# Defining a generalised version of inner product over $*$-fields

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.
• Cf. here and here – J. W. Tanner Feb 21 at 14:44
• $<x,x>$ is real-valued. You may be thinking of dot product. In a real vector space, the dot product is an inner product. – saulspatz Feb 21 at 14:45
• @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. – Sandesh Jr Feb 22 at 12:07

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.
• $\langle x|y\rangle=\frac{1}{4}\sum_{n=0}^3 i^{-n}\Vert x+i^n y\Vert^2$ (or in the real case, $\langle x|y\rangle=\frac{1}{4}(\Vert x+y\Vert^2-\Vert x-y\Vert^2)$). – J.G. Feb 22 at 12:04