- #1
Spinnor
Gold Member
- 2,216
- 430
From: http://www.sjsu.edu/faculty/watkins/spinor.htm
"Let X=(x1, x2, x3) be an element of the vector space C^3. The dot product of X with itself, X·X, is (x1x1+x2x2+x3x3). Note that if x1=a+ib then x1·x1=x1^2=a2+b2 + i(2ab), rather that a2+b2, which is x1 times the conjugate of x1.
A vector X is said to be isotropic if X·X=0. Isotropic vectors could be said to be orthogonal to themselves, but that terminology causes mental distress."
also from the same web page:
"It is impossible to visual depict isotropic vectors and spinors because three dimensional complex vectors involve six dimensions and spinors as two dimensional complex vectors involve four dimensions."
I would like to see these isotropic vectors. Does this help or work?
For an isotropic vector,
X·X = x1x1+x2x2+x3x3 = 0, so,
x1x1+x2x2 = -x3x3 (This defines a surface in C^3? What are some of its symmetries?)
Let us plot the real parts of x1 and x2 on a 3D graph with z = 0, with a red point and at the same time plot the imaginary parts of x1 and x2 with a green point. These two points lead to two pairs of solutions to x1x1+x2x2 = -x3x3? Plot the solutions on the above graph's z axis coloring real and imaginary points as above. Imagine a java program that allowed us to move the red and green plot points in the z=0 plane and have it automatically calculate solutions to,
x1x1+x2x2 = -x3x3
such a program might allow us to investigate the complex surface X·X=0 ? Is Java hard to learn? Is it expensive?
Also from the same web page:
"It can be shown that the set of isotropic vectors in C^3 form a two dimensional surface. This two dimensional surface can be parametrized by two coordinates, z0 and z1 where
z0 = [(x1-ix2)/2]1/2
z1 = i[(x1+ix2)/2]1/2.
The complex two dimensional vector Z=(z0, z1) Cartan calls a spinor. "
Since z0 and z1 are complex does this surface need four numbers to label a point of this surface?
Thank you for any thoughts.
"Let X=(x1, x2, x3) be an element of the vector space C^3. The dot product of X with itself, X·X, is (x1x1+x2x2+x3x3). Note that if x1=a+ib then x1·x1=x1^2=a2+b2 + i(2ab), rather that a2+b2, which is x1 times the conjugate of x1.
A vector X is said to be isotropic if X·X=0. Isotropic vectors could be said to be orthogonal to themselves, but that terminology causes mental distress."
also from the same web page:
"It is impossible to visual depict isotropic vectors and spinors because three dimensional complex vectors involve six dimensions and spinors as two dimensional complex vectors involve four dimensions."
I would like to see these isotropic vectors. Does this help or work?
For an isotropic vector,
X·X = x1x1+x2x2+x3x3 = 0, so,
x1x1+x2x2 = -x3x3 (This defines a surface in C^3? What are some of its symmetries?)
Let us plot the real parts of x1 and x2 on a 3D graph with z = 0, with a red point and at the same time plot the imaginary parts of x1 and x2 with a green point. These two points lead to two pairs of solutions to x1x1+x2x2 = -x3x3? Plot the solutions on the above graph's z axis coloring real and imaginary points as above. Imagine a java program that allowed us to move the red and green plot points in the z=0 plane and have it automatically calculate solutions to,
x1x1+x2x2 = -x3x3
such a program might allow us to investigate the complex surface X·X=0 ? Is Java hard to learn? Is it expensive?
Also from the same web page:
"It can be shown that the set of isotropic vectors in C^3 form a two dimensional surface. This two dimensional surface can be parametrized by two coordinates, z0 and z1 where
z0 = [(x1-ix2)/2]1/2
z1 = i[(x1+ix2)/2]1/2.
The complex two dimensional vector Z=(z0, z1) Cartan calls a spinor. "
Since z0 and z1 are complex does this surface need four numbers to label a point of this surface?
Thank you for any thoughts.
