# Hopf fiber

Basically, the Hopf fiber is a fiber structure on the $2$-sphere $S^2$. Or, it describes a decomposition of $S^3$ into $S^2$ and $S^1$.

Start from the definition of $S^3\subset \mathbb{C}^2$, it consists of points $(z,z')\in\mathbb{C}^2$ such that $|z|^2+|z'|^2=1$. On defines a projection $p:S^3\rightarrow S^2,(z,z')\mapsto (2\bar{z}z',|z|^2-|z'|^2)$ where $2\bar{z}z'$ stands for the first coordinates of points in $S^2$. Note that, the image of $p$ indeed lies in $S^2$. We will show that $p$ is surjective and for each $P\in S^2$, the fiber $p^{-1}(P)$ is a circle. And this describes a fiber structure on $S^2$.

Indeed, for any $(x+iy,w)\in S^2$, we want to find $(z,z')$ such that

$|z|^2+|z'|^2=1,|z|^2-|z'|^2=w,2\bar{z}z'=x+iy$

Yet this is almost trivial, since we have that

$|z|^2=(1+w)/2,|z'|^2=(1-w)/2,2\bar{z}z'=x+iy$

This system is always solvable. Moreover, if $p(z,z')=(x+iy,w)$, then $p^{-1}(x+iy,w)=(e^{i\theta}z,e^{i\theta}z')$. This can also be seen from the above system of equations. Note that sometimes we are likely to take this projection, $p':S^3\rightarrow S^2,(z,z')\mapsto (z,|z'|)$. One problem with this projection is that, its image is not the entire $S^2$($|z'|$ is nover negative), another point is that $p'^{-1}(p'(z,z'))=p'^{-1}(z,|z'|)=(z,e^{i\alpha}z')$, yet for the point $(z,0)\in S^2$, its fiber is only itself, only one point, not a circle.

Another interesting way to describe this fiber is to use some Lie groups. Note that, for $S^3$, the Lie group $SU(2)$ acts transitively and freely on $S^3$(because each matrix in $SU(2)$ can be written as $\begin{pmatrix}a &b \\ -\bar{b} & \bar{a} \end{pmatrix}$, in general, a matrix in $U(2)$ can be written as $e^{i\theta}\begin{pmatrix} a &b\\-\bar{b}&\bar{a}\end{pmatrix}$).

But, $SU(2)$ acts also mysteriously on $\mathbb{R}^3$! This is a rather important fact. Note that, in some sense, this is also reasonable: that is $SU(2)$ acts also on the tangent space of some point $P\in S^3$ on $S^3$, and $S^3$ has a group structure(that is exactly $SU(2)$). After giving this tangent space $T_P(S^3)$ a metric(induced from that of $S^3$, of course), $SU(2)$ acts thus on $T_P(S^3)=\mathbb{R}^3$ preserving the metric, thus acts also on $S^2\subset \mathbb{R}^3$. Yet, we know that, $SO(3)$ acts on $S^2$, we can show that the action of $SU(2)$ on $S^2$ induces a homomorphism $SU(2)\rightarrow SO(3)$, with kernel $\pm Id$. Moreover, $SO(3)$ acts transitively on $S^2$, and for each point $P\in S^2$, the stabilizer of $SO(3)$ on $P$ is a circle. Note also that, for the opposite point $-P\in S^2$, the stabilizer of $SO(3)$ on $-P$ is the same as the previous, $Stab(SO(3),P)=Stab(SO(3),-P)$. This shows that, not rigorously, that $SU(2)$ acts on $S^2$ with for each point, a stabilizer group the same as one copy of circle.

More concretely, since $SU(2)$ consists of matrices of the form $\begin{pmatrix}a&b\\-\bar{b}&\bar{a}\end{pmatrix}$ with $|a|^2+|b|^2=1$, we consider the conjugate action of $SU(2)$ on itself. This is indeed a group action, and we note this as

$Ad_g:SU(2)\rightarrow SU(2),h\mapsto ghg^{-1}$

Now take derivative with respect to the variable $h$, and transform this action to the Lie algebra

$ad_g:\mathfrak{g}(SU(2))\rightarrow \mathfrak{g}(SU(2)),H\mapsto gHg^{-1}$

Note that, the Lie algebra of $SU(2)$ consists of matrices $H\in M_{2\times 2}(\mathbb{C})$ such that $Tr(H)=0,\bar{H}^t+H=0$. More specifically, we can find a $\mathbb{R}$-basis for $\mathfrak{su}(2)=\mathfrak{g}(SU(2))$(note that, $SU(2)$ is not a complex Lie group, only a real Lie group, thus its Lie algebra is only a real vector space, not a complex one):

$e_1=\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, e_2=\begin{pmatrix}0&i\\i&0\end{pmatrix},e_3=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$

We can show that the metric induced from $SU(2)=S^3$ on $\mathfrak{su}(2)$ is preserved by $SU(2)$, and $e_1,e_2,e_3$ forms an orthonormal basis. Now, we want to show that the stabilizer of each point $P\in S^2=S^2(\mathfrak{su}(2))$ is a circle, since $SU(2)$ acts transitively on $S^2$, we can verify this for just one point, say $e_3$. Note that, if $g\in SU(2)$ preserves $e_3$, then $g=r_{\theta}=\begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} &\cos{\theta}\end{pmatrix}(\theta\in\mathbb{R})$. So, after some routine yet simple calculations, we find that, under the basis $e_1,e_2$, $ad_{r_{\theta}}=\begin{pmatrix} \cos{2\theta} & \sin{2\theta} \\ -\sin{2\theta} &\cos{2\theta}\end{pmatrix}$(why, you may wonder, there is an increase in the ‘speed’? This can be explained by taking another derivative with respect to $g$ in $ad_g$, thus we get that $ad_G(H)=[G,H]=GH-HG$. For $g=r_{\theta}$, we have that $G=\frac{dr_{\theta}}{d\theta}|_{\theta=0}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=e_3$. It is readily verified that $ad_{e_3}(e_1)=-2e_2,ad_{e_3}(e_2)=2e_1$, this little $2$ is the origin for this speed increase! And this $2$ accounts also for the fact that $\pi_1(S^3)=\pi_1(SU(2))=0,\pi_1(SO(3))=\pi_1(\mathbb{P}^3_{\mathbb{R}})=\mathbb{Z}/2$).

So, in this way, we describe another way of the Hopf fiber on $S^2$.

The second method enables us to generalize to higher dimension case.

A nice demonstration of the Hopf fiber is here.