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discussed topological space structure
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Peter McNamara
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Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The further commentary remark below applies to theaffine Grassmannian case of based loopsalso provides positive evidence.

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know.

Further commentary: Based onRegarding putting the commentclassical topology of Stephen GriffethX((t))(ℂ), if loops on G is going toone should not be homotopic toscared of the affine Grassmannianind-scheminess. ℂ((t)) has a natural structure of a topological ring, and the affine Grassmannian is a homogenous space forhence we topologise X(ℂ((t))) in the algebraic loop groupusual manner, withtaking the stabaliser ofsubspace topology using a point being topologically nonclosed embedding into affine n-trivial, then I guess the answer to the original question as I asked it is nospace for some n.

[paragraph redacted]

Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The further commentary remark below applies to the case of based loops.

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know.

Further commentary: Based on the comment of Stephen Griffeth, if loops on G is going to be homotopic to the affine Grassmannian, and the affine Grassmannian is a homogenous space for the algebraic loop group, with the stabaliser of a point being topologically non-trivial, then I guess the answer to the original question as I asked it is no.

Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The affine Grassmannian case also provides positive evidence.

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. Regarding putting the classical topology of X((t))(ℂ), one should not be scared of the ind-scheminess. ℂ((t)) has a natural structure of a topological ring, and hence we topologise X(ℂ((t))) in the usual manner, taking the subspace topology using a closed embedding into affine n-space for some n.

[paragraph redacted]

added 280 characters in body
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Peter McNamara
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Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The further commentary remark below applies to the case of based loops.

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know.

Further commentary: Based on the comment of Stephen Griffeth, if loops on G is going to be homotopic to the affine Grassmannian, and the affine Grassmannian is a homogenous space for the algebraic loop group, with the stabaliser of a point being topologically non-trivial, then I guess the answer to the original question as I asked it is no.

Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know.

Further commentary: Based on the comment of Stephen Griffeth, if loops on G is going to be homotopic to the affine Grassmannian, and the affine Grassmannian is a homogenous space for the algebraic loop group, with the stabaliser of a point being topologically non-trivial, then I guess the answer to the original question as I asked it is no.

Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The further commentary remark below applies to the case of based loops.

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know.

Further commentary: Based on the comment of Stephen Griffeth, if loops on G is going to be homotopic to the affine Grassmannian, and the affine Grassmannian is a homogenous space for the algebraic loop group, with the stabaliser of a point being topologically non-trivial, then I guess the answer to the original question as I asked it is no.

further commentary added
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Peter McNamara
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Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know.

Further commentary: Based on the comment of Stephen Griffeth, if loops on G is going to be homotopic to the affine Grassmannian, and the affine Grassmannian is a homogenous space for the algebraic loop group, with the stabaliser of a point being topologically non-trivial, then I guess the answer to the original question as I asked it is no.

Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know.

Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?

Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know.

Further commentary: Based on the comment of Stephen Griffeth, if loops on G is going to be homotopic to the affine Grassmannian, and the affine Grassmannian is a homogenous space for the algebraic loop group, with the stabaliser of a point being topologically non-trivial, then I guess the answer to the original question as I asked it is no.

it's not the restriction of scalars
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Peter McNamara
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expanded definition of X((t))
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Peter McNamara
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Peter McNamara
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