2
$\begingroup$

I'm a student studying linear algebra from S.Korea. While I was learning about inner product spaces, I came up with a question.

In inner product spaces, we have a property called conjugate symmetry.

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

If the vector space or the field is $\mathbb C$ or $\mathbb R$, we can simply think of the bar operator as changing $a+bi$ to $a-bi$. But if the vector space or the field is some unfamiliar set, what does conjugation actually mean in those cases?

In the first place, what is conjugation? Are there any other examples of conjugation than complex conjugation and $p+q \sqrt r \mapsto p-q \sqrt r$?

$\endgroup$
2
  • 4
    $\begingroup$ Inner product spaces are always spaces over $\mathbb{R}$ or $\mathbb{C}$. The product $\langle x,y\rangle$ is a scalar, i.e a complex number. (and not a vector) So conjugation is defined here, the usual complex conjugation. $\endgroup$
    – Mark
    Commented May 22, 2023 at 22:55
  • 1
    $\begingroup$ In contexts where people consider forms with values in something other than a field, "conjugation" might be replaced with a more general operation (e.g. if the set of values is a vector field over $\mathbb{C}$, a conjugate-linear mapping - one for which $L(kx) = \overline{k} L(x)$ holds for all complex scalars $k$ and all $x$ in the set). Consider e.g. $\langle a, b \rangle := b^* a$, as an operator-valued mapping on the set $\mathcal{B}(H)$ of linear operators on a Hilbert space: $\langle a, b \rangle = \langle b, a \rangle^*$ where $*$ is the adjoint operation on $\mathcal{B}(H)$. $\endgroup$
    – leslie townes
    Commented May 22, 2023 at 23:06

1 Answer 1

Reset to default
1
$\begingroup$

The word "conjugate" has multiple related and unrelated meanings (not as bad as the many uses of "regular" and "normal").

  1. On $\mathbf C$, if $z = a + bi$ then is complex-conjugate is $\overline{z} = a - bi$.

  2. In a field $E$ containing a field $F$, two elements of $E$ that are roots of the same irreducible polynomial in $F[x]$ are called $F$-conjugates. For example, $\mathbf C$ is an extension field of $\mathbf R$ and when $z = a+bi \in \mathbf C - \mathbf R$, $z$ and $\overline{z}$ are roots of the same irreducible quadratic polynomial $x^2 - 2ax + (a^2+b^2)$, so $\mathbf R$-conjugate complex numbers in the sense of field theory and complex conjugates in the first sense above.

  3. In group theory, when two elements $g$ and $g'$ in a group $G$ are related by $g' = xgx^{-1}$ for some $x$ in $G$, we call $g$ and $g'$ conjugates. (Similarly, two $n \times n$ matrices $A$ and $A'$ are called conjugate when $A' = XAX^{-1}$ for some invertible matrix $X$.) Two subgroups $H$ and $H'$ of $G$ are called conjugate when $H' = xHx^{-1}$ for some $x$ in the group.

  4. In Riemannian geometry there is a notion of conjugate points. Look here.

For a longer list, look here.

Meanings 2 and 3 occur together in the setting of Galois theory: in a finite Galois extension $E$ of $F$, two elements of $E$ are $F$-conjugates iff they are in the same orbit of ${\rm Gal}(E/F)$ and two subgroups $H$ and $H'$ of ${\rm Gal}(E/F)$ are conjugate iff their fixed fields in $E$ are $F$-isomorphic.

The concept of a complex inner product only makes sense for vector spaces over a field that has an automorphism of order $2$, and many fields don't have such an automorphism.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .