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
    24 mins ago















up vote
2
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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
    24 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 24 mins ago









Eric Wofsey

177k12202328




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zzy

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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
    24 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
    24 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
24 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
24 mins ago










3 Answers
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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.






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    up vote
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    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
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      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





















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        3 Answers
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        active

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        3 Answers
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        active

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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 30 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 28 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

























                          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























                            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 26 mins ago









                            Eric Wofsey

                            177k12202328




                            177k12202328






























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