Mathematics - 2024-04-27

12:13 AM

I still don't even know why I'm learning Clifford algebras. What if this is all pointless

Lukas Heger

@Jakobian Clifford algebras are related to K-theory

you can prove Bott periodicity with them

not sure if K-theory is enough to solve an existential crisis, though...

Thorgott

12:34 AM

sounds like an awkward approach to Bott periodicity

I think Balarka tried telling me something along these lines once upon a time

12:45 AM

@LukasHeger nah I'm trying to help my friend so its going to be something applied

I've boiled down how to derive all Clifford algebras, at least the ones over $\mathbb{R}$

$\mathbb{R}^{p, q}$ is the quadratic space (i.e. vector space with quadratic form) I've been talking about above

and $\mathbb{R}_{p, q}$ is the corresponding Clifford algebra

Then one just has to know $\mathbb{R}_{1, 0}$ and $\mathbb{R}_{0, q}$ for $0\leq q \leq 7$ as well as three formulas

1) $\mathbb{R}_{p, q}\cong \mathbb{R}_{q+1, p-1}$ for $p > 0$

2) $\mathbb{R}_{p+1, q+1}\cong \mathbb{R}_{p, q}\otimes \mathbb{R}(2)$

3) $\mathbb{R}_{p, q+8}\cong \mathbb{R}_{p, q}\otimes \mathbb{R}(2^4)$

if you know those, you can write $\mathbb{R}_{p, q}$ as an algebra of matrices over one of the $9$ algebras I've mentioned above

here $\mathbb{R}_{1, 0} = \mathbb{R}^2$, $\mathbb{R}_{0, q}$ is $\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{H}^2, \mathbb{H}(2), \mathbb{C}(4), \mathbb{R}(8)$ and $\mathbb{R}^2(8)$

oh I guess one actually only has to know $\mathbb{R}_{1, 0}$ and $\mathbb{R}_{0, q}$ for $0\leq q\leq 3$ if one apples formula 1) above

but yeah everything is an algebra of matrices over either $\mathbb{R}, \mathbb{R}^2, \mathbb{C}, \mathbb{H}$ or $\mathbb{H}^2$

@Thorgott Bott periodicity seems to be related to 3) ?

Thorgott

1:02 AM

yeah, that should be it

real K-theory is 8-periodic

1:16 AM

@Thorgott is complex K-theory 2-periodic?

this would be what I'd guess from how complex Clifford algebras only change from matrices over $\mathbb{C}$ to matrices over $\mathbb{C}^2$ and conversely

yes, it is

oh, are you knowledgeable in the topic leslie?

not really, i took a class that covered it 20 years ago

More knowledgeable than me at least

1:43 AM

Hello everyone! Some easy PT task that I can't solve, maybe someone might give me a hint...
Given $\{X_i\}$ -- uniformly distributed independent random variables. Let $X_{n} = max{X_1, X_2, ..., X_n}$. Find the limit by distribution of $n\cdot (1-X_{(n)})$

2:27 AM

Solved it, but thanks for anyone who considered it !

Lucky Chouhan

3:21 AM

@leslietownes so you're in your 40s?

i'm in at least my twenties :)

Bumblebee

Is it possible to evaluate this integral without a substitution? $$\int \frac{\sin(x) \cos(x)}{\sin^{4}(x) + \cos^{4}(x)} \, dx$$

3:36 AM

@leslietownes 20 years old genius progidy that has been listening to (and understanding!) math lectures before he was born

the change of variables is so fundamental that i'm not sure what would count as on the table if that tool is taken off. if there were bounds, at least some definite integrals involving that can be evaluated without computation (it is an odd function)

@Bumblebee What do you mean by integrating without a substitution? I don't see how it's possible, unless the function comes from the table of integration

It's certainly not the easiest one even considering the substitution. It's obvious, but seems quite technical.

Bumblebee

@MagnusAlexander I was just curios to see if it were possible.