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1. • CommentRowNumber1.
   • CommentAuthorTobias Fritz
   • CommentTimeDec 5th 2020
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   Author: Tobias Fritz
   Format: MarkdownItexDoes anyone know if there is a definition/construction of spinor bundles which starts with a plain manifold, rather than with a manifold equipped with a metric? Or to ask an equivalent question: given two Riemannian manifolds with spin structures and a diffeomorphism which preserves the spin structures, is it possible to pull back sections of the spinor bundle from one manifold to the other? And prior to that, what "should" it even mean for a diffeomorphism to preserve the spin structure? I'd appreciate any pointers to the literature or other sort of input that people here might have.

   Does anyone know if there is a definition/construction of spinor bundles which starts with a plain manifold, rather than with a manifold equipped with a metric?

   Or to ask an equivalent question: given two Riemannian manifolds with spin structures and a diffeomorphism which preserves the spin structures, is it possible to pull back sections of the spinor bundle from one manifold to the other? And prior to that, what “should” it even mean for a diffeomorphism to preserve the spin structure?

   I’d appreciate any pointers to the literature or other sort of input that people here might have.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeDec 5th 2020
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   Author: Urs
   Format: MarkdownItexI would say the following: A spinor bundle on a plain manifold is associated to a lift of the structure group of its frame bundle through the homomorphism $Spin(n) \to GL(n)$ -- but that factors, of course, as $Spin(n) \to SO(n) \to O(n) \to GL(n)$, and hence lifting the structure group to $Spin(n)$ implies/involves choosing [[G-structure]] for $G= O(n)$, which is equivalent to a Riemannian metric. So a diffeomorphism preserving spinor bundles in this sense is necessarily also an isometry. A "spin structure" is -- despite the traditional mess of terminogy -- less information than a [[G-structure]] for $G = Spin$, namely is just the homotopy class of the latter -- its "shape". Since $O(n) \to GL(n)$ is a weak homotopy equivalence/shape equivalence, a bare spin structure does not entail a Riemannian metric. But the notions of respect for all these structures run in parallel: In each case it's a choice of equivalence between the domain structure and the pullback of the codomain structure.

   I would say the following:

   A spinor bundle on a plain manifold is associated to a lift of the structure group of its frame bundle through the homomorphism $Spin(n) \to GL(n)$ – but that factors, of course, as $Spin(n) \to SO(n) \to O(n) \to GL(n)$, and hence lifting the structure group to $Spin(n)$ implies/involves choosing G-structure for $G= O(n)$, which is equivalent to a Riemannian metric. So a diffeomorphism preserving spinor bundles in this sense is necessarily also an isometry.

   A “spin structure” is – despite the traditional mess of terminogy – less information than a G-structure for $G = Spin$, namely is just the homotopy class of the latter – its “shape”. Since $O(n) \to GL(n)$ is a weak homotopy equivalence/shape equivalence, a bare spin structure does not entail a Riemannian metric.

   But the notions of respect for all these structures run in parallel: In each case it’s a choice of equivalence between the domain structure and the pullback of the codomain structure.

3. • CommentRowNumber3.
   • CommentAuthorTobias Fritz
   • CommentTimeDec 6th 2020
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   Author: Tobias Fritz
   Format: MarkdownItexThat clarifies things, thank you! Just to double-check: my takeaway from those statements is that in absence of a metric, a spinor bundle is not determined by a spin structure. The consequence for my original question is that no, we cannot expect to get any functoriality for spinor fields other than with respect to isometries. Right? (BTW I've been ignoring orientations for the purposes of this question in order to focus on the metric. I guess I should have inserted "orientation-preserving" in a few places.)

   That clarifies things, thank you!

   Just to double-check: my takeaway from those statements is that in absence of a metric, a spinor bundle is not determined by a spin structure. The consequence for my original question is that no, we cannot expect to get any functoriality for spinor fields other than with respect to isometries. Right?

   (BTW I’ve been ignoring orientations for the purposes of this question in order to focus on the metric. I guess I should have inserted “orientation-preserving” in a few places.)

4. • CommentRowNumber4.
   • CommentAuthorDmitri Pavlov
   • CommentTimeDec 7th 2020
   • (edited Dec 7th 2020)
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   Author: Dmitri Pavlov
   Format: MarkdownItex> a spinor bundle is not determined by a spin structure I think this depends on what "determined" means. Spin structures on M (when they exist) form a [[torsor]] over H^1(M,Z/2Z). If we now consider the infinite-dimensional moduli space S of Riemannian metrics on M, over each point s∈S we have an H^1(M,Z/2Z)-torsor of spin structures compatible with the given Riemannian metric s. These torsors assemble together in a principal H^1(M,Z/2Z)-bundle over S. Since S is convex, the total space of this principal bundle splits as a disjoint union of copies of S, where the disjoint union is indexed by a torsor over H^1(M,Z/2Z). Elements of this torsor can be reasonably considered spin structures on M, without a choice of a Riemannian metric.

   a spinor bundle is not determined by a spin structure

   I think this depends on what “determined” means.

   Spin structures on M (when they exist) form a torsor over H^1(M,Z/2Z).

   If we now consider the infinite-dimensional moduli space S of Riemannian metrics on M, over each point s∈S we have an H^1(M,Z/2Z)-torsor of spin structures compatible with the given Riemannian metric s. These torsors assemble together in a principal H^1(M,Z/2Z)-bundle over S.

   Since S is convex, the total space of this principal bundle splits as a disjoint union of copies of S, where the disjoint union is indexed by a torsor over H^1(M,Z/2Z).

   Elements of this torsor can be reasonably considered spin structures on M, without a choice of a Riemannian metric.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeDec 7th 2020
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   Author: Urs
   Format: MarkdownItexSo that's the difference between the homotopical and the differential geometric notion that I tried to highlight: A spin structure on $X$ is a choice of lift in $$ \array{ X && \longrightarrow && B Spin(n) \\ & {}_{\mathllap{\vdash Fr X}}\searrow &\swArrow& \swarrow \\ & & B GL(n) } $$ this happening in bare homotopy types, with $B G$ denoting the homotopy type of the classifying space of that name. A smooth spin-bundle though is a choice of lift in $$ \array{ X && \longrightarrow && \mathbf{B} Spin(n) \\ & {}_{\mathllap{\vdash Fr X}}\searrow &\swArrow& \swarrow \\ & & \mathbf{B} GL(n) } $$ where now $\mathbf{B} G$ denotes the smooth classifying stack. This latter lift is fancy notation for the usual notion of a smooth spin-bundle lifting the frame bundle $Fr(X)$, but it highlights that there is the forgetful functor from smooth spin bundles to spin structures, given by the shape unit (the "geometric realization") $$ \mathbf{B}G \longrightarrow B G $$ which forgets the smooth aspect structure, and in particular the metric implied by a smooth spin bundle-lift of the tangent bundle.

   So that’s the difference between the homotopical and the differential geometric notion that I tried to highlight:

   A spin structure on $X$ is a choice of lift in

   $$\array{ X && \longrightarrow && B Spin(n) \\ & {}_{\mathllap{\vdash Fr X}}\searrow &\swArrow& \swarrow \\ & & B GL(n) }$$

   this happening in bare homotopy types, with $B G$ denoting the homotopy type of the classifying space of that name.

   A smooth spin-bundle though is a choice of lift in

   $$\array{ X && \longrightarrow && \mathbf{B} Spin(n) \\ & {}_{\mathllap{\vdash Fr X}}\searrow &\swArrow& \swarrow \\ & & \mathbf{B} GL(n) }$$

   where now $\mathbf{B} G$ denotes the smooth classifying stack. This latter lift is fancy notation for the usual notion of a smooth spin-bundle lifting the frame bundle $Fr(X)$, but it highlights that there is the forgetful functor from smooth spin bundles to spin structures, given by the shape unit (the “geometric realization”)

   $$\mathbf{B}G \longrightarrow B G$$

   which forgets the smooth aspect structure, and in particular the metric implied by a smooth spin bundle-lift of the tangent bundle.