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Is there a canonical definition of the concept of inner products for vector spaces over arbitrary fields, i.e. other fields than ℝ or ℂ?

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up vote 4 down vote accepted

As Qiaochu wrote, the answer to the question is "not really, but..." Let me amplify on the "but" part.

Positive-definiteness inherently requires an ordering on your field. Conversely, if you have an ordered field, then the theory of inner products goes through verbatim. (The ordering does not have to be Archimedean, so this indeed gives lots more examples.)

Let's assume that you have a field K of characteristic different from 2 which cannot be ordered: by a theorem of Artin-Schreier, this is equivalent to -1 being a sum of squares in the field. Then you don't have "positive definiteness". What is more, the "standard inner product"

q(x_1,...,x_n) = x_1^2 + ... + x_n^2

will be isotropic for sufficiently large n, i.e., there will exist nonzero vectors v = (x_1,...,x_n) such that q(v) = 0. For instance, if K is finite, this occurs as soon as n >= 3.

Let me remark that "isotropic inner products" are not inherently worthless. I have a preliminary version of a wonderful book, "Linear Algebra Methods in Combinatorics" by Laszlo Babai, which indeed makes nice use of the above inner product over finite fields, even in characteristic 2.

(See http://www.cs.uchicago.edu/research/publications/combinatorics. Unfortunately it seems that the book never came to fruition. I got my copy more than 10 years ago when I took an undergraduate course in combinatorics from Babai.)

On the other hand, to any quadratic form over a field K of characteristic not 2, you can associate a symmetric bilinear form. See (among infinitely other references) p. 2 of

http://math.uga.edu/~pete/quadraticforms.pdf

As above, it is plausible that an algebraic substitute for "inner product space" is "vector space endowed with an anisotropic quadratic form", i.e., a regular quadratic form without nonzero vectors v for which q(v) = 0. Witt discovered that you can do a lot of "geometry" in this case: especially, he defined reflection through the hyperplane determined by any anisotropic vector: see (e.g....) pp. 17-18 of the above reference. More is true here than is included in my introductory notes: for instance the orthogonal group of an anistropic quadratic form has the "compactness properties" of the standard real orthogonal group O(n) (that is, it contains no nontrivial split subtorus).

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No. The axiom that fails is positive-definiteness, which doesn't mean anything for an arbitrary field. But one can still define symmetric bilinear forms.

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Thank you. What about fields which are also ordered sets? –  heiner Nov 24 '09 at 16:21
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The ordering needs to be compatible with the field operations, but yes, you can do that. –  Qiaochu Yuan Nov 24 '09 at 16:29
    
Just a trivial remark: for the quaternions, which you can think of as a "skew" field, you cannot define bilinear forms, but only sesquibilinear forms. At any rate, the important property is not so much bilinearity as non-degeneracy. –  José Figueroa-O'Farrill Nov 24 '09 at 16:53
    
I feel like there's a way to do this for fields of characteristic zero, or if not, at least real closed number fields and their algebraic closures. –  Harry Gindi Nov 25 '09 at 0:17
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To get an analogue of hermitian inner products, generalizing the inner product of complex vector spaces, one usually considers fields endowed with an involution (just as complex conjugation) and uses that in the more or less obvious way. This is how one can define the unitary groups, for example; see Dieudonné's Sur les groupes classiques.

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