Let S be a smooth cubic surface in P^3 and P a poi...

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.