Andrius Kulikauskas investigates:
What do spinors say about Bott periodicity?
I am trying to understand what are spinors. Partly, that will help me understand the physics that John Harland talks about with me, especially as regards the Dirac equation and relativity. But also, spinors manifest Bott periodicity, and I am interested to what insight they may provide.
Spin {$m$} indicates a division of everything into {$2m+1$} perspectives
In particular, I expect that spin {$m$} indicates the division of everything into {$n$} states where {$n=2m+1$}. Thus if {$m=\frac{1}{2}$} then there are two states: {$-\frac{1}{2},\frac{1}{2}$}. Whereas if {$m=1$} there are three states: {$-1,0,1$}. If {$m=0$} there is one state {$0$}.
Spinors are defined by what is done to them
Spinors are not defined by what they are. Instead, spinors are defined by what is done with them. Similarly, a "paperweight", a "piece of trash", a "sales item", a "hat" are all defined not by what they are but what is done with them.
Spinors are elements of a complex vector space
Spinors are elements of the vector spaces that are given by representations of spin groups. Currently, I am studying the Wikipedia article on Spin representation and the recipe it gives for building them.
- I am trying to understand spinors in the simplest case, the group {$\textrm{Spin}(3)\cong SU(2)$} which is the double cover of the special orthogonal group {$SO(3)$} of the rotations of the sphere.
- I should study Wikipedia: Spinors in three dimensions
- Later I should understand the spin statistics theorem
Here are my notes at Math4Wisdom: Spinors
{$\textrm{Spin}(3)\cong SU(2)$}
I am interpreting the article Wikipedia: Spin representation in terms of {$\textrm{Spin}(3)\cong SU(2)$}.
Quadratic form {$Q$} and symmetric bilinear form {$\langle\;,\;\rangle $}
{$V=\mathbb{C}^3$} with orthonormal basis {$z_1=(1,0,0),z_2=(0,1,0),z_3=(0,0,1)$} and quadratic form {$Q(c_1,c_2,c_3)=Q((c_1,c_2,c_3))=c_1^2+c_2^2+c_3^2$}.
The quadratic form {$Q$} defines a symmetric bilinear form {$\langle\;,\;\rangle $} using a polarization identity. Note that this is not sesquilinear and note also that the basis elements commute (they are not anticommmuting). The polarization identities allow us to characterize an angle between vectors in terms of the squared lengths of the two vectors, their sum and their difference. Specifically:
{$2\langle u, v\rangle = Q(u+v)-Q(u)-Q(v)$}
{$2\langle u, v\rangle = Q(u)+Q(v)-Q(u+v)$}
{$4\langle u, v\rangle = Q(u+v)-Q(u+v)$}
Isotropic spaces {$W$} and {$W^*$}
{$v\in V$} is isotropic if {$Q(v)=0$}. Subspace {$W$} of {$V$} is totally isotropic if all of its elements are isotropic.
We have isotropic elements {$z_1+iz_2$} and {$z_1-iz_2$} and isotropic spaces {$W=\{\lambda(z_1+iz_2) | \lambda\in\mathbb{C}\}$} and {$W^*=\{\gamma(z_1-iz_2) | \gamma\in\mathbb{C}\}$}. In this way, we can pair up the original basis elements.
{$V$} is odd-dimensional so we also define {$U=\{\beta x_3 | \beta\in\mathbb{C}\}$}
Note that {$\langle z_1 + iz_2, z_1 - iz_2\rangle = \langle z_1, z_1 \rangle - i\langle z_1, z_2 \rangle + i\langle z_2, z_1 \rangle - i^2\langle z_2, z_2\rangle = Q(z_1) + Q(z_2) = 1 + 1 = 2$} where two terms cancelled out because of symmetry.
Thus we have two isotropic elements {$z_1+iz_2$} and {$z_1-iz_2$} for which the bilinear form {$\langle z_1 + iz_2, z_1 - iz_2\rangle$} is not zero.
{$V=W\oplus U \oplus W^*$} where the three latter spaces are all one-dimensional.
{$V$} acts on the exterior algera {$S=\wedge^\bullet W$}
{$V$} acts on the exterior algebra {$S=\wedge^\bullet W$}, which is likewise one-dimensional. This action has {$v=w+w^*\in W\oplus W^*$} act on {$\psi\in S$} by
{$v\cdot\psi=\sqrt{2}(w\wedge\psi + \iota(w^*)\psi)$}
{$\iota(w^*)$} is defined by the bilinear form {$\langle x_1+ix_2, x_1-ix_2\rangle $} which relates {$W$} and {$W^*$}. We have {$w^*=\gamma (x_1-ix_2)$} thus {$\iota(w^*)=\iota(\gamma (x_1-ix_2))=\langle x_1+ix_2, \gamma (x_1-ix_2)\rangle $}