Is there a canonical definition of the concept of inner products for vector spaces over arbitrary fields, i.e. other fields than $\mathbb R$ or $\mathbb C$?
4 Answers
As Qiaochu wrote, the answer to the question is "not really, but..." Let me amplify on the "but" part.
Positivedefiniteness 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 ArtinSchreier, 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 $n \ge 3$.
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 https://web.archive.org/web/20100205064709/https://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://alpha.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. 1718 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).

$\begingroup$ Isotropic subspaces give rise to pointline geometries called polar spaces: en.wikipedia.org/wiki/Polar_space, and thus to several other interesting structures like strongly regular graphs (collinearity graph of almost every polar space is strongly regular), some extremal graphs, perfect codes, etc. See maths.qmul.ac.uk/~pjc/pps, and Simeon Ball's new book: cambridge.org/hr/academic/subjects/mathematics/…. $\endgroup$– AnuragMay 21, 2016 at 17:55
No. The axiom that fails is positivedefiniteness, which doesn't mean anything for an arbitrary field. But one can still define symmetric bilinear forms.

$\begingroup$ Thank you. What about fields which are also ordered sets? $\endgroup$– heinerNov 24, 2009 at 16:21

3$\begingroup$ The ordering needs to be compatible with the field operations, but yes, you can do that. $\endgroup$ Nov 24, 2009 at 16:29

$\begingroup$ 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 nondegeneracy. $\endgroup$ Nov 24, 2009 at 16:53

$\begingroup$ 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. $\endgroup$ Nov 25, 2009 at 0:17
Several approaches have been tried to conceive a wellbehaved "inner product" on nonArchimedean valued fields. Let's see some examples:
Option 1: Let $\lambda\mapsto\lambda^*$ be a field automorphism of order $2$ defined on $K$. Let $E$ be a $K$vector space. An inner product is a map $\langle,\rangle:E\times E\to K$ such that:
 $\langle x,x\rangle\neq 0$ for all $x\neq 0$, $x\in E$.
 $\langle x,y\rangle=\langle y,x\rangle^*$ for all $x,y\in E$.
 $x\mapsto \langle x,y\rangle$ is linear for each $y\in E$.
Note that if axioms 1,2,3 are assumed in the complex case then either $\langle ,\rangle$ or $\langle ,\rangle$ is positive definite.
It is proved in this book^{1} that $\langle ,\rangle$ satisfies the CauchySchwartz inequality, it induces a norm, $\langle ,\rangle$ is continuous in the topology induced by the norm and more. The down side is that there is no infinitedimensional Hilbertlike spaces over $K$.
Option 2: A similar approach was taken in the paper: A nonArchimedean inner product, L. Narici, E. Beckenstein, Contemporary Mathematics, AMS, vol.384, pp.187202 (2005). There, the "inner product" $\langle ,\rangle$ is defined by the axioms:
 $\langle x,x\rangle\neq0$ whenever $x\neq0$.
 $x\mapsto \langle x,y\rangle$ is linear for each $y\in E$.
 $\langle x,y\rangle^2\leq\langle x,x\rangle\langle y,y\rangle$.
This map defines the norm $\x\=\langle x,x\rangle^{1/2}$. Also, it is proved that in $C_0$, the sup norm is induced by $\langle x,y\rangle=\sum x_ny_n$ (for $x=(x_n)_n$ and $y=(y_n)_n$ in $C_0$) if and only if the residue class field of $K$ is real closed. Subsequent study of operators in $C_0$ over an ordered nonArchimedean valued field was developed in here^{2}, here^{3}, here^{4} and here^{5}. As in the option 1, in this case there are no infinitedimensional Hilbertlike spaces. This is shown in this other paper^{6}. However, when the valuations take values in ordered groups rather than the real numbers, the situation changes as we see in the following.
Option 3: It is possible to conceive Quadratic spaces over Krull valued fields. Under certain conditions, it is even possible to have infinitedimensional Hilbertlike spaces. For this topic, in particular, I recommend the papers:
a. On a class of orthomodular quadratic spaces, H. Gross, U.M. Künzi  Enseign. Math, 1985.
b. Banach spaces over fields with a infinite rank valuation  [H.Ochsenius A., W.H.Schikhof]  1999
c. After that see: Norm Hilbert spaces over Krull valued fields  [H. Ochsenius, W.H. Schikhof]  Indagationes Mathematicae, Elsevier  2006
^{1}PerezGarcia, C.; Schikhof, W. H. Locally convex spaces over nonArchimedean valued fields
^{2}J. Aguayo and M. Nova: NonArchimedean Hilbert like spaces, Bull. Belg. Math. Soc. Simon Stevin, Volume 14, Number 5 (2007), 787797.
^{3} José Aguayo, Miguel Nova, and Khodr Shamseddine: Characterization of compact and selfadjoint operators on free Banach spaces of countable type over the complex LeviCivita field, Journal of Mathematical Physics 54, 023503 (2013); https://doi.org/10.1063/1.4789541
^{4} José Aguayo, Miguel Nova, and Khodr Shamseddine: Inner product on $B^*$algebras of operators on a free Banach space over the LeviCivita field, Indagationes Mathematicae, Volume 26, Issue 1, January 2015, Pages 191205; https://doi.org/10.1016/j.indag.2014.09.006
^{5} José Aguayo, Miguel Nova, and Khodr Shamseddine: Positive operators on a free Banach space over the complex LeviCivita field, pAdic Numbers, Ultrametric Analysis and Applications, April 2017, Volume 9, Issue 2, pp 122–137; https://doi.org/10.1134/S2070046617020029
^{6} M. P. Solèr: Characterization of hilbert spaces by orthomodular spaces, Communications in Algebra, Volume 23, 1995  Issue 1, Pages 219243; https://doi.org/10.1080/00927879508825218

$\begingroup$ Surely you want $()^*$ to square to the identity, not have order 2, right? $\endgroup$– David Roberts ♦Sep 3 at 22:49

$\begingroup$ @DavidRoberts, re, actually, in the description of unitary groups over arbitrary fields (which one might reasonably hope to correspond to "inner products"), one usually does require that the involution be "of the second kind", i.e., genuinely of order $2$, not just of order dividing $2$. $\endgroup$– LSpiceSep 4 at 1:21

$\begingroup$ @LSpice but then this definition rules out taking K to be the reals, which definitely works here! The question is not about unitary groups, in any case. $\endgroup$– David Roberts ♦Sep 4 at 1:23

$\begingroup$ @DavidRoberts, re, but the question is about inner products, which one might expect, Erlangenly, to correspond to unitary groups. I would argue that an inner product on a real vector space is more properly viewed as a symmetric bilinear form, and its positive definiteness as a manifestation of the corresponding orthogonal group being anisotropic. But, of course, only @Chilote can say for sure what they meant. $\endgroup$– LSpiceSep 4 at 1:49

$\begingroup$ @LSpice when I studied from the mentioned references, I assumed they meant $(\lambda^*)^*=\lambda$ for every $\lambda\in K$. In this way the proofs made sense. I am not familiar with another interpretation of "automorphism of order 2". $\endgroup$– ChiloteSep 4 at 19:55
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.