I see definitions (in Wikipedia, for example) about inner product spaces over arbitrary fields. But I don't understand how positivity makes any sense for fields which are not ordered? Am I missing something?

Clarification: Let me elaborate, to make clear what I already know. An inner product $g$ is a $\mathbb F$-sesquilinear map from $V\times V\to \mathbb F$, that is conjugate symmetric, non-degenerate and positive. My qualms are regarding positivity: that $g(v,v)\geq0$ for all $v\in V$. That inequality means nothing if $\mathbb F$ has no order structure on it.

  1. A suitable solution is to actually restrict our definition to ordered fields. But, this clearly is a strong condition- even the complex numbers are not ordered.

  2. In the real case, the above definition corresponds to the familiar symmetric, non-degenerate, positive bilinear forms; as $\mathbb R$ has a trivial $*$-operation.

  3. Let us look at the complex case, where the first such non-trivial definition arises. By using conjugate symmetry, $g_{\mathbb C}(v,v)=g_{\mathbb C}(v,v)^*$, so $g_{\mathbb C}(v,v)\in \mathbb R$, and since $\mathbb R$ has a canonical order- one can talk about positivity. So, the definition seems to be consistent here. A similar argument seems to works for quaternions as well. This suggests that if your field is a normed division algebra over $\mathbb R$ (by Hurzewicz theorem, there are only four such examples), the above construction may be extendable.

  4. Does it work for $*$-fields, in general? While conjugate symmetry and sesquilinearity generalise well to $*$-fields, positivity remains an issue. For the above argument to work out in a general case, one would require the conjugate symmetric elements of a *-field to have an order structure. Any concrete counter-example to that here would be helpful.
  5. This is still a far away from general fields.
  • $\begingroup$ Cf. here and here $\endgroup$ Feb 21, 2019 at 14:44
  • $\begingroup$ $<x,x>$ is real-valued. You may be thinking of dot product. In a real vector space, the dot product is an inner product. $\endgroup$
    – saulspatz
    Feb 21, 2019 at 14:45
  • $\begingroup$ @J.W.Tanner Those answers are more along the lines of what I was thinking. So I refined the question a bit more to elucidate what I couldn't find in those answers. $\endgroup$
    – Sandesh Jr
    Feb 22, 2019 at 12:07

1 Answer 1


Because an inner product induces a norm $\|x\|=\sqrt{\langle x|x\rangle},$ and a norm, by definition, is trying to measure the "size" of something, we require the inner product of a vector with itself to be non-negative. So it makes sense in the reflexive case. If you're computing $\langle x|y\rangle,$ then all bets are off. The result could be a complex number, but that's fine. Geometrically, $\langle x|y\rangle$ is related to measuring the "angle" between $x$ and $y$, though that may not be in the Euclidean sense.

  • 1
    $\begingroup$ I'm not particularly interested in the motivation; what bugs me is the inconsistency of the definition itself. $\endgroup$
    – Sandesh Jr
    Feb 22, 2019 at 12:00
  • $\begingroup$ $\langle x|y\rangle=\frac{1}{4}\sum_{n=0}^3 i^{-n}\Vert x+i^n y\Vert^2$ (or in the real case, $\langle x|y\rangle=\frac{1}{4}(\Vert x+y\Vert^2-\Vert x-y\Vert^2)$). $\endgroup$
    – J.G.
    Feb 22, 2019 at 12:04

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