Teaching polarisation formula

Question

When teaching about Hilbert spaces, one begins with a polarisation formula, which allows us to reconstruct the scalar product from the norm: $$\langle u,v\rangle=\frac14(\|u+v\|^2-\|u-v\|^2+\imath\|u+\imath v\|^2-\imath\|u-\imath v\|^2).$$ Is there a good reason to choose this formula instead of the more symmetric one $$\langle u,v\rangle=\frac1{2\pi}\int_0^{2\pi}e^{\imath\theta}\|u+e^{\imath\theta}v\|^2d\theta,$$ or the shortest one (here $\jmath=e^{2\imath\pi/3}$) $$\langle u,v\rangle=\frac13(\|u+v\|^2+\jmath\|u+jv\|^2+\bar\jmath\|u+\bar\jmath v\|^2) \qquad?$$

Answer 1

To me it seems most natural to show that the norm determines the scalar product via the two formulas $$\Vert u + v \Vert^2 = \Vert u \Vert^2 + \Vert v \Vert^2 + 2\mathrm{Re}\langle u, v \rangle$$ and $$\mathrm{Im} \langle u, v \rangle = \mathrm{Re}\langle u, -iv \rangle.$$ Both of these are immediately obvious, in a way the polarization identities are not (IMO).

Answer 2

The polarization formula carries over to algebraic settings not directly involving real and complex numbers. Let $L/K$ be a quadratic Galois extension, where the nontrivial element of ${\rm Gal}(L/K)$ is denoted with an overline: $\overline{\alpha}$ for $\alpha \in L$. Let $V$ be a finite-dimensional $L$-vector space [edit: as nfdc23 points out, finite-dimensionality is not needed] and $B \colon V \times V \rightarrow L$ be a Hermitian form relative to the extension $L/K$: it is $K$-bilinear, $B(cv,w) = cB(v,w)$ and $B(v,cw) = \overline{c}B(v,w)$ for $c\in L$, and $B(w,v) = \overline{B(v,w)}$. Set $Q \colon V \rightarrow L$ by $Q(v) = B(v,v)$. Can we reconstruct the two-variable function $B$ from the single-variable function $Q$?

Since $$ Q(v+w) = Q(v) +Q(w) + B(v,w) + B(w,v), $$ replacing $w$ with $-w$ gives $$ Q(v-w) = Q(v) +Q(w) - B(v,w) -B(w,v), $$ so subtracting and dividing by $2$ (if we are not in characteristic $2$) gives us $$ \frac{Q(v+w)-Q(v-w)}{2} = B(v,w) + B(w,v). $$ To get a formula having only $B(v,w)$ on the right, pick $c \in L$ such that $\overline{c} \not= c$. Then $$ Q(v+cw) = Q(v) +c\overline{c}Q(w) + \overline{c}B(v,w) + cB(w,v), $$ and $$ Q(v+\overline{c}w) = Q(v) +\overline{c}cQ(w) + cB(v,w) + \overline{c}B(w,v) $$ so $$ Q(v+cw) - Q(v+\overline{c}w) = (\overline{c}-c)(B(v,w) - B(w,v)) $$ and thus $$ \frac{Q(v+cw) - Q(v+\overline{c}w)}{\overline{c}-c} = B(v,w) - B(w,v). $$ Since we have formulas for $B(v,w)+B(w,v)$ and $B(v,w)-B(w,v)$, by averaging (if we are not in characteristic $2$), $$ \frac{Q(v+w) - Q(v-w)}{4} + \frac{Q(v+cw) - Q(v+\overline{c}w)}{2(\overline{c}-c)} = B(v,w). $$ When $L = \mathbf C$, $K = \mathbf R$, $B(v,w) = \langle v,w\rangle$, $Q(v) = ||v||^2$, and $c = i$ this recovers the classical polarization formula. If in the same setting we take $c = j = e^{2\pi i/3}$, then we get a polarization formula that is different from the one you wrote down with $j$, but it fits into the general pattern described above. Since I showed how the usual polarization formula extends to the general case outside of characteristic $2$ (certainly an integration formula does not), that classical formula is not really specific to the choice $c = i$, but your version using $j$ seems specific to the choice $c=j$ since you are dividing through by $3$. I am not persuaded that a formula using $3$ terms instead of $4$ terms is genuinely simpler: the $4$-term polarization formula is essentially the result of averaging a few times. (Most people, whether by habit or otherwise, would prefer to think of $\mathbf C$ as $\mathbf R + \mathbf R{i}$ rather than as $\mathbf R + \mathbf R{j}$ for both geometric and algebraic reasons, e.g., $\overline{a+bj} = a-b-bj$ for real $a$ and $b$.)

What happens if we are using a Hermitian form in characteristic $2$? Division by $2$ in the classical polarization formula or its generalization above breaks down if we try to reconstruct $B$ from $Q$ in characteristic $2$. We can show $Q$ determines $B$ by the following argument that is valid in all characteristics: for $v$ and $w$ in $V$, and $c \in L$, we have \begin{eqnarray*} Q(cv+w) & = & c\overline{c}Q(v) + Q(w) + B(cv,w) + B(w,cv) \\ & = & c\overline{c}Q(v) + Q(w) + cB(v,w) + \overline{B(cv,w)} \\ & = & c\overline{c}Q(v) + Q(w) + {\rm Tr}_{L/K}(cB(v,w)). \end{eqnarray*} Therefore, when $v$ and $w$ are fixed in $V$, the function $f_{v,w} \colon L \rightarrow K$ given by $f_{v,w}(c) = Q(cv+w) - c\overline{c}Q(v) - Q(w)$ is completely determined by $Q$ and it is also $K$-linear since it equals ${\rm Tr}_{L/K}(cB(v,w))$. For a quadratic Galois extension $L/K$, each $K$-linear mapping $L \rightarrow K$ looks like $f(x) = {\rm Tr}_{L/K}(xy)$ for a unique $y \in L$ (this could be shown directly, or it is a consequence of non-degeneracy of the trace-pairing $L \times L \rightarrow K$, where $\langle x,y\rangle \mapsto {\rm Tr}_{L/K}(xy)$). Thus $B(v,w)$ is the unique number $y$ in $L$ such that $Q(cv+w) - c\overline{c}Q(v) - Q(w) = {\rm Tr}_{L/K}(cy)$ for all $c \in L$. There is no need for any polarization formula.

[Edit: I realized after writing this up that I wrote something essentially like this a few years ago on math.stackexchange: https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities]

Answer 3

It works for real Hilbert spaces if we remove two last terms. For two other formulas the real counterpart is not clear.