1
$\begingroup$

For the Euclidean case, we have that spinors transform under the $\mathbf{2}$ representation of (the double cover of) $SO(3,\mathbb{R})$. It would seem to me that since vectors live in $\mathbf{3}$, and have three real free components before we impose any field equations, and rank-$2$ tensors have $\mathbf{3}\otimes\mathbf{3} = \mathbf{1}\oplus\mathbf{3}\oplus\mathbf{5} \rightarrow 9$ real degrees of freedom, spinors ought to have two free real components.

Obviously a spinor $\psi$ cannot be taken to be always real, but this would imply that any one spinor could be transformed into a basis where $\psi\in\mathbb{R}^2 $. However usually the spinor is presented as having four free components: two real and two imaginary. It is not clear to me why simply because there is no real form of the $2$-dimensional representation, that we must switch to complex dimension instead of real (Indeed, $SU(2)$ is a real Lie Algebra of real dimension $3$, despite the presence of complex values; I don't see why this equivocation between real and complex dimension happens suddenly for spinors)

Improve this question
$\endgroup$
6
  • $\begingroup$ See arxiv.org/abs/hep-th/9910030 for a count of the number of real degrees of freedom for many (all?) types of spinors. $\endgroup$
    – Connor Behan
    Commented Jan 24, 2022 at 21:47
  • $\begingroup$ @Connor Behan Thanks for the suggestion, but there they seem to state '4' again without much explanation. I am surely missing something, but I am not sure what. $\endgroup$
    – Craig
    Commented Jan 24, 2022 at 22:09
  • $\begingroup$ @Craig : spinors are not vectors in physical space. The spin group of $SO(3,\mathbb R)$ is $SU(2)$ and spinors are the complex two-component vectors that the $SU(2)$ matrices act on. Thus, those spinors have four real degrees of freedom. As you say: two real and two imaginary. $\endgroup$
    – Kurt G.
    Commented Jan 25, 2022 at 15:43
  • $\begingroup$ @Kurt G. This would imply the tensor product of two spinors ought to decomposes not as $\mathbf{2}\otimes\mathbf{2} = \mathbf{1}\oplus\mathbf{3}$ but as $\mathbf{2}\otimes\mathbf{2} = \mathbf{1}_\mathbb{C} \oplus\mathbf{3}_\mathbb{C}$, no? $\endgroup$
    – Craig
    Commented Jan 25, 2022 at 17:36
  • 1
    $\begingroup$ @Craig : The tensor product of $\mathbb C^2\otimes\mathbb C^2$ (two gimmicks , sorry, spinors) is the space of complex $2\times 2$ matrices, or, if you will, complex $4$-component vectors. It sounds like by $1_{\mathbb C}\oplus 3_{\mathbb C}$ you mean the same. $\endgroup$
    – Kurt G.
    Commented Jan 25, 2022 at 18:34

1 Answer 1

Reset to default
2
$\begingroup$

For what it's worth, the defining/fundamental/spinor representation $$\bf{2}~\cong~\mathbb{C}^2~\cong~\mathbb{H}$$ of the Lie group $$Spin(3)~\cong~SU(2)~\cong~ U(1,\mathbb{H})$$ is a quaternionic/pseoudoreal representation, i.e. it is not a real representation.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Wild Guess: Are all even dimensional projective representations of $SO(3,\mathbb{R})$ pseudoreal, while the odd ones are all real? $\endgroup$
    – Craig
    Commented Jan 25, 2022 at 18:26
  • 1
    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic ♦
    Commented Jan 25, 2022 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.