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Let $B$ and $C$ be abelian groups (in additive notation). We call a function $f:B\rightarrow C$ a quadratic form if for all $x,y,z \in B$, the function $f$ satisfies the relation $$f(x+y+z)-f(x+y)-f(y+z)-f(x+z)+f(x)+f(y)+f(z)=0.$$

Is there some reference which says that it satisfies the parallelogram law $f(x+y)+f(x-y)=2f(x)+2f(y)$ and that for $B=\mathbb{Z}^n$ we can write $$f(\displaystyle\sum_{i=1}^{n} a_i e_i)=\displaystyle\sum_{i=1}^n\big(2a_i^2+\displaystyle\sum_{j=1}^n a_ia_j \big) f(e_i)+ \displaystyle\sum_{1\leq i<j\leq n}a_ia_jf(e_i+e_j),$$ where $e_i$ denote the standard basis vectors of $\mathbb{Z}^n$.

Thank you!

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Do you specifically need a reference? How about just a proof? – Scaramouche Jan 13 '12 at 22:20
Do you have one? I do not see how to do it. – Nadori Jan 13 '12 at 23:48

The parallelogram law can be proven from your relation by substituting $x=y=z=0$ to get that $f(0) = 0$; then substituting $z=-x$ to get that $$ f(y) - f(x+y) - f(y-x) + f(x) + f(y) + f(-x) =0 $$ or $$ 2 f(y) + f(x) + f(-x) = f(x+y) + f(y-x). $$ It clearly only remains to show that $f$ is even.

The second statement, your large sum expansion, follows from the first, since the parallelogram law implies the existence of an inner product defined as $$ \langle x, y \rangle := \dfrac{f(x+y) - f(x-y)}4 \text{ such that } \langle x, x \rangle = f(x). $$ (aka the polarization identity.) Thus, using this inner product and its consequential symmetry and bilinearity, the second statement immediately follows.

I'm still working on the first part, I'll put that in an edit.

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Sorry, I don't think I can show that $f$ must be even. I guess I'll leave that for you to figure out, then. – Scaramouche Jan 16 '12 at 16:13
In the literature it says that it is "well known". There should be a reference ... Maybe someone else can help us ? – Nadori Jan 17 '12 at 14:13

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