From: http://www.sjsu.edu/faculty/watkins/spinor.htm(adsbygoogle = window.adsbygoogle || []).push({});

"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.

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# Trying to understand isotropic vectors.

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