I need to know the homotopy groups of the oriented Grassmannian $\widetilde{Gr}(\infty,\infty) \cong \lim_{N \rightarrow \infty} SO(2N)/(SO(N) \times SO(N))$, and I've run into an inconsistency. It seems that the problem eventually reduces to finding the homotopy groups of the stable special orthogonal group $SO$. One can do this by exploiting Bott periodicity. I haven't found much about Bott periodicity specifically for $SO(N)$, but it's easy to find for $O(N)$: $$\pi_k(O(N)) = \begin{cases} \mathbb{Z}_2 & k = 0,1 \mod 8 \\ \mathbb{Z} & k = 3,7 \mod 8 \\ 0 & k = 2,4,5,6 \mod 8 \end{cases}$$ Since $\pi_k(SO(N)) = \pi_k(O(N))$, this gives us all the homotopy groups for $SO$.
The problem is that this tells us that for sufficiently high $N$ (i.e., $N \ge k+2$), $\pi_0(SO(N)) = \pi_0(O(N)) = \mathbb{Z}_2$. $\pi_0$ counts the path-connected components of a space, and while $O(N)$ does indeed have two of them, $SO(N)$ has only one, so $\pi_0(SO(N))$ should be $0$, not $\mathbb{Z}_2$. Therefore, something is wrong with the result from Bott periodicity.
Which of these arguments is wrong, and why?
The equality $\pi_kSO(n)=\pi_kO(n)$ is only true for $k>0$.
In fact $SO(n)$ is one of the two connected components of $O(n)$, so $\pi_0SO(n)=0$.