# How many real degrees of freedom do Euclidean spinors have in 3D Euclidean space?

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)

• See arxiv.org/abs/hep-th/9910030 for a count of the number of real degrees of freedom for many (all?) types of spinors. Commented Jan 24, 2022 at 21:47
• @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. Commented Jan 24, 2022 at 22:09
• @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. Commented Jan 25, 2022 at 15:43
• @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? Commented Jan 25, 2022 at 17:36
• @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. Commented Jan 25, 2022 at 18:34

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.
• Wild Guess: Are all even dimensional projective representations of $SO(3,\mathbb{R})$ pseudoreal, while the odd ones are all real? Commented Jan 25, 2022 at 18:26
• $\uparrow$ Yes. Commented Jan 25, 2022 at 18:29