Is the connected sum of complex manifolds also complex?












13












$begingroup$


Let $M$ and $N$ be real manifolds of dimension $n$ which happen to admit complex structures (so that necessarily $n=2k$ and both are orientable). Then does their connected sum $M# N$ also admit a complex structure?



This is true for $n=2$, because every oriented $2$-dimensional topological manifold admits a complex structure. However, this operation doesn't seem to be compatible with the information given by the structure; in particular we would want $Msqcup N$ to at least be almost-complex cobordant to $M# N$, or that these two manifolds should have the same Chern numbers. We have
$$c_1[Msqcup N]=langle c_1(tau Msqcuptau N),[Msqcup N]rangle = langle c_1(tau M),[M]rangle + langle c_1(tau N),[N]rangle = c_1[M]+c_1[N] $$
But then recall that the first Chern class of a complex rank 1 bundle $V$ is the Euler class of its realification, and so the first Chern number of a complex 1-manifold is its Euler characteristic. Thus
$$c_1[M# N] = chi(M#N)=chi(M)+chi(N)-2 neq c_1[M] + c_1[N] $$
So it seems like it's possible that the fact that complex $1$-manifolds are closed under connected sum could be a coincidence.



In higher dimensions there are no complex structures on $S^{2k}$ (except MAYBE $S^6$). But, in the positive direction, there ARE these complex Calabi-Echkmann manifolds homeomorphic to $S^{2k+1}times S^{2l+1}$. So does anyone know if, for example, the connected sum of $g$ copies of $S^5times S^5$ admits a complex structure for any $g>1$?










share|cite|improve this question









$endgroup$

















    13












    $begingroup$


    Let $M$ and $N$ be real manifolds of dimension $n$ which happen to admit complex structures (so that necessarily $n=2k$ and both are orientable). Then does their connected sum $M# N$ also admit a complex structure?



    This is true for $n=2$, because every oriented $2$-dimensional topological manifold admits a complex structure. However, this operation doesn't seem to be compatible with the information given by the structure; in particular we would want $Msqcup N$ to at least be almost-complex cobordant to $M# N$, or that these two manifolds should have the same Chern numbers. We have
    $$c_1[Msqcup N]=langle c_1(tau Msqcuptau N),[Msqcup N]rangle = langle c_1(tau M),[M]rangle + langle c_1(tau N),[N]rangle = c_1[M]+c_1[N] $$
    But then recall that the first Chern class of a complex rank 1 bundle $V$ is the Euler class of its realification, and so the first Chern number of a complex 1-manifold is its Euler characteristic. Thus
    $$c_1[M# N] = chi(M#N)=chi(M)+chi(N)-2 neq c_1[M] + c_1[N] $$
    So it seems like it's possible that the fact that complex $1$-manifolds are closed under connected sum could be a coincidence.



    In higher dimensions there are no complex structures on $S^{2k}$ (except MAYBE $S^6$). But, in the positive direction, there ARE these complex Calabi-Echkmann manifolds homeomorphic to $S^{2k+1}times S^{2l+1}$. So does anyone know if, for example, the connected sum of $g$ copies of $S^5times S^5$ admits a complex structure for any $g>1$?










    share|cite|improve this question









    $endgroup$















      13












      13








      13


      5



      $begingroup$


      Let $M$ and $N$ be real manifolds of dimension $n$ which happen to admit complex structures (so that necessarily $n=2k$ and both are orientable). Then does their connected sum $M# N$ also admit a complex structure?



      This is true for $n=2$, because every oriented $2$-dimensional topological manifold admits a complex structure. However, this operation doesn't seem to be compatible with the information given by the structure; in particular we would want $Msqcup N$ to at least be almost-complex cobordant to $M# N$, or that these two manifolds should have the same Chern numbers. We have
      $$c_1[Msqcup N]=langle c_1(tau Msqcuptau N),[Msqcup N]rangle = langle c_1(tau M),[M]rangle + langle c_1(tau N),[N]rangle = c_1[M]+c_1[N] $$
      But then recall that the first Chern class of a complex rank 1 bundle $V$ is the Euler class of its realification, and so the first Chern number of a complex 1-manifold is its Euler characteristic. Thus
      $$c_1[M# N] = chi(M#N)=chi(M)+chi(N)-2 neq c_1[M] + c_1[N] $$
      So it seems like it's possible that the fact that complex $1$-manifolds are closed under connected sum could be a coincidence.



      In higher dimensions there are no complex structures on $S^{2k}$ (except MAYBE $S^6$). But, in the positive direction, there ARE these complex Calabi-Echkmann manifolds homeomorphic to $S^{2k+1}times S^{2l+1}$. So does anyone know if, for example, the connected sum of $g$ copies of $S^5times S^5$ admits a complex structure for any $g>1$?










      share|cite|improve this question









      $endgroup$




      Let $M$ and $N$ be real manifolds of dimension $n$ which happen to admit complex structures (so that necessarily $n=2k$ and both are orientable). Then does their connected sum $M# N$ also admit a complex structure?



      This is true for $n=2$, because every oriented $2$-dimensional topological manifold admits a complex structure. However, this operation doesn't seem to be compatible with the information given by the structure; in particular we would want $Msqcup N$ to at least be almost-complex cobordant to $M# N$, or that these two manifolds should have the same Chern numbers. We have
      $$c_1[Msqcup N]=langle c_1(tau Msqcuptau N),[Msqcup N]rangle = langle c_1(tau M),[M]rangle + langle c_1(tau N),[N]rangle = c_1[M]+c_1[N] $$
      But then recall that the first Chern class of a complex rank 1 bundle $V$ is the Euler class of its realification, and so the first Chern number of a complex 1-manifold is its Euler characteristic. Thus
      $$c_1[M# N] = chi(M#N)=chi(M)+chi(N)-2 neq c_1[M] + c_1[N] $$
      So it seems like it's possible that the fact that complex $1$-manifolds are closed under connected sum could be a coincidence.



      In higher dimensions there are no complex structures on $S^{2k}$ (except MAYBE $S^6$). But, in the positive direction, there ARE these complex Calabi-Echkmann manifolds homeomorphic to $S^{2k+1}times S^{2l+1}$. So does anyone know if, for example, the connected sum of $g$ copies of $S^5times S^5$ admits a complex structure for any $g>1$?







      manifolds differential-topology complex-manifolds






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 27 '15 at 17:21









      WilliamWilliam

      2,0601120




      2,0601120






















          2 Answers
          2






          active

          oldest

          votes


















          13












          $begingroup$

          EDIT: I am greatly appreciative of Aleksandar Milivojevic for pointing out the first paragraph of my original answer is false, and therefore does not give a proof of the following two paragraphs. (For instance, $K3 # overline{K3}$ supports no complex structure: Aleksandar kindly explained to me that this has signature zero but Euler characteristic $46$, and $chi + sigma$ is always divisible by $4$ on an almost complex 4-manifold.



          I leave these here for posterity, separated from the correct content by lines. It remains an interesting question whether or not $M # N$ can support a complex structure when $M$ and $N$ are singly-even dimensional complex manifolds. There is no obstruction coming from the nonexistence of an almost complex structure, so it is not totally impossible.





          Let $M$ and $N$ be complex manifolds. Because we have a local form (there is a neighborhood of a point isomorphic to the the complex unit disc) you can verify that $M # overline N$ has a natural complex structure - just line up the copies of $Bbb C^n setminus {0}$ you're gluing together so the almost complex structures match up. Then the Nijenhuis tensor still vanishes.



          So, if $N$ supports a complex structure, does $overline N$? This would mean that $M # N$ supports a complex structure. If $text{dim}(N) = 4n+2$, then this is true: if $J$ is your (integrable almost) complex structure, then the complex conjugate $overline J$ gives a complex structure on $overline N$. The reason this doesn't work in dimension $4n$ is because $overline J$ induces the same orientation as $J$!



          So I wouldn't quite call it a fluke. It's true for every $4n+2$-dimensional complex manifold, not just $2$-dimensional ones. But it's not true in every dimension.





          For 4-manifolds, it is a theorem of Wu that $M$ admits an almost complex structure with $c_1(J) = c in H^2(M;Bbb Z)$ if and only if $c$ reduces to $w_2$ mod 2, and $c^2 = 3sigma + 2chi$. You can prove using this criterion that a most two of $M, N$, and $M # N$ can admit almost complex structures; in particular, $Bbb{CP}^2 # Bbb{CP}^2$ cannot admit an almost complex structure.



          I don't know about higher dimensions $4n$. It might be wise to think about almost complex structures instead of complex ones first, because those are easier to work with and there are no known high dimensional (complex dimension $> 2$) examples of almost complex manifolds that do not admit a complex structure.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Also, a connected sum of a nonsimply connected closed 4-manifold with itself never has a complex structure.
            $endgroup$
            – Moishe Cohen
            Aug 27 '15 at 18:10










          • $begingroup$
            Thanks, this was a very informative answer. It's hard to get information about this with Google searches. Also, your answer verifies that, for example, $#^g S^5times S^5$ should admit a complex structure for all $g$ since $10=4(2)+2$.
            $endgroup$
            – William
            Aug 28 '15 at 1:29



















          11












          $begingroup$

          $X = Bbb{CP}^4 #Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is possible in all dimensions, but probably becomes absurd very quickly. The 8-dimensional version is here. This is what we will use.



          Preliminaries: the cohomology ring of $X$ is the direct sum of two copies of the cohomology ring of $Bbb{CP}^4$, identifying the unit and volume form in each copy. Denote the two generators of $H^2(X;Bbb Z)$ as $x_1, x_2$.



          1) Wu classes. Because we know precisely what the cohomology ring is, it's easy to calculate this. In particular, $Sq^2(x_1^3) = x_1^4$, and $Sq^2(x_2^3) = x_2^4$, so $v_2 = x_1+x_2$. Similarly we see that $Sq^4(x_1^2) = x_1^4, Sq^4(x_2^2)=x_2^4$, so $v_4 = x_1^2+x_2^2$. Trivially $v_6 = 0$.



          2) Stiefel-Whitney classes. Because the total SW-class $w$ satisfies $Sq(v)$, where $Sq$ is the total Steenrod square and $v$ is the total Wu class, we obtain $w_2 = v_2 = x_1+x_2$, $w_4 = v_2^2+v_4 = 0$, $w_6 = Sq^2(v_4) = 0$.



          3) Pontryagin classes. We can give $X$ a cell structure such that its 4-skeleton is $Bbb{CP}^2 vee Bbb{CP}^2$. Because $X_4 hookrightarrow X$ induces an isomorphism on $H^4$, the Pontryagin class is determined by the restriction of the classifying map $X to BSO(8)$ to $X_4$. We can choose this classifying map to be of the form $f vee f$, where $f$ is the restriction of the classifying map of $Bbb{CP}^4$ to its 4-skeleton. Because we know $p_1(Bbb{CP}^4) = 5x_1^2$, we see that $$p_1(X) = 5x_1^2+5x_2^2.$$
          Now the Hirzebruch signature theorem tells us that $p_2 = 20x_1^4$.



          Now let's plug into the theorem. Our cohomology classes $u, v$ would have to be of the form $u = (2k+1)x_1 + (2ell+1)x_2$, and $v = 2mx_1^3 + 2nx_2^3$, where $m,n$ have the same parity (by b, c).



          Then because $chi(M) = 16$, d) takes the form $$128 = 80 + 32(km+ell n) + 16(m+n) - (2k+1)^4 - (2ell+1)^4 + 10(2k+1)^2 + 10(2ell+1)^2 - 50.$$



          Simplifying we get $$80 = 32(km + ell n - k^3 - ell^3 + k^2 + ell^2) + 16(k+ell - k^4 - ell^4 +m +n)$$
          Reducing mod 32 and remembering that $m+n$ is even we obtain $$16 equiv 16(k+ell - k^4- ell^4) mod 32$$ But $k+ell$ is odd iff $k^4 + ell^4$ is odd, so the right side is $0 mod 32$. This is a contradiction, as desired.



          That this was so much work suggests that this is, uh, the wrong approach to prove that $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ never supports a complex structure. Maybe someone more gifted with obstruction theory than I can prove this.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You're right, I can see how this would get cumbersome in higher dimensions (to say the least), but it was a nice computation. Also, I wasn't aware of these theorems giving criteria for the existence of almost-complex structures in terms of characteristic classes, that's very nice.
            $endgroup$
            – William
            Sep 23 '15 at 21:25










          • $begingroup$
            @you: I posted a mathoverflow question asking if anybody could generalize this result here. No answer yet, unfortunately.
            $endgroup$
            – Mike Miller
            Sep 23 '15 at 22:09











          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%2f1411694%2fis-the-connected-sum-of-complex-manifolds-also-complex%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          13












          $begingroup$

          EDIT: I am greatly appreciative of Aleksandar Milivojevic for pointing out the first paragraph of my original answer is false, and therefore does not give a proof of the following two paragraphs. (For instance, $K3 # overline{K3}$ supports no complex structure: Aleksandar kindly explained to me that this has signature zero but Euler characteristic $46$, and $chi + sigma$ is always divisible by $4$ on an almost complex 4-manifold.



          I leave these here for posterity, separated from the correct content by lines. It remains an interesting question whether or not $M # N$ can support a complex structure when $M$ and $N$ are singly-even dimensional complex manifolds. There is no obstruction coming from the nonexistence of an almost complex structure, so it is not totally impossible.





          Let $M$ and $N$ be complex manifolds. Because we have a local form (there is a neighborhood of a point isomorphic to the the complex unit disc) you can verify that $M # overline N$ has a natural complex structure - just line up the copies of $Bbb C^n setminus {0}$ you're gluing together so the almost complex structures match up. Then the Nijenhuis tensor still vanishes.



          So, if $N$ supports a complex structure, does $overline N$? This would mean that $M # N$ supports a complex structure. If $text{dim}(N) = 4n+2$, then this is true: if $J$ is your (integrable almost) complex structure, then the complex conjugate $overline J$ gives a complex structure on $overline N$. The reason this doesn't work in dimension $4n$ is because $overline J$ induces the same orientation as $J$!



          So I wouldn't quite call it a fluke. It's true for every $4n+2$-dimensional complex manifold, not just $2$-dimensional ones. But it's not true in every dimension.





          For 4-manifolds, it is a theorem of Wu that $M$ admits an almost complex structure with $c_1(J) = c in H^2(M;Bbb Z)$ if and only if $c$ reduces to $w_2$ mod 2, and $c^2 = 3sigma + 2chi$. You can prove using this criterion that a most two of $M, N$, and $M # N$ can admit almost complex structures; in particular, $Bbb{CP}^2 # Bbb{CP}^2$ cannot admit an almost complex structure.



          I don't know about higher dimensions $4n$. It might be wise to think about almost complex structures instead of complex ones first, because those are easier to work with and there are no known high dimensional (complex dimension $> 2$) examples of almost complex manifolds that do not admit a complex structure.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Also, a connected sum of a nonsimply connected closed 4-manifold with itself never has a complex structure.
            $endgroup$
            – Moishe Cohen
            Aug 27 '15 at 18:10










          • $begingroup$
            Thanks, this was a very informative answer. It's hard to get information about this with Google searches. Also, your answer verifies that, for example, $#^g S^5times S^5$ should admit a complex structure for all $g$ since $10=4(2)+2$.
            $endgroup$
            – William
            Aug 28 '15 at 1:29
















          13












          $begingroup$

          EDIT: I am greatly appreciative of Aleksandar Milivojevic for pointing out the first paragraph of my original answer is false, and therefore does not give a proof of the following two paragraphs. (For instance, $K3 # overline{K3}$ supports no complex structure: Aleksandar kindly explained to me that this has signature zero but Euler characteristic $46$, and $chi + sigma$ is always divisible by $4$ on an almost complex 4-manifold.



          I leave these here for posterity, separated from the correct content by lines. It remains an interesting question whether or not $M # N$ can support a complex structure when $M$ and $N$ are singly-even dimensional complex manifolds. There is no obstruction coming from the nonexistence of an almost complex structure, so it is not totally impossible.





          Let $M$ and $N$ be complex manifolds. Because we have a local form (there is a neighborhood of a point isomorphic to the the complex unit disc) you can verify that $M # overline N$ has a natural complex structure - just line up the copies of $Bbb C^n setminus {0}$ you're gluing together so the almost complex structures match up. Then the Nijenhuis tensor still vanishes.



          So, if $N$ supports a complex structure, does $overline N$? This would mean that $M # N$ supports a complex structure. If $text{dim}(N) = 4n+2$, then this is true: if $J$ is your (integrable almost) complex structure, then the complex conjugate $overline J$ gives a complex structure on $overline N$. The reason this doesn't work in dimension $4n$ is because $overline J$ induces the same orientation as $J$!



          So I wouldn't quite call it a fluke. It's true for every $4n+2$-dimensional complex manifold, not just $2$-dimensional ones. But it's not true in every dimension.





          For 4-manifolds, it is a theorem of Wu that $M$ admits an almost complex structure with $c_1(J) = c in H^2(M;Bbb Z)$ if and only if $c$ reduces to $w_2$ mod 2, and $c^2 = 3sigma + 2chi$. You can prove using this criterion that a most two of $M, N$, and $M # N$ can admit almost complex structures; in particular, $Bbb{CP}^2 # Bbb{CP}^2$ cannot admit an almost complex structure.



          I don't know about higher dimensions $4n$. It might be wise to think about almost complex structures instead of complex ones first, because those are easier to work with and there are no known high dimensional (complex dimension $> 2$) examples of almost complex manifolds that do not admit a complex structure.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Also, a connected sum of a nonsimply connected closed 4-manifold with itself never has a complex structure.
            $endgroup$
            – Moishe Cohen
            Aug 27 '15 at 18:10










          • $begingroup$
            Thanks, this was a very informative answer. It's hard to get information about this with Google searches. Also, your answer verifies that, for example, $#^g S^5times S^5$ should admit a complex structure for all $g$ since $10=4(2)+2$.
            $endgroup$
            – William
            Aug 28 '15 at 1:29














          13












          13








          13





          $begingroup$

          EDIT: I am greatly appreciative of Aleksandar Milivojevic for pointing out the first paragraph of my original answer is false, and therefore does not give a proof of the following two paragraphs. (For instance, $K3 # overline{K3}$ supports no complex structure: Aleksandar kindly explained to me that this has signature zero but Euler characteristic $46$, and $chi + sigma$ is always divisible by $4$ on an almost complex 4-manifold.



          I leave these here for posterity, separated from the correct content by lines. It remains an interesting question whether or not $M # N$ can support a complex structure when $M$ and $N$ are singly-even dimensional complex manifolds. There is no obstruction coming from the nonexistence of an almost complex structure, so it is not totally impossible.





          Let $M$ and $N$ be complex manifolds. Because we have a local form (there is a neighborhood of a point isomorphic to the the complex unit disc) you can verify that $M # overline N$ has a natural complex structure - just line up the copies of $Bbb C^n setminus {0}$ you're gluing together so the almost complex structures match up. Then the Nijenhuis tensor still vanishes.



          So, if $N$ supports a complex structure, does $overline N$? This would mean that $M # N$ supports a complex structure. If $text{dim}(N) = 4n+2$, then this is true: if $J$ is your (integrable almost) complex structure, then the complex conjugate $overline J$ gives a complex structure on $overline N$. The reason this doesn't work in dimension $4n$ is because $overline J$ induces the same orientation as $J$!



          So I wouldn't quite call it a fluke. It's true for every $4n+2$-dimensional complex manifold, not just $2$-dimensional ones. But it's not true in every dimension.





          For 4-manifolds, it is a theorem of Wu that $M$ admits an almost complex structure with $c_1(J) = c in H^2(M;Bbb Z)$ if and only if $c$ reduces to $w_2$ mod 2, and $c^2 = 3sigma + 2chi$. You can prove using this criterion that a most two of $M, N$, and $M # N$ can admit almost complex structures; in particular, $Bbb{CP}^2 # Bbb{CP}^2$ cannot admit an almost complex structure.



          I don't know about higher dimensions $4n$. It might be wise to think about almost complex structures instead of complex ones first, because those are easier to work with and there are no known high dimensional (complex dimension $> 2$) examples of almost complex manifolds that do not admit a complex structure.






          share|cite|improve this answer











          $endgroup$



          EDIT: I am greatly appreciative of Aleksandar Milivojevic for pointing out the first paragraph of my original answer is false, and therefore does not give a proof of the following two paragraphs. (For instance, $K3 # overline{K3}$ supports no complex structure: Aleksandar kindly explained to me that this has signature zero but Euler characteristic $46$, and $chi + sigma$ is always divisible by $4$ on an almost complex 4-manifold.



          I leave these here for posterity, separated from the correct content by lines. It remains an interesting question whether or not $M # N$ can support a complex structure when $M$ and $N$ are singly-even dimensional complex manifolds. There is no obstruction coming from the nonexistence of an almost complex structure, so it is not totally impossible.





          Let $M$ and $N$ be complex manifolds. Because we have a local form (there is a neighborhood of a point isomorphic to the the complex unit disc) you can verify that $M # overline N$ has a natural complex structure - just line up the copies of $Bbb C^n setminus {0}$ you're gluing together so the almost complex structures match up. Then the Nijenhuis tensor still vanishes.



          So, if $N$ supports a complex structure, does $overline N$? This would mean that $M # N$ supports a complex structure. If $text{dim}(N) = 4n+2$, then this is true: if $J$ is your (integrable almost) complex structure, then the complex conjugate $overline J$ gives a complex structure on $overline N$. The reason this doesn't work in dimension $4n$ is because $overline J$ induces the same orientation as $J$!



          So I wouldn't quite call it a fluke. It's true for every $4n+2$-dimensional complex manifold, not just $2$-dimensional ones. But it's not true in every dimension.





          For 4-manifolds, it is a theorem of Wu that $M$ admits an almost complex structure with $c_1(J) = c in H^2(M;Bbb Z)$ if and only if $c$ reduces to $w_2$ mod 2, and $c^2 = 3sigma + 2chi$. You can prove using this criterion that a most two of $M, N$, and $M # N$ can admit almost complex structures; in particular, $Bbb{CP}^2 # Bbb{CP}^2$ cannot admit an almost complex structure.



          I don't know about higher dimensions $4n$. It might be wise to think about almost complex structures instead of complex ones first, because those are easier to work with and there are no known high dimensional (complex dimension $> 2$) examples of almost complex manifolds that do not admit a complex structure.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 8:59

























          answered Aug 27 '15 at 17:57









          Mike MillerMike Miller

          37.3k472139




          37.3k472139








          • 1




            $begingroup$
            Also, a connected sum of a nonsimply connected closed 4-manifold with itself never has a complex structure.
            $endgroup$
            – Moishe Cohen
            Aug 27 '15 at 18:10










          • $begingroup$
            Thanks, this was a very informative answer. It's hard to get information about this with Google searches. Also, your answer verifies that, for example, $#^g S^5times S^5$ should admit a complex structure for all $g$ since $10=4(2)+2$.
            $endgroup$
            – William
            Aug 28 '15 at 1:29














          • 1




            $begingroup$
            Also, a connected sum of a nonsimply connected closed 4-manifold with itself never has a complex structure.
            $endgroup$
            – Moishe Cohen
            Aug 27 '15 at 18:10










          • $begingroup$
            Thanks, this was a very informative answer. It's hard to get information about this with Google searches. Also, your answer verifies that, for example, $#^g S^5times S^5$ should admit a complex structure for all $g$ since $10=4(2)+2$.
            $endgroup$
            – William
            Aug 28 '15 at 1:29








          1




          1




          $begingroup$
          Also, a connected sum of a nonsimply connected closed 4-manifold with itself never has a complex structure.
          $endgroup$
          – Moishe Cohen
          Aug 27 '15 at 18:10




          $begingroup$
          Also, a connected sum of a nonsimply connected closed 4-manifold with itself never has a complex structure.
          $endgroup$
          – Moishe Cohen
          Aug 27 '15 at 18:10












          $begingroup$
          Thanks, this was a very informative answer. It's hard to get information about this with Google searches. Also, your answer verifies that, for example, $#^g S^5times S^5$ should admit a complex structure for all $g$ since $10=4(2)+2$.
          $endgroup$
          – William
          Aug 28 '15 at 1:29




          $begingroup$
          Thanks, this was a very informative answer. It's hard to get information about this with Google searches. Also, your answer verifies that, for example, $#^g S^5times S^5$ should admit a complex structure for all $g$ since $10=4(2)+2$.
          $endgroup$
          – William
          Aug 28 '15 at 1:29











          11












          $begingroup$

          $X = Bbb{CP}^4 #Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is possible in all dimensions, but probably becomes absurd very quickly. The 8-dimensional version is here. This is what we will use.



          Preliminaries: the cohomology ring of $X$ is the direct sum of two copies of the cohomology ring of $Bbb{CP}^4$, identifying the unit and volume form in each copy. Denote the two generators of $H^2(X;Bbb Z)$ as $x_1, x_2$.



          1) Wu classes. Because we know precisely what the cohomology ring is, it's easy to calculate this. In particular, $Sq^2(x_1^3) = x_1^4$, and $Sq^2(x_2^3) = x_2^4$, so $v_2 = x_1+x_2$. Similarly we see that $Sq^4(x_1^2) = x_1^4, Sq^4(x_2^2)=x_2^4$, so $v_4 = x_1^2+x_2^2$. Trivially $v_6 = 0$.



          2) Stiefel-Whitney classes. Because the total SW-class $w$ satisfies $Sq(v)$, where $Sq$ is the total Steenrod square and $v$ is the total Wu class, we obtain $w_2 = v_2 = x_1+x_2$, $w_4 = v_2^2+v_4 = 0$, $w_6 = Sq^2(v_4) = 0$.



          3) Pontryagin classes. We can give $X$ a cell structure such that its 4-skeleton is $Bbb{CP}^2 vee Bbb{CP}^2$. Because $X_4 hookrightarrow X$ induces an isomorphism on $H^4$, the Pontryagin class is determined by the restriction of the classifying map $X to BSO(8)$ to $X_4$. We can choose this classifying map to be of the form $f vee f$, where $f$ is the restriction of the classifying map of $Bbb{CP}^4$ to its 4-skeleton. Because we know $p_1(Bbb{CP}^4) = 5x_1^2$, we see that $$p_1(X) = 5x_1^2+5x_2^2.$$
          Now the Hirzebruch signature theorem tells us that $p_2 = 20x_1^4$.



          Now let's plug into the theorem. Our cohomology classes $u, v$ would have to be of the form $u = (2k+1)x_1 + (2ell+1)x_2$, and $v = 2mx_1^3 + 2nx_2^3$, where $m,n$ have the same parity (by b, c).



          Then because $chi(M) = 16$, d) takes the form $$128 = 80 + 32(km+ell n) + 16(m+n) - (2k+1)^4 - (2ell+1)^4 + 10(2k+1)^2 + 10(2ell+1)^2 - 50.$$



          Simplifying we get $$80 = 32(km + ell n - k^3 - ell^3 + k^2 + ell^2) + 16(k+ell - k^4 - ell^4 +m +n)$$
          Reducing mod 32 and remembering that $m+n$ is even we obtain $$16 equiv 16(k+ell - k^4- ell^4) mod 32$$ But $k+ell$ is odd iff $k^4 + ell^4$ is odd, so the right side is $0 mod 32$. This is a contradiction, as desired.



          That this was so much work suggests that this is, uh, the wrong approach to prove that $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ never supports a complex structure. Maybe someone more gifted with obstruction theory than I can prove this.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You're right, I can see how this would get cumbersome in higher dimensions (to say the least), but it was a nice computation. Also, I wasn't aware of these theorems giving criteria for the existence of almost-complex structures in terms of characteristic classes, that's very nice.
            $endgroup$
            – William
            Sep 23 '15 at 21:25










          • $begingroup$
            @you: I posted a mathoverflow question asking if anybody could generalize this result here. No answer yet, unfortunately.
            $endgroup$
            – Mike Miller
            Sep 23 '15 at 22:09
















          11












          $begingroup$

          $X = Bbb{CP}^4 #Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is possible in all dimensions, but probably becomes absurd very quickly. The 8-dimensional version is here. This is what we will use.



          Preliminaries: the cohomology ring of $X$ is the direct sum of two copies of the cohomology ring of $Bbb{CP}^4$, identifying the unit and volume form in each copy. Denote the two generators of $H^2(X;Bbb Z)$ as $x_1, x_2$.



          1) Wu classes. Because we know precisely what the cohomology ring is, it's easy to calculate this. In particular, $Sq^2(x_1^3) = x_1^4$, and $Sq^2(x_2^3) = x_2^4$, so $v_2 = x_1+x_2$. Similarly we see that $Sq^4(x_1^2) = x_1^4, Sq^4(x_2^2)=x_2^4$, so $v_4 = x_1^2+x_2^2$. Trivially $v_6 = 0$.



          2) Stiefel-Whitney classes. Because the total SW-class $w$ satisfies $Sq(v)$, where $Sq$ is the total Steenrod square and $v$ is the total Wu class, we obtain $w_2 = v_2 = x_1+x_2$, $w_4 = v_2^2+v_4 = 0$, $w_6 = Sq^2(v_4) = 0$.



          3) Pontryagin classes. We can give $X$ a cell structure such that its 4-skeleton is $Bbb{CP}^2 vee Bbb{CP}^2$. Because $X_4 hookrightarrow X$ induces an isomorphism on $H^4$, the Pontryagin class is determined by the restriction of the classifying map $X to BSO(8)$ to $X_4$. We can choose this classifying map to be of the form $f vee f$, where $f$ is the restriction of the classifying map of $Bbb{CP}^4$ to its 4-skeleton. Because we know $p_1(Bbb{CP}^4) = 5x_1^2$, we see that $$p_1(X) = 5x_1^2+5x_2^2.$$
          Now the Hirzebruch signature theorem tells us that $p_2 = 20x_1^4$.



          Now let's plug into the theorem. Our cohomology classes $u, v$ would have to be of the form $u = (2k+1)x_1 + (2ell+1)x_2$, and $v = 2mx_1^3 + 2nx_2^3$, where $m,n$ have the same parity (by b, c).



          Then because $chi(M) = 16$, d) takes the form $$128 = 80 + 32(km+ell n) + 16(m+n) - (2k+1)^4 - (2ell+1)^4 + 10(2k+1)^2 + 10(2ell+1)^2 - 50.$$



          Simplifying we get $$80 = 32(km + ell n - k^3 - ell^3 + k^2 + ell^2) + 16(k+ell - k^4 - ell^4 +m +n)$$
          Reducing mod 32 and remembering that $m+n$ is even we obtain $$16 equiv 16(k+ell - k^4- ell^4) mod 32$$ But $k+ell$ is odd iff $k^4 + ell^4$ is odd, so the right side is $0 mod 32$. This is a contradiction, as desired.



          That this was so much work suggests that this is, uh, the wrong approach to prove that $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ never supports a complex structure. Maybe someone more gifted with obstruction theory than I can prove this.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You're right, I can see how this would get cumbersome in higher dimensions (to say the least), but it was a nice computation. Also, I wasn't aware of these theorems giving criteria for the existence of almost-complex structures in terms of characteristic classes, that's very nice.
            $endgroup$
            – William
            Sep 23 '15 at 21:25










          • $begingroup$
            @you: I posted a mathoverflow question asking if anybody could generalize this result here. No answer yet, unfortunately.
            $endgroup$
            – Mike Miller
            Sep 23 '15 at 22:09














          11












          11








          11





          $begingroup$

          $X = Bbb{CP}^4 #Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is possible in all dimensions, but probably becomes absurd very quickly. The 8-dimensional version is here. This is what we will use.



          Preliminaries: the cohomology ring of $X$ is the direct sum of two copies of the cohomology ring of $Bbb{CP}^4$, identifying the unit and volume form in each copy. Denote the two generators of $H^2(X;Bbb Z)$ as $x_1, x_2$.



          1) Wu classes. Because we know precisely what the cohomology ring is, it's easy to calculate this. In particular, $Sq^2(x_1^3) = x_1^4$, and $Sq^2(x_2^3) = x_2^4$, so $v_2 = x_1+x_2$. Similarly we see that $Sq^4(x_1^2) = x_1^4, Sq^4(x_2^2)=x_2^4$, so $v_4 = x_1^2+x_2^2$. Trivially $v_6 = 0$.



          2) Stiefel-Whitney classes. Because the total SW-class $w$ satisfies $Sq(v)$, where $Sq$ is the total Steenrod square and $v$ is the total Wu class, we obtain $w_2 = v_2 = x_1+x_2$, $w_4 = v_2^2+v_4 = 0$, $w_6 = Sq^2(v_4) = 0$.



          3) Pontryagin classes. We can give $X$ a cell structure such that its 4-skeleton is $Bbb{CP}^2 vee Bbb{CP}^2$. Because $X_4 hookrightarrow X$ induces an isomorphism on $H^4$, the Pontryagin class is determined by the restriction of the classifying map $X to BSO(8)$ to $X_4$. We can choose this classifying map to be of the form $f vee f$, where $f$ is the restriction of the classifying map of $Bbb{CP}^4$ to its 4-skeleton. Because we know $p_1(Bbb{CP}^4) = 5x_1^2$, we see that $$p_1(X) = 5x_1^2+5x_2^2.$$
          Now the Hirzebruch signature theorem tells us that $p_2 = 20x_1^4$.



          Now let's plug into the theorem. Our cohomology classes $u, v$ would have to be of the form $u = (2k+1)x_1 + (2ell+1)x_2$, and $v = 2mx_1^3 + 2nx_2^3$, where $m,n$ have the same parity (by b, c).



          Then because $chi(M) = 16$, d) takes the form $$128 = 80 + 32(km+ell n) + 16(m+n) - (2k+1)^4 - (2ell+1)^4 + 10(2k+1)^2 + 10(2ell+1)^2 - 50.$$



          Simplifying we get $$80 = 32(km + ell n - k^3 - ell^3 + k^2 + ell^2) + 16(k+ell - k^4 - ell^4 +m +n)$$
          Reducing mod 32 and remembering that $m+n$ is even we obtain $$16 equiv 16(k+ell - k^4- ell^4) mod 32$$ But $k+ell$ is odd iff $k^4 + ell^4$ is odd, so the right side is $0 mod 32$. This is a contradiction, as desired.



          That this was so much work suggests that this is, uh, the wrong approach to prove that $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ never supports a complex structure. Maybe someone more gifted with obstruction theory than I can prove this.






          share|cite|improve this answer









          $endgroup$



          $X = Bbb{CP}^4 #Bbb{CP}^4$ does not support an almost complex structure. I suspect this is true for $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ for any $n>0$; I can't prove it. The general idea is to obtain a "higher-dimensional" Wu theorem on which manifolds admit complex structures in terms of various characteristic classes of your manifold. I think this is possible in all dimensions, but probably becomes absurd very quickly. The 8-dimensional version is here. This is what we will use.



          Preliminaries: the cohomology ring of $X$ is the direct sum of two copies of the cohomology ring of $Bbb{CP}^4$, identifying the unit and volume form in each copy. Denote the two generators of $H^2(X;Bbb Z)$ as $x_1, x_2$.



          1) Wu classes. Because we know precisely what the cohomology ring is, it's easy to calculate this. In particular, $Sq^2(x_1^3) = x_1^4$, and $Sq^2(x_2^3) = x_2^4$, so $v_2 = x_1+x_2$. Similarly we see that $Sq^4(x_1^2) = x_1^4, Sq^4(x_2^2)=x_2^4$, so $v_4 = x_1^2+x_2^2$. Trivially $v_6 = 0$.



          2) Stiefel-Whitney classes. Because the total SW-class $w$ satisfies $Sq(v)$, where $Sq$ is the total Steenrod square and $v$ is the total Wu class, we obtain $w_2 = v_2 = x_1+x_2$, $w_4 = v_2^2+v_4 = 0$, $w_6 = Sq^2(v_4) = 0$.



          3) Pontryagin classes. We can give $X$ a cell structure such that its 4-skeleton is $Bbb{CP}^2 vee Bbb{CP}^2$. Because $X_4 hookrightarrow X$ induces an isomorphism on $H^4$, the Pontryagin class is determined by the restriction of the classifying map $X to BSO(8)$ to $X_4$. We can choose this classifying map to be of the form $f vee f$, where $f$ is the restriction of the classifying map of $Bbb{CP}^4$ to its 4-skeleton. Because we know $p_1(Bbb{CP}^4) = 5x_1^2$, we see that $$p_1(X) = 5x_1^2+5x_2^2.$$
          Now the Hirzebruch signature theorem tells us that $p_2 = 20x_1^4$.



          Now let's plug into the theorem. Our cohomology classes $u, v$ would have to be of the form $u = (2k+1)x_1 + (2ell+1)x_2$, and $v = 2mx_1^3 + 2nx_2^3$, where $m,n$ have the same parity (by b, c).



          Then because $chi(M) = 16$, d) takes the form $$128 = 80 + 32(km+ell n) + 16(m+n) - (2k+1)^4 - (2ell+1)^4 + 10(2k+1)^2 + 10(2ell+1)^2 - 50.$$



          Simplifying we get $$80 = 32(km + ell n - k^3 - ell^3 + k^2 + ell^2) + 16(k+ell - k^4 - ell^4 +m +n)$$
          Reducing mod 32 and remembering that $m+n$ is even we obtain $$16 equiv 16(k+ell - k^4- ell^4) mod 32$$ But $k+ell$ is odd iff $k^4 + ell^4$ is odd, so the right side is $0 mod 32$. This is a contradiction, as desired.



          That this was so much work suggests that this is, uh, the wrong approach to prove that $Bbb{CP}^{2n} # Bbb{CP}^{2n}$ never supports a complex structure. Maybe someone more gifted with obstruction theory than I can prove this.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 28 '15 at 20:42









          Mike MillerMike Miller

          37.3k472139




          37.3k472139












          • $begingroup$
            You're right, I can see how this would get cumbersome in higher dimensions (to say the least), but it was a nice computation. Also, I wasn't aware of these theorems giving criteria for the existence of almost-complex structures in terms of characteristic classes, that's very nice.
            $endgroup$
            – William
            Sep 23 '15 at 21:25










          • $begingroup$
            @you: I posted a mathoverflow question asking if anybody could generalize this result here. No answer yet, unfortunately.
            $endgroup$
            – Mike Miller
            Sep 23 '15 at 22:09


















          • $begingroup$
            You're right, I can see how this would get cumbersome in higher dimensions (to say the least), but it was a nice computation. Also, I wasn't aware of these theorems giving criteria for the existence of almost-complex structures in terms of characteristic classes, that's very nice.
            $endgroup$
            – William
            Sep 23 '15 at 21:25










          • $begingroup$
            @you: I posted a mathoverflow question asking if anybody could generalize this result here. No answer yet, unfortunately.
            $endgroup$
            – Mike Miller
            Sep 23 '15 at 22:09
















          $begingroup$
          You're right, I can see how this would get cumbersome in higher dimensions (to say the least), but it was a nice computation. Also, I wasn't aware of these theorems giving criteria for the existence of almost-complex structures in terms of characteristic classes, that's very nice.
          $endgroup$
          – William
          Sep 23 '15 at 21:25




          $begingroup$
          You're right, I can see how this would get cumbersome in higher dimensions (to say the least), but it was a nice computation. Also, I wasn't aware of these theorems giving criteria for the existence of almost-complex structures in terms of characteristic classes, that's very nice.
          $endgroup$
          – William
          Sep 23 '15 at 21:25












          $begingroup$
          @you: I posted a mathoverflow question asking if anybody could generalize this result here. No answer yet, unfortunately.
          $endgroup$
          – Mike Miller
          Sep 23 '15 at 22:09




          $begingroup$
          @you: I posted a mathoverflow question asking if anybody could generalize this result here. No answer yet, unfortunately.
          $endgroup$
          – Mike Miller
          Sep 23 '15 at 22:09


















          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%2f1411694%2fis-the-connected-sum-of-complex-manifolds-also-complex%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

          Plaza Victoria

          Puebla de Zaragoza

          Musa