# Prove where exp: Skew($3\times 3$) $\rightarrow SO(3)$ is local homeomorphism

The matrix exponential on skew-symmetric $3\times3$ matrices onto $SO(3)$ is not local homeomorphism everywhere. I have been instructed that one problem is with the spheres of radius $2n\pi$ ($n\in\mathbb{N}$) in the scaled Frobenius norm $\|\,\|=\frac{1}{\sqrt{2}}\|\,\|_F$. As I understand this $\exp$ is $C^\infty$ smooth and injection on the open ball of radius $\pi$.

I would like to know what is the radius of the largest open ball around zero where this matrix $\exp$ is a local homeomorphism, or maybe even diffeomorphism? Even better an explanation, a proof or a reference to the proof should be given.

In essence, I see there needs to be shown that for every $X$ in such a ball, the differential (pushforward) $\mathrm{d}_{\exp(X)}$ is linear isomorphism. I lack the knowledge of differential geometry to proceed.

#### Solutions Collecting From Web of "Prove where exp: Skew($3\times 3$) $\rightarrow SO(3)$ is local homeomorphism"

The exponential map of $SO(3)$ can be understood geometrically by looking at the double cover $S^3$, a.k.a., the group of unit quaternions under multiplication.

(Added to incorporate comment clarifications) A two-sheeted covering map $S^3 \to SO(3)$ may be defined as follows. The adjoint action of $S^3$ on its Lie algebra (the $3$-dimensional space of imaginary quaternions, hereafter denoted $\mathbf{R}^3$) is given by
$$\operatorname{ad}_{\alpha}(X) = \alpha X \alpha^{-1},\quad X \in \mathbf{R}^3.$$
The product $\alpha X \alpha^{-1}$ is an imaginary quaternion that depends linearly on $X$ and has the same length as $X$. That is, $\operatorname{ad}_{\alpha}$ is a linear isometry of $\mathbf{R}^3$, i.e., an element of $O(3)$. Since $S^3$ is connected and the identity of $SO(3)$ is in the image, $\operatorname{ad}(S^3) \subseteq SO(3)$.

Clearly, $\operatorname{ad}$ is a group homomorphism, and $\operatorname{ad}_\alpha = I$ if and only if $\alpha X \alpha^{-1} = X$ for all $X$ in $\mathbf{R}^3$, if and only if the imaginary part of $\alpha$ is zero, if and only if $\alpha = \pm 1$. That is, $\ker(\operatorname{ad}) = \{\pm1\}$, the center of $S^3$.

Let $\alpha = \alpha_0 + \alpha_1 i + \alpha_2 j + \alpha_3 k$ denote a typical element of $S^3$. Since $\operatorname{ad}_{\alpha}(X)$ is a polynomial in the components of $\alpha$, the mapping $\operatorname{ad}:S^3 \to SO(3)$ is smooth.

Direct calculation (taking $\alpha$ to be multiples of $i$, $j$, or $k$) shows that arbitrary rotations about coordinate axes of $\mathbf{R}^3$ are in the image of $\operatorname{ad}$. In particular, $\operatorname{ad}$ induces an isomorphism of Lie algebras (hence is a local diffeomorphism), $\operatorname{ad}$ is surjective (axis rotations generate $SO(3)$), and $S^3/\{\pm1\} \simeq SO(3)$ as Lie groups (isomorphic as groups, diffeomorphic as manifolds).

On $S^3$ equipped with the round (i.e., invariant) metric of unit radius, the Lie group exponential map agrees with the Riemannian exponential map, which sends a tangent vector $X$ at $1$ (i.e., a pure imaginary quaternion of norm $\|X\|$) to the geodesic
$$\gamma(t) = \exp(tX) = \cos\bigl(\|X\| t\bigr) + \frac{\sin\bigl(\|X\| t\bigr)}{\|X\|}X$$
through $1$ and tangent to $X$. The open ball of radius $\pi$ in the Euclidean norm on $T_1S^3$ clearly maps diffeomorphically into $S^3$ (missing one point, $-1$), and the open ball of radius $\pi/2$ maps diffeomorphically to the open hemisphere of $S^3$ centered at $1$. Since no two points of an open hemisphere are antipodal, the induced map from the ball of radius $\pi/2$ to $S^3/\{\pm1\}$ (i.e., to $SO(3)$ up to scaling, see below) is also a diffeomorphism. No larger open ball maps diffeomorphically into $SO(3)$ under this mapping, since the image of the closed ball of radius $\pi/2$ contains antipodal points, which project to the same point of $SO(3)$.

To get the scaling right in the preceding paragraph, let $a$, $b$, and $c$ be real (not all zero), and let
$$A = \left[\begin{array}{@{}rrr@{}} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0 \end{array}\right].$$
The vector $X = (a, b, c)$ spans the null space of $A$, and hence is a $1$-eigenvector of
$$\exp(tA) = I + tA + \dots$$
for all real $t$, i.e., spans the axis of rotation of $\exp(tA)$. The angular speed of rotation of $\exp(tA)$ is
$$\left\lVert \frac{d}{dt}\bigg|_{t=0} \exp(tA)\right\rVert = \|A\| = \frac{1}{\sqrt{2}}\sqrt{\operatorname{tr}(A^tA)} = \sqrt{a^2 + b^2 + c^2}.$$
That is, $\exp(A)$ is rotation about $X$ by oriented angle $\|A\|$ (counterclockwise if $X$ points toward your eye). Consequently, (i) the open ball of radius $\pi$ in this norm maps injectively to $SO(3)$: If $\exp(A_1) = \exp(A_2)$, then the rotation axes $X_1$ and $X_2$ coincide, as do the oriented angles; (ii) the closed ball of radius $\pi$ does not map injectively: rotating by $\pi$ or by $-\pi$ about an axis gives the same element of $SO(3)$. By the preceding paragraph, $\exp$ is a diffeomorphism on the open ball of radius $\pi$ in the scaled Frobenius norm (and a local diffeomorphism on the open ball of radius $2\pi$).

Here’s an almost purely computational answer, avoiding the double covering of the rotation group $SO(3)$ by $S^3$. Re-writing seemed likely to be easier (and possibly clearer) than retro-fitting additional explanation in service of an apparently unhelpful geometric picture. I’ve wikified the original response (with its geometric/conceptual approach) in case someone someday would like to have a go at clarifying the details.

To each unit vector $\mathbf{u} = (a, b, c)$ in $\mathbf{R}^3$, associate the skew-symmetric matrix
$$U = \left[\begin{array}{@{}rrr@{}} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0 \end{array}\right],$$
and let $S^2$ denote the set of all such matrices. The vector $\mathbf{u}$ spans the null space of $U$, and hence is a $1$-eigenvector of $\exp(tU) = I + tU + \dots$ for all real $t$, i.e., spans the axis of the rotation $\exp(tU)$. Geometrically, $U$ is the infinitesimal counterclockwise rotation about $\mathbf{u}$ of angular speed
$$\left\lVert \frac{d}{dt}\bigg|_{t=0} \exp(tU)\right\rVert = \|U\| = \frac{1}{\sqrt{2}}\sqrt{\operatorname{tr}(U^tU)} = \sqrt{a^2 + b^2 + c^2} = 1.$$

A non-zero matrix in the open ball of radius $2\pi$ is written uniquely as $tU$, with $U$ a matrix of scaled Frobenius norm one and $0 < t < 2\pi$. By the preceding calculation, $\exp(tU)$ is the counterclockwise rotation about $\mathbf{u}$ by angle $t$.

Claim 1: $\exp$ is a local differmorphism on the ball of radius $2\pi$. To prove this, it suffices to show that if $H$ is an open hemisphere of $S^2$, then the map $\Phi:H \times (0, 2\pi) \to SO(3)$ defined by $\Phi(U, t) = \exp(tU)$ is a diffeomorphism onto its image.

For longitude $\theta$ and colatitude $\phi$ between $0$ and $\pi$, consider the orthonormal basis
$$\mathbf{e}_1 = \left[\begin{matrix} \cos\theta \sin\phi \\ \sin\theta \sin\phi \\ \cos\phi \end{matrix}\right],\qquad \mathbf{e}_2 = \left[\begin{matrix} \sin\theta \\ -\cos\theta \\ 0 \end{matrix}\right],\qquad \mathbf{e}_3 = \left[\begin{matrix} \cos\theta \cos\phi \\ \sin\theta \cos\phi \\ -\sin\phi \end{matrix}\right].$$
Calling $U$ the $3 \times 3$ skew-symmetric matrix associated to the unit vector $\mathbf{e}_1$, we have
\begin{align*}
\Phi(U, t) = \exp(tU)
&= \left[\begin{matrix}
\mathbf{e}_1 &
(\cos t)\mathbf{e}_2 + (\sin t)\mathbf{e}_3 &
-(\sin t)\mathbf{e}_2 + (\cos t)\mathbf{e}_3
\end{matrix}\right] \left[\begin{matrix}
\mathbf{e}_1^t \\
\mathbf{e}_2^t \\
\mathbf{e}_3^t
\end{matrix}\right] \\
&= \mathbf{e}_1 \mathbf{e}^t
+ (\cos t)(\mathbf{e}_2 \mathbf{e}_2^t + \mathbf{e}_3 \mathbf{e}_3^t)
+ (\sin t)(\mathbf{e}_3 \mathbf{e}_2^t – \mathbf{e}_2 \mathbf{e}_3^t).
\end{align*}
It is straightforward to check algebraically (and clear geometrically) that $D\Phi(\theta, \phi, t)$ has rank $3$ for $0 < \theta < \pi$, $0 < \phi < \pi$, and $0 < t < 2\pi$. Moreover, $\Phi$ is injective, since each line through the origin of $\mathbf{R}^3$ has at most one representative in an open hemisphere, and the range of rotation angles is less than one full turn. That is, $\Phi:H \times (0, 2\pi) \to SO(3)$ is a diffeomorphism onto its image. Covering $S^2$ by open hemispheres obtained from $H$ by permuting coordinates (and/or reflecting across a coordinate plane) shows that the punctured open ball of radius $2\pi$ in the space of skew-symmetric $3 \times 3$ matrices maps locally diffeomorphically to $SO(3)$ under $\exp$. Since $\exp$ is a local diffeomorphism at the zero matrix, $\exp$ maps the ball of radius $2\pi$ locally diffeomorphically to $SO(3)$.

Claim 2: $\exp$ is injective on the open ball of radius $\pi$ (hence maps this ball diffeomorphically to its image), and not on the closed ball of radius $\pi$. This is geometrically clear: two counterclockwise rotations by an angle less than $\pi$ about distinct unit vectors are equal if and only if their axes and angles of rotation coincide, while half-turns about unit vectors $\mathbf{u}$ and $-\mathbf{u}$ are equal.

This is the sketch of differential on a smooth map between manifolds from Wikipedia:

In this case we have
$$\begin{array}{cccc} & \mbox{Domain} & \mbox{Mapping} & \mbox{Target set}\\ \mbox{Manifold} & M=\mathbb{R}_{{\rm skew}}^{3\times3} & \overset{\varphi(X)=\exp X}{\longrightarrow} & N=SO(3)\\ \mbox{Tangent space} & T_{X}M=\mathbb{R}_{{\rm skew}}^{3\times3} & \overset{\mathrm{d}_{\varphi(X)}}{\longrightarrow} & T_{\varphi(X)}N=? \end{array}$$

Differential of $\varphi$
at $X$
is a linear map $\mathrm{d}_{\varphi(X)}:T_{X}M\to T_{\varphi(X)}N$
that can be computed like this
$${\rm d}_{\varphi(X)}\underset{\begin{array}{c} \uparrow\\ \blacksquare \end{array}}{\left(\,\gamma'(0)\,\right)}=\left(\varphi\circ\gamma\right)'(0)$$
where $\gamma$
is a curve such that $\gamma(0)=X$. As a linear map, the differential is fully determined by it’s action on a basis in $T_{X}M$. Therefore the symbol $\blacksquare$ denotes a place where we want to insert one by one elements of a basis for $T_{X}M$. If $\{{\bf b}_{i}\}_{i=1}^{n}$ is a basis in $T_{X}M$, and ${\bf v}=\sum_{i=1}^{n}v_{i}{\bf b}_{i}$ where $v_{i}$ are scalars (coordinates), then we know that ${\rm d}_{\varphi\left(X\right)}{\bf v}=\sum_{i=1}^{n}v_{i}{\rm d}_{\varphi\left(X\right)}{\bf b}_{i}$. To check that ${\rm d}_{\varphi\left(X\right)}$ is an isomorphism we only need to check that $\{{\rm d}_{\varphi(X)}{\bf b}_{i}\}_{i=1}^{n}$ is a linearly independent set.

Let us use the Rodrigues’ rotation formula to ease the computation (for $B\in\mathbb{R}_{{\rm skew}}^{3\times3}$):
$$\exp B={\rm Id}+\frac{\sin\left\Vert B\right\Vert }{\left\Vert B\right\Vert }B+\frac{1-\cos\left\Vert B\right\Vert }{\left\Vert B\right\Vert ^{2}}B^{2}\in SO(3)\mbox{.}$$

My first try was to insert the elements of the canonical orthonormal basis for $\mathbb{R}_{{\rm skew}}^{3\times3}$ in $\blacksquare$. We can do that by choosing $\gamma_{i}(t)=X+tA_{i}$, so that $\gamma_{i}'(0)=A_{i}$ where
$$A_{1}=\left(\begin{array}{ccc} 0\\ & 0 & -1\\ & 1 & 0 \end{array}\right),\quad A_{2}=\left(\begin{array}{ccc} 0 & & 1\\ & 0\\ -1 & & 0 \end{array}\right),\quad A_{1}=\left(\begin{array}{ccc} 0 & -1\\ 1 & 0\\ & & 0 \end{array}\right).$$
Then, for $X=\left(\begin{array}{ccc} 0 & -x_{3} & x_{2}\\ x_{3} & 0 & -x_{1}\\ -x_{2} & x_{1} & 0 \end{array}\right)$
we have (using the Rodrigues’ formula)
$${\rm d}_{\varphi(X)}A_{i}=\frac{{\rm d}}{{\rm d}t}\exp\left(X+tA_{i}\right)_{t=0}=\frac{f(\left\Vert X\right\Vert )x_{i}}{\left\Vert X\right\Vert }X+\frac{\sin\left\Vert X\right\Vert }{\left\Vert X\right\Vert }A_{i}+\frac{g(\left\Vert X\right\Vert )x_{i}}{\left\Vert X\right\Vert }X+\frac{1-\cos\left\Vert X\right\Vert }{\left\Vert X\right\Vert ^{2}}2X.A_{i}$$
where $f(u)\overset{{\rm def}}{=}\frac{u\cos u-\sin u}{u^{2}}=\frac{\mathrm{d}}{\mathrm{d}u}\left(\frac{\sin u}{u}\right)$ and $g(u)\overset{{\rm def}}{=}\frac{u\sin u-2+2\cos u}{u^{3}}=\frac{\mathrm{d}}{\mathrm{d}u}\left(\frac{1-\cos u}{u^{2}}\right)$. Within $X.A_{i}$, the dot stands for matrix multiplication. Unfortunately, I am not able to see that $\{{\rm d}_{\varphi(X)}A_{i}\}_{i=1}^{3}$ is linearly independent, unless $X=O$. It seems as if plugging the easiest choices into $\blacksquare$
resulted with a complicated right hand side in the same formula. But at least we have proved that ${\rm d}_{\exp}$
is isomorphism at $X=O$, that is $\exp$
is local diffeomorphism at $O\in\mathbb{R}_{{\rm Skew}}^{3\times3}$.

Therefore, I started again from the beginning, this time trying to choose a basis that will result in the simplest result on the right hand side (after differentiation). Provided $X\in\mathbb{R}_{{\rm skew}}^{3\times3}$
is set, note that for each $y\in\mathbb{R}^{3}$
we have $X.y=\left(\begin{array}{c} x_{1}\\ x_{2}\\ x_{3} \end{array}\right)\times y$
(with $X.y$
I denoted matrix multiplying a vector). Provided with $X\neq O$, I choose $a=\left(\begin{array}{c} a_{1}\\ a_{2}\\ a_{3} \end{array}\right)$ and $b=\left(\begin{array}{c} b_{1}\\ b_{2}\\ b_{3} \end{array}\right)$ such that $\left\Vert a\right\Vert =1=\left\Vert b\right\Vert$ and $a\times\frac{x}{\left\Vert x\right\Vert }=b$, $\frac{x}{\left\Vert x\right\Vert }\times b=a$ (we have $a\bot b$, $a\bot x$ and $b\bot x$). Let
$$A=\left(\begin{array}{ccc} 0 & -a_{3} & a_{2}\\ a_{3} & 0 & -a_{1}\\ -a_{2} & a_{1} & 0 \end{array}\right)\mbox{ and }B=\left(\begin{array}{ccc} 0 & -b_{3} & b_{2}\\ b_{3} & 0 & -b_{1}\\ -b_{2} & b & 0 \end{array}\right)\mbox{.}$$
Then we have $A.y=a\times y$ and $B.y=b\times y$. Now I will use different $\gamma_{i}$,
$$\gamma_{1}(t)=X+tX,\gamma_{1}'(0)=X,\quad\gamma_{2}(t)=X+tA,\gamma'(0)=A,\quad\gamma_{2}(t)=X+tB,\gamma'(0)=B\mbox{.}$$
We can check that $\{X,A,B\}$ is orthogonal base in $\mathbb{R}^{3\times3}$ using the following inner product on matrices: $A:B=\frac{1}{2}{\rm Tr}\left(A^{T}.B\right)$, where ${\rm Tr}$ denotes the matrix trace. We have $A:B=a\cdot b=0$, $A:X=a\cdot x=0$ and $B:X=b\cdot x=0$ (dot denotes the ordinary vector scalar product). Therefore, $A\bot B$, $A\bot X$ and $B\bot X$.

Let us also here establish some other facts that will be used later. Let us show that $\{X,\, A,\, B,\, X^{2},\, X.A,\, X.B\}$ is linearly independent (evenmore orthogonal) set. We have
\begin{eqnarray*}
x\times\left(x\times\left(x\times y\right)\right) & = & x\times\left(x\left(x\cdot y\right)-y\left(x\cdot x\right)\right)\qquad\left\{ \begin{array}{c}
\mbox{vector}\\
\mbox{triple}\\
\mbox{product}
\end{array}\right\} \\
&=& \left(x\cdot y\right)x\times x-\left(x\cdot x\right)x\times y\qquad\left\{ \begin{array}{c}
\mbox{distributive}\\
\mbox{property}
\end{array}\right\} \\
&=& -\left\Vert x\right\Vert ^{2}\, x\times y
\end{eqnarray*}
so we conclude $X^{3}=-\left\Vert X\right\Vert ^{2}X\in\mathbb{R}_{{\rm Skew}}^{3\times3}$. Therefore $X:X^{2}=\frac{1}{2}{\rm Tr}\left(X^{T}.X^{2}\right)=\frac{1}{2}{\rm Tr}\left(-X^{3}\right)=0$. On the other hand, by using vector triple product and orthogonality between $x$, $a$ and $b$
\begin{eqnarray*}
X^{2}.A.y &=& x\times\left(x\times\left(a\times y\right)\right) =x\times\left((x\cdot y)\, a\right)=\left\Vert x\right\Vert (x\cdot y)\, b,\\
X.B.y &=& x\times\left(b\times y\right) =(x\cdot y)\, b\mbox{ .}
\end{eqnarray*}
Therefore $X^{2}.A=X.B$, so together with $X:X.A=-{\rm Tr}\left(X^{2}.A\right)=-X^{2}:A$ we get $X:X.A=-X^{2}:A=-{\rm Tr}\left(X.B\right)=X:B=0$. In the same manner we get that all pairs in $\{X,\, A,\, B,\, X^{2},\, X.A,\, X.B\}$ are orthogonal.

Back to our differential, from Rodrigues’ formula we get
\begin{eqnarray*}
{\rm d}_{\varphi(X)}X &=& \frac{{\rm d}}{{\rm d}t}\exp\left(X+tX\right)_{t=0}=\left(f(\left\Vert X\right\Vert )\left\Vert X\right\Vert +\frac{\sin\left\Vert X\right\Vert }{\left\Vert X\right\Vert }+g(\left\Vert X\right\Vert )\left\Vert X\right\Vert \right)X+2\frac{1-\cos\left\Vert X\right\Vert }{\left\Vert X\right\Vert ^{2}}X^{2},\\
{\rm d}_{\varphi(X)}A &=& \frac{{\rm d}}{{\rm d}t}\exp\left(X+tA\right)_{t=0}=\frac{\sin\left\Vert X\right\Vert }{\left\Vert X\right\Vert }A+2\frac{1-\cos\left\Vert X\right\Vert }{\left\Vert X\right\Vert ^{2}}X.A,\\
{\rm d}_{\varphi(X)}B &=& \frac{{\rm d}}{{\rm d}t}\exp\left(X+tB\right)_{t=0}=\frac{\sin\left\Vert X\right\Vert }{\left\Vert X\right\Vert }B+2\frac{1-\cos\left\Vert X\right\Vert }{\left\Vert X\right\Vert ^{2}}X.B\,\mbox{.}
\end{eqnarray*}
The final point to see is that $\{{\rm d}_{\varphi(X)}X,\,{\rm d}_{\varphi(X)}A,\,{\rm d}_{\varphi(X)}B\}$ is linearly independent set, evenmore orthogonal. Notice that each one out of $\{{\rm d}_{\varphi(X)}X,\,{\rm d}_{\varphi(X)}A,\,{\rm d}_{\varphi(X)}B\}$ is a linear combination of two different elements out of six in orthogonal set $\{X,\, A,\, B,\, X^{2},\, X.A,\, X.B\}$. Therefore, unless one of the elements in $\{{\rm d}_{\varphi(X)}X,\,{\rm d}_{\varphi(X)}A,\,{\rm d}_{\varphi(X)}B\}$ is actually a trivial combination, we have proved that ${\rm d}_{\varphi(X)}$ is isometry, that is $\exp$ is a local diffeomorphism. Note that ${\rm d}_{\varphi(X)}A=O={\rm d}_{\varphi(X)}B$ if and only if $\left\Vert X\right\Vert =n\pi,\, n\in\mathbb{N}$.
Furthermore, ${\rm d}_{\varphi(X)}X$ can not become zero unless the factor $\frac{1-\cos\left\Vert X\right\Vert }{\left\Vert X\right\Vert ^{2}}$ becomes zero, which can not happen when $\left\Vert X\right\Vert \neq n\pi,\, n\in\mathbb{N}$. Together with what we already have for $X=0$, we arrived at the conclusion that $\exp$ is a local diffeomorphism on $\mathbb{R}^{3\times 3}_{\rm Skew}$, except on the spheres of radius $2n\pi$, $n\in\mathbb{N}$.

I am inexperienced in differential geometry calculations, so please do not hesitate to be hairsplitting. I would like to get this right.