and also e1 commutes with e2 not anticommutes

some other trickery has to be done

do we know more primes or nontrivial zeros?

oh I think many of both are known

if by we you mean just you and me, i know more primes.

so if Riemann in his explicit formula used a summation over nontrivial zeros to create a sinusoidal wave to approximate the prime counting function

Ok, $Cliff_n(\Bbb C) \cong M_{2^n}(\Bbb C)$, where $\Bbb C^n$ is equipped with the inner product NOT HERMITIAN $\langle z, w\rangle = z_1 w_1 + \cdots + z_n w_n$

The rep is probably just embedding $Cliff_n(\Bbb R)$ in $Cliff_n(\Bbb C)$

the nontrivial zeroes are like the primes in the sense that they are not patterned

Er wait how can I do that

I need $Cliff_{2n}(\Bbb R)$

Oh sorry $Cliff_{2n}(\Bbb C) \cong M_{2^n}(\Bbb C)$, got my numerology wrong

So its all fine again

For example Riemann showed this: $\pi(n) = \mathrm{Li}(n) - \frac{1}{2} \mathrm{Li}\left ( \sqrt{n} \right ) - \sum_{\rho} \mathrm{Li}(x^\rho) + \text{lower order terms}$

if you could construct a function like $\pi(n)=\sum_p $ blah blah blah

where the sum is over the primes

would that be inconsistent or okay

inconsistent because you're using a summation over primes to approximate the primes themselves

you could do $\sum_p 1_{p\leq n}$

so summing up an indicator function which is 1 when $p\leq n$ and zero otherwise

@Astyx Given $\Bbb C^{2n}$ with the "form" $\langle z, w \rangle = z_1 w_1 + \cdots + z_{2n} w_{2n}$ choose an orthobasis $e_1, \cdots, e_n, e_{n+1}, \cdots, e_{2n}$ and consider $v_k = (e_k + i e_{n+k})/\sqrt{2}$, $1 \leq k \leq n$. Then $\langle v_k, v_l \rangle = 0$, the form just restricts to $0$ on the subspace $W$ spanned by $v_1, \cdots, v_k$ aka its totally isotropic.

(or $p<n$, i don't remember the specific definition of $\pi(n)$)

The form is actually nondegenerate. $\Bbb C^n \cong W^*$, by $v \mapsto \langle v, -\rangle|_W$ has kernel $W$ because totally isotropic subs have at most half the dimension, and this is also surjective, so $\Bbb C^n/W \cong W^*$, $\Bbb C^n \cong W \oplus W^*$

Let $\Bbb C^n$ act on $\Lambda^* W$ by letting $a \in W$ act by $a \cdot (w_1 \wedge \cdots \wedge w_k) = a \wedge w_1 \wedge \cdots \wedge w_k$ and $\phi \in W^*$ act by $\phi \cdot (w_1 \wedge \cdots \wedge w_k) = \sum_{i = 1}^k (-1)^i \phi(w_i) w_1\wedge \cdots \wedge w_{i-1} \wedge w_{i+1} \wedge \cdots w_k$.

seems to make sense

@Semiclassical okay I'm just trying to think. $\sum_p 1_{p\leq n}$ would be pretty much the same as $pi(n)$

$\pi(n)$

yeah

just b/c a representation is valid, doesn't mean it's useful :P

This action extends to an action of the whole tensor algebra of $\Bbb C^n$ on $\Lambda^* W$. Eg $a \otimes b$ acts by $b$ first then again by $a$. Then $(a \otimes a) \cdot (blah) = a \wedge a \wedge (blah) = 0$ and $(\phi \otimes \phi) \cdot (blah) = 0$ because of the usual boundary operation calculations and $(a \otimes \phi + \phi \otimes a) \cdot (blah) = \phi(a) (blah)$ you can check

I meant $\Bbb C^{2n}$ above

Since $a^2 = \phi^2 = 0$ and $a \otimes \phi + \phi \otimes a = \phi(a)$ in the Clifford algebra wrt this identification $\Bbb C^n \cong W \oplus W^*$ this action descends to one of $\mathrm{Cliff}_{2n}(\Bbb C)$ on $\Lambda^* W$

clifford the big red algebra

This is the $2^n$-dim irrep of the complex $2n$ dim Clifford algebra

But inside it sits the real $2n$ dim Clifford algebra so done

this whole thing makes me tensor and tensor each time i read it

@Semiclassical maybe a summation over the primes without using a indicator function could be useful?

wait, what's the tensor product you're talking about?

i mean, if you can come up with one, great

Cliff(V) = (V + (V o V) + (V o V o V) + ...)/clifford relations

Oh of course

V + (V o V) + (V o V o V) is the full tensor algebra so thats where I extend the action first

Then check the Clifford relations

sorry I'm not fully paying attention

Its all good

Are you off kneading, @Astyx?

I let the dough rest for half and hour, cut it in three portions, and let those to rest for the night

I'm on it!!

I didn't check the details, but it seems good from afar @BalarkaSen

I believe this rep $\mathrm{Cliff}_{2n}(\Bbb C) \to \mathrm{End}_{\Bbb C}(\Lambda^* W)$ is actually an iso (hence irreducible, since $V$ is an irrep of $End(V)$ always)

@BalarkaSen I'm guessing taking the "real part" doesn't change irreducibility? I haven't looked at such things in a while

It's not 100% clear to me.

the irreducibility i mean

If the real part splits, doesn't tensoring with $\Bbb C$ give you a split sub-representation of the complex representation?

That seems to work.

indeed, I should have thought of that

I don't know much.

It's not obvious that you couldn't have more to the complex representation than just tensoring the real one with $\Bbb C$; that's why I said it that way.

I don't know how to argue that the rep is an iso although I think this was a homework problem and I gave some nonsense answer

Never thought itd come back to bite me

That's when things always come back to bite you.

i just thought it was algebra

but now im reading seiberg witten theory lol

Multilinear algebra is everywhere in geometry.

I'm struggling to plot in Desmos - I can only see my plot up to x=10, it's. not powerful enough

i am ok with multilinear algebra. rep theory is when things get really dry for me

i dont understand representations

i know just enough rep theory to gauge my ignorance

is that a pun

a little bit

like, i know character theory is a thing

and character tables

rep theory is where particles come from IIRC

it's a bit part of modern particle theory, yeah

but no one talks about that. after 2 rep theory courses i dont know what representations of SO(3) are

i know SU(2) irrep theory

sorta

at least the finite-dimensional ones.

I forgot any irrep theory I knew

other than the staircase like representation of that gives meaning to spin (but it doesn't become less vague than that)

same

the algebra courses just do algebra

theres no point in learning rep theory from rep theorists

its just symbol push

In 1980 I taught enough representation theory of $SL(2,\Bbb C)$ to prove the Kähler identities. It wasn't bad at all.

$[S^j,S^k]=i\epsilon_{jkl} S_{l}$

and all that

I got a good friend who's big on rep theory

fuck rep theory, all my homies hate rep theory

tho iirc the physicist conventions for SU(2) are just different enough to be annoying

finite dimensional representations of sl(2, C) are, like, quantum harmonic oscillators

that doesn't sound right: quantum harmonic oscillator has infinitely many states

you can probably truncate the state space and ask what you get then, tho

those are the infinite dimensional reps of sl(2, C). but there's two ladder operators and a Hamiltonian

yeah, true

i dont know precisely. i asked the algebra guy and he mumbled that he forgot his physics

so

all they know is symbols

nothing stopping you from saying: when you get to the Nth state, have the raising operator kill it instead of creating the (N+1)th state

yeah thats what happens

makes sense

the word "Verma modules" might mean something to you

i've heard it before but never understood it

Peter Woit has lecture notes on it: math.columbia.edu/~woit/LieGroups-2012/vermamodules.pdf

i get repulsed by the way they write rep theory

its a string of symbols that you have to decipher every time

awful

Is the $e^x$ thing allowed or does it have to be just $x$ see here $ f(x)=\prod_{k=2}^{e^x} k$

I'm gonna say it's allowed

2) mathematicians label representations by positive integers. physicists divide that by two so even/odd -> integer/half-integer

bosons/fermions?

More likely bozos.

Okay I think I got something on Desmos that mimics $\pi(n)$

how the Latin alphabet evolved from Ancient Egyptian hieroglyphs

much like how hiragana and katakana came from Chinese characters

modern hebrew really went its own route with T