@DIRAC1930 not sure what you're trying to do, if you want to see $X = x_{\mu} \sigma^{\mu}$ written as a direct product of spinors, it's done e.g. in Ch.1 of Penrose's twistor book. It looks a bit different to what Cartan is doing and (I think) more relates to what he does when he associates a vector to a pair of spinors, e.g. in section 62, but he's starting from a more primitive idea
If you can figure out how to get the matrix of equation (5) in Section 93 of his book, i.e. the $D = 5$ matrix $X$ associated to a vector, a lot of things he's doing will make sense, it will explain the $D = 3$ matrix he uses in Section 55 as well, and you can then use it to get the $D = 4$ matrix and see how it reduces to the $2 \times 2$ matrix $X = x_{\mu} \sigma^{\mu}$, if you get stuck on this I can explain some of it
@MoreAnonymous Sure, but I don't understand what the point is. The length scale in the Ising model is an example for a renormalization scale (in that case in real-space renormalization). Renormalization scales are not observables, they're parameters that index a range of theories - more or less the theory at scale $\Lambda$ is the "best" one to predict observations made at scale $\Lambda$.
You can think about this as a continuous tower of ever more "fundamental" theories as $\Lambda$ increases (momentum-space)/decreases (real-space), but this is just a very specific instance of the general idea of approximation/coarse-graining. This isn't special to renormalization or QFT, that we can have different physical theories with different levels of detail for different use cases that are derived from each other is a rather general phenomenon
Finally, I don't understand what point you're trying to make with this about the question of whether or not there is infinite regress in the explanations of nature or a true "fundamental" theory. Such approximations are an example of "bounded" regress in that even if the regress is not technically finite there is a "most fundamental" theory in these approaches
(e.g. the UV completion in momentum-space renormalization, or the classical mechanics of individual particles in classical statistical mechanics)
This does not imply that this fundamental theory does not itself have something "more" fundamental it might be derived from, just that your current theoretical framework stops there
so this doesn't really do anything to answer the question of whether there is an infinite regress of explanations or not
and really finally, this is about physical theories explaining each other, not about "symbols" or their meaning, which is what the question that started this whole thing seemed to imply. The symbols with which we explain the "fundamental" physical theory to each other do not get their meaning from the theory; they need to already have meaning so that we can communicate - the meaning of symbols is an entirely orthogonal thing and I don't understand what you think it has to do with this
So isotropic vectors are 'really' the complex eigenvectors of a rotation matrix, e.g. a 3D rotation matrix has a real eigenvector corresponding to the direction of the 'axis' in the axis-angle decomposition of the rotation, and then it has two complex eigenvectors (which square to zero) in the plane of rotation.
What all this seems to be saying is, in 3D, visualize $F = z^2+x^2+y^2 = (x^0)^2 + (x+iy)(x-iy) = (x^0)^2+x^1x^{1'}=0$ as that (literal) cone (of vectors), and imagine $\xi_0 z + \xi_1 x + \xi_2 y = 0$ as a plane tangent to the cone at some point, and think of the $z=x^0$ axis as being the 'axis' in the axis-angle decomposition of a given rotation (with a real eigenvalue).
The plane tangent to the z-axis and tangent to the cone is full of complex conjugate eigenvectors so let's describe it, the equation of this plane is the intersection of $z = 0$ and $x^2+y^2=0$ so it's the plane containing the line $y = ix$, i.e. the equation of the plane is $0+\xi_1 x + i \xi_2 y = (\xi_1 + i \xi_2)x = 0$ which tells us $\xi_2 = i \xi_1$.
The intersection of the original tangent plane with this second tangent plane is then a line described by $\xi_0 x^0 + \xi_1 (x + i y) = 0$ i.e. $\eta_0 = \xi_0 x^0 + \xi_1 x^1 = 0$ holds.
Thus the spinor coordinates describe the intersection of a hyperplane in $D=3$ with the hyperplane orthogonal to the axis in the axis-angle decomposition of a rotation which is full of complex conjugate eigenvectors, in this case the intersection of hyperplanes gives is a line.
However, solving $F = 0$ for $x^1 = - \tfrac{(x^0)^2}{x^{1'}}$ we see there is a second way to describe this isotropic 1-line: $0 = \xi_0 x^0 - \xi_1 \tfrac{(x^0)^2}{x^{1'}}$ or $\eta_1 = \xi_0 x^{1'} - \xi_1 x^0 = 0$.
(A way of doing this step which generalizes to higher dimensions is to instead consider $\xi_0 F = 0$ and insert $\eta_0 = 0$ into this, which will again lead to $\eta_1 = 0$)
Thus $\eta_0 = 0$ and $\eta_1 = 0$ are two equations in two variables which describe the intersection of the hyperplane tangent to the null-cone with the hyperplane orthogonal to the $z$ axis in terms of coordinates $\xi_0,\xi_1$, an isotropic 1-line.
In $D = 2 \nu + 1$ we want to end up with $2^{\nu}$ equations $\eta_A = 0$ in $2^{\nu}$ variables $\xi_A$, and these equations will describe an isotropic $\nu$ hyperplane.
We're technically using complex numbers everywhere, and using the normal Euclidean inner product $F = x^2 + y^2 + z^2$ with $x,y,z$ complex, and can study $F = 0$, but we can still also use this stuff to talk about rotations when $x,y,z$ are real and use the complex numbers to talk about their eigenvectors e.g. as the pages I sent do
In $D = 5 = 2 \cdot 2 + 1$ we should end up with an isotropic $2$-plane described in terms of $2^2$ equations in $2^2$ variables by starting from $F = (x^0)^2 + x_1^2 + .. + x_4^2 = (x^0)^2 + x^1 x^{1'} + x^2 x^{2'} = 0$.
Let $\xi_0 x^0 + \xi_1 x_1 + \xi_3 x_2 + \xi_2 x_3 + \xi_4 x_4 = 0$ be tangent to the cone, then on $x^0 = 0$ we see $F = 0$ reduces to $0 = 0 + (x_1 + i x_2)(x_1 - i x_2) + (x_3 + i x_4)(x_3 - i x_4) = 0$ which contains the lines $x_2 = i x_1$, $x_4 = i x_3$ so that we first find the line $\eta_0 = \xi_0 x^0 + \xi_1 x^1 + \xi_2 x^2 = 0$, but we still need to end up with additional lines $\eta_0 = 0, \eta_1=0,\eta_2=0,\eta_{12} = 0$.
We next consider $\xi_0 F = 0$ and end up with $\xi_0 F = \sum_{i=1}^2 x^i (\xi_0 x^{i'} - \xi_i x^0) = 0$ so that we can define $\eta_i := \xi_0 x^{i'} - \xi_i x^0 + \sum_{j=1}^2 \xi_{ij} x^j = 0$ for $i=1,2$. We still need a fourth equation $\eta_{12} = 0$, Cartan shows how to get it.
From these four equations you immediately get the $X$ matrix associated to the vector $(x^0,x^1,x^2,x^{1'},x^{2'})$ in 5D. The 4D matrix immediately comes from setting $x^0=0$ and re-arranging rows/columns.
The reason I'm posting this is to try put this into simple words. What the hell is going on? How does this relate to the axis-angle stuff and complex eigenvectors? Any ideas?
I find this approach to isotropic vectors even more puzzling than Cartan's one where we put 3d vectors into 2-by-2 complex matrices :P
But I have to admit I've never tried especially hard to understand these kinds of constructions for spinors because I find the algebraic ways via representations of the covers of the rotation groups completely sufficent
This is Cartan's approach, both how he does the 3D case and how he generalizes it to higher dimensions, but he doesn't use these words he just gives you a bunch of equations and tells you at the end it's all supposed to describe an isotropic $\nu$ hyperplane in terms of $2^{\nu}$ equations $\eta_A = 0$ and $2^{\nu}$ "superfluous" parameters $\xi_A$, as the first picture I sent says, apparently Riesz at least gave some intuition in terms of (complex...) cones
Spinors appear in physics because the rotation group has all this insane behavior which is then inherited by the representation theory, but projective reps do not explain any of this
Cartan is telling you that a spinor in $D = 2 \nu + 1$ is really a $2^{\nu}$-coordinate ($\xi_A$) description of an isotropic $\nu$ plane which is defined by $2^{\nu}$ equations $\eta_A = 0$.
We're trying to figure out what a spinor really is, and this is what he says it is, that's why it's relevant. I don't (yet) see how the representation theory perspective helps explain any of this, though if it does that would be useful
...I'm not sure why I should care? Like, I also don't try to understand the gluon SU(3) adjoint rep in geometric terms, and I also don't believe there's a literal vector associated with a spin-1 massive boson, so why would a geometric interpretation for spin-1/2 matter?
Well one non-trivial practical example is that this stuff is absolutely essential in superstring theory when you get to things involving pure spinors, though on a simpler level it's like saying 'I don't need to know any Euclidean geometry, I can just learn how to work with numbers axiomatically and ignore any visual interpretations'
I've done enough superstring theory to confidently say this is not essential :P
I don't need to think about cones to work with spinors
if this kind of interpretation was necessary to work with objects, then no one would be able to work with fields transforming in arbitrary $\mathrm{SU}(N)$ representations
If you hand me some high tensor like a spin-4 $T^{\mu\nu\rho\sigma}$ I don't need to first figure out what crazy arrangement of lines or subspaces or complex eigenvectors might possibly transform like that in order to deal with it
I don't see why this should be different for spin-1/2
If there is some argument from physics why I should be looking at this way of defining what a spinor is, then that would be different. But from my perspective, the actual argument for why spinors appear in physics is because QM tells us the rotation group only needs to be represented projectively, and it so happens that the simplest projective representation is the spinors that you also get from the isotropic vector construction
but that doesn't really give me any reason to care about that construction unless for some reason I'm collecting as many mathematically equivalent descriptions of things as possible
How does the projective representation perspective, or the double cover/lack of simple connectedness, tell you how many components a spinor has in $D$ dimensions, it may but I can't remember if it does
once you've figured out that the (Dirac) spinor is the unique irrep of the Clifford algebra it's immediate that it's $2^{\lfloor n/2\rfloor}$ in $n$ dimensions
(the physics way to ask that question is to ask what size $\gamma$-matrices are in $n$ dimensions)
and the $\gamma$-matrices in turn are to my eyes well-motivated physically via the "square root of KG equation" stuff Dirac did to get to them
You're either starting from Dirac's method of introducing spinors, which is great but it's out of thin air and the question is what does any of it mean (which is the question), or you're starting from wanting reps of the orthogonal group and the question is what are these extra reps representing in the first place which leads us back to where we started
You would think that spinors have something fundamental to do with projective space if they arise from projective representations (because of the word projective...), only in 3D is there a picture that really leads to them as far as I've seen, though maybe these isotropic subspaces involve projective space somehow
we need to be careful with the adjective "projective" here - when I'm saying it I'm talking about Hilbert spaces and projective representations on them, and this is a different notion than real projective space
'A homomorphism of a group $G$ into the group PGL(V) of projective transformations of the projective space P(V) associated to a vector space V over a field k'
abstractly the notion of complex projective and real projective space is of course the same - we're looking at the space of 1d subspaces of a vector space - but complex and real projective space have rather different behaviours
QM is only about complex projective representations
A bluff (half of an iff statement maybe) is that if $X \xi = 0$ then since $X^2 = x^2 I$ we have $x^2 \xi = 0$ so it enforces $x^2 = 0$, if we don't assume that beforehand
The $2^{\nu}$ equations $\eta_A = 0$ are really $\eta = X \xi = 0$, he's defining a spinor by directly constructing this equation starting from $F = (x^0)^2 + ... = 0$ and thinking about isotropic subspaces
just because that's the way you figure out the spinor dimensions in the isotropic approach doesn't mean the same equations have to appear in other approaches
Qmechanic proves that $d$ dimensional $\gamma$-matrices are $2^{\lfloor d/2\rfloor}$ here
Yes well if you introduce spinors starting from Dirac's trick of trying to take a 'square root' of $x^2 + y^2 + z^2$ using Clifford algebras, you get $X = x^i \gamma^i$, so an obvious question is why you'd write $X \xi = 0$ and then claim it's the way to associate a spinor $\xi$ to the coordinates $(x,y,z)$ of a vector via the matrix $X$. What I just said makes sense but it's a bit too obvious and doesn't really give any picture about hyperplanes and isotropic subspaces
(In 3D the association being, explicitly, $\xi_0 = \pm \sqrt{\frac{x - i y}{2}}$ and similar for $\xi_1$, and similarly $x = \xi_0^2 - \xi_1^2$,...; in general beyond 3D...)
Well here's the $D=3$ with the matrix made explicit, without the geometry described above (and slight relabelling), $D=5$ is too big for a picture so I wont post unless people beg
Apart from $z^2 = - (x-iy)(x+iy) = 4 \frac{x - i y}{2} \frac{-x-iy}{2} = 4 (\sqrt{\frac{x - i y}{2}})^2 (\sqrt{\frac{-x-iy}{2}})^2 = 4 \xi_0^2 \xi_1^2$, and wanting to define things that let us take square roots and end up with something well-defined, you just have to be a genius to think of something like this
The above geometric picture which avoids this special 3D situation
Cut down the D=5 size:
I think that D=4 explanation is right, Cartan's book is slightly different, and other books are also slightly different in how they define say $\overline{\sigma}$ and I didn't check it all, but it just amounts to a relabelling of e.g. what is $x^1$ and what is $x^{1'}$
I think it's more that $F = 0$ (over $\mathbb{C}$) implies the existence of all these linear equations, you can infer the first one $\eta_0 = \xi_0 x^0 + .. + \xi_{\nu} x^{\nu} = 0$ abstractly from some general argument he gives, you can also see where it comes from via the geometric argument I gave earlier
Obviously you can plug any vector $\vec{x}$ into a matrix $X$, but then from $X \xi$ you have $X^2 \xi = \vec{x}^2 \xi $, if this is set equal to zero then it's because $\xi$ must be equal to zero, so...
Ah yes perhaps so because later on he doesn't constrain it to $0$. He has $X\xi = \xi'$ and only has $X \xi =0$ when he states that the matrix $X$ is associated with an isotropic vector
But then it could just be that $\xi$ is transforming to a different spinor that is associated with a different isotropic vector
The step in the $D=4$ case regarding re-ordering the indices, and the resulting zero blocks, obviously that's to do with reducibility but he gives a general (incredible) simple argument for why you need to do this
But then again, he is interested in representations of rotations which doesn't seem to be dependent on what spinor he acts on i.e. $S X S^{-1}=X'$ is true no matter what $\xi$ he is acting on
Anytime he applies say a reflection to a spinor, he's taking a non-isotropic vector, plugging it into an $X$ matrix (he calls it $A$), and gets $\xi' = A \xi$
Actually I think Iv'e figured it out! If he rotates the matrix $X$ associated with an isotropic vector through $S X S^{-1}$, the spinor must transform by $\xi\rightarrow S \xi$ since $S X S^{-1} S\xi = S (X\xi) = 0$ due to the bracketed term equalling $0$. So he has found a rep of rotations on spinors i.e. operations which leave the fundamental form invariant.
@DIRAC1930 You'll probably never get through it, he doesn't even derive the Lorentz transformation properly he just states it (as far as I remember...)
The transition to non-isotropic vectors being plugged into $X$ is immediate, it doesn't require any thought, that is something different
He plugs in non-isotropic reflection vectors into these $X$ matrices when he derives the matrix description of a reflection in Section 58 for example, without any explanation or pointing out it's not isotropic
@bolbteppa That is true but he seems to just be illustrating the representation of $\mathbb{R}^3$ on Hermitian $2\times 2$ matrices in that section (which doesn't require spinors). How it relates to the transformations on a spinor are a different story
He actually does derive it in a later chapter, after using it for half the book out of thin air, but it's not the only place where this kind of thing is going on
In Section 60 he then applies the reflection matrix $A$ (associated to a non-isotropic reflection vector) to a spinor and gets another spinor, $\xi' = A \xi$
When he studies $X' \xi'$ all he's doing is showing the whole thing is consistent and we end up seeing that if $X \xi = 0$ holds and $\xi' = A \xi$ then it implies $X' \xi' = 0$ for $X'$ the reflected version of $X$, as you'd expect, however this all only holds up to an overall scale factor which is the $m$ he's talking about. It doesn't affect this trivial issue of relating a general vector to some matrix $X$
In section 63-64 (The Case of Real Euclidean Space), he seems to first find that for a general real vector associated with a matrix $X$, operations that keep the fundamental form ($\det X$) invariant are given by $U X U^\dagger$ for $U$ being unitary.
Then he constructs the equations $D \xi = 0$ for the matrix $D$ and spinor $\xi$ associated with an isotropic vector to get the transformation law that real rotations of Euclidian space correspond to the transformation $U \xi \rightarrow \xi'$ on the spinor
This seems to be the general procedure
So there is always an isotropic vector being involved
Hopefully one day I get to the section on special relativity where I'm guessing the isotropic vector is a null vector
@bolbteppa I wonder if there is a way to relate this isotropic construction to Pauli spinors in non-rel QM
Not sure what the point is, e.g. what do you mean when you say there is always an isotropic vector involved, any time you write a spinor there is an isotropic vector involved, but so what? In section 60 when he writes $\xi' = A \xi$ the $A$ is not associated to an isotropic vector, the $\xi$ is associated to some isotropic $X$, but so what, it doesn't mean we can't associate a non-isotropic vector to $A$ and apply it to spinors.
I'm going to try and understand the $D=5$ construction. Do you think that will be enough to understand the section on SR? (I've gone through most of the next chapter with the $(p,q)$ reps used for Weyl spinors)
It'll get you started, I gave the $X$ matrix from that chapter except some things are relabelled which you can figure out to check things
To really justify where this $\eta_0 = \xi_0 x^0 + ... + \xi_{\nu} x^{\nu} = 0$ equation comes from, it involves understanding this paragraph:
I get the first sentence, the rest I don't get, where he's referring to the $(x^0,x^1,x^{\nu},x^{1'},...,x^{\nu'})$ in $F = (x^0)^2 + x^1 x^{1'} + ... + x^{\nu} x^{\nu'}$
