4
$\begingroup$

In this post John Baez states that the classical simple Lie groups "arise naturally as symmetry groups of projective spaces over $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$".

Is there some intuitive explanation for this connection?

  • I understand that $SO(n)$ is the set of all rotations about the origin of $n$-dimensional Euclidean space $\mathbb{R}^n$
  • I understand that the rotation group $O(n+1)$ is the group of isometries of the $n$-sphere. Rotations are the subset with determinant $1$, whereas reflections are the subset with determinant $-1$.

However I don't understand how projective spaces come into play here? Why are they important in this context?

$\endgroup$
  • 1
    $\begingroup$ A partial explanation is that the actions of these groups on the appropriate projective spaces is in some ways nicer than those on the real/complex/quaternionic vector spaces. For example, $SO(n)$ acts transitively on $\Bbb{RP}^{n-1} = \Bbb P(\Bbb R^n)$ but (since it preserves lengths) not on $\Bbb R^n$ itself. $\endgroup$ – Travis Apr 29 '16 at 12:22
  • 1
    $\begingroup$ One can split the difference here by looking at the $SO(n)$ action on a sphere $\Bbb S^n$ (which is a single orbit of the latter action), which we can identify with the "ray projectivization" of $\Bbb R^n$. The projection $\Bbb S^{n - 1} \to \Bbb {RP}^{n - 1}$ is a $2$-fold cover and is equivariant w.r.t. these actions. $\endgroup$ – Travis Apr 29 '16 at 12:24
2
$\begingroup$

To each lie algebra $\mathfrak{g}$ there is a unique simply connected Lie group $G$ having $\mathfrak{g}$ as its lie algebra, and furthermore any other Lie group $H$ having lie algebra $\mathfrak{g}$ is covered by the universal one $G$, in other words is a quotinet of $G$ by some discrete central subgroup $K$. (In fact, covering space theory goes on further to say that $\pi_1(H)\cong K$ canonically.) In some sense, any of the many distinct Lie groups sitting above a given lie algebra $\mathfrak{g}$ are still "approximately" the same. Indeed there is a formal sense in which they are "locally isomorphic," which is just equivalent to having the same lie algebra.

Due to this, we sometimes treat locally isomorphic Lie groups with some degree of interchangability, at least when we're speaking loosely. I've even seen physicists or physical mathematicians (which unfortunately isn't a real term) write down, in published papers and textbooks, technically false isomorphisms like $\mathrm{SU}(2)\cong\mathrm{SO}(3)$ (looking at you, Geometry of the Octonions). Anyway, the point is two spaces or groups that are the same up to covering maps are for many people more or less the same thing.

The spaces $\mathbb{R}^{n+1}$, $\mathbb{C}^{n+1}$ and $\mathbb{H}^{n+1}$ may be considered vector spaces over $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$ respectively. If you want linear transformations to act from the left by matrices, you'll want (at least for $\mathbb{H}$ since it is not commutative) scalars to be applied to vectors from the right. Each of these spaces has an inner product defined on it, essentially the dot product but with the entries of either the left or right vector conjugated first (depending on if it's a mathematician or physicist talking).

The matrix transformations which preserve the inner product, $\langle Au,Av\rangle=\langle u,v\rangle$ for all $u,v$, form the symmetry groups $\mathrm{O}(n+1)$, $\mathrm{U}(n+1)$ and $\mathrm{Sp}(n+1)$. The middle one $\mathrm{U}(n+1)$ is a little off for being simple, since it has an extra dimension coming from the fact their determinants needn't be $1$. This extra dimension can be siphoned off by passing to the projective spaces $\mathbb{RP}^n$, $\mathbb{CP}^n$, $\mathbb{HP}^n$, which are defined to be the moduli spaces of $1$-dimensional subspaces of $\mathbb{R}^{n+1}$, $\mathbb{C}^{n+1}$ or $\mathbb{H}^{n+1}$ respectively (dimension $1$ with respect to whatever the field or skew field of scalars is).

The isometry group $\mathrm{PO}(n+1)$ of $\mathbb{RP}^n$ becomes connected. If $n+1$ is odd it is $\cong\mathrm{SO}(n+1)$, and if $n+1$ is even it is $\cong\mathrm{SO}(n+1)/\{\pm\mathrm{Id}\}$. Similarly, since $e^{i\theta}\,\mathrm{Id}\,$s don't move a complex subspace of $\mathbb{C}^n$ to a different one, those disappear in the isometry group $\mathrm{PSU}(n)$ of $\mathbb{CP}^{n-1}$. It's tempting to wonder why the scalar multiples of the identity in $\mathrm{Sp}(n+1)$ don't disappear (except for $-\mathrm{Id}$) for the same reason; see if you can figure out why they act nontrivially on $\mathbb{HP}^n$.

At the infinitessimal level, the corresponding lie algebras are just $\mathfrak{so}(m)$, $\mathfrak{su}(m)$, $\mathfrak{sp}(m)$, which means we could have just introduced the simple Lie groups as $\mathrm{SO}(m)$, $\mathrm{SU}(m)$, $\mathrm{Sp}(m)$ to begin with, although keep in mind it takes some technical ingenuity to construct a uniform definition that covers all three (and generalizes to $\mathbb{O}$, but that's another story). I believe the reason Baez wants to identify the simple Lie groups as isometry groups of projective spaces, though, is because then this additionally encompasses the four exceptional lie groups $F_4,E_6,E_7,E_8$ as the isometry groups of projective planes over $\mathbb{R}\otimes\mathbb{O}$, $\mathbb{C}\otimes\mathbb{O}$, $\mathbb{H}\otimes\mathbb{O}$ and $\mathbb{O}\otimes\mathbb{O}$ respectively.

(Keep in mind there is a small caveat that $\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{s}(3)$ is not actually simple.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.