In these notes 1, the author claims on page 7 that
One can define the Lie algebra of Spin(n) in terms of quadratic elements of the Clifford algebra. This is what we will do here.
The author then proceeds with creating an exponential of bivector.
My understanding is that exponentials of bivector are rotors. And that rotors are in SO(n), not spin(n).
How is the author able to claim a rotor is representation of spins? I cannot see where (or if) he modifies the rotor in any way to get the spin.
The other element I do not understand is that I was told by a user on this site 2 that ${\rm Spin}(n)$ is not in ${\rm GL}(n,\mathbb{R})$. However, rotors are. Thus, if one represents ${\rm Spin}(n)$ with rotor, wouldn't that imply ${\rm Spin}(n)$ is in ${\rm GL}(n,\mathbb{R})$ (thus contradiction).
The thing is that "exponentials" are not well-defined in themselves. If you have a Lie group $G$ with Lie algebra $\mathfrak{g}$, you have an exponential map $\mathfrak{g}\to G$, but it depends on $G$, not just on $\mathfrak{g}$. And there are usually several groups $G$ with the same $\mathfrak{g}$ (they are of course not unrelated).
So here $\mathfrak{so}(n)$ is the Lie algebra of $SO(n)$ and of $\operatorname{Spin}(n)$ (because they are isogenous), which means there are exponential maps $\mathfrak{so}(n)\to SO(n)$ and $\mathfrak{so}(n)\to \operatorname{Spin}(n)$, which are related but different (namely, the first one is obtained from the second using the natural surjection $\operatorname{Spin}(n)\to SO(n)$).
And indeed, $\operatorname{Spin}(n)$ is not a subgroup of $GL(n)$, but this is not a contradiction.