# Why conjugate when switching order of inner product?

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 $\mathbb{C}$?

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 $\langle x,x\rangle = \overline{\langle x,x\rangle} \in \Bbb R$.

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

The first thing to understand is how the definition of the inner product is slightly different in a complex context than the one you might have learned for the reals, in that you must take the conjugate of one of the arguments, so you have the vector definition:

$$\langle x, y \rangle = x^*y$$ or $$\langle x, y \rangle = x^Hy$$ depending on your notation for the hermetian (conjugate) transpose.

Using the summation definition:

$$\langle x, y \rangle = \sum_n \overline{x[n]}y[n]$$

The reason we define it this way is that we want to be able to define the length $||u|| = \sqrt{\langle u, u \rangle}$, which should be a nonnegative real number. If we defined the inner product as $\langle x, y \rangle = \sum_n x[n]y[n]$, (or an integral for the continuous case) then any phase shift on elements of $u$ would be doubled. With the correct definition above, the phases will cancel out.

Another way to look at it (example from here) is to assume you have a non-zero vector $u$. To be a valid inner product, $\langle u, u \rangle$ should be a positive real number. So should $\langle iu, iu \rangle$, But with the normal real-valued inner product, $\langle iu, iu \rangle = i\langle u, iu \rangle = i^2\langle u, u \rangle = -\langle u, u \rangle < 0$, so this violates our condition.

so with the definitions above,$\langle y, x \rangle = y^*x = (x^*y)^* = \overline{\langle x, y \rangle}$

side note: this also this affects how you can pull coefficients out of an inner product:

$\langle cx, y \rangle = \overline c \langle x, y \rangle$

$\langle x, cy \rangle = c \langle x, y \rangle$

According to this, there's some disagreement about the argument ordering, but this seems consistent with the definition above.

$\overline{\langle x,y\rangle } = \langle y,x\rangle$ 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 $x_1^2-x_2^2-x_3^2-x_4^2$.

• The proof that that condition is enough is really a triviality, and it has nothing to do with composition algebras! – Mariano Suárez-Álvarez Jun 4 '15 at 3:57

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.