Penrose paragraph on Bundle Cross-section?

[Abhishek11235]

I am reading "Road to Reality" by Rogen Penrose. In chapter 15, Fibre and Gauge Connection ,while going through Clifford Bundle, he says:

.""""...Of course, this in itself does not tell us why the Clifford bundle has no continuous cross-sections. To understand this it will be helpful to look at the Clifford bundle in another way. In fact, it turns out that each point of our sphere $S^{3}$ can be interpreted as a unit-length ‘spinorial’ tangent vector to $S^{2}$ at one of its points. Recall from §11.3 that a spinorial object is quantity which, when completely rotated through 2$\pi$, becomes the negative of what it was originally. According to the above statement, a cross-section of our bundle $\text{B }\S^{3}$ would represent a continuous field of suchs pinorial unit vectors on $\text{M} S^{2}$. Now, it is a well-known topological fact that there is no global continuous field of ordinary unit tangent vectors on $S^{2}$. (This is the problem of combing the hair of a ‘spherical dog’! It is impossible for the hairs to lie flat in a continuous way, all over the sphere.) Making these directions ‘spinorial’ clearly does not help, so no global continuous field of unit spinorial tangent vectors can exist either. Hence our bundle B $S^{3} has no cross-sections...""""""""
Here $\text{M}$ is Manifold.

How did he conclude that Bundle B has no cross-Section?

P.S: I am new to differential Geometry.

 

[lavinia]

How did he conclude that Bundle B has no cross-Section?

He is saying that a field of non-zero spinors on the 2 sphere would determine a field of non-zero tangent vectors on the 2 sphere. But no field of non-zero tangent vectors on the 2 sphere exists.

Which would you like to see demonstrated? That there is no non-zero field of tangent vectors on the 2 sphere or a field of spinors determines a field of tangent vectors?

 

[Abhishek11235]

He is saying that a field of non-zero spinors on the 2 sphere would determine a field of non-zero tangent vectors on the 2 sphere. But no field of non-zero tangent vectors on the 2 sphere exists.

Which would you like to see demonstrated? That there is no non-zero field of tangent vectors on the 2 sphere or a field of spinors determines a field of tangent vectors?

Both

 

[lavinia]

1) Non-zero vector fields on the sphere.

Here are sketches of a couple of proofs of this theorem. I don't know how technical you want to get.

If there is a non-zero vector field then one can divide each vector by its length to get a vector field $s$ of unit length vectors. On the sphere these tangent unit vectors form a circle at each point. Across the whole sphere they form a circle bundle that is homeomorphic to the real projective space $RP^3$.

One proof shows that if $ω$ is a connection 1 form on the tangent circle bundle then the exterior derivative $dω$ is the pull back of the Gauss Bonnet form $KdV$. In the sphere this form is not exact but if there were a non-zero section then it would be. ($KdV$ is not exact because the Gauss-Bonnet theorem says that its integral is $2π$ time the Euler Characteristic which for the sphere is $2$).- If $π:RP^3→S^2$ is the bundle projection map then $π \circ s$ is the identity map on the sphere. If $ω$ is the volume form on the sphere, then $ω= s^{*}(π^{*}ω)$ where the 2-form $π^{*}ω$ is the pull back of $ω$ to $RP^3$. $π^{*}ω$ is closed and every closed form in $RP^3$ is exact. So $ω= s^{*}(π^{*}ω)$ is exact which is impossible since its integral over the sphere is the volume of the sphere.

- There is a general theorem for compact manifolds that says that if the tangent bundle has a non-zero vector field, then its Euler-Characteristic is zero. The Euler characteristic of the sphere is $2$.

2) Non-zero spin field means non-zero vector field.

There is a smooth map from $π:S^3→RP^3$ that maps antipodal points to the same point. In Penrose's description this map throws away the spin and leaves a tangent vector to the 2 sphere. So a non zero section $σ:S^2→S^3$ projects to a non-zero section of the tangent bundle of the sphere.-

 

[fresh_42]

1) Non-zero vector fields on the sphere.

Probably a silly question, but just for my understanding: the emphasis here is that the vector field has to be on the entire sphere, is that correct? And I can follow the arguments for tangent fields, but not for arbitrary vector fields, e.g. if we took a normal vector at each point of the sphere.

 

[lavinia]

Probably a silly question, but just for my understanding: the emphasis here is that the vector field has to be on the entire sphere, is that correct? And I can follow the arguments for tangent fields, but not for arbitrary vector fields, e.g. if we took a normal vector at each point of the sphere.

Yes for the entire sphere.

A vector bundle over the sphere other than the tangent bundle might have a non-zero section everywhere for instance the unit normal to the sphere is a non-zero section of the normal bundle.

 

[fresh_42]

Just a side note for amusement:

This theorem has a funny German name: "Satz vom Igel" - "Theorem of the Hedgehog". Its short version is "You can not curry the hedgehog." which in German is an alliteration "Man kann den Igel nicht striegeln."

 

[lavinia]

- All of these proofs assume that the vectors fields are smooth. But the theorem is true if they are only continuous. For this one needs different arguments.

One proof (See Milnor's,Topology from the Differentiable Viewpoint) is to show that the existence of a non-zero section of the tangent bundle of a sphere implies that the antipodal map, the map that sends each point on the sphere to its opposite pole, is homotopic to the identity map, the map that sends each point to itself. For even dimensional spheres this cannot happen because the antipodal map has degree $-1$ while the identity map has degree $+1$ and any two homotopic maps of a sphere into itself must have the same degree. For instance, a complex quadratic polynomial can not be homotopic to a complex cubic polynomial.

- A homotopy between two maps $f$ and $g$ from the topological space $X$ into the topological space $Y$ is a continuous map $H$ from the Cartesian product $X×[a,b]$ of $X$ with the interval $[a,b]$ into $Y$ such that $H(x,a) = f(x)$ and $H(x,b) = g(x)$. A homotopy is a path of maps from $f$ to $g$.

- If $H$ happens to be smooth and $X$ and $Y$ are oriented closed $n$ dimensional manifolds of the same dimension e.g. $X$ and $Y$ are both the 2 sphere, then one can use Stokes Theorem. Let $ω$ be a volume form on $Y$ with respect to some Riemannian metric. The pull back form $H^{*}ω$ is a closed $n$ form on $X×[a,b]$. By Stokes Theorem

$0 = ∫_{X×[a,b]}dH^{*}ω = ∫_{∂(X×[a,b])}H^{*}ω = ∫_{X}f^{*}ω-∫_{X}g^{*}ω$

If $f$ is orientation preserving and $g$ is orientation reversing the integrals $∫_{X}f^{*}ω$ and $∫_{X}g^{*}ω$ must have opposite sign.

 

[lavinia]

Just a side note for amusement:

This theorem has a funny German name: "Satz vom Igel" - "Theorem of the Hedgehog". Its short version is "You can not curry the hedgehog." which in German is an alliteration "Man kann den Igel nicht striegeln."

One might ask, "Who would want to comb a hedgehog?".

 

[mathwonk]

i like the "proof " in a book of lefschetz. you look at one point of the sphere where the tangent vector is non zero, and draw a small circle around that point at every point of which circle the tangent vector is nearly the same as at the given point, true by continuity. then excise the interior of the circle and consider the remaining vectors as a tangent vector field to the disc outside that circle. If you let that region outside the circle snap back flat, inside the given circle, you can "see" that at points of the boundarty circle, your tangent vector field points directly out at one point, directly in at the opposite point, and is tangent to the boundary circle at the two points equally spaced from those antipodal points. If you look carefully you can then see that the winding number of this vector field around the boundary circle is 2, hence there must be 2 zeroes inside the disc, counted with multiplicities. This proof thus shows more than just the existence of zero, it counts them.

note that originally, when you draw the little circle, that the vector field, being essentially parallel along the circle, points in at one point, say 9 o'clock, and out at the opposite point, say 3 o'clock , and is tangent to the circle at the two intermediate points, thus noon and 6 o'clock. Hence when you cut out the interior and let the exterior snap back inside, as if it were rubber, the vector that used to point in at 9 o'clock now points out at 9 o'clock, and the one that pointed out at 3 o'clock now points in at 3 o'clock, and the ones at noon and 6 o'clock still point tangent to the boundary circle in the same direction as before. It took me a while to visualize this, but it is nice eventually.

Phrased slightly differently, assume the vector at the north pole is non zeo and draw a small circle around that pole. Then the winding number of that vector field about the north pole is zero, since the direction of the vectors does not change much, and there is no zero vector inside the circle. Now note however that the winding number of that same vector field with respect to the south pole is 2, so there are 2 zeroes outside the circle, if counted with multiplicities.

aside: do hedgehogs roll up in a ball when examined closely? otherwise what does this have to do with the sphere problem?

 

[fresh_42]

aside: do hedgehogs roll up in a ball when examined closely? otherwise what does this have to do with the sphere problem?

I guess it's meant as a mnemonic and less as a biological description.

 

[lavinia]

I guess it's meant as a mnemonic and less as a biological description.

@mathwonk et al.

Hedgehogs I think do roll up in a ball in a defensive posture. I imagine their bristly fur sticks out like porcupine quills. Trying to comb them can lead to nasty bites.

 

[lavinia]

i like the "proof " in a book of lefschetz. you look at one point of the sphere where the tangent vector is non zero, and draw a small circle around that point at every point of which circle the tangent vector is nearly the same as at the given point, true by continuity. then excise the ointerior of the circle and cosider the remaining vectors as a tangent vector fild to the disc outside that circle. If you let that region outside the circle snap back flat, inside the given circle, you can "see" that at points of the boundarty circle, your tangent vector field points directly out at one point, directly in at the opposite point, and is tangent to the boundary circle at the two points equally spaced from those antipodal points. If you look carefully you can then ee that the winding number of this vector field around the boubndary circle is 2, hence there must be 2 zeroes inside the disc, counted with multiplicities. This proof thus shows more than just the existence of zero, it counts them.

Lovely proof.

 

[cosmik debris]

One might ask, "Who would want to comb a hedgehog?".

A hedgehog curry on the other hand...

Cheers

 

[lavinia]

@Abhishek11235

The Clifford algebra that Penrose is talking about is the quaternions. The Clifford bundle is a bundle over the 2 sphere whose total space is the quaternions. A non-zero section of this bundle would determine a section of its unit circle bundle by dividing each quaternion in the section by its length. One then has a section consisting of length one vectors so it lies in the unit sphere in four dimensional space( centered at the origin). This sphere is $S^3$. (In general a sphere in $n$ dimensions is a set of points that are equidistant from a central point.)

The fiber above any point in the bundle $S^3→S^2$ is a circle though not a tangent circle. Rather it is a circle of unit quaternions.

The Hopf fibration can be described explicitly in terms of unit length quaternions as follows.
If $z= a+bi+cj+dk$ is a unit quaternion then $H(a+bi+cj+dk) = (2(ad + bc), 2(bd − ac),a^2 + b^2 − c^2 − d^2)$

If $w$ is an point in a fiber circle, the the entire fiber circle is the points $e^{iθ}w$ where $e^{iθ}$ is the quaternion cosθ + sinθ$i$ + 0$j$ +0$k$.

Example of a fiber:

If $w=k$, the north pole of the three sphere, then $H(z)=(0,0,-1)$

Multiplying $w$ by $e^{iθ}$ gives -sinθ$j$+cosθ$k$.

$H(-$sinθ$j$+cosθ$k) = (0,0,-$sin$^2$θ$-$cos$^2$θ$)=(0,0,-1)$.

Conversely if $H(a+bi+cj+dk) = (0,0,-1)$ then $a^2+b^2-c^2-d^2 = -1$ while $a^2+b^2+c^2+d^2 = 1$ so $a$ and $b$ are both zero.

Quaternions and rotations of 3 space:

If one thinks of 3 dimensional space as the pure quaternions $xi+yj+zk$ (Pure means its purely real coordinate is zero.) then conjugation by an arbitrary quaternion $w = a+bi+cj+dk$ is a rotation of the pure quaternions with rotation axis the straight line through $(b,c,d)$ and rotation angle 2cos$^{-1}a/|w|$.

Two quaternions that differ by a real scalar multiple determine the same rotation so every rotation is determined by two antipodal points on the the 3 sphere $S^3$. Identifying these antipodal points with their rotation determines a 2 fold covering $S^3→SO(3)$. This interprets unit quaternions as rotations with a directed rotation axis. Forgetting the direction of the axis leaves just a rotation.

This covering map is also a group homomorphism so the unit quaternions get interpreted as the group of rotations of 3 space with directed rotation axis.

Rotations of 3 space as the unit tangent circle bundle to the 2 sphere:

If $v$ is a unit vector tangent to the 2 sphere at $p$ rotate $v$ by 90 degrees to get a positively oriented tangent frame $(v,iv)$. Add the point $p$ to get a positively oriented frame $(v,iv,p)$ in 3 space. Every positively oriented orthonormal frame is spanned in this way at some point $p$ on the sphere. So the positively oriented orthonormal frames in $R^3$ are in 1-1 correspondence with the unit tangent circle bundle to the 2 sphere. But the rotations of $R^3$ are in 1-1 correspondence with the positively oriented orthonormal frames in $R^3$ as well.

Moreover this is not just a 1-1 correspondence. Every rotation maps the unit 2 sphere into itself and its differential maps the unit tangent circle bundle into itself. From the Chain Rule it follows that $SO(3)$ actually acts on the tangent circle bundle and this action is smooth, transitive, and has no fixed points. This implies that $SO(3)$ and the tangent circle bundle to the sphere are diffeomorphic.

 

[Klystron]

My Summer reading list includes R.Penrose "The Road to Reality" after reading it back in 2005. Began reading the book this week and this quote caught my imagination, from 3.3 Real Numbers in the Physical World:

"...I tried this sort of thing [description of reality], coming up with a scheme that I referred to as the theory of 'spin networks' ... later transmogrified into 'twistor theory',..."

My impression from this section, footnotes, other Penrose books and comments in contemporary physics books is, perhaps, obvious: Roger Penrose tends to emphasize spin in his physical descriptions. Indulging in an unequal comparison, while investigating 'sparse matrices' during master's work in computer science; all my professional papers found a logical use for a sparse matrix. A vexing problem programming localized sound via a MIDI interface for human factors studies in a flight simulator -- solved via adroit manipulation of sparse matrices. Turned out to be a correct solution as processing speed and memory were constrained but also my preference from my personal study. Spin it will be!

By the time I reach the section referenced by the OP, I'm certain to appreciate Penrose's use of spin to describe reality as much as I enjoy his use of hyperbolic geometry in understanding Riemann space. I'll re-read this thread for clarity.