# Inner product spaces

## Idea

An inner product space is a vector space $V$ equipped with a (conjugate)-symmetric bilinear or sesquilinear form: a linear map from $V \otimes V$ or $\bar{V} \otimes V$ to the ground ring $k$.

One often studies 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.

## 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 conjugation; in many examples, this will simply be the identity function, but not always. An inner product on $V$ is a function

$\langle {-},{-} \rangle: V \times V \to k$

that is (1–3) sesquilinear (or bilinear when the involution is the identity) and (4) conjugate-symmetric (or symmetric when the involution is the identity). That is:

1. $\langle 0, x \rangle = 0$ and $\langle x, 0 \rangle = 0$;
2. $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z \rangle$ and $\langle x, y + z \rangle = \langle x, y \rangle + \langle x, z \rangle$;
3. $\langle c x, y \rangle = \bar{c} \langle x, y \rangle$ and $\langle x, c y \rangle = \langle x, y \rangle c$;
4. $\langle x, y \rangle = \overline{\langle y, x \rangle}$.

Here we use the physicist's convention that the inner product is antilinear (= conjugate-linear) in the first variable rather than in the second, rather than the mathematician's convention, which is the reverse. The physicist's convention fits in a little better with $2$-Hilbert spaces and is often used in a generalization for Hilbert modules. 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 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 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 real in that it equals its own conjugate, by (4).

## Definiteness

Notice that, by (1), $\langle{0, y} = 0$ for all $y$. In fact, the subset $\{ x \;|\; \forall y,\; \langle{x, y} = 0 \}$ 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:

• The inner product is semidefinite if $\{ x \;|\; {\|x\|^2} = 0 \}$ is closed under addition (and hence is a subspace); it's indefinite if there are $x$ and $y$ with ${\|x\|^2} = 0$ and ${\|y\|^2} = 0$ but ${\|x + y\|^2} \ne 0$.
• The inner product is nondegenerate if $x = 0$ whenever $\langle{x, y} = 0$ for all $y$; it's degenerate if there is $x$ with $x \ne 0$ but $\langle{x, y} = 0$ for all $y$.
• The inner product is 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 nondefinite.

(In constructive mathematics, we usually want an inequality relation relative to which the vector-space operations and the inner product are 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.)

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 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:

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

In this case, we have these theorems:

• A positive or negative inner product really is semidefinite (as the terminology implies).
• Conversely, a semidefinite inner product is either positive or negative; if $V$ is nontrivial (meaning that there exists $x$ such that $x \ne 0$), then ‘or’ strengthens to ‘xor’. (In constructive mathematics, this should only be stated for nontrivial spaces in the first place.)
• Hence, a definite inner product is either positive or negative definite. (The same remarks about ‘xor’ and constructivity apply.)
• Conversely, an inner product is indefinite if and only if some norms are positive and some are negative. (No constructive caveats here!)

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 English Wikipedia article on Krein space?s for some of it.

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

## Examples

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

• Finite-dimensional modules over $\mathbb{H}$, the quaternions.

Revised on January 25, 2014 14:12:32 by Urs Schreiber (89.204.137.78)