How can I lift a path to $\mathrm{Spin}(n)$?

Solution 1:

Nice question, Jim.

$\pi_1 SO(n) \simeq \mathbb Z_2$ provided $n \geq 3$. So your path lifts if and only if it is null homotopic. But since $\pi_1$ is order two, $\pi_1 SO(n) \simeq H_1 SO(n)$ (Hurewicz). $SO(n)$ is a closed manifold, so mod-2 Poincare duality gives you an isomorphism

$$H_i SO(n) \to H^{k - i} SO(n) \equiv Hom(H_{k-i} SO(n), \mathbb Z_2)$$

(mod-2 coefficients everywhere), the $\equiv$ on the right is universal coefficients. $k=dim(SO(n)) = {n \choose 2}$.

This says you can detect whether or not your element of $\pi_1 SO(n)$ is trivial by taking the (transverse) mod-2 intersection number with the appropriate ${n \choose 2}-1$-dimensional class in $SO(n)$.

There are many ways of describing a homology class. A useful way for this particular class would be as the subspace of $SO(n)$ where the matrix has $-1$ as an eigenvalue. First off, having $-1$ as an eigenvalue is a codimension one condition since all the eigenvalues of a matrix in $SO(n)$ are on the unit complex circle (they also appear in conjugate pairs). And if you don't have $-1$ as an eigenvalue, there's a canonical path from the matrix to $I$ in $SO(n)$ which avoids the $-1$ eigenvalue subspace -- think of perturbing the eigenvalues towards $1$.

This subspace isn't a submanifold when $n \geq 4$ but the idea of transversality still applies. To finish the argument, you'd observe that transverse intersection with this subspace gives a well-defined homomorphism $H_1 SO(n) \to \mathbb Z_2$, and since it's non-zero it has to be what you want.

In the case $n=3$, this subspace is just the rotations by $\pi$ about various axis. So it's a manifold and diffeomorphic to $\mathbb RP^2$.

When $n=4$, this space is a little more complicated. If the 2nd eigenvalue $\lambda$ of the matrix $A \in SO(4)$ is not $-1$, $A$ decomposes $\mathbb R^4$ into a 2-dimensional $(-1)$-eigenspace, and its orthogonal complement, on which $A$ acts by rotation by $\lambda$. Said another way, you can think of this space as the image of a map whose domain is $Gr_{4,2}^+ \times S^1$. $Gr_{4,2}^+$ is the Grassmannian of oriented $2$-dimensional subspaces of $\mathbb R^4$. Given a pair $(V, \lambda) \in Gr_{4,2}^+ \times S^1$, the matrix associated to $(V,\lambda)$ is the matrix where $V$ is a $(-1)$-eigenspace, and on the orthogonal complement it is rotation by $\lambda$. You need the orientations for this to make sense. This map isn't one-to-one as there is the relation $(\overline{V}, \overline{\lambda})$ and $(V,\lambda)$ are mapped to the same matrix, but also $(V,-1)$ and $(W,-1)$ are mapped to the same matrix, regardless of what $V$ and $W$ are. $\overline{V}$ is the orientation-reverse of $V$.

Another way to say this is that the subspace of matrices $A \in SO(4)$ that have $(-1)$ as an eigenvalue is the mapping cylinder of the 2:1 covering map $Gr_{4,2}^+ \to Gr_{4,2}$ with the boundary $Gr_{4,2}^+$ crushed to a point. As a stratified space, that's a 5-dimensional object which is a manifold except at a single $0$-dimensional singular point (corresponding to the matrix $-I$). So for all (our) practical purposes this is a closed manifold.

Perhaps there's a simpler description of this homology class?

For your particular example the lazy way to compute this intersection would be to sketch the eigenvalues as a function of $t$, using something like Matlab or Mathematica, and count how many times they crash through $-1$. You'd have to be careful about transversality, but your matrix does stay clear of the singular point ($-I$) so that won't cause any trouble.

I suspect you acquired your matrix as a product of simpler matrices. If that's the case, I'd write out the product and do the count by hand, on the individual factors, then add them up. This would allow for an easier check of transversality, as well.

Solution 2:

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.