# Is complex conjugation needed for valid inner product?

What are the benefits of using a conjugate linear inner product in a complex vector space vs a simple linear inner product? That is, why do we demand that $(y,x) = \overline{(x,y)}$ as opposed to $(y,x)=(x,y)$? Of course, this ensures that $(x,x)$ is real and thus makes an easy definition of norm, but is that necessary?

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We define "complex inner products" that way because the inner product we use for everything behaves that way. It is not that we use the one we use because it satisfies some a priori definition we pulled out of thin air, but the converse! – Mariano Suárez-Alvarez Sep 21 '10 at 22:08

It is in fact necessary. The inner product axioms without the conjugation are inconsistent:

(Here $u$, $v$, $w$ are vectors and $c$ is a scalar)

1. $\langle cu, v\rangle = c\langle u, v\rangle$
2. $\langle u,v\rangle = \langle v,u\rangle$
3. If $u \neq 0$, then $\langle u,u\rangle$ is a positive real number
4. $\langle u+v,w\rangle = \langle u,w\rangle + \langle v,w\rangle$

In fact, 1-3 alone are inconsistent. Indeed, let $u$ be any nonzero vector, so $\langle u,u\rangle > 0$ by condition 3. But if $i = \sqrt{-1}$, then $\langle iu, iu\rangle = i\langle u, iu\rangle$ (by 1) $= i\langle iu, u\rangle$ (by 2) $= i^2\langle u, u\rangle$ (by 1) $= -\langle u, u\rangle < 0$, contradicting condition 3.

The upshot is that you can choose: either conjugate one side of condition 2, giving you the axioms for an inner product, or get rid of condition 3, giving you the axioms for a symmetric bilinear form. You could also consider a weaker version of 3, like requiring that if $u\neq 0$, then $\langle u, v\rangle \neq 0$ for some $v$. That gives you nondegenerate symmetric bilinear forms.

Note that there's nothing wrong with bilinear forms on complex vector spaces; they're just not inner products. They're disjoint concepts, unlike in real vector spaces, where inner products are just special symmetric bilinear forms. In some ways, bilinear forms are nicer than inner products, since you don't have to worry about complex conjugation. However, bilinear forms over the complex numbers do not give rise to norms, which means they don't endow vector spaces with good geometry. Inner products do, hence their ubiquity.

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How is condition 3 contradicted? Because $-\langle u,u \rangle < 0$ means $\langle u,u \rangle > 0$, so $\langle u,u \rangle$ is a positive real number. – Cookie Jun 16 at 17:21
@lesguimauves - $\langle u, u\rangle$ is positive, but $\langle iu, iu\rangle$ is not. – Darsh Ranjan Jun 26 at 17:39
In other words, $\langle iu, iu\rangle < 0$ when $\langle u, u\rangle > 0$, which contradicts condition 3, that all $\langle x, x\rangle > 0$. – Mateen Ulhaq Nov 7 at 23:17

You need to define the inner product this way (in a complex vector space) to have a good definition of the norm of a vector. Let $V$ be this vector space, and take $X\in V$. Then, the norm of X is defined as $|X|=\sqrt{(X,X)}$. This definition is good ($|X|$ is well defined for all $X$) if $(y,x)=(x,y)^*$. If this is not the case, for example if $(y,x)=(x,y)$, then $(x,x)$ need not be positive. More: $(x,x)$ need not be real.

Aother thing, related to the above one: in metric spaces, given an inner product we define a norm (the same way we did above) and, from the norm, a distance function. This distance function determines the topology of the space (the topology induced by the inner product). To do this, we need to be able to define in a good way the norm of a vector, and the problem appears again.

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