Levi form of the unit ball

Writing $z_j=x_j+i{y}_j$, the coordinate expression $\rho ={\sum }_{j<m}(x_j^2+y_j^2)-1$ is a polynomial, so $\rho \in C^{\mathrm{\infty }}({\mathbb{C}}^m,\mathbb{R})$; applying [F3] to the partials ${\mathrm{\partial }}_{x_j}\rho =2x_j$ and ${\mathrm{\partial }}_{y_j}\rho =2y_j$ gives ${\mathrm{\partial }}_{z_j}\rho (a)=\frac{1}{2}(2x_j-2i{y}_j)=\overline{a_j}$ and ${\mathrm{\partial }}_{{\bar{z}}_j}\rho (a)=\frac{1}{2}(2x_j+2i{y}_j)=a_j$ at every $a\in {\mathbb{C}}^m$.

The set $B=\{z:\rho (z)<0\}=\{z:\parallel z\parallel <1\}=B(0,1)$ is the open ball of radius $1$ about $0$ in the sense of [F5], hence is an open subset of ${\mathbb{C}}^m$ by [F4] with $n=1$; it is nonempty because $\parallel 0\parallel =0<1$ by [F7]; and it is connected, because $\mathrm{\Phi }(B)$ is the unit ball of ${\mathbb{R}}^{2m}$ for the Euclidean norm, which is path-connected and connected by [F8], while [F6] carries the open sets of ${\mathbb{C}}^m$ onto those of ${\mathbb{R}}^{2m}$. Thus $B$ is a domain.

Differentiating the first-order expressions of step 1.1 gives ${\mathrm{\partial }}^2\rho \mathrm{/}\mathrm{\partial }{z}_j\mathrm{\partial }{\bar{z}}_k={\delta }_{jk}$ for all $j,k<m$, so [F1] yields, for every $a\in {\mathbb{C}}^m$ and every $\xi \in {\mathbb{C}}^m$, ${\mathcal{L}}_{\rho }(a;\xi )={\sum }_{j,k<m}{\delta }_{jk}{\xi }_j\overline{{\xi }_k}={\sum }_{j<m}\mathrm{\mid }{\xi }_j{\mathrm{\mid }}^2=\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{.}$

The boundary of $B$ is exactly the unit sphere $\{\parallel z\parallel =1\}$: if $\parallel p\parallel <1$ then $p\in B$ and $B$ is open by step 1.2, so $p\mathrm{\notin }\mathrm{\partial }B$ by [F9]; if $\parallel p\parallel >1$ then with $r:=(\parallel p\parallel -1)\mathrm{/}2>0$ every $z$ with $\parallel z-p\parallel <r$ satisfies $\parallel z\parallel \ge \parallel p\parallel -\parallel p-z\parallel >(\parallel p\parallel +1)\mathrm{/}2>1$ by [F7], so the ball about $p$ of radius $r$ misses $B$ and $p\mathrm{\notin }\bar{B}$, hence $p\mathrm{\notin }\mathrm{\partial }B$ by [F9]; and if $\parallel p\parallel =1$ then for $0<t<1$ the points $(1-t)p$ lie in $B$, since $\parallel (1-t)p\parallel =1-t<1$ by [F7], and converge to $p$, since $\parallel (1-t)p-p\parallel =t$, while $p\mathrm{\notin }B$; hence $p\in \bar{B}\setminus B=\bar{B}\setminus \mathrm{int}(B)=\mathrm{\partial }B$ by [F9].

At a point $p$ with $\parallel p\parallel =1$ one has $\rho (p)=0$; the complex tangent vectors at $p$ are those $\xi$ with ${\sum }_j\frac{\mathrm{\partial }\rho }{\mathrm{\partial }{z}_j}(p){\xi }_j={\sum }_j\overline{p_j}{\xi }_j=0$ by [F2] and step 1.1, and each nonzero such $\xi$ satisfies ${\mathcal{L}}_{\rho }(p;\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^2>0$ by step 2.1; also $d\rho (p)\mathrm{\ne }0$, because if $D\rho (p)=0$ then [F3] forces every Wirtinger partial ${\mathrm{\partial }}_{z_j}\rho (p)=\overline{p_j}$ and ${\mathrm{\partial }}_{{\bar{z}}_j}\rho (p)=p_j$ to vanish, whereas some $p_j\mathrm{\ne }0$ since $\parallel p\parallel =1$.

Conclusion: every boundary point of $B$ satisfies $\parallel p\parallel =1$ by step 2.2, so with $U:={\mathbb{C}}^m$ the pair $(U,\rho )$ satisfies $B\cap U=\{\rho <0\}$, $d\rho (p)\mathrm{\ne }0$ and ${\mathcal{L}}_{\rho }(p;\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^2>0$ for every nonzero complex tangent vector $\xi$ by step 3.1; this is the strict form of the condition in [F2], so the unit ball is strongly pseudoconvex, and in particular, weakening $>$ to $\ge$, it is Levi pseudoconvex in the sense of [F2]; moreover $\rho$ is strictly plurisubharmonic on all of ${\mathbb{C}}^m$ by [F1] and step 2.1, since ${\mathcal{L}}_{\rho }(a;\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^2>0$ for every $a$ and every $\xi \mathrm{\ne }0$.