On non-representability of certain hom schemes
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Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(mathbbA^1_k,mathbbA^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(mathbbA^1,mathbbA^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $mathbbA^1_Sto mathbbA^1_S$ of $S$-schemes.)
Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise.
Let X be a (positive-dimensional) variety and let $f:mathbbA^1_kto X$ be a finite morphism. How does one show that the hom-functor $mathrmHom_k(mathbbA^1_k,X)$ is not representable by an algebraic space?
What have I tried? Well, there is a natural morphism Isom$(mathbbA^1,mathbbA^1) to mathrmHom(mathbbA^1_k,X)$ which sends $g$ to $fcirc g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise?
ag.algebraic-geometry complex-geometry schemes hilbert-schemes
asked Sep 16 at 20:00
Anne F.
563
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up vote
11
down vote
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Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(mathbbA^1_k,mathbbA^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(mathbbA^1,mathbbA^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $mathbbA^1_Sto mathbbA^1_S$ of $S$-schemes.)
Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise.
Let X be a (positive-dimensional) variety and let $f:mathbbA^1_kto X$ be a finite morphism. How does one show that the hom-functor $mathrmHom_k(mathbbA^1_k,X)$ is not representable by an algebraic space?
What have I tried? Well, there is a natural morphism Isom$(mathbbA^1,mathbbA^1) to mathrmHom(mathbbA^1_k,X)$ which sends $g$ to $fcirc g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise?
ag.algebraic-geometry complex-geometry schemes hilbert-schemes
asked Sep 16 at 20:00
Anne F.
563
Probably stupid question: Isn't $textIsom(mathbbA^1_k,mathbbA^1_k)$ just $mathbbA^1_krtimesmathbbG_m,k$? That looks representable (by a scheme!) to me...?
– Will Chen
Sep 16 at 20:18
3
@WillChen What you write down is true on $k$-points. It is not however true that the scheme you write down represents the isom-functor. A proof can be found for instance after Claim 3.1 in arxiv.org/pdf/math/0602646.pdf
– Anne F.
Sep 16 at 20:21
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(mathbbA^1_k,mathbbA^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(mathbbA^1,mathbbA^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $mathbbA^1_Sto mathbbA^1_S$ of $S$-schemes.)
Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise.
Let X be a (positive-dimensional) variety and let $f:mathbbA^1_kto X$ be a finite morphism. How does one show that the hom-functor $mathrmHom_k(mathbbA^1_k,X)$ is not representable by an algebraic space?
What have I tried? Well, there is a natural morphism Isom$(mathbbA^1,mathbbA^1) to mathrmHom(mathbbA^1_k,X)$ which sends $g$ to $fcirc g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise?
ag.algebraic-geometry complex-geometry schemes hilbert-schemes
asked Sep 16 at 20:00
Anne F.
563
Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(mathbbA^1_k,mathbbA^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(mathbbA^1,mathbbA^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $mathbbA^1_Sto mathbbA^1_S$ of $S$-schemes.)
Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise.
Let X be a (positive-dimensional) variety and let $f:mathbbA^1_kto X$ be a finite morphism. How does one show that the hom-functor $mathrmHom_k(mathbbA^1_k,X)$ is not representable by an algebraic space?
What have I tried? Well, there is a natural morphism Isom$(mathbbA^1,mathbbA^1) to mathrmHom(mathbbA^1_k,X)$ which sends $g$ to $fcirc g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise?
ag.algebraic-geometry complex-geometry schemes hilbert-schemes
ag.algebraic-geometry complex-geometry schemes hilbert-schemes
asked Sep 16 at 20:00
Anne F.
563
asked Sep 16 at 20:00
Anne F.
563
asked Sep 16 at 20:00
Anne F.
563
asked Sep 16 at 20:00
Anne F.
563
asked Sep 16 at 20:00
Anne F.
563
563
Probably stupid question: Isn't $textIsom(mathbbA^1_k,mathbbA^1_k)$ just $mathbbA^1_krtimesmathbbG_m,k$? That looks representable (by a scheme!) to me...?
– Will Chen
Sep 16 at 20:18
3
@WillChen What you write down is true on $k$-points. It is not however true that the scheme you write down represents the isom-functor. A proof can be found for instance after Claim 3.1 in arxiv.org/pdf/math/0602646.pdf
– Anne F.
Sep 16 at 20:21
add a comment |Â
Probably stupid question: Isn't $textIsom(mathbbA^1_k,mathbbA^1_k)$ just $mathbbA^1_krtimesmathbbG_m,k$? That looks representable (by a scheme!) to me...?
– Will Chen
Sep 16 at 20:18
3
@WillChen What you write down is true on $k$-points. It is not however true that the scheme you write down represents the isom-functor. A proof can be found for instance after Claim 3.1 in arxiv.org/pdf/math/0602646.pdf
– Anne F.
Sep 16 at 20:21
Probably stupid question: Isn't $textIsom(mathbbA^1_k,mathbbA^1_k)$ just $mathbbA^1_krtimesmathbbG_m,k$? That looks representable (by a scheme!) to me...?
– Will Chen
Sep 16 at 20:18
Probably stupid question: Isn't $textIsom(mathbbA^1_k,mathbbA^1_k)$ just $mathbbA^1_krtimesmathbbG_m,k$? That looks representable (by a scheme!) to me...?
– Will Chen
Sep 16 at 20:18
3
3
@WillChen What you write down is true on $k$-points. It is not however true that the scheme you write down represents the isom-functor. A proof can be found for instance after Claim 3.1 in arxiv.org/pdf/math/0602646.pdf
– Anne F.
Sep 16 at 20:21
@WillChen What you write down is true on $k$-points. It is not however true that the scheme you write down represents the isom-functor. A proof can be found for instance after Claim 3.1 in arxiv.org/pdf/math/0602646.pdf
– Anne F.
Sep 16 at 20:21
add a comment |Â
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Welcome, new contributor. Let $k$ be a field. Let $Y$ be a finite type, separated $k$-scheme such that the $k$-algebra $mathcalO_Y(Y)$ is a $k$-vector space of infinite dimension. For instance, this holds for every finite type, affine $k$-scheme of positive dimension. Let $X$ be a finite type, separated $k$-scheme that is not everywhere finite, étale over $textSpec(k)$. The Hom functor $textHom_k-textSch(Y,X)$ is limit preserving because the Yoneda functor of $X$ is limit preserving. Thus, if the Hom functor is representable, it is representable by a $k$-scheme that is locally finite type. The claim is that the Hom functor is not representable by a $k$-scheme that is locally finite type.
If the Hom functor is representable by a locally finite type $k$-scheme, then after base change to the algebraic closure of $k$, also the Hom scheme of the base change is representable by a scheme that is (ed. locally) finite type over the algebraic closure. Thus, without loss of generality, assume that $k$ is algebraically closed. Since $X$ is not everywhere étale, there exists a $k$-point $x$ of $X$ at which the $k$-vector space $Omega_X/k,xotimes_mathcalO_X,x k = mathfrakm_X,x/mathfrakm_X,x^2$ has positive dimension as a $k$-vector space.
Denote by $$c_x:Yto X$$ the constant $k$-morphism with image $x$. Then the Zariski tangent space to the Hom functor at $c_x$ equals $$textHom_mathcalO_Y(c_x^*Omega_X/k,mathcalO_Y) = mathcalO_Y otimes_k left(mathfrakm_X,x/mathfrakm_X,x^2right).$$ This is a $k$-vector space of infinite dimension. If the Hom scheme were representable by a locally finite type $k$-scheme, then the Zariski tangent space at $c_x$ would be a finite-dimension $k$-vector space. Thus, the Hom scheme is not representable.
4
Thank you for this clear answer. I did not realise that the argument which proves Isom($mathbbA^1,mathbbA^1)$ is not representable actually adapts to this more general setting. Very nice. (Small typo in second paragraph: "is finite type over the algebraic closure" should be "is locally of finite type over the algebraic closure".)
– Anne F.
Sep 16 at 23:36
2
@AnneF. Thank you for catching the typo.
– Jason Starr
Sep 17 at 1:19
add a comment |Â
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Welcome, new contributor. Let $k$ be a field. Let $Y$ be a finite type, separated $k$-scheme such that the $k$-algebra $mathcalO_Y(Y)$ is a $k$-vector space of infinite dimension. For instance, this holds for every finite type, affine $k$-scheme of positive dimension. Let $X$ be a finite type, separated $k$-scheme that is not everywhere finite, étale over $textSpec(k)$. The Hom functor $textHom_k-textSch(Y,X)$ is limit preserving because the Yoneda functor of $X$ is limit preserving. Thus, if the Hom functor is representable, it is representable by a $k$-scheme that is locally finite type. The claim is that the Hom functor is not representable by a $k$-scheme that is locally finite type.
If the Hom functor is representable by a locally finite type $k$-scheme, then after base change to the algebraic closure of $k$, also the Hom scheme of the base change is representable by a scheme that is (ed. locally) finite type over the algebraic closure. Thus, without loss of generality, assume that $k$ is algebraically closed. Since $X$ is not everywhere étale, there exists a $k$-point $x$ of $X$ at which the $k$-vector space $Omega_X/k,xotimes_mathcalO_X,x k = mathfrakm_X,x/mathfrakm_X,x^2$ has positive dimension as a $k$-vector space.
Denote by $$c_x:Yto X$$ the constant $k$-morphism with image $x$. Then the Zariski tangent space to the Hom functor at $c_x$ equals $$textHom_mathcalO_Y(c_x^*Omega_X/k,mathcalO_Y) = mathcalO_Y otimes_k left(mathfrakm_X,x/mathfrakm_X,x^2right).$$ This is a $k$-vector space of infinite dimension. If the Hom scheme were representable by a locally finite type $k$-scheme, then the Zariski tangent space at $c_x$ would be a finite-dimension $k$-vector space. Thus, the Hom scheme is not representable.
4
Thank you for this clear answer. I did not realise that the argument which proves Isom($mathbbA^1,mathbbA^1)$ is not representable actually adapts to this more general setting. Very nice. (Small typo in second paragraph: "is finite type over the algebraic closure" should be "is locally of finite type over the algebraic closure".)
– Anne F.
Sep 16 at 23:36
2
@AnneF. Thank you for catching the typo.
– Jason Starr
Sep 17 at 1:19
add a comment |Â
up vote
8
down vote
accepted
Welcome, new contributor. Let $k$ be a field. Let $Y$ be a finite type, separated $k$-scheme such that the $k$-algebra $mathcalO_Y(Y)$ is a $k$-vector space of infinite dimension. For instance, this holds for every finite type, affine $k$-scheme of positive dimension. Let $X$ be a finite type, separated $k$-scheme that is not everywhere finite, étale over $textSpec(k)$. The Hom functor $textHom_k-textSch(Y,X)$ is limit preserving because the Yoneda functor of $X$ is limit preserving. Thus, if the Hom functor is representable, it is representable by a $k$-scheme that is locally finite type. The claim is that the Hom functor is not representable by a $k$-scheme that is locally finite type.
If the Hom functor is representable by a locally finite type $k$-scheme, then after base change to the algebraic closure of $k$, also the Hom scheme of the base change is representable by a scheme that is (ed. locally) finite type over the algebraic closure. Thus, without loss of generality, assume that $k$ is algebraically closed. Since $X$ is not everywhere étale, there exists a $k$-point $x$ of $X$ at which the $k$-vector space $Omega_X/k,xotimes_mathcalO_X,x k = mathfrakm_X,x/mathfrakm_X,x^2$ has positive dimension as a $k$-vector space.
Denote by $$c_x:Yto X$$ the constant $k$-morphism with image $x$. Then the Zariski tangent space to the Hom functor at $c_x$ equals $$textHom_mathcalO_Y(c_x^*Omega_X/k,mathcalO_Y) = mathcalO_Y otimes_k left(mathfrakm_X,x/mathfrakm_X,x^2right).$$ This is a $k$-vector space of infinite dimension. If the Hom scheme were representable by a locally finite type $k$-scheme, then the Zariski tangent space at $c_x$ would be a finite-dimension $k$-vector space. Thus, the Hom scheme is not representable.
4
Thank you for this clear answer. I did not realise that the argument which proves Isom($mathbbA^1,mathbbA^1)$ is not representable actually adapts to this more general setting. Very nice. (Small typo in second paragraph: "is finite type over the algebraic closure" should be "is locally of finite type over the algebraic closure".)
– Anne F.
Sep 16 at 23:36
2
@AnneF. Thank you for catching the typo.
– Jason Starr
Sep 17 at 1:19
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Welcome, new contributor. Let $k$ be a field. Let $Y$ be a finite type, separated $k$-scheme such that the $k$-algebra $mathcalO_Y(Y)$ is a $k$-vector space of infinite dimension. For instance, this holds for every finite type, affine $k$-scheme of positive dimension. Let $X$ be a finite type, separated $k$-scheme that is not everywhere finite, étale over $textSpec(k)$. The Hom functor $textHom_k-textSch(Y,X)$ is limit preserving because the Yoneda functor of $X$ is limit preserving. Thus, if the Hom functor is representable, it is representable by a $k$-scheme that is locally finite type. The claim is that the Hom functor is not representable by a $k$-scheme that is locally finite type.
If the Hom functor is representable by a locally finite type $k$-scheme, then after base change to the algebraic closure of $k$, also the Hom scheme of the base change is representable by a scheme that is (ed. locally) finite type over the algebraic closure. Thus, without loss of generality, assume that $k$ is algebraically closed. Since $X$ is not everywhere étale, there exists a $k$-point $x$ of $X$ at which the $k$-vector space $Omega_X/k,xotimes_mathcalO_X,x k = mathfrakm_X,x/mathfrakm_X,x^2$ has positive dimension as a $k$-vector space.
Denote by $$c_x:Yto X$$ the constant $k$-morphism with image $x$. Then the Zariski tangent space to the Hom functor at $c_x$ equals $$textHom_mathcalO_Y(c_x^*Omega_X/k,mathcalO_Y) = mathcalO_Y otimes_k left(mathfrakm_X,x/mathfrakm_X,x^2right).$$ This is a $k$-vector space of infinite dimension. If the Hom scheme were representable by a locally finite type $k$-scheme, then the Zariski tangent space at $c_x$ would be a finite-dimension $k$-vector space. Thus, the Hom scheme is not representable.
Welcome, new contributor. Let $k$ be a field. Let $Y$ be a finite type, separated $k$-scheme such that the $k$-algebra $mathcalO_Y(Y)$ is a $k$-vector space of infinite dimension. For instance, this holds for every finite type, affine $k$-scheme of positive dimension. Let $X$ be a finite type, separated $k$-scheme that is not everywhere finite, étale over $textSpec(k)$. The Hom functor $textHom_k-textSch(Y,X)$ is limit preserving because the Yoneda functor of $X$ is limit preserving. Thus, if the Hom functor is representable, it is representable by a $k$-scheme that is locally finite type. The claim is that the Hom functor is not representable by a $k$-scheme that is locally finite type.
If the Hom functor is representable by a locally finite type $k$-scheme, then after base change to the algebraic closure of $k$, also the Hom scheme of the base change is representable by a scheme that is (ed. locally) finite type over the algebraic closure. Thus, without loss of generality, assume that $k$ is algebraically closed. Since $X$ is not everywhere étale, there exists a $k$-point $x$ of $X$ at which the $k$-vector space $Omega_X/k,xotimes_mathcalO_X,x k = mathfrakm_X,x/mathfrakm_X,x^2$ has positive dimension as a $k$-vector space.
Denote by $$c_x:Yto X$$ the constant $k$-morphism with image $x$. Then the Zariski tangent space to the Hom functor at $c_x$ equals $$textHom_mathcalO_Y(c_x^*Omega_X/k,mathcalO_Y) = mathcalO_Y otimes_k left(mathfrakm_X,x/mathfrakm_X,x^2right).$$ This is a $k$-vector space of infinite dimension. If the Hom scheme were representable by a locally finite type $k$-scheme, then the Zariski tangent space at $c_x$ would be a finite-dimension $k$-vector space. Thus, the Hom scheme is not representable.
edited Sep 17 at 1:19
community wiki
2 revs
Jason Starr
4
Thank you for this clear answer. I did not realise that the argument which proves Isom($mathbbA^1,mathbbA^1)$ is not representable actually adapts to this more general setting. Very nice. (Small typo in second paragraph: "is finite type over the algebraic closure" should be "is locally of finite type over the algebraic closure".)
– Anne F.
Sep 16 at 23:36
2
@AnneF. Thank you for catching the typo.
– Jason Starr
Sep 17 at 1:19
add a comment |Â
4
Thank you for this clear answer. I did not realise that the argument which proves Isom($mathbbA^1,mathbbA^1)$ is not representable actually adapts to this more general setting. Very nice. (Small typo in second paragraph: "is finite type over the algebraic closure" should be "is locally of finite type over the algebraic closure".)
– Anne F.
Sep 16 at 23:36
2
@AnneF. Thank you for catching the typo.
– Jason Starr
Sep 17 at 1:19
4
4
Thank you for this clear answer. I did not realise that the argument which proves Isom($mathbbA^1,mathbbA^1)$ is not representable actually adapts to this more general setting. Very nice. (Small typo in second paragraph: "is finite type over the algebraic closure" should be "is locally of finite type over the algebraic closure".)
– Anne F.
Sep 16 at 23:36
Thank you for this clear answer. I did not realise that the argument which proves Isom($mathbbA^1,mathbbA^1)$ is not representable actually adapts to this more general setting. Very nice. (Small typo in second paragraph: "is finite type over the algebraic closure" should be "is locally of finite type over the algebraic closure".)
– Anne F.
Sep 16 at 23:36
2
2
@AnneF. Thank you for catching the typo.
– Jason Starr
Sep 17 at 1:19
@AnneF. Thank you for catching the typo.
– Jason Starr
Sep 17 at 1:19
add a comment |Â
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Probably stupid question: Isn't $textIsom(mathbbA^1_k,mathbbA^1_k)$ just $mathbbA^1_krtimesmathbbG_m,k$? That looks representable (by a scheme!) to me...?
– Will Chen
Sep 16 at 20:18
3
@WillChen What you write down is true on $k$-points. It is not however true that the scheme you write down represents the isom-functor. A proof can be found for instance after Claim 3.1 in arxiv.org/pdf/math/0602646.pdf
– Anne F.
Sep 16 at 20:21