Is it correct to state that the elements of $Spin(n)$ fulfill a Clifford algebra and that the Lie group generators of $Spin(n)$ is given by the commutator of the elements?

If not, then what is the relation of the $Spin(n)$ group, the matrices that fulfill the Clifford algebra (that can be used to construct $SO(n)$ bilinears from spinors, like the Dirac gamma matrices) and the generators of $Spin(n)$ as a Lie group?


1 Answer 1


No, elements of $Spin(n)$ don't obey the Clifford algebra. Instead, it's the gamma matrices that obey it. And no, the commutator of the $Spin(n)$ Lie algebra isn't the commutators of the elements of the group but elements of the Lie algebra.

Now positively.

The spinor representation is the representation on which the generators $J_{ij}$ (the basis of the Lie algebra) act, $s\mapsto J_{ij}\cdot s$. The elements of the $Spin(n)$ group may be obtained by exponentiation: $$ g = \exp(\sum_{i,j} i\omega_{ij}J_{ij}) $$ where $J_{ij}$ is the basis of the Lie algebra. While in the vector representation, $J_{ij}$ is given by a nearly vanishing $n\times n$ matrix with entries $\pm i$ on the $i,j$ and $j,i$ position, respectively, the matrices $J_{ij}$ have a completely different form in the spinor representation of $Spin(n)$. They may be written as $$ J_{ij} = \frac{\gamma_i \gamma_j - \gamma_j \gamma_i}{4} $$ where $\gamma_i$ are gamma matrices that do obey the Clifford algebra $$ \gamma_i \gamma_j + \gamma_j \gamma_i = 2\delta_{ij}\cdot {\bf 1}$$ So the matrices obeying this algebra may be combined to bilinear expressions, the antisymmetric tensor $J_{ij}$ with two indices, and these $J_{ij}$ obey the $Spin(n)$ Lie algebra, and as with every Lie algebra, the elements of the Lie groups may be obtained by exponentiating combinations of the Lie algebra matrices. (Equivalently, the Lie algebra is the tangent space of the Lie group manifold in the vicinity of the unit element of the group.)


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