This reduction maps to which group? General linear group reduced to spin group then squared.

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Suppose a multivector $\mathbf{u}$ of $G(\mathbb{R}^{3,1})$ (a multivector of geometric algebra 3+1 dimensions.

$$ \mathbf{u} = \exp (a + \mathbf{x} + \mathbf{f} + \mathbf{v} + \mathbf{b}) $$

where $a$ is a scalar, $\mathbf{x}$ is a vector, $\mathbf{f}$ is a bivector, $\mathbf{v}$ is a pseudovector and $\mathbf{b}$ is a pseudoscalar.

Clearly, $\mathbf{u}$ is a realization of ${\rm GL}^+(4,\mathbb{R})$.


My first question,

If I pose $\mathbf{x}\to 0, \mathbf{v} \to 0$, what group does it map to?

$$ \mathbf{u}|_{\mathbf{x}\to 0, \mathbf{v}\to 0}=\exp (a+\mathbf{f}+\mathbf{b}) $$


My second question,

If I now square the result, what group does it map to?

$$ \mathbf{u} \to \mathbf{u}^2 $$

becomes

$$ (\mathbf{u}|_{\mathbf{x}\to 0, \mathbf{v}\to 0})^2=\exp 2(a+\mathbf{f}+\mathbf{b}) $$