Let $P:SO(n) \to SO(n)$ be the square map $P(A)=A^2$.

Does $P$ define a covering map structure?

If yes, is there an action of a finite group $G$ on $SO(n)$ such that each fiber of the covering space is a $G$-orbit and each orbit is a fiber??

If yes, can the tangent bundles of $SO(n)$ be acted by $G$ such that we obtain a $G$ bundle structure?


2 Answers 2


It can't be a covering map for $n>2$, since the fundamental group of $SO(n)$ for $n>2$ is finite, hence it can't cover itself with more than one sheet. But note that $P(I)=I$ and $P(A)=I$, where $$A=\begin{pmatrix} -1 & 0 &0 & \cdots & 0 \\ 0 &-1 & 0 &\cdots &0 \\ 0 &0 & 1 &\cdots & 0 \\ 0 &0 &0 &\cdots &1 \end{pmatrix}.$$

For $n=2$, $SO(n)$ is $S^1$, so the map is a covering map.

For $n=1$, it is the identity, so the map is a covering map.

  • 3
    $\begingroup$ This is technically correct but kinda non-constructive. It is interesting to understand where this map fails to be a covering. $\endgroup$
    – Wolfram
    Mar 3, 2017 at 18:40
  • $\begingroup$ ...for $n \geq 3$. $SO(2)$ is a circle. $\endgroup$
    – Stephen
    Mar 3, 2017 at 18:54
  • $\begingroup$ @Wolfram I admit this answer very much leaves me wondering that. Is part of the issue that it's not 2-to-1 but actually infinity-to-one for $n\gt 2$? (For instance, there are infinitely many $M\in SO(3)$ with $M^2=I$: take the rotation of 180 degrees about any axis; then this fixes exactly two antipodal points and is uniquely determined by the points it fixes.) $\endgroup$ Mar 3, 2017 at 19:08
  • 1
    $\begingroup$ @StevenStadnicki Indeed. If we take the rotations with close axes we get very close elements from the preimage of $I$, so the preimage is not discrete $\Rightarrow$ not a covering. $SO(n)$ can be handled just padding the matrices for $SO(3)$ with $1$s on the diagonal. $\endgroup$
    – Wolfram
    Mar 3, 2017 at 19:16

There are several obstructions to this map being a covering map for $n \ge 3$. A simple one is that its fibers are not discrete: for example, $P^{-1}(I)$ contains all $180^{\circ}$ rotations about every axis.

Alternatively, you can use the fact that a map between closed manifolds is a covering map iff it's a local diffeomorphism, and then compute the differential of $P$ to see where it fails to be a local diffeomorphism. The tangent space at an element $X \in SO(n)$ can be identified with the set of matrices of the form $X(I + \epsilon Y)$ where $Y \in \mathfrak{so}(n)$ is a skew-symmetric matrix and $\epsilon^2 = 0$. Squaring such a matrix gives

$$\left( X + \epsilon XY \right)^2 = X^2 + \epsilon (X^2 Y + XYX) = X^2 \left( I + \epsilon (Y + X^{-1} Y X) \right).$$

So the differential $P_X$ can be identified with the linear map $Y \mapsto Y + X^{-1} Y X$. This is an isomorphism iff it has nontrivial kernel, and its kernel consists of matrices $Y$ satisfying $Y + X^{-1} Y X = 0$.

The matrices $Y$ such that some $X$ exists with this property are matrices conjugate to $-Y$ by an element of $SO(n)$. Every skew-symmetric matrix $Y$ is conjugate to $-Y$, even by an element of $O(n)$, as follows: $Y$ is conjugate via an element of $O(n)$ to a block diagonal matrix whose blocks have the form

$$\left[ \begin{array}{cc} 0 & \theta \\ - \theta & 0 \end{array} \right]$$

for some $\theta$, and any such block is conjugate to its negation via a reflection. The extra condition that we need $Y$ to be conjugate to $-Y$ via an element of $SO(n)$ means we need to do an even number of reflections, which is always possible when $n \ge 4$. When $n = 3$ one of the eigenvalues of $Y$ is zero and so we can just stick a $-1$ into $X$ as necessary.

Hence for any nonzero value of $Y$ we can find $X$ with the above property, and so there are many points at which the differential fails to be surjective and hence when $P$ fails to be a local diffeomorphism.

  • $\begingroup$ Thanks for the answer. So my question was a trivial question since I did not pay attention to the (infinitness ) of level sets. But how can we compare two typical level sets(Are they the same topologicaly?). Do we have a (at least topological) fiber bundle? $\endgroup$ Mar 5, 2017 at 16:08
  • $\begingroup$ @Ali: no, you can also find some fibers that are discrete (take the preimage of something with distinct eigenvalues). $\endgroup$ Mar 5, 2017 at 19:03
  • $\begingroup$ What about the square map on $S^{3}$?Are there two different fubers(topologically)? $\endgroup$ Mar 6, 2017 at 8:37
  • $\begingroup$ @Ali: yes, the situation is basically the same as for $SO(3)$, you can consider e.g. the preimage of the identity and of an element with distinct eigenvalues. $\endgroup$ Mar 6, 2017 at 18:47
  • $\begingroup$ Thanks again for your answer. I am sorry that I can not accept two answers. Just a question on your statement "...map between closed manifolds is a covering map iff it's a local diffeomorphism". Can not one modify $x\mapsto x^3$ on $\mathbb{R}$ to a simillar example on a closed manifold? $\endgroup$ Aug 7, 2018 at 14:32

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