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Using the laguage of derived category, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

can be described as a Fourier-Mukai transformation $\Phi_{\mathcal F}$.

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

Here, $x \in X$ is a point of a smooth projective variety over an algebraically closed field.

$\mathcal F$ is a coherent sheaf on $X \times X$ which is flat over the first factor, and $\mathcal F_x$ means a $\mathcal {i}^*_{x\times X} \mathcal F$.

(which is equal to $\Phi_{\mathcal F} (k(x))$)

Of course, KS map cannot beis not defined as a differential of some morphism, but please consider the following argument ;

Suppose $\mathcal F_x$ is concentrated at $x$ for all $x \in X$. It means the map $f:x \mapsto \mathcal F_x$ is injective. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Here, the definition of $f$ is a nonsense. It is not even clear where the image of $f$ lives. But the argument is quite persuasive. And it is even geometrically intuitive, because it describes a KS map as a differential of some "function".

I think this must be a shadow of rigorous mathematical contents(probably a deformation theory), but I failed to make it so. Could someone explain to me what's going on? Thanks in advance.

Using the laguage of derived category, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

can be described as a Fourier-Mukai transformation $\Phi_{\mathcal F}$.

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

Here, $x \in X$ is a point of a smooth projective variety over an algebraically closed field.

$\mathcal F$ is a coherent sheaf on $X \times X$ which is flat over the first factor, and $\mathcal F_x$ means a $\mathcal {i}^*_{x\times X} \mathcal F$.

(which is equal to $\Phi_{\mathcal F} (k(x))$)

Of course, KS map cannot be a differential of some morphism, but please consider the following argument ;

Suppose $\mathcal F_x$ is concentrated at $x$ for all $x \in X$. It means the map $f:x \mapsto \mathcal F_x$ is injective. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Here, the definition of $f$ is a nonsense. It is not even clear where the image of $f$ lives. But the argument is quite persuasive. And it is even geometrically intuitive, because it describes a KS map as a differential of some "function".

I think this must be a shadow of rigorous mathematical contents(probably a deformation theory), but I failed to make it so. Could someone explain to me what's going on? Thanks in advance.

Using the laguage of derived category, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

can be described as a Fourier-Mukai transformation $\Phi_{\mathcal F}$.

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

Here, $x \in X$ is a point of a smooth projective variety over an algebraically closed field.

$\mathcal F$ is a coherent sheaf on $X \times X$ which is flat over the first factor, and $\mathcal F_x$ means a $\mathcal {i}^*_{x\times X} \mathcal F$.

(which is equal to $\Phi_{\mathcal F} (k(x))$)

Of course, KS map is not defined as a differential of some morphism, but please consider the following argument ;

Suppose $\mathcal F_x$ is concentrated at $x$ for all $x \in X$. It means the map $f:x \mapsto \mathcal F_x$ is injective. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Here, the definition of $f$ is a nonsense. It is not even clear where the image of $f$ lives. But the argument is quite persuasive. And it is even geometrically intuitive, because it describes a KS map as a differential of some "function".

I think this must be a shadow of rigorous mathematical contents(probably a deformation theory), but I failed to make it so. Could someone explain to me what's going on? Thanks in advance.

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Intuitive argument for the Kodaira-Spencer map as a "differential"

I met this problem inUsing the D.Huybrechts' booklaguage of derived category, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

can be described as a Fourier-Mukai transformation Fourier-Mukai transform in algebraic geometry$\Phi_{\mathcal F}$.

Let$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

Here, $X$ be$x \in X$ is a point of a smooth projective variety over an algebraically closed field.

And let $\mathcal F$ beis a coherent sheaf on $X \times X$ which is flat over the first factor.

Futhermore, assumeand $\mathcal F_x$ is concentrated in $x$ for every $x$.

Since $\mathcal F$ is flat over $X$, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

is nothing butmeans a Fourier-Mukai transformation $\Phi_{\mathcal F}$

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $$\mathcal {i}^*_{x\times X} \mathcal F$.

( For notational convenience, I wrote $\mathcal F_x := \Phi_{\mathcal F}(k(x)) =$ $\mathcal {Li}^*_{x\times X} \mathcal F$which is equal to $\Phi_{\mathcal F} (k(x))$)

At this pointOf course, author proved $\kappa(x)$ is injective for the generic point $x \in X$ byKS map cannot be a differential of some morphism, but please consider the non-rigourous shortcut;following argument ;

TheSuppose $\mathcal F_x$ is concentrated at $x$ for all $x \in X$. It means the map $f:x \mapsto \mathcal F_x$ is injective by assumptions. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Even though itHere, the definition of $f$ is a nonsense. It is not even clear where the image of $f$ lives,. But the argument is quite persuasive and. And it is even geometrically intuitivegeometrically intuitive, because it describes a KS map as a differential of some "function". But

I think this must be a shadow of rigorous mathematical contents(probably a deformation theory), but I failed to make it rigourousso. Could someone explain to me what's going on? Thanks in advance.

Intuitive argument for the Kodaira-Spencer map

I met this problem in the D.Huybrechts' book Fourier-Mukai transform in algebraic geometry

Let $X$ be a smooth projective variety over an algebraically closed field.

And let $\mathcal F$ be a coherent sheaf on $X \times X$ which is flat over the first factor.

Futhermore, assume $\mathcal F_x$ is concentrated in $x$ for every $x$.

Since $\mathcal F$ is flat over $X$, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

is nothing but a Fourier-Mukai transformation $\Phi_{\mathcal F}$

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

( For notational convenience, I wrote $\mathcal F_x := \Phi_{\mathcal F}(k(x)) =$ $\mathcal {Li}^*_{x\times X} \mathcal F$ )

At this point, author proved $\kappa(x)$ is injective for the generic point $x \in X$ by the non-rigourous shortcut;

The map $f:x \mapsto \mathcal F_x$ is injective by assumptions. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Even though it is not even clear where the image of $f$ lives, the argument is quite persuasive and even geometrically intuitive. But I failed to make it rigourous. Could someone explain to me what's going on? Thanks in advance

Kodaira-Spencer map as a "differential"

Using the laguage of derived category, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

can be described as a Fourier-Mukai transformation $\Phi_{\mathcal F}$.

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

Here, $x \in X$ is a point of a smooth projective variety over an algebraically closed field.

$\mathcal F$ is a coherent sheaf on $X \times X$ which is flat over the first factor, and $\mathcal F_x$ means a $\mathcal {i}^*_{x\times X} \mathcal F$.

(which is equal to $\Phi_{\mathcal F} (k(x))$)

Of course, KS map cannot be a differential of some morphism, but please consider the following argument ;

Suppose $\mathcal F_x$ is concentrated at $x$ for all $x \in X$. It means the map $f:x \mapsto \mathcal F_x$ is injective. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Here, the definition of $f$ is a nonsense. It is not even clear where the image of $f$ lives. But the argument is quite persuasive. And it is even geometrically intuitive, because it describes a KS map as a differential of some "function".

I think this must be a shadow of rigorous mathematical contents(probably a deformation theory), but I failed to make it so. Could someone explain to me what's going on? Thanks in advance.

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I met this problem in the D.Huybrechts' book Fourier-Mukai transform in algebraic geometry

Let $X$ be a smooth projective variety over an algebraically closed field.

And let $\mathcal F$ be a coherent sheaf on $X \times X$ which is flat over the first factor.

Futhermore, assume $\mathcal F_x$ is concentrated in $x$ for every $x$.

Since $\mathcal F$ is flat over $X$, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

is nothing but a Fourier-Mukai transformation $\Phi_{\mathcal F}$

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

( For notational convenience, writeI wrote $\mathcal F_x := \Phi_{\mathcal F}(k(x)) =$ $\mathcal {Li}^*_{x\times X} \mathcal F$ )

At this point, author used the following shortcut to prove thatproved $\kappa(x)$ is injective atfor the generic point of $X$;$x \in X$ by the non-rigourous shortcut;

The map $f:x \mapsto \mathcal F_x$ is injective by assumptions. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Even though it is not even clear where the image of $f$ lives, the argument is quite persuasive and even geometrically intuitive. But I failed to make it rigourous. Could someone explain to me what's going on? Thanks in advance

I met this problem in the D.Huybrechts' book Fourier-Mukai transform in algebraic geometry

Let $X$ be a smooth projective variety over an algebraically closed field.

And let $\mathcal F$ be a coherent sheaf on $X \times X$ which is flat over the first factor.

Futhermore, assume $\mathcal F_x$ is concentrated in $x$ for every $x$.

Since $\mathcal F$ is flat over $X$, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

is nothing but a Fourier-Mukai transformation $\Phi_{\mathcal F}$

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

( For notational convenience, write $\mathcal F_x := \Phi_{\mathcal F}(k(x)) =$ $\mathcal {Li}^*_{x\times X} \mathcal F$ )

At this point, author used the following shortcut to prove that $\kappa(x)$ is injective at generic point of $X$;

The map $f:x \mapsto \mathcal F_x$ is injective by assumptions. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Even though it is not even clear where the image of $f$ lives, the argument is quite persuasive and even geometrically intuitive. But I failed to make it rigourous. Could someone explain to me what's going on? Thanks in advance

I met this problem in the D.Huybrechts' book Fourier-Mukai transform in algebraic geometry

Let $X$ be a smooth projective variety over an algebraically closed field.

And let $\mathcal F$ be a coherent sheaf on $X \times X$ which is flat over the first factor.

Futhermore, assume $\mathcal F_x$ is concentrated in $x$ for every $x$.

Since $\mathcal F$ is flat over $X$, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

is nothing but a Fourier-Mukai transformation $\Phi_{\mathcal F}$

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

( For notational convenience, I wrote $\mathcal F_x := \Phi_{\mathcal F}(k(x)) =$ $\mathcal {Li}^*_{x\times X} \mathcal F$ )

At this point, author proved $\kappa(x)$ is injective for the generic point $x \in X$ by the non-rigourous shortcut;

The map $f:x \mapsto \mathcal F_x$ is injective by assumptions. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Even though it is not even clear where the image of $f$ lives, the argument is quite persuasive and even geometrically intuitive. But I failed to make it rigourous. Could someone explain to me what's going on? Thanks in advance

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