A question on Lorentzian matrices from “Foundations of Hyperbolic Manifolds” by John G. Ratcliffe

83 Views Asked by At

I’m stuck understanding the implication from questions 3 and 4 in Exercise 3.1 of the book Foundations of Hyperbolic Manifolds, hoping someone can help! I have provided some details at the bottom explaining terminology used here and in the book.

Q3. Let $A = (a_{ij})$ be a matrix in $O(1, n − 1)$. Show that A is positive (negative) if and only if $a_{11} > 0$ ($a_{11} < 0$).

Q4. Let $A = (a_{ij})$ be a matrix in $O^{+}(1, n − 1)$. Prove that $a_{11} ≥ 1$ with equality if and only if A is orthogonal.

By Q3, a matrix $A$ in $O(1, n − 1)$ with $a_{11}=1/2$ is positive. I.e. it’s in $O^{+}(1, n − 1)$. But by Q4, such a matrix must have $a_{11} \geq 1$. A contradiction. I must be missing something, hoping someone can advise.

Some more detail on notation and terminology from the book:

  • $O(1, n − 1)$ is the set of Lorentzian $n x n$ matrices.
  • A Lorentzian matrix A is said to be positive (resp. negative) if and only if A transforms positive time-like vectors into positive (resp. negative) time-like vectors.
  • $O^{+}(1, n − 1)$ is the set of all positive matrices in $O(1, n − 1)$.

enter image description here

1

There are 1 best solutions below

1
On BEST ANSWER

Let $q(x)=x_1^2-x_2^2-... -x_n^2$ be the quadratic form invariant under the group $G=O(1,n-1)$. The hyperboloid $H=\{x: q(x)=1\}$ is $G$-invariant and has two components: One (let's call it $H_+$) is contained in the half-space $\{x: x_1>0\}$ and the other (let's call it $H_-$) is contained in the half-space $\{x: x_1< 0\}$. The key observation (which is an elementary exercise in algebra) is that the intersection of $H$ with the open slab $$ \{x: -1< x_1<1\} $$ is empty and the intersection of $H$ with the boundary of the slab consists of just two points: $\pm e_1$. In particular,

The next observation is that the first component of the vector $Ae_1$ is $a_{11}$. In particular, $a_{11}$ cannot belong to the interval $(-1,1)$.

Thus, $Ae_1\in H_+$ if and only if $a_{11}\ge 1$, equivalently, $a_{11}>0$. Similarly, $Ae_1\in H_-$ if and only if $a_{11}\le -1$, equivalently, $a_{11}<0$.

Since $H_\pm$ are connected, it follows that $A\in O^{+}(1,n-1)$ if and only if for each "positive" vector $x$ (i.e. a vector satisfying $q(x)>0$ if and only if $Ae_1\in H_+$, if and only if $a_{11}\ge 1$. Similarly, for negative matrices $A$ in $O(1,n-1)$.

This answers affirmatively your question Q3.

To answer Q4 you should think about the vector $Ae_1$ assuming that $a_{11}=1$. Then use the fact that $A$ preserves the quadratic form $q$ to verify orthogonality of $A$. To check the converse, assume that $A\in O(n)\cap O^+(1,n-1)$ and note that it preserves both the unit sphere (with respect to the Euclidean metric) and the upper hyperboloid $H_+$. Then compute $Ae_1$.

All in all, there are no contradictions in the statements of the textbook you are reading.