Lie group structure on the complex projective space











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There is a famous theorem about when $S^n$ has the structure of a Lie group. What about the complex projective space $mathbb CP^n$? For example, why $mathbb CP^2$ is not a Lie group (without using classification for low dimension compact Lie groups)?






algebraic-topology lie-groups projective-space







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  • It is an amusing observation that a polynomial $f$ over $Bbb C$ that does not split into linear factors gives a non-trivial finite field extension $F/Bbb C$, and $Bbb P(F) = Bbb CP^{dim F - 1}$ is then a commutative positive-dimensional Lie group, and so the proofs below give as a corollary the fundamental theorem of algebra.
    – Mike Miller
    33 mins ago















up vote
2
down vote

favorite












There is a famous theorem about when $S^n$ has the structure of a Lie group. What about the complex projective space $mathbb CP^n$? For example, why $mathbb CP^2$ is not a Lie group (without using classification for low dimension compact Lie groups)?






algebraic-topology lie-groups projective-space







share|cite|improve this question
























  • It is an amusing observation that a polynomial $f$ over $Bbb C$ that does not split into linear factors gives a non-trivial finite field extension $F/Bbb C$, and $Bbb P(F) = Bbb CP^{dim F - 1}$ is then a commutative positive-dimensional Lie group, and so the proofs below give as a corollary the fundamental theorem of algebra.
    – Mike Miller
    33 mins ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











There is a famous theorem about when $S^n$ has the structure of a Lie group. What about the complex projective space $mathbb CP^n$? For example, why $mathbb CP^2$ is not a Lie group (without using classification for low dimension compact Lie groups)?






algebraic-topology lie-groups projective-space







share|cite|improve this question















There is a famous theorem about when $S^n$ has the structure of a Lie group. What about the complex projective space $mathbb CP^n$? For example, why $mathbb CP^2$ is not a Lie group (without using classification for low dimension compact Lie groups)?






algebraic-topology lie-groups projective-space




algebraic-topology lie-groups projective-space






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edited 33 mins ago









Eric Wofsey

177k12202328




177k12202328










asked 59 mins ago









zzy

2,3051419




2,3051419












  • It is an amusing observation that a polynomial $f$ over $Bbb C$ that does not split into linear factors gives a non-trivial finite field extension $F/Bbb C$, and $Bbb P(F) = Bbb CP^{dim F - 1}$ is then a commutative positive-dimensional Lie group, and so the proofs below give as a corollary the fundamental theorem of algebra.
    – Mike Miller
    33 mins ago


















  • It is an amusing observation that a polynomial $f$ over $Bbb C$ that does not split into linear factors gives a non-trivial finite field extension $F/Bbb C$, and $Bbb P(F) = Bbb CP^{dim F - 1}$ is then a commutative positive-dimensional Lie group, and so the proofs below give as a corollary the fundamental theorem of algebra.
    – Mike Miller
    33 mins ago
















It is an amusing observation that a polynomial $f$ over $Bbb C$ that does not split into linear factors gives a non-trivial finite field extension $F/Bbb C$, and $Bbb P(F) = Bbb CP^{dim F - 1}$ is then a commutative positive-dimensional Lie group, and so the proofs below give as a corollary the fundamental theorem of algebra.
– Mike Miller
33 mins ago




It is an amusing observation that a polynomial $f$ over $Bbb C$ that does not split into linear factors gives a non-trivial finite field extension $F/Bbb C$, and $Bbb P(F) = Bbb CP^{dim F - 1}$ is then a commutative positive-dimensional Lie group, and so the proofs below give as a corollary the fundamental theorem of algebra.
– Mike Miller
33 mins ago










3 Answers
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2
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$mathbb{CP}^n$ has Euler characteristic $n+1$, but a compact (positive-dimensional) Lie group has Euler characteristic $0$, for example by the Lefschetz fixed point theorem.



Alternatively, you can show in various ways that the rational cohomology ring of a compact Lie group must be an exterior algebra on a finite number of odd generators. But $H^{bullet}(mathbb{CP}^n, mathbb{Q})$ is concentrated in even degrees, so doesn't admit odd generators.






share|cite|improve this answer




























    up vote
    1
    down vote













    $pi_2(mathbb{C}P^n)$ is $mathbb{Z}$ and $pi_2(G)$ is trivial where $G$ is a connected Lie group.



    https://en.wikipedia.org/wiki/Complex_projective_space#Homotopy_groups



    https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups






    share|cite|improve this answer




























      up vote
      0
      down vote













      A complex projective space (of positive dimension) never admits a Lie group structure. There are lots of ways to prove this. For instance, the rational cohomology ring of any Lie group is a graded Hopf algebra (the comultiplication coming from the group operation) but the cohomology ring $mathbb{Q}[x]/(x^n)$ of $mathbb{CP}^n$ does not admit a Hopf algebra structure. Indeed, for reasons of degree, $Delta(x)$ would have to be $xotimes 1+1otimes x$ but then $Delta(x^n)=Delta(x)^n=sum_{k=0}^n binom{n}{k} x^kotimes x^{n-k}$ would be nonzero (all the terms except $k=0$ and $k=n$ are nonzero), which is a contradiction.






      share|cite|improve this answer





















      • Can you talk a bit about the Hopf structure on the cohomology ring? The "standard" way I've seen to get a Hopf algebra out of a Lie group is to consider the universal enveloping algebra of its Lie algebra, but this is obviously coarser. Are the two related in any way?
        – Ashwin Trisal
        6 mins ago










      • It's very simple: if $G$ is a topological group, the multiplication $mu:Gtimes Gto G$ gives a map $mu^*: H^*(G)to H^*(Gtimes G)cong H^*(G)otimes H^*(G)$ (here cohomology is with coefficients in a field to get the latter isomorphism), and the group axioms for $mu$ say exactly that $mu^*$ is the comultiplication of a Hopf algebra structure on $H^*(G)$. I don't know of any connection to the universal enveloping algebra (there may be one but it would have to be fairly indirect).
        – Eric Wofsey
        1 min ago











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      3 Answers
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      3 Answers
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      up vote
      2
      down vote













      $mathbb{CP}^n$ has Euler characteristic $n+1$, but a compact (positive-dimensional) Lie group has Euler characteristic $0$, for example by the Lefschetz fixed point theorem.



      Alternatively, you can show in various ways that the rational cohomology ring of a compact Lie group must be an exterior algebra on a finite number of odd generators. But $H^{bullet}(mathbb{CP}^n, mathbb{Q})$ is concentrated in even degrees, so doesn't admit odd generators.






      share|cite|improve this answer

























        up vote
        2
        down vote













        $mathbb{CP}^n$ has Euler characteristic $n+1$, but a compact (positive-dimensional) Lie group has Euler characteristic $0$, for example by the Lefschetz fixed point theorem.



        Alternatively, you can show in various ways that the rational cohomology ring of a compact Lie group must be an exterior algebra on a finite number of odd generators. But $H^{bullet}(mathbb{CP}^n, mathbb{Q})$ is concentrated in even degrees, so doesn't admit odd generators.






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          $mathbb{CP}^n$ has Euler characteristic $n+1$, but a compact (positive-dimensional) Lie group has Euler characteristic $0$, for example by the Lefschetz fixed point theorem.



          Alternatively, you can show in various ways that the rational cohomology ring of a compact Lie group must be an exterior algebra on a finite number of odd generators. But $H^{bullet}(mathbb{CP}^n, mathbb{Q})$ is concentrated in even degrees, so doesn't admit odd generators.






          share|cite|improve this answer












          $mathbb{CP}^n$ has Euler characteristic $n+1$, but a compact (positive-dimensional) Lie group has Euler characteristic $0$, for example by the Lefschetz fixed point theorem.



          Alternatively, you can show in various ways that the rational cohomology ring of a compact Lie group must be an exterior algebra on a finite number of odd generators. But $H^{bullet}(mathbb{CP}^n, mathbb{Q})$ is concentrated in even degrees, so doesn't admit odd generators.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 39 mins ago









          Qiaochu Yuan

          276k32579917




          276k32579917






















              up vote
              1
              down vote













              $pi_2(mathbb{C}P^n)$ is $mathbb{Z}$ and $pi_2(G)$ is trivial where $G$ is a connected Lie group.



              https://en.wikipedia.org/wiki/Complex_projective_space#Homotopy_groups



              https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups






              share|cite|improve this answer

























                up vote
                1
                down vote













                $pi_2(mathbb{C}P^n)$ is $mathbb{Z}$ and $pi_2(G)$ is trivial where $G$ is a connected Lie group.



                https://en.wikipedia.org/wiki/Complex_projective_space#Homotopy_groups



                https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  $pi_2(mathbb{C}P^n)$ is $mathbb{Z}$ and $pi_2(G)$ is trivial where $G$ is a connected Lie group.



                  https://en.wikipedia.org/wiki/Complex_projective_space#Homotopy_groups



                  https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups






                  share|cite|improve this answer












                  $pi_2(mathbb{C}P^n)$ is $mathbb{Z}$ and $pi_2(G)$ is trivial where $G$ is a connected Lie group.



                  https://en.wikipedia.org/wiki/Complex_projective_space#Homotopy_groups



                  https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 37 mins ago









                  Tsemo Aristide

                  55.2k11444




                  55.2k11444






















                      up vote
                      0
                      down vote













                      A complex projective space (of positive dimension) never admits a Lie group structure. There are lots of ways to prove this. For instance, the rational cohomology ring of any Lie group is a graded Hopf algebra (the comultiplication coming from the group operation) but the cohomology ring $mathbb{Q}[x]/(x^n)$ of $mathbb{CP}^n$ does not admit a Hopf algebra structure. Indeed, for reasons of degree, $Delta(x)$ would have to be $xotimes 1+1otimes x$ but then $Delta(x^n)=Delta(x)^n=sum_{k=0}^n binom{n}{k} x^kotimes x^{n-k}$ would be nonzero (all the terms except $k=0$ and $k=n$ are nonzero), which is a contradiction.






                      share|cite|improve this answer





















                      • Can you talk a bit about the Hopf structure on the cohomology ring? The "standard" way I've seen to get a Hopf algebra out of a Lie group is to consider the universal enveloping algebra of its Lie algebra, but this is obviously coarser. Are the two related in any way?
                        – Ashwin Trisal
                        6 mins ago










                      • It's very simple: if $G$ is a topological group, the multiplication $mu:Gtimes Gto G$ gives a map $mu^*: H^*(G)to H^*(Gtimes G)cong H^*(G)otimes H^*(G)$ (here cohomology is with coefficients in a field to get the latter isomorphism), and the group axioms for $mu$ say exactly that $mu^*$ is the comultiplication of a Hopf algebra structure on $H^*(G)$. I don't know of any connection to the universal enveloping algebra (there may be one but it would have to be fairly indirect).
                        – Eric Wofsey
                        1 min ago















                      up vote
                      0
                      down vote













                      A complex projective space (of positive dimension) never admits a Lie group structure. There are lots of ways to prove this. For instance, the rational cohomology ring of any Lie group is a graded Hopf algebra (the comultiplication coming from the group operation) but the cohomology ring $mathbb{Q}[x]/(x^n)$ of $mathbb{CP}^n$ does not admit a Hopf algebra structure. Indeed, for reasons of degree, $Delta(x)$ would have to be $xotimes 1+1otimes x$ but then $Delta(x^n)=Delta(x)^n=sum_{k=0}^n binom{n}{k} x^kotimes x^{n-k}$ would be nonzero (all the terms except $k=0$ and $k=n$ are nonzero), which is a contradiction.






                      share|cite|improve this answer





















                      • Can you talk a bit about the Hopf structure on the cohomology ring? The "standard" way I've seen to get a Hopf algebra out of a Lie group is to consider the universal enveloping algebra of its Lie algebra, but this is obviously coarser. Are the two related in any way?
                        – Ashwin Trisal
                        6 mins ago










                      • It's very simple: if $G$ is a topological group, the multiplication $mu:Gtimes Gto G$ gives a map $mu^*: H^*(G)to H^*(Gtimes G)cong H^*(G)otimes H^*(G)$ (here cohomology is with coefficients in a field to get the latter isomorphism), and the group axioms for $mu$ say exactly that $mu^*$ is the comultiplication of a Hopf algebra structure on $H^*(G)$. I don't know of any connection to the universal enveloping algebra (there may be one but it would have to be fairly indirect).
                        – Eric Wofsey
                        1 min ago













                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote









                      A complex projective space (of positive dimension) never admits a Lie group structure. There are lots of ways to prove this. For instance, the rational cohomology ring of any Lie group is a graded Hopf algebra (the comultiplication coming from the group operation) but the cohomology ring $mathbb{Q}[x]/(x^n)$ of $mathbb{CP}^n$ does not admit a Hopf algebra structure. Indeed, for reasons of degree, $Delta(x)$ would have to be $xotimes 1+1otimes x$ but then $Delta(x^n)=Delta(x)^n=sum_{k=0}^n binom{n}{k} x^kotimes x^{n-k}$ would be nonzero (all the terms except $k=0$ and $k=n$ are nonzero), which is a contradiction.






                      share|cite|improve this answer












                      A complex projective space (of positive dimension) never admits a Lie group structure. There are lots of ways to prove this. For instance, the rational cohomology ring of any Lie group is a graded Hopf algebra (the comultiplication coming from the group operation) but the cohomology ring $mathbb{Q}[x]/(x^n)$ of $mathbb{CP}^n$ does not admit a Hopf algebra structure. Indeed, for reasons of degree, $Delta(x)$ would have to be $xotimes 1+1otimes x$ but then $Delta(x^n)=Delta(x)^n=sum_{k=0}^n binom{n}{k} x^kotimes x^{n-k}$ would be nonzero (all the terms except $k=0$ and $k=n$ are nonzero), which is a contradiction.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 35 mins ago









                      Eric Wofsey

                      177k12202328




                      177k12202328












                      • Can you talk a bit about the Hopf structure on the cohomology ring? The "standard" way I've seen to get a Hopf algebra out of a Lie group is to consider the universal enveloping algebra of its Lie algebra, but this is obviously coarser. Are the two related in any way?
                        – Ashwin Trisal
                        6 mins ago










                      • It's very simple: if $G$ is a topological group, the multiplication $mu:Gtimes Gto G$ gives a map $mu^*: H^*(G)to H^*(Gtimes G)cong H^*(G)otimes H^*(G)$ (here cohomology is with coefficients in a field to get the latter isomorphism), and the group axioms for $mu$ say exactly that $mu^*$ is the comultiplication of a Hopf algebra structure on $H^*(G)$. I don't know of any connection to the universal enveloping algebra (there may be one but it would have to be fairly indirect).
                        – Eric Wofsey
                        1 min ago


















                      • Can you talk a bit about the Hopf structure on the cohomology ring? The "standard" way I've seen to get a Hopf algebra out of a Lie group is to consider the universal enveloping algebra of its Lie algebra, but this is obviously coarser. Are the two related in any way?
                        – Ashwin Trisal
                        6 mins ago










                      • It's very simple: if $G$ is a topological group, the multiplication $mu:Gtimes Gto G$ gives a map $mu^*: H^*(G)to H^*(Gtimes G)cong H^*(G)otimes H^*(G)$ (here cohomology is with coefficients in a field to get the latter isomorphism), and the group axioms for $mu$ say exactly that $mu^*$ is the comultiplication of a Hopf algebra structure on $H^*(G)$. I don't know of any connection to the universal enveloping algebra (there may be one but it would have to be fairly indirect).
                        – Eric Wofsey
                        1 min ago
















                      Can you talk a bit about the Hopf structure on the cohomology ring? The "standard" way I've seen to get a Hopf algebra out of a Lie group is to consider the universal enveloping algebra of its Lie algebra, but this is obviously coarser. Are the two related in any way?
                      – Ashwin Trisal
                      6 mins ago




                      Can you talk a bit about the Hopf structure on the cohomology ring? The "standard" way I've seen to get a Hopf algebra out of a Lie group is to consider the universal enveloping algebra of its Lie algebra, but this is obviously coarser. Are the two related in any way?
                      – Ashwin Trisal
                      6 mins ago












                      It's very simple: if $G$ is a topological group, the multiplication $mu:Gtimes Gto G$ gives a map $mu^*: H^*(G)to H^*(Gtimes G)cong H^*(G)otimes H^*(G)$ (here cohomology is with coefficients in a field to get the latter isomorphism), and the group axioms for $mu$ say exactly that $mu^*$ is the comultiplication of a Hopf algebra structure on $H^*(G)$. I don't know of any connection to the universal enveloping algebra (there may be one but it would have to be fairly indirect).
                      – Eric Wofsey
                      1 min ago




                      It's very simple: if $G$ is a topological group, the multiplication $mu:Gtimes Gto G$ gives a map $mu^*: H^*(G)to H^*(Gtimes G)cong H^*(G)otimes H^*(G)$ (here cohomology is with coefficients in a field to get the latter isomorphism), and the group axioms for $mu$ say exactly that $mu^*$ is the comultiplication of a Hopf algebra structure on $H^*(G)$. I don't know of any connection to the universal enveloping algebra (there may be one but it would have to be fairly indirect).
                      – Eric Wofsey
                      1 min ago


















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