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Consider the spin group, we know it is a double cover with the map:

$\rho: Spin(n,n)\longrightarrow SO(n,n)$ s.t $\rho(x)(v)= xvx^{-1}$ where $v$ is an element of 2n dimensional vector space V and $x$ is an element of spin group (multiplications are Clifford multiplication). I read that this map induces a lie algebra representation given by: $d\rho:so(n,n) \longrightarrow so(n,n)$ s.t $d \rho_{x}(v)=xv-vx$ here $x$ is an element of $so(n,n)$ and $v$ is again an element of V.

I cannot understand the derivation of this lie algebra representation. Can anyone help me? :)

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up vote 4 down vote accepted

This basically comes from the product rule of differentiation. Recall that for a general Lie algebra homomorphism $\rho : G\to H$ you can compute its derivative $d\rho: Lie(G) \to Lie(H)$ by the formula $$ d\rho(X) = \frac{d}{dt}\vert_{t=0} \rho(\exp(tX)). $$ In the case you have, this gives $$ d\rho(x) (v) = \frac{d}{dt}\vert_{t=0} \exp(tx) v \exp(-tx), $$ where $x \in so(n,n)$, which is identified with the second filtration of the Clifford algebra. Now you can think of the right hand side is taking place in the Clifford algebra and the product rule of differentiation holds so you get $$ d\rho(x)(v) = \frac{d}{dt} \vert_{t=0} \exp(tx) v \exp(0x) + \exp(0x) v \frac{d}{dt}\vert_{t=0} \exp(-tx) = xv +v(-x). $$

Notice that this formula does not make sense if you think of $so(n,n)$ in terms of $2n \times 2n$ matrices since right multiplying a column vector by a matrix doesn't make sense. The $xv - vx$ is taking place in the Clifford algebra.

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Very neat and clear. Thank you very much. – muratguner Jul 15 '12 at 15:01

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