# Do the spinor transformation matrices form a matrix representation of the corresponding Lorentz group?

Suppose $$\Psi$$ is a Dirac spinor, then let the transformation matrix $$S$$ be defined as usual: $$\Psi'=S(\Lambda)\Psi$$, where $$\Lambda$$ is the Lorentz transformation matrix.

Then the questions is: for two Lorentz transformations $$\Lambda_1$$ and $$\Lambda_2$$, do $$S(\Lambda_1)$$ and $$S(\Lambda_2)$$ obey $$S(\Lambda_1)S(\Lambda_2)=S(\Lambda_1\Lambda_2)$$ and $$S(\Lambda)S(\Lambda^{-1})=I~?$$

They form a complex projective representation in which

$$S(\Lambda_1)S(\Lambda_2)=c(\Lambda_1,\Lambda_2)S(\Lambda_1\Lambda_2)$$

where $$c(\Lambda_1,\Lambda_2)$$ is a number.

For an arbitrary group, the $$c(\Lambda_1,\Lambda_2)$$ for its complex projective representations will in general be complex numbers. But in the case of the Lorentz group, complex projective representations can always be chosen to have $$c(\Lambda_1,\Lambda_2)=\pm 1$$.

See the Wikipedia articles on Projective Hilbert space and Wigner’s Theorem to begin to understand why projective representations (and not just ordinary representations) are relevant in quantum mechanics.

In some cases the projective representations of a group are equivalent to the ordinary representations of its universal covering group. For more about this, see this question.

No. For example, if $$Λ$$ is a 180-degree spatial rotation, then $$S(Λ)^2 = -I \ne I = S(Λ^2)$$.