Let $S\subset \mathbf P^3$ be a smooth cubic surface and let $P\in S$. Let

$$H=T_P S$$

be the tangent plane at $P$. Then

$$C:=H\cap S$$

is a plane cubic curve, because it is cut out on the plane $H\simeq \mathbf P^2$ by the cubic equation of $S$.

The key point is that $H$ is tangent to $S$ at $P$, so the plane section $C$ is singular at $P$.

Indeed, choose affine coordinates near $P$ so that $P=(0,0,0)$ and the tangent plane is

$$H=\{z=0\}.$$

Since $S$ is smooth at $P$, locally its equation can be written in the form

$$z+q_2(x,y,z)+q_3(x,y,z)=0,$$

where $q_2$ and $q_3$ have degree $2$ and $3$. Restricting to $H$, i.e. setting $z=0$, gives

$$q_2(x,y,0)+q_3(x,y,0)=0.$$

There is no linear term. Therefore $P$ is a singular point of the plane cubic $C$.

Now use the assumption that there is no line on $S$ passing through $P$.

If $C=H\cap S$ were reducible, then since $C$ is a plane cubic, its components would have degrees adding to $3$. The possible reducible cases are:

$$C=L+Q$$

with $L$ a line and $Q$ a conic, or

$$C=L_1+L_2+L_3$$

as a sum of lines, possibly with multiplicities.

But every component of $C$ is contained in both $H$ and $S$. Hence any line component of $C$ is a line on $S$.

Because $P\in C$ and $C$ is singular at $P$, any reducible decomposition would force at least one line component through $P$. That contradicts the hypothesis that no line on $S$ passes through $P$.

Therefore $C$ cannot be reducible.

So

$$H\cap S$$

is an irreducible plane cubic curve singular at $P$.

Since an irreducible singular plane cubic has arithmetic genus $1$ and one singularity of $\delta$-invariant $1$, its singularity is either a node or a cusp. Thus:

$$\boxed{H\cap S \text{ is an irreducible singular plane cubic with singular point } P.}$$

More precisely,

$$\boxed{H\cap S \text{ is either a nodal cubic or a cuspidal cubic, singular at } P.}$$

Geometrically, the tangent plane cuts the cubic surface in a cubic curve having a double point at $P$. The absence of lines through $P$ rules out the familiar reducible tangent-plane sections such as a line plus a conic or three lines.

The involution on $S'$ is the Geiser involution. It is constructed from projection away from $P$.

Assume for simplicity that the ground field has characteristic $\neq 2$, for example $S$ is over $\mathbf C$.

Let

$$\pi:S'\to S$$

be the blowup of $S$ at $P$, and let $E=\pi^{-1}(P)$ be the exceptional curve.

The hyperplanes in $\mathbf P^3$ passing through $P$ form a projective plane. Restricting them to $S$, they give the linear system of hyperplane sections of $S$ passing through $P$. After blowing up $P$, this becomes the base-point-free linear system

$$|\pi^*\mathcal O_S(1)-E|.$$

But since $S\subset \mathbf P^3$ is a cubic surface, $-K_S=\mathcal O_S(1)$, so

$$-K_{S'}=\pi^*(-K_S)-E=\pi^*\mathcal O_S(1)-E.$$

Thus the system is exactly

$$|-K_{S'}|.$$

It defines a morphism

$$\varphi:S'\longrightarrow \mathbf P^2.$$

Geometrically, $\varphi$ is just the projection of $S$ from the point $P$, with the indeterminacy at $P$ resolved by the blowup.

Now take a general line $\ell\subset \mathbf P^3$ passing through $P$. Since no line on $S$ passes through $P$, $\ell$ is not contained in $S$. Hence

$$\ell\cap S$$

is a degree $3$ scheme on $\ell$, and since $P\in S$, we can write

$$\ell\cap S=P+Q+R$$

counting multiplicities.

The map $\varphi:S'\to \mathbf P^2$ sends both $Q$ and $R$ to the same line $\ell$ through $P$. Therefore a general fiber of $\varphi$ consists of two points, $Q$ and $R$.

So $\varphi$ is a finite morphism of degree $2$. Indeed,

$$(\pi^*\mathcal O_S(1)-E)^2=3-1=2.$$

Therefore $\varphi:S'\to \mathbf P^2$ is a double cover. The involution is the deck transformation of this double cover:

$$\iota:S'\to S'.$$

Concretely, for a general point $Q\in S\setminus\{P\}$, define

$$\iota(Q)=R,$$

where $R$ is the third residual point determined by

$$\overline{PQ}\cap S=P+Q+R.$$

Thus the involution exchanges the two residual intersection points of $S$ with a line through $P$.

On the exceptional divisor $E$, the construction extends as follows. A point of $E$ corresponds to a tangent direction at $P$, equivalently to a line

$$\ell\subset T_P S$$

through $P$. Since $\ell\subset T_P S$, the intersection multiplicity of $\ell$ with $S$ at $P$ is at least $2$, so

$$\ell\cap S=2P+R_\ell.$$

Then the involution sends the point of $E$ corresponding to the direction $\ell$ to the residual point $R_\ell$. Conversely, $R_\ell$ is sent back to that point of $E$.

In particular, if

$$C=T_P S\cap S,$$

then on $S'$,

$$\varphi^{-1}(\text{line corresponding to }T_PS)=E+\widetilde C,$$

and the involution exchanges $E$ with the strict transform $\widetilde C$. Their intersection points are fixed points of the involution.

Equivalently, one can describe the construction in coordinates. Put

$$P=[0:0:0:1],$$

and write the cubic equation of $S$ as

$$F=x_3^2L(x_0,x_1,x_2)+x_3Q(x_0,x_1,x_2)+C(x_0,x_1,x_2),$$

where $L,Q,C$ are homogeneous of degrees $1,2,3$. Projection from $P$ is

$$[x_0:x_1:x_2:x_3]\mapsto [x_0:x_1:x_2].$$

For fixed $[x_0:x_1:x_2]$, the fiber is given by the quadratic equation

$$Lx_3^2+Qx_3+C=0.$$

The involution exchanges the two roots of this quadratic. Thus, where $L\neq 0$, it is given by

$$x_3\longmapsto -\frac{Q}{L}-x_3.$$

After blowing up $P$, this formula extends regularly everywhere on $S'$.

So the desired involution is:

$$\boxed{\text{the deck involution of the double cover }S'\to \mathbf P^2\text{ induced by projection from }P.}$$

It is the classical Geiser involution associated to the point $P$.

No.

$$\mathbf P^1\times \mathbf P^1$$

and

$$\mathbb F_1 \simeq \operatorname{Bl}_p \mathbf P^2$$

are not deformation equivalent as smooth complex projective surfaces.

A clean way to see this is by comparing their intersection forms.

For

$$\mathbf P^1\times \mathbf P^1,$$

let $f_1,f_2$ be the two rulings. Then

$$f_1^2=0,\qquad f_2^2=0,\qquad f_1\cdot f_2=1.$$

So the intersection form is

$$\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}.$$

This is an even form: every divisor class has even self-intersection, because

$$(af_1+bf_2)^2=2ab.$$

By contrast, on

$$\mathbb F_1=\operatorname{Bl}_p\mathbf P^2,$$

let $H$ be the pullback of a line in $\mathbf P^2$, and let $E$ be the exceptional curve. Then

$$H^2=1,\qquad E^2=-1,\qquad H\cdot E=0.$$

Thus the intersection form is

$$\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}.$$

This is an odd form, since $H^2=1$.

So:

$$\mathbf P^1\times \mathbf P^1$$

has even intersection form, while

$$\mathbb F_1$$

has odd intersection form.

If two smooth projective varieties are deformation equivalent, then by Ehresmann’s theorem their underlying smooth manifolds are diffeomorphic. In particular, their intersection forms must be isomorphic. But an even unimodular rank-two form cannot be isomorphic to an odd one.

Therefore

$$\boxed{\mathbf P^1\times \mathbf P^1 \text{ and } \mathbb F_1 \text{ are not deformation equivalent}.}$$

Equivalently, among Hirzebruch surfaces $\mathbb F_n$, one has the rule:

$$\mathbb F_n \sim_{\mathrm{def}} \mathbb F_m \quad\Longleftrightarrow\quad n\equiv m \pmod 2.$$

Since

$$\mathbf P^1\times \mathbf P^1=\mathbb F_0$$

and

$$\operatorname{Bl}_p\mathbf P^2=\mathbb F_1,$$

they lie in different deformation classes.

Use the description of Hirzebruch surfaces as projectivized rank-two bundles over $\mathbf P^1$:

$$\mathbb F_n \simeq \mathbf P_{\mathbf P^1}\bigl(\mathcal O_{\mathbf P^1}\oplus \mathcal O_{\mathbf P^1}(n)\bigr).$$

The key step is:

$$\boxed{\mathbb F_n \text{ deforms to } \mathbb F_{n-2} \quad \text{for } n\ge 2.}$$

Then, if $m-n$ is even, repeatedly applying this step gives a deformation chain

$$\mathbb F_m \rightsquigarrow \mathbb F_{m-2} \rightsquigarrow \mathbb F_{m-4} \rightsquigarrow \cdots \rightsquigarrow \mathbb F_n.$$

Here is the standard construction.

Consider extensions on $\mathbf P^1$

$$0\longrightarrow \mathcal O_{\mathbf P^1} \longrightarrow E \longrightarrow \mathcal O_{\mathbf P^1}(n) \longrightarrow 0.$$

These extensions are classified by

$$\operatorname{Ext}^1(\mathcal O(n),\mathcal O) \simeq H^1(\mathbf P^1,\mathcal O(-n)).$$

For $n\ge 2$, this space is nonzero.

The zero extension gives

$$E_0\simeq \mathcal O\oplus \mathcal O(n),$$

hence

$$\mathbf P(E_0)\simeq \mathbb F_n.$$

Now take a suitable nonzero extension class. One can realize it explicitly by the exact sequence

$$0\longrightarrow \mathcal O \longrightarrow \mathcal O(1)\oplus \mathcal O(n-1) \longrightarrow \mathcal O(n) \longrightarrow 0.$$

For example, choose homogeneous coordinates $[u:v]$ on $\mathbf P^1$, and embed

$$\mathcal O \hookrightarrow \mathcal O(1)\oplus \mathcal O(n-1)$$

by

$$1\longmapsto (u, v^{n-1}).$$

The two sections $u$ and $v^{n-1}$ have no common zero, so the quotient is a line bundle. Its degree is

$$\deg(\mathcal O(1)\oplus \mathcal O(n-1))-\deg(\mathcal O)=n,$$

so the quotient is $\mathcal O(n)$. Thus this gives a non-split extension.

Now take a one-parameter family of extension classes

$$t\xi\in H^1(\mathbf P^1,\mathcal O(-n)), \qquad t\in \mathbf A^1,$$

where $\xi\neq 0$ is the class above. This gives a family of vector bundles $E_t$ such that

$$E_0\simeq \mathcal O\oplus \mathcal O(n),$$

while for $t\neq 0$,

$$E_t\simeq \mathcal O(1)\oplus \mathcal O(n-1).$$

Projectivizing gives a smooth family

$$\mathbf P(E_t)\longrightarrow \mathbf A^1.$$

The central fiber is

$$\mathbf P(E_0) \simeq \mathbf P(\mathcal O\oplus \mathcal O(n)) \simeq \mathbb F_n.$$

For $t\neq 0$, we get

$$\mathbf P(E_t) \simeq \mathbf P(\mathcal O(1)\oplus \mathcal O(n-1)).$$

But projectivization is unchanged by tensoring the bundle with a line bundle, so

$$\mathbf P(\mathcal O(1)\oplus \mathcal O(n-1)) \simeq \mathbf P\bigl((\mathcal O(1)\oplus \mathcal O(n-1))\otimes \mathcal O(-1)\bigr).$$

Hence

$$\mathbf P(\mathcal O(1)\oplus \mathcal O(n-1)) \simeq \mathbf P(\mathcal O\oplus \mathcal O(n-2)) \simeq \mathbb F_{n-2}.$$

Therefore

$$\boxed{\mathbb F_n \text{ degenerates to } \mathbb F_{n-2},}$$

or equivalently $\mathbb F_n$ and $\mathbb F_{n-2}$ lie in the same deformation class.

Iterating,

$$\mathbb F_m \sim_{\mathrm{def}} \mathbb F_{m-2} \sim_{\mathrm{def}} \mathbb F_{m-4} \sim_{\mathrm{def}}\cdots.$$

So if

$$m-n\in 2\mathbf Z,$$

then

$$\boxed{\mathbb F_m \text{ and } \mathbb F_n \text{ are deformation equivalent}.}$$

This also explains why there are exactly two deformation classes of Hirzebruch surfaces: the even ones

$$\mathbb F_0,\mathbb F_2,\mathbb F_4,\dots$$

and the odd ones

$$\mathbb F_1,\mathbb F_3,\mathbb F_5,\dots.$$

Let $n\ge 2$. We want a family of rank-two bundles

$$E_t \quad \text{on } \mathbf P^1$$

such that

$$E_0\simeq \mathcal O\oplus \mathcal O(n),$$

but for $t\neq 0$,

$$E_t\simeq \mathcal O(1)\oplus \mathcal O(n-1).$$

Then projectivizing gives

$$\mathbf P(E_0)\simeq \mathbb F_n,$$

whereas

$$\mathbf P(E_t)\simeq \mathbf P(\mathcal O(1)\oplus \mathcal O(n-1)) \simeq \mathbf P(\mathcal O\oplus \mathcal O(n-2)) \simeq \mathbb F_{n-2}.$$

So the main point is to build $E_t$.

---

Take the standard open cover of $\mathbf P^1$:

$$U_0=\{u\neq 0\},\qquad U_\infty=\{v\neq 0\},$$

with affine coordinate

$$x=\frac vu$$

on $U_0$. On the overlap $U_0\cap U_\infty\simeq \mathbf G_m$, the line bundle $\mathcal O(d)$ has transition function

$$x^{-d}.$$

Now define a rank-two bundle $E_t$ by gluing trivial bundles on $U_0$ and $U_\infty$ using the transition matrix

$$G_t(x)= \begin{pmatrix} 1 & t x^{-(n-1)}\\ 0 & x^{-n} \end{pmatrix}.$$

That is, over $U_0\cap U_\infty$, the two local frames are related by this matrix.

Because

$$\det G_t=x^{-n}$$

is invertible on $U_0\cap U_\infty$, this indeed defines a vector bundle.

This matrix gives an extension

$$0\longrightarrow \mathcal O \longrightarrow E_t \longrightarrow \mathcal O(n) \longrightarrow 0.$$

The extension class is

$$t\cdot [x^{-(n-1)}]\in H^1(\mathbf P^1,\mathcal O(-n)).$$

When $t=0$, the matrix becomes diagonal:

$$G_0(x)= \begin{pmatrix} 1 & 0\\ 0 & x^{-n} \end{pmatrix}.$$

Therefore

$$E_0\simeq \mathcal O\oplus \mathcal O(n).$$

For $t\neq 0$, multiplying the extension class by a nonzero scalar does not change the isomorphism class of the middle bundle, so

$$E_t\simeq E_1.$$

It remains to identify $E_1$.

The extension class represented by $x^{-(n-1)}$ is realized by the exact sequence

$$0 \longrightarrow \mathcal O \longrightarrow \mathcal O(1)\oplus \mathcal O(n-1) \longrightarrow \mathcal O(n) \longrightarrow 0.$$

Explicitly, the injection is

$$\mathcal O\longrightarrow \mathcal O(1)\oplus \mathcal O(n-1),$$

given by

$$1\longmapsto (u, v^{n-1}),$$

and the quotient map is

$$\mathcal O(1)\oplus \mathcal O(n-1) \longrightarrow \mathcal O(n),$$

given by

$$(a,b)\longmapsto v^{n-1}a-u b.$$

Indeed,

$$v^{n-1}\cdot u-u\cdot v^{n-1}=0,$$

so the image of the first map lies in the kernel of the second. The second map is surjective because $u$ and $v^{n-1}$ have no common zero on $\mathbf P^1$. Hence the kernel is exactly $\mathcal O$.

Thus the middle term of the nonzero extension is

$$E_t\simeq \mathcal O(1)\oplus \mathcal O(n-1) \qquad \text{for } t\neq 0.$$

So the desired family is produced by the transition matrix

$$\boxed{ G_t(x)= \begin{pmatrix} 1 & t x^{-(n-1)}\\ 0 & x^{-n} \end{pmatrix}. }$$

Equivalently, one can say:

$$\boxed{ E_t \text{ is the extension of } \mathcal O(n) \text{ by } \mathcal O \text{ with class } t\xi\in H^1(\mathbf P^1,\mathcal O(-n)). }$$

Here $\xi\neq 0$ is chosen so that the middle term is $\mathcal O(1)\oplus \mathcal O(n-1)$.

Projectivizing the relative bundle over $\mathbf P^1\times \mathbf A^1$ gives a smooth family

$$\mathbf P(E_t)\longrightarrow \mathbf A^1$$

with

$$\mathbf P(E_0)\simeq \mathbb F_n,$$

and for $t\neq 0$,

$$\mathbf P(E_t)\simeq \mathbb F_{n-2}.$$

That is the explicit deformation step

$$\mathbb F_n \rightsquigarrow \mathbb F_{n-2}.$$

Yes. A standard example is given by two smooth cubic surfaces in $\mathbf P^3$.

Let

$$S_0=\{x_0^3+x_1^3+x_2^3+x_3^3=0\}\subset \mathbf P^3$$

be the Fermat cubic surface, and let

$$S_1=\{x_0^3+x_1^3+x_2^3+x_3^3+x_0x_1x_2=0\}\subset \mathbf P^3.$$

Both are smooth cubic surfaces. Since a smooth cubic surface satisfies

$$K_S=(K_{\mathbf P^3}+S)|_S =(-4H+3H)|_S =-H|_S,$$

we get

$$-K_S=H|_S,$$

which is ample. Hence every smooth cubic surface is a Fano surface, more precisely a del Pezzo surface of degree $3$.

Now consider the one-parameter family

$$\mathcal S_t = \{x_0^3+x_1^3+x_2^3+x_3^3+t x_0x_1x_2=0\} \subset \mathbf P^3\times \mathbf A^1.$$

The fiber over $t$ is

$$S_t=\{x_0^3+x_1^3+x_2^3+x_3^3+t x_0x_1x_2=0\}.$$

For all but finitely many values of $t$, this cubic surface is smooth. In fact, the singular values occur when

$$t^3=-27.$$

Thus over the connected base

$$B=\mathbf A^1\setminus \{t:t^3=-27\},$$

we obtain a smooth family

$$\mathcal S\to B.$$

Both $0$ and $1$ lie in $B$, so $S_0$ and $S_1$ occur as fibers of the same smooth connected family. Therefore

$$\boxed{S_0 \text{ and } S_1 \text{ are deformation equivalent Fano manifolds}.}$$

More conceptually: the space of smooth cubic surfaces is a connected Zariski-open subset of the projective space of all cubic equations in $\mathbf P^3$. Hence any two smooth cubic surfaces are deformation equivalent, and they are all Fano.

Yes. A standard example is the Pasquier–Perrin / Kuznetsov degeneration

$$\operatorname{OGr}(2,7) \rightsquigarrow X_5,$$

where $X_5$ is the $G_2$-horospherical variety of Picard number $1$.

More precisely, there exists a smooth projective family

$$\pi:\mathcal X\longrightarrow \mathbb A^1$$

such that

$$\mathcal X_t \simeq \operatorname{OGr}(2,7) \qquad \text{for every } t\neq 0,$$

while

$$\mathcal X_0 \simeq X_5.$$

Kuznetsov gives explicit geometric constructions of exactly such a smooth family: central fiber the $G_2$-horospherical variety, and all nonzero fibers the isotropic orthogonal Grassmannian $\operatorname{OGr}(2,7)$. (arXiv)

Here

$$\operatorname{OGr}(2,7)$$

is the homogeneous variety parametrizing $2$-dimensional isotropic subspaces in a $7$-dimensional quadratic vector space. Since it is a rational homogeneous space $G/P$, it is Fano.

The central fiber $X_5$ is also Fano. It is a smooth $7$-fold of Picard number $1$, index $4$, and is the exceptional nonhomogeneous $G_2$-horospherical variety appearing in Pasquier–Perrin’s work. (grassmannian.info)

So this gives the desired pattern:

$$\boxed{ \mathcal X_t \cong \operatorname{OGr}(2,7)\quad (t\neq 0), \qquad \mathcal X_0\cong X_5, }$$

with both fibers smooth Fano manifolds.

They are not isomorphic: $\operatorname{OGr}(2,7)$ is homogeneous, while $X_5$ is horospherical but nonhomogeneous. Thus this is a genuine Fano-to-Fano specialization.

The standard classification is the Kulikov classification of semistable degenerations of K3 surfaces.

Suppose

$$\pi:\mathcal X\to \Delta$$

is a degeneration over a disk, with general fiber

$$X_t=\pi^{-1}(t)$$

a smooth K3 surface for $t\neq 0$. After a finite base change and birational modifications, one may assume $\pi$ is a Kulikov model, meaning:

$$\mathcal X_0$$

is a reduced simple normal crossings divisor and

$$K_{\mathcal X}\simeq \mathcal O_{\mathcal X}.$$

Then the central fiber falls into exactly one of three types.

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Type I: smooth central fiber

In Type I, the central fiber is itself a smooth K3 surface:

$$\mathcal X_0 \text{ is a smooth K3 surface.}$$

So the degeneration is actually not very singular after choosing the right model. The family extends smoothly across the central point.

Equivalently, the monodromy around $0$ is finite. After base change, it becomes trivial:

$$N=0,$$

where

$$N=\log T$$

is the logarithm of the unipotent monodromy operator on $H^2(X_t,\mathbf Q)$.

So Type I means:

$$\boxed{\mathcal X_0 \text{ is a smooth K3 surface, and } N=0.}$$

Geometrically, this is the mildest possible degeneration.

---

Type II: chain of surfaces with elliptic double curves

In Type II, the central fiber is reducible:

$$\mathcal X_0=V_0\cup V_1\cup\cdots\cup V_r.$$

The components form a chain:

$$V_0 - V_1 - \cdots - V_r.$$

The double curves

$$D_i=V_i\cap V_{i+1}$$

are smooth elliptic curves. The end components $V_0,V_r$ are rational surfaces, while the middle components $V_1,\dots,V_{r-1}$ are elliptic ruled surfaces.

Schematically:

$$\boxed{ \text{rational surface} \;-\; \text{elliptic ruled surfaces} \;-\; \text{rational surface} }$$

with all intersections along elliptic curves.

The dual complex is an interval:

$$[0,1].$$

The monodromy satisfies

$$N\neq 0,\qquad N^2=0.$$

So Type II means:

$$\boxed{N\neq 0,\quad N^2=0.}$$

Geometrically, Type II degenerations are intermediate: the surface breaks into pieces arranged in a chain, and an elliptic curve appears as the essential limiting object.

---

Type III: union of rational surfaces with sphere-like dual complex

In Type III, the central fiber is a union of rational surfaces:

$$\mathcal X_0=\bigcup_i V_i,$$

where each $V_i$ is a rational surface.

The double curves are rational curves, and the components fit together in a much more complicated two-dimensional pattern. The dual complex is a triangulation of the sphere:

$$\Delta(\mathcal X_0)\simeq S^2.$$

So the combinatorial shape of the central fiber is spherical.

The monodromy satisfies

$$N^2\neq 0,\qquad N^3=0.$$

Since $H^2$ is the relevant cohomology group for K3 surfaces, this is the most degenerate possible unipotent monodromy.

Thus Type III means:

$$\boxed{N^2\neq 0,\quad N^3=0.}$$

Geometrically, Type III is the maximally degenerate case. The K3 surface collapses into a configuration of rational surfaces whose dual intersection complex is a sphere.

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Summary table

$$\begin{array}{c|c|c|c} \text{Type} & \mathcal X_0 & \text{Dual complex} & \text{Monodromy} \\ \hline \text{I} & \text{smooth K3 surface} & \text{point} & N=0 \\ \text{II} & \text{chain of surfaces, elliptic double curves} & \text{interval} & N\neq 0,\ N^2=0 \\ \text{III} & \text{union of rational surfaces} & S^2 & N^2\neq 0,\ N^3=0 \end{array}$$

So the three types can be remembered as:

$$\boxed{ \begin{aligned} \text{Type I} &: \text{smooth K3 limit},\\ \text{Type II} &: \text{chain with elliptic curves},\\ \text{Type III} &: \text{maximal degeneration into rational surfaces}. \end{aligned} }$$

A useful slogan is:

$$\text{Type I} \leftrightarrow \text{finite monodromy},$$ $$\text{Type II} \leftrightarrow \text{elliptic curve in the limit},$$ $$\text{Type III} \leftrightarrow \text{sphere-like rational surface decomposition}.$$

Let $X$ be a variety or smooth manifold, and let

$$Z\subset X$$

be a closed subvariety. The deformation to the normal cone is a canonical one-parameter degeneration in which the general fiber is $X$, while the special fiber remembers the first-order normal geometry of $Z\subset X$.

The slogan is:

$$\boxed{ X \rightsquigarrow C_{Z/X}, }$$

where $C_{Z/X}$ is the normal cone of $Z$ in $X$. If $Z\subset X$ is smooth and regularly embedded, then

$$C_{Z/X}=N_{Z/X},$$

the normal bundle.

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1. Blowup construction

Start with the trivial family

$$X\times \mathbb A^1 \longrightarrow \mathbb A^1.$$

Inside the central fiber $X\times\{0\}$, put the copy of $Z$:

$$Z\times \{0\}\subset X\times \mathbb A^1.$$

Now blow it up:

$$\mathcal Y:=\operatorname{Bl}_{Z\times\{0\}}(X\times \mathbb A^1).$$

There is a natural morphism

$$\mathcal Y\to \mathbb A^1.$$

For $t\neq 0$, the center of the blowup does not meet $X\times\{t\}$. Therefore

$$\mathcal Y_t\simeq X \qquad t\neq 0.$$

The interesting part is the central fiber.

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2. The central fiber

The central fiber of $\mathcal Y\to \mathbb A^1$ has two pieces:

$$\mathcal Y_0 = \operatorname{Bl}_Z X \cup \mathbf P(C_{Z/X}\oplus \mathcal O_Z).$$

If $Z\subset X$ is smooth, this becomes

$$\mathcal Y_0 = \operatorname{Bl}_Z X \cup \mathbf P(N_{Z/X}\oplus \mathcal O_Z).$$

These two components meet along the exceptional divisor

$$\mathbf P(C_{Z/X})$$

or, in the smooth case,

$$\mathbf P(N_{Z/X}).$$

So the compactified deformation looks like

$$\boxed{ \mathcal Y_t\simeq X \text{ for } t\neq 0, }$$

and

$$\boxed{ \mathcal Y_0 = \operatorname{Bl}_Z X \cup_{\mathbf P(N_{Z/X})} \mathbf P(N_{Z/X}\oplus \mathcal O_Z) }$$

when $Z\subset X$ is smooth.

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3. The open version: actual deformation to the normal cone

Often one removes the strict transform of $X\times\{0\}$. That is, define

$$\mathcal D_{Z/X} := \operatorname{Bl}_{Z\times\{0\}}(X\times \mathbb A^1) \setminus \operatorname{Bl}_Z X.$$

Then

$$\mathcal D_{Z/X}\to \mathbb A^1$$

has fibers

$$(\mathcal D_{Z/X})_t\simeq X \qquad t\neq 0,$$

and

$$(\mathcal D_{Z/X})_0\simeq C_{Z/X}.$$

If $Z\subset X$ is smooth, then

$$(\mathcal D_{Z/X})_0\simeq N_{Z/X}.$$

This is the reason for the name:

$$\boxed{ \text{deformation to the normal cone} }$$

or, in the smooth case,

$$\boxed{ \text{deformation to the normal bundle}. }$$

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4. Local affine description

Suppose

$$X=\operatorname{Spec} A$$

and

$$Z=V(I)$$

for an ideal $I\subset A$. Then the normal cone is

$$C_{Z/X} = \operatorname{Spec} \left( \bigoplus_{k\ge 0} I^k/I^{k+1} \right).$$

The deformation space is controlled by the Rees algebra

$$\mathcal R(I) = A[It] = A\oplus It\oplus I^2t^2\oplus \cdots \subset A[t].$$

One convenient form is

$$\operatorname{Spec}\left(A[t,I/t]\right) \longrightarrow \mathbb A^1.$$

For $t\neq 0$, $t$ is invertible, so the fiber is just

$$\operatorname{Spec} A=X.$$

At $t=0$, the fiber becomes

$$\operatorname{Spec} \left( \bigoplus_{k\ge 0} I^k/I^{k+1} \right) = C_{Z/X}.$$

Thus algebraically, deformation to the normal cone replaces $A$ by the associated graded algebra of $I$ at the central fiber.

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5. Basic example: a point in affine space

Let

$$X=\mathbb A^n, \qquad Z=\{0\}.$$

Then

$$N_{Z/X}\simeq T_0\mathbb A^n\simeq \mathbb A^n.$$

The deformation to the normal cone is a family whose general fiber is $\mathbb A^n$, and whose central fiber is the tangent space at the origin:

$$\mathbb A^n \rightsquigarrow T_0\mathbb A^n.$$

For affine space this looks trivial, because the tangent space is again $\mathbb A^n$. But for a subvariety $Z\subset X$, the construction is nontrivial because the central fiber remembers only the normal directions to $Z$ inside $X$.

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6. Geometric intuition

Near $Z$, imagine zooming in on $X$ transverse to $Z$. As the parameter $t\to 0$, the variety $X$ is replaced by its first-order transverse approximation along $Z$.

So:

$$X \quad\text{degenerates to}\quad N_{Z/X}.$$

The tangent directions along $Z$ stay inside $Z$, while the transverse directions become the fibers of the normal bundle.

This is analogous to how a smooth manifold looks like its tangent space under infinite magnification at a point.

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7. Why the compactified version has two components

The blowup

$$\operatorname{Bl}_{Z\times\{0\}}(X\times\mathbb A^1)$$

does not only produce the normal cone. It also keeps track of the strict transform of the original central fiber $X\times\{0\}$, which is

$$\operatorname{Bl}_Z X.$$

The exceptional divisor is

$$\mathbf P(N_{Z/X}\oplus \mathcal O_Z).$$

Hence the central fiber is

$$\operatorname{Bl}_Z X \cup \mathbf P(N_{Z/X}\oplus\mathcal O_Z).$$

If one removes the $\operatorname{Bl}_Z X$ component, one obtains the noncompact deformation whose central fiber is exactly $N_{Z/X}$.

So there are two related objects:

$$\boxed{ \text{open DNC: central fiber } N_{Z/X} }$$

and

$$\boxed{ \text{compactified DNC: central fiber } \operatorname{Bl}_Z X \cup \mathbf P(N_{Z/X}\oplus\mathcal O_Z). }$$

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8. Example relevant to projective geometry

Let $X$ be smooth projective and $Z\subset X$ smooth. The compactified deformation to the normal cone gives a degeneration

$$\mathcal Y\to \mathbb A^1$$

with

$$\mathcal Y_t\simeq X \qquad t\neq 0,$$

and

$$\mathcal Y_0 = \operatorname{Bl}_Z X \cup_{\mathbf P(N_{Z/X})} \mathbf P(N_{Z/X}\oplus\mathcal O_Z).$$

This construction is very useful in intersection theory, Gromov-Witten theory, and K-stability. In K-stability, for instance, one often applies deformation to the normal cone to a subscheme $Z\subset X$ in order to produce a test configuration.

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9. Relation to degeneration of Fano manifolds

If $X$ is Fano and $Z\subset X$ is a subvariety, the deformation to the normal cone gives a natural degeneration of $X$. However, the central fiber is usually reducible:

$$\operatorname{Bl}_Z X \cup \mathbf P(N_{Z/X}\oplus \mathcal O_Z).$$

So it is typically not a smooth Fano manifold. Rather, it is a controlled degeneration whose components are explicitly understood.

This is different from the earlier type of example you asked for, where

$$\mathcal X_t\simeq X \quad t\neq 0,$$

and

$$\mathcal X_0\simeq Y$$

is another smooth Fano manifold. Deformation to the normal cone usually produces a singular or reducible central fiber, but it is extremely useful because it is canonical and computable.

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The key formula to remember is:

$$\boxed{ \operatorname{Bl}_{Z\times\{0\}}(X\times\mathbb A^1) \to \mathbb A^1 }$$

with

$$\boxed{ \mathcal Y_t\simeq X \text{ for } t\neq 0 }$$

and, when $Z\subset X$ is smooth,

$$\boxed{ \mathcal Y_0= \operatorname{Bl}_Z X \cup_{\mathbf P(N_{Z/X})} \mathbf P(N_{Z/X}\oplus\mathcal O_Z). }$$

Yes—but not in general. There are two different operations:

1. Removing the strict transform of $X$, which always produces the nonproper family with central fiber $N_{Z/X}$.
2. Contracting the strict transform inside a proper family, which requires additional positivity. The cleanest case is when $Z$ is a smooth ample divisor.

Let

$$\mathcal Y=\operatorname{Bl}_{Z\times\{0\}}(X\times \mathbb A^1) \longrightarrow \mathbb A^1.$$

Write

$$\widetilde X=\operatorname{Bl}_Z X$$

for the strict transform of $X\times\{0\}$, and

$$F=\mathbf P_Z\!\left(N_{Z/X}\oplus\mathcal O_Z\right)$$

for the exceptional component. Then

$$\mathcal Y_0=\widetilde X\cup_E F, \qquad E=\mathbf P_Z(N_{Z/X}).$$

1. The normal bundle of $\widetilde X$

Because $\mathcal Y_0=\widetilde X+F$ is a principal divisor—the fiber over $0$—we have

$$\mathcal O_{\mathcal Y}(\widetilde X+F)\simeq \mathcal O_{\mathcal Y}.$$

Restricting to $\widetilde X$, and noting that

$$F|_{\widetilde X}=E,$$

gives

$$N_{\widetilde X/\mathcal Y} = \mathcal O_{\mathcal Y}(\widetilde X)|_{\widetilde X} \simeq \mathcal O_{\widetilde X}(-E).$$

Thus any contraction of $\widetilde X$ is governed by the positivity of $E$ on $\widetilde X$.

For a contraction of all of $\widetilde X$ to a point, one expects the dual normal bundle

$$N_{\widetilde X/\mathcal Y}^{\vee} \simeq \mathcal O_{\widetilde X}(E)$$

to be ample.

That condition usually fails when $\operatorname{codim}_X Z\geq 2$. Indeed, if

$$\ell\subset E$$

is a line in a fiber of

$$E=\mathbf P(N_{Z/X})\to Z,$$

then

$$\mathcal O_{\widetilde X}(E)|_\ell\simeq \mathcal O_{\mathbf P^1}(-1).$$

Hence

$$E\cdot \ell=-1,$$

so $\mathcal O_{\widetilde X}(E)$ is not ample. Therefore there is generally no contraction of $\widetilde X$ to a point.

There can be special higher-codimensional examples in which $\widetilde X$ admits another Mori contraction, but this depends on the geometry of $X$ and $Z$; it is not part of the general deformation-to-the-normal-cone construction.

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The important positive case: $Z$ is an ample divisor

Suppose now that

$$Z=D\subset X$$

is a smooth Cartier divisor. Then blowing up $X$ along $D$ does nothing, so

$$\widetilde X\simeq X.$$

The intersection divisor is simply

$$E=D,$$

and therefore

$$N_{\widetilde X/\mathcal Y} \simeq \mathcal O_X(-D).$$

If $D$ is ample, this is an anti-ample normal bundle. In this situation $\widetilde X$ can be contracted to a point.

The resulting construction is commonly called the deformation to the projective cone.

2. The contracting line bundle

Assume, for clarity, that

$$L=\mathcal O_X(D)$$

is very ample and $D$ is cut out by a section

$$s\in H^0(X,L).$$

Let

$$b:\mathcal Y\to X\times\mathbb A^1$$

be the blowup map, and let $F$ denote the exceptional divisor. Consider

$$\mathcal M = b^*p_X^*L\otimes \mathcal O_{\mathcal Y}(-F).$$

On the strict transform $\widetilde X\simeq X$,

$$\mathcal M|_{\widetilde X} \simeq L\otimes \mathcal O_X(-D) \simeq \mathcal O_X.$$

Thus every section of the relevant relative linear system is constant on $\widetilde X$, so the associated morphism contracts $\widetilde X$ to a point.

For $t\neq 0$, the exceptional divisor does not meet the fiber, so

$$\mathcal M|_{\mathcal Y_t}\simeq L.$$

Consequently, on every nonzero fiber the morphism is the embedding of $X$ defined by $L$. In particular, the general fiber remains isomorphic to $X$.

Using a sufficiently large tensor power if necessary, $\mathcal M$ defines a morphism

$$c:\mathcal Y\longrightarrow \overline{\mathcal Y}$$

over $\mathbb A^1$ such that

$$c(\widetilde X)=\{v\}$$

is one point, and $c$ is an isomorphism away from $\widetilde X$, up to the chosen projective embedding.

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3. What happens to the other component?

In the divisor case,

$$F=\mathbf P_D\!\left(\mathcal O_D\oplus N_{D/X}\right).$$

The intersection

$$D=\widetilde X\cap F$$

is one of the two natural sections of this $\mathbf P^1$-bundle. Its normal bundle inside $F$ is

$$N_{D/F}\simeq N_{D/X}^{-1}.$$

Since $D$ is ample in $X$,

$$N_{D/X}=\mathcal O_D(D)$$

is ample, so this section has negative normal bundle. Under the contraction $c$, that section is contracted to the same point $v$.

Thus the central fiber after contraction is the image of $F$:

$$\overline{\mathcal Y}_0=c(F).$$

This is the projective cone over $D$, polarized by

$$N_{D/X}=\mathcal O_D(D).$$

Symbolically,

$$\boxed{ \overline{\mathcal Y}_t\simeq X \quad (t\neq 0), }$$

while

$$\boxed{ \overline{\mathcal Y}_0 \simeq \operatorname{Cone}\bigl(D,\mathcal O_D(D)\bigr). }$$

The central fiber is now irreducible, but it is generally singular at the cone vertex.

Schematically, the operation is

$$X\cup_D \mathbf P_D(\mathcal O_D\oplus N_{D/X}) \quad\longrightarrow\quad \operatorname{Cone}(D,N_{D/X}),$$

where the entire $X$-component and the attaching section $D$ are collapsed to the vertex.

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4. Why the divisor case is special

For a divisor $D$,

$$\widetilde X=X$$

and

$$N_{\widetilde X/\mathcal Y}=\mathcal O_X(-D).$$

Thus positivity of $D$ gives negativity of the normal bundle, precisely the geometry needed for contraction.

For a center of codimension at least two,

$$\widetilde X=\operatorname{Bl}_Z X$$

and

$$N_{\widetilde X/\mathcal Y} = \mathcal O_{\widetilde X}(-E).$$

But $E$ has negative degree on the fibers of $E\to Z$, so $E$ is not ample. The simple contraction-to-a-point mechanism therefore fails.

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5. Three possibilities summarized

Always possible, but nonproper

Delete the strict transform:

$$\mathcal Y\setminus\widetilde X\to\mathbb A^1.$$

Then

$$(\mathcal Y\setminus\widetilde X)_t\simeq X \quad(t\neq0),$$

and

$$(\mathcal Y\setminus\widetilde X)_0\simeq N_{Z/X}.$$

The central fiber is irreducible, but the family is generally nonproper.

Proper and reducible

Keep the whole blowup:

$$\mathcal Y_0 = \operatorname{Bl}_Z X \cup \mathbf P(N_{Z/X}\oplus\mathcal O_Z).$$

This works for every smooth center $Z$.

Proper and irreducible after contraction

This requires extra geometry. The standard situation is:

$$Z=D\subset X$$

a smooth ample divisor. Then one contracts the $X$-component and obtains

$$X\rightsquigarrow \operatorname{Cone}\bigl(D,\mathcal O_D(D)\bigr).$$

So the answer is:

$$\boxed{\text{Yes in important special cases, especially for a smooth ample divisor; not for a general center.}}$$