- #1
Spinnor
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Shape defined by x∈ℂ^3 and x⋅x = 0.
From, http://www.sjsu.edu/faculty/watkins/spinor.htm
"The concept of spinor is now important in theoretical physics but it is a difficult topic to gain acquaintance with. Spinors were defined by Elie Cartan, the French mathematician, in terms of three dimensional vectors whose components are complex. The vectors which are of interest are the ones such that their dot product with themselves is zero.
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 x=a+ib then x·x=x^2=a^2+b^2 + i(2ab), rather that a^2+b^2, which is x times the conjugate of x.
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.
It can be shown that the set of isotropic vectors in C3 form a two dimensional surface."
Are there some simple ways to try and understand a bit about the shape of this surface?
From, http://www.sjsu.edu/faculty/watkins/spinor.htm
"The concept of spinor is now important in theoretical physics but it is a difficult topic to gain acquaintance with. Spinors were defined by Elie Cartan, the French mathematician, in terms of three dimensional vectors whose components are complex. The vectors which are of interest are the ones such that their dot product with themselves is zero.
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 x=a+ib then x·x=x^2=a^2+b^2 + i(2ab), rather that a^2+b^2, which is x times the conjugate of x.
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.
It can be shown that the set of isotropic vectors in C3 form a two dimensional surface."
Are there some simple ways to try and understand a bit about the shape of this surface?