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\begin{document}

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\section*{inner product space}

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

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

[[!include homotopy - contents]]

\hypertarget{inner_product_spaces}{}\section*{{Inner product spaces}}\label{inner_product_spaces}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak
\noindent\hyperlink{definite}{Definiteness}\dotfill \pageref*{definite} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


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

An \textbf{inner product space} (``scalar product'') is a [[vector space]] $V$ equipped with a (conjugate)-symmetric [[bilinear form|bilinear]] or [[sesquilinear form]]: a [[linear map]] from the [[tensor product]] $V \otimes V$ of $V$ with itself, or of $V$ with its [[dual module]] $\bar{V} \otimes V$ to the [[ground ring]] $k$.

One often studies \emph{positive-definite} inner product spaces; for these, see [[Hilbert space]]. Here we do not assume positivity (positive semidefiniteness) or definiteness (nondegeneracy). See also [[bilinear form]].

The group of automorphisms of an inner product space is the [[orthogonal group of an inner product space]].

\hypertarget{definitions}{}\subsection*{{Definitions}}\label{definitions}

Let $V$ be a [[vector space]] over the [[field]] (or more generally a [[ring]]) $k$. Suppose that $k$ is equipped with an [[involution]] $r \mapsto \bar{r}$, called \emph{conjugation}; in many examples, this will simply be the [[identity function]], but not always. An \textbf{inner product} on $V$ is a function

\begin{displaymath}
\langle {-},{-} \rangle: V \times V \to k
\end{displaymath}
that is (1--3) \emph{sesquilinear} (or \emph{bilinear} when the involution is the identity) and (4) \emph{conjugate-symmetric} (or \emph{symmetric} when the involution is the identity). That is:

\begin{enumerate}%
\item $\langle 0, x \rangle = 0$ and $\langle x, 0 \rangle = 0$;
\item $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z \rangle$ and $\langle x, y + z \rangle = \langle x, y \rangle + \langle x, z \rangle$;
\item $\langle c x, y \rangle = \bar{c} \langle x, y \rangle$ and $\langle x, c y \rangle = \langle x, y \rangle c$;
\item $\langle x, y \rangle = \overline{\langle y, x \rangle}$.

\end{enumerate}
Here we use the \emph{physicist's convention} that the inner product is antilinear (= conjugate-linear) in the first variable rather than in the second, rather than the \emph{mathematician's convention}, which is the reverse. The physicist's convention fits in a little better with $2$-[[2-Hilbert space|Hilbert space]]s and is often used in a generalization for [[Hilbert module]]s. Note that we use the same ring as values of the inner product as for [[scalars]]. Notice that $\langle x, c y \rangle = \langle x, y \rangle c$ is written with $c$ on the right for the case that we deal with noncommutative division ring.

Are the two conventions really equivalent when $k$ is noncommutative? ---Toby

(The axiom list above is rather redundant. First of all, (1) follows from (3) by setting $c = 0$; besides that, (1--3) come in pairs, only one of which is needed, since each half follows from the other using (4). It is even possible to derive (3) from (2) under some circumstances.)

An \textbf{inner product space} is simply a vector space equipped with an inner product.

We define a function ${\|{-}\|^2}\colon V \to k$ by ${\|x\|^2} = \langle x, x \rangle$; this is called the \textbf{norm} of $x$. As the notation suggests, it is common to take the norm of $x$ to be the square root of this expression in contexts where that makes sense, but for us ${\|{-}\|^2}$ is an atomic symbol. The norm of $x$ is \textbf{real} in that it equals its own conjugate, by (4).

\hypertarget{definite}{}\subsection*{{Definiteness}}\label{definite}

Notice that, by (1), $\langle 0, y \rangle = 0$ for all $y$. In fact, the [[subset]] $\{ x \;|\; \forall y,\; \langle x, y \rangle = 0 \}$ is a [[linear subspace]] of $V$. Of course, we also have ${\|0\|^2} = 0$, but $\{ x \;|\; {\|x\|^2} = 0 \}$ may not be a subspace. These observations motivate some possible conditions on the inner product:

\begin{itemize}%
\item The inner product is \textbf{semidefinite} if $\{ x \;|\; {\|x\|^2} = 0 \}$ \emph{is} closed under addition (and hence is a subspace); it's \textbf{indefinite} if there are $x$ and $y$ with ${\|x\|^2} = 0$ and ${\|y\|^2} = 0$ but ${\|x + y\|^2} \ne 0$.
\item The inner product is \textbf{nondegenerate} if $x = 0$ whenever $\langle x, y \rangle = 0$ for all $y$; it's \textbf{degenerate} if there is $x$ with $x \ne 0$ but $\langle x, y \rangle = 0$ for all $y$.
\item The inner product is \textbf{definite} if $x = 0$ whenever ${\|x\|^2} = 0$; there ought to be a term for the condition that there is some $x \ne 0$ with ${\|x\|^2} = 0$, so let's call it \textbf{nondefinite}.

\end{itemize}
(In [[constructive mathematics]], we usually want an [[inequality relation]] relative to which the vector-space operations and the inner product are [[strongly extensional function|strongly extensional]], to make sense of the conditions with $\ne$ in them. We can also use [[contrapositives]] to put $\ne$ in the other conditions, which makes them stronger if the inequality relation is [[tight relation|tight]].)

An inner product is definite iff it's both semidefinite and nondegenerate. Semidefinite inner products behave very much like definite ones; you can mod out by the elements with norm $0$ to get a [[quotient vector space|quotient space]] with a definite inner product. In a similar way, every inner product space has a nondegenerate quotient.

Now suppose that $k$ is equipped with a [[partial order]]. (Note that the [[complex numbers]] are standardly so equipped, with $a \leq b$ iff $b - a$ is a nonnegative real.) Then we can consider other conditions on the inner product:

\begin{itemize}%
\item The inner product is \textbf{positive semidefinite}, or simply \textbf{positive}, if ${\|x\|^2} \geq 0$ always.
\item The inner product is \textbf{positive definite} if it is both positive and definite, in other words if ${\|x\|^2} \gt 0$ whenever $x \ne 0$.
\item The inner product is \textbf{negative semidefinite}, or simply \textbf{negative}, if ${\|x\|^2} \leq 0$ always.
\item The inner product is \textbf{negative definite} if it is both positive and definite, in other words if ${\|x\|^2} \lt 0$ whenever $x \ne 0$.

\end{itemize}
In this case, we have these theorems:

\begin{itemize}%
\item A positive or negative inner product really is semidefinite (as the terminology implies).
\item Conversely, a semidefinite inner product is either positive or negative, at least if we use [[classical logic]]. (In [[constructive mathematics]], we can say that a semidefinite inner product is either positive or negative, indeed is positive [[xor]] negative, as long as it is nonzero in the sense that at least some $\lang{x,y}\rang$ is [[apart]] from zero. And of course, the zero inner product is the only one that is both positive and negative.)
\item Hence, a definite inner product is either positive or negative definite. (We can strengthen this to `xor' and state it constructively if we assume that $V$ is nonzero in that there exists $x$ with $x \ne 0$.)
\item Conversely, an inner product is indefinite if and only if some norms are positive and some are negative. (No constructive caveats here!)

\end{itemize}
Negative (semi)definite inner products behave very much like positive (semi)definite ones; you can turn one into the other by multiplying all inner products by $-1$.

The study of positive definite inner product spaces (hence essentially of all semidefinite inner product spaces over partially ordered fields) is essentially the study of [[Hilbert spaces]]. (For Hilbert spaces, one usually uses a [[topological field]], typically $\mathbb{C}$, and requires a completeness condition, but this does not effect the algebraic properties much.) The study of indefinite inner product spaces is very different; see the \href{http://secure.wikimedia.org/wikipedia/en/wiki/Krein_space}{English Wikipedia article} on [[Krein space]]s for some of it.

All of this definiteness terminology may now be applied to an \emph{[[linear operator|operator]]} $T$ on $V$, since $(x, y) \mapsto \langle{x, T y}\rangle$ is another inner product (on $\dom T$, if necessary). See [[positive operator]].

\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\begin{itemize}%
\item [[Euclidean space]];


\item [[Hilbert spaces]] (over $\mathbb{C}$), for example $\mathbb{C}^n$;


\item Finite-dimensional modules over $\mathbb{H}$, the [[quaternions]].



\end{itemize}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[orthogonal structure]]


\item [[self-dual object]]


\item [[inner product of vector bundles]]



\end{itemize}
[[!redirects inner product]] [[!redirects inner products]] [[!redirects inner product space]] [[!redirects inner product spaces]]

[[!redirects scalar product]] [[!redirects scalar products]]



\end{document}