3
$\begingroup$

In undergraduate linear algebra, the concept of a dot product, generalized to the inner product on an inner product space, is introduced fairly early as a way to multiply 2 vectors together to get a scalar.

As one continues through an undergraduate curriculum to physics, however, this notion gets largely replaced by that of a dual space, and the inner product of vectors becomes replaced by the notion of multiplying vectors with a member of their dual space.

This seems to me to be two realizations of the same motivation: multiplying 2 vectors together to get a scalar. I understand that there are formal mathematical differences between the two, and the Riesz representation theorem gives a map between them, but I'm struggling with why we need both to exist to begin with.

To help me understand the fundamental conceptual differences between the two, in physics, why does "multiplying 2 vectors together" take the form of the inner product in some circumstances, and dual spaces in others?

E.g., I've seen bra/ket multiplication represented both by an inner product and dual space formalism, and it seems the structures are somewhat redundant if they have the same utility.

General relativity seems to prefer using dual spaces, but it's not clear to me why the extra structure is necessary and, for example, why we couldn't inner product vectors at a point with other members of the same vector space to achieve the same physical results rather than formalizing the physics in a dual space construct.

I'm sure there are more examples, where physics needs a concept of "multiplying 2 vectors together" and the community decides to say the second operand lives in its own dual vector space, rather than having it "share" an inner product space with the first operand. Why is this natural? In what cases is one representation of this idea more "powerful" or "natural" than the other?

$\endgroup$
3
$\begingroup$

In Hilbert Spaces, you are correct that the Riesz Representation Theorem (RRT) allows you to use inner products and dual spaces interchangeably, however there are some reasons the dual space can be preferable, or even necessary.

Example 1: In Quantum Mechanics one can associate a pure state with an element $|\phi\rangle$ of a Hilbert Space $\mathcal H$, and its naturally corresponding yes/no observable with an element $\langle \phi |$ of the dual $\mathcal H^*$. The probability of a positive outcome for a measurement of $\langle \chi |$ in the state $|\psi\rangle$ is given by $|\langle\chi|\psi\rangle|^2$. Without referencing the dual, we could alternatively just use the symbols $\chi,\psi\in \mathcal H$, and the same calculation would be represented by using the inner product $|(\chi,\psi)|^2$. Now, however, we have to explain what this means since both of the symbols in the inner product are naturally interpreted as states, and there appear to be no measurements involved. In other words, we have two vectors which each have different physical interpretations and the use of the dual distinguishes them perfectly. Not to mention all the other symbolic advantages of the Dirac notation, which are mathematically grounded in the use of the dual space.

Example 2: In an infinite dimensional inner product space $V$ (that is not a Hilbert Space, so that we can't invoke the RRT) the dual $V^*$ can actually be larger than the original space, $V\subset V^*$ (in the sense of an isomorphic embedding). In Quantum Mechanics this fact is used to build something called a Rigged Hilbert Space in which certain states live in the dual but do not have counterparts in the inner product space itself. One can then only calculate probabilities by evaluating a dual element. (This is how Dirac Delta functions can be handled rigorously.)

Example 3: In relativity theory physical quantities are general tensors, which are defined properly as multilinear functionals of the form: $$\tau:V^*\times V^* \cdots \times V^* \times V\times V \cdots \times V \rightarrow \mathbb R$$Whilst a bilinear form corresponding to a 'dot product' of vectors is certainly a tensor (and an important one at that!), most tensors do not reduce to such simple rules. Sooner or later, you have to get used to working with duals.

$\endgroup$
3
$\begingroup$

An inner product space is more than just the space and its dual in the following way:

  • In an inner product space, one can indeed unambiguously multiply two vectors to form a scalar using the inner product. This provides a canonical map from the space to its dual, $$V\ni v\longmapsto \omega=\left<v,\cdot\right> \in V^*\,.$$
  • On the other hand, if you just have a space and its dual, there is no such identification (e.g. the dual basis does not provide one). Hence, all you can do is apply elements of $V^*$ to elements of $V$ to get scalars, but you cannot take to elements of $V$ to form a scalar.

(As usual, there are subtleties when dealing with infinte-dimensional spaces, which I gloss over here.)

$\endgroup$
1
$\begingroup$

On one hand, a VS has always an algebraic dual VS, and a topological VS has always a continuous dual VS, but is not necessarily an inner product space.

On the other hand, the norm of an inner product space may be a physical observable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.