How Do Spinors Fit in With Differential Geometry

[stevendaryl]

When I studied General Relativity using Misner, Thorne and Wheeler's "Gravitation", it was eye-opening to me to learn the geometric meanings of vectors, tensors, etc. The way such objects were taught in introductory physics classes were heavily dependent on coordinates: "A vector is a collection of 3 (or 4) numbers that transform in such and such way under rotations..." or whatever. I much prefer the geometric definition of a vector as a "tangent" to a smooth parametrized path, and a covector as a linear operator on vectors, etc.

But once you introduce spinors, it seems that the geometric way of understanding things fails. People are reduced to saying "a spinor is something that transforms in such-and-such a way under rotations" again.The mathematically sophisticated way of describing them in terms of representations of the group of rotations (or Lorentz boosts) don't really provide much more understanding of what they really "are".

Is there some geometric way of understanding spinors that is more in line with the "tangent to a parametrized path" understanding of a vector?

 

[DEvens]

It's hard to know what would be a satisfying explanation when "irreducible rep of the rotation group" is not satisfying. In a sense, this definition is very physical. It just may be that the physical nature of the definition did not come through.

If you think about something like an electron, a spin 1/2 thing, and how it behaves under rotation, you may see it. Under rotation an electron is still an electron. It does not turn into an electron and a photon, for example. It does not become an electron and another electron-positron pair. That is, under rotations, the electron goes onto another appearance of the same electron. That is, it is an irreducible object under rotation. In this sense, the mathematical definition is very physical.

If it was reducible, then rotations could split off the electron into smaller parts of some kind. So, a photon and an electron would be, in some sense, in a spin 1 and 1/2 rep, which would be reducible. The irreps are the fundamental particles.

From the standpoint of explaining spinors, the usual mathematical concept is that they are a mapping of space-time onto some kind of extended manifold. Often you get into principal fibre bundles and structures of that nature. In a sense, spinors are a way to represent and transmit the symmetries of the physical situation.

Heh. But that is as far as I got in that area of study. I still remember the point at which my brain's circuit breakers tripped. It was when we started studying "Lie algebra valued one-forms."

 

[stevendaryl]

It's hard to know what would be a satisfying explanation when "irreducible rep of the rotation group" is not satisfying. In a sense, this definition is very physical. It just may be that the physical nature of the definition did not come through.

Well, it's very different in nature than the geometric definition of "vector" and "1-form".

 

[Ben Niehoff]

How much do you understand about the geometrical meaning of spinors in flat space? That would help before you go putting them on curved spaces.

E.g., if I give you a 2-component spinor in $\mathbb{R}^3$, say

$$\begin{pmatrix} \xi \\ \zeta \end{pmatrix}$$
can you tell me which direction it points, and how it is "rotated" by the action of $SU(2) \sim \mathrm{Spin}(3)$? That would be a start.

 

[dextercioby]

There was an entire 2-volume book written about spinors in space-time by Penrose and Rindler, but I'd go with Wald, Chapter 13 anytime.

 

[stevendaryl]

How much do you understand about the geometrical meaning of spinors in flat space? That would help before you go putting them on curved spaces.

E.g., if I give you a 2-component spinor in $\mathbb{R}^3$, say

$$\begin{pmatrix} \xi \\ \zeta \end{pmatrix}$$
can you tell me which direction it points, and how it is "rotated" by the action of $SU(2) \sim \mathrm{Spin}(3)$? That would be a start.

Well, there is a canonical way, using Pauli matrices, to convert a two-component spinor into a vector representing the way that it "points":

$\psi = \left( \begin{array}\\ \xi \\ \zeta \end{array} \right)$

$\psi^\dagger = \left( \begin{array}\\ \xi^* & \zeta^* \end{array} \right)$

$\psi \psi^\dagger = \left( \begin{array}\\ |\xi|^2 & \xi \zeta^* \\ \xi^* \zeta & |\zeta|^2 \end{array} \right)$

Then you write $\psi \psi^\dagger = \frac{1}{2}(1 + \hat{r} \cdot \vec{\sigma})$, where $\hat{r}$ is a unit vector in the direction the spin is "pointing".
Working all this out gives the following relationships between $\xi, \zeta$ and the usual spherical angles $\theta$ and $\phi$:

$\xi = cos(\frac{\theta}{2}) e^{-i \frac{\phi}{2}}$
$\zeta = sin(\frac{\theta}{2}) e^{+i \frac{\phi}{2}}$

But that kind of monkeying with components is the opposite of what I was wanting. In the same way that a tangent vector can be defined without reference to components, it seems to me that a spinor should similarly be definable without reference to components.

 

[bolbteppa]

My vague incomplete intuition is that we can take, say, a 3 x 3 matrix A and 'represent' it using addition (tensor products) or multiplication (spinors).

The additive decomposition is described here (page 10) nicely, basically take A = A_s + A_a (sum of symmetric and anti-symmetric) then take the trace out of the symmetric part to decompose a 3x3 A into A = A_ts + A_s + A_a which is the sum of a 1, 5 & 3 component matrix respectively.

However, for a similar matrix B, we know that A = U^-1 B U gives a multiplicative decomposition of A into A = CD = [U^{-1}B]U, where the matrices U are rotation matrices. Since any rotation is the product of 2 reflections, http://www.euclideanspace.com/maths/geometry/rotations/theory/as2reflections/ , and reflections live in a plane, we can represent the vector we're rotating as a 2x2 matrix acting on planes then use similarity transformations to rotate by the reflections, e.g. http://en.wikipedia.org/wiki/Spinors_in_three_dimensions#Formulation . These matrices are going to have to act on some vectors and we call those vectors in this new space spinors. Further note that A = U^-1 B U = (- U)^{-1} B (- U), which is the geometric origin of the double-cover-edness of the spinor representation.

So basically the unitary change of basis A = U^-1 B U along with a decomposition into the product of reflections is the intuitive origin of spinors in 3 dimensions as far as I can see, it makes sense because this represents changing your coordinate system basis, and spinors arise in QM when you rotate the z-axis and get your wave function to be a linear combination of 2l + 1 wave functions.

Would like to see someone develop my intuition a bit better and explain the 4-d Lorentz case via this product of reflections intuition :)

 

[samalkhaiat]

Spinors can be introduced only in a manifold admitting of the existence of null vectors. Indeed, if we consider the null cone (i.e., the set of null vectors) in 3D Euclidean complex manifold, then it is easy to show that a plane tangent to the null cone describes spinors. In fact, 2 out of the 3 coordinates of a current point in such a plane correspond to the two components of the spinor. Try to do it yourself, the math is very easy.

Sam

 

[pervect]

When I studied General Relativity using Misner, Thorne and Wheeler's "Gravitation", it was eye-opening to me to learn the geometric meanings of vectors, tensors, etc. The way such objects were taught in introductory physics classes were heavily dependent on coordinates: "A vector is a collection of 3 (or 4) numbers that transform in such and such way under rotations..." or whatever. I much prefer the geometric definition of a vector as a "tangent" to a smooth parametrized path, and a covector as a linear operator on vectors, etc.

But once you introduce spinors, it seems that the geometric way of understanding things fails. People are reduced to saying "a spinor is something that transforms in such-and-such a way under rotations" again.The mathematically sophisticated way of describing them in terms of representations of the group of rotations (or Lorentz boosts) don't really provide much more understanding of what they really "are".

Is there some geometric way of understanding spinors that is more in line with the "tangent to a parametrized path" understanding of a vector?

When you rotate an electron (or any fermion) through 360 degrees, the phase of its wave function inverts. The group of standard rotations in 3 dimensions, which mathematicians call SO(3), doesn't model this behavior of inverting the sign of the wave function. SO(3) means something like "special orthogonal group", see for instance http://en.wikipedia.org/wiki/Rotation_group_SO(3)

What does match the behavior of inverting the wavefunction is the group SU(2), see for instance http://en.wikipedia.org/wiki/Special_unitary_group. It's a 2 dimensonal complex matrix, a group of unitary transformations via paulii matrices. It's not particularly intuitive, but it's what works for spin 1/2 particles.

The mathematically relevant statement is that SU(2) is a double cover of SO(3). This is precisely the group structure you need to have a rotation through 360 degrees change the sign.

It's not particularly intuitive, but if you are used to rotations being matrices, they still are matrices, it's just that they are 2x2 complex matrices, the Paulii matrices, rather than 3x3 real matrices. It also fits in with the quantum mechanical idea that spin is represented as a complex amplitude for the |up> state, and another complex amplitude for the |down> state, said states being the only possible states you can observe. The Stern-Gerlach experiment demonstrates the quantitization of spin, the only two possible results of a spin measurement for a fermion are |up> and |down>, rather than the infinite number of results you'd get with a classical spinning particle.

From the standpoint of classical theory, the fact that SU(2) is a double cover of SO(3) is more of an annoyance than anything else, as you treat classical objects as having a classical 3-d spin axis. It's closely related (in my mind, at least) to the idea of using quaternions for representing rotations (this is used by some 3d modelling computer programs for efficiency reasons). The double-cover is more of an annoyance than a feature. But it turns out the double-cover structure of SU(2) to SO(3) is useful idea when you want to extend GR to handle particles with quantum spin,like Einstein–Cartan–Sciama–Kibblen theories. SU(2) is really the natural representation of the spin of a fermion. The whole idea is abstract and not particularly intuitive by my view, but it turns out to be useful. I believe there are some purely classical problems that the spinor approach makes a lot easier, though I couldn't provide any details.

 

[Ben Niehoff]

So, to return to the OP, I have thought a little bit about this, and frankly getting a fully geometrical view of spinors is hard. Spinors do not arise from any easily-visualized construction, such as vectors arising from equivalence classes of curves. However, I can assemble a few useful facts.

First of all, consider the 2-component spinor in $\mathbb{R}^3$ that I wrote about earlier:

$$\psi = \begin{pmatrix} \xi \\ \zeta \end{pmatrix}.$$
I asked if you knew an easy way to map this into a direction in 3-space, and while the construction you gave is correct, there is a more intuitive, geometrical way to think of it: Simply define the complex number

$$z = \frac{\xi}{\zeta}.$$
This complex number marks a point in the complex plane $\mathbb{C} \simeq \mathbb{R}^2$, which you can think of as the x-y plane. To map this onto some direction in 3-space, just use stereographic projection to map it up onto the unit sphere. You will find that SU(2) rotations of $\psi$ correspond to fractional linear transformations of $z$:

$$z \mapsto \frac{a z + b}{c z+ d},$$
where $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is a matrix in SU(2).

The use of stereographic projection here is important: it means we are discussing a projective representation of the rotation group. My knowledge of representation theory is a bit weak, but apparently for $n \geq 3$, the projective representations of SO(n) are on a one-to-one correspondence with linear representations of Spin(n):

http://en.wikipedia.org/wiki/Projective_representation

This works because for $n \geq 3$, Spin(n) is the universal cover of SO(n). For SO(2) there are an infinite tower of covering groups.

This would seem to imply that there is a similar interpretation of spinors in higher dimensions, but I'm not sure how straightforward it is. I haven't yet thought much about specifically how projective reps of SO(n) relate to Clifford algebras.

Another bit of a useful information comes from Stiefel-Whitney classes. It turns out that not every manifold actually admits spinors, and the obstruction to putting a spin bundle on a manifold is the second Stiefel-Whitney class. Stiefel-Whitney classes measure the obstruction to having a non-zero section of some exterior power of the tangent bundle. For example, if the first Stiefel-Whitney class vanishes, then you can have a globally non-zero n-vector; hence it measures the obstruction to your manifold being orientable. If the second Stiefel-Whitney class vanishes, then you can have a globally non-zero (n-1)-vector; and if you recall, the space of lines in $\mathbb{R}^n$ is the projective space $\mathbb{RP}^{n-1}$. Thus a globally non-zero (n-1)-vector gives you enough coordinates to discuss projective reps of the rotation group.

 

[marmoset]

If I remember rightly there is a description in Pertti Lounesto's book Clifford Algebras and Spinors (http://www.amazon.com/dp/0521005515/?tag=pfamazon01-20) of how a Pauli spinor field defines a basis at each point in three dimensional space and how a Dirac Spinor field is almost characterized by its bilinear covariants (four current, stress tensor etc. from Bjorken and Drell for example). I think the chapter on 'Fierz identities and boomerangs' shows something like that if a certain combination of the bilinear covariants isn't zero then they uniquely define a Dirac spinor or something like that - I can't remember and I don't have access to the book at the moment. If I've remembered right then this might be one way to make spinors intuitive, since if they are completely specified by more familiar objects (four currents, stress tensors, spin tensors etc.) then the coordinate independent understanding of the simpler fields would kind of carry over to the spinor field as a way of compactly encoding the simpler fields.

 

[Geometry_dude]

OP, I know exactly what you mean. The work of David Hestenes is the closest I have got so far in answering this question. Here's an article. (You probably find a free pdf somewhere online).

 

[haael]

@samalkhaiat is right. Spinors are built on complex numbers.

Speaking of geometrical intuitions:
@OP, what you know about vectors is false. The object you think of as a vector is in fact a spinor. Spinors are quite simple, it's the vectors that are bizzare.

Spinor is an object that carries the following information:
1. Direction.
2. Magnitude.
3. Information regarding number of times it has been rotated (odd/even).
I imagine it as "an arrow attached to a wall". Think of an arrow (say a pencil) attached to a wall using a ribbon. One edge of the ribbon is connected to the pencil, the opposite one is connected to the wall. The geometrical object that can be used to describe this system is a spinor. You have the direction the pencil is pointing, the length of the pencil and the number of wraps of the ribbon (only the parity).

You might have thought that this system (save for the ribbon) is in fact a vector. The concept of a vector in school is taught as something that has direction and magnitude. This is not true. A vector can describe much more. Something with direction and magnitude (plus parity number of wraps) is a spinor. A vector is a much more complex object.

The "parity of wraps" feature of a spinor is a reflection of the fact that certain systems pointing in some directions may be realized in more than one way. Imagine you want to build a robotic arm that can point (say a gun) in any direction. You can build it with solid blocks of any shape and rotating joints. However the joints can rotate only 360 degrees at maximum. They are similar to human joints in this respect.
How many joints do you need to build such an arm? Three. When you have the 3 joints, how many combinations of the joints angles are there to select one direction? There are two. You can encode any direction in two ways. A spinor is a full mathematical description of such a system. You may think of the joints rotations as of the Pauli matrices.

I must say one thing here, that spinors have nothing to do with quantum mechanics. There are purely classical objects. In fact, they need to be quantized to admit properties needed in QM. So we should talk about two kinds of objects: classical spinors and quantum spinors. Unfortunately, spinors are usually introduced in the quantum flavor, without the more simple classical case.

Let's drag the robotic arm analogy a bit further.
The arm is pointing in some direction. Its joints are set to some angle to encode that direction. Now:
1. If the joints have restricted angle range (say 0-360 deg) their rotations can be thought of as Pauli matrices and the setup is described by a spinor.
2. If the joints are unrestricted (they can rotate as many times as you wish) their rotations can be thought of as normal rotation matrices. You can build more devices using a limited count of such joints than with restricted ones and it reflects the fact that vectors carry more information than spinors.
3. If you have joints that can be set only in a discrete angle increments (i.e. with a ratchet), you get quantized rotations. The system is described by a quantum spin vector and the joints angles correspond to quantum spin numbers in certain direction.
As you may have guessed, an electron would correspond to a combination of 3 restricted joints with ratchets.

The most important thing to change in popularization of quantum mechanics in my opinion is the introduction of classical spinors. We should talk more about them, instead of introducing quantum spinors directly.

 

[Ben Niehoff]

@samalkhaiat is right. Spinors are built on complex numbers.

Samalkhaiat is talking about Cartan's "isotropic vectors", which is a very algebraic way to think about spinors. I'm not sure how to interpret such an idea geometrically, in real space.

Spinor is an object that carries the following information:
1. Direction.
2. Magnitude.
3. Information regarding number of times it has been rotated (odd/even).

This is a way of saying that the spinor transforms under a rep of the covering group. For SO(3) and higher, the universal cover is a double cover, hence why one needs to remember parity information.

I think the OP's real question is why one needs the covering group, and what this means geometrically. A spinor lives in a vector space at a point, so I don't really see the "attached by a ribbon" model being realized mathematically; there's no object sitting around to attach ribbons to. My post above is pointing out that the reps of the covering group are in fact related to projective reps of the group we started with, and it's really projective reps that we're interested in.

3. If you have joints that can be set only in a discrete angle increments (i.e. with a ratchet), you get quantized rotations. The system is described by a quantum spin vector and the joints angles correspond to quantum spin numbers in certain direction.
As you may have guessed, an electron would correspond to a combination of 3 restricted joints with ratchets.

This is not how quantized rotations work. Actually, this is a pretty bad way to think of quantization in general. Quantization is NOT the same thing as discretization.

 

[Ben Niehoff]

Spinors can be introduced only in a manifold admitting of the existence of null vectors.

Also, this is false. Spinors can be defined on any spin manifold (i.e., manifold with vanishing second Stiefel-Whitney class).

Indeed, if we consider the null cone (i.e., the set of null vectors) in 3D Euclidean complex manifold, then it is easy to show that a plane tangent to the null cone describes spinors.

The question is how to interpret these complex null vectors in the original, real tangent space.

 

[bolbteppa]

I think you guys are focusing on the wrong thing for intuition about spinors, as I explained in my post the intuition comes from thinking about the operators acting on the spinors not the spinors themselves. In fact I think it naturally motivates the whole reason for defining isotropic vectors in the first place also, and once you see this the space of spinors actually makes complete intuitive sense. If you re-read it and think about rotating forward then back to the original point I think that explains spinors completely, at least in the 3-d case (I hope!). Intuitively spinors themselves are just vectors parametrized by the components of another vector as it is rotated and are constructed using the idea that when you rotate the vector then rotate it back to it's original point nothing should happen therefore the operator representing this rotation acting on the spinors should do nothing. Yeah!

 

[stevendaryl]

I think you guys are focusing on the wrong thing for intuition about spinors, as I explained in my post the intuition comes from thinking about the operators acting on the spinors not the spinors themselves. In fact I think it naturally motivates the whole reason for defining isotropic vectors in the first place also, and once you see this the space of spinors actually makes complete intuitive sense. If you re-read it and think about rotating forward then back to the original point I think that explains spinors completely, at least in the 3-d case (I hope!). Intuitively spinors themselves are just vectors parametrized by the components of another vector as it is rotated and are constructed using the idea that when you rotate the vector then rotate it back to it's original point nothing should happen therefore the operator representing this rotation acting on the spinors should do nothing. Yeah!

I don't disagree with what you're saying, but my original point is that this way of thinking about spinors is very different in flavor from the differential geometry approach to thinking about vectors. If you a manifold $\mathcal{M}$, two natural things that you would like to do with that manifold are:

1. Define a "parametrized path", which is a function $\mathcal{P}(\lambda)$ of type $R \rightarrow \mathcal{M}$.
2. Define a "scalar field", which is a function $\phi(\mathcal{P})$ of type $\mathcal{M} \rightarrow R$.

(Tangent) vectors and 1-forms are very natural ways to characterized the local behavior of parametrized paths and scalar fields, respectively.

In contrast, a spinor, defined as something like "an irreducible rep of the rotation group" seems out of left field. I can, with work, understand what that means, mathematically, but unlike vectors and 1-forms, it's not clear what the motivation would be for studying such things. I mean, in hindsight, it's clear that they are useful in quantum mechanics in describing Fermions, and they are very handy for representing rotations (much nicer than Euler angles). But without the benefit of hindsight, it's hard to see how to motivate spinors.

 

[bolbteppa]

Yeah I agree but I think we just don't fully understand what's going on. Introducing a manifold makes it interesting, because you cannot exploit the isomorphism between R^n and it's tangent space in any way! My attempt at making things intuitive on a manifold is to say the following:

in QM we want to measure things, so let's measure something happening at a point of our manifold, with the wave function being our scalar field. The measurement is going to depend on how I set up my coordinate system to do the measurement, in other words the scalar wave function output will be dependent on how my tangent space basis looks, how it is rotated.

Remember, the way I described spinors above was in terms of an active rotation of a vector, right? But you can equivalently talk about passive rotations, in terms of rotating the coordinate system. So a spinor is just a way to parametrize how we've set up a coordinate system, the value of our scalar function will depend on the orientation of the tangent space basis at the point of measurement.

So, the natural loose intuitive way of motivating spinors is to say that if the value of a scalar function depends on how our coordinate system looks then we end up using spinors to describe that property.

This is heuristic and potentially wrong, but nice, can we build on it?

 

[Haelfix]

There is a whole section in MTF which describes Spinors using little 'flags' that rotate around but don't come back to the identity under 360 degree rotations, which they justify with some extra conditions. In some sense the obstruction to giving spinors a straightforward geometric interpretation is precisely the necessity for these extra conditions in order to make sense of the physics.

Mathematically, its easy to see that there is an extra phase that is required to be 'bundled' over each point and one is forced to impose at the very least an orthonormal set of basis vectors (so you don't need coordinates, but you at the very least need a Vielbein), b/c there is a convention that one is forced to choose and this has ramifications for the type of manifold that is allowed (as Ben points out, the manifold is required to be a spin manifold, eg a manifold with a vanishing 2nd Stiefel Whitney class).

Often with these things, it's probably possible in principle to generalize the mathematical objects in such a way as to make spinors natural geometric objects, (perhaps something along the lines of what the Clifford Algebra people do) but the downside is the base space will no longer have a natural physical interpretation.

 

[samalkhaiat]

Samalkhaiat is talking about Cartan's "isotropic vectors", which is a very algebraic way to think about spinors.

What is it that is “algebraic” about the statement “a plane tangent to the isotropic cone”? In fact, that description for spinor is not “less geometric” than describing vector as “tangent to a parameterized curve”.

I'm not sure how to interpret such an idea geometrically, in real space.

Theoretical physics does not need real space to interpret geometrical object.

 

[samalkhaiat]

Also, this is false.

Okay, to make you happy, the manifold I was talking about is one with trivial topology. In such manifold, the existence of isotropic vectors guarantees a unique spin structure [*].

Spinors can be defined on any spin manifold

Yeah, cucumbers can also be defined on any cucumber manifold.

(i.e., manifold with vanishing second Stiefel-Whitney class).

Since you mentioned it, I will give you an exercise to think about:
It is well-known that an oriented bundle $E$ with Euclidean (compact) structure group $SO(n)$ is spinable, i.e., admits a spin structure, if and only if the second Stiefel-Whitney class vanishes, i.e.$w_{2}(E) = 0$.
Of course, the situation for bundles with metrics of indefinite signature is quite different. So, let $E$ be an oriented bundle with (non-compact) structure group $SO(n,m)$.
What is/are the necessary and sufficient condition(s) for $E$ to admit spin structure?
If the base manifold $\mathcal{M}$ is not simply-connected, the spin structure will, in general, not be unique, why is that?
In this case, how many different spin structures are there? And, what cohomology group would you use to classify these different spin structures?
And finally, what is the relation between number of elements of this group and the number of different Dirac-type Lagrangians?

The question is how to interpret these complex null vectors in the original, real tangent space.

See page 48-56 in [*]

Sam

[*] R. Penrose & W. Rindler, “Spinors and space-time” Vol 1, Cambridge University Press 1990.
See also:
R. Geroch (1968), J. Math. Phys. 9, 1739.
G. Wiston (1974), Int. J. Theor. Phys. 11, 341.

 

[Ben Niehoff]

What is it that is “algebraic” about the statement “a plane tangent to the isotropic cone”? In fact, that description for spinor is not “less geometric” than describing vector as “tangent to a parameterized curve”.

Yes, I thought you might object for that reason.

I say this construction is algebraic because of the following: I hand you a real vector space, say $\mathbb{R}^n$, and ask you how to interpret spinor representations of SO(n). You respond with, "Well, let's look at what happens when SO(n) acts on the complexification of your vector space, $\mathbb{C}^n$". Who ordered that? And how does it relate to my question?

While it's certainly a way of arriving at spinor reps, from the perspective of real vector spaces, it doesn't seem any less algebraic than the usual route involving Clifford algebras. After all, the "tangent plane to the isotropic cone" is not a thing that exists in $\mathbb{R}^n$ with Euclidean inner product. Since a spinor is essentially a "kind of thing that can be rotated", one hopes to find some intuitive way to interpret them without resorting to such constructions.

Even in vector spaces with indefinite quadratic form, I don't think you mean to imply that spinors are literally the same thing as null vectors. You would still have to look at the null cone in the complexification $\mathbb{C}^{m+n}$.

Since you mentioned it, I will give you an exercise to think about:
It is well-known that an oriented bundle $E$ with Euclidean (compact) structure group $SO(n)$ is spinable, i.e., admits a spin structure, if and only if the second Stiefel-Whitney class vanishes, i.e.$w_{2}(E) = 0$.
Of course, the situation for bundles with metrics of indefinite signature is quite different. So, let $E$ be an oriented bundle with (non-compact) structure group $SO(n,m)$.
What is/are the necessary and sufficient condition(s) for $E$ to admit spin structure?
If the base manifold $\mathcal{M}$ is not simply-connected, the spin structure will, in general, not be unique, why is that?
In this case, how many different spin structures are there? And, what cohomology group would you use to classify these different spin structures?
And finally, what is the relation between number of elements of this group and the number of different Dirac-type Lagrangians?

Interesting things to think about, thanks. I'm pretty sure the obstruction to having a spin structure in indefinite signature is still $w_2(E)$, though.

 

[Spinnor]

If I might pester the experts, spinors exist in space time where null vectors are defined but from readings spinors also "live" in Calabi Yau manifolds? Could the experts here give an example of simple closed spaces where spinors "live" (what is the right term, defined, exist?).

Thanks.

 

[Haelfix]

Spin structure in indefinite signature is significantly more complicated, if I recall some classes in string theory from grad school. I think the trick was that one had to split the orientable bundle E into a diirect sum of a positive definite part and a negative definite part, and then the obstruction to having a spin structure was that the difference of the second Stieffel Whitney class applied to both pieces had to vanish. I would be incapable of proving this piece of memory though.

 

[Spinnor]

Spin structure in indefinite signature is significantly more complicated, if I recall some classes in string theory from grad school. I think the trick was that one had to split the orientable bundle E into a diirect sum of a positive definite part and a negative definite part, and then the obstruction to having a spin structure was that the difference of the second Stieffel Whitney class applied to both pieces had to vanish. I would be incapable of proving this piece of memory though.

Thank you. I guess all those buzz words don't apply to the humble torus, no spinors on a torus?

 

[Ben Niehoff]

Thank you. I guess all those buzz words don't apply to the humble torus, no spinors on a torus?

You can have spinors on the torus, as well as every orientable Riemann surface.

 

[Spinnor]

You can have spinors on the torus, as well as every orientable Riemann surface.

That is cool! Thank you. You can't know too many facts about spinors and now I know 2 more!

The Wiki article on spinors has a table summary that lists "objects" that exist for a given signature of Euclidean space (which I guess are different types of spinors)?

Can there be different signatures on a torus that also give rise to different types of spinors?

http://en.wikipedia.org/wiki/Spinor#Summary_in_low_dimensions

Thanks for the help!.

 

[lavinia]

Spin structure in indefinite signature is significantly more complicated, if I recall some classes in string theory from grad school. I think the trick was that one had to split the orientable bundle E into a diirect sum of a positive definite part and a negative definite part, and then the obstruction to having a spin structure was that the difference of the second Stieffel Whitney class applied to both pieces had to vanish. I would be incapable of proving this piece of memory though.

The reason, I think, is that if the SO(n,m) bundle is (n,m)-orientable, that is, the structure group can be reduced to the connected component of the identity, then the bundle splits into a Whitney sum of two oriented sub-bundles and a Spin(n,m) structure reduces to a Spin structure on each summand. It follows that the second Stiefel Whitney class of each summand must be zero.

 

[lavinia]

While I don't know anything about Spin geometry, in order to answer the original question which was how do Spinors play a role in Differential Geometry, perhaps this may be helpful to you. I am paraphrasing a section of the introduction to the book "Spin Geometry" by Lawson and Michelson.

The tangent bundle of a manifold has structure group the general linear group. If the structure group can be lifted to the universal covering group, the manifold is a Spin manifold. Sadly, the universal covering group has no finite dimensional representations other than those obtained by projecting onto the general linear group and then following the projection with a representation of GL(n). So from this point of view Spin structures do not provide any new geometry.

However with a Riemannian metric on an orientable manifold, one can reduce the structure group to SO(n), and the universal cover, Spin(n), does in fact have independent finite dimensional representations. This means that the metric allows one to construct new vector bundles which do not exist on manifolds that do not have a Spin structure.

These "bundles of Spinors" depend on the choice of metric and thus should be expected to be linked with the geometry of the manifold.

Interestingly, the authors say that mathematicians are trying to formulate a "spinor calculus" analogous to tensor calculus but at the time of the writing of the book, it has not been well defined.

I imagine that since the Spin(n) principle bundle covers the principle SO(n) bundle that a connection on the SO(n) bundle lifts to a connection on the principal Spin(n) bundle. This connection together with a finite dimensional representation of the Spinor group will produce a covariant differentiation on the induced Spinor bundle.

 

[samalkhaiat]

Spin structure in indefinite signature is significantly more complicated, if I recall some classes in string theory from grad school. I think the trick was that one had to split the orientable bundle E into a diirect sum of a positive definite part and a negative definite part, and then the obstruction to having a spin structure was that the difference of the second Stieffel Whitney class applied to both pieces had to vanish. I would be incapable of proving this piece of memory though.

The complicated detail was worked out by Karoubi in 1968[*] I summarize his results here:
If an orientable bundle $E$ has structure group $SO(n,m)$, one can always reduce the structure group to its maximal compact subgroup $S(O(n) × O(m))$. This reduction is equivalent to the choice of a maximal positive definite sub-bundle $E^{+} < E$, which allows us to write $E$ as the direct sum, $E = E^{+} \oplus E^{-}$ , where $E^{-} = (E^{+})^{\perp}$ is negative definite. Then, one can show that the $SO(n, m)$ bundle $E$ admits $\mbox{Spin} (n, m)$ structure if and only if $w_{2} (E^{+}) = w_{2} (E^{-})$.

[*] M. Karoubi (1968). "Algèbres de Clifford et K-théorie". Ann. Sci. Éc. Norm. Sup.1 (2): 161–270.

 

[lavinia]

The complicated detail was worked out by Karoubi in 1968[*] I summarize his results here:
If an orientable bundle $E$ has structure group $SO(n,m)$, one can always reduce the structure group to its maximal compact subgroup $S(O(n) × O(m))$. This reduction is equivalent to the choice of a maximal positive definite sub-bundle $E^{+} < E$, which allows us to write $E$ as the direct sum, $E = E^{+} \oplus E^{-}$ , where $E^{-} = (E^{+})^{\perp}$ is negative definite. Then, one can show that the $SO(n, m)$ bundle $E$ admits $\mbox{Spin} (n, m)$ structure if and only if $w_{2} (E^{+}) = w_{2} (E^{-})$.

I would like to see the proof. It seems that the condition only guarantees that the second Stiefel-Whitney class of the bundle is zero which means that it has a Spin(p+q) structure not a Spin(p,q) structure. For that I would think the second Whitney class of each summand must be zero.

* Roughly, the universal classifying space for SO(p,q) structures ought to be BSO(p) x BSO(q) up to homotopy equivalence so that its second Z2 cohomology would be generated by the two universal Stiefel-Whitney classes(Kunneth formula) of the two factors and the condition that the manifold admits a Spin(p,q) structure should imply that the induced map on the second Z2 cohomology group is zero.

One can define a Spin structure as a cohomology class in $H^1$(SO(p) bundle;Z2) whose restriction to $H^1$(Fiber;Z2) is surjective for each fiber,F. (For details see http://retro.seals.ch/cntmng?pid=ensmat-001:1963:9::66 )

It follows that the transgression mapping $t: H^1$(Fiber) $-> H^2$(Manifold) is zero.
However, in the universal bundle the transgression mapping is injective. So by a diagram chasing argument the classifying map for the bundle must map the universal second Stiefel-Whitney class to zero.

For an SO(p,q) bundle the argument is the same except the cohomology coefficients are Z2xZ2 rather than Z2 and there are now two universal second Stiefel Whitney classes, one for each factor.

 

[pellis]

.
You might have thought that this system (save for the ribbon) is in fact a vector. The concept of a vector in school is taught as something that has direction and magnitude. This is not true. A vector can describe much more. Something with direction and magnitude (plus parity number of wraps) is a spinor. A vector is a much more complex object.
.

Interesting thread on a perennial topic.

Could you please expand on the ways in which you reckon that a vector "is a much more complex object"? (I realize that your use of "complex" here is in the natural language rather than mathematical sense).

 

[pellis]

Could you please expand on the ways in which you reckon that a vector "is a much more complex object"? .

Further: Having looked at the various uses of the term vector in maths ( http://en.wikipedia.org/wiki/Vector_(mathematics_and_physics) ) it seems that for almost every entry one can replace the word vector by the word spinor, e.g. vector field, vector space, vector calculus, vector bundle etc. all have their equivalents as spinor field, spinor space, spinor calculus, spinor bundle. The more specific terms like Darboux vector or wave vector refer to physical situations represented by vectors, just as there are specific applications of spinors, e.g pulse synthesis in medical NMR spectroscopy.

Can you still support your contention that a vector "is a much more complex object" than a spinor, please, haael?

Regards - pellis

 

[lavinia]

But once you introduce spinors, it seems that the geometric way of understanding things fails. People are reduced to saying "a spinor is something that transforms in such-and-such a way under rotations" again.The mathematically sophisticated way of describing them in terms of representations of the group of rotations (or Lorentz boosts) don't really provide much more understanding of what they really "are".

Is there some geometric way of understanding spinors that is more in line with the "tangent to a parametrized path" understanding of a vector?

While this doesn't really answer your question I find these ideas helpful.

- First of all, on a general vector bundle, there is no natural geometric way to interpret the vectors other than algebraically as elements of a vector space at each point of the manifold.

When one wants to take directional derivatives, or covariant derivatives, one must have tangent vectors and it is the tangent bundle which is naturally geometric. One can have some generalized ideas of geometry on vectors bundles using a connection but intuition goes away quickly and differentiation is still done with respect to tangent vectors.

- When a manifold is oriented its structure group can be reduced to SO(n) from O(n). An orientation is not a property of a vector but rather of the entire vector space.

By analogy one can think of a Spin structure as an "orientation" of an oriented manifold.

A manifold is orientable when its first Stiefel-Whitney class is zero and then it has two orientations ,one for each element of $H^0(M;Z2)$

Similarly, if an oriented manifold has a zero second Stiefel Whitney class then it has one Spin structure for each element of $H^1(M;Z2)$.
This can be seen from the exact cohomology sequence (with Z2 coefficients),

$0 ->H^1$(Manifold) ->$H^1$(SO(p) Principal bundle) ->$H^1$(SO(p)) ->$H^2$(Manifold)

In the case that the second Stiefel-Whitney class is zero the last term on the right may be taken to be zero. Since $H^1(SO(p)) = Z2$
the sequence becomes

$0 ->H^1$(Manifold) -> $H^1$(SO(p) Principal bundle) -> Z2 -> 0

So each element of $H^1$(Manifold) corresponds to a Spin structure on the bundle of oriented orthonormal frames.