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\begin{document}

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\section*{polarization identity}

\begin{quote}%
This entry is about the notion in [[linear algebra]] relating bilinear and quadratic forms. For the notion in [[symplectic geometry]] see at \emph{[[polarization]]}. For polarization of light, see \emph{[[wave polarization]]} (if we ever write it).


\end{quote}
\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{linear_algebra}{}\paragraph*{{Linear algebra}}\label{linear_algebra}

[[!include homotopy - contents]]

\hypertarget{the_polarization_identity}{}\section*{{The polarization identity}}\label{the_polarization_identity}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{higher_order}{Higher order}\dotfill \pageref*{higher_order} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

Any [[symmetric bilinear form]] $(-) \cdot (-)$ defines a [[quadratic form]] $(-)^2$. The \emph{polarization identity} reconstructs the bilinear form from the quadratic form. More generally, starting from any [[bilinear form]], the polarization identity reconstructs its symmetrization. A slight variation applies this also to [[sesquilinear form]]s. The whole business actually applies to [[bilinear maps]], not just forms (that is, taking arbitrary values, not just values in the [[base field]] or some other [[line]]). The linearity is crucial: polarization doesn't work without addition (and subtraction); we must also be able to divide by $2$. It is also possible to generalize from quadratic forms to higher-order [[homogenous form]]s.

\hypertarget{statement}{}\subsection*{{Statement}}\label{statement}

Let $R$ be a [[commutative ring]]. Let $V$ and $W$ be $R$-[[modules]], and let $m\colon V \times V \to W$ be a [[bilinear map]]; that is, we have an $R$-module [[homomorphism]] $V \otimes V \to W$. Let $Q\colon V \to W$ be the [[quadratic map]] given by $Q(v) = m(v,v)$; this is \emph{not} an $R$-module homomorphism.

Then we have:

\begin{itemize}%
\item the \emph{[[parallelogram law]]}: $2 Q(x) + 2 Q(y) = Q(x + y) + Q(x - y)$, and
\item the \textbf{polarization identity}: $2 m(x,y) + 2 m(y,x) = Q(x + y) - Q(x - y)$.

\end{itemize}
Writing $Q(x)$ as $x^2$ and $m(x,y)$ as $x y$, these read:

\begin{itemize}%
\item $2 x^2 + 2 y^2 = (x +y )^2 + (x - y)^2$,
\item $2 x y + 2 y x = (x + y)^2 - (x - y)^2$.

\end{itemize}
The polarization identity also has these alternative forms:

\begin{itemize}%
\item $x y + y x = (x + y)^2 - x^2 - y^2$,
\item $x y + y x = x^2 + y^2 - (x - y)^2$;

\end{itemize}
these may derived by adding or subtracting the parallelogram law and the first polarization identity; although the derivation requires cancelling $2$, the alternative polarization identities remain valid regardless of whether $2$ is cancellable in $W$.

Now suppose that $m$ is [[symmetric bilinear map|symmetric]], so that $x y = y x$. And suppose that $2 \coloneqq 1 + 1$ is (not merely cancellable but also) invertible in $W$. Then the polarization identities read:

\begin{itemize}%
\item $x y = \frac{1}{4} (x + y)^2 - \frac{1}{4} (x - y)^2$,
\item $x y = \frac{1}{2} (x + y)^2 - \frac{1}{2} x^2 - \frac{1}{2} y^2$,
\item $x y = \frac{1}{2} x^2 + \frac{1}{2} y^2 - \frac{1}{2} (x - y)^2$;

\end{itemize}
That is, we may recover $m$ from $Q$ (in any of these ways). Regardless of the original symmetry of $m$, we may recover its symmetrization $x \circ y = (x y + y x)/2$.

We can go the other direction: given a quadratic map $Q$, if $2$ is invertible, then any polarization identity defines a symmetric bilinear map $m$; these all agree if $Q$ obeys the parallelogram law, and then $Q$ may be recovered from this $m$ once more.

If $R$ is a $*$-[[star-ring|ring]], then $m$ could be conjugate-symmetric ([[Hermitian map|Hermitian]]). Then $Q$ would satisfy $Q(t x) = t^* t Q(x)$ instead of $Q(t x) = t^2 Q(x)$ as for a quadratic map. Since everything is still bilinear or quadratic over the [[integers]], the parallelogram identity still follows, as does the polarization identity in its general forms (before assuming that $m$ is symmetric). We can still recover $m$ from $Q$ if $R$ has an [[imaginary unit]]: an element $i$ such that $i + i^* = 0$ and $i i^* = 1$; we do this as follows:

\begin{displaymath}
x y = \frac{1}{4} Q(x + y) - \frac{1}{4} Q(x - y) + \frac{1}{4} i Q(x + i y) - \frac{1}{4} i Q(x - i y) .
\end{displaymath}
\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

This is best known in the case of bilinear and quadratic \emph{forms}, where $W$ is the [[ground ring]] $R$. Here, $m$ is an [[inner product]], making $V$ into an [[inner product space]], and $Q$ is (the square of) the norm, making $V$ into a [[normed space]].

This also applies to [[commutative algebras]], where $W$ is $V$. Actually, there is no need for $m$ to be associative; although one rarely studies commutative but [[non-associative algebras]], we have an exception with [[Jordan algebras]]. Although the Jordan identity is simpler to express in terms of the multiplication operation (as usual), the application to [[quantum mechanics]] may be more easily motivated through the squaring operation (since the square of an [[observable]] has a more obvious meaning than the Jordan product of two observables), and the polarization identities allow us to recover multiplication from squaring (and subtraction).

\hypertarget{higher_order}{}\subsection*{{Higher order}}\label{higher_order}

Let $R, V, W$ be as before. Let $p$ be a [[natural number]], and let $Q\colon V \to W$ be homogeneous of degree $p$; that is,

\begin{displaymath}
Q(t v) = t^p Q(v) .
\end{displaymath}
We wish to turn $Q$ into a symmetric [[multilinear map]] $m$ of rank $p$ (so $m\colon \Sym_p V \to W$ is linear) as follows:

\begin{itemize}%
\item If $p = 0$, $m() = Q(0)$;
\item If $p = 1$, $m(v_1) = Q(v_1)$;
\item If $p = 2$, $m(v_1, v_2) = Q(v_1 + v_2) - Q(v_1) - Q(v_2)$;
\item If $p = 3$, $m(v_1, v_2, v_3) = Q(v_1 + v_2 + v_3) - Q(v_1 + v_2) - Q(v_1 + v_3) - Q(v_2 + v_3) + Q(v_1) + Q(v_2) + Q(v_3)$;
\item If $p = 4$, $m(v_1, v_2, v_3, v_4) = Q(v_1 + v_2 + v_3 + v_4) - Q(v_1 + v_2 + v_3) - Q(v_1 + v_2 + v_4) - Q(v_1 + v_3 + v_4) - Q(v_2 + v_3 + v_4) + Q(v_1 + v_2) + Q(v_1 + v_3) + Q(v_1 + v_4) + Q(v_2 + v_3) + Q(v_2 + v_4) + Q(v_3 + v_4) - Q(v_1) - Q(v_2) - Q(v_3) - Q(v_4)$;
\item etc.

\end{itemize}
So defined, $m$ is manifestly symmetric, but it might \emph{not} be multilinear just because $Q$ is homogeneous (except for $p = 1$); instead, we \emph{define} a \textbf{homogeneous polynomial} of degree $p$ from $V$ to $W$ be such a homogeneous $Q$ such that $m$ \emph{is} multilinear. (Then a \textbf{[[polynomial]]} from $V$ to $W$ is a sum of homogeneous polynomials of various degrees.) Note that if $V$ is the [[free module]] $R^n$, then this is the usual notion of a polynomial with $n$ variables.

Morally, the expressions on the right-hand side of each item above should end with $\pm Q(0)$, but this ends up not mattering for the definition of a homogeneous polynomial (except when $p = 0$), in which case $Q(0) = 0$ (including when $p = 0$). If one were ever to consider the polarization of a non-polynomial homogeneous map, however, then presumably this term would need to be restored.

To get the same $m$ as in earlier sections of this text, the expressions here must be divided by the [[factorial]] $p!$. Of course, this only works if $p!$ is invertible in $R$; even if $R$ is a [[field]], we need its [[characteristic]] to be greater than $p$ (or $0$). The expressions above work regardless of this to define what a homogeneous polynomial is. However, if you wish to recover $Q$ from $m$, then you do need to divide by the polynomial; otherwise, the only rule that holds in general is that

\begin{displaymath}
m(v,\ldots,v) = p! Q(v) .
\end{displaymath}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[cubical structure on a line bundle]]

\end{itemize}
[[!redirects polarization identity]] [[!redirects polarization identities]] [[!redirects polarisation identity]] [[!redirects polarisation identities]]



\end{document}