Odd tangent bundle is the same as mapping stack from shifted affine line?
$begingroup$
Fix a field $k$ of characteristic $0$.
Let $X = operatorname{Spec} A$ be an affine derived scheme of finite type, i.e., $A$ is a cdga such that $H^0(A)$ is a finitely generated over $k$ and $H^{i}(A)$ is finitely generated over $H^0(A)$ for $i<0$.
Assume $A$ is quasi-free so the cotangent complex $mathbb{L}_A$ can be computed by the K"{a}her differential $Omega_A$ as in the classical case.
The odd tangent bundle $mathbb{T}X[-1]$ is defined to be the relative affine scheme $operatorname{Spec}_{A} (operatorname{Sym}^bullet Omega_A[1])$.
Denote $mathbb{A}[1]$ the derived stack defined by
$mathbb{A}[1](S) = Hom_{DGA_k}(k[eta],S)$ where $eta$ is of $deg 1$.
I'm looking for a proof of the isomorphism:
$$ Map(mathbb{A}[1],X) cong mathbb{T}X[-1]$$
where $Map$ denotes the mapping stack.
I know one can show that they have the same fiber over each $S$-point $x : operatorname{Spec}(S) rightarrow X$ by using the universal property of the cotangent complex. But I don't understand how one can get a global map in the first place? Reference or explanations are both welcome.
abstract-algebra geometry algebraic-geometry homological-algebra derived-categories
asked Dec 3 '18 at 5:12
Chris KuoChris Kuo
570210
$endgroup$
add a comment |
$begingroup$
Fix a field $k$ of characteristic $0$.
Let $X = operatorname{Spec} A$ be an affine derived scheme of finite type, i.e., $A$ is a cdga such that $H^0(A)$ is a finitely generated over $k$ and $H^{i}(A)$ is finitely generated over $H^0(A)$ for $i<0$.
Assume $A$ is quasi-free so the cotangent complex $mathbb{L}_A$ can be computed by the K"{a}her differential $Omega_A$ as in the classical case.
The odd tangent bundle $mathbb{T}X[-1]$ is defined to be the relative affine scheme $operatorname{Spec}_{A} (operatorname{Sym}^bullet Omega_A[1])$.
Denote $mathbb{A}[1]$ the derived stack defined by
$mathbb{A}[1](S) = Hom_{DGA_k}(k[eta],S)$ where $eta$ is of $deg 1$.
I'm looking for a proof of the isomorphism:
$$ Map(mathbb{A}[1],X) cong mathbb{T}X[-1]$$
where $Map$ denotes the mapping stack.
I know one can show that they have the same fiber over each $S$-point $x : operatorname{Spec}(S) rightarrow X$ by using the universal property of the cotangent complex. But I don't understand how one can get a global map in the first place? Reference or explanations are both welcome.
abstract-algebra geometry algebraic-geometry homological-algebra derived-categories
asked Dec 3 '18 at 5:12
Chris KuoChris Kuo
570210
$endgroup$
add a comment |
$begingroup$
Fix a field $k$ of characteristic $0$.
Let $X = operatorname{Spec} A$ be an affine derived scheme of finite type, i.e., $A$ is a cdga such that $H^0(A)$ is a finitely generated over $k$ and $H^{i}(A)$ is finitely generated over $H^0(A)$ for $i<0$.
Assume $A$ is quasi-free so the cotangent complex $mathbb{L}_A$ can be computed by the K"{a}her differential $Omega_A$ as in the classical case.
The odd tangent bundle $mathbb{T}X[-1]$ is defined to be the relative affine scheme $operatorname{Spec}_{A} (operatorname{Sym}^bullet Omega_A[1])$.
Denote $mathbb{A}[1]$ the derived stack defined by
$mathbb{A}[1](S) = Hom_{DGA_k}(k[eta],S)$ where $eta$ is of $deg 1$.
I'm looking for a proof of the isomorphism:
$$ Map(mathbb{A}[1],X) cong mathbb{T}X[-1]$$
where $Map$ denotes the mapping stack.
I know one can show that they have the same fiber over each $S$-point $x : operatorname{Spec}(S) rightarrow X$ by using the universal property of the cotangent complex. But I don't understand how one can get a global map in the first place? Reference or explanations are both welcome.
abstract-algebra geometry algebraic-geometry homological-algebra derived-categories
asked Dec 3 '18 at 5:12
Chris KuoChris Kuo
570210
$endgroup$
Fix a field $k$ of characteristic $0$.
Let $X = operatorname{Spec} A$ be an affine derived scheme of finite type, i.e., $A$ is a cdga such that $H^0(A)$ is a finitely generated over $k$ and $H^{i}(A)$ is finitely generated over $H^0(A)$ for $i<0$.
Assume $A$ is quasi-free so the cotangent complex $mathbb{L}_A$ can be computed by the K"{a}her differential $Omega_A$ as in the classical case.
The odd tangent bundle $mathbb{T}X[-1]$ is defined to be the relative affine scheme $operatorname{Spec}_{A} (operatorname{Sym}^bullet Omega_A[1])$.
Denote $mathbb{A}[1]$ the derived stack defined by
$mathbb{A}[1](S) = Hom_{DGA_k}(k[eta],S)$ where $eta$ is of $deg 1$.
I'm looking for a proof of the isomorphism:
$$ Map(mathbb{A}[1],X) cong mathbb{T}X[-1]$$
where $Map$ denotes the mapping stack.
I know one can show that they have the same fiber over each $S$-point $x : operatorname{Spec}(S) rightarrow X$ by using the universal property of the cotangent complex. But I don't understand how one can get a global map in the first place? Reference or explanations are both welcome.
abstract-algebra geometry algebraic-geometry homological-algebra derived-categories
abstract-algebra geometry algebraic-geometry homological-algebra derived-categories
asked Dec 3 '18 at 5:12
Chris KuoChris Kuo
570210
asked Dec 3 '18 at 5:12
Chris KuoChris Kuo
570210
asked Dec 3 '18 at 5:12
Chris KuoChris Kuo
570210
asked Dec 3 '18 at 5:12
Chris KuoChris Kuo
570210
asked Dec 3 '18 at 5:12
Chris KuoChris Kuo
570210
570210
add a comment |
add a comment |
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