What separates Hilbert space from other spaces?

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Please forgive me if I repeat things that have been said before.
A Banach space is complete and has a norm, so distance and convergence is defined. Every Hilbert space is a Banach space that additionally has a dot product which defines the norm and angles. So the extra things that you get with a Hilbert space are angles, orthogonal decomposition, Pythagorean Theorem, parallelogram law, etc. If you are working in QM with a space that is a Hilbert space, then you would want to take advantage of those extra things that give you so much more geometric intuition.
 
chiro
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I actually meant continuous. Everything has to have continuity when it comes to analysis of these operators and continuous things are also measurable [no infinity terms or divergence].

Continuity is easy to think about when you have to have every mapping being draw with a pencil [or filled in if it is a region] by taking a pencil and putting it on some paper and doing the mapping without taking it off the page.

Also - finite dimensional spaces are not studied because they don't need to be. We assume in a hilbert space that the coefficients of the vector in l^2 are real or complex numbers of any kind and in L^2 we assume the integral to always be finite when it comes to norms and metrics.

It's easy to prove because summing and multiplying numbers always gives finite result if all are finite. The infinite part is what makes hilbert space and operator algebra theory hard.

You might want to consider how hard it is to find a way to get multiple infinite series to converge all at the same time to know why the generalization to infinite dimensions is not trivial.
 
fresh_42
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I actually meant continuous.
... which is still a property of functions and not space.
Everything has to have continuity when it comes to analysis
... which is wrong in this generality. E.g. step functions are frequently used in analysis or with the integrals in functional analysis. If we consider linear operators, we have continuity for them, but these aren't the only ones which are studied and says nothing about the elements these operator act on. It is not part of the definition of a Hilbert or Banach space. It doesn't even have to be functions or sequences! The square integrable functions which are often used in physics are simply an example, not the necessity.
Also - finite dimensional spaces are not studied because they don't need to be.
... which doesn't change the fact that ##\mathbb{R}^n## is a Hilbert space and thus also a Banach space: real and finite dimensional.
 
PeroK
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Continuity is easy to think about when you have to have every mapping being draw with a pencil [or filled in if it is a region] by taking a pencil and putting it on some paper and doing the mapping without taking it off the page.

Also - finite dimensional spaces are not studied because they don't need to be. We assume in a hilbert space that the coefficients of the vector in l^2 are real or complex numbers of any kind and in L^2 we assume the integral to always be finite when it comes to norms and metrics.
This is an A level thread. It's hard to accept "drawing a line without lifting a pencil off the page" as an adequate description of continuity of linear operators on function spaces!

The theory of quantum spin utilises finite dimensional Hilbert spaces.
 
chiro
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There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?
 
fresh_42
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There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.
This is a debatable point of view. In any case these examples have still to hold, if theorems on arbitrary Hilbert spaces are proven. They are a good test object.
You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.
... since you can have infinite dimension. Whether a geometric intuition is needed is a very personal decision. I think no intuition and merely formulas are still better than a wrong intuition. And to draw a line with a pencil without interruption is wrong since the days of Cantor.
How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?
I work with the definitions and finite dimensional examples. A couple of things are different in finite and infinite dimensional vector spaces, but this wan't the question here. The answer to the OP's question is: inner product and completeness, not dimensionality.
 
FactChecker
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There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.
It may be better to say that our knowledge of Rn or Cn is abstracted by identifying their Banach and Hilbert properties and applying them to other spaces. Then we can have Rn or Cn motivated insight in those spaces.
You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.
The Dirac delta function, associated comb functions and other generalized functions are very important in Fourier analysis. Continuity is not required for a function to be measurable and Lebesgue integration and measure can handle very strange functions.
How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?
Integration or convolution with an appropriate kernel function is the infinite-dimensional equivalent of matrices.
 
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Just elaborating a bit on what FactChecker said.

Hilbert spaces were invented by David Hilbert and others to help in areas like Fourier Analysis, Partial Differential Equations and eventually QM.

Nowadays, especially for applications to QM and Fourier Transforms, IMHO the easiest way to naturally motivate infinite Hilbert spaces is via what's called Rigged Hilbert spaces of great value in understanding mathematically difficult/unusual things like the Dirac Delta Function - which is not part of a Hilbert space.

What you do is take the set of all complex (just to be definite row) vectors of finite length then consider its dual. Its in fact any infinite sequence at all. Well that leads you naturally to look at dual's of subspaces of that huge and very general space. What you notice is sometimes that dual contains the original space and sometimes its the other way around. There is one very special space whose dual is basically itself - that is called the Hilbert space. This is expressed in the so called Gelfland Triple:

S* ⊃ H ⊃ S - where H is a Hilbert space (ie the space such that the sum of the squares of the sequences absolute value exists) - S is some subspace of H and S* is dual.

This view is especially useful in studying distribution theory that makes Fourier transform's a snap - otherwise you quickly run into issues of convergence and functions that do not belong to a Hilbert space eg that dreaded Dirac Delta Function mentioned before. It bedeviled the greats like Von-Neumann who I believe actually invented the name - Hilbert Space - in honor of David Hilbert who he was assistant to (wow what a pair that would be - like Wheeler and Feynman).

That's just a way of motivating why its interesting to study it - you can of course treat it purely mathematically with no motivation at all.

Added Later:
In light of Marks response - note the original space of finite vectors. Finite vectors are Hilbert spaces ie they all can be placed in one to one correspondence with its dual. We see in that light infinite Hilbert spaces are simply a generalization of that interesting property to infinite spaces. The Dirac Delta function belongs to the dual of a subset of a Hilbert space - the space of what are called good functions is usually used because it has the interesting property of the Fourier transform of a good function is a good function. Its dual is called a Schwartz space - and its easy to define the Fourier Transform of any member of that space because of the Fourier Transform property of good functions. The Dirac Delta function is a member of the Schwartz space. It took quite a while for mathematicians to figure it all out and solve what the great Von-Neumann couldn't.

Thanks
Bill
 
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There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.
A Hilbert space can be of infinite dimension, but doesn't have to be.

Emphasis added in both quotes.
From Wikipedia (https://en.wikipedia.org/wiki/Hilbert_space):
(A Hilbert space) extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions.
From Wolfram MathWorld (http://mathworld.wolfram.com/HilbertSpace.html)
Examples of finite-dimensional Hilbert spaces include
1. The real numbers ##\mathbb R^n## with <v, u> the dot product of v and u.
2. The complex numbers ##\mathbb C^n## with <v, u> the dot product of v and the complex conjugate of u.
 
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SemM
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Please forgive me if I repeat things that have been said before.
A Banach space is complete and has a norm, so distance and convergence is defined. Every Hilbert space is a Banach space that additionally has a dot product which defines the norm and angles. So the extra things that you get with a Hilbert space are angles, orthogonal decomposition, Pythagorean Theorem, parallelogram law, etc. If you are working in QM with a space that is a Hilbert space, then you would want to take advantage of those extra things that give you so much more geometric intuition.
Does this mean that Banach spaces can be more suitable for classical mechanics, diffusion problems etc?
 
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If you are working in QM with a space that is a Hilbert space, then you would want to take advantage of those extra things that give you so much more geometric intuition.
Does this mean that Banach spaces can be more suitable for classical mechanics, diffusion problems etc?
I don't believe it means that at all.
It appears to me that you are committing a logical error by starting from an implication that is assumed true, and then negating both the hypothesis and conclusion.
FactChecker: If (you're working in QM) then (a Hilbert space is preferred)
SemM: If (you're not working in QM) then (you should not use a Hilbert space (i.e., use a Banach space))
The second implication is faulty, because nerely negating the hypothesis and conclusion isn't valid.

Here's a simple mathematical example that shows why this negation is invalid.
If x = 2, then x2 = 4
This is clearly a true statement.

The implication "If x ≠ 2, then x2 ≠ 4" is untrue, as is easily seen by the counterexample x = -2.
 
SemM
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Take for instance curvilinear phenomena as gravity and relativistic time/space asymmetry phenomena, having no orthogonality and having no need for inner product, would Banach space be more suitable than Hilbert space , for these ?
 
PeroK
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Take for instance curvilinear phenomena as gravity and relativistic time/space asymmetry phenomena, having no orthogonality and having no need for inner product, would Banach space be more suitable than Hilbert space , for these ?
No need for inner product in GR?
 
SemM
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No need for inner product in GR?
That is what I am saying, its curvilinear, and Banach space has no inner product...so Banach space OK for GR?
 
Orodruin
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That is what I am saying, its curvilinear, and Banach space has no inner product...so Banach space OK for GR?
No. Even asking that question shows a fundamental misunderstanding of how GR works and the central role played by the metric.
 
SemM
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No. Even asking that question shows a fundamental misunderstanding of how GR works and the central role played by the metric.
Please explain, that would be the meaning of this thread... when Banach? When Hilbert (QM).. When RIenman? When others? That may not be the original question, but it helps in understanding and answering the original question. Lets be pedagogic here shall we, so we can make Physicsforum pleasant and informative!
 
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Orodruin
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Please explain, that would be the meaning of this thread... when Banach? When Hilbert (QM).. When RIenman? When others? That may not be the original question, but it helps in understanding and answering the original question. Lets be pedagogic here shall we, so we can make Physicsforum pleasant and informative!
GR is based upon differential geometry and the description of space-time as a Lorentzian manifold, i.e., a manifold equipped with a non-degenerate symmetric bilinear form with signature (1,3). This defines a pseudo-inner product on all tangent spaces.
 
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Curvature alone does not rule out an inner product. A space can have curvature but when you zoom in to a tiny neighborhood of any point, then there are inner products, angles, right angles and orthogonal vectors. They are local to that point. As long as you have them, you should (judiciously) use them where they apply. Because they are defined differently at every point, they are trickier to use. Moving continuously from one point to another, the definitions of the angles and coordinates also change continuously.
 
SemM
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Curvature alone does not rule out an inner product. A space can have curvature but when you zoom in to a tiny neighborhood of any point, then there are inner products, angles, right angles and orthogonal vectors. They are local to that point. As long as you have them, you should (judiciously) use them where they apply. Because they are defined differently at every point, they are trickier to use. Moving continuously from one point to another, the definitions of the angles and coordinates also change continuously.
Thanks, I thought curvature implied non-orthogonality of it elements.
 
PeroK
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Thanks, I thought curvature implied non-orthogonality of it elements.
If you don't have an inner product, what does "orthogonal" mean?
 
SemM
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Nothing, that is what I am saying here:

"I thought curvature implied non-orthogonality of it elements." No orthogonality, no inner product.
 
fresh_42
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Nothing, that is what I am saying here:

"I thought curvature implied non-orthogonality of it elements." No orthogonality, no inner product.
The curvature of a space is usually defined by its behavior of tangents (second derivative), i.e. linear approximations. Usually here means, I'm not sure whether there are any exotic possibilities to do it otherwise. But with tangents of different directions at a point, we have an angle between them, an angle between intersecting lines. This is the reason why manifolds are considered. They behave locally, i.e. in small open neighborhoods of points like a Euclidean space, which means for arbitrary close approximations we have all geometric advantages.
 
Orodruin
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To add to what has been said, curvature is a priori completely unrelated to the inner product as it determines the failure of parallel transport around a closed loop to return the same tangent vector. This depends only on the affine connection that is placed on the manifold and relates nearby tangent spaces. It does not even require a metric tensor on the manifold. It is related to the metric only if the connection is assumed to be metric compatible, which is the case, e.g., for the Levi-Civita connection (which is the unique torsion free metric compatible connection).
 
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A familiar example is the surface of the Earth. At any point, p, on Earth (except the North and South poles), one can talk about the angles between lines of latitude and lines of longitude. They are at right angles. An inner product, orthogonal vectors, and angles are defined at that point. Yet the surface of the Earth has curvature at that point.
 
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