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I ##GL(n,\mathbb{R})## Spin Reps

  1. Jun 13, 2017 #1
    Motivated by a statement in GSW (discussed here and here), in this closed thread

    https://www.physicsforums.com/threads/why-there-are-no-spinors-for-gl-n.240240/

    a discussion asking 'Why there are no spinors for GL(n)' is had.

    In trying to follow the proof in one of the references they mention in the pf thread, from here, sec. 177, we want the prove the theorem at the bottom of the page:

    eW8y0.png

    but exclusively for ##x_i## in ##\mathbb{R}##, as is needed in GSW (the proof above does it over ##\mathbb{C}##). Most of the proof, mimicking the above as much as possible, seems to go as follows, for ##D > 2##:

    If ##A^{\mu} \mapsto A'^{\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}} A^{\nu}## where the matrix ##\frac{\partial x'^{\mu}}{\partial x^{\nu}}## is an element of ##GL(D,\mathbb{R})## admits tensor representations, we want to ask if the spin (projective) representations of ##SO(D)## come out of restrictions from ##GL(D,\mathbb{R})##. Were we able to obtain operators in the spin representation by elements ##\frac{\partial x'^{\mu}}{\partial x^{\nu}} \in GL(D,\mathbb{R})##, it would mean we could restrict the element corresponding to ##x_1' = x_1 \cos(\theta) - x_1 \sin(\theta)##, ##x_2' = x_1 \sin(\theta) + x_2 \cos(\theta)## to an element of the spin representation. But we can see this restriction is double-valued from

    $$\begin{aligned}
    \zeta_0 &= \pm \sqrt{\frac{x_1-ix_2}{2}} \\
    \zeta_1 &= \pm \sqrt{ \frac{-x_1-ix_2}{2} }
    \end{aligned},$$
    where the components change sign after a rotation by ##2 \pi##
    $$\begin{aligned}
    \zeta_0' &= \pm \sqrt{\frac{x_1'-ix_2'}{2}} = \pm \sqrt{\frac{e^{i 2 \pi}(x_1-ix_2)}{2}} = \mp \sqrt{\frac{x_1-ix_2}{2}} \\
    \zeta_1' &= \pm \sqrt{ \frac{-x'_1-ix'_2}{2} } = \mp \sqrt{ \frac{-x_1-iyx_2}{2} }
    \end{aligned}$$
    showing that the ##\frac{\partial x'^{\mu}}{\partial x^{\nu}} \in GL(D,\mathbb{R})## must restrict to both ##(\zeta_0,\zeta_1)## and ## (-\zeta_0,-\zeta_1)##, that is, the group of transformations ##x_1' = x_1 \cos(\theta) - x_1 \sin(\theta)##, ##x_2' = x_1 \sin(\theta) + x_2 \cos(\theta)## would have a double-valued spin representation.

    Now comes the iffy part.

    (If we follow the proof over ##\mathbb{C}##, at this stage we finish the proof with a topological argument - we have found a multivalued representation of the matrix
    $$ \begin{bmatrix} a && b \\ c && d \end{bmatrix} = \begin{bmatrix} \cos(\theta) && -\sin(\theta) \\ \sin(\theta) && \cos(\theta) \end{bmatrix}$$
    by the spinors ##(zeta_1,zeta_2)## and ##(-\zeta_1,-\zeta_2)##, but multi-valued representations of complex matrices of this form with non-zero determinant are impossible, as the group of matrices is simply connected, so that following a closed path so that the final point is represented by a different spinor to the initial one, and then we shrink the path down to a point at the origin without modifying the final spinor, we obtain a contradiction.)

    We need to modify this over ##\mathbb{R}## as the group is no longer simply connected. The jist of the proof seems to be in showing we cannot set up a faithful representation of the group of matrices, by using real projective transformations
    $$z' = \frac{az+b}{cz+d},$$
    setting ##z = \tan(x), z' = \tan(x')## and then modifying ##a,b,c,d## so that we end up with ##z' \pm \pi##, but I am not clear on this part of the proof and how it actually finishes the proof.

    Any help on finishing this proof clearly, really appreciate it!? The proof is given here, sec. 86:

    asa.jpg

    asb.jpg
     
    Last edited: Jun 13, 2017
  2. jcsd
  3. Jun 18, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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