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I was studying this paper and on page 8 the author defines the orthogonal group associated to the Minkowski linear form in the usual way ( $A \in O(1,n) \Longleftrightarrow \langle Ax | Ay \rangle = \langle x|y \rangle \ \forall x, y \in \mathbb{R}^{1,n}$, where $\mathbb{R}^{1,n}$ is $\mathbb{R}^{n+1}$ with the Minkowski linear form). $\mathbb{H}^n$ is defined as the $x : \langle x | x \rangle = -1$ and $x_0 > 0$.

The part that I can't prove is that the stabilizer of $\mathbb{H}^n$ is equal to the subset of $O(1,n)$ such that the first column belongs to $\mathbb{H}^n$. I did prove that if $A \in Stab_{O(1,n)} \ \mathbb{H}^n$ then $A_0$ must belong to $\mathbb{H}^n$, but the reverse is escaping me.

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Werner Germán Busch
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1 Answers1

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Intuitively speaking you have this: you know that any $A\in O(1,n)$ will map the point $x_0=(1,0,0,\dots,0)$ to an element of the two-sheeted hyperboloid. It could be the upper sheet or the lower sheet. With the extra condition of $A_{00}>0$ you know it will be mapped to the upper sheet. So one point from the upper sheet goes to a point on the upper sheet. From this you want to conclude that all points on the upper sheet stay on the upper sheet.

You will need some kind of continuity argument here. Something that says that if a point $x$ stays on the upper sheet, then a point $y$ close by will stay there, too. Assume this were not the case. Then you would have a continuous path from $x$ to $y$ made of points on the upper sheet, which gets mapped to something that lies partially on the upper sheet and partially on the lower sheet. This cannot be done without a discontinuity, and linear functions are continuous.

This is just a proof idea. You may need to flesh this out to make it rigorous enough for your formalism, but it should be enough to guide your approach.

MvG
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