I am reading Thurston's Three-dimensional Geometry and Topology and in particular the sections on isometries of hyperbolic space. I have read (and mostly understood) the qualitative/geometric proofs that Thurston gives and would like to understand the algebraic proof, which is outlined on page 97, problem 2.5.24. It seems like "just" linear algebra, but I am stuck on some parts. I am looking for answers or hints to parts d,e, and g. I'll write some of what I have for the other parts, but will skip some detail (this is already long; sorry).
We are using the hyperbolic model of $\mathbb{H}^n$—let $Q$ be the quadratic form $Q(x_1,\ldots,x_{n+1}) = x_1^2 + \cdots + x_{n}^2 - x_{n+1}^2$ on $\mathbb{R}^{n+1}$ and set $\mathbb{H}^n$ to be the upper sheet of $Q^{-1}(-1)$. This is a smooth manifold, and inherits a Riemannian metric from $Q$. The linear maps which preserve $Q$, denoted $SO(n,1)$ restrict to isometries of $\mathbb{H}^n$, and and moreover every isometry of $\mathbb{H}^n$ is given by such a linear map. (This is all background)
a. As a warm-up exercise, let $V$ be a two dimensional real vector space, $Q$ a possibly degenerate form on $V$ and $A: V \to V$ a linear map preserving $Q$. Describe the relationship between $Q$ and the eigenvectors and eigenvalues of $A$.
I'm guessing he wants me to say that for an eigenvector $v$, we have $$Q(v) = Q(Av) = Q(\lambda v) = \lambda^2Q(v),$$ so $\lambda = \pm 1$ whenever $Q(v) \neq 0$.
b. (skipping)
c. Let $A \in SO(n,1)$. If $W$ is a minimal invariant subspace of $A$, show that $W$ has dimension 1 or 2. If the dimension is two, $Q$ is positive definite on $W$.
If $W$ has a (real) eigenvector, we're done. It must have a complex eigenvector, $\lambda$ and an eigenvector $v = v_1+ iv_2 \in W + iW$ in the complexification of $W$. We also get the conjugate eigen-pair $(\overline{\lambda}, \overline{v})$. $A$ must preserve the subspace $span_{\mathbb{C}}(v) \oplus span_{\mathbb{C}}(\overline{v})$, which intersects with $W$ in $span_{\mathbb{R}}(v_1,v_2)$, a two-dimensional invariant subspace.
If $Q$ were not positive definite on $W$, $W$ would intersect the cone $Q^{-1}(0)$ in a line. Since $A$ preserves both $W$ and the cone, this line is an eigenspace, meaning $W$ is one dimensional.
d. Factor the characteristic polynomial $p$ of $A$ into irreducible quadratic and linear factors over $\mathbb{R}$. For any irreducible quadratic factor $q$ of $p$, the subspace annihilated by $q(A)$ must be positive definite, so the roots of $q$ are on the unit circle.
I'm a little uneasy on this one. Write $q(A) = (I - \lambda)(I - \overline{\lambda})$ over $\mathbb{C}$. I think the subspace annihilated by $q(A)$ is the real part of $E_\lambda^A \oplus E_{\overline{\lambda}}^A$ (these are eigenspaces in the complexification), but in general this needn't be 2-dimensional—for instance, a block sum of rotations through the same angle (is this the only way? presumably not, as these have trivial jordan blocks). Assuming it is 2-dimensional, we can apply the previous part to deduce that $Q$ is positive definite. I then want to define a "complexified" quadratic form by $Q'(v) = v^*Bv$, where $Q$ is given by $Q(v) = v^T Bv$. Hopefully this is also positive definite. Then we have $$Q'(v) = Q'(Av) = \overline{\lambda}\lambda Q(v)$$ so $|\lambda| = 1$ or $Q'(v) = 0$, which can't happen since $Q'$ is(?) positive definite.
e. If $p$ has any roots that are not on the unit circle, there are precisely two such roots, $\lambda, \lambda^{-1}$, and the isometry of hyperbolic space induced by $A$ is hyperbolic.
I don't really know how to approach this. From part 1, I know that if $Av = \lambda v$ for $\lambda \neq \pm 1$, then $Q(v) = 0$. Also from $A^TBA = B$ that $\text{det}(A) = \pm 1$, but I can't quite figure out why there can only be two non-1 eigenvectors.
Even if I assume that there are only two, I don't know how to show that this gives a hyperbolic isometry of $\mathbb{H}^n$, (ie the isometry has a unique axis which is acted on by translation). This seems eminently plausible to me (I can see the picture), but I don't know how to prove it!
f. If $A$ fixes some vector $v$ with $Q(v) < 0$, its induced isometry is elliptic; it fixes some totally geodesic subspace $P \subset \mathbb{H}^n$ isometric to $\mathbb{H^k}$, and "rotates" the normal space of $P$.
Since $A$ fixes $v$ with $Q(v) < 0$, we can rescale to find a fixed a point on $\mathbb{H^n}$. By possibly conjugating, we may assume that this point is $(0,\ldots, 0,1)$. Then the $1$-eigenspace is a $k$-dimensional subspace including the vertical line, which corresponds to a totally geodesic submanifold isometric to $\mathbb{H}^{k-1}$ (in particular, this means $A$ is elliptic). The action on the normal space must be orthogonal, since $A$ preserves $Q$.
g. If all characteristic roots are on the unit circle and $A$ fixes no vectors $v$ with $Q(v) < 0$, $A$ fixes a non-zero vector $v$ with $Q(v) = 0$. All such vectors are multiples of one another. The isometry induced by $A$ is parabolic [by which we mean the infimum of the translation distance is 0, but never attained]. The hyperplane $P = \{w \mid v \cdot w = 1\}$, where $\cdot$ is the inner product associated with $Q$ is invariant by $A$. The quadratic form $Q$ is degenerate on the tangent space of $P$ and induces a Euclidean metric on the quotient space $P/\mathbb{R}v$. The transformation induced by $A$ on $P/\mathbb{R}v$ is a Euclidean isometry without fixed points.
For the first part, I don't even know where to begin, much less translate the linear algebra into geometry. Also, it appears the "hyperplane" $P$ isn't a subspace? Clearly I'm missing something.