@bolbteppa In your $D=5$ text, how did you get to equation 4? What is $\xi_{ij}$ and where does the antisymmetry condition come from?

Look at the last equality in the third equation, the first two terms there are the first two terms in the fourth equation, the last part of the fourth equation can be added to the last part of the third equation and it will automatically vanish if $\xi_{ij}$ is anti-symmetric so it's a valid term to add

The last part of the third equation can be written as $\sum_{i=1}^2 x^i \eta_i = 0$

Ah okay thanks

Ill play around with it tommorow

Is there a word for when one doesn't understand a single word an author is talking about and consequently gets an uncomfortable feeling in their stomach?

anxiety

Stress

Panic

Single word requests are frowned upon at English Language & Usage.SE; but, you could try.

the robot has spoken

@nickbros123 I think it's just called "feeling stupid" :P

I'm not saying you (or anyone else) is stupid or that actually being stupid (whatever that means) feels like that, just that that's what people mean when they say something makes them feel stupid

mathwithbaddrawings.com/2017/01/25/…

You can try to check here

I don't feel stupid when I'm still trying to understand, I do when I make mistakes I shouldn't do

Sometimes I freak out about a single mistake for a couple of days

I resonate with this stick figure

@Slereah explanagony is terrible

it's terrible if it happens in front of other people, which is why you use a rubber duck :P

Even more terrible if it happens here and it will remain forever

I've heard programmers use that, does it really work?

the one emotion there I think is debatable is elegance envy - I get where it's coming from, but I think plenty of people actually enjoy seeing proofs from "the book" that are slicker than their own

@ACuriousMind i can't help it but think "why didn't I think of that"

@user85795 I mean, few people use a literal rubber duck, but yes, trying to express a problem in your own words in a coherent manner that someone else could understand often helps yourself understanding what the actual problem is

Hmm, good point.

As long as my proof works I'm fine. Math is not programming :P

@nickbros123 sure - but the next thought should be "I'm putting this into my toolbox for next time" ;)

Also, quicker$\neq$easier

I see it more like friendly rivalry than envy - "next time I'm gonna be the one who writes a proof like that!"

@ACuriousMind yeah agreed. Sometimes it so happens someone comes up with a truly elegant proof that for some reason would make my day

For me it also depends on who wrote the proof: if someone with a better knowledge of the field than mine I'm fine but if it's someone in my class, envy could kill me :P

Well, maybe that's not simply envy but a mix of rage and respect

I stopped envying people when my 7th grade cousin solved an IIT entrance exam problem I was stuck in, using a rather slick method I couldn't have possibly thought of

:O

Hello

I have a question on Thermodynamics.

@bolbteppa Are the signs of the terms in $\xi \eta_j - \xi_j \eta_i + \dots$ and $\xi_{ij}\eta_k + \xi_{jk}\eta_i + \dots$ just an ansatz or is there a reason behind them?

It needs to be anti-symmetric in the $ij$ indices, i.e. we formed $\xi_i \eta_j - \xi_j \eta_i = 2 \xi_{[i} \eta_{j]}$ so that we could add it to the anti-symmetric quantity $\xi_{ij} \eta_0$, similarly the $ijk$ indices are anti-symmetric etc...

Ah okay

Also, he defines $\eta_i=0$ when it is really $\sum_i x^i \eta_i =0$. What allows him to do that? He seems to pull the expression $\xi_{[i} \eta_{j]} + \xi_{ij}\eta_0$ out of thin air. Where does it come from?

$\sum_i x^i \eta_i = 0$ must hold for all choices of $x^i$ which is only true if the $\eta_i$ are each separately zero

Apart from noticing that the indices are anti-symmetric in $2 \xi_{[i} \eta_{j]} + \xi_{ij} \eta_0 = 0$, so that it's a well-defined tensor equation, and the idea that expanding it will lead to something new, I can't really add more than that unfortunately

Do you know why the $\eta$'s define an equation for the $v$ plane? e.g. what does $\eta_0 = \xi_i x^i=0$ represent geometrically?

Looking at your $D=3$ case, we have what he says are the 'equations for the isotropic directions' but what exactly does this mean for a simple example

I tried to give a picture here yesterday in 3D which shows what $\eta_0 = 0$ means visually

Ah okay thanks. I'm going to think about what you wrote

I think I need to understand sections 9 & 10 fully first

That's section 9

@bolbteppa I think that he uses the word direction in the term 'isotropic direction' because $F$ is invariant under the scaling of isotropic vectors i.e. $X^i \rightarrow \gamma X^i$ still keeps the fundamental form invariant since $F= X_i X^i = 0 \rightarrow F'=\gamma^2 X_i X^i = 0$. Therefore it's only the direction that matters

At the end he shows a reflection is a reversal (i.e. has determinant $-1$), then you can see any rotation is a product of reflections from the fact that the determinant of a rotation is $+1$ and a reversal has $-1$, the only thing left is to show a rotation is a product of $\leq n$ reversals, not just any old number of them. In the first part of 10 he proves most of it quickly, then a technicality makes the rest of it complicated

A vector has magnitude and direction, however since $F = 0$ we see the vector really only has a (non-trivial) direction, the vector is thus the direction

@bolbteppa How can I have a plane tangent to the cone and also tangent to the $z$ axis?

Nevermind

I was trying to visualize it but it's impossible visualizing a cone in 3d complex space lol

Referring to page 62 (the first page I sent yesterday), the plane $\xi X + \eta Y + \zeta Z = 0$ that is tangent to the cone is perpendicular to the z-axis if it has a fixed value for $Z$, if we choose $Z = 0$ then if $X^2 + Y^2 = 0$ also holds we have $Y = \pm i X$ thus this plane that is tangent to the cone also contains an isotropic straight line $Y = i X$ and it is perpendicular to the $Z$ axis

Okay I think I've understood the idea

So the intersection of the two hyperplanes is given by a line $\eta_0=0$

But then why do we need the extra $\eta_1$? What extra information does it give?

In the $3 = 2 \cdot 1 + 1$ case there are $2^1$ equations for $2^1$ 'superfluous' variables $\xi_0,\xi_1$. Only one equation means one of the two $\xi_0,\xi_1$ is free, but two equations completely fixes them right

Cartan doesn't seem to explain anything lol

Or often at most offer a sentence or so explanation for something complicated

What does he mean by dimension of the subspace?

Where

Well mainly the first page of chapter V

He says "any isotropic subspace (i.e., a subspace in which all the vectors are isotropic) has dimension at most $\nu$", so if it's $p$ there are $p$ basis vectors $\mathbf{e}_1,...,\mathbf{e}_p$ for this subspace squaring to $0$. Another point he makes is: the space orthogonal to this space has dimension $n-p$, however since the $\mathbf{e}_i$, $i=1,...,p$, are also orthogonal to themselves they live among these $n-p$ vectors so there are only $n-2p$ basis vectors remaining in the total space.

Clearly we must have $n - 2p \geq 0$ which means $2p \leq 2 \nu + 1$ so that $p \leq \nu$.

Okay thanks

If you re-write $F = (x^0)^2 + x^1 x^{1'} = 0$ in 3D as $F = (x^0,x^1) \cdot (x^0,x^{1'}) = 0$ it seems to be saying that in a smaller 2D space there are vectors orthogonal to $(x^0,x^1)$, if we denote a general vector as $(\xi_0,\xi_1)$ then we have $(x^0,x^1) \cdot (\xi_0,\xi_1) = \xi_0 x^0 + \xi_1 x^1 := \eta_0 = 0$. In other words the existence of this initial $\eta_0 = 0$ equation in all $D = 2 \nu + 1$ dimensions would follow immediately if this make sense. Not sure how valid this is...

I think perhaps there might be a way to think about this more formally in terms of group theory (I'm not sure however). For the 2d case the quantity $\xi_0/\xi_1$ is actually the map of the Reimann sphere onto the complex plane and you can show that as you rotate the sphere, the quantity $\xi_0/\xi_1$ transforms homographically therefore the spinor $\xi=(\xi_0,\xi_1)$ transforms through $\xi \rightarrow U \xi$.

There must be a similar way to show how the intersection of the isotropic planes transform under a transformation of the cone

Not sure what you're trying to do, looks like you are still trying to appreciate why a spinor should transform as $\xi' = A \xi$ under a reflection

If $\eta = X \xi = 0$ holds, and $X$ transforms as $X' = - A X A$ under a reflection, then if we perform a reflection on a given system, $X \xi = 0$ should reflect into an equation like $X' \xi' = 0$, i.e. $- A X A \xi' = 0$, how is $\xi'$ defined in the reflected system so that the whole thing remains consistent? Clearly $\xi' = A \xi$ will lead to $X' \xi' = - A X \xi = A 0 = 0$ so that $X' \xi' = 0$ indeed holds, so it's a good definition.

Taking this definition however leads to a technicality in that all this only constrains the definition of $\xi'$ up to an overall scalar, we could equivalently define $\xi' = m A \xi$ with $m$ a scalar, but from wanting to get the original spinor back (up to the overall $\pm$ indeterminacy) under the reflection done twice gives $\pm \xi = \xi'' = m^2 \xi$ we see $m^2 = \pm 1$ should hold so we get two definitions. This is really important in QM, especially it's analog in 4D.

No, I've always understood that

I'm saying that by definition that there is a representation of the rotation group being represented on the representation space of intersecting isotropic planes that you can construct explicitly that could perhaps give a better idea on how this construction works in higher dimensions