Element of the spin group

Question

I've got the following question: why it true, 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 real vector space.

Answer 1

You use a non-standard definition of the Clifford algebra (up to sign), but it matters nothing with respect to Spin groups.

If $x\mapsto u\,x\,u^{-1}$ is a special orthogonal operator on V and u is known to be even, then $u\in\mathrm{Spin}(V)$. But if parity of u is unknown, then we can’t be sure u belongs to Spin(V). BTW, you are also confused about adjoint – we don’t need any involutions to define Spin if u is even.