Intersection number and cup product












1












$begingroup$


It is known that for a closed oriented smooth manifold $M$, if A and B are oriented submanifolds of $M$, and if A and B intersect transversely, then the Poincare dual of A ∩ B is the cup product of the Poincare duals of A and B(see Ref 1).



When applying to the case where $M$ is a 4 dimensional(also closed, oriented, smooth), and $A$ and $B$ are two dimensional (oriented) submanifolds, we can have



begin{align}
([A]^* cup [B]^*)cap M=([A cap B]^*cap M
end{align}

where $[A]^*,[B]^*$ denote the Poincare dual of $A, B$, and similarly for $[Acap B]^*$ is the Poincare dual to the intersection of A and B, namely $A cap B$, which in general is a zero dimensional submanifold of $M$.



In understanding the 4 dimensional case, I meet some basic problems:



(1) What is the definition of Poincare dual of $A$ if $A$ is not a closed submanifold of $M$? For example, if $A$ is the seifert surface of a line in $S^4$, what is the Poincare dual of this seifert surface?



(2) If I want to calculate the selfintersection number of a open oriented 2-dimensional submanifold $A$, it seems that I need to take two submanifolds $A_1$ and $A_2$ which are two representatives of the cohomology class [A] and also intersect transversely, and then apply the above formula
begin{align}
([A_1]^*cup [A_2]^*)cap M=[A_1cap A_2]^*cap M
end{align}

Now if I consider $M=S^4$, $A$ is the Seifert surface of a line(an embedded one dimensional submanifold) in $S^4$, is it possible that the selfintersection number equal to one? Why, or why not?



I am not very sure whether I have made some mistake in describing my puzzles. If you guys have any idea or suggestion or references, please just post, and I am really grateful!










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    It is known that for a closed oriented smooth manifold $M$, if A and B are oriented submanifolds of $M$, and if A and B intersect transversely, then the Poincare dual of A ∩ B is the cup product of the Poincare duals of A and B(see Ref 1).



    When applying to the case where $M$ is a 4 dimensional(also closed, oriented, smooth), and $A$ and $B$ are two dimensional (oriented) submanifolds, we can have



    begin{align}
    ([A]^* cup [B]^*)cap M=([A cap B]^*cap M
    end{align}

    where $[A]^*,[B]^*$ denote the Poincare dual of $A, B$, and similarly for $[Acap B]^*$ is the Poincare dual to the intersection of A and B, namely $A cap B$, which in general is a zero dimensional submanifold of $M$.



    In understanding the 4 dimensional case, I meet some basic problems:



    (1) What is the definition of Poincare dual of $A$ if $A$ is not a closed submanifold of $M$? For example, if $A$ is the seifert surface of a line in $S^4$, what is the Poincare dual of this seifert surface?



    (2) If I want to calculate the selfintersection number of a open oriented 2-dimensional submanifold $A$, it seems that I need to take two submanifolds $A_1$ and $A_2$ which are two representatives of the cohomology class [A] and also intersect transversely, and then apply the above formula
    begin{align}
    ([A_1]^*cup [A_2]^*)cap M=[A_1cap A_2]^*cap M
    end{align}

    Now if I consider $M=S^4$, $A$ is the Seifert surface of a line(an embedded one dimensional submanifold) in $S^4$, is it possible that the selfintersection number equal to one? Why, or why not?



    I am not very sure whether I have made some mistake in describing my puzzles. If you guys have any idea or suggestion or references, please just post, and I am really grateful!










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      It is known that for a closed oriented smooth manifold $M$, if A and B are oriented submanifolds of $M$, and if A and B intersect transversely, then the Poincare dual of A ∩ B is the cup product of the Poincare duals of A and B(see Ref 1).



      When applying to the case where $M$ is a 4 dimensional(also closed, oriented, smooth), and $A$ and $B$ are two dimensional (oriented) submanifolds, we can have



      begin{align}
      ([A]^* cup [B]^*)cap M=([A cap B]^*cap M
      end{align}

      where $[A]^*,[B]^*$ denote the Poincare dual of $A, B$, and similarly for $[Acap B]^*$ is the Poincare dual to the intersection of A and B, namely $A cap B$, which in general is a zero dimensional submanifold of $M$.



      In understanding the 4 dimensional case, I meet some basic problems:



      (1) What is the definition of Poincare dual of $A$ if $A$ is not a closed submanifold of $M$? For example, if $A$ is the seifert surface of a line in $S^4$, what is the Poincare dual of this seifert surface?



      (2) If I want to calculate the selfintersection number of a open oriented 2-dimensional submanifold $A$, it seems that I need to take two submanifolds $A_1$ and $A_2$ which are two representatives of the cohomology class [A] and also intersect transversely, and then apply the above formula
      begin{align}
      ([A_1]^*cup [A_2]^*)cap M=[A_1cap A_2]^*cap M
      end{align}

      Now if I consider $M=S^4$, $A$ is the Seifert surface of a line(an embedded one dimensional submanifold) in $S^4$, is it possible that the selfintersection number equal to one? Why, or why not?



      I am not very sure whether I have made some mistake in describing my puzzles. If you guys have any idea or suggestion or references, please just post, and I am really grateful!










      share|cite|improve this question









      $endgroup$




      It is known that for a closed oriented smooth manifold $M$, if A and B are oriented submanifolds of $M$, and if A and B intersect transversely, then the Poincare dual of A ∩ B is the cup product of the Poincare duals of A and B(see Ref 1).



      When applying to the case where $M$ is a 4 dimensional(also closed, oriented, smooth), and $A$ and $B$ are two dimensional (oriented) submanifolds, we can have



      begin{align}
      ([A]^* cup [B]^*)cap M=([A cap B]^*cap M
      end{align}

      where $[A]^*,[B]^*$ denote the Poincare dual of $A, B$, and similarly for $[Acap B]^*$ is the Poincare dual to the intersection of A and B, namely $A cap B$, which in general is a zero dimensional submanifold of $M$.



      In understanding the 4 dimensional case, I meet some basic problems:



      (1) What is the definition of Poincare dual of $A$ if $A$ is not a closed submanifold of $M$? For example, if $A$ is the seifert surface of a line in $S^4$, what is the Poincare dual of this seifert surface?



      (2) If I want to calculate the selfintersection number of a open oriented 2-dimensional submanifold $A$, it seems that I need to take two submanifolds $A_1$ and $A_2$ which are two representatives of the cohomology class [A] and also intersect transversely, and then apply the above formula
      begin{align}
      ([A_1]^*cup [A_2]^*)cap M=[A_1cap A_2]^*cap M
      end{align}

      Now if I consider $M=S^4$, $A$ is the Seifert surface of a line(an embedded one dimensional submanifold) in $S^4$, is it possible that the selfintersection number equal to one? Why, or why not?



      I am not very sure whether I have made some mistake in describing my puzzles. If you guys have any idea or suggestion or references, please just post, and I am really grateful!







      algebraic-topology intersection-theory poincare-duality






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 31 '18 at 20:00









      wlnwln

      255




      255






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          (1) If $Sigma$ is a submanifold with boundary in $M$, so that $Sigma cap partial M = partial Sigma$, then $Sigma$ has a fundamental class in relative homology which you may push forward and dualizing to obtain a class in $H^2(M)$. I don't know what you would mean by the Poincare dual of something whose boundary does whatever it feels. I doubt it's what you want.



          (2) If $Sigma$ and $Sigma'$ are submanifolds of complementary dimension so that $partial Sigma cap Sigma' = varnothing$ and similarly $Sigma cap partial Sigma' = varnothing$, you may still define an intersection product (make them transverse without moving the boundary). In cohomological language (assuming the ambient manifold is closed so I don't need to say "compactly supported cohomology"), you are taking the cup product of cycles in $H^2(M, Sigma)$ and $H^2(M, Sigma')$ to obtain a class in $H^4(M, Sigma cup Sigma') cong Bbb Z$, the last isomorphism determined by an orientation on $M$. These intersection numbers will not be invariant under arbitrary homotopy. Instead, one is allowed to homotope $f_t: Sigma to M$ under the rule that $partial Sigma$ stays fixed, and $f_t(Sigma) cap partial Sigma' = varnothing$ for all $t in [0,1]$. Similarly with $Sigma'$. If you allow more general homotopies the result is unlikely to be invariant.



          If $Sigma cap Sigma'$ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore, since to undo this intersection point you need to push the boundaries off of one another. This breaks the grand rule of the previous paragraph: you can see in practice this is not a well-defined notion.



          In particular, no self-intersections.



          On a related note, if $M$ is noncompact you can define intersection numbers of properly embedded submanifolds, which are invariant under proper isotopy of each. If the pieces have boundary, again you need to fix a neighborhood if that boundary.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Actually I don't assume $Sigmacap partial M=partial Sigma$. In fact, I assume $M$ to be closed. I see in some reference(see 1) where Corollary 2.2 says for a compact s-dimensional oriented submanifold $S$(not necessary closed) of the smooth, finite type oriented manifold $M$ of dimension m, we can find a tubular neighborhood $N$ of $S$ in $M$, and define the Poincare dual of $S$ in $N$, then promote it to be the Poincare dual of $S$ in M by some extension. Actually I don't understand much about this.
            $endgroup$
            – wln
            Jan 1 at 4:11












          • $begingroup$
            (2) Sorry, what do you mean by complementary boundary? And why do you say "If Σ∩Σ′ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore." I do not understand this, could you say it in more details.
            $endgroup$
            – wln
            Jan 1 at 4:22










          • $begingroup$
            @wln Nowhere in that text is M or S allowed to have boundary, though the author doesn't state it explicitly. (Much does not make sense if they did.) Whenever the author says "compact manifold" they mean compact without boundary, and whenever they say "closed submanifold" they mean S is closed as a subset of M (this is equivalent to what I call being a "properly embedded submanifold).
            $endgroup$
            – user98602
            Jan 1 at 4:23










          • $begingroup$
            That should have said "complientary dimension". Fixed now. As for (2), already draw what happens if you try to take intersection numbers of [0,2] on the x-axis with the unit circle, as you move the arc outside the unit circle. The intersection numbers changed because you allowed one manifold to move over the boundary of the other.
            $endgroup$
            – user98602
            Jan 1 at 4:25










          • $begingroup$
            There were a few typos in the response (2) concerning what the boundary is allowed to intersect or not. Sorry about that. It's fixed now.
            $endgroup$
            – user98602
            Jan 1 at 4:30












          Your Answer








          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%2f3058003%2fintersection-number-and-cup-product%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












          $begingroup$

          (1) If $Sigma$ is a submanifold with boundary in $M$, so that $Sigma cap partial M = partial Sigma$, then $Sigma$ has a fundamental class in relative homology which you may push forward and dualizing to obtain a class in $H^2(M)$. I don't know what you would mean by the Poincare dual of something whose boundary does whatever it feels. I doubt it's what you want.



          (2) If $Sigma$ and $Sigma'$ are submanifolds of complementary dimension so that $partial Sigma cap Sigma' = varnothing$ and similarly $Sigma cap partial Sigma' = varnothing$, you may still define an intersection product (make them transverse without moving the boundary). In cohomological language (assuming the ambient manifold is closed so I don't need to say "compactly supported cohomology"), you are taking the cup product of cycles in $H^2(M, Sigma)$ and $H^2(M, Sigma')$ to obtain a class in $H^4(M, Sigma cup Sigma') cong Bbb Z$, the last isomorphism determined by an orientation on $M$. These intersection numbers will not be invariant under arbitrary homotopy. Instead, one is allowed to homotope $f_t: Sigma to M$ under the rule that $partial Sigma$ stays fixed, and $f_t(Sigma) cap partial Sigma' = varnothing$ for all $t in [0,1]$. Similarly with $Sigma'$. If you allow more general homotopies the result is unlikely to be invariant.



          If $Sigma cap Sigma'$ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore, since to undo this intersection point you need to push the boundaries off of one another. This breaks the grand rule of the previous paragraph: you can see in practice this is not a well-defined notion.



          In particular, no self-intersections.



          On a related note, if $M$ is noncompact you can define intersection numbers of properly embedded submanifolds, which are invariant under proper isotopy of each. If the pieces have boundary, again you need to fix a neighborhood if that boundary.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Actually I don't assume $Sigmacap partial M=partial Sigma$. In fact, I assume $M$ to be closed. I see in some reference(see 1) where Corollary 2.2 says for a compact s-dimensional oriented submanifold $S$(not necessary closed) of the smooth, finite type oriented manifold $M$ of dimension m, we can find a tubular neighborhood $N$ of $S$ in $M$, and define the Poincare dual of $S$ in $N$, then promote it to be the Poincare dual of $S$ in M by some extension. Actually I don't understand much about this.
            $endgroup$
            – wln
            Jan 1 at 4:11












          • $begingroup$
            (2) Sorry, what do you mean by complementary boundary? And why do you say "If Σ∩Σ′ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore." I do not understand this, could you say it in more details.
            $endgroup$
            – wln
            Jan 1 at 4:22










          • $begingroup$
            @wln Nowhere in that text is M or S allowed to have boundary, though the author doesn't state it explicitly. (Much does not make sense if they did.) Whenever the author says "compact manifold" they mean compact without boundary, and whenever they say "closed submanifold" they mean S is closed as a subset of M (this is equivalent to what I call being a "properly embedded submanifold).
            $endgroup$
            – user98602
            Jan 1 at 4:23










          • $begingroup$
            That should have said "complientary dimension". Fixed now. As for (2), already draw what happens if you try to take intersection numbers of [0,2] on the x-axis with the unit circle, as you move the arc outside the unit circle. The intersection numbers changed because you allowed one manifold to move over the boundary of the other.
            $endgroup$
            – user98602
            Jan 1 at 4:25










          • $begingroup$
            There were a few typos in the response (2) concerning what the boundary is allowed to intersect or not. Sorry about that. It's fixed now.
            $endgroup$
            – user98602
            Jan 1 at 4:30
















          2












          $begingroup$

          (1) If $Sigma$ is a submanifold with boundary in $M$, so that $Sigma cap partial M = partial Sigma$, then $Sigma$ has a fundamental class in relative homology which you may push forward and dualizing to obtain a class in $H^2(M)$. I don't know what you would mean by the Poincare dual of something whose boundary does whatever it feels. I doubt it's what you want.



          (2) If $Sigma$ and $Sigma'$ are submanifolds of complementary dimension so that $partial Sigma cap Sigma' = varnothing$ and similarly $Sigma cap partial Sigma' = varnothing$, you may still define an intersection product (make them transverse without moving the boundary). In cohomological language (assuming the ambient manifold is closed so I don't need to say "compactly supported cohomology"), you are taking the cup product of cycles in $H^2(M, Sigma)$ and $H^2(M, Sigma')$ to obtain a class in $H^4(M, Sigma cup Sigma') cong Bbb Z$, the last isomorphism determined by an orientation on $M$. These intersection numbers will not be invariant under arbitrary homotopy. Instead, one is allowed to homotope $f_t: Sigma to M$ under the rule that $partial Sigma$ stays fixed, and $f_t(Sigma) cap partial Sigma' = varnothing$ for all $t in [0,1]$. Similarly with $Sigma'$. If you allow more general homotopies the result is unlikely to be invariant.



          If $Sigma cap Sigma'$ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore, since to undo this intersection point you need to push the boundaries off of one another. This breaks the grand rule of the previous paragraph: you can see in practice this is not a well-defined notion.



          In particular, no self-intersections.



          On a related note, if $M$ is noncompact you can define intersection numbers of properly embedded submanifolds, which are invariant under proper isotopy of each. If the pieces have boundary, again you need to fix a neighborhood if that boundary.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Actually I don't assume $Sigmacap partial M=partial Sigma$. In fact, I assume $M$ to be closed. I see in some reference(see 1) where Corollary 2.2 says for a compact s-dimensional oriented submanifold $S$(not necessary closed) of the smooth, finite type oriented manifold $M$ of dimension m, we can find a tubular neighborhood $N$ of $S$ in $M$, and define the Poincare dual of $S$ in $N$, then promote it to be the Poincare dual of $S$ in M by some extension. Actually I don't understand much about this.
            $endgroup$
            – wln
            Jan 1 at 4:11












          • $begingroup$
            (2) Sorry, what do you mean by complementary boundary? And why do you say "If Σ∩Σ′ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore." I do not understand this, could you say it in more details.
            $endgroup$
            – wln
            Jan 1 at 4:22










          • $begingroup$
            @wln Nowhere in that text is M or S allowed to have boundary, though the author doesn't state it explicitly. (Much does not make sense if they did.) Whenever the author says "compact manifold" they mean compact without boundary, and whenever they say "closed submanifold" they mean S is closed as a subset of M (this is equivalent to what I call being a "properly embedded submanifold).
            $endgroup$
            – user98602
            Jan 1 at 4:23










          • $begingroup$
            That should have said "complientary dimension". Fixed now. As for (2), already draw what happens if you try to take intersection numbers of [0,2] on the x-axis with the unit circle, as you move the arc outside the unit circle. The intersection numbers changed because you allowed one manifold to move over the boundary of the other.
            $endgroup$
            – user98602
            Jan 1 at 4:25










          • $begingroup$
            There were a few typos in the response (2) concerning what the boundary is allowed to intersect or not. Sorry about that. It's fixed now.
            $endgroup$
            – user98602
            Jan 1 at 4:30














          2












          2








          2





          $begingroup$

          (1) If $Sigma$ is a submanifold with boundary in $M$, so that $Sigma cap partial M = partial Sigma$, then $Sigma$ has a fundamental class in relative homology which you may push forward and dualizing to obtain a class in $H^2(M)$. I don't know what you would mean by the Poincare dual of something whose boundary does whatever it feels. I doubt it's what you want.



          (2) If $Sigma$ and $Sigma'$ are submanifolds of complementary dimension so that $partial Sigma cap Sigma' = varnothing$ and similarly $Sigma cap partial Sigma' = varnothing$, you may still define an intersection product (make them transverse without moving the boundary). In cohomological language (assuming the ambient manifold is closed so I don't need to say "compactly supported cohomology"), you are taking the cup product of cycles in $H^2(M, Sigma)$ and $H^2(M, Sigma')$ to obtain a class in $H^4(M, Sigma cup Sigma') cong Bbb Z$, the last isomorphism determined by an orientation on $M$. These intersection numbers will not be invariant under arbitrary homotopy. Instead, one is allowed to homotope $f_t: Sigma to M$ under the rule that $partial Sigma$ stays fixed, and $f_t(Sigma) cap partial Sigma' = varnothing$ for all $t in [0,1]$. Similarly with $Sigma'$. If you allow more general homotopies the result is unlikely to be invariant.



          If $Sigma cap Sigma'$ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore, since to undo this intersection point you need to push the boundaries off of one another. This breaks the grand rule of the previous paragraph: you can see in practice this is not a well-defined notion.



          In particular, no self-intersections.



          On a related note, if $M$ is noncompact you can define intersection numbers of properly embedded submanifolds, which are invariant under proper isotopy of each. If the pieces have boundary, again you need to fix a neighborhood if that boundary.






          share|cite|improve this answer











          $endgroup$



          (1) If $Sigma$ is a submanifold with boundary in $M$, so that $Sigma cap partial M = partial Sigma$, then $Sigma$ has a fundamental class in relative homology which you may push forward and dualizing to obtain a class in $H^2(M)$. I don't know what you would mean by the Poincare dual of something whose boundary does whatever it feels. I doubt it's what you want.



          (2) If $Sigma$ and $Sigma'$ are submanifolds of complementary dimension so that $partial Sigma cap Sigma' = varnothing$ and similarly $Sigma cap partial Sigma' = varnothing$, you may still define an intersection product (make them transverse without moving the boundary). In cohomological language (assuming the ambient manifold is closed so I don't need to say "compactly supported cohomology"), you are taking the cup product of cycles in $H^2(M, Sigma)$ and $H^2(M, Sigma')$ to obtain a class in $H^4(M, Sigma cup Sigma') cong Bbb Z$, the last isomorphism determined by an orientation on $M$. These intersection numbers will not be invariant under arbitrary homotopy. Instead, one is allowed to homotope $f_t: Sigma to M$ under the rule that $partial Sigma$ stays fixed, and $f_t(Sigma) cap partial Sigma' = varnothing$ for all $t in [0,1]$. Similarly with $Sigma'$. If you allow more general homotopies the result is unlikely to be invariant.



          If $Sigma cap Sigma'$ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore, since to undo this intersection point you need to push the boundaries off of one another. This breaks the grand rule of the previous paragraph: you can see in practice this is not a well-defined notion.



          In particular, no self-intersections.



          On a related note, if $M$ is noncompact you can define intersection numbers of properly embedded submanifolds, which are invariant under proper isotopy of each. If the pieces have boundary, again you need to fix a neighborhood if that boundary.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 1 at 4:29

























          answered Dec 31 '18 at 21:48







          user98602



















          • $begingroup$
            Thanks. Actually I don't assume $Sigmacap partial M=partial Sigma$. In fact, I assume $M$ to be closed. I see in some reference(see 1) where Corollary 2.2 says for a compact s-dimensional oriented submanifold $S$(not necessary closed) of the smooth, finite type oriented manifold $M$ of dimension m, we can find a tubular neighborhood $N$ of $S$ in $M$, and define the Poincare dual of $S$ in $N$, then promote it to be the Poincare dual of $S$ in M by some extension. Actually I don't understand much about this.
            $endgroup$
            – wln
            Jan 1 at 4:11












          • $begingroup$
            (2) Sorry, what do you mean by complementary boundary? And why do you say "If Σ∩Σ′ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore." I do not understand this, could you say it in more details.
            $endgroup$
            – wln
            Jan 1 at 4:22










          • $begingroup$
            @wln Nowhere in that text is M or S allowed to have boundary, though the author doesn't state it explicitly. (Much does not make sense if they did.) Whenever the author says "compact manifold" they mean compact without boundary, and whenever they say "closed submanifold" they mean S is closed as a subset of M (this is equivalent to what I call being a "properly embedded submanifold).
            $endgroup$
            – user98602
            Jan 1 at 4:23










          • $begingroup$
            That should have said "complientary dimension". Fixed now. As for (2), already draw what happens if you try to take intersection numbers of [0,2] on the x-axis with the unit circle, as you move the arc outside the unit circle. The intersection numbers changed because you allowed one manifold to move over the boundary of the other.
            $endgroup$
            – user98602
            Jan 1 at 4:25










          • $begingroup$
            There were a few typos in the response (2) concerning what the boundary is allowed to intersect or not. Sorry about that. It's fixed now.
            $endgroup$
            – user98602
            Jan 1 at 4:30


















          • $begingroup$
            Thanks. Actually I don't assume $Sigmacap partial M=partial Sigma$. In fact, I assume $M$ to be closed. I see in some reference(see 1) where Corollary 2.2 says for a compact s-dimensional oriented submanifold $S$(not necessary closed) of the smooth, finite type oriented manifold $M$ of dimension m, we can find a tubular neighborhood $N$ of $S$ in $M$, and define the Poincare dual of $S$ in $N$, then promote it to be the Poincare dual of $S$ in M by some extension. Actually I don't understand much about this.
            $endgroup$
            – wln
            Jan 1 at 4:11












          • $begingroup$
            (2) Sorry, what do you mean by complementary boundary? And why do you say "If Σ∩Σ′ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore." I do not understand this, could you say it in more details.
            $endgroup$
            – wln
            Jan 1 at 4:22










          • $begingroup$
            @wln Nowhere in that text is M or S allowed to have boundary, though the author doesn't state it explicitly. (Much does not make sense if they did.) Whenever the author says "compact manifold" they mean compact without boundary, and whenever they say "closed submanifold" they mean S is closed as a subset of M (this is equivalent to what I call being a "properly embedded submanifold).
            $endgroup$
            – user98602
            Jan 1 at 4:23










          • $begingroup$
            That should have said "complientary dimension". Fixed now. As for (2), already draw what happens if you try to take intersection numbers of [0,2] on the x-axis with the unit circle, as you move the arc outside the unit circle. The intersection numbers changed because you allowed one manifold to move over the boundary of the other.
            $endgroup$
            – user98602
            Jan 1 at 4:25










          • $begingroup$
            There were a few typos in the response (2) concerning what the boundary is allowed to intersect or not. Sorry about that. It's fixed now.
            $endgroup$
            – user98602
            Jan 1 at 4:30
















          $begingroup$
          Thanks. Actually I don't assume $Sigmacap partial M=partial Sigma$. In fact, I assume $M$ to be closed. I see in some reference(see 1) where Corollary 2.2 says for a compact s-dimensional oriented submanifold $S$(not necessary closed) of the smooth, finite type oriented manifold $M$ of dimension m, we can find a tubular neighborhood $N$ of $S$ in $M$, and define the Poincare dual of $S$ in $N$, then promote it to be the Poincare dual of $S$ in M by some extension. Actually I don't understand much about this.
          $endgroup$
          – wln
          Jan 1 at 4:11






          $begingroup$
          Thanks. Actually I don't assume $Sigmacap partial M=partial Sigma$. In fact, I assume $M$ to be closed. I see in some reference(see 1) where Corollary 2.2 says for a compact s-dimensional oriented submanifold $S$(not necessary closed) of the smooth, finite type oriented manifold $M$ of dimension m, we can find a tubular neighborhood $N$ of $S$ in $M$, and define the Poincare dual of $S$ in $N$, then promote it to be the Poincare dual of $S$ in M by some extension. Actually I don't understand much about this.
          $endgroup$
          – wln
          Jan 1 at 4:11














          $begingroup$
          (2) Sorry, what do you mean by complementary boundary? And why do you say "If Σ∩Σ′ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore." I do not understand this, could you say it in more details.
          $endgroup$
          – wln
          Jan 1 at 4:22




          $begingroup$
          (2) Sorry, what do you mean by complementary boundary? And why do you say "If Σ∩Σ′ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore." I do not understand this, could you say it in more details.
          $endgroup$
          – wln
          Jan 1 at 4:22












          $begingroup$
          @wln Nowhere in that text is M or S allowed to have boundary, though the author doesn't state it explicitly. (Much does not make sense if they did.) Whenever the author says "compact manifold" they mean compact without boundary, and whenever they say "closed submanifold" they mean S is closed as a subset of M (this is equivalent to what I call being a "properly embedded submanifold).
          $endgroup$
          – user98602
          Jan 1 at 4:23




          $begingroup$
          @wln Nowhere in that text is M or S allowed to have boundary, though the author doesn't state it explicitly. (Much does not make sense if they did.) Whenever the author says "compact manifold" they mean compact without boundary, and whenever they say "closed submanifold" they mean S is closed as a subset of M (this is equivalent to what I call being a "properly embedded submanifold).
          $endgroup$
          – user98602
          Jan 1 at 4:23












          $begingroup$
          That should have said "complientary dimension". Fixed now. As for (2), already draw what happens if you try to take intersection numbers of [0,2] on the x-axis with the unit circle, as you move the arc outside the unit circle. The intersection numbers changed because you allowed one manifold to move over the boundary of the other.
          $endgroup$
          – user98602
          Jan 1 at 4:25




          $begingroup$
          That should have said "complientary dimension". Fixed now. As for (2), already draw what happens if you try to take intersection numbers of [0,2] on the x-axis with the unit circle, as you move the arc outside the unit circle. The intersection numbers changed because you allowed one manifold to move over the boundary of the other.
          $endgroup$
          – user98602
          Jan 1 at 4:25












          $begingroup$
          There were a few typos in the response (2) concerning what the boundary is allowed to intersect or not. Sorry about that. It's fixed now.
          $endgroup$
          – user98602
          Jan 1 at 4:30




          $begingroup$
          There were a few typos in the response (2) concerning what the boundary is allowed to intersect or not. Sorry about that. It's fixed now.
          $endgroup$
          – user98602
          Jan 1 at 4:30


















          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%2f3058003%2fintersection-number-and-cup-product%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

          mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

          How to change which sound is reproduced for terminal bell?

          Can I use Tabulator js library in my java Spring + Thymeleaf project?