00:00 - 20:0020:00 - 00:00

12:00 AM
Like, lacking vocabulary ?

a "value" of $\rho$ is what you get when you plug any specific $p_0,q_0,t_0$ into some specific function $\rho$, i.e. $\rho(p_0,q_0,t_0)$

yes
that's what I mean
and different values, represent different positions in phase space
for this function /rho
\rho*

$\rho(p_0,q_0,t_0)$ is just the probability (density) to be in the microstate $(q_0,p_0)$ at time $t_0$ for this particular macrostate

one second

a different macrostate would assign a different value to at least one $q_0,p_0$
i.e. it would be a different function $\rho_2$

12:03 AM
but ofc, to find this probability you would need to integrate over some region in phase space
around the point $(q_0,p_0)$

Okay, i worked it out in component form and it's invariant

for example

yes, sure
a different macrostate is just a different function

of '\rho

yes

12:04 AM
see, we are in the same mind ahhaha

I'm just saying that saying it's a "different value" is strange, because a "value" is what you get when you plug points into these different choices for $\rho$

yes
but to me it makes more sense
let me clarify
In phase space, you take 2 different points in t_0, and you integrate \rho in the volume surrounding these 2 points, which are far away from each other. Now \rho, will have different values for these 2 points, which physically would translate to two different macro states, each with a multiplicity,which corrsponds to the nr. of points in each volume.
What did I say wrong, and what did I say right

uhhh
$\rho$ is the macrostate

yes

you don't need to integrate it to get the macrostate

12:07 AM
no
I am integrating to get the probabilities of the microstate

that it has different values around different points is all part of it, not an indication of "different macrostates"
"Now \rho, will have different values for these 2 points, which physically would translate to two different macro states" I don't really get what you're imagining here, but it's either wrong or wrongly expressed

Didn't you say that integrating \rho (probability density function /macrostate) for some volume, gives you the probability of the system being in a microstate?

it gives you the probability of the system being in a microstate in that volume, yes

and where all the microstates in that volume have same proability
right?
1/V

that's an equilibrium assumption again

12:11 AM
and if not equilibrium?

then it depends on what function $\rho$ is!

but you have multiples microstates

also, note that it's not "every microstate in this volume has probability 1/V", it's "the probability density over this volume is 1/v"

how can you find the proability for each of them

the probability for each microstate (which is a point!) is 0
that's how probability densities work :P

12:13 AM
yeah

you only get non-zero probabilities for volumes, not points

Ok wait
I get it
in the normal distribution when you integrate the pdf in some region, you find the probability that the function, which has a normal distribution, has a value in this region
and the same logic is being applied in phase space
where you find the probability that the microstate is located in this volume

yes, exactly
the concepts are identical

and in the case of non equilibrium
the expression of \rho
gives you insight about the proability of the microstates being in the volume
since here the microstates don't have same probability
but that varies

yup

12:17 AM
that's insane
I hate statistical mechanics, but it's to damn awesome
So in the case of the simple harmonic osciallator
where the trajectory is a ellipse, meaning energy conserved
along it
I don't think the concepts of micro and macro state apply here, in this example
or do thhey?
maybe when you have multiple of them

they don't apply in the sense of thermodynamics, no
you can of course be uncertain about what state a SHO is in and consider some phase space densities that model your uncertainty
but it doesn't make sense to talk about equilbrium or canonical ensembles or anything like that
the point of the "macrostates" of thermodynamics is that the particular microscopic dynamics of your system are irrelevant because you have so many particles
you don't care what complicated Hamiltonian would model the collision of two gas particles

and for SHO you have countable amount
and a small one at that

well, you can do thermodynamics on a collection of a lot of SHOs

yes

but then it doesn't really matter anymore they're SHOs

12:23 AM
and?

as I said, you don't care what the individual particles are doing

aha

or, well, you will, later on

this is what you meant

e.g. it's relevant when you want to compute stuff like entropy

12:24 AM
you mean like einstein solid/

but it's not relevant for the concepts of microstate, macrostate, equilbrium, etc. we've been discussing so far

I have 2 last questions,
just for extra knowledge
We have a system, it has a phase space with all the possible microstates, and initially it's in non equilibrium. Then later on it reaches equlibrium. Now how does the phase space "change". What I mean by that, and I need your confirmation is that when the system ain't in eq. the proability density function is non-zero around phase space
then when the sysstem reaches stability, the \rho is non-zero constant in a region and zero elsewhere
is this correct to imagine
assume?

sure

Ok
2nd:
We have a system in eq with multiplicity /omega
the entropy is S=kln \Omega
this is what my professor did
If we assume that the system is in a mixed state, hence density matrix

...why are we suddenly doing QM again?

12:30 AM
because this is what the professor did
jumped from classic to qm

???

without caring to much :)

no wonder you're confused :P

Trust me
quantum statistical mechanics
is 1000 easier then classic
so anyway you have the density matrix
and the above equation
S=kln \Omega which
according to him (magic)

Why can you build invariants that don't seem to make sense in covariant notation?

12:32 AM
S=-k \Sigma_i P_i |i><i|
sorry
\rho = \Sigma_i P_i|i><i| is the density matrix of a mixed state
S=kln\Omega
and then he does
S=-k\Sigma P_i ln P_i
how? does he come to this result
ofc you know, definitely
P_i = 1/\Omega

ln(1/V) = -ln(V)

yes

and the sum over all the $p_i$ is 1 by definition of how a probability works

yes
but where does it come from
you simply multiply with 1
and then transform it?

well, it's true, isn't it?
why does it need to "come" from anywhere?
if I start with 4 and write 4=2+2, where did the 2 "come from"?

12:39 AM
Because it has no physics basis
it's pure math

so?

If it has no physics basis
why would I ever think of doing that?
the equation is perfectly fine without that 1
there should be some reasoning as to why we made this transformation

I disagree

Why would you needlessly add stuff ?

not every mathematical step in the course of a derivation needs a "reason"
you're just trying stuff
it looks like magic because of course they don't show you all the cases where researchers tried random manipulations on equations and they didn't lead anywhere

12:41 AM
Yeah I guess so
But I think it's just normal
Wouldn't anyone think of asking why?
That's why I got confused

you've never done a proper math class with proofs, have you? :P
random manipulations drop from the sky all the time
2

Not really
And I know that math class with proof, just proving stuff is extremely difficult, but it develops your brain insanely
I remember asking my math professor about how is it possible to prove theories
And he told me

it's just part of the process - you have an equation, like here that involves $S$ and $\Omega$, and you have the expression for $\rho$ that involves the $P_i$, and some relation between $P_i$ and $\Omega$ and you're just trying random stuff on to get something that relates $S$ to the $P_i$

"Well, those who prove mathematical theories, are demigods"
2
I see

there's no secret that tells you "how" to do it, but once you've seen a dozen proofs like this, you'll get used to just multiplying by cleverly expressed 1s or adding/subtracting a 0 being common techniques
and then next time you'll try it, too

12:46 AM
Anyway I will continue to do physics for like 2 -3 hours, going 12 now
But you helped me immensely, yet again
in order to fundamentally understand more about the abstract phase space

you're welcome, and take a break :P

Thank you, and don't worry, in a couple of days time, I'll be here again, sorry in advance :D

1:00 AM
@ACuriousMind One thing, that I forgot. I want to understand more about the free energy/enthalpy/ free enthaly, I want to understand what do they represent, their meaning and how they differ from each other, and the reason as to why we had to come up with these potentials, do you have any pdf/ script or w/e where I can read about them, understand them and see the difference between them ?

2 hours later…
3:22 AM
Is there any difference between $(\psi^*)^\alpha$ and $(\psi^\alpha)^*$?

2 hours later…
5:48 AM
@DIRAC1930 Well the fundamental quantity that you have is the spinor $\psi$
on it you have some function $^*$
and then you can project that spinor onto a basis
In which case you have like $$\psi^* = (\psi^a e_a)^* = (\psi^a)^* (e_a)^*$$
I think
$(e_a)^*$ projecting it onto whatever conjugate spaced

6:21 AM
Hello @JohnRennie Sir.
Three forces P , Q , R act perpendicularly to the sides of a triangle at their middle points and are proportional to the sides. The forces are in equilibrium...
My question is that, these three forces in equilibrium means triangle wont move any where, triangle also in equilibrium or not?
I am confused because if these three equilibrium forces act on a particle or concurrent, it means their resultant is zero. But in this case we have three different sides of triangle suppose triangle as has three uniform rods.
Here forces are concurrent the triangle is not a uniform lamina, also centroid/COM of triangle is at different place from the point of concurrency of triangle, in second picture of attachment i have shown that.
Here forces are concurrent the triangle is not a uniform lamina, also centroid/COM of triangle is at different place from the point of concurrency of triangle, in second picture of attachment i have shown that.
Here forces are concurrent the triangle is not a uniform lamina, also centroid/COM of triangle is at different place from the point of concurrency of triangle, in second picture of attachment i have shown that.

@ACuriousMind Is there a term for a vector field normal to a foliation defining a connection

5 hours later…
11:10 AM
I hope this chatroom is an appropriate place to ask this. I sometimes get this error (?) that there's a red dot next to one of the review queues while it actually is empty. It looks like this: imgur.com/a/AiBlseH
Is this something that happens to other users too, sometimes or is it a possible mistake on my side?

@Jonas This is a known problem.

I see, thanks for the reply!

14

Short explanation This is a known issue for some set of users (particularly active users, unfortunately). We're working a major review queue overhaul (Yaakov is heading up this effort). Longer explanation Review queues are an exercise in trade-offs. Mainly, with respect to performance and spee...

11:26 AM
interesting, I don't think I've ever seen a red-lighted review queue with not zero review tasks

@Slereah a symmetrized form of the 'norm' $\psi^{\dagger} \gamma^0 \psi$

11:50 AM
Question is
What is $\gamma_0$ generically
I know it's actual $\gamma_0$ for Dirac and $\sigma_2$ for Pauli
But how do you pick it

You don't pick it, it's an abstract element of an algebra
You can construct a representation of the algebra where it takes a certain form, but any other equivalent representation is fine too

Is it always a member of the Dirac algebra though
and if so, is it a specific one or can you pick any of them

You mean could you do $\psi^{\dagger} \gamma^3 \psi$ instead
Basically one can prove that it's always possible to choose the $\gamma_a$'s associated to the Minkowski metric $\eta_{ab}$ so that $\gamma_0$ is Hermitian and $\gamma_i$ is anti-Hermitian, sometimes that's given in the definition instead, so the $\gamma_0$ is just the Hermitian one

12:06 PM
I see
Is that true generally?
Does any basis of the spin algebra have a hermitian generator and a bunch of anti hermitian ones
Although $\gamma^i$ are just big $\sigma^i$ matrices, so I suspect not

Yeah it generalizes to higher $D$ and to more general $(+,+,...,-,-,...)$ metrics
It's an abstract algebra, but you can construct irreducible representations of it with these properties, remember some theorem about reps being equivalent to unitary reps, it's along that line of thinking

12:26 PM
Hm
Seems to be the opposite, but I guess it's the physicist convention of using $\sigma^i$ instead of $i \sigma^i$
I'm afraid that for my measurement nonsense, I'm gonna have to bring up the Cartan connection
Because for some reason, it's very fashionable in that field to use Weyl gravity
So might be a good idea to keep the geometry a bit fluid

@Jonas that just mans you're an exemplary reviewer and if we had more users like you the red indicators would be much rarer ;)

fortunately I don't think anyone has done it for supergravity yet
idk what kind of holonomy lets you measure the Rarita Schwinger field
is there even a solder form for Weyl gravity, I think you don't reduce the structure group at all
You just sit there with your full GL group

12:58 PM
but then again I know there is a "metric" for the affine structure
and it's just the Lorentz metric up to equivalence of geodesics
idk what G-structure or whatever that is
"Geometry of affine manifolds is essentially a network of longstanding conjectures"
Marvelous

1:10 PM
@Qmechanic Responding to a message from november 19. About the youtube show PBS Space Time, written and presented by Matt O'Dowd. A week earlier the show was titled: 'Is ACTION The Most Fundamental Property in Physics?' In that show Matt O'Down asserted that in General Relativity there is a principle of Least Proper Time. That is, according to Matt O'Dowd motion along a geodesic is a path of least proper time. i.stack.imgur.com/SpV0J.jpg

1:31 PM
@Cleonis : Ha. Matt O'Dowd probably meant to say maximal proper time :)

@ACuriousMind reviewing is a great way to procrastinate :)

1:57 PM
I remember that well, yes :P

It's a good thing the causal structure is mostly the same no matter the G-structure you're considering
At least for globally hyperbolic spacetimes

@Qmechanic Well, he had his production team go through creating an image with those words, to be placed in the background. I took a screenshot. i.stack.imgur.com/SpV0J.jpg So it's definitely not a simple slip. Here is my hypothesis: Matt wrote the script for that show together with a collaborator. They wanted to see a connection between Fermat's least time and motion of matter. And then a thought occurred: hey, GR has time dilation. Fermat least time <=> Principle of Least Proper time.

@Cleonis well, we also speak of the principle of least action when it's really the principle of extremal action
so it's not an unusual simplification/inaccuracy

@ACuriousMind It could be the most action!

I mean, you can just put a minus sign in front of the action to turn least into most :P

2:14 PM
@ACuriousMind There is a difference. 'Least action' and 'stationary action' are not opposites. 'least action' says: it must be a minimum. 'Stationary action' says: whether it is a minimum or a maximum is immaterial. Hypothesis: Matt O'Down wanted really badly to unify Fermat least time and motion of matter. That led Matt O'Down to believe the opposite of what is in reality the case.

@Cleonis it can also be an inflection point

So, $\psi^1 \psi^{*1} + \psi^2 \psi^{*2}$ is invariant but not manifestly invariant. So in general, all invariants cannot be built just using the usual tensor contraction stuff. Why is that?

idk my dude

Are you talking to me or someone else?

I am

2:19 PM
@Cleonis I agree there's a difference, I'm just saying this isn't the only place where someone doesn't pay much attention to the difference

I assume something weird probably arrises from the structure of the group i.e. the determinant of the elements equalling $0$
You can't really control accidents like that happening

I should look into how conformal gravity works
I don't rly know how you swap the metric to the conformal metric in a rigorous way

2:35 PM
I must be missing something. People in non rel stuff write expressions like $\psi^\dagger \psi \rightarrow \psi^\dagger U^\dagger U \psi = \psi^\dagger \psi$ all the time but you can't really write it in spinor notation

@ACuriousMind The point is, claiming that GR features a Principle of Least Proper Time is not a simplification/inaccuracy. It's a blunder. If you go to that youtube video and you scroll through the latest comments, there are all these people responding: what a great insight: Fermat's least time <=> GR Principle of Least Proper Time. The view count is currently at 608.000 views

I don't really see the harm - it's the same level of wrongness as saying classical mechanics has a principle of least action

... and what does view counts have to do with accuracy?

you find many resources talking about least action, and they're all wrong, technically. But I think it's an "acceptable" level of wrongness, since explaining why it's really stationary points and not extrema that matter is a technicality that doesn't matter if you're not actually doing the math
pop-sci and entry level physics is full of these sorts of tricks - I don't generally like them either and would prefer if we were much more explicit about when we're sweeping technical stuff under the rug, but that doesn't seem to be the majority opinion

communicative effectiveness requires the sender and the receiver to agree on what makes a difference and what doesn't.

2:48 PM
as so often, that platitude doesn't apply here - the average receivers don't know anything about the topic and therefore cannot rationally agree to a level of discourse in any way

$\psi'^{\dagger} \psi' = \psi^{\dagger} U^{\dagger} U \psi = \psi^{\dagger} \psi$ is fine. If $\psi = \begin{bmatrix} \psi^1 \\ \psi^2 \end{bmatrix}$ then $\psi^{\dagger} = \begin{bmatrix} \psi^{1*} & \psi^{2*} \end{bmatrix}$ so $\psi^{\dagger} \psi = \psi^1 \psi^{1*} + \psi^2 \psi^{2*}$ is 'manifestly invariant' under $\mathrm{SU}(2)$.

But I can't write that like $M_{\alpha \beta}\psi^{*\alpha}\psi^\beta$

$\psi^{\dagger} \psi = \delta_{\alpha \beta} \psi^{* \alpha} \psi^{\beta}$

I've tried that, it doesn't work

2:55 PM

$$\psi^{\dagger} \psi = \psi^1 \psi^{1*} + \psi^2 \psi^{2*} = \delta_{11}\psi^1 \psi^{1*} + \delta_{22} \psi^2 \psi^{2*} = \delta_{\alpha 1}\psi^{\alpha} \psi^{1*} + \delta_{\alpha 2} \psi^{\alpha} \psi^{2*} = \delta_{\alpha \beta} \psi^{\alpha} \psi^{* \beta}$$

Yes but it transforms into minus itself
$F{}_\alpha{}^\varphi{} F{}_\beta{}^\xi{} (F^{-1})^*{}_\gamma{}^\alpha{} (F^{-1})_\delta{}^\beta{}\delta_{\varphi \xi}\psi^*{}^\gamma \psi^\delta$
That is the transformation
It will give you $-\delta_{\alpha \beta}\psi^{*\alpha} \psi^\beta$ if I recall correctly

It doesn't, you know the left-hand side transforms as $\psi'^{\dagger} \psi' = \psi^{\dagger} U^{\dagger} U \psi = \psi^{\dagger} \psi$. So now you need to re-derive the same result when written as $\psi'^{\dagger} \psi' = \delta_{\alpha \beta} \psi'^{\alpha} \psi'^{\beta *}$. It's really the exact same thing, just written in matrix notation with the indices explicit vs. matrix notation with them left out as 'obvious'

There are too many G-structures on a spacetime
How do they work
I can only find people who mention them but not how they work or people who mention how they work but not what they are

Does a dagger operation lower an index?

3:05 PM
"The study of causal relations in spaces equipped with a field of convex null cones in TM received a major contribution from the Alexandrov school in Russia under the name of chronogeometry"
It's like I never run out of new things to learn

But $\psi^*$ doesn't transform the same as $\psi^\dagger$

"The projective and conformal structures of a metric space determine uniquely its metric."

What is N=2 supersymmetry? What means N=2 here? Anybody knows it?

"A fully rigorous formalization [of our axioms and proofs] has not yet been achieved, but we nevertheless hope that the main line of reasoning will be intelligible and convincing to the sympathetic reader."
Oh no
@peterh The number of SUSY generators

Hi All...

3:09 PM
If you work it out explicitly it works with the transformations I am using so I think they are right

Or the odd dimension of the superspace, alternatively

When I took the dagger I didn't write them with the indices down

Also means that there are two spinor fields associated with the whatever you're doing

How $\tau = r \times F = I \alpha$ , because $r \times F$ does not take care of rotational inertia.

The classic example is a scalar field associated with two spinor fields

3:12 PM
In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object around a specific axis. Another definition of torque is the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically...

Firstly your $\delta_{\alpha \beta}$ is not frame independent

$\delta_{\alpha \beta}$ is invariant under $\mathrm{SU}(2)$, $U^{\dagger} I U = I$
The wiki derives $\tau = I \alpha$

that would be $\delta^{\alpha}{}_\beta$

My confusion is that, How $r \times F$ take care of rotational inertia of body?

3:17 PM
You've defined the lowered one to be $\delta_{11}=1,\delta_{22}=1$
which is not invariant
$\delta_{\alpha \beta}$ is usually the epsilon tensor $\epsilon_{\alpha \beta}$
Let me just check I'm doing my determinants correctly

@ACuriousMind Well, as we know: in the most prevalent cases the point of stationary action corresponds to a minimum of the action. Cases with minimum action are: inverse square force, inverse force, and constant force. When the force increases with the square of displacement (hence potential energy increases with cube of displacement) the true trajectory corresponds to a maximum of Hamilton's action. Harmonic oscillation is at the cusp: both kinetic energy and potential energy are quadratic.

projective structure and causal structure seem to both be modeled on the projective group
I am guessing there is a nuance
Also you can do a G-structure from the time function, it seems?
Question is, what does the connection of the causal structure look like!
stick around to find out

3:41 PM
Is it not the case that if we use the index notation, then $U^{\alpha}_{\ \ \beta} \in \mathrm{SU}(2)$ leads to $(U^{\alpha}_{\ \ \beta})^T = U^{\beta}_{\ \ \alpha}$ and similarly $(U^{\alpha}_{\ \ \beta})^{\dagger} = U^{* \beta}_{\ \ \ \ \ \alpha}$, so that the relation $U U^{\dagger} = I$ now reads as $\delta^{\alpha}_{\ \ \beta} = U^{\alpha}_{\ \ \gamma} (U^{\dagger})^{\gamma}_{\ \ \beta} = U^{\alpha}_{\ \ \gamma} U^{* \beta}_{\ \ \ \ \ \gamma} = \delta^{\alpha \beta}$,
where I wrote the last delta with indices up because the indices were all raised in the term before it, and you can imagine what $U^{\dagger} U = I$ is going to lead to. So breaking these conventions is unavoidable, but it's fine because $\epsilon$ is the metric not $\delta$?

So to get the result you have to break something
If you do the transformation $\psi^1 \psi^{*1} + \psi^2 \psi^{*2}$ explicitly in index notation it is invariant but I can't manage to get it through just manipulating indices in the usual way

I suspect that the measurement business I am doing can probably be boiled down to the holonomies of the various structures of a manifold
and hopefully each one can be determined by various data from such holonomies
I probably need to order those various structures into each other I think
They seem to flow into one another
Something like metric → conformal → affine → projective → causal
Probably involving a bunch of diagram chasing
also idk where the structure induced by the time function fits in there but I'm pretty sure it's the one you can actually measure properly

4:00 PM
Can we write the spinor field explicitly like $\psi_R^\alpha + i \psi_I^\alpha$ where now $\psi_R, \psi_I$ are real spinor fields?
Maybe that wouldn't help

I mean you can, sure
it's just complex numbers
You can always split them

Does this index notation stuff not come up with all the invariants in general?

I agree it's confusing, I think the best thing to do is to ignore the positioning of the indices on the matrices unless you're using the $\epsilon$ to properly raise and lower like L&L do, for example $\psi'^{\alpha} = (U \psi)^{\alpha}$ and $(\psi'^{\alpha})^{\dagger}= [(U \psi)^{\alpha}]^{\dagger} = (\psi^{\dagger} U^{\dagger})^{\alpha}$ combine as
$$\psi'^{\dagger} \psi = \delta_{\alpha \beta} \psi'^{\dagger \alpha} \psi'^{\beta} = (\psi^{\dagger} U^{\dagger})^{\alpha} \delta_{\alpha \beta} (U \psi)^{\beta} = (\psi^{\dagger} U^{\dagger} I U \psi) = (\psi^{\dagger} I \psi).$$

I was using the $\epsilon$ to raise and lower indicies
Hang on, doesn't when I got $(-)$ after the transformation imply that the whole thing is $0$ in the first place, because if you make the continuous transformation infinitesimal, the only way you can really get a minus sign is if the quantity to begin with is $0$
Man this is so confusing because spinors are derived from isotropic vectors like in that Cartan book
So is $\psi^1 \psi^{*1} + \psi^2 \psi^{*2}$ invariant but also equal to $0$?

4:20 PM
It's not equal to zero

I wish it were

Running into the old problem of "problem is too big for a single paper" where I am running out of symbols to use
Too many things I want to call $\sigma$

I had to resort to using $\psi$ as an index so I was writing $\psi^\psi$ lol

also check out this cool paper

@DIRAC1930 gross

4:32 PM
I wonder how generally one can use that bundle model for spacetime
Trivially true for globally hyperbolic spacetimes, probably alright for causal spacetimes generally
I don't think it would work for say those weird non-distinguishing spacetimes
or with topology changing spacetimes
that paper calls the musical isomorphism the "Riesz operator"
makes sense but weird

4:52 PM
mama mia

5:30 PM
In general, can you have invariants other than the ones with fully contracted indices?

6:02 PM
I honestly think it doesnt work because you need $\delta_{\alpha \dot{\alpha}}$
If you do that, then the transformations cancel

Not sure what you're talking about

Well with $SL(2,\mathbb{C})$ you never mix the two spaces up
Well doesnt $\overline{V}^*$ act on $\overline{V}$ and $V^*$ act on $V$?
We are trying to make $V^* \times V^*$ act on $\overline{V} V$
instead of $\overline{V}^* \times V^*$ act on $\overline{V} V$
excuse my sloppy notation. I don't know how to write it properly

6:18 PM
Don't know what that means, are they vector spaces acting on each other?

$*$ is the dual space
Overlined is the conjugate space
Let me just write it properly
$(\overline{V}^* \times V ^*) \times (\overline{V} \times V) \rightarrow \mathbb{C}$ is what we want
We are trying to do $({V}^* \times V ^*) \times (\overline{V} \times V) \rightarrow$ which I'm not sure what it transforms to

Why are you acting on vector spaces with other vector spaces

Because thats what tensors do write?
$\psi_\alpha \psi^\alpha$ is $V^* \times V \rightarrow \mathbb{C}$ right?

Lets say $\psi \cdot \psi = \psi_{\alpha} \psi^{\alpha}$ is a dot product, what you're talking about is the dot product as a map $\cdot : V^* \times V \to \mathbb{C}$ which evaluates as $\cdot(\psi,\psi) = \psi \cdot \psi = \psi_{\alpha} \psi^{\alpha}$, but it's not vector spaces acting on vector spaces

How do I write $\delta_{\dot{\alpha} \alpha }\overline{\psi}{}^\dot{\alpha} \psi{}^\alpha$ in that notation?

6:33 PM
The same way, just replace that by the one I wrote with indices up and down

That's what I did though?
here $(\overline{V}^* \times V ^*) \times (\overline{V} \times V) \rightarrow \mathbb{C}$
But then what exactly is the difference between $\delta_{\alpha \beta}$ and $\delta_{\alpha \dot{\alpha}}$ other than how it transforms

6:48 PM
You're not explaining what any of this means and this notation is unreadable

7:18 PM
It's standard notation?
It's in every book on SUSY and $SL(2,\mathbb{C})$
I'm just using the same notation
$\psi^\dagger U^\dagger U \psi$ implies that you are using $\delta_{\alpha \dot{\alpha}}$. The question is if anyone has an explanation of what exactly is going on
Can someone help me?

7:38 PM
@ACuriousMind

7:51 PM
@fqq
I think page 5 of this is relevant
So yes $M^*$ is related to $M$ by a similarity transformation with the matrix $\epsilon$
So they are equivalent which 2 people mentioned here the other day
So if the $\overline{\psi}$ and $\psi$ representation are equivalent, why does that allow us to write $\delta_{\alpha \dot{\alpha}}$?

00:00 - 20:0020:00 - 00:00