Mock scalar product's nonlinearity as a derivation

Question

Context. I was reading about the converse of the parallelogram law (a norm satisfying the law defines an inner product), and in an answer on MO it is stated that this is impossible without the norm's inequality axiom (yielding continuity). Ivanov constructs an explicit counterexample, and makes a claim (below) which I am unable to verify.

Definition. A mock scalar product $\langle-,-\rangle$ on a vector space $V$ over a field $\mathbb{F}$ satisfies

• homogeneity: $\langle tu,tv\rangle=t^2\langle u,v\rangle$
• symmetry: $\langle u,v\rangle=\langle v,u\rangle$
• biadditivity: $\langle u,v+v'\rangle=\langle u,v\rangle+\langle u,v'\rangle$

(Homogeneity is equivalent to $\langle tu,tu\rangle=t^2\langle u,u\rangle$ via polarization unless $\mathrm{char}\,\mathbb{F}=2$.)

Claim. $D_{u,v}(t):=\langle u,vt\rangle-\langle u,v\rangle t$ is a derivation of $t\in\mathbb{F}$ (for any $u,v\in V$).

(Recall a derivation $D$ satisfies $D(s+t)=D(s)+D(t)$ and $D(st)=D(s)t+sD(t)$ for any $s,t\in\mathbb{F}$. I wrote $t$ on the right of the $v$ only so $D_{u,v}$ looks more like an operator.)

Question. How do we verify the claim from the MO thread?

Work. Writing out the product rule for $D_{u,v}$, we could add $2\langle u,v\rangle st$ to both sides but I don't see how this helps. Presumably there is a way to move scalars around using homogeneity but I don't see it. When I use homogeneity to verify the special case $D_{u,v}(1/t)=-D_{u,v}(t)/t^2$ I instead get

$$ \begin{array}{cl} D_{u,v}(1/t) & = \langle u,v/t\rangle-\langle u,v\rangle/t \\ & = \big(\langle tu,v\rangle-\langle u,v\rangle t\big)/t^2 \\ & = D_{v,u}(t)/t^2 \end{array} $$

When $u=v$ this seems to be the opposite of what it ought to be if $D_{u,v}$ is a derivation.

Answer 1

Here is a proof assuming the characteristic is different from $2$. For convenience of typing I will write $\langle u,v\rangle$ as $(u,v)$. Note first that homogeneity with respect to the scalar $s+t$ gives $$(su,sv)+(su,tv)+(tu,sv)+(tu,tv)=(s^2+2st+t^2)(u,v)$$ and hence $$(su,tv)+(tu,sv)=2st(u,v). \quad(*)$$

(By the way, when $t=1$, this implies $D_{u,v}=-D_{v,u}$ so $D_{u,u}=0$ if the characteristic is different from $2$, which explains the computation that you did.)

Now the idea is to use $(*)$ repeatedly to see what you get for an expression like $(stu,v)$ where you multiply by $st$ inside the scalar product. Two applications of $(*)$ gives $$(stu,v)+(su,tv)+2t(u,vs)=2t(su,v)+2t(u,sv)=4st(u,v).$$

Adding this equation to the same equation with $s$ and $t$ swapped gives $$2(stu,v)+(su,tv)+(tu,sv) + 2t(u,sv)+2s(u,tv)= 8st(u,v)$$ and combining this with $(*)$ and dividing by $2$ yields $$(stu,v)+t(u,sv)+s(u,tv)=3st(u,v).$$

Now applying $(*)$ with $st$ in place of $s$ and $1$ in place of $t$ gives $(stu,v)+(u,stv)=2st(u,v)$ so the equation above gives $$-st(u,v)+t(u,sv)+s(u,tv)=(u,stv).$$

Now note that $t(u,sv)=tD_{u,v}(s)-st(u,v)$, $s(u,tv)=sD_{u,v}(t)-st(u,v)$, and $(u,stv)=D_{u,v}(st)-st(u,v)$. Thus adding $st(u,v)$ to both sides of the previous equation gives $$tD_{u,v}(s)+sD_{u,v}(t)=D_{u,v}(st).$$

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On the other hand, here is a counterexample in characteristic $2$. Let $V=\mathbb{F}$ be a field of characteristic $2$ and let $f:\mathbb{F}\to \mathbb{F}$ be an $\mathbb{F}^2$-linear map such that $f(1)=0$, and define $$(x,y)=f(xy).$$ This is a mock scalar product, with homogeneity following from the fact that $f$ is $\mathbb{F}^2$-linear. The function $D_{1,1}$ is then just $f$, which need not be a derivation. For instance, if $\mathbb{F}=\mathbb{F}_2(x,y)$, then $1,x,y,$ and $xy$ are linearly independent over $\mathbb{F}^2$, so $f$ can take arbitrary values on $x,y,$ and $xy$.

(Conversely, in characteristic $2$, $(*)$ implies that $(su,v)=(u,sv)$ so $(s^2u,v)=(su,sv)=s^2(u,v)$ and thus $D_{u,v}$ must be $\mathbb{F}^2$-linear.)