How does a section of a bundle with fibre $mathbb{P}^N$ give us a spin structure?












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In "Spin Structures on Manifolds" Milnor says that for a principal $SO(n)$ bundle $pi:Eto X$ a spin structure can be defined as follows: Let $Spin(n)$ act on a sphere $S^N$ in such a way that the kernel $mathbb{Z}_2$ of the cover $Spin(n) to SO(n)$ acts freely, thus giving an action of the quotient $SO(n) $ on the quotient $mathbb{P}^N$.



He considers the bundle with fibre $mathbb{P}^N$ associated to the principal bundle $pi:Eto X$ and says that a spin structure can be seen as a section of this bundle but he doesn't give any detail. How can we get a spin structure on $E$ from such a section?






differential-geometry algebraic-topology differential-topology







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$endgroup$

















    4












    $begingroup$


    In "Spin Structures on Manifolds" Milnor says that for a principal $SO(n)$ bundle $pi:Eto X$ a spin structure can be defined as follows: Let $Spin(n)$ act on a sphere $S^N$ in such a way that the kernel $mathbb{Z}_2$ of the cover $Spin(n) to SO(n)$ acts freely, thus giving an action of the quotient $SO(n) $ on the quotient $mathbb{P}^N$.



    He considers the bundle with fibre $mathbb{P}^N$ associated to the principal bundle $pi:Eto X$ and says that a spin structure can be seen as a section of this bundle but he doesn't give any detail. How can we get a spin structure on $E$ from such a section?






    differential-geometry algebraic-topology differential-topology







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      2



      $begingroup$


      In "Spin Structures on Manifolds" Milnor says that for a principal $SO(n)$ bundle $pi:Eto X$ a spin structure can be defined as follows: Let $Spin(n)$ act on a sphere $S^N$ in such a way that the kernel $mathbb{Z}_2$ of the cover $Spin(n) to SO(n)$ acts freely, thus giving an action of the quotient $SO(n) $ on the quotient $mathbb{P}^N$.



      He considers the bundle with fibre $mathbb{P}^N$ associated to the principal bundle $pi:Eto X$ and says that a spin structure can be seen as a section of this bundle but he doesn't give any detail. How can we get a spin structure on $E$ from such a section?






      differential-geometry algebraic-topology differential-topology







      share|cite|improve this question









      $endgroup$




      In "Spin Structures on Manifolds" Milnor says that for a principal $SO(n)$ bundle $pi:Eto X$ a spin structure can be defined as follows: Let $Spin(n)$ act on a sphere $S^N$ in such a way that the kernel $mathbb{Z}_2$ of the cover $Spin(n) to SO(n)$ acts freely, thus giving an action of the quotient $SO(n) $ on the quotient $mathbb{P}^N$.



      He considers the bundle with fibre $mathbb{P}^N$ associated to the principal bundle $pi:Eto X$ and says that a spin structure can be seen as a section of this bundle but he doesn't give any detail. How can we get a spin structure on $E$ from such a section?






      differential-geometry algebraic-topology differential-topology




      differential-geometry algebraic-topology differential-topology






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 28 at 10:51









      Eugenio LandiEugenio Landi

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          Let $lambda colon Spin(n) to SO(n)$ be the usual double-cover; let $rhocolon Spin(n) to SO(N+1)$ be a spinor representation, so that $ker(lambda)cong mathbb{Z}/2$ acts freely on $S^N$, and suppose that $N> dim X$; let $(E,B)$ be your favourite classifying space functor; let $P_{SO}$ be a principal $SO(n)$-bundle over $X$, classified by $ccolon Xto BSO(n)$; finally let $mathbb{P}_{SO}$ be the $mathbb{P}^N$ bundle associated to $P_{SO}$, via the chosen spinor representation.



          The Short Answer to "How can we get a spin structure on $P_{SO}$ from a section of $mathbb{P}_{SO}$?"



          There is a model of $BSpin(n)$ as a $mathbb{P}^infty sim B(mathbb{Z}/2)$ bundle over $BSO(n)$, given by replacing the $SO(n)$ fibres of $ESO(n)$ with copies of $mathbb{P}^infty$ in a way induced by $rho$, and the inclusion $mathbb{P}^Ntomathbb{P}^infty$ induces an inclusion $mathbb{P}_{SO} to c^*BSpin(n)$. From a section $sigmacolon Xto mathbb{P}_{SO}$ we can form the composition



          $$ tilde{c}colon X to mathbb{P}_{SO} to c^*BSpin(n) to BSpin(n) $$



          which is a lift of $c$ along $Blambda$. Then $tilde{c}^* ESpin(n)$ is a spin structure for $P_{SO}$.



          The Long Answer to "How are isomorphism classes of Spin structures on $P_{SO}$ the same as homotopy classes of sections of $mathbb{P}_{SO}$?"



          I found a way of seeing it in two steps using classifying spaces: first going from Spin structures in the sense of Milnor to lifts of classifying maps along $Blambdacolon BSpin(n) to BSO(n)$, and then from lifts of classifying maps to sections of $mathbb{P}_{SO}$. I think Milnor would've know about this stuff when he wrote the paper, so this may (or may not) be what he had in mind.



          Step 1: Spin structures correspond to lifts of classifying maps.



          The first step is to translate the paper's definition of a spin structure into a homotopy class of lifts of a classifying map along $Blambdacolon BSpin(n) to BSO(n)$ (this is a reasonably common interpretation). Specifically let $ccolon Xto BSO(n)$ be a classifying map for the principal bundle $P_{SO} = c^* ESO(n) to X$ and consider the diagram
          $require{AMScd}$
          begin{CD}
          P_{SO} @>>> ESO(n) @< Elambda << ESpin(n) \
          @V V V @VV pi_{SO} V @VV pi_{Spin} V\
          X @>c>> BSO(n) @< Blambda << BSpin(n)
          end{CD}



          The functorial map $Elambdacolon ESpin(n) to ESO(n)$ is a Spin structure for $ESO(n)$ in the sense of Milnor.



          Proposition 1. Isomorphism classes of Spin structures on $P_{SO}$ correspond to homotopy classes of lifts of $c$ along $Blambda$.



          Proof Sketch. If $tilde{c}colon Xto BSpin(n)$ is a lift of $c$ along $Blambda$, then the pulled-back principal $Spin(n)$ bundle $P_{Spin} = tilde{c}^*ESpin(n)$ has a canonical map to $P_{SO}$ due to the universal property of the pullback, and moreover is a Spin structure. The converse is a bit trickier and I'm not 100% but I think it works: if $Pto P_{SO}$ is a Spin structure in the sense of Milnor then in particular it is a principal $Spin(n)$ bundle so it is classified by some map $tilde{c}colon Xto BSpin(n)$ and we need to show that $c = Blambdacirctilde{c}$, at least up to homotopy. The claim is that $tilde{c}^*ESpin(n)$ double-covers $P_{SO}$ in the same way as $P$, but by uniqueness of double covers of $SO(n)$ it should follow that $tilde{c}^*ESpin(n)cong P$ and then $Blambda circ tilde{c} sim c$ by classifying space theory.



          Step 2: Lifts of classifying maps correspond to sections of the projective space bundle



          Let $mathbb{P}_{SO}$ be the bundle with fibre $mathbb{P}^N$ associated to $P_{SO}$ via the chosen spinor representation, where $N>> dim X$.



          Proposition 2. Homotopy classes of lifts of $c$ along $Blambda$ correspond to homotopy classes of sections of $mathbb{P}_{SO}$.



          Proof sketch. This is essentially due to the following fact: If $Kto G to H$ is a short exact sequence of groups then $BK to BG to BH$ is a fibration sequence. In fact, using the Borel Construction you can explicitly describe $BG to BH$ as the universal principal $H$-bundle with fibres replaced by $BK$ (I will do this below). Then $BSpin(n)$ is the universal $SO(n)$-bundle with $B(mathbb{Z}/2)sim mathbb{P}^infty$ plugged in (via countably many copies of the chosen spinor representation), so by naturality of replacing fibres $c^*BSpin(n)$ is $P_{SO}$ with fibres replaced by $mathbb{P}^infty$ in the same way. The inclusion $mathbb{P}^N to mathbb{P}^infty$ induces a fibre-wise inclusion $mathbb{P}_{SO} to c^*BSpin(n)$, and since $N$ is large it is sufficiently highly connected so that sections of $mathbb{P}_{SO}$ correspond to sections of $c^*BSpin(n)$ (up to homotopy). Now you can check that the space of sections of $c^*BSpin(n)$ is actually homeomorphic to the space of lifts of $c$ along $Blambda$, finishing the proof sketch.



          (Supplemental: The Borel Construction)



          Definition. Let $G$ be a topological group, $X$ a right $G$-space, and $Y$ a left $G$-space. The diagonal action of $G$ on $Xtimes Y$ is defined by $(x, y)cdot g = (xg, g^{-1}y)$. The Borel Construction of $G$ acting on these spaces is defined as



          $$Xtimes_{G} Y := (X times Y)/G.$$



          (Equivalently you could take the quotient by the relation $(xg,y) sim (x,gy)$ but it is useful in our case to express it as the quotient of an action.) For a first exercise, check that the Borel Construction comes with natural maps to $X/G$ and $Gbackslash Y$ (the "projections" to $X$ and $Y$ will typically not be well-defined). In particular if $X$ is a principle $G$ bundle, the Borel Construction has the effect of replacing the $G$ fibres with $Y$; in fact if $E$ is a bundle with fibre $F$ and structure group $G$, part of expressing $G$ as the structure group involves choosing a representation $varphicolon Gto Homeo(F)$, and if $P$ is the underlying principal $G$-bundle then
          $$Econg Ptimes_G F.$$
          I'm going to use the following lemma to switch from a $(SO(n),mathbb{P}^N)$ context to a $(Spin(n), S^N)$ context. I don't have a citation for it off hand, but it can be checked directly as an exercise:



          Lemma. Let $G$, $X$, $Y$ be as above, and suppose $K$ is a normal subgroup of $G$ which acts trivially on $X$. Then $Xtimes_G Y$ is homeomorphic to $X_{G/K}(Y/K)$.



          (The idea is that whatever acts trivially on the left side can be eaten up on the right side.)



          Now suppose $Kto G to H$ is a short exact sequence of groups. $G$ acts on $EH$ via the homomorphism $Gto H$, and the kernel of this action is the normal subgroup $K$. Now suppose $E$ is a contractible $G$-space on which $K$ acts freely, so that the diagonal action of $G$ on $EHtimes E$ is free. Then



          $$ BG = EH times_{G} E cong EHtimes_{G/K} (E/K) cong EH times_H BK$$



          That is, $BG$ is the the universal $H$-bundle with fibres replaced by $BK$ with $H$ acting via the quotient action of $G/K$ on $E/K$.



          In the spin case, we can add together countably many copies of our spinor representation to get an action of $Spin(n)$ on $S^infty$ where $mathbb{Z}/2$ acts freely, so then



          $$ BSpin(n) = ESO(n)times_{Spin(n)} S^infty cong ESO(n) times_{SO(n)} mathbb{P}^infty $$



          Moreover the map $ESO(n)times_{SO(n)} mathbb{P}^N to BSpin(n)$ is as connected as the map $mathbb{P}^N to mathbb{P}^infty$






          share|cite|improve this answer











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            $begingroup$

            Let $lambda colon Spin(n) to SO(n)$ be the usual double-cover; let $rhocolon Spin(n) to SO(N+1)$ be a spinor representation, so that $ker(lambda)cong mathbb{Z}/2$ acts freely on $S^N$, and suppose that $N> dim X$; let $(E,B)$ be your favourite classifying space functor; let $P_{SO}$ be a principal $SO(n)$-bundle over $X$, classified by $ccolon Xto BSO(n)$; finally let $mathbb{P}_{SO}$ be the $mathbb{P}^N$ bundle associated to $P_{SO}$, via the chosen spinor representation.



            The Short Answer to "How can we get a spin structure on $P_{SO}$ from a section of $mathbb{P}_{SO}$?"



            There is a model of $BSpin(n)$ as a $mathbb{P}^infty sim B(mathbb{Z}/2)$ bundle over $BSO(n)$, given by replacing the $SO(n)$ fibres of $ESO(n)$ with copies of $mathbb{P}^infty$ in a way induced by $rho$, and the inclusion $mathbb{P}^Ntomathbb{P}^infty$ induces an inclusion $mathbb{P}_{SO} to c^*BSpin(n)$. From a section $sigmacolon Xto mathbb{P}_{SO}$ we can form the composition



            $$ tilde{c}colon X to mathbb{P}_{SO} to c^*BSpin(n) to BSpin(n) $$



            which is a lift of $c$ along $Blambda$. Then $tilde{c}^* ESpin(n)$ is a spin structure for $P_{SO}$.



            The Long Answer to "How are isomorphism classes of Spin structures on $P_{SO}$ the same as homotopy classes of sections of $mathbb{P}_{SO}$?"



            I found a way of seeing it in two steps using classifying spaces: first going from Spin structures in the sense of Milnor to lifts of classifying maps along $Blambdacolon BSpin(n) to BSO(n)$, and then from lifts of classifying maps to sections of $mathbb{P}_{SO}$. I think Milnor would've know about this stuff when he wrote the paper, so this may (or may not) be what he had in mind.



            Step 1: Spin structures correspond to lifts of classifying maps.



            The first step is to translate the paper's definition of a spin structure into a homotopy class of lifts of a classifying map along $Blambdacolon BSpin(n) to BSO(n)$ (this is a reasonably common interpretation). Specifically let $ccolon Xto BSO(n)$ be a classifying map for the principal bundle $P_{SO} = c^* ESO(n) to X$ and consider the diagram
            $require{AMScd}$
            begin{CD}
            P_{SO} @>>> ESO(n) @< Elambda << ESpin(n) \
            @V V V @VV pi_{SO} V @VV pi_{Spin} V\
            X @>c>> BSO(n) @< Blambda << BSpin(n)
            end{CD}



            The functorial map $Elambdacolon ESpin(n) to ESO(n)$ is a Spin structure for $ESO(n)$ in the sense of Milnor.



            Proposition 1. Isomorphism classes of Spin structures on $P_{SO}$ correspond to homotopy classes of lifts of $c$ along $Blambda$.



            Proof Sketch. If $tilde{c}colon Xto BSpin(n)$ is a lift of $c$ along $Blambda$, then the pulled-back principal $Spin(n)$ bundle $P_{Spin} = tilde{c}^*ESpin(n)$ has a canonical map to $P_{SO}$ due to the universal property of the pullback, and moreover is a Spin structure. The converse is a bit trickier and I'm not 100% but I think it works: if $Pto P_{SO}$ is a Spin structure in the sense of Milnor then in particular it is a principal $Spin(n)$ bundle so it is classified by some map $tilde{c}colon Xto BSpin(n)$ and we need to show that $c = Blambdacirctilde{c}$, at least up to homotopy. The claim is that $tilde{c}^*ESpin(n)$ double-covers $P_{SO}$ in the same way as $P$, but by uniqueness of double covers of $SO(n)$ it should follow that $tilde{c}^*ESpin(n)cong P$ and then $Blambda circ tilde{c} sim c$ by classifying space theory.



            Step 2: Lifts of classifying maps correspond to sections of the projective space bundle



            Let $mathbb{P}_{SO}$ be the bundle with fibre $mathbb{P}^N$ associated to $P_{SO}$ via the chosen spinor representation, where $N>> dim X$.



            Proposition 2. Homotopy classes of lifts of $c$ along $Blambda$ correspond to homotopy classes of sections of $mathbb{P}_{SO}$.



            Proof sketch. This is essentially due to the following fact: If $Kto G to H$ is a short exact sequence of groups then $BK to BG to BH$ is a fibration sequence. In fact, using the Borel Construction you can explicitly describe $BG to BH$ as the universal principal $H$-bundle with fibres replaced by $BK$ (I will do this below). Then $BSpin(n)$ is the universal $SO(n)$-bundle with $B(mathbb{Z}/2)sim mathbb{P}^infty$ plugged in (via countably many copies of the chosen spinor representation), so by naturality of replacing fibres $c^*BSpin(n)$ is $P_{SO}$ with fibres replaced by $mathbb{P}^infty$ in the same way. The inclusion $mathbb{P}^N to mathbb{P}^infty$ induces a fibre-wise inclusion $mathbb{P}_{SO} to c^*BSpin(n)$, and since $N$ is large it is sufficiently highly connected so that sections of $mathbb{P}_{SO}$ correspond to sections of $c^*BSpin(n)$ (up to homotopy). Now you can check that the space of sections of $c^*BSpin(n)$ is actually homeomorphic to the space of lifts of $c$ along $Blambda$, finishing the proof sketch.



            (Supplemental: The Borel Construction)



            Definition. Let $G$ be a topological group, $X$ a right $G$-space, and $Y$ a left $G$-space. The diagonal action of $G$ on $Xtimes Y$ is defined by $(x, y)cdot g = (xg, g^{-1}y)$. The Borel Construction of $G$ acting on these spaces is defined as



            $$Xtimes_{G} Y := (X times Y)/G.$$



            (Equivalently you could take the quotient by the relation $(xg,y) sim (x,gy)$ but it is useful in our case to express it as the quotient of an action.) For a first exercise, check that the Borel Construction comes with natural maps to $X/G$ and $Gbackslash Y$ (the "projections" to $X$ and $Y$ will typically not be well-defined). In particular if $X$ is a principle $G$ bundle, the Borel Construction has the effect of replacing the $G$ fibres with $Y$; in fact if $E$ is a bundle with fibre $F$ and structure group $G$, part of expressing $G$ as the structure group involves choosing a representation $varphicolon Gto Homeo(F)$, and if $P$ is the underlying principal $G$-bundle then
            $$Econg Ptimes_G F.$$
            I'm going to use the following lemma to switch from a $(SO(n),mathbb{P}^N)$ context to a $(Spin(n), S^N)$ context. I don't have a citation for it off hand, but it can be checked directly as an exercise:



            Lemma. Let $G$, $X$, $Y$ be as above, and suppose $K$ is a normal subgroup of $G$ which acts trivially on $X$. Then $Xtimes_G Y$ is homeomorphic to $X_{G/K}(Y/K)$.



            (The idea is that whatever acts trivially on the left side can be eaten up on the right side.)



            Now suppose $Kto G to H$ is a short exact sequence of groups. $G$ acts on $EH$ via the homomorphism $Gto H$, and the kernel of this action is the normal subgroup $K$. Now suppose $E$ is a contractible $G$-space on which $K$ acts freely, so that the diagonal action of $G$ on $EHtimes E$ is free. Then



            $$ BG = EH times_{G} E cong EHtimes_{G/K} (E/K) cong EH times_H BK$$



            That is, $BG$ is the the universal $H$-bundle with fibres replaced by $BK$ with $H$ acting via the quotient action of $G/K$ on $E/K$.



            In the spin case, we can add together countably many copies of our spinor representation to get an action of $Spin(n)$ on $S^infty$ where $mathbb{Z}/2$ acts freely, so then



            $$ BSpin(n) = ESO(n)times_{Spin(n)} S^infty cong ESO(n) times_{SO(n)} mathbb{P}^infty $$



            Moreover the map $ESO(n)times_{SO(n)} mathbb{P}^N to BSpin(n)$ is as connected as the map $mathbb{P}^N to mathbb{P}^infty$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Let $lambda colon Spin(n) to SO(n)$ be the usual double-cover; let $rhocolon Spin(n) to SO(N+1)$ be a spinor representation, so that $ker(lambda)cong mathbb{Z}/2$ acts freely on $S^N$, and suppose that $N> dim X$; let $(E,B)$ be your favourite classifying space functor; let $P_{SO}$ be a principal $SO(n)$-bundle over $X$, classified by $ccolon Xto BSO(n)$; finally let $mathbb{P}_{SO}$ be the $mathbb{P}^N$ bundle associated to $P_{SO}$, via the chosen spinor representation.



              The Short Answer to "How can we get a spin structure on $P_{SO}$ from a section of $mathbb{P}_{SO}$?"



              There is a model of $BSpin(n)$ as a $mathbb{P}^infty sim B(mathbb{Z}/2)$ bundle over $BSO(n)$, given by replacing the $SO(n)$ fibres of $ESO(n)$ with copies of $mathbb{P}^infty$ in a way induced by $rho$, and the inclusion $mathbb{P}^Ntomathbb{P}^infty$ induces an inclusion $mathbb{P}_{SO} to c^*BSpin(n)$. From a section $sigmacolon Xto mathbb{P}_{SO}$ we can form the composition



              $$ tilde{c}colon X to mathbb{P}_{SO} to c^*BSpin(n) to BSpin(n) $$



              which is a lift of $c$ along $Blambda$. Then $tilde{c}^* ESpin(n)$ is a spin structure for $P_{SO}$.



              The Long Answer to "How are isomorphism classes of Spin structures on $P_{SO}$ the same as homotopy classes of sections of $mathbb{P}_{SO}$?"



              I found a way of seeing it in two steps using classifying spaces: first going from Spin structures in the sense of Milnor to lifts of classifying maps along $Blambdacolon BSpin(n) to BSO(n)$, and then from lifts of classifying maps to sections of $mathbb{P}_{SO}$. I think Milnor would've know about this stuff when he wrote the paper, so this may (or may not) be what he had in mind.



              Step 1: Spin structures correspond to lifts of classifying maps.



              The first step is to translate the paper's definition of a spin structure into a homotopy class of lifts of a classifying map along $Blambdacolon BSpin(n) to BSO(n)$ (this is a reasonably common interpretation). Specifically let $ccolon Xto BSO(n)$ be a classifying map for the principal bundle $P_{SO} = c^* ESO(n) to X$ and consider the diagram
              $require{AMScd}$
              begin{CD}
              P_{SO} @>>> ESO(n) @< Elambda << ESpin(n) \
              @V V V @VV pi_{SO} V @VV pi_{Spin} V\
              X @>c>> BSO(n) @< Blambda << BSpin(n)
              end{CD}



              The functorial map $Elambdacolon ESpin(n) to ESO(n)$ is a Spin structure for $ESO(n)$ in the sense of Milnor.



              Proposition 1. Isomorphism classes of Spin structures on $P_{SO}$ correspond to homotopy classes of lifts of $c$ along $Blambda$.



              Proof Sketch. If $tilde{c}colon Xto BSpin(n)$ is a lift of $c$ along $Blambda$, then the pulled-back principal $Spin(n)$ bundle $P_{Spin} = tilde{c}^*ESpin(n)$ has a canonical map to $P_{SO}$ due to the universal property of the pullback, and moreover is a Spin structure. The converse is a bit trickier and I'm not 100% but I think it works: if $Pto P_{SO}$ is a Spin structure in the sense of Milnor then in particular it is a principal $Spin(n)$ bundle so it is classified by some map $tilde{c}colon Xto BSpin(n)$ and we need to show that $c = Blambdacirctilde{c}$, at least up to homotopy. The claim is that $tilde{c}^*ESpin(n)$ double-covers $P_{SO}$ in the same way as $P$, but by uniqueness of double covers of $SO(n)$ it should follow that $tilde{c}^*ESpin(n)cong P$ and then $Blambda circ tilde{c} sim c$ by classifying space theory.



              Step 2: Lifts of classifying maps correspond to sections of the projective space bundle



              Let $mathbb{P}_{SO}$ be the bundle with fibre $mathbb{P}^N$ associated to $P_{SO}$ via the chosen spinor representation, where $N>> dim X$.



              Proposition 2. Homotopy classes of lifts of $c$ along $Blambda$ correspond to homotopy classes of sections of $mathbb{P}_{SO}$.



              Proof sketch. This is essentially due to the following fact: If $Kto G to H$ is a short exact sequence of groups then $BK to BG to BH$ is a fibration sequence. In fact, using the Borel Construction you can explicitly describe $BG to BH$ as the universal principal $H$-bundle with fibres replaced by $BK$ (I will do this below). Then $BSpin(n)$ is the universal $SO(n)$-bundle with $B(mathbb{Z}/2)sim mathbb{P}^infty$ plugged in (via countably many copies of the chosen spinor representation), so by naturality of replacing fibres $c^*BSpin(n)$ is $P_{SO}$ with fibres replaced by $mathbb{P}^infty$ in the same way. The inclusion $mathbb{P}^N to mathbb{P}^infty$ induces a fibre-wise inclusion $mathbb{P}_{SO} to c^*BSpin(n)$, and since $N$ is large it is sufficiently highly connected so that sections of $mathbb{P}_{SO}$ correspond to sections of $c^*BSpin(n)$ (up to homotopy). Now you can check that the space of sections of $c^*BSpin(n)$ is actually homeomorphic to the space of lifts of $c$ along $Blambda$, finishing the proof sketch.



              (Supplemental: The Borel Construction)



              Definition. Let $G$ be a topological group, $X$ a right $G$-space, and $Y$ a left $G$-space. The diagonal action of $G$ on $Xtimes Y$ is defined by $(x, y)cdot g = (xg, g^{-1}y)$. The Borel Construction of $G$ acting on these spaces is defined as



              $$Xtimes_{G} Y := (X times Y)/G.$$



              (Equivalently you could take the quotient by the relation $(xg,y) sim (x,gy)$ but it is useful in our case to express it as the quotient of an action.) For a first exercise, check that the Borel Construction comes with natural maps to $X/G$ and $Gbackslash Y$ (the "projections" to $X$ and $Y$ will typically not be well-defined). In particular if $X$ is a principle $G$ bundle, the Borel Construction has the effect of replacing the $G$ fibres with $Y$; in fact if $E$ is a bundle with fibre $F$ and structure group $G$, part of expressing $G$ as the structure group involves choosing a representation $varphicolon Gto Homeo(F)$, and if $P$ is the underlying principal $G$-bundle then
              $$Econg Ptimes_G F.$$
              I'm going to use the following lemma to switch from a $(SO(n),mathbb{P}^N)$ context to a $(Spin(n), S^N)$ context. I don't have a citation for it off hand, but it can be checked directly as an exercise:



              Lemma. Let $G$, $X$, $Y$ be as above, and suppose $K$ is a normal subgroup of $G$ which acts trivially on $X$. Then $Xtimes_G Y$ is homeomorphic to $X_{G/K}(Y/K)$.



              (The idea is that whatever acts trivially on the left side can be eaten up on the right side.)



              Now suppose $Kto G to H$ is a short exact sequence of groups. $G$ acts on $EH$ via the homomorphism $Gto H$, and the kernel of this action is the normal subgroup $K$. Now suppose $E$ is a contractible $G$-space on which $K$ acts freely, so that the diagonal action of $G$ on $EHtimes E$ is free. Then



              $$ BG = EH times_{G} E cong EHtimes_{G/K} (E/K) cong EH times_H BK$$



              That is, $BG$ is the the universal $H$-bundle with fibres replaced by $BK$ with $H$ acting via the quotient action of $G/K$ on $E/K$.



              In the spin case, we can add together countably many copies of our spinor representation to get an action of $Spin(n)$ on $S^infty$ where $mathbb{Z}/2$ acts freely, so then



              $$ BSpin(n) = ESO(n)times_{Spin(n)} S^infty cong ESO(n) times_{SO(n)} mathbb{P}^infty $$



              Moreover the map $ESO(n)times_{SO(n)} mathbb{P}^N to BSpin(n)$ is as connected as the map $mathbb{P}^N to mathbb{P}^infty$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Let $lambda colon Spin(n) to SO(n)$ be the usual double-cover; let $rhocolon Spin(n) to SO(N+1)$ be a spinor representation, so that $ker(lambda)cong mathbb{Z}/2$ acts freely on $S^N$, and suppose that $N> dim X$; let $(E,B)$ be your favourite classifying space functor; let $P_{SO}$ be a principal $SO(n)$-bundle over $X$, classified by $ccolon Xto BSO(n)$; finally let $mathbb{P}_{SO}$ be the $mathbb{P}^N$ bundle associated to $P_{SO}$, via the chosen spinor representation.



                The Short Answer to "How can we get a spin structure on $P_{SO}$ from a section of $mathbb{P}_{SO}$?"



                There is a model of $BSpin(n)$ as a $mathbb{P}^infty sim B(mathbb{Z}/2)$ bundle over $BSO(n)$, given by replacing the $SO(n)$ fibres of $ESO(n)$ with copies of $mathbb{P}^infty$ in a way induced by $rho$, and the inclusion $mathbb{P}^Ntomathbb{P}^infty$ induces an inclusion $mathbb{P}_{SO} to c^*BSpin(n)$. From a section $sigmacolon Xto mathbb{P}_{SO}$ we can form the composition



                $$ tilde{c}colon X to mathbb{P}_{SO} to c^*BSpin(n) to BSpin(n) $$



                which is a lift of $c$ along $Blambda$. Then $tilde{c}^* ESpin(n)$ is a spin structure for $P_{SO}$.



                The Long Answer to "How are isomorphism classes of Spin structures on $P_{SO}$ the same as homotopy classes of sections of $mathbb{P}_{SO}$?"



                I found a way of seeing it in two steps using classifying spaces: first going from Spin structures in the sense of Milnor to lifts of classifying maps along $Blambdacolon BSpin(n) to BSO(n)$, and then from lifts of classifying maps to sections of $mathbb{P}_{SO}$. I think Milnor would've know about this stuff when he wrote the paper, so this may (or may not) be what he had in mind.



                Step 1: Spin structures correspond to lifts of classifying maps.



                The first step is to translate the paper's definition of a spin structure into a homotopy class of lifts of a classifying map along $Blambdacolon BSpin(n) to BSO(n)$ (this is a reasonably common interpretation). Specifically let $ccolon Xto BSO(n)$ be a classifying map for the principal bundle $P_{SO} = c^* ESO(n) to X$ and consider the diagram
                $require{AMScd}$
                begin{CD}
                P_{SO} @>>> ESO(n) @< Elambda << ESpin(n) \
                @V V V @VV pi_{SO} V @VV pi_{Spin} V\
                X @>c>> BSO(n) @< Blambda << BSpin(n)
                end{CD}



                The functorial map $Elambdacolon ESpin(n) to ESO(n)$ is a Spin structure for $ESO(n)$ in the sense of Milnor.



                Proposition 1. Isomorphism classes of Spin structures on $P_{SO}$ correspond to homotopy classes of lifts of $c$ along $Blambda$.



                Proof Sketch. If $tilde{c}colon Xto BSpin(n)$ is a lift of $c$ along $Blambda$, then the pulled-back principal $Spin(n)$ bundle $P_{Spin} = tilde{c}^*ESpin(n)$ has a canonical map to $P_{SO}$ due to the universal property of the pullback, and moreover is a Spin structure. The converse is a bit trickier and I'm not 100% but I think it works: if $Pto P_{SO}$ is a Spin structure in the sense of Milnor then in particular it is a principal $Spin(n)$ bundle so it is classified by some map $tilde{c}colon Xto BSpin(n)$ and we need to show that $c = Blambdacirctilde{c}$, at least up to homotopy. The claim is that $tilde{c}^*ESpin(n)$ double-covers $P_{SO}$ in the same way as $P$, but by uniqueness of double covers of $SO(n)$ it should follow that $tilde{c}^*ESpin(n)cong P$ and then $Blambda circ tilde{c} sim c$ by classifying space theory.



                Step 2: Lifts of classifying maps correspond to sections of the projective space bundle



                Let $mathbb{P}_{SO}$ be the bundle with fibre $mathbb{P}^N$ associated to $P_{SO}$ via the chosen spinor representation, where $N>> dim X$.



                Proposition 2. Homotopy classes of lifts of $c$ along $Blambda$ correspond to homotopy classes of sections of $mathbb{P}_{SO}$.



                Proof sketch. This is essentially due to the following fact: If $Kto G to H$ is a short exact sequence of groups then $BK to BG to BH$ is a fibration sequence. In fact, using the Borel Construction you can explicitly describe $BG to BH$ as the universal principal $H$-bundle with fibres replaced by $BK$ (I will do this below). Then $BSpin(n)$ is the universal $SO(n)$-bundle with $B(mathbb{Z}/2)sim mathbb{P}^infty$ plugged in (via countably many copies of the chosen spinor representation), so by naturality of replacing fibres $c^*BSpin(n)$ is $P_{SO}$ with fibres replaced by $mathbb{P}^infty$ in the same way. The inclusion $mathbb{P}^N to mathbb{P}^infty$ induces a fibre-wise inclusion $mathbb{P}_{SO} to c^*BSpin(n)$, and since $N$ is large it is sufficiently highly connected so that sections of $mathbb{P}_{SO}$ correspond to sections of $c^*BSpin(n)$ (up to homotopy). Now you can check that the space of sections of $c^*BSpin(n)$ is actually homeomorphic to the space of lifts of $c$ along $Blambda$, finishing the proof sketch.



                (Supplemental: The Borel Construction)



                Definition. Let $G$ be a topological group, $X$ a right $G$-space, and $Y$ a left $G$-space. The diagonal action of $G$ on $Xtimes Y$ is defined by $(x, y)cdot g = (xg, g^{-1}y)$. The Borel Construction of $G$ acting on these spaces is defined as



                $$Xtimes_{G} Y := (X times Y)/G.$$



                (Equivalently you could take the quotient by the relation $(xg,y) sim (x,gy)$ but it is useful in our case to express it as the quotient of an action.) For a first exercise, check that the Borel Construction comes with natural maps to $X/G$ and $Gbackslash Y$ (the "projections" to $X$ and $Y$ will typically not be well-defined). In particular if $X$ is a principle $G$ bundle, the Borel Construction has the effect of replacing the $G$ fibres with $Y$; in fact if $E$ is a bundle with fibre $F$ and structure group $G$, part of expressing $G$ as the structure group involves choosing a representation $varphicolon Gto Homeo(F)$, and if $P$ is the underlying principal $G$-bundle then
                $$Econg Ptimes_G F.$$
                I'm going to use the following lemma to switch from a $(SO(n),mathbb{P}^N)$ context to a $(Spin(n), S^N)$ context. I don't have a citation for it off hand, but it can be checked directly as an exercise:



                Lemma. Let $G$, $X$, $Y$ be as above, and suppose $K$ is a normal subgroup of $G$ which acts trivially on $X$. Then $Xtimes_G Y$ is homeomorphic to $X_{G/K}(Y/K)$.



                (The idea is that whatever acts trivially on the left side can be eaten up on the right side.)



                Now suppose $Kto G to H$ is a short exact sequence of groups. $G$ acts on $EH$ via the homomorphism $Gto H$, and the kernel of this action is the normal subgroup $K$. Now suppose $E$ is a contractible $G$-space on which $K$ acts freely, so that the diagonal action of $G$ on $EHtimes E$ is free. Then



                $$ BG = EH times_{G} E cong EHtimes_{G/K} (E/K) cong EH times_H BK$$



                That is, $BG$ is the the universal $H$-bundle with fibres replaced by $BK$ with $H$ acting via the quotient action of $G/K$ on $E/K$.



                In the spin case, we can add together countably many copies of our spinor representation to get an action of $Spin(n)$ on $S^infty$ where $mathbb{Z}/2$ acts freely, so then



                $$ BSpin(n) = ESO(n)times_{Spin(n)} S^infty cong ESO(n) times_{SO(n)} mathbb{P}^infty $$



                Moreover the map $ESO(n)times_{SO(n)} mathbb{P}^N to BSpin(n)$ is as connected as the map $mathbb{P}^N to mathbb{P}^infty$






                share|cite|improve this answer











                $endgroup$



                Let $lambda colon Spin(n) to SO(n)$ be the usual double-cover; let $rhocolon Spin(n) to SO(N+1)$ be a spinor representation, so that $ker(lambda)cong mathbb{Z}/2$ acts freely on $S^N$, and suppose that $N> dim X$; let $(E,B)$ be your favourite classifying space functor; let $P_{SO}$ be a principal $SO(n)$-bundle over $X$, classified by $ccolon Xto BSO(n)$; finally let $mathbb{P}_{SO}$ be the $mathbb{P}^N$ bundle associated to $P_{SO}$, via the chosen spinor representation.



                The Short Answer to "How can we get a spin structure on $P_{SO}$ from a section of $mathbb{P}_{SO}$?"



                There is a model of $BSpin(n)$ as a $mathbb{P}^infty sim B(mathbb{Z}/2)$ bundle over $BSO(n)$, given by replacing the $SO(n)$ fibres of $ESO(n)$ with copies of $mathbb{P}^infty$ in a way induced by $rho$, and the inclusion $mathbb{P}^Ntomathbb{P}^infty$ induces an inclusion $mathbb{P}_{SO} to c^*BSpin(n)$. From a section $sigmacolon Xto mathbb{P}_{SO}$ we can form the composition



                $$ tilde{c}colon X to mathbb{P}_{SO} to c^*BSpin(n) to BSpin(n) $$



                which is a lift of $c$ along $Blambda$. Then $tilde{c}^* ESpin(n)$ is a spin structure for $P_{SO}$.



                The Long Answer to "How are isomorphism classes of Spin structures on $P_{SO}$ the same as homotopy classes of sections of $mathbb{P}_{SO}$?"



                I found a way of seeing it in two steps using classifying spaces: first going from Spin structures in the sense of Milnor to lifts of classifying maps along $Blambdacolon BSpin(n) to BSO(n)$, and then from lifts of classifying maps to sections of $mathbb{P}_{SO}$. I think Milnor would've know about this stuff when he wrote the paper, so this may (or may not) be what he had in mind.



                Step 1: Spin structures correspond to lifts of classifying maps.



                The first step is to translate the paper's definition of a spin structure into a homotopy class of lifts of a classifying map along $Blambdacolon BSpin(n) to BSO(n)$ (this is a reasonably common interpretation). Specifically let $ccolon Xto BSO(n)$ be a classifying map for the principal bundle $P_{SO} = c^* ESO(n) to X$ and consider the diagram
                $require{AMScd}$
                begin{CD}
                P_{SO} @>>> ESO(n) @< Elambda << ESpin(n) \
                @V V V @VV pi_{SO} V @VV pi_{Spin} V\
                X @>c>> BSO(n) @< Blambda << BSpin(n)
                end{CD}



                The functorial map $Elambdacolon ESpin(n) to ESO(n)$ is a Spin structure for $ESO(n)$ in the sense of Milnor.



                Proposition 1. Isomorphism classes of Spin structures on $P_{SO}$ correspond to homotopy classes of lifts of $c$ along $Blambda$.



                Proof Sketch. If $tilde{c}colon Xto BSpin(n)$ is a lift of $c$ along $Blambda$, then the pulled-back principal $Spin(n)$ bundle $P_{Spin} = tilde{c}^*ESpin(n)$ has a canonical map to $P_{SO}$ due to the universal property of the pullback, and moreover is a Spin structure. The converse is a bit trickier and I'm not 100% but I think it works: if $Pto P_{SO}$ is a Spin structure in the sense of Milnor then in particular it is a principal $Spin(n)$ bundle so it is classified by some map $tilde{c}colon Xto BSpin(n)$ and we need to show that $c = Blambdacirctilde{c}$, at least up to homotopy. The claim is that $tilde{c}^*ESpin(n)$ double-covers $P_{SO}$ in the same way as $P$, but by uniqueness of double covers of $SO(n)$ it should follow that $tilde{c}^*ESpin(n)cong P$ and then $Blambda circ tilde{c} sim c$ by classifying space theory.



                Step 2: Lifts of classifying maps correspond to sections of the projective space bundle



                Let $mathbb{P}_{SO}$ be the bundle with fibre $mathbb{P}^N$ associated to $P_{SO}$ via the chosen spinor representation, where $N>> dim X$.



                Proposition 2. Homotopy classes of lifts of $c$ along $Blambda$ correspond to homotopy classes of sections of $mathbb{P}_{SO}$.



                Proof sketch. This is essentially due to the following fact: If $Kto G to H$ is a short exact sequence of groups then $BK to BG to BH$ is a fibration sequence. In fact, using the Borel Construction you can explicitly describe $BG to BH$ as the universal principal $H$-bundle with fibres replaced by $BK$ (I will do this below). Then $BSpin(n)$ is the universal $SO(n)$-bundle with $B(mathbb{Z}/2)sim mathbb{P}^infty$ plugged in (via countably many copies of the chosen spinor representation), so by naturality of replacing fibres $c^*BSpin(n)$ is $P_{SO}$ with fibres replaced by $mathbb{P}^infty$ in the same way. The inclusion $mathbb{P}^N to mathbb{P}^infty$ induces a fibre-wise inclusion $mathbb{P}_{SO} to c^*BSpin(n)$, and since $N$ is large it is sufficiently highly connected so that sections of $mathbb{P}_{SO}$ correspond to sections of $c^*BSpin(n)$ (up to homotopy). Now you can check that the space of sections of $c^*BSpin(n)$ is actually homeomorphic to the space of lifts of $c$ along $Blambda$, finishing the proof sketch.



                (Supplemental: The Borel Construction)



                Definition. Let $G$ be a topological group, $X$ a right $G$-space, and $Y$ a left $G$-space. The diagonal action of $G$ on $Xtimes Y$ is defined by $(x, y)cdot g = (xg, g^{-1}y)$. The Borel Construction of $G$ acting on these spaces is defined as



                $$Xtimes_{G} Y := (X times Y)/G.$$



                (Equivalently you could take the quotient by the relation $(xg,y) sim (x,gy)$ but it is useful in our case to express it as the quotient of an action.) For a first exercise, check that the Borel Construction comes with natural maps to $X/G$ and $Gbackslash Y$ (the "projections" to $X$ and $Y$ will typically not be well-defined). In particular if $X$ is a principle $G$ bundle, the Borel Construction has the effect of replacing the $G$ fibres with $Y$; in fact if $E$ is a bundle with fibre $F$ and structure group $G$, part of expressing $G$ as the structure group involves choosing a representation $varphicolon Gto Homeo(F)$, and if $P$ is the underlying principal $G$-bundle then
                $$Econg Ptimes_G F.$$
                I'm going to use the following lemma to switch from a $(SO(n),mathbb{P}^N)$ context to a $(Spin(n), S^N)$ context. I don't have a citation for it off hand, but it can be checked directly as an exercise:



                Lemma. Let $G$, $X$, $Y$ be as above, and suppose $K$ is a normal subgroup of $G$ which acts trivially on $X$. Then $Xtimes_G Y$ is homeomorphic to $X_{G/K}(Y/K)$.



                (The idea is that whatever acts trivially on the left side can be eaten up on the right side.)



                Now suppose $Kto G to H$ is a short exact sequence of groups. $G$ acts on $EH$ via the homomorphism $Gto H$, and the kernel of this action is the normal subgroup $K$. Now suppose $E$ is a contractible $G$-space on which $K$ acts freely, so that the diagonal action of $G$ on $EHtimes E$ is free. Then



                $$ BG = EH times_{G} E cong EHtimes_{G/K} (E/K) cong EH times_H BK$$



                That is, $BG$ is the the universal $H$-bundle with fibres replaced by $BK$ with $H$ acting via the quotient action of $G/K$ on $E/K$.



                In the spin case, we can add together countably many copies of our spinor representation to get an action of $Spin(n)$ on $S^infty$ where $mathbb{Z}/2$ acts freely, so then



                $$ BSpin(n) = ESO(n)times_{Spin(n)} S^infty cong ESO(n) times_{SO(n)} mathbb{P}^infty $$



                Moreover the map $ESO(n)times_{SO(n)} mathbb{P}^N to BSpin(n)$ is as connected as the map $mathbb{P}^N to mathbb{P}^infty$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 2 at 19:28

























                answered Jan 31 at 4:58









                WilliamWilliam

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