Why conjugate when switching order of inner product?

Question

There is an axiom of inner product spaces that states:

• $\overline{\langle x,y\rangle } = \langle y,x\rangle$

Basically (without any conceptual understanding) it seems like all you have to do when you swap the order of the arguments in an inner product space is take their conjugate.

How does this make any sense? I know if we are dealing with an inner product space over $\mathbb{R}$ then the conjugate of a real number is just the real number itself so there is no change. But how does this make sense over the field $\mathbb{C}$?

Answer 1

The conjugate is necessary because you want to define a norm $\|\cdot\|: V \to \Bbb R_{\geq 0}$ by using that inner product, putting $$\|x\| = \sqrt{\langle x,x\rangle},$$ and for this you need $\langle x,x\rangle$ to be real. The conjugation gives $\langle x,x\rangle = \overline{\langle x,x\rangle} \in \Bbb R$.

Answer 2

$\overline{\langle x,y\rangle } = \langle y,x\rangle$ is used to ensure the norm is real. The proof that this condition is enough to prove $\langle x,x\rangle$ is real is part of Hurwitz's proof that there are only four composition algebras, and a proof of this (in a book!) is here.

Without the condition you give, the complex norm is not necessarily real. Further examples and details on non-real norms can be found in this Wikipedia page on Minkowski spaces, which considers the norms of say $x_1^2-x_2^2-x_3^2-x_4^2$.

Answer 3

Conjugation is there to make sure the signs work out. If you don't conjugate, then you'll find that $\langle ix, ix \rangle = -\langle x, x\rangle$.

Answer 4

At least in the case of finite-dimensional vectors, if you agree that $\langle a, b\rangle = a^\dagger b \equiv \overline{a}^\top b$, then $$\langle a, b\rangle = a^\dagger b = \overline{a^\top} b = \overline{a \overline{b^\top}} = \overline{\overline{b}^\top a} = \overline{b^\dagger a} = \overline{\langle b, a\rangle}$$

If you don't agree with the conjugation in the premise, note that we need a conjugation to ensure $$\langle a, a\rangle = \lVert a\rVert^2 = \overline{a}^\top a$$ is real.