I've got the following question: why is it true (if it really is?), that if I have a unitary element $u$ in the (real) Clifford algebra $Cl(V,g)$ which is even and the operator $\varphi(u)$ defined via $\varphi(u)(x)=uxu^{-1}$ is an element of $SO(V,g)$ then actually $u \in Spin(V)$? Here $g$ is assumed to be positive definite, Clifford algebra is defined via identifying $xy+yx$ wich $2g(x,y)$ and the adjoint is therefore $x^*=x$ for $x \in V$, $V$ denoting the underlying real vector space.
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$\begingroup$ For me this is the definition of $\mathrm{Spin}(V)$. What is yours? $\endgroup$– abxSep 7, 2014 at 18:03
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$\begingroup$ $Spin(V)$ is a subgroup of invertible elements in the even part of Clifford algebra, generated by unit vectors in $V$. Each element in $Spin(V)$ is a unitary, is contained in $Cl(V,g)$ and one shows that $\varphi(u)$ is therefore in $SO(V)$. The above question is about the converse. $\endgroup$– truebaranSep 7, 2014 at 18:38
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$\begingroup$ Good question. This is the group called $SP(V,g)$ in Lawson+Michelsohn's Spin Geometry. I don't recall it being shown to be precisely the Spin group. $\endgroup$– José Figueroa-O'FarrillSep 7, 2014 at 19:21
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2$\begingroup$ If you look at the Wikipedia page on Clifford algebras, I think that you will see that you are asking about the even unitary elements in the Clifford group. $\endgroup$– Robert BryantSep 7, 2014 at 21:08
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$\begingroup$ Thank You for Your comment. So, as I understood correctly (in general setting): if $u$ is such that $\varphi(u) in SO(V)$ then $u \in \Gamma$ where $\Gamma$ is so called Clifford group. If $n=2m$ and $g$ is nondegenerate one can show that $\Gamma$ is the group generated by nonzero vectors. If $n=2m+1$ there is only an inclusion but for the even parts still we have equality. Therefore $u=v_1...v_{2k}$ and if we assume that $u$ is unitary, then we easily see that $u \in Spin(V)$? $\endgroup$– truebaranSep 8, 2014 at 20:34
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