TITLE: How can I lift a path to $\mathrm{Spin}(n)$?
QUESTION [21 upvotes]: Suppose I am given an explicit differentiable path $\gamma\colon[a,b]\to SO(n)$, with $\gamma(a)=\gamma(b)=I$.  Then $\gamma$ either does or does not lift to a closed loop in $\mathrm{Spin}(n)$.  How do I tell which is which?
For example, let $\gamma\colon[0,2\pi]\to SO(4)$ be the function
$$
\gamma(t) \;=\; \begin{bmatrix}
\cos t \cos 2t & \sin t \cos 2t & 0 & \sin 2t \\[6.5pt]
-\sin t\cos 3t & \cos t \cos 3t & \sin 3t & 0 \\[6.5pt]
\sin t \sin 3t & -\cos t \sin 3t & \cos 3t & 0 \\[6.5pt]
-\cos t \sin 2t & -\sin t \sin 2t & 0 & \cos 2t
\end{bmatrix}
$$
Does this lift to a loop in $\mathrm{Spin}(4)$, or not?  What calculation would I need to do to figure this out?
Edit: Following Ryan's suggestion, I made a plot of the eigenvalues of the above matrix in Mathematica.  Only the two eigenvalues with $\mathrm{Im}(\lambda)>0\text{ }$are shown, but the other two are simply the complex conjugates:

Note that a $-1$ eigenvalue corresponds to the top edge of the rectangle.  There are four obvious values of $t$ where one of the eigenvalues passes through $-1$ transversely, but it also appears that one of the eigenvalues is tangent to $-1$ at $t=\pi$.
To resolve the tangency, I perturbed the path $\gamma(t)$ by a path $\delta(t) \in SO(4)$ that stays close to the identity matrix.  Here are the eignevalues for the product $\delta(t)\gamma(t)$:

The tangency has now resolved itself into two transverse intersections, making a total of six transverse intersections.  Since six is even, this path lifts to a loop in $\mathrm{Spin}(4)$.

REPLY [3 votes]: This is a minor modification to the last part of Ryan Budney's great answer. Instead of finding the eigenvalues of $A$ and looking to see where they become $1$, plot $\sqrt{\det(A+\mathrm{Id})}$. Here is Jim Belk's example:

The function $\det(A+\mathrm{Id})$ vanishes to order $2$ on the divisor Ryan describes (and nowhere else), so it locally has a smooth square root. It has a global smooth square root up on the double cover $Spin(n)$, but not on $SO(n)$, because Ryan Budney's divisor doesn't disconnect $SO(n)$, so we can't choose a side on which to take the positive square root. We can tell whether $\gamma(t)$ lifts to a closed path in $Spin(n)$ by seeing whether $\sqrt{\det(\gamma(t)+\mathrm{Id})}$ wants to lift to a smooth function with $f(0) = f(1)$ or $f(0) = -f(1)$. 
Of course, if you actually tell Mathematica to plot the square root, it will plot the positive square root. But if you look at it, you can see when it wants to cross the $x$-acis and when it wants to be tangent to it (as happened in Jim Belk's computation). In this case, it wants to cross 4 times and to be tangent once, so the path lifts.
This isn't much different the previous answer, but I would expect computing determinants to be a little less taxing than computing eigenvalues.