### The Hopf Bundles

Consider a complex column vector \begin{equation} v = \begin{pmatrix}b\cr c\cr\end{pmatrix} \label{column} \end{equation} where $b,c\in\CC$. What can we do with $v$? Well, we can take its Hermitian conjugate, defined as before, namely \begin{equation} v^\dagger = \begin{pmatrix}\bar{b}& \bar{c}\cr\end{pmatrix} \end{equation} We can now multiply $v$ and $v^\dagger$ in two quite different ways. We can produce a complex number \begin{equation} v^\dagger v = |b|^2 + |c|^2 \label{scalar} \end{equation} and, if we multiply in the opposite order, a matrix \begin{equation} v v^\dagger = \begin{pmatrix} |b|^2& b\bar{c}\cr \noalign{\smallskip} c\bar{b}& |c|^2 \end{pmatrix} \end{equation}

One way to visualize $v$ is to think of it as a point in $\CC^2=\RR^4$, in which case ($\ref{scalar}$) is just the usual (squared) vector norm. Suppose we normalize $v$ by requiring that $v^\dagger v = 1$. Then $v$ lies on the unit sphere in $\RR^4$, which is 3-dimensional, and hence denoted by $\SS^3$.

What about $vv^\dagger$? This matrix is Hermitian! We can therefore identify it with a vector in spacetime, via \begin{align} t+z &= |b|^2 \\ t-z &= |c|^2 \\ x+iy &= c\bar{b} \end{align} We can solve this for the coordinates themselves, obtaining 1) \begin{align} t &= \frac12\, (|b|^2 + |c|^2) \\ z &= \frac12\, (|b|^2 - |c|^2) \\ x &= \frac12\, (c\bar{b} + b\bar{c}) \\ iy &= \frac12\, (c\bar{b} - b\bar{c}) \end{align} Note however that the normalization ($\ref{scalar}$) of $v$ tells us that $t$ is constant. This also follows immediately from \begin{equation} 2t = \tr(vv^\dagger) = v^\dagger v = 1 \end{equation} So the trace of $vv^\dagger$ is special; what about its determinant? We have \begin{equation} \det(vv^\dagger) = |b|^2 |c|^2 - (b\bar{c})(c\bar{b}) = |b|^2 |c|^2 - |b\bar{c}|^2 = |b|^2 |c|^2 - |bc|^2 = 0 \label{division} \end{equation} Thus, $vv^\dagger$ represents a null vector! Putting this all together, we have \begin{equation} x^2 + y^2 + z^2 = t^2 = \hbox{constant} \end{equation} so that we can think of $vv^\dagger$ as a point on a particular sphere, which we denote $\SS^2$.

We have constructed a famous map between $\SS^3$ and $\SS^2$ known as the *Hopf map*, and which for us is simply given by \begin{align} \SS^3 &\longrightarrow \SS^2 \nonumber\\ v &\longmapsto vv^\dagger \label{Hopf} \end{align} This map is not one-to-one. After all, it takes 3 angles to specify a point in $\SS^3$, but only 2 to specify a point in $\SS^2$.

What information have we lost in “squaring” $v$ to get $vv^\dagger$? If we multiply $v$ by a phase, that is, a complex number of the form $e^{i\theta}$, then not only is the norm ($\ref{scalar}$) unchanged, but also the square! Explicitly, if \begin{equation} w = v e^{i\theta} \label{phase} \end{equation} then \begin{equation} ww^\dagger = vv^\dagger \label{square} \end{equation} This phase freedom is also associated with a sphere, namely the circle $\SS^1$ of unit complex numbers — which includes antipodal points as a special case. This $\SS^1$ is in fact also a group, denoted $U(1)$, since the product of any two unit complex numbers is another such number. The Hopf map is also referred to as the *Hopf bundle*, because it describes $\SS^3$ as a *fibre bundle* with base space $\SS^2$ and fibre $\SS^1$; roughly speaking, there is a copy of $\SS^1$, given by the phase freedom, at each point of $\SS^2$.

Not surprisingly, the above construction works over any of the division algebras. Note the crucial use of the division algebra property in ($\ref{division}$) to conclude that $vv^\dagger$ is a null vector! Thus, allowing the components of $v$ in turn to be in $\RR$, $\CC$, $\HH$, and $\OO$ results in $vv^\dagger$ being a null vector in 3, 4, 6, or 10 spacetime dimensions, respectively. Imposing the normalization condition ($\ref{scalar}$) allows $vv^\dagger$ to be thought of as a point on the spheres $\SS^1$, $\SS^2$, $\SS^4$, and $\SS^8$, respectively, while $v$ can be interpreted as a point on the spheres $\SS^1$, $\SS^3$, $\SS^7$, and $\SS^{15}$, respectively.

There are a couple of details to check. Over $\HH$, it is essential that the phase freedom in ($\ref{phase}$) be written on the right. And of course the phase is given by an arbitrary unit quaternion, corresponding to $\SS^3$. Over $\OO$, we must be even more careful in identifying the phase freedom; there appear to be associativity issues in the transition from ($\ref{phase}$) to ($\ref{square}$). However, there is a way around this: $vv^\dagger$ is really complex, since there it contains only one octonionic direction. It therefore admits a complex “square root” $u$. If we apply the phase freedom to $u$, rather than $v$, the computation becomes quaternionic; the associativity problems disappear. Explicitly, choosing 2) \begin{equation} u = \begin{pmatrix} {b\bar{c}\over|c|}\cr \noalign{\smallskip} |c|\cr \end{pmatrix} \label{octohopf} \end{equation} we have \begin{equation} uu^\dagger = vv^\dagger \end{equation} and the most general column vector $w$ satisfying ($\ref{square}$) is \begin{equation} w = u\xi \end{equation} where the phase $\xi$ is any unit octonion, which can be thought of as an element of $\SS^7$. Note that we can recover $v$ by putting $\xi={c\over|c|}$.

Putting this all together, we obtain the four Hopf bundles \begin{align} \SS^{15} &\longrightarrow \SS^8 \nonumber\\ \SS^7 &\longrightarrow \SS^4 \nonumber\\ \SS^3 &\longrightarrow \SS^2 \\ \SS^1 &\longrightarrow \SS^1 \nonumber \end{align} (The last of these is pretty simple; the fibres are the 0-sphere $\SS^0$, which consists of the two points $\pm1$.) It is remarkable that there are precisely four of these bundles. It is even more remarkable that they are related to the four division algebras.

There is another way to view $vv^\dagger$, namely as an element in projective space. Consider a pair of real numbers $(b,c)$, and identify points on the same line through the origin. This can be though of as introducing an equivalence relation of the form \begin{equation} (b,c) \sim (b\chi,c\chi) \label{equiv} \end{equation} where $0\ne\chi\in\RR$. The resulting space can be identified with the (unit) circle of all possible directions in $\RR^2$, with antipodal points identified. This is the *real projective space* $\RR\PP^1$. But this space can also be identified with the squares of *normalized* column vector ($\ref{column}$) (with $b,c\in\RR$). In other words, \begin{equation} \RR\PP^1 = \{vv^\dagger: v\in\RR^2, v^\dagger v=1\} \end{equation}

The normalization condition can be written in terms of the trace of $vv^\dagger$, since \begin{equation} \tr(vv^\dagger) = v^\dagger v \end{equation} There is yet another way to write this, since \begin{equation} (vv^\dagger)(vv^\dagger) = v(v^\dagger v)v^\dagger = \left(\tr(vv^\dagger)\right) \, (vv^\dagger) \label{trID} \end{equation} Putting the pieces together, we obtain a matrix description of $\RR\PP^1$ in terms of $2\times2$ real matrices ($GL(2,\RR)$), namely 3) \begin{equation} \RR\PP^1 = \{\XX\in GL(2,\RR): \XX^\dagger=\XX, \XX^2 = \XX\} \label{projective} \end{equation}

Not surprisingly, all of this works over the other division algebras as well; ($\ref{trID}$) holds even over $\OO$ since $v$ has only 2 components, so that the computation takes place in a quaternionic subalgebra. Thus, ($\ref{projective}$) can be used to *define* the projective spaces $\RR\PP^1$, $\CC\PP^1$, $\HH\PP^1$, and $\OO\PP^1$.

However, the traditional definition, in terms of ($\ref{equiv}$), requires modification over the octonions, along the lines of the disucssion leading up to ($\ref{octohopf}$). One possible choice would be \begin{equation} \OO\PP^1 = \{(b,c)\in\OO^2:(b,c)\sim\left((bc^{-1})\chi,\chi\right),0\ne\chi\in\OO\} \end{equation} with $c=0$ handled as a special case.