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A vector space $V$ makes sense over any field $F$, or even a division ring. So why does adding an inner product suddenly not make sense without taking the $F=\Bbb R$ or $\Bbb C$? What are the primary properties we want in our scalar field that forces it to be one of these two? (In particular I don't see why completeness of the scalar field is necessary in a pre-Hilbert space, but even in a Hilbert space completeness of the vector space does not imply completeness of the scalar field.)

  • Do any problems arise when taking $F$ to be an involutive field with an absolute value satifying $|x^*|=|x|$ and defining $||x||=\sqrt{|\langle x,x\rangle|}$?

  • Or, if we stick to the standard definition $||x||=\sqrt{\langle x,x\rangle}$ (which only makes sense when $F$ has a subfield $K$, identified with a subfield of $\Bbb R$, such that $\langle x,x\rangle\in K$ for all $x\in V$), what problems arise if $F$ is not complete, or at least quadratically complete? (Note that the expression $\sqrt{\langle x,x\rangle}$ is evaluated in $\Bbb R$, not $K$.)

Mario Carneiro
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  • Look into the first answer to this question: http://math.stackexchange.com/questions/49348/inner-product-spaces-over-finite-fields – balddraz Oct 08 '15 at 23:25
  • @anon What about proper subfields of $\mathbb{R}$? – Ian Oct 08 '15 at 23:26
  • @ZeroXLR Although that is helpful for explaining why we want characteristic zero, there is a big gap between that and $\Bbb R$, and I'd like to understand how to close that gap. – Mario Carneiro Oct 08 '15 at 23:47
  • You could define an inner product space on a field with generalized conjugation ... But is it really more interesting than the complex numbers? Asking why some concepts are not common, is like asking why we haven't found live outside earth. Totally meaningless. – user251257 Oct 09 '15 at 00:35
  • @user251257 I have a practical reason for asking this question: I want to define inner product spaces and Hilbert spaces in a way which allows for both $\Bbb R$ and $\Bbb C$, but without explicitly listing them, because that would not be invariant under isomorphisms. Even if in the end it turns out to only allow $\Bbb R$ and $\Bbb C$, I want to have a set of characterizing "niceness" properties which make it clear exactly what makes $\Bbb C$ a good choice. (Also I think Hilbert space can be done on $\Bbb H$ as well, with the only loss being commutativity.) – Mario Carneiro Oct 09 '15 at 00:50
  • Just list some involutions (think as identity and conjugation). – user251257 Oct 09 '15 at 00:51
  • @user251257 Only if you assume that we are starting from a subfield of $\Bbb C$ (that is, if we demand that $F$ is a complete subfield of $\Bbb C$ then we already have obtained $\Bbb R$ or $\Bbb C$). But I don't see how that eliminates larger fields. – Mario Carneiro Oct 09 '15 at 01:00
  • @MarioCarneiro I would start with some extension of $\mathbb Q$ (for positive definite) or extension of $\mathbb R$ (for completeness) – user251257 Oct 09 '15 at 01:04
  • The point is, there are few justifications to extend $\mathbb C$, neither from an algebraic nor an analytic point of view. – user251257 Oct 09 '15 at 01:06
  • It is not about coming up with some nice concepts but concepts which produce nice results. If you have no use case for an inner product space over an other field, why bother? – user251257 Oct 09 '15 at 01:10
  • @user251257 (1) "There are few justifications..." Field theory, or even abstract algebra in general, is all about finding generalizations of $\Bbb C$. And there are plenty of extensions of $\Bbb Q$ larger than $\Bbb C$, for example $\Bbb C(x)$. (2) "why bother": Because I don't want to do all my proofs under case analysis. I'd much rather prove that the field must be isomorphic to $\Bbb R$ or $\Bbb C$, and in doing so establish what are the essential properties of the field that I need for the proof. – Mario Carneiro Oct 09 '15 at 01:15
  • @MarioCarneiro what do you need to prove explicitly for the reals? If you prove a result for the complex numbers, it usually includes the real case. Anyway you are the one who want to start with a subfield of $\mathbb C$. There is not need for it. – user251257 Oct 09 '15 at 01:23
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    Do you mean something like that? Hilbert C*-module – user251257 Oct 09 '15 at 01:41
  • @user251257 A theorem about complex Hilbert spaces does not imply a theorem on real Hilbert spaces; the relation between them is more complex than simple subset (because we are changing the relations between vectors). I just want a way to treat both possibilities with the same language. – Mario Carneiro Oct 09 '15 at 01:46
  • You could replace $\mathbb R$ and $\mathbb C$ with a (commutative) $\mathbb R$ - * - algebra. It would include both cases – user251257 Oct 09 '15 at 02:07

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One of the properties of an inner product is positive-definiteness, which requires the field of scalars to contain an ordered sub-field; in particular, finite fields, and fields of finite characteristic will not work, as it is not possible to define an order for them compatible with the field operations.

If we wish the inner product to define a norm via $\|x\| = \sqrt{\langle x,x\rangle}$, and we want this to return a scalar in our field, $F$ needs to contain a Euclidean sub-field (that is, an ordered field in which every non-negative number has a square root).

It is possible to, say, define an inner-product space over the field of all real algebraic numbers, or the field of constructible real numbers, from a purely algebraic point of view, but such inner-product spaces lack nice topological properties (a complete metric).

These topological properties become important when studying function spaces-typically "arbitrary" functions are too bizarre to study in any great detail, so we limit ourselves to collections of "nice" functions (for example, "smooth", or perhaps only "continuous" ones).

David Wheeler
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  • A norm defines an expression in $\Bbb R$, not the scalar field; I see no reason a priori to assume that the two coincide. (Or is this a nonstandard view? A field with an absolute value generally is a function $|\cdot|:F\to\Bbb R$ - we don't try to use the norm as an element of the field.) – Mario Carneiro Oct 09 '15 at 00:55
  • @MarioCarneiro That's a different kind of norm don't you think? David was talking about norm on a vector space (over a scalar field $F$) with the usual properties. He actually was trying to argue that the $F$ should be an extension of $\mathbb R$ with some "absolute value" function you mentioned. – Quang Hoang Oct 09 '15 at 01:23
  • @QuangHoang For me a normed vector space is a vector space over a normed field (i.e. the field also has an absolute value) and $||rx||=|r|,||x||$, so there are really three algebraic spaces in play, $V$, $F$, and $\Bbb R$. It might help to know in what circumstances we need a norm to "pretend" as if it is a member of $F$; I usually think of it as only describing "sizes" of vectors/scalars and so existing in a different space than either the vectors or the scalars. (con't) – Mario Carneiro Oct 09 '15 at 01:33
  • One could generalize the $\Bbb R$ to some other ordered field, but even in this case it need not relate to $F$ in any way other than through the norm itself. – Mario Carneiro Oct 09 '15 at 01:38
  • @MarioCarneiro yes, it is possible to define a norm that has nothing to do with the inner product. However, it is harder to pretend this if our norm is induced by the inner product, no? For some reason, mathematicians prefer multi-structures with "cross-structure compatibility". On another note, another reason we consider complete fields, is we want to regard an integral linear functional, as a kind of limit (as our dimensions go to infinity) of a weighted dot-product linear functional (the "weight coefficients" become a "measure distribution"). – David Wheeler Oct 09 '15 at 01:58