Past exam of the short options course of the University of Oxford / 2017 / 4

The map is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatEquivalently, $f(x+h)=f(x)+Lh+o(\Vert h\Vert )$. This linear map is unique and is the Fréchet derivative $D{f}_x$. Its matrix in the standard bases is the Jacobian matrix, whose $(i,j)$ entry is $\partial f_i/\partial x_j$ at $x$. A function is differentiable on its domain if this condition holds at every point.

A sufficient condition is that the derivative $D{f}_x:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is surjective, equivalently that its Jacobian matrix has rank $m$. Then the implicit function theorem makes the zero set near $x$ a $C^1$ embedded submanifold of dimensionThe rank condition persists nearby because the derivative is continuous. This is the local form of the regular level set theorem; only the rank at the specified point is needed for this local conclusion.

Expand the increment of the quadratic form, using $A^T=A$:The last term obeys $|h^{T}Ah|\le \Vert A\Vert \Vert h{\Vert }^2=o(\Vert h\Vert )$. Thus $q$ is differentiable andThis is the quadratic form gradient. The derivative is a linear functional of the increment $h$, whereas the gradient is its representing vector under the Euclidean inner product.

Since $A$ is invertible, the quadratic form gradient vanishes exactly at the origin. At every nonzero point of $C_{\alpha }$, the scalar derivative is surjective, so the regular level set theorem gives a smooth submanifold of dimension $n-1$.

If $\alpha \ne 0$, the origin is absent. Therefore the largest relatively open smooth subset is all of $C_{\alpha }$.

If $\alpha =0$, the nontriviality assumption forces the form to be an indefinite quadratic form: an invertible definite form would have zero set consisting only of the origin. Every nonzero point is regular. To show that the origin really fails to be a manifold point, use Sylvester's law of inertia to reduce to $|u{|}^2-|w{|}^2=0$. With positive and negative coordinate vectors $e_i,f_j$, the null vectors $e_i+f_j$ and $e_i-f_j$ span ${\mathbb{R}}^n$. Each line in one of these directions lies in the cone.

If a relatively open manifold neighborhood of the origin existed, its tangent space would contain all these line directions, so would have dimension $n$. Such an embedded manifold would be open in ${\mathbb{R}}^n$, impossible for the zero set of a nonzero quadratic form. ConsequentlyThis gives the smooth locus of a nondegenerate real quadratic level set, including maximality at the singular vertex.

Writing the coordinates as $(x,y,z)$, the level equation is $x^2+y^2-z^2=\alpha$. Therefore:

The cone vertex is an instance of the double cone singularity. The tips of the two-sheeted hyperboloid are regular because the gradient $(2x,2y,-2z)$ is nonzero there.