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A closed analytic subset of a complex manifold is nonsingular when every point has a neighbourhood on which the subset is the zero set of a holomorphic map of full rank along that zero set. The holomorphic implicit function theorem then makes it a local coordinate subspace. This includes a closed complex submanifold, whereas an arbitrary open piece of a complex submanifold need not be globally closed in its ambient manifold.

At $c=0$ the left-hand root is $-2$, and the derivative of the polynomial with respect to $z$ is $3(-2{)}^2=12\ne 0$. The holomorphic implicit function theorem states that if $P(z,c)$ is holomorphic near $(z_0,c_0)$, $P(z_0,c_0)=0$ and $P_z(z_0,c_0)\ne 0$, there is a unique holomorphic function $z=\gamma (c)$ near $c_0$, with $\gamma (c_0)=z_0$ and $P(\gamma (c),c)=0$. It therefore supplies an analytic root branch $\gamma (c)$ near zero. Differentiate its equation:It follows that ${\gamma }^{\prime }(0)=i/6$ andFor sufficiently small positive $c$ its imaginary part is $c/6+O(c^2)>0$. By the unique-root localization from the preceding part, this is the root in the left half-plane. Thus that root lies above the real axis for all sufficiently small positive $c$.

A function element is a pair $(f,D)$ where $D$ is a nonempty disk and $f$ is holomorphic there; the germ at its center captures its local information. Elements continue one another when their functions agree on an overlapping connected neighborhood. A complete analytic function is a maximal collection reachable from one element by chains of analytic continuation, including all further continuations of its members.

If $f^n=p$ in the initial disk, the identity theorem propagates this identity across every overlapping disk in a continuation chain. Induction along the chain proves every element of the complete analytic function also satisfies ${\tilde{f}}^n=p$.

Put $F(z,w)=w^n-p(z)$. Its gradient is $(-p^{\prime }(z),n{w}^{n-1})$. At a point of the curve with $w\ne 0$, the second entry is nonzero; at $w=0$, the root of $p$ is simple and the first entry is nonzero. Thus the holomorphic implicit function theorem supplies charts. Its precise local statement is that a holomorphic equation with nonzero derivative in one variable can locally be solved uniquely and holomorphically for that variable as a function of the other. For $w\ne 0$, use $z$ as coordinate and write $w=w(z)$; at a simple root use $w$ as coordinate and write $z=z(w)$. The transition maps are holomorphic with nonzero derivatives, giving a Riemann surface. The two projections are holomorphic because they are holomorphic expressions in either kind of chart.

Away from the zeros of $p$, the projection onto $z$ is an unramified $n$-sheeted covering. A germ of a root corresponds to $(z,f(z))$ and its analytic continuations follow this covering. Near a simple root, $z-z_0$ is a nonzero analytic factor times $w^n$; a circuit about $z_0$ cyclically permutes all $n$ branches. For nonconstant square-free $p$, these circuits connect all sheets. Hence the germ surface $S(F)$ is the curve with the points over the roots removed, and $C$ is its branched completion obtained by adding those ramification points. An ordinary function element in the variable $z$ cannot include such a point: the zero order of a holomorphic $n$th power is divisible by $n$, whereas $p$ has zero order one. If a convention already includes these branch points in the Riemann surface of a multivalued function, that completed surface is $C$ itself. For a nonzero constant $p$, the curve has $n$ separate components, and the chosen element determines just one of them. The identically zero polynomial is excluded by the non-singularity premise.

For a nonsingular plane curve $F(s,t)=0$, at a point with $F_t\ne 0$ the holomorphic implicit function theorem gives $t=t(s)$ and the chart coordinate $s$. Where $F_s\ne 0$, use $t$ and express $s=s(t)$. Nonsingularity ensures at least one choice is available, and the coordinate transition functions and both restricted projections are holomorphic.

Here $F=t^4-P(s)$ with $P(s)=(s^2-1)(s-4)$. A singular point would require $t=0$, $P(s)=0$ and $P^{\prime }(s)=0$. But $P$ has three distinct roots $-1,1,4$, so this is impossible. Hence the curve is nonsingular.

If $t\ne 0$, $F_t=4t^3\ne 0$, so $s$ is a local coordinate and the projection to $s$ is unramified. At $(s_0,0)$ for $s_0\in \{-1,1,4\}$, use $t$ as coordinate. Since $P^{\prime }(s_0)\ne 0$, solving $P(s)=t^4$ gives $s-s_0=t^4/P^{\prime }(s_0)+O(t^8)$. ThusThese are all the ramification points on the stated affine curve; no points at infinity belong to $Y\subset {\mathbb{C}}^2$.

The assertion is false for $k\ge 2$, but true for $k=1$. For $k\ge 2$, take the jointly holomorphic right-hand side $A_{\alpha }(z,w)=k{z}^{k-1}+\alpha$. Its solution with zero initial value isAt $\alpha =0$, the value zero has multiplicity $k$ at the origin. For $\alpha \ne 0$, however, every zero is a simple zero: $f_{\alpha }^{\prime }(0)=\alpha$, and at a nonzero zero $z^{k-1}=-\alpha$,Thus there is no zero of multiplicity $k$ anywhere, even though the $k$ simple zeros stay close to zero for small $\alpha$. When $k=1$, part (a) gives exactly one zero counted with multiplicity in a small disc, so that zero is simple. Alternatively, the holomorphic implicit function theorem gives its holomorphic location as a function of $\alpha$.

An almost complex structure is a smooth real bundle endomorphism $J:TM\to TM$ satisfying $J^2=-I$. In particular the real dimension is even, say $2m$. After complexifying, the type decomposition of the complexified tangent bundle isThe corresponding differential forms of type (p, q) are smooth sections of ${\Lambda }^p(T^{1,0}M{)}^{\ast }\otimes {\Lambda }^q(T^{0,1}M{)}^{\ast }$, regarded as alternating forms by the wedge product of differential forms. A form of type $(p,q)$ vanishes unless its arguments include $p$ vectors of type $(1,0)$ and $q$ of type $(0,1)$.

Besides vanishing of the Nijenhuis tensor, two equivalent descriptions of an integrable almost complex structure are: the smooth sections of $T^{0,1}M$ are closed under Lie bracket; and every point has complex coordinates in which $J$ is multiplication by $i$, with holomorphic changes of coordinates. The equivalence with existence of these coordinates is the Newlander-Nirenberg theorem for smooth $J$. A further equivalent condition is that the exterior derivative has only the two type components $d=\partial +\bar{\partial }$, with no components of bidegree $(2,-1)$ or $(-1,2)$. Indeed evaluation of $d$ of a $(1,0)$-form on two $(0,1)$ vector fields detects the $(1,0)$ component of their bracket; complex conjugation detects the other component.

For the almost complex structure induced by a complex atlas, write $z^j=x^j+i{y}^j$ and put $J{\partial }_{x^j}={\partial }_{y^j}$, $J{\partial }_{y^j}=-{\partial }_{x^j}$. The differentials of holomorphic transition maps are complex linear, so these local definitions agree. In these coordinates $J$ has constant coefficients and all coordinate vector fields commute. The expression defining the Nijenhuis tensor is tensorial, so evaluating it on this frame proves it is zero everywhere. Thus the induced structure is integrable.

Here is a direct proof for the smooth submanifolds with invariant complex tangent spaces are complex assertion. At a point of $Y$, its tangent space is a complex-linear subspace, of real dimension $2k$. Make a complex-linear change of ambient holomorphic coordinates so that this tangent space is ${\mathbb{C}}^k\times 0$. Projection onto the first $k$ coordinates has invertible real differential on $Y$. The real inverse function theorem states that a smooth map between equal-dimensional manifolds with invertible differential is a diffeomorphism on sufficiently small neighbourhoods. It therefore writes $Y$ locally as a smooth graph $w=F(z)$ over an open set in ${\mathbb{C}}^k$.

A tangent vector to this graph is $(v,dF(v))$. Its image under $J$ is $(iv,idF(v))$. Tangent-space invariance impliesso $F$ satisfies the Cauchy-Riemann equations. Smooth functions satisfying these equations are holomorphic: restrict to each coordinate complex line, use the one-variable criterion, and apply the iterated Cauchy integral formula on a small polydisc to get jointly holomorphic coefficients. Hence the graph parametrization is holomorphic. Restrictions of ambient holomorphic coordinates supply holomorphic transition maps between these graph charts. The restricted structure on $Y$ is an integrable complex structure. This also proves the induced structure agrees with the original restriction, rather than merely producing some complex structure on $Y$.

A nonsingular analytic subvariety is a closed analytic subset which near each of its points is the zero set of a holomorphic submersion of constant rank; equivalently, it is locally a coordinate subspace in holomorphic coordinates. The holomorphic implicit function theorem says that if a holomorphic map to ${\mathbb{C}}^r$ has differential of rank $r$ at a zero, its nearby zero set is a complex submanifold of codimension $r$.

The homogeneous equation defines a subset of Complex projective space because multiplying a vector by $\lambda \ne 0$ multiplies its quadratic value by ${\lambda }^2$. Its gradient on ${\mathbb{C}}^4$ is $2Qz$, which is nonzero for $z\ne 0$. More explicitly, in an affine chart set $z_i=1$. If all derivatives of the restricted equation vanished at one of its zeros, all components of $2Qz$ except possibly the $i$th would vanish. The Euler homogeneous function theorem gives ${\sum }_j{z}_j{\partial }_{j}p=2p=0$, forcing the remaining component to vanish as well. This contradicts invertibility of $Q$. The holomorphic implicit function theorem thus proves $S$ is a smooth complex surface.

To see its geometry, reduce the nondegenerate symmetric bilinear form to diagonal form over $\mathbb{C}$. There is a vector of nonzero self-pairing, since otherwise polarization would make the entire form zero. Normalize such a vector using a square root of its self-pairing. Its orthogonal complement remains nondegenerate, so induction gives an orthonormal basis. Apply the same argument to the nondegenerate form $w_0{w}_3-w_1{w}_2$: the two forms are equivalent under an invertible linear change of variables. The allowed projective linear transformation therefore reduces $S$ toView its coordinates as the nonzero rank-one matrix $A=\left(\begin{array}{cc}{w}_0& w_1\\ w_2& w_3\end{array}\right)=u{v}^T$. At $[A]$, holding $[u]$ fixed and varying $[v]$ gives a projective line, and holding $[v]$ fixed and varying $[u]$ gives another. They are distinct and meet at $[A]$. In fact these are the only lines through it: if a second rank-one matrix is $a{b}^T$, the entire pencil has determinant zero only if $\det (u,a)\det (v,b)=0$, so one projective factor is fixed. These are the two rulings of a smooth quadric surface.

The Segre embedding is the holomorphic mapEvery nonzero rank-one matrix has such a factorization, unique up to reciprocal rescaling, so $\Phi$ is bijective. Its inverse is holomorphic on each open set where a matrix entry $a_{ij}\ne 0$: the projective column $[a_{0j}:a_{1j}]$ recovers $[u]$, and the projective row $[a_{i0}:a_{i1}]$ recovers $[v]$. These formulas agree on overlaps. Thus the Segre description of a smooth quadric surface gives the promised stronger conclusion

Let $H$ denote a hyperplane-section divisor of a projective plane curve. The adjunction formula for a smooth degree-$d$ projective plane curve statesSince $\mathrm{deg}(H{|}_C)=d$, this gives the canonical degree of a smooth plane curveCombining this with $\mathrm{deg}{K}_C=2g-2$ gives the genus of a smooth plane curve

Homogenization of the affine equation gives the projective completionThis is the plane model y plus x cubed plus xy cubed equals zero of the Klein quartic: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^{3}B+B^{3}C+C^{3}A=0$.

Its first partial derivatives areIf one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in projective space. If $XYZ\ne 0$, multiplying the three derivative equations givesagain a contradiction. The Jacobian criterion therefore proves that $C$ is smooth.

The rational function $y=Y/Z$ defines a rational map of projective varieties $C⇢{\mathbb{P}}^1$. Because $C$ is a smooth projective curve, it extends uniquely to a morphismFor a generic finite value of $y$, its fibre is given by the cubicso the degree of a holomorphic map is $3$. The discriminant of a depressed cubic is

At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the holomorphic implicit function theorem givesso $x$ is a local coordinate and $\pi$ has ramification index of a holomorphic map $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.

It remains to inspect the points at infinity. They areNear $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^{3}v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a simple pole. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+s{t}^3=0$, so $t=-s^3+O(s^{10})$ andThus $y$ has a double pole at $P$ and $e_P=2$. These calculations are the ramification of the x-coordinate on the Klein quartic after the coordinate relabelling above.

The total ramification contribution isThe Riemann-Hurwitz formula for the degree-three map to the projective line now givesand therefore

LetIts complex partial derivatives areThere is no point of $X^{\prime }$ at which both vanish. Indeed, if $F=F_x=F_y=0$, then $F_y=0$ and $F=0$ giveNeither $x$ nor $y$ can be zero because $F_x=0$. Substitution into $F_x=0$ gives $12y^7+1=0$, whereas substitution into $F_y=0$ gives $8y^7+3=0$; these equations are incompatible.

For every $p=(x_0,y_0)\in X^{\prime }$, at least one partial derivative is nonzero. If $F_y(p)\ne 0$, the holomorphic implicit function theorem expresses the curve near $p$ as $y=g(x)$, and the restriction of the $x$-projection is a holomorphic coordinate chart. If $F_x(p)\ne 0$, use the $y$-projection instead. On overlaps, the transition functions are restrictions of holomorphic coordinate functions and are holomorphic. These charts define an atlas making $X^{\prime }$ a one-dimensional complex manifold, hence a Riemann surface.