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Stream: deprecated: mathematics

Topic: hermitian matrix over normed division algebra

Avi Levy (Sep 26 2020 at 21:53):

Sorry for the somewhat off-topic question, but I figured some people here might be familiar with Baez' paper "The Octonions". I'm having trouble when I try to be pedantic with definitions in the first paragraph of this page https://math.ucr.edu/home/baez/octonions/node9.html where the space of 2x2 hermitian matrices over a normed division algebra is considered. My interpretation is that a priori the definition doesn't strictly make sense, until one uses a classification theorem to recognize the normed division algebra as one of R, C, H, O and then equip it with its canonical star-algebra structure. Or maybe this is not necessary - is there a direct way of equipping an abstract normed division algebra with a conjugation operation?

David Michael Roberts (Sep 26 2020 at 23:51):

Hmm, interesting point. Can a normed division algebra be shown to have a $*$-algebra structure before using the classification? I suppose it must, else "hermitian" is not that good an adjective. I guess for $x\neq 0$ in a normed division algebra, we can define $x^* = ||x||^2/ x$ (and $0^*=0$), and then check this works.

Avi Levy (Sep 27 2020 at 00:10):

David Michael Roberts said:

Hmm, interesting point. Can a normed division algebra be shown to have a $*$-algebra structure before using the classification? I suppose it must, else "hermitian" is not that good an adjective. I guess for $x\neq 0$ in a normed division algebra, we can define $x^* = ||x||^2/ x$ (and $0^*=0$), and then check this works.

Before the classification, how do you show that the normed division algebra has inverses? There's some discussion here https://math.ucr.edu/home/baez/octonions/node2.html that shows things are a little more subtle than they may seem at first (since we are not assuming associativity of the algebra).

[EDIT] This follows from the construction of the imaginary part of any normed division algebra, outlined here https://math.ucr.edu/home/baez/octonions/node6.html#clifford with a key technical detail given here https://math.ucr.edu/home/baez/octonions/proof.html

John Baez (Sep 28 2020 at 21:06):

It sounds like you figured out the answer to your question, but I'll summarize. With some work (that "key technical detail" you mentioned) one can prove that for any finite-dimensional normed division algebra $\mathbb{K}$ over $\mathbb{R}$ the norm comes from an inner product. Then one can define the space of imaginary elements to be the orthogonal complement of the real multiples of the identity $\mathbb{R} \subseteq \mathbb{K}$ and define $\ast \colon \mathbb{K} \to \mathbb{K}$ to be the linear map that's the identity on $\mathbb{R} \subseteq \mathbb{K}$ and minus the identity on the imaginary elements. Then one can prove $\ast$ is an anti-automorphism.

John Baez (Sep 28 2020 at 21:09):

It might be just as efficient to classify the finite-dimensional normed division algebras and then work case-by-case, but it's nice to know that conjugation is something that makes sense "in general", i.e. in all 4 cases, without reference to the individual case.

John Baez (Sep 28 2020 at 21:14):

Oh, and by the way: you can show that in any finite-dimensional normed division algebra over $\mathbb{R}$ you have

$x^{-1} = x^*/\|x\|^2$

for $x \ne 0$.

John Baez (Sep 28 2020 at 21:15):

What I mean by this is

$x x^* = x^* x = 1$

John Baez (Sep 28 2020 at 21:19):

There are a lot of "obvious facts" like this to prove, and they all tend to involve calculations that rely on other "obvious facts", so you need to work a few days to organize all the arguments into a chain with no circular arguments.