What is the difference between the Lie algebra of the orthogonal group and the spin group?

Question

I'm going to restrict my question to $\mathbb{R}^n$. I'm reading that the Lie algebra of $O(n)$ is the algebra of bivectors on $\mathbb{R}^n$. I'm also reading that if you write the Clifford algebra $$Cl(n)=Cl^0(n)\oplus Cl^1(n)\oplus Cl^2(n)\oplus ...$$ That the Lie algebra of $Spin(n)$ is $Cl^2$. But isn't $Cl^2$ the space of bivectors on $\mathbb{R}^n$? Or am I mixing it up with the exterior algebra? If $Spin(n)$ and $O(n)$ have the same Lie algebra, is there a difference in the exponential map?

Answer 1

The spin group is a cover of the orthogonal group, therefore they have the same Lie algebra