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This morning I heard the awful news that Bill Thurston died last night. Many of us knew that Bill was very ill, but we all hoped (or imagined?) that he would still be with us for a while yet, and the suddenness of this is very harsh. As Sarah Koch put it in an email to me, “Although this was not unexpected, it is still shocking.” On the other hand, I am glad to hear that he was surrounded by family, and died peacefully.

I counted Bill as my friend, as well as my mentor, and I have many vivid and happy memories of time I spent with him. I hope that writing down a few of these reminiscences will be cathartic for me, and for others who are coping with this loss.

I am Alden, one of Danny’s students. Error/naivete that may (will) be found here is mine. In these posts, I will attempt to give notes from Danny’s class on hyperbolic geometry (157b). This first post covers some models for hyperbolic space.

1. Models

We have a very good natural geometric understanding of ${\mathbb{E}^3}$, i.e. 3-space with the euclidean metric. Pretty much all of our geometric and topological intuition about manifolds (Riemannian or not) comes from finding some reasonable way to embed or immerse them (perhaps locally) in ${\mathbb{E}^3}$. Let us look at some examples of 2-manifolds.

• Example (curvature = 1) ${S^2}$ with its standard metric embeds in ${\mathbb{E}^2}$; moreover, any isometry of ${S^2}$ is the restriction of (exactly one) isometry of the ambient space (this group of isometries being ${SO(3)}$). We could not ask for anything more from an embedding.
• Example (curvature = 0) Planes embed similarly.
• Example (curvature = -1) The pseudosphere gives an example of an isometric embedding of a manifold with constant curvature -1. Consider a person standing in the plane at the origin. The person holds a string attached to a rock at ${(0,1)}$, and they proceed to walk due east dragging the rock behind them. The movement of the rock is always straight towards the person, and its distance is always 1 (the string does not stretch). The line traced out by the rock is a tractrix. Draw a right triangle with hypotenuse the tangent line to the curve and vertical side a vertical line to the ${x}$-axis. The bottom has length ${\sqrt{1-y^2}}$, which shows that the tractrix is the solution to the differential equation$\displaystyle \frac{-y}{\sqrt{1-y^2}} = \frac{dy}{dx}$

The Tractrix

The surface of revolution about the ${x}$-axis is the pseudosphere, an isometric embedding of a surface of constant curvature -1. Like the sphere, there are some isometries of the pseudosphere that we can understand as isometries of ${\mathbb{E}^3}$, namely rotations about the ${x}$-axis. However, there are lots of isometries which do not extend, so this embeddeding does not serve us all that well.

• Example (hyperbolic space) By the Nash embedding theorem, there is a ${\mathcal{C}^1}$ immersion of ${\mathbb{H}^2}$ in ${\mathbb{E}^3}$, but by Hilbert, there is no ${\mathcal{C}^2}$ immersion of any complete hyperbolic surface.That last example is the important one to consider when thinking about hypobolic spaces. Intuitively, manifolds with negative curvature have a hard time fitting in euclidean space because volume grows too fast — there is not enough room for them. The solution is to find (local, or global in the case of ${\mathbb{H}^2}$) models for hyperbolic manfolds such that the geometry is distorted from the usual euclidean geometry, but the isometries of the space are clear.

2. 1-Dimensional Models for Hyperbolic Space

While studying 1-dimensional hyperbolic space might seem simplistic, there are nice models such that higher dimensions are simple generalizations of the 1-dimensional case, and we have such a dimensional advantage that our understanding is relatively easy.

2.1. Hyperboloid Model

Parameterizing ${H}$

Consider the quadratic form ${\langle \cdot, \cdot \rangle_H}$ on ${\mathbb{R}^2}$ defined by ${\langle v, w \rangle_A = \langle v, w \rangle_H = v^TAw}$, where ${A = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]}$. This doesn’t give a norm, since ${A}$ is not positive definite, but we can still ask for the set of points ${v}$ with ${\langle v, v \rangle_H = -1}$. This is (both sheets of) the hyperbola ${x^2-y^2 = -1}$. Let ${H}$ be the upper sheet of the hyperbola. This will be 1-dimensional hyperbolic space.

For any ${n\times n}$ matrix ${B}$, let ${O(B) = \{ M \in \mathrm{Mat}(n,\mathbb{R}) \, | \, \langle v, w \rangle_B = \langle Mv, Mw \rangle_B \}}$. That is, matrices which preserve the form given by ${A}$. The condition is equivalent to requiring that ${M^TBM = B}$. Notice that if we let ${B}$ be the identity matrix, we would get the regular orthogonal group. We define ${O(p,q) = O(B)}$, where ${B}$ has ${p}$ positive eigenvalues and ${q}$ negative eigenvalues. Thus ${O(1,1) = O(A)}$. We similarly define ${SO(1,1)}$ to be matricies of determinant 1 preserving ${A}$, and ${SO_0(1,1)}$ to be the connected component of the identity. ${SO_0(1,1)}$ is then the group of matrices preserving both orientation and the sheets of the hyperbolas.

We can find an explicit form for the elements of ${SO_0(1,1)}$. Consider the matrix ${M = \left[ \begin{array}{cc} a & b \\ c& d \end{array} \right]}$. Writing down the equations ${M^TAM = A}$ and ${\det(M) = 1}$ gives us four equations, which we can solve to get the solutions

$\displaystyle \left[ \begin{array}{cc} \sqrt{b^2+1} & b \\ b & \sqrt{b^2+1} \end{array} \right] \textrm{ and } \left[ \begin{array}{cc} -\sqrt{b^2+1} & b \\ b & -\sqrt{b^2+1} \end{array} \right].$

Since we are interested in the connected component of the identity, we discard the solution on the right. It is useful to do a change of variables ${b = \sinh(t)}$, so we have (recall that ${\cosh^2(t) - \sinh^2(t) = 1}$).

$\displaystyle SO_0(1,1) = \left\{ \left[ \begin{array}{cc} \cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t) \end{array} \right] \, | \, t \in \mathbb{R} \right\}$

These matrices take ${\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]}$ to ${\left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right]}$. In other words, ${SO_0(1,1)}$ acts transitively on ${H}$ with trivial stabilizers, and in particular we have parmeterizing maps

$\displaystyle \mathbb{R} \rightarrow SO_0(1,1) \rightarrow H \textrm{ defined by } t \mapsto \left[ \begin{array}{cc} \cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t) \end{array} \right] \mapsto \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right]$

The first map is actually a Lie group isomorphism (with the group action on ${\mathbb{R}}$ being ${+}$) in addition to a diffeomorphism, since

$\displaystyle \left[ \begin{array}{cc} \cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t) \end{array} \right] \left[ \begin{array}{cc} \cosh(s) & \sinh(s) \\ \sinh(s) & \cosh(s) \end{array} \right] = \left[ \begin{array}{cc} \cosh(t+s) & \sinh(t+s) \\ \sinh(t+s) & \cosh(t+s) \end{array} \right]$

Metric

As mentioned above, ${\langle \cdot, \cdot \rangle_H}$ is not positive definite, but its restriction to the tangent space of ${H}$ is. We can see this in the following way: tangent vectors at a point ${p \in H}$ are characterized by the form ${\langle \cdot, \cdot \rangle_H}$. Specifically, ${v\in T_pH \Leftrightarrow \langle v, p \rangle_H}$, since (by a calculation) ${\frac{d}{dt} \langle p+tv, p+tv \rangle_H = 0 \Leftrightarrow \langle v, p \rangle_H}$. Therefore, ${SO_0(1,1)}$ takes tangent vectors to tangent vectors and preserves the form (and is transitive), so we only need to check that the form is positive definite on one tangent space. This is obvious on the tangent space to the point ${\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]}$. Thus, ${H}$ is a Riemannian manifold, and ${SO_0(1,1)}$ acts by isometries.

Let’s use the parameterization ${\phi: t \mapsto \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right]}$. The unit (in the ${H}$ metric) tangent at ${\phi(t) = \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right]}$ is ${\left[ \begin{array}{c} \cosh(t) \\ \sinh(t) \end{array} \right]}$. The distance between the points ${\phi(s)}$ and ${\phi(t)}$ is

$\displaystyle d_H(\phi(s), \phi(t)) = \left| \int_s^t\sqrt{\langle \left[ \begin{array}{c} \cosh(t) \\ \sinh(t) \end{array} \right], \left[ \begin{array}{c} \cosh(t) \\ \sinh(t) \end{array} \right] \rangle_H dv } \right| = \left|\int_s^tdv \right| = |t-s|$

In other words, ${\phi}$ is an isometry from ${\mathbb{E}^1}$ to ${H}$.

1-dimensional hyperbollic space. The hyperboloid model is shown in blue, and the projective model is shown in red. An example of the projection map identifying ${H}$ with ${(-1,1) \subseteq \mathbb{R}\mathrm{P}^1}$ is shown.

2.2. Projective Model

Parameterizing

Real projective space ${\mathbb{R}\mathrm{P}^1}$ is the set of lines through the origin in ${\mathbb{R}^2}$. We can think about ${\mathbb{R}\mathrm{P}^1}$ as ${\mathbb{R} \cup \{\infty\}}$, where ${x\in \mathbb{R}}$ is associated with the line (point in ${\mathbb{R}\mathrm{P}^1}$) intersecting ${\{y=1\}}$ in ${x}$, and ${\infty}$ is the horizontal line. There is a natural projection ${\mathbb{R}^2 \setminus \{0\} \rightarrow \mathbb{R}\mathrm{P}^1}$ by projecting a point to the line it is on. Under this projection, ${H}$ maps to ${(-1,1)\subseteq \mathbb{R} \subseteq \mathbb{R}\mathrm{P}^1}$.

Since ${SO_0(1,1)}$ acts on ${\mathbb{R}^2}$ preserving the lines ${y = \pm x}$, it gives a projective action on ${\mathbb{R}\mathrm{P}^1}$ fixing the points ${\pm 1}$. Now suppose we have any projective linear isomorphism of ${\mathbb{R}\mathrm{P}^1}$ fixing ${\pm 1}$. The isomorphism is represented by a matrix ${A \in \mathrm{PGL}(2,\mathbb{R})}$ with eigenvectors ${\left[ \begin{array}{c} 1 \\ \pm 1 \end{array} \right]}$. Since scaling ${A}$ preserves its projective class, we may assume it has determinant 1. Its eigenvalues are thus ${\lambda}$ and ${\lambda^{-1}}$. The determinant equation, plus the fact that

$\displaystyle A \left[ \begin{array}{c} 1 \\ \pm 1 \end{array} \right] = \left[ \begin{array}{c} \lambda^{\pm 1} \\ \pm \lambda^{\pm 1} \end{array} \right]$

Implies that ${A}$ is of the form of a matrix in ${SO_0(1,1)}$. Therefore, the projective linear structure on ${(-1,1) \subseteq \mathbb{R}\mathrm{P}^1}$ is the “same” (has the same isometry (isomorphism) group) as the hyperbolic (Riemannian) structure on ${H}$.

Metric

Clearly, we’re going to use the pushforward metric under the projection of ${H}$ to ${(-1,1)}$, but it turns out that this metric is a natural choice for other reasons, and it has a nice expression.

The map taking ${H}$ to ${(-1,1) \subseteq \mathbb{R}\mathrm{P}^1}$ is ${\psi: \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right] \rightarrow \frac{\sinh(t)}{\cosh(T)} = \tanh(t)}$. The hyperbolic distance between ${x}$ and ${y}$ in ${(-1,1)}$ is then ${d_H(x,y) = |\tanh^{-1}(x) - \tanh^{-1}(y)|}$ (by the fact from the previous sections that ${\phi}$ is an isometry).

Recall the fact that ${\tanh(a\pm b) = \frac{\tanh(a) \pm \tanh(b)}{1 \pm \tanh(a)\tanh(b)}}$. Applying this, we get the nice form

$\displaystyle d_H(x,y) = \frac{y-x}{1 - xy}$

We also recall the cross ratio, for which we fix notation as ${ (z_1, z_2; z_3, z_4) := \frac{(z_3 -z_1)(z_4-z_2)}{(z_2-z_1)(z_4-z_3)}}$. Then

$\displaystyle (-1, x;y,1 ) = \frac{(y+1)(1-x)}{(x+1)(1-y)} = \frac{1-xy + (y-x)}{1-xy + (x-y)}$

Call the numerator of that fraction by ${N}$ and the denominator by ${D}$. Then, recalling that ${\tanh(u) = \frac{e^{2u}-1}{e^{2u}+1}}$, we have

$\displaystyle \tanh(\frac{1}{2} \log(-1,x;y,1)) = \frac{\frac{N}{D} -1}{\frac{N}{D} +1} = \frac{N-D}{N+D} = \frac{2(y-x)}{2(1-xy)} = \tanh(d_H(x,y))$

Therefore, ${d_H(x,y) = \frac{1}{2}\log(-1,x;y,-1)}$.

3. Hilbert Metric

Notice that the expression on the right above has nothing, a priori, to do with the hyperbolic projection. In fact, for any open convex body in ${\mathbb{R}\mathrm{P}^n}$, we can define the Hilbert metric on ${C}$ by setting ${d_H(p,q) = \frac{1}{2}\log(a,p,q,b)}$, where ${a}$ and ${b}$ are the intersections of the line through ${a}$ and ${b}$ with the boundary of ${C}$. How is it possible to take the cross ratio, since ${a,p,q,b}$ are not numbers? The line containing all of them is projectively isomorphic to ${\mathbb{R}\mathrm{P}^1}$, which we can parameterize as ${\mathbb{R} \cup \{\infty\}}$. The cross ratio does not depend on the choice of parameterization, so it is well defined. Note that the Hilbert metric is not necessarily a Riemannian metric, but it does make any open convex set into a metric space.

Therefore, we see that any open convex body in ${\mathbb{R}\mathrm{P}^n}$ has a natural metric, and the hyperbolic metric in ${H = (-1,1)}$ agrees with this metric when ${(-1,1)}$ is thought of as a open convex set in ${\mathbb{R}\mathrm{P}^1}$.

4. Higher-Dimensional Hyperbolic Space

4.1. Hyperboloid

The higher dimensional hyperbolic spaces are completely analogous to the 1-dimensional case. Consider ${\mathbb{R}^{n+1}}$ with the basis ${\{e_i\}_{i=1}^n \cup \{e\}}$ and the 2-form ${\langle v, w \rangle_H = \sum_{i=1}^n v_iw_i - v_{n+1}w_{n+1}}$. This is the form defined by the matrix ${J = I \oplus (-1)}$. Define ${\mathbb{H}^n}$ to be the positive (positive in the ${e}$ direction) sheet of the hyperbola ${\langle v,v\rangle_H = -1}$.

Let ${O(n,1)}$ be the linear transformations preserving the form, so ${O(n,1) = \{ A \, | \, A^TJA = J\}}$. This group is generated by ${O(1,1) \subseteq O(n,1)}$ as symmetries of the ${e_1, e}$ plane, together with ${O(n) \subseteq O(n,1)}$ as symmetries of the span of the ${e_i}$ (this subspace is euclidean). The group ${SO_0(n,1)}$ is the set of orientation preserving elements of ${O(n,1)}$ which preserve the positive sheet of the hyperboloid (${\mathbb{H}^n}$). This group acts transitively on ${\mathbb{H}^n}$ with point stabilizers ${SO(n)}$: this is easiest to see by considering the point ${(0,\cdots, 0, 1) \in \mathbb{H}^n}$. Here the stabilizer is clearly ${SO(n)}$, and because ${SO_0(n,1)}$ acts transitively, any stabilizer is a conjugate of this.

As in the 1-dimensional case, the metric on ${\mathbb{H}^n}$ is ${\langle \cdot , \cdot \rangle_H|_{T_p\mathbb{H}^n}}$, which is invariant under ${SO_0(n,1)}$.

Geodesics in ${\mathbb{H}^n}$ can be understood by consdering the fixed point sets of isometries, which are always totally geodesic. Here, reflection in a vertical (containing ${e}$) plane restricts to an (orientation-reversing, but that’s ok) isometry of ${\mathbb{H}^n}$, and the fixed point set is obviously the intersection of this plane with ${\mathbb{H}^n}$. Now ${SO_0(n,1)}$ is transitive on ${\mathbb{H}^n}$, and it sends planes to planes in ${\mathbb{R}^{n+1}}$, so we have a bijection

{Totally geodesic subspaces through ${p}$} ${\leftrightarrow}$ ${\mathbb{H}^n \cap}$ {linear subspaces of ${\mathbb{R}^{n+1}}$ through ${p}$ }

By considering planes through ${e}$, we can see that these totally geodesic subspaces are isometric to lower dimensional hyperbolic spaces.

4.2. Projective

Analogously, we define the projective model as follows: consider the disk ${\{v_{n+1} \,| v_{n+1} = 1, \langle v,v \rangle_H < 0\}}$. I.e. the points in the ${v_{n+1}}$ plane inside the cone ${\langle v,v \rangle_H = 0}$. We can think of ${\mathbb{R}\mathrm{P}^n}$ as ${\mathbb{R}^n \cup \mathbb{R}\mathrm{P}^{n-1}}$, so this disk is ${D^\circ \subseteq \mathbb{R}^n \subseteq \mathbb{R}\mathrm{P}^n}$. There is, as before, the natural projection of ${\mathbb{H}^n}$ to ${D^\circ}$, and the pushforward of the hyperbolic metric agrees with the Hilbert metric on ${D^\circ}$ as an open convex body in ${\mathbb{R}\mathrm{P}^n}$.

Geodesics in the projective model are the intersections of planes in ${\mathbb{R}^{n+1}}$ with ${D^\circ}$; that is, they are geodesics in the euclidean space spanned by the ${e_i}$. One interesting consequence of this is that any theorem which is true in euclidean geometry which does not reply on facts about angles is still true for hyperbolic space. For example, Pappus’ hexagon theorem, the proof of which does not use angles, is true.

4.3. Projective Model in Dimension 2

In the case that ${n=2}$, we can understand the projective isomorphisms of ${\mathbb{H}^2 = D \subseteq \mathbb{R}\mathrm{P}^2}$ by looking at their actions on the boundary ${\partial D}$. The set ${\partial D}$ is projectively isomorphic to ${\mathbb{R}\mathrm{P}^1}$ as an abstract manifold, but it should be noted that ${\partial D}$ is not a straight line in ${\mathbb{R}\mathrm{P}^2}$, which would be the most natural way to find ${\mathbb{R}\mathrm{P}^1}$‘s embedded in ${\mathbb{R}\mathrm{P}^2}$.

In addition, any projective isomorphism of ${\mathbb{R}\mathrm{P}^1 \cong \partial D}$ can be extended to a real projective isomorphism of ${\mathbb{R}\mathrm{P}^2}$. In other words, we can understand isometries of 2-dimensional hyperbolic space by looking at the action on the boundary. Since ${\partial D}$ is not a straight line, the extension is not trivial. We now show how to do this.

The automorphisms of ${\partial D \cong \mathbb{R}\mathrm{P}^1}$ are ${\mathrm{PSL}(2,\mathbb{R}}$. We will consider ${\mathrm{SL}(2,\mathbb{R})}$. For any Lie group ${G}$, there is an Adjoint action ${G \rightarrow \mathrm{Aut}(T_eG)}$ defined by (the derivative of) conjugation. We can similarly define an adjoint action ${\mathrm{ad}}$ by the Lie algebra on itself, as ${\mathrm{ad}(\gamma '(0)) := \left. \frac{d}{dt} \right|_{t=0} \mathrm{Ad}(\gamma(t))}$ for any path ${\gamma}$ with ${\gamma(0) = e}$. If the tangent vectors ${v}$ and ${w}$ are matrices, then ${\mathrm{ad}(v)(w) = [v,w] = vw-wv}$.

We can define the Killing form ${B}$ on the Lie algebra by ${B(v,w) = \mathrm{Tr}(\mathrm{ad}(v)\mathrm{ad}(w))}$. Note that ${\mathrm{ad}(v)}$ is a matrix, so this makes sense, and the Lie group acts on the tangent space (Lie algebra) preserving this form.

Now let’s look at ${\mathrm{SL}(2,\mathbb{R})}$ specifically. A basis for the tangent space (Lie algebra) is ${e_1 = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]}$, ${e_2 = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]}$, and ${e_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]}$. We can check that ${[e_1,e_2] = e_3}$, ${[e_1,e_3] = -2e_1}$, and ${[e_2, e_3]=2e_2}$. Using these relations plus the antisymmetry of the Lie bracket, we know

$\displaystyle \mathrm{ad}(e_1) = \left[ \begin{array}{ccc} 0 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \qquad \mathrm{ad}(e_2) = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ -1 & 0 & 0 \end{array}\right] \qquad \mathrm{ad}(e_3) = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{array}\right]$

Therefore, the matrix for the Killing form in this basis is

$\displaystyle B_{ij} = B(e_i,e_j) = \mathrm{Tr}(\mathrm{ad}(e_i)\mathrm{ad}(e_j)) = \left[ \begin{array}{ccc} 0 & 4 & 0 \\ 4 & 0 & 0 \\ 0 & 0 & 8 \end{array}\right]$

This matrix has 2 positive eigenvalues and one negative eigenvalue, so its signature is ${(2,1)}$. Since ${\mathrm{SL}(2,\mathbb{R})}$ acts on ${T_e(\mathrm{SL}(2,\mathbb{R}))}$ preserving this form, we have ${\mathrm{SL}(2,\mathbb{R}) \cong O(2,1)}$, otherwise known at the group of isometries of the disk in projective space ${\mathbb{R}\mathrm{P}^2}$, otherwise known as ${\mathbb{H}^2}$.

Any element of ${\mathrm{PSL}(2,\mathbb{R})}$ (which, recall, was acting on the boundary of projective hyperbolic space ${\partial D}$) therefore extends to an element of ${O(2,1)}$, the isometries of hyperbolic space, i.e. we can extend the action over the disk.

This means that we can classify isometries of 2-dimensional hyperbolic space by what they do to the boundary, which is determined generally by their eigevectors (${\mathrm{PSL}(2,\mathbb{R})}$ acts on ${\mathbb{R}\mathrm{P}^1}$ by projecting the action on ${\mathbb{R}^2}$, so an eigenvector of a matrix corresponds to a fixed line in ${\mathbb{R}^2}$, so a fixed point in ${\mathbb{R}\mathrm{P}^1 \cong \partial D}$. For a matrix ${A}$, we have the following:

• ${|\mathrm{Tr}(A)| < 2}$ (elliptic) In this case, there are no real eigenvalues, so no real eigenvectors. The action here is rotation, which extends to a rotation of the entire disk.
• ${|\mathrm{Tr}(A)| = 2}$ (parabolic) There is a single real eigenvector. There is a single fixed point, to which all other points are attracted (in one direction) and repelled from (in the other). For example, the action in projective coordinates sending ${[x:y]}$ to ${[x+1:y]}$: infinity is such a fixed point.
• ${|\mathrm{Tr}(A)| > 2}$ (hyperbolic) There are two fixed point, one attracting and one repelling.
•

5. Complex Hyperbolic Space

We can do a construction analogous to real hyperbolic space over the complexes. Define a Hermitian form ${q}$ on ${\mathbb{C}^{n+1}}$ with coordinates ${\{z_1,\cdots, z_n\} \cup \{w\}}$ by ${q(x_1,\cdots x_n, w) = |z_1|^2 + \cdots + |z_n|^2 - |w|^2}$. We will also refer to ${q}$ as ${\langle \cdot, \cdot \rangle_q}$. The (complex) matrix for this form is ${J = I \oplus (-1)}$, where ${q(v,w) = v^*Jw}$. Complex linear isomorphisms preserving this form are matrices ${A}$ such that ${A^*JA = J}$. This is our definition for ${\mathrm{U}(q) := \mathrm{U}(n,1)}$, and we define ${\mathrm{SU}(n,1)}$ to be those elements of ${\mathrm{U}(n,1)}$ with determinant of norm 1.

The set of points ${z}$ such that ${q(z) = -1}$ is not quite what we are looking for: first it is a ${2n+1}$ real dimensional manifold (not ${2n}$ as we would like for whatever our definition of “complex hyperbolic ${n}$ space” is), but more importantly, ${q}$ does not restrict to a positive definite form on the tangent spaces. Call the set of points ${z}$ where ${q(z) = -1}$ by ${\bar{H}}$. Consider a point ${p}$ in ${\bar{H}}$ and ${v}$ in ${T_p\bar{H}}$. As with the real case, by the fact that ${v}$ is in the tangent space,

$\displaystyle \left. \frac{d}{dt} \right|_{t=0} \langle p + tv, p+tv\rangle_q = 0 \quad \Rightarrow \quad \langle v, p \rangle_q + \langle p,v \rangle_q = 0$

Because ${q}$ is hermitian, the expression on the right does not mean that ${\langle v,p\rangle_q = 0}$, but it does mean that ${\langle v,p \rangle_q}$ is purely imaginary. If ${\langle v,p \rangle_q = ik}$, then ${\langle v,v\rangle_q < 0}$, i.e. ${q}$ is not positive definite on the tangent spaces.

However, we can get rid of this negative definite subspace. ${S^1}$ as the complex numbers of unit length (or ${\mathrm{U}(1)}$, say) acts on ${\mathbb{C}^{n+1}}$ by multiplying coordinates, and this action preserves ${q}$: any phase goes away when we apply the absolute value. The quotient of ${\bar{H}}$ by this action is ${\mathbb{C}\mathbb{H}^n}$. The isometry group of this space is still ${\mathrm{U}(n,1)}$, but now there are point stabilizers because of the action of ${\mathrm{U}(1)}$. We can think of ${\mathrm{U}(1)}$ inside ${\mathrm{U}(n,1)}$ as the diagonal matrices, so we can write

$\displaystyle \mathrm{SU}(n,1) \times \mathrm{U}(1) \cong U(n,1)$

And the projectivized matrices ${\mathrm{PSU}(n,1)}$ is the group of isometries of ${\mathbb{C}\mathbb{H}^n \subseteq \mathbb{C}^n \subseteq \mathbb{C}\mathrm{P}^n}$, where the middle ${\mathbb{C}^n}$ is all vectors in ${\mathbb{C}^{n+1}}$ with ${w=1}$ (which we think of as part of complex projective space). We can also approach this group by projectivizing, since that will get rid of the unwanted point stabilizers too: we have ${\mathrm{PU}(n,1) \cong \mathrm{PSU}(n,1)}$.

5.1. Case ${n=1}$

In the case ${n=1}$, we can actually picture ${\mathbb{C}\mathrm{P}^1}$. We can’t picture the original ${\mathbb{C}^4}$, but we are looking at the set of ${(z,w)}$ such that ${|z|^2 - |w|^2 = -1}$. Notice that ${|w| \ge 1}$. After projectivizing, we may divide by ${w}$, so ${|z/w| - 1 = -1/|w|}$. The set of points ${z/w}$ which satisfy this is the interior of the unit circle, so this is what we think of for ${\mathbb{C}\mathbb{H}^1}$. The group of complex projective isometries of the disk is ${\mathrm{PU}(1,1)}$. The straight horizontal line is a geodesic, and the complex isometries send circles to circles, so the geodesics in ${\mathbb{C}\mathbb{H}^1}$ are circles perpendicular to the boundary of ${S^1}$ in ${\mathbb{C}}$.

Imagine the real projective model as a disk sitting at height one, and the geodesics are the intersections of planes with the disk. Complex hyperbolic space is the upper hemisphere of a sphere of radius one with equator the boundary of real hyperbolic space. To get the geodesics in complex hyperbolic space, intersect a plane with this upper hemisphere and stereographically project it flat. This gives the familiar Poincare disk model.

5.2. Real ${\mathbb{H}^2}$‘s contained in ${\mathbb{C}\mathbb{H}^n}$

${\mathbb{C}\mathbb{H}^2}$ contains 2 kinds of real hyperbolic spaces. The subset of real points in ${\mathbb{C}\mathbb{H}^n}$ is (real) ${\mathbb{H}^n}$, so we have a many ${\mathbb{H}^2 \subseteq \mathbb{H}^n \subseteq \mathbb{C}\mathbb{H}^n}$. In addition, we have copies of ${\mathbb{C}\mathbb{H}^1}$, which, as discussed above, has the same geometry (i.e. has the same isometry group) as real ${\mathbb{H}^2}$. However, these two real hyperbolic spaces are not isometric. the complex hyperbolic space ${\mathbb{C}\mathbb{H}^1}$ has a more negative curvature than the real hyperbolic spaces. If we scale the metric on ${\mathbb{C}\mathbb{H}^n}$ so that the real hyperbolic spaces have curvature ${-1}$, then the copies of ${\mathbb{C}\mathbb{H}^1}$ will have curvature ${-4}$.

In a similar vein, there is a symplectic structure on ${\mathbb{C}\mathbb{H}^n}$ such that the real ${\mathbb{H}^2}$ are lagrangian subspaces (the flattest), and the ${\mathbb{C}\mathbb{H}^1}$ are symplectic, the most negatively curved.

An important thing to mention is that complex hyperbolic space does not have constant curvature(!).

6. Poincare Disk Model and Upper Half Space Model

The projective models that we have been dealing with have many nice properties, especially the fact that geodesics in hyperbolic space are straight lines in projective space. However, the angles are wrong. There are models in which the straight lines are “curved” i.e. curved in the euclidean metric, but the angles between them are accurate. Here we are interested in a group of isometries which preserves angles, so we are looking at a conformal model. Dimension 2 is special, because complex geometry is real conformal geometry, but nevertheless, there is a model of ${\mathbb{R}\mathbb{H}^n}$ in which the isometries of the space are conformal.

Consider the unit disk ${D^n}$ in ${n}$ dimensions. The conformal automorphisms are the maps taking (straight) diameters and arcs of circles perpendicular to the boundary to this same set. This model is abstractly isomorphic to the Klein model in projective space. Imagine the unit disk in a flat plane of height one with an upper hemisphere over it. The geodesics in the Klein model are the intersections of this flat plane with subspaces (so they are straight lines, for example, in dimension 2). Intersecting vertical planes with the upper hemisphere and stereographically projecting it flat give geodesics in the Poincare disk model. The fact that this model is the “same” (up to scaling the metric) as the example above of ${\mathbb{C}\mathbb{H}^1}$ is a (nice) coincidence.

The Klein model is the flat disk inside the sphere, and the Poincare disk model is the sphere. Geodesics in the Klein model are intersections of subspaces (the angled plane) with the flat plane at height 1. Geodesics in the Poincare model are intersections of vertical planes with the upper hemisphere. The two darkened geodesics, one in the Klein model and one in the Poincare, correspond under orthogonal projection. We get the usual Poincare disk model by stereographically projecting the upper hemisphere to the disk. The projection of the geodesic is shown as the curved line inside the disk

The Poincare disk model. A few geodesics are shown.

Now we have the Poincare disk model, where the geodesics are straight diameters and arcs of circles perpendicular to the boundary and the isometries are the conformal automorphisms of the unit disk. There is a conformal map from the disk to an open half space (we typically choose to conformally identify it with the upper half space). Conveniently, the hyperbolic metric on the upper half space ${d_H}$ can be expressed at a point ${(x,t)}$ (euclidean coordinates) as ${d_H = d_E/t}$. I.e. the hyperbolic metric is just a rescaling (at each point) of the euclidean metric.

One of the important things that we wanted in our models was the ability to realize isometries of the model with isometries of the ambient space. In the case of a one-parameter family of isometries of hyperbolic space, this is possible. Suppose that we have a set of elliptic isometries. Then in the disk model, we can move that point to the origin and realize the isometries by rotations. In the upper half space model, we can move the point to infinity, and realize them by translations.

Geometric group theory is not a coherent and unified field of enquiry so much as a collection of overlapping methods, examples, and contexts. The most important examples of groups are those that arise in nature: free groups and fundamental groups of surfaces, the automorphism groups of the same, lattices, Coxeter and Artin groups, and so on; whereas the most important properties of groups are those that lend themselves to applications or can be used in certain proof templates: linearity, hyperbolicity, orderability, property (T), coherence, amenability, etc. It is natural to confront examples arising in one context with properties that arise in the other, and this is the source of a wealth of (usually very difficult) problems; e.g. do mapping class groups have property (T)? (no, by Andersen) or: is every lattice in $\text{PSL}(2,\mathbb{C})$ virtually orderable?

As remarked above, it is natural to formulate these questions, but not necessarily productive. Gromov, in his essay Spaces and Questions remarks that

often the mirage of naturality lures us into featureless desert with no clear perspective where the solution, even if found, does not quench our thirst for structural mathematics . . . Another approach . . . has a better chance for a successful outcome with questions following (rather than preceding) construction of new objects.

A famous question of the kind Gromov warns against is the following:

Question: Is Thompson’s group $F$ amenable?

Recall that Thompson’s group is the group of (orientation-preserving) PL homeomorphisms of the unit interval with breakpoints at dyadic rationals (i.e. rational numbers of the form $p/2^q$ for integers $p,q$) and derivatives all powers of $2$. This group is a rich source of examples/counterexamples in geometric group theory: it is finitely presented (in fact $FP_\infty$) but “looks like” a transformation group; it contains no nonabelian free group (by Brin-Squier), but obeys no law. It is not elementary amenable (i.e. it cannot be built up from finite or abelian groups by elementary operations — subgroups, quotients, extensions, directed unions), so it is “natural” to wonder whether it is amenable at all, or whether it is one of the rare examples of nonamenable groups without nonabelian free subgroups (see this post for a discussion of amenability versus the existence of free subgroups, and von Neumann’s conjecture). This question has attracted a great deal of attention, possibly because of its long historical pedigree, rather than because of the potential applications of a positive (or negative) answer.

Recently, two papers were posted on the arXiv, promising competing resolutions of the question. In February, Azer Akhmedov posted a preprint claiming to show that the group $F$ is not amenable. This preprint was revised, withdrawn, then revised again, and as of the end of April, Akhmedov continues to press his claim. Akhmedov’s argument depends on a new geometric criterion for nonamenability, roughly speaking, the existence of a $2$-generator subgroup and a subadditive non-negative function on the group whose values grow at a definite rate on words in the subgroup whose exponents satisfy suitable parity conditions and inequalities. The non-negative function (Akhmedov calls it a “height function”) certifies the existence of a sufficiently “bushy” subset of the group to violate Folner’s criterion for amenability. Akhmedov’s paper reads like a “conventional” paper in geometric group theory, using methods from coarse geometry, careful combinatorial and counting arguments to establish the existence of a geometric object with certain large-scale properties, and an appeal to a standard geometric criterion to obtain the desired result. Akhmedov’s paper is part of a series, relating (non)amenability to certain other interesting geometric properties, some related to the so-called “traveling salesman” property, introduced earlier by Akhmedov.

On the other hand, in May, E. Shavgulidze posted a preprint claiming to show that the group $F$ is amenable. Interestingly enough, Shavgulidze’s argument does not apply to the slightly more general class of Stein-Thompson groups in which slopes and denominators of break points can be divisible by an arbitrary (but prescribed) finite set of prime numbers. Moreover, his methods are very unlike any that one would expect to find in the typical geometric group theory paper. The argument depends on the construction, going back (in some sense) to a paper of Shavgulidze from 1978, of a measure on the space $C(I)$ of continuous functions on the interval which is quasi-invariant under the natural action of the group of diffeomorphisms of the interval of regularity at least $C^3$. In more detail, let $D^n$ denote the group of diffeomorphisms of the interval of regularity at least $C^n$ for each $n$, and let $C$ denote the Banach space of continuous functions on the interval that vanish at the origin. Define $A:D^1 \to C$ by the formula $A(f)(t) = \log(f'(t)) - \log(f'(0))$. The space $C$ can be equipped with a natural measure — the Wiener measure $w_\sigma$ of variance $\sigma$, and this measure can be pulled back to $D^1$ by $A$, which is thought of as a topological space with the $C^1$ topology. Shavgulidze shows that the left action of $D^3$ on $D^1$ quasi-preserves this measure. Here the Wiener measure on $C$ is the probability measure associated to Brownian motion (with given variance). A “sample” trajectory $W_t$ from $C$ is characterized by three properties: that it starts at the origin (i.e. $W_0=0$), that is it continuous almost surely (this is already implicit in the fact that the measure is supported on the space $C$ and not some more general space), and that increments are independent, with the distribution of $W_t - W_s$ equal to a Gaussian with mean zero and variance $(t-s)\sigma$. Shavgulidze’s argument depends first on an argument of Ghys-Sergiescu that shows Thompson’s group is conjugate (by a homeomorphism) to a discrete subgroup of the group of $C^\infty$ diffeomorphisms of the interval. A bounded function $f$ on $F$ determines a continuous bounded function $\pi_\delta(f)$ on $D^{1+\delta}$ (for $\delta<1/2$) by a certain convolution trick, using both the group structure of $F$, and its discreteness in $D^3$. Roughly, given an element $g \in D^{1+\delta}$, the set of elements of $F$ whose (group) composition with $g$ is uniformly bounded in the $C^{1+\delta}$ norm is finite; so the value of $\pi_\delta(f)$ is obtained by taking a suitable average of the value of $f$ on this finite subset of $F$. This reduces the problem of the amenability of $F$ to the existence of a suitable functional on the space of bounded continuous functions on $D^{1+\delta}$, which is constructed via the pulled back Wiener measure as above.

There are several distinctive features of Shavgulidze’s preprint. One of the most striking is that it depends on very delicate analytic features of the Wiener measure, and the way it transforms under the action of $D^3$ on $D^1$ — a transformation law involving the Schwartzian derivative — and suggesting that certain parts of the argument could be clarified (at least from the point of view of a topologist?) by using projective geometry and Sturm-Liouville theory. Another is that the crucial analytic quality — namely differentiability of class $C^{1+1/2}$ — is also crucial for many other natural problems in $1$-dimensional analysis and geometry, from regularity estimates in the thin obstacle problem, to Navas’ work on actions of property (T) groups on the circle. At least one of the preprints by Akhmedov and Shavgulidze must be in error (in fact, a real skeptic’s skeptic such as Michael Aschbacher is not even willing to concede that much . . .) but even if wrong, it is possible that they contain things more valuable than a resolution of the question that prompted them.

Update (7/6): Azer Akhmedov sent me a construction of a (nonabelian) free subgroup of $D^1$ that is discrete in the $C^1$ topology. This is not quite enough regularity to intersect with Shavgulidze’s program, but it is interesting, and worth explaining. This is my (minor) modification of Azer’s construction (any errors are due to me):

Proposition: The group $D^1$ contains a discrete nonabelian free subgroup.

Sketch of Proof: First, decompose the interval $[0,1]$ into countably many disjoint subintervals accumulating only at the endpoints. Choose a free action on two generators by doing something generic on each subinterval, in such a way that the derivative is equal to $1$ at the endpoints. This can certainly be accomplished; for concreteness, choose the action so that for each subinterval $I_i$ there is a point $x_i$ in the interior of $I_i$ whose stabilizer is trivial.

Second, for each pair of distinct words in the generators, choose a subinterval and modify the action there so that the derivatives of those words in that subinterval differ by at least some definite constant $C$ at some point. In more detail: enumerate the pairs of words somehow $p_1, p_2, p_3$ where each $p_i$ is a pair of words $(w_{i1}, w_{i2})$ in the generators, and modify the action on the subinterval $I_i$ so the words in $p_i$ differ by at least $C$ in the $C^1$ norm on the interval $I_i$. Since we are modifying the generators infinitely many times, but in such a way that the support of the modification exits any compact subset of the interior, we just need to check that the modifications are $C^1$. Since there are only finitely many pairs of words, both of which are of bounded length (for any given bound), when $i$ is sufficiently big, one of the words $w_{i1}$,  $w_{i2}$ has length at least $n(i)$ where $n(i)$ goes to infinity as $i$ goes to infinity. Without loss of generality, we can order the pairs so that $w_{i1}$ is the “long” word.

Now this is how we modify the action in $I_i$. Recall that the point $x_i$ has trivial stabilizer, so the translates $y_{ij}$ of $x_i$ under the suffixes of $w_{i1}$ are distinct. Take disjoint intervals about the $y_{ij}$ and observe that each $y_{ij}$ is taken to $y_{ij+1}$ by one of the generators. Modify this generator inside this disjoint neighborhood so that $y_{ij}$ is still taken to $y_{ij+1}$, but the derivative at that point is multiplied by $1+ C/n(i)$, and the derivative at nearby points is not multiplied by more than $1+C/n(i)$. Since the neighborhoods of the $y_{ij}$ are disjoint, these modifications are all compatible, and the derivative of the generators does not change by more than $1+C/n(i)$ at any point. Since $n(i)$ goes to infinity as $i$ goes to infinity, we can perform such modifications for each $i$, and the resulting action is still $C^1$. But now the derivative of $w_{i1}$ at $x_i$ has been multiplied by $1+C$, so $w_{i1}$ and $w_{i2}$ differ by at least $C$ in the $C^1$ norm.  qed.

It is interesting to observe that this construction, while $C^1$, is not $C^{1+\epsilon}$ for any $\epsilon>0$. For big $i$, we have $n(i) \sim \log(i)$ whereas $|I_i| = o(1/i)$. Introducing a “bump” which modifies the derivative by $1/\log(i)$ in a subinterval of size $o(1/i)$ will blow up every Holder norm.

(Update 8/10): Mark Sapir has created a webpage to discuss Shavgulidze’s paper here. Also, Matt Brin has posted notes on Shavgulidze’s paper here. The notes are very nice, and go into great detail, as far as they go. Matt promises to update the notes periodically.

(Update 11/18): Matt Brin has let me know by email that a significant gap has emerged in Shavgulidze’s argument. He writes:

Lemma 5 is still unproven. It claims a property about the distributions $u_n$ on the simplexes $D_n$ that is needed for the second part of the paper. The main result does not need the particular distributions $u_n$ given in the paper, but does need distributions on the $D_n$ that satisfy the properties claimed by Lemmas 5, 6 and that cooperate with Lemma 9. Ufe Haagerup claims an argument that the $u_n$ in the paper does not satisfy the conclusion of Lemma 5. Another distribution was said to be suggested by Shavgulidze, but at last report, it did not seem to be working out.

In light of this, it would seem to be reasonable to consider the question of whether $F$ is amenable as wide open.

(Update 9/21/2012): Justin Moore has posted a preprint on the arXiv claiming to prove amenability of $F$. It is too early to suggest that there is expert consensus on the correctness of the proof, but certainly everything I have heard is promising. I have not had time to look carefully at the argument yet, but hope to get a chance to do so before too long.

(Update 10/2/2012): Justin has withdrawn his claim of a proof. A gap was found by Akhmedov.

A few days ago, Joel Friedman posted a paper on the arXiv purporting to give a proof of the (strengthened) Hanna Neumann conjecture, a well-known problem in geometric group theory.

Simply stated, the problem is as follows.

Conjecture (Hanna Neumann): Let $F$ be a free group, and let $G$ and $H$ be finitely generated subgroups. For a subgroup $E$ of $F$, let $\rho(E) = \max(\text{rank}(E)-1,0)$. Then there is an inequality $\rho(G \cap H) \le \rho(G)\rho(H)$.

This conjecture was further strengthened by Walter Neumann (her son):

Conjecture (strengthened Hanna Neumann): With notation above, there is an inequality $\sum_x \rho(G \cap xHx^{-1}) \le \rho(G)\rho(H)$ where the sum is taken over $x \in H\backslash F / G$, i.e. the double coset representatives.

Notice by the way that since any free group embeds into $F_2$, the free group of rank $2$, one can assume that $F$ has rank $2$ above. This fact is implicit in the discussion below.

Friedman’s paper seems to be very carefully written, and contains some new ideas (which I do not yet really understand), namely an approach using sheaf theory. But in this post I want to restrict myself to some simple (and probably well-known) geometric observations.

The first step is to reduce the problem to a completely graph-theoretic one, following Stallings; in fact, Benson Farb tells me that he thinks this reduction was known to Stallings, or at least to Dicks/Formanek (and in any case is very close to some ideas Stallings and Gersten introduced to study the problem; more on that in a later post). Friedman makes the following definition:

Definition: Let $\mathcal{G}$ be a finite group and $g_1,g_2 \in \mathcal{G}$ be two elements (that do not necessarily generate $\mathcal{G}$). The directed Cayley graph $C$ is the graph with vertex set $\mathcal{G}$ and with a directed edge from $v$ to $vg_i$ labeled $i$ for each $v \in \mathcal{G}$ and $i=1,2$.

In other words, $C$ is a graph whose edges are oriented and labeled with either $1$ or $2$ in such a way that each vertex has at most one outgoing and one incoming edge with each label, and such that there is a transitive (on the vertices) free action of a group $\mathcal{G}$ on $C$. (Note: for some reason, Friedman wants his group to act on the right, and therefore has directed edges from $v$ to $g_iv$, but this is just a matter of convention).

For any finite graph $K$, not necessarily connected, let $\rho(K) = \sum_j \max(0,-\chi(K_j))$; i.e. $\rho(K) = \sum_j \rho(\pi_1(K_j))$ where the sum is taken over the connected components $K_j$ of $K$. Friedman shows (but this reduction is well-known) that the SHNC is equivalent to the following graph-theoretic inequality:

Theorem: The SHNC is equivalent to the following statement. For any graph $C$ as above, and any two subgraphs $K,K'$ we have $\sum_{g \in \mathcal{G}} \rho(K \cap gK') \le \rho(K)\rho(K')$.

The purpose of this blog entry is to show that there is a very simple proof of this inequality when $\rho$ is replaced with $-\chi$. This is not such a strange thing to do, since $\rho$ and $-\chi$ are equal for graphs without acyclic components (i.e. without components that are trees), and for “random” graphs $K,K'$ one does not expect the difference between $\rho$ and $-\chi$ to be very big. The argument proceeds as follows. Suppose $K$ has $v$ vertices and $e_1,e_2$ edges of kind $1,2$ respectively, and define $v',e_1',e_2'$ similarly for $K'$. Then

• $(-\chi(K))(-\chi(K')) = (v-e_1-e_2)(v'-e_1'-e_2')$

On the other hand, since Euler characteristic is local, we just need to count how many vertices and edges of each kind turn up in each $K \cap gK'$. But this is easy: every vertex of $K$ is equal to exactly one translate of every vertex of $K'$, and similarly for edges of each kind. Hence

• $\sum_g -\chi(K \cap gK') = e_1e_1' + e_2e_2' - vv'$

So the inequality one wants to show is $e_1e_1' + e_2e_2' - vv' \le (v-e_1-e_2)(v'-e_1'-e_2')$ which simplifies to

• $v(e_1' + e_2') + v'(e_1 + e_2) \le 2vv' + e_1e_2' + e_2 e_1'$

On the other hand, each graph $K,K'$ has at most two edges at any vertex with either label, and therefore we have inequalities $0 \le e_1,e_2 \le v, 0 \le e_1',e_2' \le v'$. Subject to these constraints, the inequality above is straightforward to prove. To see this, first fix some non-negative values of $v,v'$ and let $X$ be the four-dimensional cube of possible values of $e_1,e_2,e_1',e_2'$. Since both sides of the inequality are linear as a function of each $e_i$ or $e_i'$, if the inequality is violated at any point in $X$ one may draw a straight line in $X$ corresponding to varying one of the co-ordinates (e.g. $e_1$) while keeping the others fixed, and deduce that the inequality must be violated on one of the faces of $X$. Inductively, if the inequality is violated at all, it is violated at a vertex of $X$, which may be ruled out by inspection; qed.

This argument shows that the whole game is to understand the acyclic components of $K \cap gK'$; i.e. those which are topologically trees, and therefore contribute $0$ to $\rho$, but $-1$ to $-\chi$.

Incidentally, for all I know, this simple argument is explicitly contained in either Stallings’ or Gersten’s paper (it is surely not original in any case). If a reader can verify this, please let me know!

Update: Walter Neumann informs me that this observation (that the inequality is true with $-\chi$ in place of $\rho$) is in his paper in which he introduces the SHNC! He further shows in that paper that for “most” $G$, the SHNC is true for all $H$.

Update (6/29): Warren Dicks informs me that he was not aware of the reduction of SHNC to the graph-theoretic formulation described above. Friedman’s webpage acknowledges the existence of an error in the paper, and says that he is working to correct it. One problem that I know of (discovered mostly by my student Steven Frankel) concerns the commutativity of the diagram on page 10.

Update (10/22): It has been a few months since I last edited this page, and Joel Friedman has not updated either the arXiv paper, or the statement on his webpage that he is “trying to fix the error”. Since wikipedia mentions Friedman’s announcement, I thought it would be worth going on record at this point to say that Friedman’s arXiv paper (version 1 — the only version at the point I write this) is definitely in error, and that I believe the error is fundamental, and cannot be repaired (this is not to say that the paper does not contain some things of interest (it does), or that Friedman does not acknowledge the error (he does), just that it is worth clearing up any possible ambiguity about the situation for readers who are wondering about the status of the SHNC). The problem is the “not entirely standard” (quote from Friedman’s paper) diagrams, like the one on page 10. In particular, the claimed proof of Theorem 5.6, that the projections constructed in Lemma 5.5 (by a very general dimension counting argument) fit into a diagram with the desired properties is false. Any construction of projections satisfying the desired properties must be quite special. Nevertheless, one can certainly still define Friedman’s sheaf $\mathcal{K}$, and ask whether it has $\rho(\mathcal{K})=0$ (in Friedman’s sense); this would, as far as I can tell, prove SHNC; however, I do not know of any reason why it should hold (or whether there are any counterexamples, which might exist even if SHNC is true).