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
edited 33 mins ago
Eric Wofsey
177k12202328
asked 59 mins ago
zzy
2,3051419
add a comment |
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
edited 33 mins ago
Eric Wofsey
177k12202328
asked 59 mins ago
zzy
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
add a comment |
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
edited 33 mins ago
Eric Wofsey
177k12202328
asked 59 mins ago
zzy
2,3051419
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
edited 33 mins ago
Eric Wofsey
177k12202328
asked 59 mins ago
zzy
2,3051419
edited 33 mins ago
Eric Wofsey
177k12202328
asked 59 mins ago
zzy
2,3051419
edited 33 mins ago
Eric Wofsey
177k12202328
edited 33 mins ago
Eric Wofsey
177k12202328
edited 33 mins ago
Eric Wofsey
177k12202328
177k12202328
asked 59 mins ago
zzy
2,3051419
asked 59 mins ago
zzy
2,3051419
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
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.
answered 39 mins ago
Qiaochu Yuan
276k32579917
add a comment |
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
answered 37 mins ago
Tsemo Aristide
55.2k11444
add a comment |
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.
answered 35 mins ago
Eric Wofsey
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
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 39 mins ago
Qiaochu Yuan
276k32579917
add a comment |
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.
answered 39 mins ago
Qiaochu Yuan
276k32579917
add a comment |
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.
answered 39 mins ago
Qiaochu Yuan
276k32579917
$mathbb{CP}^n$ 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.
answered 39 mins ago
Qiaochu Yuan
276k32579917
answered 39 mins ago
Qiaochu Yuan
276k32579917
answered 39 mins ago
Qiaochu Yuan
276k32579917
answered 39 mins ago
Qiaochu Yuan
276k32579917
276k32579917
add a comment |
add a comment |
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
answered 37 mins ago
Tsemo Aristide
55.2k11444
add a comment |
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
answered 37 mins ago
Tsemo Aristide
55.2k11444
add a comment |
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
answered 37 mins ago
Tsemo Aristide
55.2k11444
$pi_2(mathbb{C}P^n)$ 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
answered 37 mins ago
Tsemo Aristide
55.2k11444
answered 37 mins ago
Tsemo Aristide
55.2k11444
answered 37 mins ago
Tsemo Aristide
55.2k11444
answered 37 mins ago
Tsemo Aristide
55.2k11444
55.2k11444
add a comment |
add a comment |
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.
answered 35 mins ago
Eric Wofsey
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
add a comment |
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.
answered 35 mins ago
Eric Wofsey
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
add a comment |
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.
answered 35 mins ago
Eric Wofsey
177k12202328
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.
answered 35 mins ago
Eric Wofsey
177k12202328
answered 35 mins ago
Eric Wofsey
177k12202328
answered 35 mins ago
Eric Wofsey
177k12202328
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
add a comment |
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
add a comment |
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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