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11:11 AM
ugh
If I have a generic coordinate change like $\tilde x^\mu=x^\mu+\epsilon^\mu$ and I need to calculate $\partial x^\alpha/\partial \tilde x^\mu$ am I forced to break the index notation?
 
what do you mean by "break"?
 
i.e. it feels like I have to write $x^\alpha=\tilde x^\mu-\epsilon^\alpha$
 
your feeling is wrong :P
what's so bad about $\frac{\partial}{\partial\bar{x}^\mu} (\bar{x}^\alpha - \epsilon^\alpha)$?
there's no reason to "mix" the indices in the bracket, you can just give both an $\alpha$
 
oh yeh
I see, that was pretty duh, thanks :P
That actually raises a related question, I feel like we would have to be very careful about which indices are in which coordinate system, eg. the above gives $$\delta^{\tilde\alpha}_{\tilde\mu}-\partial_{\tilde \mu}\epsilon^\alpha$$
actually nvm I just need to think for a bit
Ok no I'm definitely not seeing it, if you take a Jacobian thingy like $\partial \tilde x^\alpha/\partial \tilde x^\mu$ you're going to get a Kronecker delta $\delta^{\tilde \alpha}_{\tilde \mu}$ in the tilde coordinates but it feels like we're never careful about putting tildes (or primes) on indices in Kronecker deltas for eg.
even though it seems like that should matter, even if the Kronecker delta is basically just the identity matrix in any coordinate system
 
11:44 AM
The indices on coordinates are a bit of an abuse of notation
you have to be careful about this when the $x^\mu$ occur directly in your equations, but in general, the indices in equations that only involve vectors, covectors, etc. are either all w.r.t. a single coordinate system or they're "abstract"
 
Like I can see how the things I'm looking at are derived if I look the other way a bit on primed vs unprimed indices, but when I write it out explicitly it seems like I run into strange looking expressions
alright, I won't waste too much time worrying atm then
 
 
2 hours later…
1:33 PM
anyone has some resource on madgraph5?
 
2:08 PM
Nicolaescu's notes on Seiberg-Witten theory are once again top-notch
(by the way - he's actually running for MathOverflow mod right now)
 
3:06 PM
@ACuriousMind As a sheaf, the points of a spacetime are an equivalence of the maximal ideal of the function algebra, is that correct?
 
@Slereah what's the sheaf supposed to be a sheaf on there?
 
It's too early for philosophical questions
 
usually you have the space(time) given and then you can consider various sheafs on it
 
Well
 
but the points of the spacetime aren't "sheafy", it's just a topological space
 
3:15 PM
Is there a bijection between the maximal ideals of the function algebra and the space points
Also unrelatedly, what's going on with the parity reversion functor
How does it work, the two vector spaces aren't even necessarily of the same dimension
 
@Slereah What function algebra? That of continuous functions?
 
$C^\infty(M)$, yes
 
that's the ring of smooth functions, but okay :P
in that case, yes, the maximal ideals of $C^\infty(M)$ are the points of the spacetime if $M$ is compact
 
the ring module whatever
Hm, why the compactness?
what's a counterexample
Divergent functions at infinity maybe?
 
@Slereah for non-compact manifolds you can have maximal ideals that aren't points
 
3:21 PM
Like I would guess positive functions on $\mathbb{R}$ such that $\lim_{x \to \infty} = \infty$ are a maximal ideal
 
take the ideal of all compactly supported functions - it is contained in some maximal ideal, but that ideal can't be the ideal corresponding to a point
 
22
Q: Maximal ideals in the ring of continuous real-valued functions on R

Alon AmitFor a compact space $K$, the maximal ideals in the ring $C(K)$ of continuous real-valued functions on $K$ are easily identified with the points of $K$ (a point defines the maximal ideal of functions vanishing at that point). Now take $K=\mathbb{R}$. Is there a useful characterization of the set...

"For any manifold M, the maximal ideals of C(M) whose residue field is ℝ is exactly in bijection with the points of M."
God I hate algebra
 
1
Q: Sheaves of analytic functions were once called multi-valued functions.

R. EmeryI understand the Taylor series and I understand what a germ is. "Any vector $g = (z_0, α_0, α_1, ...)$ is a germ if it represents a power series of an analytic function around $z_0$" And a sheaf is a set of germs. But then I read "Sheaves of analytic functions were once called multi-valued f...

 
@Slereah do you like the topological characterization of the maximal ideals being the points of the Stone-Čech compactification better? :P
 
"Let me now explain why you have heard of sheaves as "special functions". Indeed, suppose you want to define a "multifunction","
3
A: What is an intuitive Geometrical explanation of a "sheaf?"

Andrea MarinoA sheaf $F$ on a topological space is something that associates to every open set an object $F(U)$, e.g. an abelian group. Elements of $F(U)$ are usually called sections on U. Furthermore,whenever you have $U \subset V$ there is an operator of restriction $| U: F(V) \to F(U)$. As the name sugges...

 
3:31 PM
I hate everything
 
@bolbteppa why are you linking sheaf stuff, Slereah said "sheaf" but this really just about properties of the ring of continuous/smooth functions
 
I guess the parity reversal functor only acts on the spaces, not the vectors themselves?
 
it doesn't actually have anything to do with sheaves
@Slereah why is parity reversal a functor?
 
So instead of $\mathbb{R} \oplus \mathbb{Whatever}$ it's $\mathbb{Whatever} \oplus \mathbb{R}$
 
shouldn't the parity operator just be the action of the parity inversion in $O(p,q)$ under the representation?
 
3:34 PM
Apparently the functor that switches the degree of vector spaces in $\mathbb{Z}_2$ graded spaces
 
I always thought parity was just so annoying because physicists usually just look at representing the $\mathfrak{so}(p,q)$ algebra, which doesn't tell you what parity does, so every time it becomes relevant you have to go back and figure out which representation of parity you actually have to choose for all your stuff
I don't see why it would be a functor, or on what category
 
"parity" can mean many different things
it just means "two"
That parity is more for SUSY I think
if we go for the physics metaphore
Turning fermions to bosons
 
have you been reading nlab again :P
3
 
currently this
people talking superspaces tend to be a bit stingy about actual examples
Or if they do it's usually gonna be like $V \oplus \{*\}$
 
in that case, yes, the functor just acts on the spaces - it turns $\mathbb{R}^{p|q}$ into $\mathbb{R}^{q|p}$
 
3:43 PM
ah yes
It seemed odd because there's no reason that there would be a unique map otherwise
especially if the graded spaces don't have the same dimension
 
It makes some sense to say the biggest set of functions which are products with $f$ such that $f(c) = 0$ is 'associated' with the point $c$, why would you bring in sheaves and the whole multivalued thing
 
I don't know about all those shenanigans
I only know that if you drop a ball it falls down
a more interesting example
 
In the intro to those notes it complains about path integrals because one has to mimic the finite dimensional case and says the BV stuff offers an alternative, then says the idea of the rest of the notes are about setting up the finite dimensional integral BV case :p
 
nobody's perfect
I'm sure nlab has the perfect solution but I am too scared to look
no proof of that example though
hm
what's a not too scary place to look into
I have DeWitt on supermanifolds but it's a scary book
 
@Slereah what's there to prove - the first equality is just by definition, and the correspondence is just sending products $\theta^i \theta^j$ to $e^i\wedge e^j$ in some basis $e^i$
 
3:53 PM
Oh that's a definition?
I mean I guess there is an actual proof in DeWitt if you start with Grassmann algebras, but
that is part of the scary place
 
There seems to be three good reasons to look at bv, a Lagrangian perspective on the brst stuff that applies even to classical mechanics theoretically, extending brst beyond the cases of constant 'structure-constants', and also when the algebra includes extra eom-dependent terms
 
@Slereah yeah, I mean, you have to define what you mean by "smooth functions" on a purely odd algebra, and that's what one usually takes as definition
 
hi everyone
 
DeWitt defines a whole topology and such on supernumbers, but
I am willing to accept this on faith
 
I am sorry if I come across as desperate, but I am really approaching a deadline
 
3:56 PM
if it means I don't have to look back into it
 
I need help with a thermodynamics question,may I ask?
 
there's a specific channel for homework questions, I think?
 
in chat?
 

 Problem Solving Strategies

General chat for high school physics. For MathJax see [here](m...
Possibly this one
 
that room isnt of use on sunday evenings
thats why I am trying my luck here :p
 
3:59 PM
The people who invented this stuff would probably never have invented it if they started worrying about the abstract topology of the maximal ideals of the flabby perverse sheaves of smooth functions on an odd algebra :p
 
Yeah I remember the physics version of the proof
"Well, if we assume that they have a Taylor series, then only those bits are non-zero"
But what if things are not analytic???
Can that happen???
 
Assuming that the relationship between the pressure and specific volume of air in the
atmosphere is given by $pv^{n}$ = constant. If the pressure and specific volume on the
surface of earth are 1 bar and 0.8 m3
/kg respectively, determine the height of the
atmosphere in meters. The value of ‘n’ may be taken as 1.4 and acceleration due to
gravity may be assumed to be constant as 9.8 m/s2
.
 
In supergravity you're going to end up with black holes in superspace
 
I have an approach
 
Once in a while I try to read DeWitt to lay my fears to rest, but then when I look inside it looks like this :
I'd rather live in fear
 
4:03 PM
$$dp/dh= -\rho g$$
$$pv^{1/n}= c=(0.8)^{1/n} $$
$$v=1/\rho$$
so,we get
 
Maybe that huge Witten book has an easier rigorous superspace approach than that one
 
ohh wait i typed 1/n instead of n
but anyways, this was the approach, but i am getting a really low value for height...
 
I looked around and all superspace stuff is either like DeWitt's approach or nlab
Gotta pick my poison
 
There's a good few susy books that do superspace normally
 
well there's also the physicist approach
"Here's some $Q$ operators, don't worry about where they came from"
 
4:09 PM
Wait so is Minkowski space not actually a representation of the SUSY algebra, you need a superspace?
 
"Also they anticommute"
Minkowski space is a superspace, I guess?
Of dimension $(n, 0)$
Trivially
 
@Charlie it is a representation, but one where the supercharges act trivially (there's no fermions they could send the bosonic coordinates to)
 
super Minkowski space is a thing apparently
oh
do you need some kind of ungodly graded manifold, however that would work, to get fermions
you know what I don't even want to know it'll only make me sad
 
I mean you can get fermions the usual way, with fiber bundles and all
but yeah you can have graded manifolds also
which are locally homeomorphic to $\mathbb{R}^n \times \mathbb{R}_a^m$
 
what's the a subscript there?
 
4:12 PM
anticommuting
 
what structure actually anticommutes there?
 
Well, if you have, say, $a, b \in \mathbb{R}_a$
Then their product is $ab = -ba$
 
oh like grassman numbers
 
It differs as an algebra
they are Grassmann numbers yeah
Except the actual explicit construction is super weird
 
of superspace?
 
4:14 PM
of $\mathbb{R}_a$
 
I feel like we're going to run out of fancy sounding names for new things in physics one day
 
if you take anticommuting numbers as a given, superspaces aren't too hard
Only physicists call it superX, tbh
Mathematicians just call them graded whatever
 
I still don't really know what the principle of graded Lie algebras is, not that I've really looked far into it, just seems like a Lie algebra with two Lie brackets
 
yeah once you're familiar with USp and OSp then superspace is just a coset space
 
the basics of graded algebras is that your vector space is split in two
so a vector is like $v = (a,b)$
The first part commutes, the second part anticommutes, as an algebra
simple enough
 
4:19 PM
Superspace is basically just the space on which a function like $f(x,\theta)$ makes sense, you can then define in this space a 'translation' like $\delta x^a = - i \overline{\varepsilon} \gamma^a \theta$ and so build up susy transformations. We can play games like generalizing Minkowski = Poincare/Lorentz to define it using cosets, it just means adding these $\theta$'s to Poincare (super-Poincare) and quotienting by Lorentz
 
although actually the construction of superPoincaré is really annoying
I'm pretty sure you have to do a Inonu-Wigner contraction somewhere
 
unfortunately the actual construction of anticommuting numbers is either short but abstract, or straightforward and super long
I don't think it's technically too important for actual physics but it does bother me
 
@NiharKarve i.e. you can't just use $OSp(4|N)$, you need to Inonu-Wigner contract it to embed all the generators exactly
 
That comes up through a DeSitter isomorphism in the orthosymplectic group if you want to define a supergroup
 
Ironically since the space of smooth functions is isomorphic to the exterior algebra, it's much more straightforward
 
4:25 PM
@NiharKarve but yeah once that's done you can just quotient by $SO(1, 3)$ to get superspace
or at least you definitely can for $\mathcal N=1$ (who uses the other ones anyway)
 
I think it's weird that most SUSY people deal with the ring of smooth functions directly and I wonder if at first they tried the DeWitt way and said "Let's never to that again"
"However one should anticipate that some of the geometric intuition failsand we cannot think in terms of points due to the presence of the odd coordinates"
dang it
"The supermanifold is defined by gluing superdomains. However, the gluing should be done with some care and for the rigorous treatment we need to usethe sheaf theory."
Nooooooooooooooo
 
4:50 PM
Have I understood correctly that the supersymmetric "superpartners" of the particles in the standard model are added by hand in supersymmetric theories to preserve CPT symmetry?
 
Not a clue!
as far as I know you can just do SUSY theories from scratch
 
@Charlie no I don't think so
 
hmm ok
 
I mean, once you posit supersymmetry, something has to fill the other space in the multiplet (for N=1)
 
you do need another field to transform into I suppose
unless it's a trivial SUSY space
 
4:56 PM
although to preserve CPT invariance, you can't just have a (0, 1/2) multiplet for example
you add in the CPT conjugate to make (-1/2,0) $\oplus$ (0,1/2) in total
because CPT flips the helicity
 
Yeah I think this is roughly what I was looking at
And these "multiplets", these are just labelling certain representations of the SUSY algebra like the $j$'s labelling representations of $\mathfrak{su}(2)$ and stuff right?
 
oki, makes sense
I'm not actually completely happy with the idea of a vector space which contains both fermions (spinors) and a scalar (or whatever), how do you actually construct the representations spaces of the SUSY algebra from these Lorentz reps?
 
the susy representation is the direct sum of the bosonic and fermionic reps of the Lorentz algeblra
 
For instance if I'm able to act with the $Q$ generators in some representation and turn fermions into bosons, my vector space has to contain something like a spinor and something like a scalar and have some kind of well-defined map between them
ok it's just the direct sum, nice
 
5:06 PM
pretty much
 
actually one small further point, are the Majorana spinors (in dimensions that admit them) a subspace of the entire $\Bbb C^n$ space?
I feel like if you're going to have SUSY Majorana theories you shouldn't be able to turn Majorana's into Dirac spinors by the action of the SUSY algebra so they should form an invariant subspace wrt the SUSY algebra no?
Although maybe I have just typed a meaningless load of jargon there, who knows
Wait they are literally an invariant subspace of the Lorentz reps so they should be a vector subspace at least surely
 
@Charlie I'm not sure what you're saying, at least :P
 
Maybe I can give a better example, if you have a theory containing Majoranas (and whatever else), surely you don't want to be able to turn Majoranas into Dirac spinors
I was wondering if the subset of spinors that satisfy the Majorana reality condition thingy form an invariant subspace of the Lorentz algebra or better yet the SUSY algebra
If that doesn't make sense then I surrender my question for now :P
 
they're not an invariant subspace for SUSY because SUSY turns them into some bosons
but Majoranas are a proper representation of the Lorentz algebra
 
oh yeah ofc I didn't think about turning them into other stuff, just losing their Majorana property
 
5:15 PM
if $V_D$ is the Dirac representation, then if Majoranas exist then they're a real invariant subspace $V_M\subset V_D$
same dimension as the Dirac representation, but real, not complex
 
oh
are they isomorphic to some $R^n$ then?
 
I'm actually slightly surprised that they are a real subspace, I thought the catch of Majorana's is that it isn't enough to simply impose $\psi^*=\psi$, you had to have that they are related by some charge conjugation matrix thingy
but I guess those aren't incompatible, just that those spinors for which $\psi^*=\psi$ are not actually a vector subspace
 
of course they are a vector subspace
but the physics literature is terrible in really saying what they're doing :P
note that $\psi^\ast = \psi$ is a basis-dependent formula, you have to be careful what you really mean by it
 
Ok then in that case I don't really know what is gained by defining Majorana's like that is
oh it is?
hmm maybe I just don't have enough experience with complex vector spaces
 
5:25 PM
Take $\mathbb{C}^2$, and pick some orthonormal basis $e_1,e_2$. Then the complex conjugate of the vector $v = \mathrm{i}e_2$ is $-\mathrm{i}e_2$. But if you pick the basis $\mathrm{i}e_1,\mathrm{i}e_2$ instead, then the conjugate is just $v$ itself
 
ah, I see
 
both conjugation and transpose are basis-dependent operations, which is really annoying if you try to use them to define Majoranas, which is why I asked this question a while back
 
"Let us discuss a slightly more complicated example of gradedcotangent bundleover cotangent bundle"
it is quite harrowing but I think I am nearing the point where he explains what the hell the BV formalism is
it involves the Fourier transform for some reason?
 
As far as I can tell your question is asking whether or not it is possible to prove that you can have Majoranas in $d$ dimensions without resorting to choosing a basis, I guess proving this would require defining a basis independent Majorana condition, the answer given is quite involved, is the conclusion that you can find a basis-independent Majorana constraint in allowed dimensions?
I've actually stumbled onto your question there a few times while looking for various bits of information, I never actually understood what you were asking properly before, pretty interesting though
 
@Charlie the "constraint" is easy - you ask for a real invariant subspace to exist, or equivalently for the existence of a real form
 
5:37 PM
oh, sure
 
the question is how to prove these actually exist without choosing a basis
 
presumably you can have several real invariant subspaces of so(1,3) in arbitrary dimensions
 
For a lorentz boost in 2 dimensions, this can be thought of as hyperbolic rotation for a hyperbola 1/x. What if you use 1/x^2 instead of 1/x? 1/x^2 is not a hyperbola
 
unless not, in which case Majoranas would be unique, then again if they aren't unique surely you need to choose a Majorana rep when you build a theory
 
@Charlie Majoranas are specifically real versions of the Dirac spinors
there might be more than one way to choose a real invariant subspace, but they will be isomorphic as representations
 
5:44 PM
um
actually I'm a bit thrown by that qualifier, Dirac spinors being spinor field solutions of the dirac equation
and actually also why any real invariant subspaces are necessarily isomorphic as representations lol
 
when I say "Dirac spinor" I mean the spinors that arise from the irreducible representation of the Clifford algebra
 
oh
 
@Charlie they all have the same $(j_1,j_2)$, which they directly inherit from the Dirac representation
 
I had never really considered that an irreducible complex representation might have real invariant subspaces
 
That's a hefty Stokes theorem
"Before discussing our main topic, let us remind the reader of some facts about theChevalley-Eilenberg complex for the Lie algebras."
Please, I just wish to die
 
6:04 PM
In mathematics, Lie algebra cohomology is a cohomology theory for Lie algebras. It was first introduced in 1929 by Élie Cartan to study the topology of Lie groups and homogeneous spaces by relating cohomological methods of Georges de Rham to properties of the Lie algebra. It was later extended by Claude Chevalley and Samuel Eilenberg (1948) to coefficients in an arbitrary Lie module. == Motivation == If G {\displaystyle G} is a compact simply connected Lie group, then it is determined by its Lie algebra, so it should be possible to calculate its cohomology...
The idea behind it is pretty interesting
 
I think physicists would probably see it in the context of BRST
 
I have skimmed the rest of the paper and he basically never applies the formalism to physics
I think I need another paper
A man cannot live on cohomologies alone
 

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