kind of thing is what Fleisch is referring to with "traditional". It's wrong and causes mathematicians to shout at physicists a lot, but you can get away with it.

 

[fresh_42]

I'm not really happy with Fleisch's presentation as it mixes two different concepts of what a dual basis is. He discusses two different bases of the same vector space, connected by a transformation $T$. One basis is $B=\{\vec{e}_1,\vec{e}_2\}$ and the other one is $B'=\{\vec{e}^1,\vec{e}^2\}.$ In this sense, we have
$$
\vec{A}=A^x\vec{e}_1+A^y\vec{e}_2= \left(A^x,A^y\right)_B=\left(A_x,A_y\right)_{B'}=A_x\vec{e}^1+A_y\vec{e}^2 \quad\quad (*)
$$
the same vector $\vec{A}$ represented in two different bases. This has priorily nothing to do with duality and can be done for any basis transformation.

Then he uses the word duality to describe that the two bases are related by an inner product $g$ and
$$
g(\vec{e}_i,\vec{e}^j) = \delta_{ij}\quad\quad (**).
$$
The standard inner product in this context would be $g=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$

This is a purely geometric perspective that takes place in the same vector space so that we get two different expressions for the same vector $\vec{A}.$ Angles and lengths are defined by the metric $g$ which is an inner product.

There is also an algebraic point of view that uses the word duality. In this case, a dual vector is a linear transformation, a dual form. We start with a vector $\vec{e}_1$ and consider the linear transformation
$$
\vec{e}^1\, : \,\vec{A}\longmapsto g\left(\vec{A},\vec{e}_1\right)
$$
from the vector space $V$ into the (one-dimensional) vector space of real numbers $\mathbb{R}.$ This linear transformation is written as $\vec{e}^1$ and is also called dual vector since it is defined by $g$ and $\vec{e}_1.$ (The same is done with $\vec{e}_2$ to obtain $\vec{e}^2.$) In this algebraic case, we identify the dual vectors with linear transformations. In my one-dimensional example, we would identify $\vec{2}$ with the dual vector, a linear transformation, $\vec{2}^*\, : \,\vec{x}\longmapsto \dfrac{1}{2}\cdot \vec{x}$ so that $g\left(\vec{2},\vec{2}^*\right)=1.$

Both concepts, the geometrical and the algebraical, are basically the same since they are defined by the same equation $(**)$ of the metric. Moreover, the vector space $V$ and the vector space $V^*$ of linear transformations $V\rightarrow \mathbb{R}$ are isomorphic and can be identified. However, this isomorphism isn't a natural one. It depends on the equation $(**)$ and is as such dependent on $g.$ This isomorphism allows us to write the equation $(*).$ However, it is algebraically problematic since it identifies two objects that are not in the same vector space: $\vec{A}\in V=\operatorname{span}\{\vec{e}_1,\vec{e}_2\}$ and $\vec{A}^*\in V^*=\operatorname{span}\{\vec{e}^1,\vec{e}^2\}.$ The identification via $g$ makes them "the same" and phrased in the language of projections, they are the same. Phrased in the language of linear algebra, they do not live in the same vector space, only in an isomorphic copy. Hopefully, Fleisch will explain this duality on the next pages.

 

[Rick16]

He discusses two different bases of the same vector space

This is apparently the crux of the matter. Fleisch repeatedly stresses the point that we always deal with one and the same object, and I really got this idea anchored in my head. On page 133 he writes again:

"So vector $\vec A$ represents the same entity whether it is expressed using contravariant components $A^i$ or covariant components $A_i$: $$\vec A=A^i \vec e_i = A_i \vec e^i,$$ where $\vec e_i$ represents a covariant basis vector and $\vec e^i$ represents a contravariant basis vector."

If I understand correctly, this is true when both entities are in the same vector space, whereas in general, vectors and covectors are in different vector spaces and are therefore different objects.

This now leaves me with the question about the notation. Is it generally so that the dot notation is only used for the inner product between two vectors or two covectors ($\vec A\cdot \vec B$ or $\tilde A \cdot \tilde B$), whereas the bracket notation is only used for the inner product between a vector and a covector ($\vec A(\tilde B)$ or $\langle \vec A, \tilde B\rangle$)?

If this is so, wouldn't it be more systematic to respect this notation also for inner products between bases? It seems to me that in the case of bases we can use whatever notation we want, i.e. $\vec e^i\cdot \vec e_j = \langle \vec e^i, \vec e_j\rangle = \vec e^i(\vec e_j)$. Is this correct?

 

[fresh_42]

This is apparently the crux of the matter. Fleisch repeatedly stresses the point that we always deal with one and the same object, and I really got this idea anchored in my head. On page 133 he writes again:

"So vector $\vec A$ represents the same entity whether it is expressed using contravariant components $A^i$ or covariant components $A_i$: $$\vec A=A^i \vec e_i = A_i \vec e^i,$$ where $\vec e_i$ represents a covariant basis vector and $\vec e^i$ represents a contravariant basis vector."

If I understand correctly, this is true when both entities are in the same vector space, whereas in general, vectors and covectors are in different vector spaces and are therefore different objects.

The inner product is defined as a bilinear transformation of the same vector space $V\times V$ into the scalar field. But as soon as we write vectors by their coordinates, we have $\mathbb{R}^n\times \mathbb{R}^n$ and it doesn't matter anymore what $\mathbb{R}^n$ has represented before. This is particularly true for $V\cong \mathbb{R}^n\cong V^*.$

We can also formally define an inner product $V\times V^*$ by $\bigl\langle v,w^* \bigr\rangle =w^*(v).$

They are all the same if written in coordinates. An inner product space is normally given by an inner product $V\times V$ on the same vector space for both components. Mixing all of them and even hiding them behind projections isn't very enlightening in my opinion. That's my criticism of Fleisch's presentation.

This now leaves me with the question about the notation. Is it generally so that the dot notation is only used for the inner product between two vectors or two covectors ($\vec A\cdot \vec B$ or $\tilde A \cdot \tilde B$), whereas the bracket notation is only used for the inner product between a vector and a covector ($\vec A(\tilde B)$ or $\langle \vec A, \tilde B\rangle$)?

As already mentioned, if at all, then the dot refers to the same vector space, and the bracket allows the consideration of $\bigl\langle V,V^* \bigr\rangle.$ But there is not really a difference. It is a notation and you have to figure out which way an author uses them in each case. I think it is also often used to distinguish real and complex cases since in the complex case, we no longer have bilinear, but sesquilinear products:
$$
\bigl\langle \alpha\vec{v},\beta\vec{w} \bigr\rangle = \alpha \bar{\beta}\bigl\langle \vec{v},\vec{w} \bigr\rangle.
$$
You wouldn't expect this from a dot product. On the other side, we write $z\cdot \overline{z}$ for complex numbers, i.e. with a dot. But do not really on such conventions. They might differ from author to author.

We here call it simply a scalar product since the result is a scalar. That covers all cases and doesn't require a certain notation. The downside is, that you could confuse it with a scalar multiplication. The solution is that we call one a product and the other one a multiplication.

If this is so, wouldn't it be more systematic to respect this notation also for inner products between bases? It seems to me that in the case of bases we can use whatever notation we want, i.e. $\vec e^i\cdot \vec e_j = \langle \vec e^i, \vec e_j\rangle = \vec e^i(\vec e_j)$. Is this correct?

Yes, see my comments above. I think it even has a historical dimension. I use keyboard shortcuts and $\bigl\langle \vec{v},\vec{w} \bigr\rangle$ took me 10 keys (including the 4 for the tags) to write it. Not so long ago, people had to change the typeball of their typewriters to do the same! I would certainly have preferred the dot over the bracket!

The culmination of all is the total differential
$$
df(v)=\sum_{k=1}^n v_k \dfrac{\partial f}{\partial x_k}\,dx_k= v^k\,\partial^k f \,dx_k= (df)\cdot \vec{v}=\nabla f\cdot v=\operatorname{grad}f\cdot v=\bigl\langle \nabla f,v \bigr\rangle
$$
and to make the notational confusion perfect with Einstein notation. Now figure out what the components are, what the summation index is, what the basis vectors are, and of which vector spaces, the tangent space or its dual space, the cotangent space?

 

[Rick16]

Thank you very much for all the answers. This has been very enlightening. I will have to rethink everything I thought I knew, but I should be able to make more sense of it now.

 

[mathwonk]

I am not a physicist, so take this with a grain of salt. But I'll try to give a math guy's view of this.

A vector represents a physical arrow in a vector space, say a position (with respect to a given origin), or (more intrinsically) a velocity. A covector represents an operator on vectors that spits out a number. A vector plus a metric yields a covector. I.e. if A is a vector in V, then A.( ) is the corresponding covector in V*.

Some objects in physics are naturally vectors and some are naturally covectors. If we have a basis e1,...,en of our vector space V, we use it to represent our vectors, i.e. to know where they are. Thus we write A in terms of the e1,...,en to locate A. If we have a covector in V* such as the dot product operator A.( ), we would like to write it in terms of a nice basis of V*.

Now if e1,...,en is a basis of V, then the dot products e1.( ),......, en( ), will be a basis of of V*, but not a particularly useful one unless {ej} is orthonormal. I.e. if we are given a vector A, written in terms of the ej, then A.( ) can be written in terms of the basis ej.( ), but this won't give us much information about the operation A.( ). I.e. we would like to be able to evaluate A.v on any v, if we know the expansion of v as a1e1+...+anen. But this is not so easy, since when {ej} is not orthonormal, we don't even know how e1 acts on itself, or any of the other ej. Thus writing A.( ) in terms of the ej.( ), is unhelpful, not even to compute A.e1. Of course we can do it, but the answer will be complicated, and we want it to be simple.

Thus if we are really interested in the operator A.( ) rather than the physical vector A, we want to write A.( ) in terms of a nicer basis of V*. That means finding a basis of V* whose action on the basis {ej} is very simple. This is given by the basis {e^j.( )} for V*. I.e. given the vector basis e1,...en, and a vector A = a1e1+...+anen, to understand the operator A.( ), we find the "dual basis" e^1,...,e^n for V, and rewrite A in terms of this basis, A = c1e^1+...+cne^n.

Then we have the expansion A.( ) = c1(e^1.( )) +....+ cn(e^n.( )), and this is very useful for computing the operator A.( ) on vectors. I.e. given a vector B = b1e1+...+bnen, A.B is just c1b1+...cnbn.

For comparison, using the expansion A = a1e1+...+anen, we have
A.B = a1b1(e1.e1) + a1b2(e1.e2) +....+ a1bn(e1.en)
+ a2b1(e2.e1) + a2b2(e2.e2)+......+a2bn(e2.en)
......
+anb1(en.e1) +......+anbn(en.en). So kind of a long mess, n^2 terms.

It's the difference between using an arbitrary matrix or a diagonal matrix to compute the dot product; or in fact, just the identity matrix. I.e. given a basis {ej} for V, the dot product on V is given by a matrix, and finding the dual basis {e^j} essentially diagonalizes that matrix, and even makes all diagonal elements equal to 1.

To repeat, given A = a1e1+...+anen in V, if what we are really interested in is computing dot products A.B for all B in V, then we want to first rewrite A in terms of the dual basis e^1,...,e^n, as A = c1e^1+...+cne^n.
Then writing A in terms of the "dual basis" {e^j}, and writing B in terms of the original basis {ej}, makes it easy to compute A.B.

So indeed a1e1+...+anen, and c1e^1+...+cne^n are the same object, namely both equal the vector A in V, but the second representation is of interest for studying not so much the physical object A, but its role as an operator A.( ).

On the other hand, even though we may be interested more in the operation of dotting with a given vector, there may still be some meaning in the physical position of that vector, since of course that position does determine its action when dotting. E.g. although the gradient of a function of two variables is, in modern terms, an operator on tangent vectors, telling you the derivative of the function in that direction, still the physical vector whose dot product gives you that directional derivative can be visualized as a vector pointing in the direction where the function increases fastest.

I.e. given a (curve C with) velocity vector v at p, in the domain of f(x,y) in the plane, then grad_pf is the covector whose value at v is grad_pf(v) = the slope of the graph at the point above p, and in the direction of v, (and multiplied by the length of v). But it also equals the dot product del_pf.v, where del_pf is the vector in the plane whose coordinates are the partials of f. The vector del_pf lives in the plane V, and the covector grad_pf lives in the dual plane V*. This is all confusing of course since we are used to calling the vector del_pf also "grad f at p", i.e. we speak of "the gradient vector". Thus the gradient as a covector, is the operation of dotting with the gradient vector.

So always keep in mind there are two different concepts: w a vector, and w.( ) a covector, the operation of dotting with that vector. The confusion arises by dropping the notation w.( ) for the dot product, and thinking of w and w.( ) as the same object, which they are not.

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*.

In modern terms, it is the operators e^1.( ), ..., e^n( ) which are properly called the dual basis for e1,...,en. I.e. in modern language, the dual basis of a basis for V, is a basis for V*. This states the relation of the bases {ej} and {e^j} this way: given a basis {ej} for V, find the basis {e^j} for V whose dot products yield the correct, intrinsic, dual basis for {ej} in V*.

Note: if you keep the basis {ej} but change the metric, the basis {e^j} will also change, (since the correspondence between V and V* will change), but the intrinsic dual basis for V* given by the operators {e^j.( )} will not change. I.e. if we change the metric, the "dual basis" of {ej} in V will change to some other basis {d^j} for V, but the resulting basis for V*, given by the operators {d^j.( )} will be the same as the old basis {e^j.( )}.

 

[Rick16]

I am not a physicist

And I am not a mathematician, so please excuse my ignorance. I am just a guy who got a bachelor in physics years ago and is now trying to learn general relativity on his own, in his (very limited) spare time.

A covector represents an operator on vectors that spits out a number.

On the other hand, a vector can also be an operator on a covector. If I am not mistaken, $\vec A(\tilde B)=\tilde B(\vec A)$. But this detail is not even so confusing. I just find the whole approach very unexpected. The confusion sets in later.

the dot product operator A.( )

This is just a little detail, but one of my issues was with notation and terminology. It seems that the terms "dot product", "inner product", and "scalar product" can really be used interchangeably, and even regardless of the notation used.

So indeed a1e1+...+anen, and c1e^1+...+cne^n are the same object, namely both equal the vector A in V, but the second representation is of interest for studying not so much the physical object A, but its role as an operator A.( ).

Now it gets really confusing. I just thought that I had understood that they are not the same object, because they live in different vector spaces. Isn't it rather so that they both represent the same object, but are not the same object?

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*.

This is even more confusing. Is this just meant to clarify the meaning of the basis e1,...,en? You also write:

The confusion arises by dropping the notation w.( ) for the dot product

So, when you wrote "c1e^1+...+cne^n", did you drop the notation for the dot product and did you mean "c1e^1.( ), ..., cne^n( )", or should I see these two as different objects? I feel that I am beginning to lose it, and I must stop for now.

In any case, thank you very much for this post. I think it will keep me busy for a while.

 

[mathwonk]

I meant exactly what I said in every instance you cite. in particular, a covector is an operator on vectors. a vector cannot be viewed as operating on vectors without a dot product. a vector is not a covector. a vector, plus a dot product does allow the construction of a corresponding covector, namely the operation of dotting with the vector. If A is a vector, then A.( ) is the corresponding covector. In particular, these should be seen as two different objects. This is what fresh42 meant in post#3 by the rhetorical question: "are 2 and the function sending x to 2x, the same object?" [answer: no.]

Thus the presence of a dot product every vector A in V does define a covector in V*, namely the operation A.( ) = A*, which lives in V*. A lives in V; it is the operator A.( ) = A* that lives in V*.

This is what people mean when they say that a vector "can be viewed as a covector" in the presence of a dot product. They do not mean that A itself shifts from living in V to living in V*. They just mean that one can focus on the covector A.( ) instead of on the vector A. I think it is confusing when people (including me) say things like "one can look at the vector A as a covector". That is not strictly true, A is a vector no matter how you look at it. They mean, instead of the vector A, one can consider the associated covector A.( ) = A*.

Thus when one writes A = a1e1+...+anen = c1e^1 +...+ cne^n, both these vectors do live in V and are equal to the same object namely A. But when one is more interested in the covector A.( ) = A*, which lives in V*, the second expansion is more convenient. I.e. the first expansion for A leads to the expansion a1e1.( ) +....+ anen.( ) = a1e1*+...+anen*, for A*, which is not ideal since in the general case, i.e. the non orthonormal case, {ej.( )} = {ej*} is not a nice basis of V*. I.e. the basis {ej*} is not usually the correct dual basis for {ej} since it need not satisfy ej*(ej) = 1, and ej*(ei) = 0 for I≠j.

So the whole question is this: given a basis {ej} for V, find another basis {e^j} for V, such that the corresponding basis {e^j*} of covectors, will be the correct dual basis (of V*) to {ej}. For this one should clearly understand the meaning of the "correct" dual basis.

Here is the modern "algebraic" definition:

If {ej} is a basis for V, then the dual basis is a basis for V*, consisting of covectors {E^j} such that E^j(ej) = 1, and E^j(ei) = 0 for j≠i.

And the definition in your book is this:
In the presence of a dot product for V, the dual basis of {ej}, is another basis {e^j} for V, such that e^j.ej = 1, and e^j.ei = 0 for j≠i.

The relation between these definitions is that E^j = e^j.( ) = e^j*, is the covector associated to e^j by the dot product.
We also have E^j = ej*, if {ej} is orthonormal, but not otherwise.

Fleisch calls {e^j} the dual basis (a basis of V), and the moderns call {e^j*} the dual basis (a basis of V*).

In all of this I am using the notation e^j as it is used in your book by Fleisch.
Let me try to relate it to the discussion in post #8. In that post, there are two different notations in different parts of the discussion; the meaning of the symbol e^j changes from one paragraph to the next. At the beginning, a1e1+...+anen and c1e^1+...+cne^n are indeed the same object, and this is stated below equation (*).

Then the "algebraic" point of view is explained, and to do so, the meaning of e^j is changed. The symbol e^j is redefined in the equation 8 lines below equation (**), by saying that now e^j = the operator ej.( ), which I have called ej*. With this new meaning of e^j, as ej*, equation (*) post #8 now would say A = A*, which would not be strictly true, since the two objects are in different spaces.

I.e. now that the meaning of the symbol e^j has changed, indeed a linear combination of the new "e^j" is no longer an element of V. This is however not the meaning it has in your book. If you want to answer the question of which space e^j belongs to, you have to say what e^j means once and for all. If it means a vector such that e^j.ej = 1, and e^j.ei = 0 for I≠j, then it belongs to V, as Fleisch intends. If it means instead the operator ej.( ) = ej*, as it is changed to mean in the middle of post #8, then it belongs to V*.

More importantly, note that even with the new definition of e^j as an element of V*, it would not equal my E^j, unless the basis {ej} was orthonormal. In particular the basis {e^j} defined 8 lines below (**) in post #8, will not be the correct dual basis for V* unless the basis {ej} was orthonormal.

Now importantly, note that right below equation (**) in post #8, it is stated that "the standard inner product in this context is g = [Id]". I believe this means that from here on in post #8, one is assuming the basis {ej} to be orthonormal. Thus with this assumption, the e^j in the rest of post #8 is equal to my E^j.

I.e. there are two possible inner product matrices; one is in terms of the one basis {ej}, which is the matrix [ei.ej], which is usually what is meant by the matrix of g. The other is in terms of two bases {ej} and {e^i}, which is [e^i.ej], which is displayed in equation (**).

I think the term "inner product matrix" for g in post #8, refers to the first matrix, so in post #8, it seems the identity dot product matrix is meant to be [ei.ej], so the basis is being assumed orthonormal, and thus taking e^j = ej* does give the correct dual basis.

The whole point of my post was to understand the non- orthonormal case.

In case {ej} is not orthonormal, we must define the corresponding intrinsic dual basis of V* as {e^j*}, where {e^j} is Fleisch's "dual basis" for V, defined to be the unique basis satisfying equation (**) post#8, i.e. the unique basis for V such that the dot product matrix [e^i.ej] is the identity matrix.

In this approach, equation (*) in post #8 still holds true and both sides equal A, since the e^j still belong to V.

you certainly have my sympathy as to how confusing this is. as usual, most confusion stems from people assigning different meanings to the same symbol or the same word. it is also especially confusing that in general, the dual basis does not consist of the "duals", i.e. the associated covectors, of the vectors in the original basis. I.e. "the basis of duals" differs from the "dual of the basis".

 

[Rick16]

I meant exactly what I said in every instance you cite.

I don't doubt it. The problem is that it is often not clear to me what you meant.

This paragraph is right at the beginning of your explanation, and it is already highly unclear to me:

A vector represents a physical arrow in a vector space, say a position (with respect to a given origin), or (more intrinsically) a velocity. A covector represents an operator on vectors that spits out a number. A vector plus a metric yields a covector. I.e. if A is a vector in V, then A.( ) is the corresponding covector in V*.

If A is a vector, then I would read A.( ) as the vector A operating on an argument. Is it understood in your notation that if $\vec A$ is a vector in V, then $\tilde A(~)$ is the corresponding covector? This I would understand, but I don't understand why I should identify A.( ) as a covector, when A is a vector. When A is a vector, I would rather think that A.( ) is still a vector. You write that a covector represents an operator on vectors that spits out a number. But a vector also represents an operator on covectors that spits out a number. Maybe I misunderstand the dot in your notation. I see the dot as being part of the notation for the inner product. Is it possible that you use the dot as an identifier for covectors, similar to the tilde?

 

[mathwonk]

Re-reading your post, its seems you are focussed on notation. You seem to assume that using different notation makes a concept different.

You say " Is it understood in your notation that if A is a vector in V, then Atilda( ) is the corresponding covector? This I would understand, but I don't understand why I should identify A.( ) as a covector, when A is a vector. "

By your definition in line 9 of post #1, ATilda and A.( ), or Vtilda and V.( ) are exactly the same object. I.e. A covector is a function, and two functions are the same if they have the same values at every argument. The definition you give there of VTilda is that it is a function whose value VTilda(A) at A is V.A. Therefore VTilda = VTilda( ) = V.( ).

I.e. a covector is a function, and therefore it is determined by its values. It does not matter what notation you use to write it, as long as you define what that notation means. You seem to assume that an object is a covector if and only if it is denoted with a Tilda. that is irrelevant. What matters is whether it is a function from vectors to numbers. No matter what notation is used, if they both represent functions, and if the same input always yields the same output, then the two functions are the same, in particular they are the same covector.

I must apologize however for not reading your posts closely enough to realize that Fleisch has already introduced a notation for A.( ), namely he calls it ATilda. So he is not guilty of using the same symbol for two different things. When he uses A, (with an arrow over it), he means the vector A, and when he is thinking of the covector A.( ) defined by A, he writes it as ATilda. Wonderful.

 

[mathwonk]

Forgive me for ignoring your questions about notation. The key is to realize what are the concepts, and that different people use different notation to represent the same concept. Fresh42 discussed this. (E.g. in this case the two notations Fresh42 and fresh_42 denote the same person.)

An inner product on V is a function of 2 variables VxV-->R (real numbers), with certain properties, (symmetry, positivity, bilinearity). Some people call it ( ).( ), some call it < , >, and apparently some call it g( , ).

Since it is determined by its values on a basis, if {ej} is a basis of n vectors, it is entirely determined by the nxn matrix [ei.ej], which some also call g.

Therefore the covector determined by a vector A may be written as A.( ), or as <A, >, or as < ,A> (by symmetry), or as g(A, ), or as g( ,A), etc.....

Since these notations are long and clunky, you may also introduce a new symbol for the covector, as Fleisch apparently does, namely just say that "ATilda" = A.( ) = <A, > = g(A, ) = g( ,A), etc....

Of course in my notation, the symbol ATilda is also clunky, since I don't have the software to put the wiggly Tilda sign over the A, and that's why I don't use it.

To understand any notation, you must use the definition of that notation. You wrote in line 9 of post #1 that VTilda(A) = V.A. You said later you could not understand why V.( ) meant the same as VTilda( ), but to me this is answered by your own statement that they are the same function.
I.e. the confusion perhaps stemmed from my thinking they are the same because they are both functions on the same domain, with the same values, and your question over why they use different symbols. In math, if two symbols are related by an equals sign, they are interchangeable. does this help?

 

[Rick16]

Thank you for your answers. I don't have the time now to get into details. I just want to mention the part that still confuses me the most. For one thing, I do not think that a covector must be written with a tilde. I am aware that can it be represented as a function, although I have never seen your notation before and that confused me too.

I am mostly confused by the fact that you only present covectors as functions. Given what I have learned so far, it seems to me that both vectors and covectors can be treated as simple objects (i.e. not as functions) and as functions. Here is again a quote from Fleisch (Fleisch is not my only source, but it so happens that I find many relevant passages in his book):

"Authors using the modern approach often place strong emphasis on vectors and one-forms as operators (or rules), so you're likely to encounter statements that vectors "take" one-forms and produce scalars, just as one-forms "take" vectors and produce scalars."

Given this statement, I don't understand why you look only at covectors as functions, and you do not consider vectors to be functions. As if the fact that covectors can be functions were the defining characteristic of a covector.

 

[mathwonk]

I am puzzled when you say:

"For one thing, I do not think that a covector must be written with a tilde. I am aware that can it be represented as a function, although I have never seen your notation before and that confused me too."

Why do you say you have never seen my notation before when you used it yourself in line 9 of post #1? [see edit below] In fact you use it again in lines 11 and 12, when you propose just writing A.B instead of ATilda(B). You also propose writing A(B), which is perhaps ok if you mean here that ATilda= A( ), since then the notation, including the parenthesis, makes clear you are meaning an operator. Indeed that is almost exactly what both you and I suggest writing for ATilda, namely A.( ), as you suggest in line 12. To be clear, when you say that ATilda(B) = A.B, this is the same as saying that ATilda = A.( ). I.e. both are equal to the function whose value at B is A.B.

edit: Oh, my apologies, I think this is what you are saying was not clear, namely that A.( ) denotes the function whose value at B is A.B. Sorry, this just seemed "obvious" to me. Obviously, I am not as precise as I pretend to be.

Now for this sentence from Fleisch:

"Authors using the modern approach often place strong emphasis on vectors and one-forms as operators (or rules), so you're likely to encounter statements that vectors "take" one-forms and produce scalars, just as one-forms "take" vectors and produce scalars."

This sentence introduces another new wrinkle on vectors and covectors and duality, (called "double duality"), which I had not yet discussed, given how much trouble we were having with the more basic stuff.

I.e. when we write a function (or functional value) as f(x), and think of f as a function acting on x, we could also "think of x as a function acting on f"! I.e. if we have a real vector space V, we can also look at the real vector space V* of all linear real functions on V. Since we can add functions and multiply them by scalars, the function space V* satisfies the axioms for an abstract vector space. Then a vector A in V does yield a linear function A**, on elements f of V*, defined by A**(f) = f(A).

Strictly speaking however, this function A** is not the same object as A, since it is a function from V* to R, it lives not in V, but in (V*)* = V**. However, since the association of A** to A, defines a one-one correspondence between V and V**, at least in case V is finite dimensional, these spaces are often identified. They are not the same however, and If V is infinite dimensional, the mapping V-->V** is not onto, i.e. there will exist "functions on one forms" in V** that do not arise from evaluating at vectors in V. So in infinite dimensions one cannot identify V with V**. There is still a map V-->V** however, so vectors A in V will still yield certain special functions A** in V**. So one can always consider vectors A as functions A** operating on one forms, even in infinite dimensions, but in that case not all operators on one forms arise that way.

But to be precise, and not sloppy, one should not say "the vector A" takes the one form f to the number f(A)", rather the precise correct statement is "the function: 'evaluation at A' " takes the one form f to the number f(A)".

So while Fleisch is right that "you're likely to encounter statements that vectors "take" one-forms and produce scalars", those statements are not precisely correct. Notice that Fleisch did not say they were, just that you were likely to encounter them, namely by people who are not as precise as I am trying to be. That is why he wrote quotation marks around the word "take", since it is to be understood with a grain of salt. But I admit that once they begin to understand each other, people do frequently use such imprecise statements, to save time and words.

Note however that if someone is willing to sometimes think of A as in V, and sometimes in V**, i.e. as A**, but without changing the notation, then the question "is A a vector or an operator on covectors?" has no precise answer.

I.e. in our discussion, we have considered the objects A, ATilda, and A**. These are all distinct objects, but since in context there is only one way to have A operate on vectors (via inner product), or on covectors (via evaluation), people often say A when they really mean one of the other objects, and they just expect you to understand what they mean.

Just to confuse things more, note that as above, given vectors A and B in V, and an inner product ( ).( ) on V, (where I am writing ( ).( ) for the function taking A,B to A.B), if the operation on V taking A to B.A is called B*, then we can write the innocent inner product A.B also as A**(B*) = B*(A) = B.A = A.B.

Given this sort of mess, people do often just say: "let A act on B, (meaning think of A as A* = ATilda taking B to A.B)", or "let A act on the one form f, (meaning think of A as A** taking f to f(A))". But then they don't ask whether A is inherently a vector or a covector or whatever.

Or if they do, they say fuzzy stuff like "let A act on B as a covector (via the inner product, i.e. as A* if you must ask)".

Or they may say "let A act on the one form f, (as evaluation of course, i.e. as the double dual A**)", where the parenthetical part is silent, unless you ask.

This has me rethinking my whole world. I.e. I am going out walking and taking a big walking stick with me, which I also use for protection. So is it a walking stick or a club? I suspect it is really just a piece of wood.

 

[Rick16]

To be clear, when you say that ATilda(B) = A.B, this is the same as saying that ATilda = A.( ). I.e. both are equal to the function whose value at B is A.B.

edit: Oh, my apologies, I think this is what you are saying was not clear, namely that A.( ) denotes the function whose value at B is A.B. Sorry, this just seemed "obvious" to me. Obviously, I am not as precise as I pretend to be.

Yes, this was not obvious at all to me. You use a dot AND a bracket, and I have so far only seen using either a dot OR a bracket. And I still find it strange that you write $\tilde A=\vec A\cdot(~)$. How to use the dot and the bracket was part of my original question. For this point I have to leave Fleisch, because Fleisch does not go into this. My major reference is here Giancarlo Bernacchi: Tensors made easy. Schutz (A First Course in General Relativity) uses the same notation, but Schutz's explanations are much less clear to me than Bernacchi's.

Bernacchi distinguishes between the homogeneous inner product (between two vectors or between two covectors) and the heterogenous inner product (between a vector and a covector). He uses the dot exclusively for the homogeneous product and the bracket exclusively for the heterogeneous product. It makes sense to me to distinguish between the two, because if we do not make this distinction we would have $\vec A \cdot \vec B$ = $\vec A \cdot \tilde B$, as I already wrote in post #1. This would only be true if $\vec B=\tilde B$, and this is how this entire thread started. When you write A.( ) it looks like you mix two different notations that should be mutually exclusive. On the other hand, when I now look at it written as $\vec A \cdot (~)$, I begin to see it differently. I begin to see that the bracket is not actually the sign of an operator, but simply a placeholder. When I write "sign of an operator" I mean that if $\vec A$ is a vector, then $\vec A(~)$ is a vector written as an operator. The bracket shows that I am dealing with an operator. So I always saw A.( ) as A followed by two operator symbols. It looks like I have finally solved the mystery. You may find it strange that I get hung up on such details, but I am new to this, and it is easy to throw me off the track.

By the way, I don't have any software either to write equations on this site. I just use the built-in Latex editor. Doesn't it work for you?

Here is another thought about A.( ). If you write $\tilde A=\vec A \cdot (~)$, you could also write $\vec A=\tilde A \cdot (~)$. Written as A.( ), how do you know which is which? Where is the information in A.( ) that identifies it as a covector? You later write:

But then they don't ask whether A is inherently a vector or a covector or whatever.

At my current level, I don't want to go into the details about vector spaces V, V*, and V**. Bernacchi treats vectors and covectors and their related operations as equals. He writes "We define covector $\tilde P$ ... a linear scalar function of the vector $\vec V$." And two pages later he writes "In this "reversed" interpretation a vector can be seen as a linear scalar function of a covector $\tilde P$." This is for now good enough for me, but the knowledge that the operation $\vec A=\tilde A \cdot (~)$ exists prevents me from seeing A.( ) as an unambiguous expression for a covector.

In any case, I now finally understand why you write a covector as A.( ). It was a long way to a small result, but this is how it often goes. I sometimes wonder why I don't find an easier hobby. It just does not let me go.

 

[mathwonk]

well we got somewhere. communication is hard, but possible.

yes for me parentheses are a place holder to show that the object described is a function whose inputs should be inserted between the parentheses.

For that reason for me. your equation A = ATilda.( ) is hard to understand. The right side is apparently a function, and since ATilda is an element of V*, namely a function on V, I don't want to equate it with the left side, which is A, which is not a function on V but an element of V. I would be happy with A.( ) = ATilda = ATilda( ).

Also I am assuming the dot notation means the inner product on VxV, so it is not defined for ATilda, which lives in V*. Of course an author is free to define dot notation on V* if he says what it means. There is a natural pairing, or inner product on VxV*, namely evaluation, taking the pair (A,f) to f(A). Thus it takes the pair (B,ATilda) to ATilda(B). This means there is no need to introduce a dot for this pairing, since the evaluation notation is already defined.

Nonetheless even if we use ATilda.( ) to represent the already well defined object ATilda = ATilda( ), it still does not live in V, so can't equal A. Language like "A can be seen to live in V*", just confuses matters for me, and I try not to use it. It would be slightly better if they said "A can be viewed as living in V* via the inner product", since that translates as "A corresponds via the inner product to the operator A* = ATilda, where A*(B) = A.B". But I only understand it if I can make that translation precisely.

edit: [I confess that I also use this imprecise language when it suits me, but since we are trying here to understand the difference between vectors and convectors, I feel it would be unhelpful to be imprecise. I guess one goal is to be able to read imprecise language, and yet to have a precise understanding of what is meant.]

In the last part of your post you quote Bernacchi:
"In this "reversed" interpretation a vector A can be seen as a linear scalar function of a covector ATilda" , which is equivalent to "A can be viewed as living in V**". Now in this case, there is only one way to view A as living in V**, namely by equating A with A** = evaluation at A.

I would say that this is more understandable, since there is only one way to do this, (which does not depend on a choice of inner product), namely by letting the vector A define the double dual operator A**, but it is not precise language to say "can be seen as" even if it is very common.

Of course your quote includes the words ""In this "reversed" interpretation ..." which suggests you have left out the part where he explains what he means, i.e. how to view A as an operator on covectors, namely by pretending that ATilda(B) denotes a function B acting on the input ATilda. So he is saying that sometimes he is going to write B but think of B*.

This causes no trouble as long as I do know what he means. But if someone asks "is A a vector or an operator on covectors?", the only answer can be "sometimes I think of it as one, and sometimes as the other". My reply to this is "but how am I supposed to know when you are doing which?" In my opinion the most satisfactory response to that is "Ok I won't jump around in what my notation means, I will write A* when I am thinking of A as an element of V* and will write A** when I am thinking of it as in V**".

But we don't live entirely in that precise world, so we have to get used to statements that we will encounter, that may be ambiguous.

My advice is again to focus on the concepts. Then no matter what notation we encounter, it has to be trying to represent one of the basic concepts. I.e. there is only one way to pair a vector and a covector, namely by evaluating the covector at the vector. So no matter what they write, if it pairs a vector and a covector, (or a covector and a vector), it has to be evaluation, i.e. A in V and f in V*, can only be paired as f(A), so A in V and BTilda in V* can only be paired as BTilda(A). So that is what is meant even if they write it as BTilda.A or BTilda.(A), or A(BTilda), or A.BTilda or A.(BTilda),.....

So just ask: "what does this have to mean?"

Notice if ( ).( ) and < , > are two different inner products, and I say I want to think of A as a covector, you don't know whether I mean
A.( ) or <A, >. So in my opinion I should include the notation for the inner product to make clear not only that I am thinking of A as a covector, but how I am doing so.

 

[mathwonk]

I am also becoming consumed by this, and trying to find a good explanation for why one should not think of A and ATilda as the same object. I am also trying to understand this stuff without reading the book, which may be making it harder.

One observation: when we choose a basis {ej} for V we automatically get a "dual" basis for V*, namely the covectors f1,...,fn such that fi(ej) = 0 if I≠j, and fj(ej) = 1. When we also choose an inner product for V we get a second basis for V*, namely {e*j}. Which one should we use?

I.e. if we introduce an inner product in V, we get an isomorphism of V with V*, (making A correspond to A* = ATilda = A.( )), and hence the basis {ej} defines a basis {ej*} = {ejTilda} of V*, but this is not the same as the natural dual basis {fj} above. I.e. unless {ej} happens to be orthonormal, we do not have ei*(ei) = 1, nor ei*(ej) = 0 for I≠j.

If we identify V with V* by the inner product, then the natural dual basis {fj} above for V* corresponds to another basis for V which we call {e^j}. This is not the same as the original basis {ej}, but is "dual" to it in the sense that e^j.ei = kronecker delta (ij). So given a basis as well as an inner product for V, we have two associated bases for V*, and two for V. This may introduce confusion when using coordinates, since a basis determines a coordinate system, and different bases determine different coordinate systems.

[I almost trapped myself here by using imprecise language and saying a basis is a coordinate system, or can be thought of as a coordinate system!]

So we have two different ways to identify V with R^n, using either the basis {ej} or the dual basis {e^j}. The upshot seems to be that one wants to use one identification when we want to think of a vector A as a vector in V, and the other way when we want to think of it as a covector A* in V*.

E.g., if we use the basis {ej} to identify V with R^n, we may want to draw vectors in R^n representing vectors in V using their {ej} coordinates. But then A and B will not look perpendicular in R^n just because A.B = 0, since the basis identification using {ej} does not preserve inner products since {ej} is not orthonormal.

Here is one solution: To picture A.B in R^n, write A in terms of the basis {e^j} and write B in terms of the basis {ej}. I.e. thinking of A as A* = ATilda, express it in V* in terms of the dual basis {fj} corresponding naturally to {ej}, instead of the basis {ej*} corresponding to {ej} by means of the inner product.

This means we depict A in R^n using the coordinates of the expansion in terms of the dual basis {e^j} instead of the basis {ej}. I.e. since we are interested in depicting A* rather than A, we use the natural coordinates for V*. Thus (I think in Fleisch's notation) we use the coordinates {Aj} rather than the coordinates {A^j}. This will represent the vector A by a different arrow in R^n from the one representing A in the basis {ej}. Let's call that new arrow A* (coming from the dual basis {e^j}).

Now given two vectors A,B in V, graph them in R^n using different bases, i.e. different coordinate systems. Graph B using the basis {ej}, and graph A as A* using the basis {e^j}. I.e. graph A using the coordinates (A1,...,An), and graph B using the coordinates (B^1,...,B^n).

Then in R^n, the arrows A* and B will look perpendicular if and only if A.B = 0 back in V.

The problem was that the non orthonormal basis {ej} identifies V with R^n in such a way that the inner product on V is identified with a non standard inner product on R^n defined by the matrix Q = [ei.ej].
This is also the "change of basis" matrix that transforms the {ej} coordinates of A into the {e^j} coordinates of A.
I.e. if A = A^1e1 +...+A^nen = A1e^1 +...+Ane^n, then the coordinate vector (A1,...,An) is obtained by multiplying the coordinate vector (A^1,...,A^n) by the inner product matrix Q,
i.e.[A1,...,An] = Q.[A^1,...,A^n].

I.e. in {ej} coordinates A.B = [A^1,...,A^n]^t . Q . [B^1,...,B^n] (where the dots on the right represent matrix multiplication), and Q is symmetric,
so [A^1,...,A^n]^t . Q = [A^1,...,A^n]^t . Q^t = [Q. [A^1,...,A^n]]^t, so

A.B = [A^1,...,A^n]^t . Q . [B^1,...,B^n]
= [Q. [A^1,...,A^n]]^t . [B^1,...,B^n]
= [A1,...,An]^t . [B^1,...,B^n], now using {ej} coordinates for B and {e^j} coordinates for A.

I.e. representing A by its {e^j} coordinates (or A* by its {fj} coordinates), and B by its {ej} coordinates, transforms the inner product Q into the usual one for R^n, and hence makes the arrows A* and B depicted in R^n, look perpendicular whenever A.B = 0 in V.

(Unfortunately I am myself committing the cardinal sin of using A*, (and B), for more than one thing.)

I don't know if this is a motivation of your author until I get around to reading his book, but it is something I thought of as to why he might have introduced the dual basis {e^j}.

By the way, if you change bases for V, from {ej} to {uj}, with "change of basis" matrix M, which means the new u- coordinates for A are obtained by multiplying the old e-coordinates by M, then apparently the new dual {u^j} coordinates for A* will be obtained by multiplying the old {e^j} coordinates by "M^(-t)" = (M^t)^(-1) = "inverse transpose" of M.

This is another reason for distinguishing ATilda = A*, i.e. A thought of as a covector, from A thought of as a vector, since their natural coordinates transform differently. This is also apparently what is meant by calling one set of coordinates contra- variant, and the other one covariant.

 

[Rick16]

Nonetheless even if we use ATilda.( ) to represent the already well defined object ATilda = ATilda( ), it still does not live in V, so can't equal A.

Then what about the equation $\tilde A=\vec A\cdot (~)$? Couldn't you critisize it for the same reason? Actually, I wrote the equation $\vec A =\tilde A \cdot (~)$ this way, because you wrote above "To be clear, when you say that ATilda(B) = A.B, this is the same as saying that ATilda = A.( )." Consequently $\vec A =\tilde A \cdot (~)$ should be the same as saying that $\vec A(\tilde B) =\tilde A \cdot \tilde B$. Here both sides are scalars, so there should not be any incompatibility problems.

Beyond that, I cannot really comment on what you wrote. Your posts contain so many details that I cannot process so quickly. I need time to think about all of this.

Back in post #7 Ibix wrote:

"Vector and covector aren't the same object. They live in different vector spaces, for a start."
​

That is a good start for me, and I will for now rely on that and fill in the details later. But I thank you very much for your detailed posts.

 

[mathwonk]

I am assuming that the symbol "dot" or "." has only been defined for vectors in V. Hence in your first sentence, what I wrote, namely A.( ) makes sense as a function on V, since if you fill in the space between the parentheses with a vector from V, the resulting dot product is defined.

However what you wrote in the third sentence, namely ATilda.( ), is not defined since ATilda lives in V* and the symbol "." has not been defined on V*. Perhaps you wish to define it on V* by saying that ATilda.BTilda = A.B, but I was not aware you had done so. I would also prefer not to use the same symbol, " . ", for two different things, namely products on two different spaces.

 

[mathwonk]

I think I have a better idea of what is bothering you, namely my use of functional notation. So here is one more shot:

If f is a function (of one variable), then it is possible to write it as f or, less commonly, as f( ). In more old fashioned notation one used to also write f(x), where x represents the variable.

Thus in our case, if we have a dot product “.” defined on our vector space V, and if we have defined, for each vector A in V, the function ATilda:V—>R, whose value at the vector B is A.B, then we can write this function either as ATilda, or ATilda( ), or even ATilda(x), I guess, and possibly you can write it as ATilda.( ) if you insist on also using a dot to represent evaluation. However since A is not a function, but a vector, it does not make sense to write A( ). In particular A does not mean the same as "A( )", which actually does not mean anything.

Now whenever we have a function of two variables say F, we may write it as
F( , ), and then when we fix one variable, as say A, we can plug in A and leave the other space free, obtaining a function of one variable, namely F(A, ). The function here is not A, but F, with one variable fixed as A. It would be possible here to say that A becomes a function of one variable by means of F, but just writing A for the function F(A, ) would be pretty mysterious.

Now in our case, the dot product is a function of two variables. “.”, from VxV to R. As above, we can also write this dot function as ( ).( ). In all these cases the parentheses are just place holders for variables.

Now if we fix one variable as A, we get a function of one variable, namely
(A).( ), which I have also written as A.( ). Note the function here is not A, but “dot”, with one variable fixed. Thus saying that A is a covector, any means mod the dot product, is true but pretty much conceals a lot of information as to how that is done. Just as it is more revealing to say that the function above is not A but F(A, ), here also we should say that the covector is not A but A.( ).

E.g. what if we have more than one function of two variables, say F and G, then we get more than one function of one variable, namely F(A, ) and G(A, ), and we don't want to call them both A.

Tada, drumroll…. thus we can write A.( ) = ATilda = ATilda( ), since both sides denote a function, and both have the same value at B, namely A.B. But we cannot write A = ATilda( ), since one side is a vector and the other side is a function on vectors.
I.e. A is not the function, the dot is, so without the dot, A does not clearly denote a function.

sorry if this is just too much.

 

[Rick16]

Thanks for trying again to get this through my thick skull. You go very much into details, which are doubtlessly necessary to be precise. However, at this point I just want to understand the basic idea: why do dual basis vectors live in a different vector space? You spent a lot of time explaining this, but I got lost in the details. Back in post # 14 you wrote

it is also especially confusing that in general, the dual basis does not consist of the "duals", i.e. the associated covectors, of the vectors in the original basis. I.e. "the basis of duals" differs from the "dual of the basis".

You are right, this is especially confusing.

From my naive perspective, I don't even see why we need more than one vector space. We have different sets of vectors, so why don't we let them all live happily together? When Fleisch writes $\vec A = A^i\vec e_i=A_i\vec e^i$, it is not clear at all why the entities $A^i\vec e_i$ and $A_i\vec e^i$ should not be in the same vector space. It has been pointed out before that in Fleisch's presentation they do live in the same space, but that this is for some reason not generally the case.

Here are some other points that are not clear to me:

I don't know why I should not use the same symbol for the same operation, if the operation takes place in different vector spaces.

I don't know why I should define an operation in a vector space. If I see an equation in a book, I assume that it is valid. Here is equation 4.13 from Bernacchi: $\vec A \cdot \vec B=\tilde A \cdot \tilde B$. I assume it is not enough for you to just state an equation like this, because you see a whole theorical framework behind it, which I do not see. By the way, Bernacchi's book is a text for non-mathematicians. That's why I got it in the first place. I don't want to get to the bottom of everything, I have much more modest goals.

The whole construction with vectors as functions is completely new to me, and I am not convinced of the necessity of this approach. I already quoted Fleisch about this before: "Authors using the modern approach often place strong emphasis on vectors and one-forms as operators...". Apparently there is a different (according to Fleisch "traditional") approach that achieves the same results without operators/functions. This whole function idea probably makes things more precise for mathematicians, but for me it looks like overcomplicating everything.

 

[PeroK]

From my naive perspective, I don't even see why we need more than one vector space. We have different sets of vectors, so why don't we let them all live happily together? When Fleisch writes $\vec A = A^i\vec e_i=A_i\vec e^i$, it is not clear at all why the entities $A^i\vec e_i$ and $A_i\vec e^i$ should not be in the same vector space. It has been pointed out before that in Fleisch's presentation they do live in the same space, but that this is for some reason not generally the case.

I don't have Schutz, but if he uses the more mathematically flexible approach of having a dual vector space, then Fleisch is not much good to you. Hartle's book, which is the most accessible serious introduction to GR, takes the simplified approach. One option is to give Hartle a try - which might not be a bad idea if you have limited time.

I also have Sean Carroll's book and he takes the more sophisticated approach.

There is also a series of MIT graduate lectures online. The first lecture is here:



Professor Hughes' lectures are compatible with Carroll and Schutz (he discusses these books at the start of the course).

You could even try these lectures as an alternative to a textbook. I really liked his treatment of vectors and one-forms, so perhaps his presentation of the material might help you to grasp it. And it is a bona fide MIT graduate course!

My recommendation would be to supplement Schutz with Prof Hughes' lectures; or, bail out and try Hartle.

 

[Rick16]

My recommendation would be to supplement Schutz with Prof Hughes' lectures; or, bail out and try Hartle.

Thank you for the tips. I will certainly try out different sources. I have for the moment given up on Schutz and I am now reading Foster & Nightingale, which I like lot (for now, but I am only in chapter 1).

 

[jbergman]

TL;DR Summary: confused about the inner product

I posted a follow-up question on an earlier question, but did not get a response. So I will start a new thread.

A vector $\vec V$ and a covector/one-form $\tilde V$ represent the same object, right? $\vec V=V^i \vec e_i = V_i \vec e^i = \tilde V$.

We also have $\mathbf g(\vec V)=\tilde V$, where $\mathbf g$ is the metric tensor.

Putting these two together, I get $\mathbf g(\vec V)=\vec V$. That's odd. How can I make sense of this?

And when I look at equations like $\tilde V(\vec A):= \mathbf g(\vec V, \vec A)=\vec V \cdot \vec A$ (Schutz, A First Course in General Relativity, 3rd edtion, page 68), I wonder if it is really necessary to do it like this. Couldn't I simply write $\tilde V(\vec A)=\vec V(\vec A)$, or $\tilde V \cdot \vec A=\vec V \cdot \vec A$, since $\vec V$ and $\tilde V$ are the same object?

Is the distinction between bracket notation and dot notation purely conventional, is it always done like this, is it necessary to do it like this?

Unfortunately, the way things are taught can be imprecise at times and introduce confusion.
First of $\tilde V$ is not equal to $V$. If we have a vector space, $W$ there is it's dual vector space that covectors belong to, $W^{*}$, where a covector is a function which takes a vector and returns a number. $W$ and $W^*$ are the same dimension hence are isomorphic as vector spaces.

A metric or an inner product gives you a canonical isomorphism between $W$ and $W^*$. In other words as your wrote, we can define $\tilde{V} = g(V)$.

Regarding the $\cdot$ notation versus the parentheses notation, $\tilde{V}(A)$ is applying the covector to a vector $A$ since a covector is a function. The $\cdot$ notation is the way to take an inner product between two vectors $V \cdot A = g(V, A)$ makes sense but $\tilde{V} \cdot A = g(\tilde{V}, A)$ doesn't make sense because you can't take an inner product between a covector and a vector. Likewise $V(A)$ doesn't make sense.

However since we have the canonical isomorphism given a metric between a vector space and its dual people tend to play fast and loose with these things.

 

[Rick16]

Unfortunately, the way things are taught can be imprecise at times and introduce confusion.
First of $\tilde V$ is not equal to $V$. If we have a vector space, $W$ there is it's dual vector space that covectors belong to, $W^{*}$, where a covector is a function which takes a vector and returns a number. $W$ and $W^*$ are the same dimension hence are isomorphic as vector spaces.

A metric or an inner product gives you a canonical isomorphism between $W$ and $W^*$. In other words as your wrote, we can define $\tilde{V} = g(V)$.

Regarding the $\cdot$ notation versus the parentheses notation, $\tilde{V}(A)$ is applying the covector to a vector $A$ since a covector is a function. The $\cdot$ notation is the way to take an inner product between two vectors $V \cdot A = g(V, A)$ makes sense but $\tilde{V} \cdot A = g(\tilde{V}, A)$ doesn't make sense because you can't take an inner product between a covector and a vector. Likewise $V(A)$ doesn't make sense.

However since we have the canonical isomorphism given a metric between a vector space and its dual people tend to play fast and loose with these things.

All of this would make sense if I understood this:

If we have a vector space, W there is it's dual vector space that covectors belong to, W∗,

Why does the covector belong to W*? fresh_42 writes in post #8:

I'm not really happy with Fleisch's presentation as it mixes two different concepts of what a dual basis is. He discusses two different bases of the same vector space, connected by a transformation T. One basis is B={e→1,e→2} and the other one is B′={e→1,e→2}. In this sense, we have
A→=Axe→1+Aye→2=(Ax,Ay)B=(Ax,Ay)B′=Axe→1+Aye→2(∗)
the same vector A→ represented in two different bases. This has priorily nothing to do with duality and can be done for any basis transformation.

So in Fleisch's presentation we have two different bases of the same vector space, i.e. vector and covector both belong to the same vector space. What do covectors look like that do not belong to the same vector space?

 

[martinbn]

So in Fleisch's presentation we have two different bases of the same vector space, i.e. vector and covector both belong to the same vector space. What do covectors look like that do not belong to the same vector space?

I think he identifies the two spaces using the metric and considers them as if they are the same. Either stick with this book and not worry about these things or try another.

 

[Rick16]

I think he identifies the two spaces using the metric and considers them as if they are the same. Either stick with this book and not worry about these things or try another.

Does this basically mean that covectors belong to a different (dual) vector space by definition? Without any additional geometrical considerations?

 

[martinbn]

Does this basically mean that covectors belong to a different (dual) vector space by definition? Without any additional geometrical considerations?

Yes, they belong to the dual space, which is different. This was explained may times in this thread. In infinite dimension they are not isomorphic, you cannot identify them.

 

[PeroK]

Does this basically mean that covectors belong to a different (dual) vector space by definition? Without any additional geometrical considerations?

Yes. You can associate them isomorphically with vectors in the original space. The original vector is the dual of its dual vector.

However, in more general vector spaces, this does not always hold.

 

[Rick16]

Yes, they belong to the dual space, which is different. This was explained may times in this thread. In infinite dimension they are not isomorphic, you cannot identify them.

Okay, thank you. Then everything seems to be pretty clear now.

 

[Rick16]

However, in more general vector spaces, this does not always hold.

And here comes the next confusing point. But I don't think that I have to go into this at this stage.

 

[fresh_42]

So in Fleisch's presentation we have two different bases of the same vector space, i.e. vector and covector both belong to the same vector space. What do covectors look like that do not belong to the same vector space?

The vector spaces $V$ and $\tilde{V}=V^*$ are formally not the same. One consists of arrows, and the other one consists of linear functions. In the case of finite dimension $n$ they are isomorphic. I wrote it as
$$
V\cong \mathbb{R}^n\cong \tilde{V}=V^* \quad {(*)}.
$$
This means that both - arrows and functions - can be described by $n$-tuples of real numbers. That doesn't make them the same object, but it makes them have the same sort of representation, namely by $n$ real coordinates. It also allows us (in the cases $n=0,1,2,3$) to draw them on the same sheet of paper: the origin of this entire discussion here and my criticism of Fleisch's treatment. Fleisch interprets those linear functions as the lengths of their projections on those arrows, i.e. a linear function (projection) that eats arrows and spits out a length, a real number.

Those isomorphisms in ${(*)}$ are not natural. They depend (as in our case here) on the choice of the metric $g.$ This makes them sort of "artificial". Nevertheless, you can make this geometric interpretation at the expense of confusing two initially different concepts: arrows and linear functions.
$$
v\stackrel{1:1}{\longleftrightarrow } \left(w\stackrel{\tilde{v}}{\longmapsto} \bigl\langle w,v \bigr\rangle =w\cdot v =g(w,v)=w^\tau\cdot v \right)
$$
You can see that this linear function on the right (eating an arrow $w$ and spitting out the real number $g(w,v)$) can be written in many ways $(\tilde{v}=v^*\, , \,g(w,v)\, , \,w\cdot v\, , \,w^\tau \cdot v\, , \,\bigl\langle w,v \bigr\rangle )$ adding a notational mess to what has already been confused.

The best advice is therefore:

Either stick with this book and not worry about these things or try another.

Here is a text that explains these "dualities" a bit better:
https://arxiv.org/pdf/1205.5935
but it's not an easy read in my opinion.

 

[fresh_42]

And here comes the next confusing point. But I don't think that I have to go into this at this stage.

That's easy. If the vector space consists e.g. of sequences as vectors, then it isn't usually of finite dimension any longer. This means we cannot express them as $n$-tuples of real numbers anymore. And then the identification (isomorphisms) $V\cong \mathbb{R}^n\cong \tilde{V}$ breaks down.

Vector spaces in general are a set of objects that allow addition, stretching, and compressing. The vectors do not always have to be arrows. Functions, sequences, series etc. can all also be added and stretched.

 

[Rick16]

One last remark: My confusion was obviously caused by the fact that I interpreted $\vec A=A^i \vec e_i = A_i \vec e^i$ from Fleisch as meaning $\vec A=\tilde A$, but Fleisch never writes $A_i \vec e^i=\tilde A$. He writes that $A_i\vec e^i$ are the covariant components of vector $\vec A$. Later he writes "The first thing to understand is that the traditional approach tends to treat contravariant and covariant components as representations of the same object, whereas in the modern approach objects are classified either as "vectors" or as "one-forms" (also called "covectors"). In the modern terminoloy, vectors transform as contravariant quantities, and one-forms transform as covariant quantities." He never writes that vectors with covariant components are the same as one-forms. One must really read these texts very closely. I am sorry that I took up so much of everybody's time with this.

 

[mathwonk]

[I have not read today's posts, and am following up on your question as to why I did not understand the equation A = ATilda.( ).]

Forgive me, I seem to have misunderstood your notation. I want to reassure you that your instincts are excellent, and that, once we agree what the notation means, there is indeed an interpretation in which the equation A = ATilda.( ) is true and meaningful, namely as follows ( where I will write A* instead of ATilda): So I want to justify the claim that indeed, properly understood, “A = A*.( )”.

Le V be a vector space with a real-valued inner product (positive, symmetric, bilinear function of two variables) denoted by A.B for A,B in V. Let V* denote the vector space of real valued linear functions on V, and if A is in V, write A* for the element of V* whose value at B in V is A.B.

Then we can write the function A* as A* = A*( ) = A.( ).

If V is finite dimensional, this correspondence, taking A to A*, is an isomorphism from V to V*, (because the inner product is assumed positive definite), and we can use it to carry the inner product from V over to V*. Namely given elements f,g in V*, then f = A*, and g = B*, for some unique elements A,B of V, and we can define f.g = A*.B* = A.B.

Note there is always a natural “evaluation” pairing between V and V*, not dependent on the inner product, taking the function F = F( ) in V* and the vector B in V, to the value F(B), of F at B.

Thus if A is a vector in V, then A* = A*( ) denotes the linear function on V whose value at B is A*(B) = A.B.

With our definitions above, then A*.( ) denotes however the linear function on V* whose value at B* is A*.B* = A.B.

Now let V** = (V*)* denote the vector space of all real valued linear functions on V*. Thus the function A*.( ) belongs to V**, (whereas A*( ) belongs to V*).

Then for each A in V, we can define the “evaluation” function on V* whose value at f is f(A), and we denote this linear function on V*, as A** = evaluation at A. In particular, A**(B*) = B*(A) = B.A = A.B. I.e. A** is the unique linear function on V* whose value at B* is A.B.

Since A*.( ) is also a linear function on V* whose value at B* is A*.(B*) = A*.B* = A.B, then A** and A*.( ) are the same function on V*. Hence. A*.( ) = A**, i.e. they denote the same element of V**.

Finally, since the mapping from V to V** taking A to A** is an isomorphism when V is finite dimensional, and does not depend on any choice of inner product, we may identify V naturally with V**, and under this identification, indeed the element A*.( ) of V** corresponds to A.

In this sense, “A = A*.( )”, i.e. although the two sides are in different spaces, they do correspond under the natural isomorphism between those spaces. (Note however the fact they define the same function depended on the symmetry of the inner product.)

 

[Rick16]

And one final last remark: Bernacchi writes: "Even a covector can then be expressed as an expansion on its own basis: $\tilde P=\tilde e^\alpha P_\alpha$". So Bernacchi's expansion of a covector looks the same as Fleisch's expansion of a vector with covariant components. Which further added to my confusion.

 

[fresh_42]

And one final last remark: Bernacchi writes: "Even a covector can then be expressed as an expansion on its own basis: $\tilde P=\tilde e^\alpha P_\alpha$". So Bernacchi's expansion of a covector looks the same as Fleisch's expansion of a vector with covariant components. Which further added to my confusion.

See my post #37.

The origin of the confusion is that a vector is $v=(v_1,\ldots,v_n)$ and $\tilde{v}= \operatorname{ times }\;(v_1,\ldots,v_n).$

 

[Rick16]

See my post #37.

The origin of the confusion is that a vector is $v=(v_1,\ldots,v_n)$ and $\tilde{v}= \operatorname{ times }\;(v_1,\ldots,v_n).$

Which means that a vector with covariant components is not the same as a covector, doesn't it?

 

[mathwonk]

I think dealing with a lot of this just boils down to: when someone says something equals A, ask your self: "well, maybe he means A*, or A**; would it make sense then?"

 

[Rick16]

[I have not read today's posts, and am following up on your question as to why I did not understand the equation A = ATilda.( ).]

Forgive me, I seem to have misunderstood your notation. I want to reassure you that your instincts are excellent, and that, once we agree what the notation means, there is indeed an interpretation in which the equation A = ATilda.( ) is true and meaningful, namely as follows ( where I will write A* instead of ATilda): So I want to justify the claim that indeed, properly understood, “A = A*.( )”.

Le V be a vector space with a real-valued inner product (positive, symmetric, bilinear function of two variables) denoted by A.B for A,B in V. Let V* denote the vector space of real valued linear functions on V, and if A is in V, write A* for the element of V* whose value at B in V is A.B.

Then we can write the function A* as A* = A*( ) = A.( ).

If V is finite dimensional, this correspondence, taking A to A*, is an isomorphism from V to V*, (because the inner product is assumed positive definite), and we can use it to carry the inner product from V over to V*. Namely given elements f,g in V*, then f = A*, and g = B*, for some unique elements A,B of V, and we can define f.g = A*.B* = A.B.

Note there is always a natural “evaluation” pairing between V and V*, not dependent on the inner product, taking the function F = F( ) in V* and the vector B in V, to the value F(B), of F at B.

Thus if A is a vector in V, then A* = A*( ) denotes the linear function on V whose value at B is A*(B) = A.B.

With our definitions above, then A*.( ) denotes however the linear function on V* whose value at B* is A*.B* = A.B.

Now let V** = (V*)* denote the vector space of all real valued linear functions on V*. Thus the function A*.( ) belongs to V**, (whereas A*( ) belongs to V*).

Then for each A in V, we can define the “evaluation” function on V* whose value at f is f(A), and we denote this linear function on V*, as A** = evaluation at A. In particular, A**(B*) = B*(A) = B.A = A.B. I.e. A** is the unique linear function on V* whose value at B* is A.B.

Since A*.( ) is also a linear function on V* whose value at B* is A*.(B*) = A*.B* = A.B, then A** and A*.( ) are the same function on V*. Hence. A*.( ) = A**, i.e. they denote the same element of V**.

Finally, since the mapping from V to V** taking A to A** is an isomorphism when V is finite dimensional, and does not depend on any choice of inner product, we may identify V naturally with V**, and under this identification, indeed the element A*.( ) of V** corresponds to A.

In this sense, “A = A*.( )”, i.e. although the two sides are in different spaces, they do correspond under the natural isomorphism between those spaces. (Note however the fact they define the same function depended on the symmetry of the inner product.)

I am glad that you acknowledge this equation. I am new to this whole subject, but my acquaintance with Fleisch's presentation is a little older than the functional approach. I learned about the functional approach from Bernacchi, and Bernacchi takes a "symmetric" approach, i.e. he treats covectors as functions of vectors, and vectors as functions of covectors. I always wondered why you only considered covectors as functions, and vectors just as vectors, which seems like an "asymmetric" approach. I have since seen other posts on this thread with the same presentation like yours, and Schutz seems to do it this way, too. This is apparently something that I have to get acquainted with. In Bernacchi's approach it seems that vectors and covectors are objects on an equal footing, but it seems that I have to get away from this idea and start seeing them as different objects.

 

[fresh_42]

Which means that a vector with covariant components is not the same as a covector, doesn't it?

I'm not particularly familiar with that co-contra-variant babble. It means something completely different in mathematics than physicists use it for. I like to speak of vectors (like tangents) and their coordinates, and dual vectors (co-vectors, like cotangents) and their coordinates. A component of a vector is a coordinate is a number. What should a covariant number be? We have only one kind of numbers.

 

[Rick16]

What should a covariant number be?

I hope you are not asking me? I am just trying to learn this.

 

[Rick16]

I think dealing with a lot of this just boils down to: when someone says something equals A, ask your self: "well, maybe he means A*, or A**; would it make sense then?"

This could make sense in a certain way, but in another way it could also confuse things further.

 

[mathwonk]

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

 

[mathwonk]

Ok I have found Bernacchi and read pages 1-11. In these pages he is discussing only V and V*. There is no inner product on V, and he calls the natural evaluation pairing VxV*-->R, taking v,f to f(v), the "heterogeneous inner product".

There is at first no way to identify V with V*, so vectors cannot yet be considered as covectors. Then he chooses a basis e1,...,en for V, which gives a way to identify V with coordinate space R^n, by associating to each vector in V, its coordinates in R^n.

But then there is always a unique associated "dual" basis for V*, namely the basis of functions f1,...,fn in V* such that fi(ej) = 1 iff I=j, and = 0 otherwise. Then using this basis for V*, the space V* can also be identified with R^n, using the coordinates in terms of this basis to represent functions in V*.

Since both V and V* are identified with R^n by the choice of a basis of V, this choice of basis gives a way to identify V with V*, i.e. to view vectors as covectors. Namely if A is a vector with coordinates a^1,...,a^n, and B is another vector with coordinates b^1,...,b^n, we view A as the covector whose value at B is a^1b^1+...+a^nb^n.

What is going on here is that R^n has a standard inner product, and once the basis is used to identify V with R^n, we can transfer that inner product to V. Then with this inner product on V, call it <A,B>, we are just getting the familiar identification of V with V*, via A corresponds to <A, >. This is the unique inner product on V for which the chosen basis e1,...,en is orthonormal.

Later however, when he starts over with V, with no basis, and introduces first a "homogeneous" inner product g( , ) for V, and then chooses a basis u1,...,un for V that is not orthonormal for g( , ), we will have two different ways to identify V with V*, and hence two different ways to identify V with R^n, one by means of the basis, as above, and another by means of the inner product g.

I.e. the basis u1,...,un for V defines a dual basis f1,...,fn for V* such that fi(uj) = kronecker delta ij. Using the inner product g, we can transfer this basis to a basis for V, by choosing u^j to be the unique vector in V such that for all B in V, we have g(u^j,B) = fj(B). If we write a "dot" for g, we have u^j.B = fj(B).

Now if A is any vector in V, we get an element A* of V* where A* is the function A*(B) = A.B. In terms of the dual basis f1,...,fn for V*, this A* has an expansion as A* = c1f1+...+cnfn, for some c1,...,cn.

Now that we have two bases for V, namely {uj} and {u^j}, each vector A has two expansions, A = a^1u1+...+a^nun = a1u^1+...+anu^n.

I claim aj = cj for all j. I.e. the coefficients cj of A* in terms of the natural "dual basis" {fj}, are equal to the coefficients aj of A in terms of the basis {u^j} for V. (This is why Fleisch calls {u^j} the dual basis for V.)

E.g. note that A.u1 = (a1u^1+...+anu^n).u1 = a1, by definition of the u^j. But also A.u1 = A*(u1) = (c1f1+...+cnfn)(u1) = c1, by definition off the fj. Thus a1 = A.u1 = c1.

Similarly, aj = cj, for all j.

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 Disney's "Ratatouille" said, when the gourmet rat asked him what he was eating, I don't really know.)