The cone of positive definite 2×2 matrices is a light cone

Idea. The determinant of a $2\times2$ symmetric matrix is a Lorentzian quadratic form in suitable coordinates, and $\mathrm{PD}$ is the "future light cone".

Part 1: criterion. Let $S=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ and $u=(u_1,u_2)\ne0$. If $a>0$, completing the square,
$$u^TSu=au_1^2+2bu_1u_2+cu_2^2=a\Bigl(u_1+\frac bau_2\Bigr)^2+\Bigl(c-\frac{b^2}a\Bigr)u_2^2.$$
If $ac>b^2$ then $c-b^2/a>0$ and the form is positive for $u\ne0$ (both squares vanish only if $u_2=0$ and then $u_1=0$). Conversely, if $S$ is positive definite then $a=e_1^TSe_1>0$ and, taking $u=(-b,a)$, $u^TSu=a(ac-b^2)>0$ gives $ac>b^2$. (Equivalently: Sylvester's criterion.)

Part 2: cone, convexity, homeomorphism. Put $x=\frac{a+c}2$, $y=\frac{a-c}2$, so $a=x+y$, $c=x-y$, and $(a,b,c)\mapsto(x,y,b)$ is a linear isomorphism. Then $ac-b^2=x^2-y^2-b^2$.
- If $x>\sqrt{y^2+b^2}$, then $x>0$ and $x^2>y^2+b^2$; then $a=x+y\ge x-|y|>0$ and $ac-b^2>0$, so the matrix is PD.
- Conversely, if $a>0$ and $ac>b^2\ge0$ then $c>0$, so $x=\frac{a+c}2>0$, and $x^2>y^2+b^2$ gives $x>\sqrt{y^2+b^2}$.

Hence $\mathrm{PD}=\{(x,y,b):x>\|(y,b)\|\}$, where $\|\cdot\|$ is the Euclidean norm of $\mathbb{R}^2$. It is *open* (the strict inequality between continuous functions), a *cone* (stable under multiplication by $\lambda>0$) and *convex*: if $x_i>\|(y_i,b_i)\|$ then for $\lambda\in[0,1]$, by the triangle inequality $\|\lambda(y_1,b_1)+(1-\lambda)(y_2,b_2)\|\le\lambda\|(y_1,b_1)\|+(1-\lambda)\|(y_2,b_2)\|<\lambda x_1+(1-\lambda)x_2$.

*Homeomorphism with $\mathbb{R}^3$.* The map $(x,y,b)\mapsto\bigl(y,b,\ln(x-\|(y,b)\|)\bigr)$ is a continuous bijection from $\mathrm{PD}$ onto $\mathbb{R}^2\times\mathbb{R}$ with continuous inverse $(y,b,s)\mapsto(\|(y,b)\|+e^s,y,b)$.

Part 3: boundary. The set $F=\{x\ge\|(y,b)\|\}$ is closed. Every point of $\{x=\|(y,b)\|\}$ is the limit of the points $(x+\frac1k,y,b)\in\mathrm{PD}$. Hence $\overline{\mathrm{PD}}=F$, and since $\mathrm{PD}$ is open, $\partial\,\mathrm{PD}=F\setminus\mathrm{PD}=\{x=\|(y,b)\|\}$. In the original coordinates, $x=\sqrt{y^2+b^2}$ means $x\ge0$ and $x^2-y^2=b^2$, i.e. $ac=b^2$ and $a+c\ge0$; together with $ac=b^2\ge0$ this means $a,c\ge0$ (they have the same sign and a nonnegative sum). So
$$\partial\,\mathrm{PD}=\{(a,b,c):ac=b^2,\ a,c\ge0\},$$
the set of positive semidefinite singular matrices, a circular cone with vertex $0$.

Check. $S=I$ has $(x,y,b)=(1,0,0)$ with $x>0$. $S=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ has $(x,y,b)=(1,0,1)$, $x=\|(y,b)\|$: on the boundary (it is PSD with $\det=0$).

Remark. In the Minkowski space $\mathbb{R}^{2,1}$ with the form $x^2-y^2-b^2=\det S$, $\mathrm{PD}$ is the open future light cone. This is the picture behind the hyperbolic plane $\mathrm{PD}\cap\{\det=1\}\cong\mathbb{H}^2$ and its link with $SL_2(\mathbb{R})$.

Lesson. Choose coordinates in which the defining quadratic form is diagonal; cones become norm inequalities.