“Free” Hopf algebraWhen does a certain natural construction on monoidal categories yield a Hopf...
“Free” Hopf algebra
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I am reading through the Dyer and Lashof's paper "Homology of Iterated Loop Spaces". They are quoting the following theorem:
If $A$ is a connected, free, associative, commutative, primitively generated mod $p$ Hopf algebra of finite type, then $A$ is isomorphic as a Hopf algebra to a tensor product of free monogenic Hopf algebras, i.e. to a tensor product of exterior and polynomial Hopf algebras each having one generator.
I can find respective theorems in Milnor-Moore's paper, but I cannot understand the "freeness" assumption, as it does not appear elsewhere.
So my questions are:
1) What "free Hopf algebra" might mean in this context? Is it free as an algebra?
2) Is it known that the only "free" Hopf algebras with one generator are external and polynomial ones?
at.algebraic-topology hopf-algebras
asked 3 hours ago
Igor SikoraIgor Sikora
3558
$endgroup$
add a comment |
$begingroup$
I am reading through the Dyer and Lashof's paper "Homology of Iterated Loop Spaces". They are quoting the following theorem:
If $A$ is a connected, free, associative, commutative, primitively generated mod $p$ Hopf algebra of finite type, then $A$ is isomorphic as a Hopf algebra to a tensor product of free monogenic Hopf algebras, i.e. to a tensor product of exterior and polynomial Hopf algebras each having one generator.
I can find respective theorems in Milnor-Moore's paper, but I cannot understand the "freeness" assumption, as it does not appear elsewhere.
So my questions are:
1) What "free Hopf algebra" might mean in this context? Is it free as an algebra?
2) Is it known that the only "free" Hopf algebras with one generator are external and polynomial ones?
at.algebraic-topology hopf-algebras
asked 3 hours ago
Igor SikoraIgor Sikora
3558
$endgroup$
add a comment |
$begingroup$
I am reading through the Dyer and Lashof's paper "Homology of Iterated Loop Spaces". They are quoting the following theorem:
If $A$ is a connected, free, associative, commutative, primitively generated mod $p$ Hopf algebra of finite type, then $A$ is isomorphic as a Hopf algebra to a tensor product of free monogenic Hopf algebras, i.e. to a tensor product of exterior and polynomial Hopf algebras each having one generator.
I can find respective theorems in Milnor-Moore's paper, but I cannot understand the "freeness" assumption, as it does not appear elsewhere.
So my questions are:
1) What "free Hopf algebra" might mean in this context? Is it free as an algebra?
2) Is it known that the only "free" Hopf algebras with one generator are external and polynomial ones?
at.algebraic-topology hopf-algebras
asked 3 hours ago
Igor SikoraIgor Sikora
3558
$endgroup$
I am reading through the Dyer and Lashof's paper "Homology of Iterated Loop Spaces". They are quoting the following theorem:
If $A$ is a connected, free, associative, commutative, primitively generated mod $p$ Hopf algebra of finite type, then $A$ is isomorphic as a Hopf algebra to a tensor product of free monogenic Hopf algebras, i.e. to a tensor product of exterior and polynomial Hopf algebras each having one generator.
I can find respective theorems in Milnor-Moore's paper, but I cannot understand the "freeness" assumption, as it does not appear elsewhere.
So my questions are:
1) What "free Hopf algebra" might mean in this context? Is it free as an algebra?
2) Is it known that the only "free" Hopf algebras with one generator are external and polynomial ones?
at.algebraic-topology hopf-algebras
at.algebraic-topology hopf-algebras
asked 3 hours ago
Igor SikoraIgor Sikora
3558
asked 3 hours ago
Igor SikoraIgor Sikora
3558
asked 3 hours ago
Igor SikoraIgor Sikora
3558
asked 3 hours ago
Igor SikoraIgor Sikora
3558
asked 3 hours ago
Igor SikoraIgor Sikora
3558
3558
add a comment |
add a comment |
1 Answer
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$begingroup$
1) Yes, free means free as an algebra. (I don't think it could possibly mean anything else. "Free Hopf algebras" and "free coalgebras" as left adjoints to the forgetful functor don't exist.)
2) is elementary, I think. Let $H$ be a connected free monogenic Hopf algebra. Let $x$ be a generator. Thanks to the connectivity assumption, $deg x > 0$. The algebra $H$ is the free graded-commutative algebra on $x$, either $Bbbk[x]$ if $deg x$ is even or $p = 2$, or $Lambda(x)$ otherwise. Then by counitality and degree arguments, $Delta(x) = 1 otimes x + x otimes 1$, so $H$ is indeed either a polynomial or exterior Hopf algebra.
Note that I used the connectivity assumption above. If you don't assume that $H$ is connected (or at the very least locally conilpotent) then the result becomes false. Consider for example the Hopf algebra of dimension $2$, with basis $(1,x)$, both in degree zero, $x cdot x = 1$, $epsilon(x) = 1$ and $Delta(x) = x otimes x$. In other words, the group algebra of $mathbb{Z}/2mathbb{Z}$.
answered 2 hours ago
Najib IdrissiNajib Idrissi
1,98911027
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
1) Yes, free means free as an algebra. (I don't think it could possibly mean anything else. "Free Hopf algebras" and "free coalgebras" as left adjoints to the forgetful functor don't exist.)
2) is elementary, I think. Let $H$ be a connected free monogenic Hopf algebra. Let $x$ be a generator. Thanks to the connectivity assumption, $deg x > 0$. The algebra $H$ is the free graded-commutative algebra on $x$, either $Bbbk[x]$ if $deg x$ is even or $p = 2$, or $Lambda(x)$ otherwise. Then by counitality and degree arguments, $Delta(x) = 1 otimes x + x otimes 1$, so $H$ is indeed either a polynomial or exterior Hopf algebra.
Note that I used the connectivity assumption above. If you don't assume that $H$ is connected (or at the very least locally conilpotent) then the result becomes false. Consider for example the Hopf algebra of dimension $2$, with basis $(1,x)$, both in degree zero, $x cdot x = 1$, $epsilon(x) = 1$ and $Delta(x) = x otimes x$. In other words, the group algebra of $mathbb{Z}/2mathbb{Z}$.
answered 2 hours ago
Najib IdrissiNajib Idrissi
1,98911027
$endgroup$
add a comment |
$begingroup$
1) Yes, free means free as an algebra. (I don't think it could possibly mean anything else. "Free Hopf algebras" and "free coalgebras" as left adjoints to the forgetful functor don't exist.)
2) is elementary, I think. Let $H$ be a connected free monogenic Hopf algebra. Let $x$ be a generator. Thanks to the connectivity assumption, $deg x > 0$. The algebra $H$ is the free graded-commutative algebra on $x$, either $Bbbk[x]$ if $deg x$ is even or $p = 2$, or $Lambda(x)$ otherwise. Then by counitality and degree arguments, $Delta(x) = 1 otimes x + x otimes 1$, so $H$ is indeed either a polynomial or exterior Hopf algebra.
Note that I used the connectivity assumption above. If you don't assume that $H$ is connected (or at the very least locally conilpotent) then the result becomes false. Consider for example the Hopf algebra of dimension $2$, with basis $(1,x)$, both in degree zero, $x cdot x = 1$, $epsilon(x) = 1$ and $Delta(x) = x otimes x$. In other words, the group algebra of $mathbb{Z}/2mathbb{Z}$.
answered 2 hours ago
Najib IdrissiNajib Idrissi
1,98911027
$endgroup$
add a comment |
$begingroup$
1) Yes, free means free as an algebra. (I don't think it could possibly mean anything else. "Free Hopf algebras" and "free coalgebras" as left adjoints to the forgetful functor don't exist.)
2) is elementary, I think. Let $H$ be a connected free monogenic Hopf algebra. Let $x$ be a generator. Thanks to the connectivity assumption, $deg x > 0$. The algebra $H$ is the free graded-commutative algebra on $x$, either $Bbbk[x]$ if $deg x$ is even or $p = 2$, or $Lambda(x)$ otherwise. Then by counitality and degree arguments, $Delta(x) = 1 otimes x + x otimes 1$, so $H$ is indeed either a polynomial or exterior Hopf algebra.
Note that I used the connectivity assumption above. If you don't assume that $H$ is connected (or at the very least locally conilpotent) then the result becomes false. Consider for example the Hopf algebra of dimension $2$, with basis $(1,x)$, both in degree zero, $x cdot x = 1$, $epsilon(x) = 1$ and $Delta(x) = x otimes x$. In other words, the group algebra of $mathbb{Z}/2mathbb{Z}$.
answered 2 hours ago
Najib IdrissiNajib Idrissi
1,98911027
$endgroup$
1) Yes, free means free as an algebra. (I don't think it could possibly mean anything else. "Free Hopf algebras" and "free coalgebras" as left adjoints to the forgetful functor don't exist.)
2) is elementary, I think. Let $H$ be a connected free monogenic Hopf algebra. Let $x$ be a generator. Thanks to the connectivity assumption, $deg x > 0$. The algebra $H$ is the free graded-commutative algebra on $x$, either $Bbbk[x]$ if $deg x$ is even or $p = 2$, or $Lambda(x)$ otherwise. Then by counitality and degree arguments, $Delta(x) = 1 otimes x + x otimes 1$, so $H$ is indeed either a polynomial or exterior Hopf algebra.
Note that I used the connectivity assumption above. If you don't assume that $H$ is connected (or at the very least locally conilpotent) then the result becomes false. Consider for example the Hopf algebra of dimension $2$, with basis $(1,x)$, both in degree zero, $x cdot x = 1$, $epsilon(x) = 1$ and $Delta(x) = x otimes x$. In other words, the group algebra of $mathbb{Z}/2mathbb{Z}$.
answered 2 hours ago
Najib IdrissiNajib Idrissi
1,98911027
edited 2 hours ago
answered 2 hours ago
Najib IdrissiNajib Idrissi
1,98911027
answered 2 hours ago
Najib IdrissiNajib Idrissi
1,98911027
answered 2 hours ago
Najib IdrissiNajib Idrissi
1,98911027
1,98911027
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