= Smooth locus of a nondegenerate real quadratic level set
{title2=$C_\alpha=\{x:x^TAx=\alpha\},\quad A=A^T,\quad\det A\ne0$}
The <quadratic form gradient> gives $Dq_x(h)=2x^TAh$. Since $A$ is invertible, this <linear map> is nonzero exactly when $x\ne0$. The <regular level set theorem> therefore makes every nonzero point of $C_\alpha$ a smooth point of dimension $n-1$. If $\alpha\ne0$, the entire level set is smooth.
If $\alpha=0$ and the level set contains a nonzero point, the form is an <indefinite quadratic form>. Its only singular point is the origin. To prove this rather than merely observe a vanishing derivative, use <Sylvester's law of inertia> to write $q=|u|^2-|v|^2$ with both sign spaces nonzero. The null vectors $e_i\pm f_j$ span the whole ambient space. Their straight-line curves through the origin lie in the cone. If a relatively open neighborhood of the origin in the cone were an <embedded submanifold>, its <tangent space> would contain all these directions and hence have dimension $n$. It would then be open in the ambient space, impossible for the zero set of this nonzero quadratic form. Thus the smooth locus is $C_0\setminus\{0\}$.
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2017-quadratic-level-surfaces.png]
{title=Three levels of an indefinite quadratic form in three dimensions}
{description=Original diagrams and generic model calculations generated by the adjacent Python script.}
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