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 4


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 $n \ge 3$.

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


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).


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.

  • $\begingroup$ Thank you. What about fields which are also ordered sets? $\endgroup$
    – heiner
    Nov 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 non-degeneracy. $\endgroup$ 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 well-behaved "inner product" on non-Archimedean 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:

  1. $\langle x,x\rangle\neq 0$ for all $x\neq 0$, $x\in E$.
  2. $\langle x,y\rangle=\langle y,x\rangle^*$ for all $x,y\in E$.
  3. $x\mapsto \langle x,y\rangle$ 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 book1 that $\langle ,\rangle$ satisfies the Cauchy-Schwartz 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 infinite-dimensional Hilbert-like spaces over $K$.

Option 2: A similar approach was taken in the paper: A non-Archimedean inner product, L. Narici, E. Beckenstein, Contemporary Mathematics, AMS, vol.384, pp.187--202 (2005). There, the "inner product" $\langle ,\rangle$ is defined by the axioms:

  1. $\langle x,x\rangle\neq0$ whenever $x\neq0$.
  2. $x\mapsto \langle x,y\rangle$ is linear for each $y\in E$.
  3. $|\langle x,y\rangle|^2\leq|\langle x,x\rangle||\langle y,y\rangle|$.

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 non-Archimedean valued field was developed in here2, here3, here4 and here5. As in the option 1, in this case there are no infinite-dimensional Hilbert-like spaces. This is shown in this other paper6. 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 infinite-dimensional Hilbert-like 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

1Perez-Garcia, C.; Schikhof, W. H. Locally convex spaces over non-Archimedean valued fields
2J. Aguayo and M. Nova: Non-Archimedean Hilbert like spaces, Bull. Belg. Math. Soc. Simon Stevin, Volume 14, Number 5 (2007), 787-797.
3 José Aguayo, Miguel Nova, and Khodr Shamseddine: Characterization of compact and self-adjoint operators on free Banach spaces of countable type over the complex Levi-Civita 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 Levi-Civita field, Indagationes Mathematicae, Volume 26, Issue 1, January 2015, Pages 191-205; 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 Levi-Civita field, p-Adic 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 219-243; 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 $2$. $\endgroup$
    – LSpice
    Sep 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, Erlangen-ly, 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. $\endgroup$
    – LSpice
    Sep 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". $\endgroup$
    – Chilote
    Sep 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.


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