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

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 $\mathrm{Spin}(4)$.

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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.

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Thanks for the great answer! The specific path I mentioned turns out to be tangent to the $-1$ eigenvalue subspace at $t = \pi$ (see above), but I was able to determine the intersection number by perturbing the path a bit. – Jim Belk Apr 21 '12 at 16:48

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.

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I have decided that all my interactions with Jim have been awesome, so I'm going through his old questions to see if I have anything useful to say about them. – David Speyer Nov 22 '13 at 19:46
Well, I certainly have no objection to that! Actually, I haven't asked very many questions, but I'm glad that you've enjoyed my recent ones. – Jim Belk Nov 23 '13 at 2:27
By the way, I like this determinant method (making one plot is certainly easier than making several), but is there some way to tell that the tangency in your graph has order two? I had to perturb the loop in order to resolve the tangency, because I couldn't tell by looking at it whether the order was two or three (or higher). Shouldn't the same thing be necessary here? – Jim Belk Nov 23 '13 at 2:31
I think, but haven't proved, this works: Choose a portion $(t_0, t_1)$ of the path on which $n-2$ of the eigenvalues remain bounded away from $-1$. Let $V_i$ (for $i=1$, $2$) be the $2$-plane corresponding to the most negative eigenvalues of $\gamma(t_i)$. Choose an orientation of $V_i$ so that $\gamma(t_i)$ rotates it clockwise by slightly less than $\pi$. If the orthogonal projection $V_0 \to V_1$ reverses these orientations, I think you have crossed Budney's barrier; if it respects them, I think you have bounced off. – David Speyer Nov 24 '13 at 17:40