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For nonzero real $a_i$, scale the ambient coordinates by $q_i=a_i{r}_i$. On the half where $\epsilon q_i>0$, omitting $q_i$ is a manifold chart onto the open unit ball; its inverse restores $q_i=\epsilon \sqrt{1-{\sum }_{j\ne i}{q}_j^2}$. These $2(n+1)$ charts form a smooth atlas of the ellipsoid, including for negative $a_i$.

Rescale the ambient coordinates by $q_i=a_i{r}_i$. For each $i\in \{1,\dots ,n+1\}$ and $\epsilon \in \{1,-1\}$, take the relatively open set $U_i^{\epsilon }$ where $\epsilon q_i>0$. Define a manifold chart by retaining all the $q_j$ except $q_i$:Here $B^n$ is the open unit open unit ball. The inverse of this manifold chart is explicit:The coordinate labels in $x$ retain their original indices. All the $a_i$ are nonzero, so this inverse exists; the strictly positive radicand makes it smooth on the whole open open unit ball. Projection and this inverse are continuous, so each manifold chart is a homeomorphism onto an open subset of ${\mathbb{R}}^n$.

These $2(n+1)$ manifold charts cover the ellipsoid, because at every point at least one $q_i$ is nonzero. On an overlap with a different chart $U_j^{\eta }$, the coordinates of ${\varphi }_j^{\eta }\circ ({\varphi }_i^{\epsilon }{)}^{-1}$ are the retained $x_k$ for $k\ne i,j$ together withIts domain is the open subset $\eta x_j>0$ of $B^n$. The displayed functions, and the reverse transition functions obtained by exchanging $i,j$, are smooth. Transitions between identical charts are the identity; opposite-sign charts with the same omitted coordinate do not overlap. The ambient subspace topology is Hausdorff and second countable, as it is inherited from Euclidean space. Thus this coordinate-projection atlas of an ellipsoid is a smooth atlas, proving the ellipsoid is a smooth manifold of dimension $n$. The rescaling $r\mapsto q$ also gives a diffeomorphism with the unit sphere $S^n$.

For the three-sphere, use its identification with SU(2) rather than separate local constructions. With the Pauli matrices ${\tau }_i$, put $T_i=-i{\tau }_i$. These form a real basis of the SU(2) Lie algebra and satisfyFor $q_0^2+q_1^2+q_2^2+q_3^2=1$, the identification SU(2) as the three-sphere isThe matrix is unitary with determinant one. Conversely, every SU(2) matrix has this form, and both the correspondence and its inverse are smooth.

The left translations $L_g(h)=gh$ are diffeomorphisms. Define three left-invariant vector fields byMatrix multiplication makes these vector fields smooth globally. Since $(d{L}_g{)}_e$ is an isomorphism from the SU(2) Lie algebra to $T_g\mathrm{SU}(2)$, the three vectors are linearly independent at every point, and each is nowhere zero. This is the parallelization of a Lie group by left translations.

Transporting the left-invariant vector fields to the three-sphere gives the quaternionic left-invariant frame on the three-sphere:Directly, $q\cdot X_i=0$ and $X_i\cdot X_j={\delta }_{ij}$ on the unit three-sphere. Thus the vectors are tangent and form a global orthonormal frame of a vector bundle. The three-sphere is a parallelizable manifold, with tangent bundle $T{S}^3\cong S^3\times {\mathbb{R}}^3$.

The interior of the diamond is $B=\{(x_1,x_2):|x_1|+|x_2|<1\}$. Its Minkowski functional, or the uniqueness argument already proved, gives the L1 norm:The final unheaded set $C$ cannot be an open unit ball of a norm, because every norm ball is convex by homogeneity and the triangle inequality. Both $p=(9/10,0)$ and $q=(0,9/10)$ lie in $C$: their squared-distance expression is $101/100>1$. Their midpoint has that expressionso $(p+q)/2\notin C$. Thus $\boxed{\text{no norm has }C\text{ as its open unit ball}}$. The strict inequalities in the definition are respected by this explicit counterexample.

The unheaded preliminary requests are addressed before the first example. A norm is a function $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly for $x=0$, $p(ax)=|a|p(x)$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d(x,y)=p(x-y)$. Translation $T_v$ preserves this distance, and scalar multiplication satisfies $d(ax,ay)=|a|d(x,y)$, so both are continuous; multiplication by zero is constant.

The open unit ball determines the norm uniquely through its Minkowski functional:because $x/t\in B$ is equivalent to $p(x)<t$. This also works for $x=0$.

Under the given radial and convexity hypotheses, set $p_B(0)=0$ and $p_B(v)=1/\lambda (v)$ for $v\ne 0$. The finite symmetric segment along each line gives positivity, absolute homogeneity and $v\in B⟺p_B(v)<1$. If $s>p_B(x)$ and $t>p_B(y)$, then $x/s,y/t\in B$, and the stated convexity condition yields $(x+y)/(s+t)\in B$. Hence $p_B(x+y)<s+t$; taking $s\downarrow p_B(x)$ and $t\downarrow p_B(y)$ proves the triangle inequality. Thus this is a norm determined by a convex radial unit ball. The case $V=\{0\}$ is immediate.

For the coordinate box, the Minkowski functional is the supremum norm:Indeed $x/t\in B$ exactly when $t>{\max }_i|x_i|$.

Write $B$ for the open unit ball. Because $T$ is a linear map, the hypothesis scales toFix $y\in B$. Choose numbers $r$ and $\theta$ such thatthen $r/(1-\theta )<1$.

Set $z_0=y$. Inductively, if $\Vert z_{j-1}\Vert <r{\theta }^{j-1}$, the scaled density statement lets us chooseand we put $z_j=z_{j-1}-T{x}_j$. The geometric series givesBecause $X$ is a Banach space, $x={\sum }_{j\ge 1}{x}_j$ exists and lies in $B$. Since $T$ is a bounded linear operator, it is continuous, whileTherefore $Tx=y$. As $y\in B$ was arbitrary,This is the dense open-unit-ball image criterion.