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I Shape defined by x∈ℂ^3 and x⋅x = 0.

  1. Mar 6, 2016 #1
    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?
  2. jcsd
  3. Mar 6, 2016 #2


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    Since a and b are real, the above requires that every coordinate of a point in the set ##\{X\in\mathbb{C}^3 | X \cdot X=0\}## is zero. So the set is simply the origin of ##\mathbb{C}^3##.

    When I went web-searching for isotropic vectors I didn't come up with any definition for them except for the thread you posted here in 2009 asking about the same topic.

    'Isotropic' is a term that can be applied to bilinear forms. Metrics in spacetime are isotropic in that sense. In that context, we can call null vectors, which are lightlike vectors, isotropic.

    But the standard metric on ##\mathbb{C}^3## that you have described above is not isotropic, so the set of all isotropic vectors in ##\mathbb{C}^3## with that metric is just the zero vector. There is no 2D surface.
  4. Mar 7, 2016 #3
    I am sorry, there is an important typo in the link above. As written,

    "x·x=x^2=a^2+b^2 + i(2ab), rather that a^2+b^2"

    should be,

    x·x=x^2=a^2-b^2 + i(2ab), rather that a^2+b^2. The minus sign is important.

    So we have,

    x⋅x = (x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2 = 0

    I hope the above is clear as the original quote was not..

  5. Mar 7, 2016 #4
    So a post of andrewkirk's disappeared but think I remember most of it. So as for the physics content of my question I agree that this is probably a dead end, but still maybe an interesting math question. The link I gave of Thayer Watkins website was material that he probably got from the book of the famous mathematician Elie Cartan, The Theory of Spinors? I butchered the question, sorry. Here is a relevant page from that book, (which Google lets you read a bit)

    https://books.google.com/books?id=f-_DAgAAQBAJ&printsec=frontcover&dq=the+theory+of+spinors&hl=en&sa=X&ved=0ahUKEwi9063Y-K_LAhWLFx4KHWFvCGAQ6AEIHTAA#v=onepage&q=the theory of spinors&f=false


    So what is the shape of the surfacec defined by x∈ℂ^3 and x⋅x = 0? My first guess is that it is a
    boring 4 dimensional infinite space. My second guess is that because it is complex maybe something interesting is going on. We have,

    x⋅x = (x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2 = 0 or

    (x_1^2 + x_2^2 + x_3^2) - (y_1^2 + y_2^2 + y_3^2) = 0 and

    (x_1y_1 + x_2y_2 + x_3y_3) = 0 So what?

    x⋅x = (x_1 + iy_1)^2 + (x_2 + iy_2)^2 + (x_3 + iy_3)^2 = z_1^2 + z_2^2 + z_3^2 = 0 or

    z_1^2 + z_2^2 = - z_3^2 So what?

    Is it true that 2z_1dz_1 + 2z_2dz_2 = - 2z_3dz_3 ? Can this help to understand the surface or do we need more powerful mathematics? Looking for simple ways at first to understand this surface.

    Last edited: Mar 7, 2016
  6. Mar 7, 2016 #5
    I don't really know about spinors, but this is an interesting math question.

    Let's say a typical point in ℂ3 is called (u, v, w), so we can avoid subscripts. Then according to the Watkins notes*, a spinor is the locus S of the equation

    u2 + v2 + w2 = 0,​

    or in other words, the set

    S = {(u, v, w) ∈ ℂ3 | u2 + v2 + w2 = 0}.​

    Note that this is a homogeneous polynomial, meaning that each monomial has the same degree (in this case, 2). Homogeneity implies that if (u, v, w) ∈ S, then the same is true of any complex multiple of it:

    K(u, v, w) = (Ku, Kv, Kw) ∈ S​

    where K is an arbitrary complex number, which is easy to verify. Now let's ignore the point (0, 0, 0) of S, which means that we will also exclude the complex number K from taking the value 0. Then the set

    S* = S - {0, 0, 0}​

    is the disjoint union of sets of the form

    P(a, b, c) = {K(a, b, c) | K ∈ ℂ*},​

    where ℂ* denotes ℂ - {0} and (a, b, c) is any arbitrary but fixed point of ℂ3.

    Now, each such set P(a, b, c) is just a copy of ℂ*, which may be thought of as determining a "complex line" (actually a 2-dimensional real plane) in ℂ3.

    The set of all complex lines in ℂ3 is a 4-dimensional manifold (i.e., it has 2 complex dimensions) known as the "complex projective plane" and denote by the symbol ℂℙ2. It turns out to be the most natural space to study the locus of a homogeneous polynomial in 3 complex variables.

    A beautiful fact is that the topology of the set of points in ℂℙ2 that occur as complex lines P(a, b, c) that form part of the locus S is a compact orientable surface of genus g (i.e., a "g-holed torus") — and that the genus g is a simple function of the degree d of the original homogeneous polynomial, as follows:

    g = (d-1)(d-2)/2.

    In our case, d = 2 and so g = 0. This means that topologically, the locus in ℂℙ2 is a 2-dimensional sphere S2.

    Backing up to the original locus S, this means that if we remove (0, 0, 0) from S to get what we called S*, then S* is the union of one copy of C* for each point of a 2-sphere S2.

    * Of course, u2 + v2 + w2 is not the usual "Hermitian" inner product on ℂ3 of (u, v, w) with itself, which would be u conj(u) + v conj(v) + w conj(w) = |u|2 + |v|2 + |w|2.
  7. Mar 8, 2016 #6


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    Yes. I wrote something, then decided after a few minutes that it was actually rubbish, so I deleted it.

    That happens sometimes :biggrin:
  8. Mar 8, 2016 #7
    Thank you Zinq! Much to digest this evening.
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