clifford-algebragroup-theorylie-algebramathematical physicsrepresentation-theory
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?
It's pretty annoying that P&S just give you $$S^{\mu \nu} = \frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]$$ from thin air, here is a way to derive it similar to Bjorken-Drell's derivation (who start from the Dirac equation) but from the Clifford algebra directly, assuming that products of the gamma matrices form a basis. Given a Clifford algebra of $\gamma^{\mu}$'s satisfying \begin{align} \{ \gamma^{\mu} , \gamma^{\mu} \} = 2 \eta^{\mu \nu} I \end{align} we note that for an invertible transformation $S$ we have \begin{align} 2 \eta^{\mu \nu} I &= 2 \eta^{\mu \nu} S^{-1} S \\ &= S^{-1}(2 \eta^{\mu \nu}) S \\ &= S^{-1}\{ \gamma^{\mu} , \gamma^{\mu} \} S \\ &= \{ S^{-1} \gamma^{\mu} S, S^{-1}\gamma^{\mu} S \} \\ &= \{ \gamma'^{\mu} , \gamma'^{\mu} \} \end{align} showing us that the Clifford algebra of matrices $$\gamma'^{\mu} = S^{-1} \gamma^{\mu} S$$ also satisfies the Clifford algebra, hence any set of matrices satisfying the Clifford algebra can be obtained from a given set $\gamma^{\mu}$ using a non-singular transformation $S$. Since the anti-commutation relations involve the metric $\eta_{\mu \nu}$, and we know the metric is left invariant under Lorentz transformations $$\eta^{\mu \nu} = \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} \eta^{\rho \sigma} $$ this immediately implies \begin{align} 2 \eta^{\mu \nu} I &= \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} 2 \eta^{\rho \sigma} I \\ &= \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} \{ \gamma^{\rho} , \gamma^{\sigma} \} \\ &= \{ \Lambda^{\mu} \, _{\rho} \gamma^{\rho} , \Lambda^{\nu} \, _{\sigma} \gamma^{\sigma} \} \\ &= \{ \gamma'^{\mu} , \gamma'^{\mu} \} \end{align} which shows that the Lorentz transformation of a gamma matrix also satisfies the Clifford algebra, and so is itself a gamma matrix, and hence can be expressed in terms of some non-singular transformation $S$ \begin{align} \gamma'^{\mu} &= \Lambda^{\mu} \, _{\nu} \gamma^{\nu} \\ &= S^{-1} \gamma^{\mu} S \end{align} where $S$ is to be determined. Since the operators $S$ represent performing a Lorentz transformation on $\gamma^{\mu}$, and Lorentz transformations on fields expand as $I - \frac{i}{2}\omega_{\mu \nu} M^{\mu \nu}$, we expand $\Lambda$ and $S$ as \begin{align} \Lambda^{\mu} \, _{\nu} &= \delta ^{\mu} \, _{\nu} + \omega^{\mu} \, _{\nu} \\ S &= I - \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu} \end{align} where $\Sigma^{\mu \nu}$ must be anti-symmetric and constructed from a basis of gamma matrices, hence from \begin{align} \gamma'^a &= \Lambda^a \, _{\mu} \gamma^{\mu} \\ &= (\delta^a \, _{\mu} + \omega^a \, _{\mu})\gamma^{\mu} \\ &= \gamma^a + \omega^a \, _{\mu} \gamma^{\mu} \\ &= \gamma^a + \omega_{b \mu} \eta^{a b} \gamma^{\mu} \\ &= \gamma^a + \omega_{b \mu} \eta^{a [b} \gamma^{\mu]} \\ &= \gamma^a + \frac{1}{2} \omega_{b \mu} (\eta^{a b} \gamma^{\mu} - \eta^{a \mu} \gamma^b) \\ &= \gamma^a + \frac{1}{2} \omega_{\nu} (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) \\ &= S^{-1} \gamma^a S \\ &= (I - \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}) \gamma^a (I + \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}) \\ &= \gamma^a - \frac{i}{2} \omega_{\mu \nu} [\gamma^a, \Sigma^{\mu \nu}] \end{align} we have the relation (which can be interpreted as saying that $\gamma^a$ transforms as a vector under spinor representations of Lorentz transformations, as e.g. in Tong's QFT notes) \begin{align} i (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) = [\gamma^a, \Sigma^{\mu \nu}] \end{align} and we know $\Sigma^{\mu \nu}$, since it is anti-symmetric, must involve a product's of $\gamma$ matrices (because of the 16-dimensional basis formed from Clifford algebra elements), only two by the left-hand side, and from \begin{align} \gamma^{\mu} \gamma^{\nu} &= - \gamma^{\nu} \gamma^{\mu} , \ \ \ \mu \neq \nu, \\ \gamma^{\mu} \gamma^{\mu} &= \gamma^{\nu} \gamma^{\mu} , \ \ \ \mu = \nu, \end{align} we expect that \begin{align} \Sigma^{\mu \nu} &= c [\gamma^{\mu},\gamma^{\nu}] \\ &= c (\gamma^{\mu} \gamma^{\nu} - \gamma^{\nu} \gamma^{\mu}) \\ &= 2 c ( \gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu}) \end{align} for some $c$ which we constrain by the (vector) relation above \begin{align} i (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) &= [\gamma^a, \Sigma^{\mu \nu}] \\ &= c [\gamma^a, 2( \gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu})] \\ &= 2 c [\gamma^a, \gamma^{\mu} \gamma^{\nu}] \\ &= 2 c ( \gamma^{\mu} [\gamma^a,\gamma^{\nu}] + [\gamma^a, \gamma^{\mu}] \gamma^{\nu}) \\ &= 2 c [ \gamma^{\mu} 2( \gamma^a \gamma^{\nu} - \eta^{a \nu}) + 2( \gamma^a \gamma^{\mu} - \eta^{a \mu}) \gamma^{\nu}] \\ &= 4 c [ \gamma^{\mu} ( \gamma^a \gamma^{\nu} - \eta^{a \nu}) + ( \gamma^{\mu} \gamma^{a} + 2 \eta^{a \mu} - \eta^{a \mu}) \gamma^{\nu}] \\ &= 4 c (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}). \end{align} This gives the result $c = i/4$. The generator of Lorentz transformations of gamma matrices is \begin{align} \Sigma^{\mu \nu} &= \dfrac{i}{4} [\gamma^{\mu},\gamma^{\nu}] \\ &= \dfrac{i}{2}(\gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu}) \ \ \text{i.e.} \\ S &= I - \frac{i}{2} \omega_{\mu \nu} (\frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]) \\ &= I + \dfrac{1}{8} \omega_{\mu \nu} [\gamma^{\mu},\gamma^{\nu}]. \end{align} Using the fact that the gamma matrices transform as a vector under the spinor representation of an infinitesimal Lorentz transformation, \begin{align} [\Sigma^{\mu \nu}, \gamma^{\rho}] = i (\gamma^{\mu} \eta^{\nu \rho} - \gamma^{\nu} \eta^{\mu \rho}) \end{align} we can show the spinor representation of a Lorentz transformation satisfies the Lorentz algebra commutation relations, since for $\rho \neq \sigma$ \begin{align} [\Sigma^{\mu \nu},\Sigma^{\rho \sigma}] &= \frac{i}{2}[\Sigma^{\mu \nu},\gamma^{\rho} \gamma^{\sigma}] \\ &= \frac{i}{2}([\Sigma^{\mu \nu},\gamma^{\rho} ] \gamma^{\sigma} + \gamma^{\rho} [\Sigma^{\mu \nu}, \gamma^{\sigma}]) \\ &= \frac{i}{2}\{ i (\gamma^{\mu} \eta^{\nu \rho} - \gamma^{\nu} \eta^{\mu \rho}) \gamma^{\sigma} + \gamma^{\rho} i (\gamma^{\mu} \eta^{\nu \sigma} - \gamma^{\nu} \eta^{\mu \sigma}) \} \\ &= - \frac{1}{2}\{ \gamma^{\mu} \eta^{\nu \rho} \gamma^{\sigma} - \gamma^{\nu} \eta^{\mu \rho} \gamma^{\sigma} + \gamma^{\rho} \gamma^{\mu} \eta^{\nu \sigma} - \gamma^{\rho} \gamma^{\nu} \eta^{\mu \sigma} \} \\ &= \frac{i}{2}\{ \eta^{\nu \rho} (2 \Sigma^{\mu \sigma} + \eta^{\mu \sigma}) - \eta^{\mu \rho} (2 \Sigma^{\nu \sigma} - \eta^{\nu \sigma}) + (2 \Sigma^{\rho \mu} - \eta^{\rho \mu}) \eta^{\nu \sigma} - (2 \Sigma^{\rho \nu}) - \eta^{\rho \nu}) \eta^{\mu \sigma} \} \\ &= i ( \eta^{\nu \rho} \Sigma^{\mu \sigma} - \eta^{\mu \rho} \Sigma^{\nu \sigma} + \Sigma^{\rho \mu} \eta^{\nu \sigma} - \Sigma^{\rho \nu} \eta^{\mu \sigma} ). \end{align} This method generalizes from $SO(3,1)$ to $SO(N)$, see e.g. Kaku QFT Sec. 2.6, and the underlying reason for doing any of this in the first place is that one seeks to find projective representations which arise due to the non-simple-connectedness of these orthogonal groups. Regarding your question about arbitrary metrics $g_{\mu \nu}$, this method applies to, and arises due to the non-simple-connectedness of, special orthogonal groups, you can't generalize to arbitrary metrics, this is a problem which can be circumvented in supergravity and superstring theory using veilbein's.
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Best Answer
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.)