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Let $\mathbb{F}$ be an arbitrary field of characteristic $\mathrm{char}(\mathbb{F})\neq 2$ and $V$ a $\mathbb{F}$-vector space. The definition of a quadratic form I am used to is a map $\varphi\colon V\to\mathbb{F}$ with the property $\varphi(\lambda v)=\lambda^{2}\varphi(v)$ such that $$\beta_{\varphi}(v,w):=\frac{1}{2}(\varphi(v+w)-\varphi(v)-\varphi(w))$$ is bilinear. Now, clearly, $\beta_{\varphi}$ is symmetric and $\beta_{\varphi}(v,v)=\varphi(v)$, by definition. In the case $\mathbb{F}=\mathbb{R}$, I have seen the following claim:

Proposition 1: If $\varphi\colon V\to\mathbb{R}$ with the property $\varphi(\lambda v)=\lambda^{2}\varphi(v)$ satisfies the paralellogram law $$\varphi(v+w)+\varphi(v-w)=2\varphi(v)+2\varphi(w)$$ then it is a quadratic form, i.e. there exists a symmetric bilinear form $\beta_{\varphi}$ such that $\beta_{\varphi}(v,v)=\varphi(v)$.

This is essentially a similar statement as the Jordan-von Neumann theorem, which asserts that a norm $\Vert\cdot\Vert$ on $V$ admits a inner product $\langle\cdot,\cdot\rangle$ inducing it if and only if the paralellogram law $$\Vert v+w\Vert^{2}+\Vert v-w\Vert^{2}=2\Vert v\Vert^{2}+2\Vert w\Vert^{2}$$ holds. Now, my question is the following:

Is Proposition 1 also true for arbitrary fields (excluding $\mathrm{char}(\mathbb{F})=2$ of course)?

I was not able to find a proof of this statement. I tried to mimik the proof of Jordan-von Neumann, but all the proofs I know of this result are somehow specific to $\mathbb{R}$. The idea of the proof is usually to define $$\beta_{\varphi}(v,w):=\frac{1}{2}(\varphi(v+w)-\varphi(v)-\varphi(w))$$ Then clearly $\beta_{\varphi}(v,v)=\varphi(v)$ and $\beta_{\varphi}$ is symmetric. The only thing to show is that $\beta_{\varphi}$ actually defines a bilinear form.

Any literature on this is appreciated.

G. Blaickner
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2 Answers2

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Without positivity, this is not even true for $\mathbb F=\mathbb R$. See this wonderful post.

We show some more general results. It has been shown by P. Quinton that $(\cdot, \cdot):=\beta_{\varphi}$ is bi-additive. In particular it follows $$(-u, v)=(u,-v)=-(u,v)$$ Consider $$\begin{cases} (\lambda(u+v), \lambda(u+v)=\lambda^2(u+v, u+v) & (1)\\ (\lambda(u-v), \lambda(u-v))=\lambda^2(u-v,u-v) & (2)\end{cases}$$

Consider $(1)-(2)$, and after additive expansion, we get $4(\lambda u, \lambda v) =4 \lambda^2(u, v)$ that is $$(\lambda u, \lambda v)=\lambda^2(u, v)$$

where $u, v$ are not necessarily equal.

Now we show $(\lambda u, u)=\lambda (u, u)$. This follows from $$((1+\lambda)u, (1+\lambda)u)=(1+\lambda)^2(u, u)$$

After the expansion and cancellation, $2(\lambda u, u)=2\lambda (u, u)$.

Now we can show that in the very special case $\dim V = 1$, $\beta_{\varphi}$ is bilinear. And since the post has constructed examples for $\dim V=2$, we know the statement also fails for all $\dim V\ge 2$.

Let $V=\text{Span}\{e\}$, then assume $a\not=0$, $$(ae,be)=a^2(e, \frac{b}{a}e)=a^2(\frac{b}{a}e, e)=a^2\frac{b}{a}(e,e)=ab(e,e)$$ In other words, $(\cdot, \cdot)$ is completely determined by $(e,e)$ alone.

Just a user
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    Thank you very much for your answer. Thats interesting. I have seen resources defining real quadratic forms to be maps $Q:V\to\mathbb{R}$ with the properties $Q(\lambda v)=\lambda^{2}Q(v)$ and the paralellgram law. I was assuming that this is an equivalent definition to the usual one (the one in my post), but appaerently not! – G. Blaickner Mar 08 '24 at 09:11
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This needs a total of 9 applications of Proposition 1, sorry about that, maybe one can do better. I can prove that $\beta_\varphi(u+v,w)=\beta_\varphi(u,w)+\beta_\varphi(v,w)$, I don't know if $\beta_\varphi(av,w)=a\beta_\varphi(v,w)$.

We can apply Propposition 1 three times to get \begin{align*} \varphi(-u+v+w)&=2\varphi(w)+2\varphi(-u+v)-\varphi(-u+v-w)\\ &=2\varphi(w)+2\varphi(-u+v)-\varphi(u-v+w)\\ \varphi(u-v+w)&=2\varphi(u)+2\varphi(-v+w)-\varphi(-u-v+w)\\ &=2\varphi(u)+2\varphi(-v+w)-\varphi(u+v-w)\\ \varphi(u+v-w)&=2\varphi(v)+2\varphi(u-w)-\varphi(u-v-w)\\ &=2\varphi(v)+2\varphi(u-w)-\varphi(-u+v+w)\\ \end{align*} summing the three equations, rearanging and dividing by $2$ yields \begin{align*} &\varphi(-u+v+w)+\varphi(u-v+w)+\varphi(u+v-w)\\ =&\varphi(u)+\varphi(v)+\varphi(w)+\varphi(-u+v)+\varphi(-v+w)+\varphi(u-w) \end{align*}

Now we can apply Proposition 1 three more times to the term $\varphi(u+v+w)$ to get \begin{align*} \varphi(u+v+w) &=2\varphi(u+v)+2\varphi(w)-\varphi(u+v-w)\\ &=2\varphi(u+w)+2\varphi(v)-\varphi(u-v+w)\\ &=2\varphi(v+w)+2\varphi(u)-\varphi(-u+v+w) \end{align*} summing the three and plugging in the previous equation gives \begin{align*} 3\varphi(u+v+w)&=2\varphi(u+v)+2\varphi(u+w)+2\varphi(v+w)+2\varphi(u)+2\varphi(v)+2\varphi(w)\\ &-\varphi(u)-\varphi(v)-\varphi(w)-\varphi(-u+v)-\varphi(-v+w)-\varphi(u-w) \end{align*} Finally applying three last times Proposition 1 \begin{align*} \varphi(-u+v)&=2\varphi(u)+2\varphi(v)-\varphi(u+v)\\ \varphi(-v+w)&=2\varphi(v)+2\varphi(w)-\varphi(v+w)\\ \varphi(u-w)&=2\varphi(u)+2\varphi(w)-\varphi(u+w) \end{align*}

and plugging in as well as dividing by $3$ yields \begin{align*} \varphi(u+v+w)&=\varphi(u+v)+\varphi(u+w)+\varphi(v+w)-\varphi(u)-\varphi(v)-\varphi(w) \end{align*}

With this we can prove that \begin{align*} 2\beta_\varphi(u+v,w)&=\varphi(u+v+w)-\varphi(u+v)-\varphi(w)\\ &=\varphi(u+w)+\varphi(v+w)-\varphi(u)-\varphi(v)-2\varphi(w)\\ &=2\beta_\varphi(u,w)+2\beta_\varphi(v,w) \end{align*}

P. Quinton
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  • Hi @P.Quinton. Thank you very much for your detailed reply! This is indeed the same proof as for inner products of the Jordan-Neumann theorem. My problem was more to show $\beta_{\varphi}(\lambda v,w)=\lambda\beta_{\varphi}(v,w)$. Because, in the case of Jordan-Neumann and inner products in $\mathbb{R}$, this part of the proof uses usually methods specific for $\mathbb{R}$... – G. Blaickner Mar 04 '24 at 15:51
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    @G.Blaickner I would believe this is not true in general, it feels like you would have to do things like $\beta_\varphi(n u, w)=n\beta_\varphi(u,w)$ for $n\in\mathbb N$ and $nu=u+u+u+...$ $n$ times. Then you can even have a notion of rationals, but I think this would work for $\mathbb F_p$, $\mathbb Q$, $\mathbb R$ and such. – P. Quinton Mar 05 '24 at 08:01