Projective Lines Next: OP1 and Bott Periodicity Up: Octonionic Projective Geometry Previous: Octonionic Projective Geometry

3.1 Projective Lines

A one-dimensional projective space is called a projective line. Projective lines are not very interesting from the viewpoint of axiomatic projective geometry, since they have only one line on which all the points lie. Nonetheless, they can be geometrically and topologically interesting. This is especially true of the octonionic projective line. As we shall see, this space has a deep connection to Bott periodicity, and also to the Lorentzian geometry of 10-dimensional spacetime.

Suppose $\K$ is a normed division algebra. We have already defined $\KP^1$ when $\K$ is associative, but this definition does not work well for the octonions: it is wiser to take a detour through Jordan algebras. Let $\h _2(\K)$ be the space of $2 \times 2$ hermitian matrices with entries in $\K$. It is easy to check that this becomes a Jordan algebra with the product $a \circ b = {1\over 2}(ab + ba)$. We can try to build a projective space from this Jordan algebra using the construction in the previous section. To see if this succeeds, we need to ponder the projections in $\h _2(\K)$. A little calculation shows that besides the trivial projections 0 and 1, they are all of the form

% latex2html id marker 1582
\left( \begin{array}{c} x^* \\...
x^* x & x^* y \\  y^* x & y^* y
\end{array} \right)

where $(x,y) \in \K^2$ has

% latex2html id marker 1583
\Vert x\Vert^2 + \Vert y\Vert^2 = 1. \end{displaymath}

These nontrivial projections all have rank 1, so they are the points of our would-be projective space. Our would-be projective space has just one line, corresponding to the projection 1, and all the points lie on this line. It is easy to check that the axioms for a projective space hold. Since this projective space is 1-dimensional, we have succeeded in creating the projective line over $\K$. We call the set of points of this projective line $\KP^1$.

Given any nonzero element $(x,y) \in \K^2$, we can normalize it and then use the above formula to get a point in $\KP^1$, which we call $[(x,y)]$. This allows us to describe $\KP^1$ in terms of lines through the origin, as follows. Define an equivalence relation on nonzero elements of $\K^2$ by

% latex2html id marker 1584
(x,y) \sim (x',y') \; \iff \; [(x,y)] = [(x',y')] .\end{displaymath}

We call an equivalence class for this relation a line through the origin in $\K^2$. We can then identify points in $\KP^1$ with lines through the origin in $\K^2$.

Be careful: when $\K$ is the octonions, the line through the origin containing $(x,y)$ is not always equal to

% latex2html id marker 1585
\{(\alpha x, \alpha y)\; \colon \; \alpha \in \K\}. \end{displaymath}

This is only true when $\K$ is associative, or when $x$ or $y$ is $1$. Luckily, we have $(x,y) \sim (y^{-1}x,1)$ when $y \ne 0$ and $(x,y) \sim (1,x^{-1}y)$ when $x \ne 0$. Thus in either case we get a concrete description of the line through the origin containing $(x,y)$: when $x \ne 0$ it equals

% latex2html id marker 1586
\{(\alpha(y^{-1}x), \alpha)\; \colon \; \alpha \in \K\} , \end{displaymath}

and when $y \ne 0$ it equals

% latex2html id marker 1587
\{(\alpha, \alpha(x^{-1}y)\; \colon \; \alpha \in \K\} . \end{displaymath}

In particular, the line through the origin containing $(x,y)$ is always a real vector space isomorphic to $\K$.

We can make $\KP^1$ into a manifold as follows. By the above observations, we can cover it with two coordinate charts: one containing all points of the form $[(x,1)]$, the other containing all points of the form $[(1,y)]$. It is easy to check that $[(x,1)] = [(1,y)]$ iff $y =
x^{-1}$, so the transition function from the first chart to the second is the map $x \mapsto x^{-1}$. Since this transition function and its inverse are smooth on the intersection of the two charts, $\KP^1$ becomes a smooth manifold.

When pondering the geometry of projective lines it is handy to visualize the complex case, since $\CP^1$ is just the familiar 'Riemann sphere'. In this case, the map

\begin{displaymath}x \mapsto [(x,1)] \end{displaymath}

is given by stereographic projection:

% latex2html id marker 519

where we choose the sphere to have diameter 1. This map from $\C$ to $\CP^1$ is one-to-one and almost onto, missing only the point at infinity, or 'north pole'. Similarly, the map

\begin{displaymath}y \mapsto [(1,y)] \end{displaymath}

misses only the south pole. Composing the first map with the inverse of the second, we get the map $x \mapsto x^{-1}$, which goes by the name of 'conformal inversion'. The southern hemisphere of the Riemann sphere consists of all points $[(x,1)]$ with % latex2html id marker 3597
$\Vert x\Vert \le 1$, while the northern hemisphere consists of all $[(1,y)]$ with % latex2html id marker 3601
$\Vert y\Vert \le 1$. Unit complex numbers $x$ give points $[(x,1)] = [(1,x^{-1})]$ on the equator.

All these ideas painlessly generalize to $\KP^1$ for any normed division algebra $\K$. First of all, as a smooth manifold $\KP^1$ is just a sphere with dimension equal to that of $\K$:

% latex2html id marker 1590
\RP^1 &\i...
...S^2 \\  \HP^1 &\iso & S^4 \\  \OP^1 &\iso & S^8.

We can think of it as the one-point compactification of $\K$. The 'southern hemisphere', 'northern hemisphere', and 'equator' of $\K$ have descriptions exactly like those given above for the complex case. Also, as in the complex case, the maps $x \mapsto [(x,1)]$ and $y \mapsto [(1,y)]$ are angle-preserving with respect to the usual Euclidean metric on $\K$ and the round metric on the sphere.

One of the nice things about $\KP^1$ is that it comes equipped with a vector bundle whose fiber over the point $[(x,y)]$ is the line through the origin corresponding to this point. This bundle is called the canonical line bundle, $L_\K$. Of course, when we are working with a particular division algebra, 'line' means a copy of this division algebra, so if we think of them as real vector bundles, $L_\R, L_\C,
L_\H$ and $L_\O$ have dimensions 1,2,4, and 8, respectively.

These bundles play an important role in topology, so it is good to understand them in a number of ways. In general, any $k$-dimensional real vector bundle over $S^n$ can be formed by taking trivial bundles over the northern and southern hemispheres and gluing them together along the equator via a map $f \maps S^{n-1} \to \OO (k)$. We must therefore be able to build the canonical line bundles $L_\R, L_\C,
L_\H$ and $L_\O$ using maps

% latex2html id marker 1591
f_\R \ma... & \OO (4) \\  f_\O \maps &S^7& \to & \OO (8).

What are these maps? We can describe them all simultaneously. Suppose $\K$ is a normed division algebra of dimension $n$. In the southern hemisphere of $\KP^1$, we can identify any fiber of $L_\K$ with $\K$ by mapping the point $(\alpha x, \alpha)$ in the line $[(x,1)]$ to the element $\alpha \in \K$. This trivializes the canonical line bundle over the southern hemisphere. Similarly, we can trivialize this bundle over the northern hemisphere by mapping the point $(\beta,\beta y)$ in the line $[(1,y)]$ to the element $\beta \in \K$. If $x \in \K$ has norm one, $[(x,1)] = [(1,x^{-1})]$ is a point on the equator, so we get two trivializations of the fiber over this point. These are related as follows: if $(\alpha x, \alpha) = (\beta, \beta x^{-1})$ then $\beta =
\alpha x$. The map $\alpha \mapsto \beta$ is thus right multiplication by $x$. In short,

\begin{displaymath}f_\K \maps S^{n-1} \to \OO (n) \end{displaymath}

is just the map sending any norm-one element $x \in \K$ to the operation of right multiplication by $x$.

The importance of the map $f_\K$ becomes clearest if we form the inductive limit of the groups $\OO (n)$ using the obvious inclusions $\OO (n) \hookrightarrow \OO (n+1)$, obtaining a topological group called $\OO (\infty)$. Since $\OO (n)$ is included in $\OO (\infty)$, we can think of $f_\K$ as a map from $S^{n-1}$ to $\OO (\infty)$. Its homotopy class $[f_\K]$ has the following marvelous property, mentioned in the Introduction:

Another nice perspective on the canonical line bundles $L_\K$ comes from looking at their unit sphere bundles. Any fiber of $L_\K$ is naturally an inner product space, since it is a line through the origin in $\K^2$. If we take the unit sphere in each fiber, we get a bundle of $(n-1)$-spheres over $\KP^1$ called the Hopf bundle:

\begin{displaymath}p_\K \maps E_\K \to \KP^1 \end{displaymath}

The projection $p_\K$ is called the Hopf map. The total space $E_\K$ consists of all the unit vectors in $\K^2$, so it is a sphere of dimension $2n-1$. In short, the Hopf bundles look like this:

% latex2html id marker 1594
\K = \R:...
...O: \qquad &S^7 \hookrightarrow &S^{15}& \to &S^8

We can understand the Hopf maps better by thinking about inverse images of points. The inverse image $p_\K^{-1}(x)$ of any point $x \in S^n$ is a $(n-1)$-sphere in $S^{2n-1}$, and the inverse image of any pair of distinct points is a pair of linked spheres of this sort. When $\K =
\C$ we get linked circles in $S^3$, which form the famous Hopf link:

% latex2html id marker 548

When $\K = \O$, we get a pair of linked 7-spheres in $S^{15}$.

To quantify this notion of linking, we can use the 'Hopf invariant'. Suppose for a moment that $n$ is any natural number greater than one, and let $f \maps S^{2n-1} \to S^n$ be any smooth map. If $\omega$ is the normalized volume form on $S^n$, then $f^* \omega$ is a closed $n$-form on $S^{2n-1}$. Since the $n$th cohomology of $S^{2n-1}$ vanishes, $f^\ast \omega = d \alpha$ for some $(n-1)$-form $\alpha$. We define the Hopf invariant of $f$ to be the number

\begin{displaymath}H(f) = \int_{S^{2n-1}} \alpha \wedge d\alpha .\end{displaymath}

This is easily seen to be invariant under smooth homotopies of the map $f$.

To see how the Hopf invariant is related to linking, we can compute it using homology rather than cohomology. If we take any two regular values of $f$, say $x$ and $y$, the inverse images of these points are compact oriented $(n-1)$-dimensional submanifolds of $S^{2n-1}$. We can always find an oriented $n$-dimensional submanifold $X \subset
S^{2n-1}$ that has boundary equal to $f^{-1}(x)$ and that intersects $f^{-1}(y)$ transversely. The dimensions of $X$ and $f^{-1}(y)$ add up to $2n-1$, so their intersection number is well-defined. By the duality between homology and cohomology, this number equals the Hopf invariant $H(f)$. This shows that the Hopf invariant is an integer. Moreover, it shows that when the Hopf invariant is nonzero, the inverse images of $x$ and $y$ are linked.

Using either of these approaches we can compute the Hopf invariant of $p_\C$, $p_\H$ and $p_\O$. They all turn out to have Hopf invariant 1. This implies, for example, that the inverse images of distinct points under $p_\O$ are nontrivially linked 7-spheres in $S^{15}$. It also implies that $p_\C$, $p_\H$ and $p_\O$ give nontrivial elements of $\pi_{2n-1}(S^n)$ for $n = 2, 4$, and $8$. In fact, these elements generate the torsion-free part of $\pi_{2n-1}(S^n)$.

A deep study of the Hopf invariant is one way to prove that any division algebra must have dimension 1, 2, 4 or 8. One can show that if there exists an $n$-dimensional division algebra, then $S^{n-1}$ must be parallelizable: it must admit $n-1$ pointwise linearly independent smooth vector fields. One can also show that for $n > 1$, $S^{n-1}$ is parallelizable iff there exists a map $f \maps S^{2n-1} \to S^n$ with $H(f) = 1$. The hard part is the final step: showing that a map from $S^{2n-1}$ to $S^n$ can have Hopf invariant 1 only if $n = 2, 4$, or $8$. Proving this requires algebraic topology that goes far beyond the scope of this paper [2,9,52]. There is just one thing we wish to note about this proof: it involves Bott periodicity, which we describe in the next section. As we shall see, Bott periodicity has a natural explanation in terms of the canonical line bundle over $\OP^1$. There is thus a sense in which the existence of the octonions is 'responsible' for the nonexistence of division algebras in dimensions other than 1, 2, 4, and 8!

Next: OP1 and Bott Periodicity Up: Octonionic Projective Geometry Previous: Octonionic Projective Geometry

© 2001 John Baez