$W$ spin implies $partial W$ spin












5












$begingroup$


Let $M$ be a compact orientable manifold with the first two Stieffel-Whitney numbers equal to zero (this is my definition of SPIN manifold). Let $B$ be the boundary of $M$; I want to show that $B$ is spin (in the sense that I've already stated).



I know that there is a theorem of Pontryagin that holds in a more general context and gives a stronger result (if $M$ is a compact manifold with boundary, than the Stieffel-Whitney numbers $w_i$ with $i geq 1$ of $partial M$ vanish) but I was wandering if in this case there exists a simpler proof of this fact.



I know for example that if $TM$ is orientable (as bundle, since as manifold is always orientable) than $M$ is spin iff the restriction of the tangent bundle to the 2-skeleton of $M$ is trivial (given a cellularization of M as CW-complex). The problem is that I don't know how to use this, since I don't know how to relate the $2$-skeleton of $partial M$ to the $2$-skeleton of $M$.










share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    Let $M$ be a compact orientable manifold with the first two Stieffel-Whitney numbers equal to zero (this is my definition of SPIN manifold). Let $B$ be the boundary of $M$; I want to show that $B$ is spin (in the sense that I've already stated).



    I know that there is a theorem of Pontryagin that holds in a more general context and gives a stronger result (if $M$ is a compact manifold with boundary, than the Stieffel-Whitney numbers $w_i$ with $i geq 1$ of $partial M$ vanish) but I was wandering if in this case there exists a simpler proof of this fact.



    I know for example that if $TM$ is orientable (as bundle, since as manifold is always orientable) than $M$ is spin iff the restriction of the tangent bundle to the 2-skeleton of $M$ is trivial (given a cellularization of M as CW-complex). The problem is that I don't know how to use this, since I don't know how to relate the $2$-skeleton of $partial M$ to the $2$-skeleton of $M$.










    share|cite|improve this question









    $endgroup$















      5












      5








      5


      2



      $begingroup$


      Let $M$ be a compact orientable manifold with the first two Stieffel-Whitney numbers equal to zero (this is my definition of SPIN manifold). Let $B$ be the boundary of $M$; I want to show that $B$ is spin (in the sense that I've already stated).



      I know that there is a theorem of Pontryagin that holds in a more general context and gives a stronger result (if $M$ is a compact manifold with boundary, than the Stieffel-Whitney numbers $w_i$ with $i geq 1$ of $partial M$ vanish) but I was wandering if in this case there exists a simpler proof of this fact.



      I know for example that if $TM$ is orientable (as bundle, since as manifold is always orientable) than $M$ is spin iff the restriction of the tangent bundle to the 2-skeleton of $M$ is trivial (given a cellularization of M as CW-complex). The problem is that I don't know how to use this, since I don't know how to relate the $2$-skeleton of $partial M$ to the $2$-skeleton of $M$.










      share|cite|improve this question









      $endgroup$




      Let $M$ be a compact orientable manifold with the first two Stieffel-Whitney numbers equal to zero (this is my definition of SPIN manifold). Let $B$ be the boundary of $M$; I want to show that $B$ is spin (in the sense that I've already stated).



      I know that there is a theorem of Pontryagin that holds in a more general context and gives a stronger result (if $M$ is a compact manifold with boundary, than the Stieffel-Whitney numbers $w_i$ with $i geq 1$ of $partial M$ vanish) but I was wandering if in this case there exists a simpler proof of this fact.



      I know for example that if $TM$ is orientable (as bundle, since as manifold is always orientable) than $M$ is spin iff the restriction of the tangent bundle to the 2-skeleton of $M$ is trivial (given a cellularization of M as CW-complex). The problem is that I don't know how to use this, since I don't know how to relate the $2$-skeleton of $partial M$ to the $2$-skeleton of $M$.







      algebraic-topology vector-bundles characteristic-classes manifolds-with-boundary spin-geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 2 '18 at 16:46









      Giuseppe BargagnatiGiuseppe Bargagnati

      1,094514




      1,094514






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          There are some confusions here I should clear up before answering.



          1) This is a definition of spinnable manifold, not of a spin manifold. (Similar to the difference between oriented manifold and orientable manifold.) When spin structures exist, there are $H^1(W;Bbb Z/2)$-many of them, and a spin manifold requires one such choice.



          2) If $M$ is the boundary of another manifold $W$, then it is not necessarily true that the Stiefel-Whitney classes $w_i(M) in H^i(M;Bbb Z/2)$ vanish. For instance, if $M$ has nontrivial $w_i(M)$ for $i < dim M$, then so does $M # M$, but the latter is always null-bordant. What instead is the case is that the Stiefel-Whitney numbers vanish. These are the results of all products of $w_i(M)$ that land in $H^{dim M}(M; Bbb Z/2) cong Bbb Z/2$; they are labeled by partitions of $dim M$.



          3) A vector bundle $E$ of rank at least 3 is spinnable if and only if it is trivializable over the 2-skeleton. This is not true for bundles of rank 2: there are spinnable bundles which are not trivial over the 2-skeleton. Think $TS^2$.





          As for your actual question, the point is that we have a natural isomorphism $T(partial W) oplus Bbb R cong TWbig|_{partial W}$. The easy claim is that if $W$ is spinnable, then $partial W$ is spinnable: naturality of Stiefel-Whitney classes implies that $j^*(w_i(W)) = w_i(partial W)$, where $j: partial W to W$ is the inclusion. If you already know that $w_1(W) = w_2(W) = 0$, you hence know that as well for $partial W$. This argument applies for any condition defined by the vanishing of some set of Stiefel-Whitney classes.



          To actually pin down a specific spin structure, we need to be able to argue that a spin structure on $E oplus Bbb R$ induces a natural spin structure on $E$. (This is true for orientations!) This amounts to the inverse claim that the map sending spin structures on $E$ to spin structures on $E oplus Bbb R$ (via the natural map $text{Spin}(n) to text{Spin}(n+1)$) is a bijection; and this may be verified using that spin structures are affine over the group of isomorphism classes of real line bundles, aka $H^1(W;Bbb Z/2)$.



          This argument doesn't work for arbitrary sorts of structure on the tangent bundle. For instance, one that famously doesn't work is "$W$ parallelizable implies $partial W$ parallelizable". Every disc $D^n$ is parallelizable, but the only spheres which are are $S^0, S^1, S^3$, and $S^7$. In the argument above, the way that showed up was in our descriptions of spin structures as affine over $H^1(W;Bbb Z/2)$, which is true for spin structures on rank $n$ bundles for all $n$; trivializations are affine over $[W, SO(n)]$, which depends on $n$ until $n$ is quite large.






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022871%2fw-spin-implies-partial-w-spin%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            There are some confusions here I should clear up before answering.



            1) This is a definition of spinnable manifold, not of a spin manifold. (Similar to the difference between oriented manifold and orientable manifold.) When spin structures exist, there are $H^1(W;Bbb Z/2)$-many of them, and a spin manifold requires one such choice.



            2) If $M$ is the boundary of another manifold $W$, then it is not necessarily true that the Stiefel-Whitney classes $w_i(M) in H^i(M;Bbb Z/2)$ vanish. For instance, if $M$ has nontrivial $w_i(M)$ for $i < dim M$, then so does $M # M$, but the latter is always null-bordant. What instead is the case is that the Stiefel-Whitney numbers vanish. These are the results of all products of $w_i(M)$ that land in $H^{dim M}(M; Bbb Z/2) cong Bbb Z/2$; they are labeled by partitions of $dim M$.



            3) A vector bundle $E$ of rank at least 3 is spinnable if and only if it is trivializable over the 2-skeleton. This is not true for bundles of rank 2: there are spinnable bundles which are not trivial over the 2-skeleton. Think $TS^2$.





            As for your actual question, the point is that we have a natural isomorphism $T(partial W) oplus Bbb R cong TWbig|_{partial W}$. The easy claim is that if $W$ is spinnable, then $partial W$ is spinnable: naturality of Stiefel-Whitney classes implies that $j^*(w_i(W)) = w_i(partial W)$, where $j: partial W to W$ is the inclusion. If you already know that $w_1(W) = w_2(W) = 0$, you hence know that as well for $partial W$. This argument applies for any condition defined by the vanishing of some set of Stiefel-Whitney classes.



            To actually pin down a specific spin structure, we need to be able to argue that a spin structure on $E oplus Bbb R$ induces a natural spin structure on $E$. (This is true for orientations!) This amounts to the inverse claim that the map sending spin structures on $E$ to spin structures on $E oplus Bbb R$ (via the natural map $text{Spin}(n) to text{Spin}(n+1)$) is a bijection; and this may be verified using that spin structures are affine over the group of isomorphism classes of real line bundles, aka $H^1(W;Bbb Z/2)$.



            This argument doesn't work for arbitrary sorts of structure on the tangent bundle. For instance, one that famously doesn't work is "$W$ parallelizable implies $partial W$ parallelizable". Every disc $D^n$ is parallelizable, but the only spheres which are are $S^0, S^1, S^3$, and $S^7$. In the argument above, the way that showed up was in our descriptions of spin structures as affine over $H^1(W;Bbb Z/2)$, which is true for spin structures on rank $n$ bundles for all $n$; trivializations are affine over $[W, SO(n)]$, which depends on $n$ until $n$ is quite large.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              There are some confusions here I should clear up before answering.



              1) This is a definition of spinnable manifold, not of a spin manifold. (Similar to the difference between oriented manifold and orientable manifold.) When spin structures exist, there are $H^1(W;Bbb Z/2)$-many of them, and a spin manifold requires one such choice.



              2) If $M$ is the boundary of another manifold $W$, then it is not necessarily true that the Stiefel-Whitney classes $w_i(M) in H^i(M;Bbb Z/2)$ vanish. For instance, if $M$ has nontrivial $w_i(M)$ for $i < dim M$, then so does $M # M$, but the latter is always null-bordant. What instead is the case is that the Stiefel-Whitney numbers vanish. These are the results of all products of $w_i(M)$ that land in $H^{dim M}(M; Bbb Z/2) cong Bbb Z/2$; they are labeled by partitions of $dim M$.



              3) A vector bundle $E$ of rank at least 3 is spinnable if and only if it is trivializable over the 2-skeleton. This is not true for bundles of rank 2: there are spinnable bundles which are not trivial over the 2-skeleton. Think $TS^2$.





              As for your actual question, the point is that we have a natural isomorphism $T(partial W) oplus Bbb R cong TWbig|_{partial W}$. The easy claim is that if $W$ is spinnable, then $partial W$ is spinnable: naturality of Stiefel-Whitney classes implies that $j^*(w_i(W)) = w_i(partial W)$, where $j: partial W to W$ is the inclusion. If you already know that $w_1(W) = w_2(W) = 0$, you hence know that as well for $partial W$. This argument applies for any condition defined by the vanishing of some set of Stiefel-Whitney classes.



              To actually pin down a specific spin structure, we need to be able to argue that a spin structure on $E oplus Bbb R$ induces a natural spin structure on $E$. (This is true for orientations!) This amounts to the inverse claim that the map sending spin structures on $E$ to spin structures on $E oplus Bbb R$ (via the natural map $text{Spin}(n) to text{Spin}(n+1)$) is a bijection; and this may be verified using that spin structures are affine over the group of isomorphism classes of real line bundles, aka $H^1(W;Bbb Z/2)$.



              This argument doesn't work for arbitrary sorts of structure on the tangent bundle. For instance, one that famously doesn't work is "$W$ parallelizable implies $partial W$ parallelizable". Every disc $D^n$ is parallelizable, but the only spheres which are are $S^0, S^1, S^3$, and $S^7$. In the argument above, the way that showed up was in our descriptions of spin structures as affine over $H^1(W;Bbb Z/2)$, which is true for spin structures on rank $n$ bundles for all $n$; trivializations are affine over $[W, SO(n)]$, which depends on $n$ until $n$ is quite large.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                There are some confusions here I should clear up before answering.



                1) This is a definition of spinnable manifold, not of a spin manifold. (Similar to the difference between oriented manifold and orientable manifold.) When spin structures exist, there are $H^1(W;Bbb Z/2)$-many of them, and a spin manifold requires one such choice.



                2) If $M$ is the boundary of another manifold $W$, then it is not necessarily true that the Stiefel-Whitney classes $w_i(M) in H^i(M;Bbb Z/2)$ vanish. For instance, if $M$ has nontrivial $w_i(M)$ for $i < dim M$, then so does $M # M$, but the latter is always null-bordant. What instead is the case is that the Stiefel-Whitney numbers vanish. These are the results of all products of $w_i(M)$ that land in $H^{dim M}(M; Bbb Z/2) cong Bbb Z/2$; they are labeled by partitions of $dim M$.



                3) A vector bundle $E$ of rank at least 3 is spinnable if and only if it is trivializable over the 2-skeleton. This is not true for bundles of rank 2: there are spinnable bundles which are not trivial over the 2-skeleton. Think $TS^2$.





                As for your actual question, the point is that we have a natural isomorphism $T(partial W) oplus Bbb R cong TWbig|_{partial W}$. The easy claim is that if $W$ is spinnable, then $partial W$ is spinnable: naturality of Stiefel-Whitney classes implies that $j^*(w_i(W)) = w_i(partial W)$, where $j: partial W to W$ is the inclusion. If you already know that $w_1(W) = w_2(W) = 0$, you hence know that as well for $partial W$. This argument applies for any condition defined by the vanishing of some set of Stiefel-Whitney classes.



                To actually pin down a specific spin structure, we need to be able to argue that a spin structure on $E oplus Bbb R$ induces a natural spin structure on $E$. (This is true for orientations!) This amounts to the inverse claim that the map sending spin structures on $E$ to spin structures on $E oplus Bbb R$ (via the natural map $text{Spin}(n) to text{Spin}(n+1)$) is a bijection; and this may be verified using that spin structures are affine over the group of isomorphism classes of real line bundles, aka $H^1(W;Bbb Z/2)$.



                This argument doesn't work for arbitrary sorts of structure on the tangent bundle. For instance, one that famously doesn't work is "$W$ parallelizable implies $partial W$ parallelizable". Every disc $D^n$ is parallelizable, but the only spheres which are are $S^0, S^1, S^3$, and $S^7$. In the argument above, the way that showed up was in our descriptions of spin structures as affine over $H^1(W;Bbb Z/2)$, which is true for spin structures on rank $n$ bundles for all $n$; trivializations are affine over $[W, SO(n)]$, which depends on $n$ until $n$ is quite large.






                share|cite|improve this answer









                $endgroup$



                There are some confusions here I should clear up before answering.



                1) This is a definition of spinnable manifold, not of a spin manifold. (Similar to the difference between oriented manifold and orientable manifold.) When spin structures exist, there are $H^1(W;Bbb Z/2)$-many of them, and a spin manifold requires one such choice.



                2) If $M$ is the boundary of another manifold $W$, then it is not necessarily true that the Stiefel-Whitney classes $w_i(M) in H^i(M;Bbb Z/2)$ vanish. For instance, if $M$ has nontrivial $w_i(M)$ for $i < dim M$, then so does $M # M$, but the latter is always null-bordant. What instead is the case is that the Stiefel-Whitney numbers vanish. These are the results of all products of $w_i(M)$ that land in $H^{dim M}(M; Bbb Z/2) cong Bbb Z/2$; they are labeled by partitions of $dim M$.



                3) A vector bundle $E$ of rank at least 3 is spinnable if and only if it is trivializable over the 2-skeleton. This is not true for bundles of rank 2: there are spinnable bundles which are not trivial over the 2-skeleton. Think $TS^2$.





                As for your actual question, the point is that we have a natural isomorphism $T(partial W) oplus Bbb R cong TWbig|_{partial W}$. The easy claim is that if $W$ is spinnable, then $partial W$ is spinnable: naturality of Stiefel-Whitney classes implies that $j^*(w_i(W)) = w_i(partial W)$, where $j: partial W to W$ is the inclusion. If you already know that $w_1(W) = w_2(W) = 0$, you hence know that as well for $partial W$. This argument applies for any condition defined by the vanishing of some set of Stiefel-Whitney classes.



                To actually pin down a specific spin structure, we need to be able to argue that a spin structure on $E oplus Bbb R$ induces a natural spin structure on $E$. (This is true for orientations!) This amounts to the inverse claim that the map sending spin structures on $E$ to spin structures on $E oplus Bbb R$ (via the natural map $text{Spin}(n) to text{Spin}(n+1)$) is a bijection; and this may be verified using that spin structures are affine over the group of isomorphism classes of real line bundles, aka $H^1(W;Bbb Z/2)$.



                This argument doesn't work for arbitrary sorts of structure on the tangent bundle. For instance, one that famously doesn't work is "$W$ parallelizable implies $partial W$ parallelizable". Every disc $D^n$ is parallelizable, but the only spheres which are are $S^0, S^1, S^3$, and $S^7$. In the argument above, the way that showed up was in our descriptions of spin structures as affine over $H^1(W;Bbb Z/2)$, which is true for spin structures on rank $n$ bundles for all $n$; trivializations are affine over $[W, SO(n)]$, which depends on $n$ until $n$ is quite large.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 19:15









                Mike MillerMike Miller

                36.6k470137




                36.6k470137






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022871%2fw-spin-implies-partial-w-spin%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Ellipse (mathématiques)

                    Quarter-circle Tiles

                    Mont Emei