Inner products and the dual space

In summary: If ##T: X \to Y## is a bounded operator, then the adjoint of ##T##, ##T^{\star}: B' \to A'## is defined by ##(T^{\star}v)(w)=v(Tw)##. Note that ##T^{\star}## is defined using the inner product on ##B##.Also, one can show that if ##X, Y## are Hilbert spaces and ##T: X \to Y## is a bounded operator, then ##T^{\star}## is the adjoint of ##T## in the sense of Hilbert spaces.In summary, the inner product is defined between vectors in the same space instead of between a
  • #1
orion
93
2
I searched the forums, and I can't find anywhere where someone asked this question point blank. I may be completely off base so I apologize in advance if that is the case.

In quantum mechanics, an inner product is formed as the bra-ket <φ|ψ>. We are told that vectors are represented by the kets and the bras represent dual vectors. However, after reading linear algebra texts, I understand that inner products are not generally formed from a vector and a dual vector, but from two vectors in the same space. Now let V, V* and F be a vector space, dual space, and field respectively. I am familiar with the Riesz representation theorem used to justify the bra-ket formalism, however the fact that the dual basis is defined as ei(ej) = δij and the fact that the inner product is a linear functional mapping elements in V to the field F are very suggestive that a natural way to define the inner product should be between a vector in V and a dual vector in V*.

Also, the use of the Einstein summation convention in relativity is suggestive because it sums over a vector with an “upstairs” index and a vector with a “downstairs” index. If two vectors with an “upstairs” index are to be summed over, one has to use the metric to lower the indices of one of the vectors.

Consider a vector space over the reals. Yet another example would be a representation of a vector as a column vector. By the ordinary rules of matrix multiplication, a scalar product can only be formed with a row vector. Often column vectors are identified as vectors and row vectors as dual vectors.

My question is why are inner products defined in linear algebra texts as between vectors in the same space instead of between a vector and a dual vector when in practice it seems the opposite is true? Is there a reason why the inner product is not defined as between a vector and a dual vector? Why is the inner product defined as between two vectors of the same space?

Given the Riesz representation theorem, perhaps my questions are moot, but I would still like to conceptually understand this.
 
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  • #2
orion said:
I am familiar with the Riesz representation theorem used to justify the bra-ket formalism, however the fact that the dual basis is defined as ei(ej) = δij and the fact that the inner product is a linear functional mapping elements in V to the field F are very suggestive that a natural way to define the inner product should be between a vector in V and a dual vector in V*.

There is always a natural pairing between a vector space and its dual as you note above. The trick, however, is knowing just what vector each (continuous) linear functional corresponds to. That is the role of the inner product and different choices of inner product will change the vector representing this functional.

My question is why are inner products defined in linear algebra texts as between vectors in the same space instead of between a vector and a dual vector when in practice it seems the opposite is true?

In practice inner products are defined between vectors. The reasons why we define things this way are numerous. It gives a convenient dictionary between the vector space and its dual. It generalizes notions like length. You get the idea.

Is there a reason why the inner product is not defined as between a vector and a dual vector?

Because there is already a pairing between vectors and covectors! Besides there is no a priori notion of which vector corresponds to which covector. We need an inner product to determine that.
 
  • #3
Algebraically, the inner product should be thought of as a map that takes vectors in V to dual vectors in V*. The vector v gets sent to the element in the dual space which is <v,.>.

Geometrically however in finite dimensional space the inner product measures angles. <u,v>/(||u|| ||v||) is cosine of the angle between u and v, and this statement only makes sense if u and v are both elements of V. Since it is the geometry of the space that people are most often interested in, the inner product is usually thought of as a bilinear form on V.

Basically you have your priorities backwards. If we define the inner product as a map VxV* to F, then we can prove lots of cool things, but it's not clear why you care about them. However if it is defined on VxV then there are lots of geometric results that people really want to know, and the Riesz Representation Theorem is simply a powerful tool to study these geometry problems.
 
  • #4
Part of the difficulty is in terminology. For mathematicians, Euclidean and Hilbert spaces (inner product spaces) are self dual. Physicists make a distinction between the space and its dual..
 
  • #5
mathman said:
Part of the difficulty is in terminology. For mathematicians, Euclidean and Hilbert spaces (inner product spaces) are self dual. Physicists make a distinction between the space and its dual..

So what if they were not self dual: the inner product would be between what and what? Vectors and covectors or vectors and vectors?
 
  • #6
jgens said:
It gives a convenient dictionary between the vector space and its dual.

Could you elaborate on that?
 
  • #7
Office_Shredder said:
Algebraically, the inner product should be thought of as a map that takes vectors in V to dual vectors in V*. The vector v gets sent to the element in the dual space which is <v,.>.

Yes, this is what I am thinking.
 
  • #8
orion said:
So what if they were not self dual: the inner product would be between what and what? Vectors and covectors or vectors and vectors?

Comes up in the study of adjoint operators

Let ##{B}## be a normed space (which might not to be Hilbert in general). Define ##(\cdot, \cdot ): B \times B' \to \mathbb{K}## by ##(v, v' ):= v'(v)##. This a bilinear form which satisfies
##| (v,v')| \leq \| v\| \, \|v'\|##.

Let ##X## and ##Y## are normed spaces, and ##T \in \mathcal{B}(X,Y)##. A operator ##T' \in \mathcal{B}(X', Y')## is a adjoint of ##T## if
##(Tx, y') = (x, T' y')## for all ##x \in X## and ##y' \in Y'##.

I've forgotten what adjoint operators are used for.
Edit: note that what I wrote above is different from the Hermition adjoint on Wikipedia.

However, this is not the inner product. Inner products exist on inner product spaces. An arbitrary normed space does not have an inner product. If the inner product space is not complete then it is not self-dual in general. Nonetheless, the inner product is defined as between two vectors.
 
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  • #9
orion said:
So what if they were not self dual: the inner product would be between what and what? Vectors and covectors or vectors and vectors?

It is a question of point of view:

(x,y) can be thought of as a representation of an operator (y) acting on an element (x).

Alternatively it describes a mapping of HxH to the scalars, where (x,x) is the square of the length of x.
 
  • #10
mathman said:
It is a question of point of view:

(x,y) can be thought of as a representation of an operator (y) acting on an element (x).

Alternatively it describes a mapping of HxH to the scalars, where (x,x) is the square of the length of x.

Thanks, mathman, this is actually quite helpful to know that it is a question of point of view and not that one way is right and one way is wrong. I thought this was the case, but I wasn't sure.
 
  • #11
I have another question which I hope that someone in this forum can answer (or else I can post it in the relativity forum).

Why in relativity do we consider a dot product as "naturally" occurring between a covector and a vector? For example, we have

A[itex]\cdot[/itex]B = AiBi

and not

A[itex]\cdot[/itex]B = AiBi

I suspect the answer lies in the fact that by definition ei[itex]\cdot[/itex]ejij whereas ei[itex]\cdot[/itex]ej is not necessarily = δij. (i.e. non-orthogonal coordinate system).

Is this correct?
 
  • #12
orion said:
Thanks, mathman, this is actually quite helpful to know that it is a question of point of view and not that one way is right and one way is wrong. I thought this was the case, but I wasn't sure.

This is an annoying detail, but in the infinite dimensional case, the usual definition of inner product is forced (unless of course you know the continuous dual a priori). So there is a reason to prefer the bilinear form version.

orion said:
A[itex]\cdot[/itex]B = AiBi

and not

A[itex]\cdot[/itex]B = AiBi

In the first case there is just a hidden use of the musical isomorphisms combined with contraction. The second case violates the summation convention which is (one reason) why the procedure above is preferred.
 
  • #13
Thanks for your quick reply.

jgens said:
This is an annoying detail, but in the infinite dimensional case, the usual definition of inner product is forced (unless of course you know the continuous dual a priori). So there is a reason to prefer the bilinear form version.

Point taken, but I don't think it comes into play in the spaces I'm interested in.
jgens said:
The second case violates the summation convention which is (one reason) why the procedure above is preferred.

I agree it violates the summation convention, but violating the summation convention seems like a superficial reason to me.

Also, it was stated explicitly on another thread that a dot product in relativity is taken between a vector and a covector which was my understanding as well. I'm just trying to reconcile what I learned in relativity with what I learned in linear algebra and trying to understand the underlying mathematical base.

I will look into the musical isomorphism, but from first glance it is just saying that there is an isomorphism between the tangent bundle and the cotangent bundle for a Riemannian manifold. Am I understanding that correctly? Does it follow from this that the tangent space at a particular point on a Riemannian manifold is isomorphic to the cotangent space?
 
  • #14
orion said:
Point taken, but I don't think it comes into play in the spaces I'm interested in.

If your spaces are finite-dimensional, then absolutely. But the point still stands that there is a reason to prefer starting with bilinear maps VxV→F instead of maps V→V*.

I agree it violates the summation convention, but violating the summation convention seems like a superficial reason to me.

It may seem superficial, but it really is not. Things violating the summation convention more often than not fail to be tensorial, suitably coordinate invariant, etc.

Also, it was stated explicitly on another thread that a dot product in relativity is taken between a vector and a covector which was my understanding as well.

A dot product is between two vectors. In the event it was, for some unknown reason, not clear the first time around: since we like equations satisfying the summation convention and we have a Riemannian/Lorentzian metric lying around we can apply a musical isomorphism to the first vector and then contract to get the desired dot product formula.
 
  • #15
I should have supplied the link (that the inner product is between a vector and covector):

https://www.physicsforums.com/showthread.php?t=735287&highlight=covariant+contravariant

Check out post #4. It is also stated in physics textbooks.

So what I gather from your post is that the musical isomorphism is the key to why we can associate a vector with a covector in a Riemannian space. I must admit that this is the first I've heard of the musical isomorphism.
 
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  • #16
In that post PeterDonis either means the musical isomorphism + contraction trick mentioned earlier (and just is not very explicit about it) or is just plain mistaken. I am fairly certain it is the former. If you want some motivation for that, then consider the following: supposing the dot product were the contraction of a vector and covector, the result would have absolutely no dependence on our choice of metric, which is certainly problematic.

Since the musical isomorphism thing seems to be causing confusion I will note it is essentially just the Riesz representation theorem. Basically it allows us to transform vector fields into covector fields (1-forms).
 
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  • #17
jgens said:
In that post PeterDonis either means the musical isomorphism + contraction trick mentioned earlier (and just is not very explicit about it) or is just plain mistaken. I am fairly certain it is the former. If you want some motivation for that, then consider the following: supposing the dot product were the contraction of a vector and covector, the result would have absolutely no dependence on our choice of metric, which is certainly problematic.

Since the musical isomorphism thing seems to be causing confusion I will note it is essentially just the Riesz representation theorem. Basically it allows us to transform vector fields into covector fields (1-forms).

Thanks. That clarifies things ... I think. However textbooks clearly state that the dot product is taken between a vector and covector. I don't have mine available at the moment, but a quick google search turned up the following:

If the inner product is taken of two vectors, one must be a contravariant vector and the other a covariant vector. The inner product of two covariant or two contravariant vectors is not defined.

Spherical Astronomy, Robin Michael Green page 495.

I'm not advocating that book but just using it as an example of what is being taught in physics classes. In fact, any search on Googlebooks will return many, many books saying essentially the same thing.

Now I get that you are saying that they are employing the musical isomorphism and contraction, but as mentioned in the above quote they are also saying that any other type of inner product is undefined. So this brings me back full circle to my original question: Why do linear algebra texts define the dot product as between two vectors and not a vector and a dual vector? How to reconcile these two points of view?
 
  • #18
orion said:
Why do linear algebra texts define the dot product as between two vectors and not a vector and a dual vector?

First the contraction of vectors and covectors does not recover the information from an inner product so there is one reason. In the finite-dimensional case a linear map V→V* does recover this information, but as mentioned before, the usual definition in linear algebra is the only one that generalizes appropriately in the infinite-dimensional case. Inner products are useful for much more than simply representing linear functionals, however, and that provides yet more reasons to prefer the definition in your linear algebra books.

How to reconcile these two points of view?

Simple. If one takes an uncharitable view, then just assert the definitions passed around in these physics books are wrong. It is possible that relativists have their own conventions about inner products that differ from the mathematical conventions, but in this case, I highly doubt it. More likely is these books/authors are being lazy.
 
  • #19
I have to tell you, you are destroying my worldview here and going against everything that was drilled in my head throughout graduate school. Wow, this is a major paradigm shift.


Getting a little geometrical, here is some insight as to why physicists define the dot product as they do:

(The traditional dot product) is the product of the magnitudes of the two vectors multiplied by the cosine of the angle between them. Quite obviously, such a definition (stated here in terms of arrows, since arrows are the traditional way in which vectors are represented) cannot possibly be invariant to a general coordinate transformation since it involves measurement both of angles and of numerical lengths. Yet a different definition, equivalent to the first if one restricts oneself to rigid rotations, can be constructed, provided that a dot product is always taken of one arrow and one stack.

Geometrical Vectors, Gabriel Weinreich, pages 22-23.
 
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  • #20
I have to tell you, you are destroying my worldview here and going against everything that was drilled in my head throughout graduate school. Wow, this is a major paradigm shift.

That conventions in math and physics differ should be altogether unsurprising. You see small differences everywhere in QM and relativity.
orion said:
Getting a little geometrical, here is some insight as to why physicists define the dot product as they do

As mentioned in the last post I am not convinced they honestly define dot products any differently than mathematicians*. More likely is the authors are saying one thing but meaning another. Saying that inner products are the contraction of vectors and covectors is obviously wrong, since the contraction has zero dependence on the underlying metric (including its signature). But since relativists like to hide all their implicit uses of musical isomorphisms, it is unsurprising they "define" inner products that way, while meaning the standard math definition.

*There is another terminological issue here that merits note. Technically an inner product should be positive definite (i.e. <v,v> ≥ 0 for all vectors v), but relativists often require the slightly weaker condition that inner products need only be non-degenerate. Some mathematicians have adopted the physics notion of inner product, but many have not. Terminology is always a mess. Fortunately so long as one is aware of the differences, it usually poses no real barrier.
 
  • #21
Thank you so much, jgens. I did a lot of thinking about this the last few days and you are absolutely right.

Every vector in the tangent bundle of a Riemannian manifold has an associated covector in the cotangent bundle through the musical isomorphism. So when the dot product is taken in relativity, it is not merely a contraction, but a real scalar product. The raising or lowering of indices through the metric is an invocation of the musical isomorphism.

I didn't see this before because it is never discussed in physics. Physicists often just "do what works" and ignore the details.

Thanks again. I finally feel like I understand something that was greatly troubling me before.
 

1. What is an inner product?

An inner product is a mathematical operation that takes in two vectors and produces a scalar quantity. It is also known as a dot product or scalar product. It is commonly used in linear algebra and functional analysis to measure the angle between two vectors and the length of a vector.

2. What is the significance of an inner product?

An inner product allows us to define important concepts such as orthogonality, distance, and angle between vectors in a vector space. It also provides a way to generalize the concept of length and angle to abstract vector spaces, making it a fundamental tool in many areas of mathematics and physics.

3. What is the dual space?

The dual space of a vector space is the set of all linear functionals on that vector space. In other words, it is the space of all linear maps that take a vector as input and produce a scalar as output. It is denoted by V* and is itself a vector space.

4. How are inner products and the dual space related?

The dual space of a vector space can be thought of as the space of all possible inner products on that vector space. This means that every element in the dual space is a linear functional that takes in two vectors and produces a scalar, which is essentially an inner product. Therefore, inner products play a crucial role in defining and characterizing the dual space.

5. Can inner products and the dual space be extended to infinite-dimensional vector spaces?

Yes, inner products and the dual space can be extended to infinite-dimensional vector spaces. In fact, many important applications of inner products and dual spaces, such as Fourier series and Hilbert spaces, involve infinite-dimensional vector spaces. In this case, the inner product is defined using integrals or summations, and the dual space is often referred to as the continuous dual space.

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