# Inner products and the dual space

1. Jan 27, 2014

### orion

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.

2. Jan 27, 2014

### jgens

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.

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.

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. Jan 27, 2014

### Office_Shredder

Staff Emeritus
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. Jan 27, 2014

### mathman

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. Jan 27, 2014

### orion

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. Jan 27, 2014

### orion

Could you elaborate on that?

7. Jan 27, 2014

### orion

Yes, this is what I am thinking.

8. Jan 27, 2014

### pwsnafu

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.

Last edited: Jan 28, 2014
9. Jan 28, 2014

### mathman

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. Jan 29, 2014

### orion

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. Jan 29, 2014

### orion

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$\cdot$B = AiBi

and not

A$\cdot$B = AiBi

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

Is this correct?

12. Jan 29, 2014

### jgens

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.

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. Jan 29, 2014

### orion

Point taken, but I don't think it comes into play in the spaces I'm interested in.

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. Jan 29, 2014

### jgens

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*.

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.

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. Jan 29, 2014

### orion

I should have supplied the link (that the inner product is between a vector and covector):

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.

Last edited: Jan 29, 2014
16. Jan 29, 2014

### jgens

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).

Last edited: Jan 29, 2014
17. Jan 29, 2014

### orion

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:

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. Jan 29, 2014

### jgens

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.

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. Jan 30, 2014

### orion

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:

Geometrical Vectors, Gabriel Weinreich, pages 22-23.

Last edited: Jan 30, 2014
20. Jan 30, 2014

### jgens

That conventions in math and physics differ should be altogether unsurprising. You see small differences everywhere in QM and relativity.

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.