Differential 1-forms on an irreducible projective variety
Let $k$ be an algebraically closed field and let $X$ be an irreducible projective variety over $k$. I am wondering what the module of differential 1-forms on $X$ is.
Since $X$ is a projective variety, its ring of regular functions $mathcal{O}_X (X)$ is precisely the ring of locally constant functions on X. Now since $X$ is irreducible, it is connected, hence $mathcal{O}_X (X)$ is the ring of constant functions on $X$. By identifying a function with its image we find $mathcal{O}_X (X) cong k$, thus $mathcal{O}_X (X)$ is a $1$-dimensional $k$-vector space with as a basis the constant function $1$.
Now the $mathcal{O}_X (X)$-module $Omega^1(X)$ of differential 1-forms on $X$ and the Kahler differential $dcolon mathcal{O}_X (X) to Omega^1(X)$ are defined as follows. $Omega^1(X)$ and $d$ are such that for any $mathcal{O}_X (X)$-module $M$ and derivation, i.e. a $k$-linear map satisfying the Leibniz rule, $Dcolon mathcal{O}_X (X) to M$ there is a unique $k$-linear map $varphicolon Omega^1(X) to M$ such that $D=varphi circ d$.
Firstly, since $D$ and $d$ are derivations and $mathcal{O}_X (X)$ is a $k$-vector space with basis $1$, we have $D(mathcal{O}_X (X))=d(mathcal{O}_X (X))={0}$. This seems to imply that there is no restriction on what $Omega^1(X)$ is other than a $k$-vector space, since I can always take $varphi$ to be the zero map.
Now on the one hand, in the construction of $Omega^1$ for a general $k$-algebra $A$, thus replacing $mathcal{O}_X (X)$ with $A$ in the previous definition, it follows that all elements are of the form $adb$ with $a,bin A$, implying that $Omega^1(X)=0$.
On the other hand, I am trying to do an exercise in which I calculate $Omega^1(X)$ for a smooth irreducible projective curve $X$ from a presentation of $X$ by glueing the local $Omega^1(U_i )$ to the global set. However I found that the local $Omega^1(U_i )$ are 1-dimensional and they glue to a 1-dimensional vector space.
In conlusion I have three different ways in which I can construct $Omega^1(X)$ and they all seem to imply that $Omega^1(X)$ is something else. Which method is correct?
algebraic-geometry differential-forms projective-schemes projective-varieties
asked Nov 26 '18 at 19:48
Grimp0w
414
add a comment |
Let $k$ be an algebraically closed field and let $X$ be an irreducible projective variety over $k$. I am wondering what the module of differential 1-forms on $X$ is.
Since $X$ is a projective variety, its ring of regular functions $mathcal{O}_X (X)$ is precisely the ring of locally constant functions on X. Now since $X$ is irreducible, it is connected, hence $mathcal{O}_X (X)$ is the ring of constant functions on $X$. By identifying a function with its image we find $mathcal{O}_X (X) cong k$, thus $mathcal{O}_X (X)$ is a $1$-dimensional $k$-vector space with as a basis the constant function $1$.
Now the $mathcal{O}_X (X)$-module $Omega^1(X)$ of differential 1-forms on $X$ and the Kahler differential $dcolon mathcal{O}_X (X) to Omega^1(X)$ are defined as follows. $Omega^1(X)$ and $d$ are such that for any $mathcal{O}_X (X)$-module $M$ and derivation, i.e. a $k$-linear map satisfying the Leibniz rule, $Dcolon mathcal{O}_X (X) to M$ there is a unique $k$-linear map $varphicolon Omega^1(X) to M$ such that $D=varphi circ d$.
Firstly, since $D$ and $d$ are derivations and $mathcal{O}_X (X)$ is a $k$-vector space with basis $1$, we have $D(mathcal{O}_X (X))=d(mathcal{O}_X (X))={0}$. This seems to imply that there is no restriction on what $Omega^1(X)$ is other than a $k$-vector space, since I can always take $varphi$ to be the zero map.
Now on the one hand, in the construction of $Omega^1$ for a general $k$-algebra $A$, thus replacing $mathcal{O}_X (X)$ with $A$ in the previous definition, it follows that all elements are of the form $adb$ with $a,bin A$, implying that $Omega^1(X)=0$.
On the other hand, I am trying to do an exercise in which I calculate $Omega^1(X)$ for a smooth irreducible projective curve $X$ from a presentation of $X$ by glueing the local $Omega^1(U_i )$ to the global set. However I found that the local $Omega^1(U_i )$ are 1-dimensional and they glue to a 1-dimensional vector space.
In conlusion I have three different ways in which I can construct $Omega^1(X)$ and they all seem to imply that $Omega^1(X)$ is something else. Which method is correct?
algebraic-geometry differential-forms projective-schemes projective-varieties
asked Nov 26 '18 at 19:48
Grimp0w
414
Note that global sections of a sheaf, in particular $Omega^1(X)$ can be zero but still the sheaf might be non-trivial. For example $mathcal{O}(-1)$.
– random123
Nov 27 '18 at 5:08
Your definition of $Omega^1(X)$ is incorrect. The definition you write is correct for affine schemes only.
– Asal Beag Dubh
Nov 27 '18 at 10:36
add a comment |
Let $k$ be an algebraically closed field and let $X$ be an irreducible projective variety over $k$. I am wondering what the module of differential 1-forms on $X$ is.
Since $X$ is a projective variety, its ring of regular functions $mathcal{O}_X (X)$ is precisely the ring of locally constant functions on X. Now since $X$ is irreducible, it is connected, hence $mathcal{O}_X (X)$ is the ring of constant functions on $X$. By identifying a function with its image we find $mathcal{O}_X (X) cong k$, thus $mathcal{O}_X (X)$ is a $1$-dimensional $k$-vector space with as a basis the constant function $1$.
Now the $mathcal{O}_X (X)$-module $Omega^1(X)$ of differential 1-forms on $X$ and the Kahler differential $dcolon mathcal{O}_X (X) to Omega^1(X)$ are defined as follows. $Omega^1(X)$ and $d$ are such that for any $mathcal{O}_X (X)$-module $M$ and derivation, i.e. a $k$-linear map satisfying the Leibniz rule, $Dcolon mathcal{O}_X (X) to M$ there is a unique $k$-linear map $varphicolon Omega^1(X) to M$ such that $D=varphi circ d$.
Firstly, since $D$ and $d$ are derivations and $mathcal{O}_X (X)$ is a $k$-vector space with basis $1$, we have $D(mathcal{O}_X (X))=d(mathcal{O}_X (X))={0}$. This seems to imply that there is no restriction on what $Omega^1(X)$ is other than a $k$-vector space, since I can always take $varphi$ to be the zero map.
Now on the one hand, in the construction of $Omega^1$ for a general $k$-algebra $A$, thus replacing $mathcal{O}_X (X)$ with $A$ in the previous definition, it follows that all elements are of the form $adb$ with $a,bin A$, implying that $Omega^1(X)=0$.
On the other hand, I am trying to do an exercise in which I calculate $Omega^1(X)$ for a smooth irreducible projective curve $X$ from a presentation of $X$ by glueing the local $Omega^1(U_i )$ to the global set. However I found that the local $Omega^1(U_i )$ are 1-dimensional and they glue to a 1-dimensional vector space.
In conlusion I have three different ways in which I can construct $Omega^1(X)$ and they all seem to imply that $Omega^1(X)$ is something else. Which method is correct?
algebraic-geometry differential-forms projective-schemes projective-varieties
asked Nov 26 '18 at 19:48
Grimp0w
414
Let $k$ be an algebraically closed field and let $X$ be an irreducible projective variety over $k$. I am wondering what the module of differential 1-forms on $X$ is.
Since $X$ is a projective variety, its ring of regular functions $mathcal{O}_X (X)$ is precisely the ring of locally constant functions on X. Now since $X$ is irreducible, it is connected, hence $mathcal{O}_X (X)$ is the ring of constant functions on $X$. By identifying a function with its image we find $mathcal{O}_X (X) cong k$, thus $mathcal{O}_X (X)$ is a $1$-dimensional $k$-vector space with as a basis the constant function $1$.
Now the $mathcal{O}_X (X)$-module $Omega^1(X)$ of differential 1-forms on $X$ and the Kahler differential $dcolon mathcal{O}_X (X) to Omega^1(X)$ are defined as follows. $Omega^1(X)$ and $d$ are such that for any $mathcal{O}_X (X)$-module $M$ and derivation, i.e. a $k$-linear map satisfying the Leibniz rule, $Dcolon mathcal{O}_X (X) to M$ there is a unique $k$-linear map $varphicolon Omega^1(X) to M$ such that $D=varphi circ d$.
Firstly, since $D$ and $d$ are derivations and $mathcal{O}_X (X)$ is a $k$-vector space with basis $1$, we have $D(mathcal{O}_X (X))=d(mathcal{O}_X (X))={0}$. This seems to imply that there is no restriction on what $Omega^1(X)$ is other than a $k$-vector space, since I can always take $varphi$ to be the zero map.
Now on the one hand, in the construction of $Omega^1$ for a general $k$-algebra $A$, thus replacing $mathcal{O}_X (X)$ with $A$ in the previous definition, it follows that all elements are of the form $adb$ with $a,bin A$, implying that $Omega^1(X)=0$.
On the other hand, I am trying to do an exercise in which I calculate $Omega^1(X)$ for a smooth irreducible projective curve $X$ from a presentation of $X$ by glueing the local $Omega^1(U_i )$ to the global set. However I found that the local $Omega^1(U_i )$ are 1-dimensional and they glue to a 1-dimensional vector space.
In conlusion I have three different ways in which I can construct $Omega^1(X)$ and they all seem to imply that $Omega^1(X)$ is something else. Which method is correct?
algebraic-geometry differential-forms projective-schemes projective-varieties
algebraic-geometry differential-forms projective-schemes projective-varieties
asked Nov 26 '18 at 19:48
Grimp0w
414
asked Nov 26 '18 at 19:48
Grimp0w
414
asked Nov 26 '18 at 19:48
Grimp0w
414
asked Nov 26 '18 at 19:48
Grimp0w
414
asked Nov 26 '18 at 19:48
Grimp0w
414
414
Note that global sections of a sheaf, in particular $Omega^1(X)$ can be zero but still the sheaf might be non-trivial. For example $mathcal{O}(-1)$.
– random123
Nov 27 '18 at 5:08
Your definition of $Omega^1(X)$ is incorrect. The definition you write is correct for affine schemes only.
– Asal Beag Dubh
Nov 27 '18 at 10:36
add a comment |
Note that global sections of a sheaf, in particular $Omega^1(X)$ can be zero but still the sheaf might be non-trivial. For example $mathcal{O}(-1)$.
– random123
Nov 27 '18 at 5:08
Your definition of $Omega^1(X)$ is incorrect. The definition you write is correct for affine schemes only.
– Asal Beag Dubh
Nov 27 '18 at 10:36
Note that global sections of a sheaf, in particular $Omega^1(X)$ can be zero but still the sheaf might be non-trivial. For example $mathcal{O}(-1)$.
– random123
Nov 27 '18 at 5:08
Note that global sections of a sheaf, in particular $Omega^1(X)$ can be zero but still the sheaf might be non-trivial. For example $mathcal{O}(-1)$.
– random123
Nov 27 '18 at 5:08
Your definition of $Omega^1(X)$ is incorrect. The definition you write is correct for affine schemes only.
– Asal Beag Dubh
Nov 27 '18 at 10:36
Your definition of $Omega^1(X)$ is incorrect. The definition you write is correct for affine schemes only.
– Asal Beag Dubh
Nov 27 '18 at 10:36
add a comment |
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Note that global sections of a sheaf, in particular $Omega^1(X)$ can be zero but still the sheaf might be non-trivial. For example $mathcal{O}(-1)$.
– random123
Nov 27 '18 at 5:08
Your definition of $Omega^1(X)$ is incorrect. The definition you write is correct for affine schemes only.
– Asal Beag Dubh
Nov 27 '18 at 10:36