Quasimorphisms and Bounded Cohomology: Quantitative Version?
$begingroup$
Consider maps from a discrete group $Gamma$ to the additive group $mathbb{R}$. A function $f:Gamma to mathbb{R}$ is called a quasimorphism if it is locally close to being a group homomorphism. More precisely, $f$ is a quasimorphism if there exists some constant $C$ such that for every $x,y in Gamma$,
$$|f(xy)-f(x)-f(y)|<C$$
In other words, $f(xy)$ is at bounded distance from $f(x)+f(y)$.
A natural question is whether any quasimorphism is simply a homomorphism perturbed by a bounded function. In this regard, we can define the bounded cohomology of $Gamma$ with coefficients in $mathbb{R}$, and consider the comparison homomorphism
$$c:H_b^2(Gamma,mathbb{R}) to H^2(Gamma,mathbb{R})$$
and it is known that the kernel of this map is precisely the space of non-trivial quasimorphisms. More precisely
$$ker(c) = QM(Gamma)/(Hom(Gamma,mathbb{R}) oplus C_b(Gamma,mathbb{R}))$$
where $QM(Gamma)$ is the space of quasimorphisms, $C_b(Gamma,mathbb{R})$ is the space of all bounded functions, and $Hom(Gamma,mathbb{R})$ is the space of homomorphisms.
So when $ker(c)$ is trivial, every quasimorphism is trivial: it is obtained by perturbing a homomorphism by a bounded function.
Is there a quantitative version of this statement? That is, suppose $C$ is the uniform bound for the quasimorphism $f$ as above, and suppose $ker(c)$ is trivial. Then can we actually give a concrete bound for the distance of $f$ from the homomorphism in $Hom(Gamma,mathbb{R})$ in terms of $C$?
All I find in references is that $f$ can be written as a sum of a homomorphism and a bounded function, but I have not come across any actual bound on that function in terms of the original bound on $f(xy)-f(x)-f(y)$. This seems like a very natural question, so do help out with directions and references if available.
*UPDATE: So in one direction it seems to be easy. Suppose $f=phi+g$ where $phi in Hom(Gamma,mathbb{R})$ and $g:Gamma to mathbb{R}$ is a bounded function with $|g(x)|<D$ for every $x in Gamma$. Then
$$|g(xy)-g(x)-g(y)|<3D$$
by the triangle inequality, and
$$|f(xy)-f(x)-f(y)|leq |phi(xy)-phi(x)-phi(y)|+|g(xy)-g(x)-g(y)|$$
and so
$$|f(xy)-f(x)-f(y)|< 3D$$
So if $f$ is globally $D$-close to a homomorphism, then it is locally $3D$-close to being a homomorphism, or a $3D$-quasimorphism. But I am interested more in the other direction.
gr.group-theory group-cohomology geometric-group-theory
$endgroup$
add a comment |
$begingroup$
Consider maps from a discrete group $Gamma$ to the additive group $mathbb{R}$. A function $f:Gamma to mathbb{R}$ is called a quasimorphism if it is locally close to being a group homomorphism. More precisely, $f$ is a quasimorphism if there exists some constant $C$ such that for every $x,y in Gamma$,
$$|f(xy)-f(x)-f(y)|<C$$
In other words, $f(xy)$ is at bounded distance from $f(x)+f(y)$.
A natural question is whether any quasimorphism is simply a homomorphism perturbed by a bounded function. In this regard, we can define the bounded cohomology of $Gamma$ with coefficients in $mathbb{R}$, and consider the comparison homomorphism
$$c:H_b^2(Gamma,mathbb{R}) to H^2(Gamma,mathbb{R})$$
and it is known that the kernel of this map is precisely the space of non-trivial quasimorphisms. More precisely
$$ker(c) = QM(Gamma)/(Hom(Gamma,mathbb{R}) oplus C_b(Gamma,mathbb{R}))$$
where $QM(Gamma)$ is the space of quasimorphisms, $C_b(Gamma,mathbb{R})$ is the space of all bounded functions, and $Hom(Gamma,mathbb{R})$ is the space of homomorphisms.
So when $ker(c)$ is trivial, every quasimorphism is trivial: it is obtained by perturbing a homomorphism by a bounded function.
Is there a quantitative version of this statement? That is, suppose $C$ is the uniform bound for the quasimorphism $f$ as above, and suppose $ker(c)$ is trivial. Then can we actually give a concrete bound for the distance of $f$ from the homomorphism in $Hom(Gamma,mathbb{R})$ in terms of $C$?
All I find in references is that $f$ can be written as a sum of a homomorphism and a bounded function, but I have not come across any actual bound on that function in terms of the original bound on $f(xy)-f(x)-f(y)$. This seems like a very natural question, so do help out with directions and references if available.
*UPDATE: So in one direction it seems to be easy. Suppose $f=phi+g$ where $phi in Hom(Gamma,mathbb{R})$ and $g:Gamma to mathbb{R}$ is a bounded function with $|g(x)|<D$ for every $x in Gamma$. Then
$$|g(xy)-g(x)-g(y)|<3D$$
by the triangle inequality, and
$$|f(xy)-f(x)-f(y)|leq |phi(xy)-phi(x)-phi(y)|+|g(xy)-g(x)-g(y)|$$
and so
$$|f(xy)-f(x)-f(y)|< 3D$$
So if $f$ is globally $D$-close to a homomorphism, then it is locally $3D$-close to being a homomorphism, or a $3D$-quasimorphism. But I am interested more in the other direction.
gr.group-theory group-cohomology geometric-group-theory
$endgroup$
add a comment |
$begingroup$
Consider maps from a discrete group $Gamma$ to the additive group $mathbb{R}$. A function $f:Gamma to mathbb{R}$ is called a quasimorphism if it is locally close to being a group homomorphism. More precisely, $f$ is a quasimorphism if there exists some constant $C$ such that for every $x,y in Gamma$,
$$|f(xy)-f(x)-f(y)|<C$$
In other words, $f(xy)$ is at bounded distance from $f(x)+f(y)$.
A natural question is whether any quasimorphism is simply a homomorphism perturbed by a bounded function. In this regard, we can define the bounded cohomology of $Gamma$ with coefficients in $mathbb{R}$, and consider the comparison homomorphism
$$c:H_b^2(Gamma,mathbb{R}) to H^2(Gamma,mathbb{R})$$
and it is known that the kernel of this map is precisely the space of non-trivial quasimorphisms. More precisely
$$ker(c) = QM(Gamma)/(Hom(Gamma,mathbb{R}) oplus C_b(Gamma,mathbb{R}))$$
where $QM(Gamma)$ is the space of quasimorphisms, $C_b(Gamma,mathbb{R})$ is the space of all bounded functions, and $Hom(Gamma,mathbb{R})$ is the space of homomorphisms.
So when $ker(c)$ is trivial, every quasimorphism is trivial: it is obtained by perturbing a homomorphism by a bounded function.
Is there a quantitative version of this statement? That is, suppose $C$ is the uniform bound for the quasimorphism $f$ as above, and suppose $ker(c)$ is trivial. Then can we actually give a concrete bound for the distance of $f$ from the homomorphism in $Hom(Gamma,mathbb{R})$ in terms of $C$?
All I find in references is that $f$ can be written as a sum of a homomorphism and a bounded function, but I have not come across any actual bound on that function in terms of the original bound on $f(xy)-f(x)-f(y)$. This seems like a very natural question, so do help out with directions and references if available.
*UPDATE: So in one direction it seems to be easy. Suppose $f=phi+g$ where $phi in Hom(Gamma,mathbb{R})$ and $g:Gamma to mathbb{R}$ is a bounded function with $|g(x)|<D$ for every $x in Gamma$. Then
$$|g(xy)-g(x)-g(y)|<3D$$
by the triangle inequality, and
$$|f(xy)-f(x)-f(y)|leq |phi(xy)-phi(x)-phi(y)|+|g(xy)-g(x)-g(y)|$$
and so
$$|f(xy)-f(x)-f(y)|< 3D$$
So if $f$ is globally $D$-close to a homomorphism, then it is locally $3D$-close to being a homomorphism, or a $3D$-quasimorphism. But I am interested more in the other direction.
gr.group-theory group-cohomology geometric-group-theory
$endgroup$
Consider maps from a discrete group $Gamma$ to the additive group $mathbb{R}$. A function $f:Gamma to mathbb{R}$ is called a quasimorphism if it is locally close to being a group homomorphism. More precisely, $f$ is a quasimorphism if there exists some constant $C$ such that for every $x,y in Gamma$,
$$|f(xy)-f(x)-f(y)|<C$$
In other words, $f(xy)$ is at bounded distance from $f(x)+f(y)$.
A natural question is whether any quasimorphism is simply a homomorphism perturbed by a bounded function. In this regard, we can define the bounded cohomology of $Gamma$ with coefficients in $mathbb{R}$, and consider the comparison homomorphism
$$c:H_b^2(Gamma,mathbb{R}) to H^2(Gamma,mathbb{R})$$
and it is known that the kernel of this map is precisely the space of non-trivial quasimorphisms. More precisely
$$ker(c) = QM(Gamma)/(Hom(Gamma,mathbb{R}) oplus C_b(Gamma,mathbb{R}))$$
where $QM(Gamma)$ is the space of quasimorphisms, $C_b(Gamma,mathbb{R})$ is the space of all bounded functions, and $Hom(Gamma,mathbb{R})$ is the space of homomorphisms.
So when $ker(c)$ is trivial, every quasimorphism is trivial: it is obtained by perturbing a homomorphism by a bounded function.
Is there a quantitative version of this statement? That is, suppose $C$ is the uniform bound for the quasimorphism $f$ as above, and suppose $ker(c)$ is trivial. Then can we actually give a concrete bound for the distance of $f$ from the homomorphism in $Hom(Gamma,mathbb{R})$ in terms of $C$?
All I find in references is that $f$ can be written as a sum of a homomorphism and a bounded function, but I have not come across any actual bound on that function in terms of the original bound on $f(xy)-f(x)-f(y)$. This seems like a very natural question, so do help out with directions and references if available.
*UPDATE: So in one direction it seems to be easy. Suppose $f=phi+g$ where $phi in Hom(Gamma,mathbb{R})$ and $g:Gamma to mathbb{R}$ is a bounded function with $|g(x)|<D$ for every $x in Gamma$. Then
$$|g(xy)-g(x)-g(y)|<3D$$
by the triangle inequality, and
$$|f(xy)-f(x)-f(y)|leq |phi(xy)-phi(x)-phi(y)|+|g(xy)-g(x)-g(y)|$$
and so
$$|f(xy)-f(x)-f(y)|< 3D$$
So if $f$ is globally $D$-close to a homomorphism, then it is locally $3D$-close to being a homomorphism, or a $3D$-quasimorphism. But I am interested more in the other direction.
gr.group-theory group-cohomology geometric-group-theory
gr.group-theory group-cohomology geometric-group-theory
edited 6 hours ago
BharatRam
asked 8 hours ago
BharatRamBharatRam
369213
369213
add a comment |
add a comment |
1 Answer
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$begingroup$
Suppose that $f : Gamma to mathbf{R}$ is a trivial quasi-isomorphism and write $f = phi + g$ where $phi in Hom(Gamma,mathbf{R})$ and $g : Gamma to mathbf{R}$ is a bounded function as you did. Here we prove that if
$$|f(x) + f(y) - f(xy)| le C,$$
for all $x,y in Gamma$, then $g$ is bounded by $C$.
Suppose that $g$ is bounded by $D$ and that the bound is optimal, namely $D = sup_{x in Gamma} |g(x)|$. For simplicity, let us assume for the moment that there exists $x in Gamma$ such that $|g(x)| = D$. First of all, as $phi$ is a group homomorphism, we have
$$|g(x) + g(y) - g(xy)|= |f(x) + f(y) - f(xy)|.$$
Since
$$2D - |g(x^2)| leleft|2|g(x)| - |g(x^2)|right| le|2g(x) - g(x^2)| = |2f(x) - f(x^2)| le C,$$
we have $|g(x^2)| ge 2D - C$. On the other hand $D ge |g(x^2)|$ by assumption, so $C ge D$. Therefore $g$ is bounded by $C$.
In the general case, that is if we don't assume that $|g(x)| = D$ for some $x in Gamma$, then for every $varepsilon > 0$ we can still find $x in Gamma$ such that $|g(x)| ge D - varepsilon$. A similar argument shows that $C + 2varepsilon ge D$ for every $varepsilon > 0$. Therefore $C$ is a bound of $g$.
$endgroup$
add a comment |
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$begingroup$
Suppose that $f : Gamma to mathbf{R}$ is a trivial quasi-isomorphism and write $f = phi + g$ where $phi in Hom(Gamma,mathbf{R})$ and $g : Gamma to mathbf{R}$ is a bounded function as you did. Here we prove that if
$$|f(x) + f(y) - f(xy)| le C,$$
for all $x,y in Gamma$, then $g$ is bounded by $C$.
Suppose that $g$ is bounded by $D$ and that the bound is optimal, namely $D = sup_{x in Gamma} |g(x)|$. For simplicity, let us assume for the moment that there exists $x in Gamma$ such that $|g(x)| = D$. First of all, as $phi$ is a group homomorphism, we have
$$|g(x) + g(y) - g(xy)|= |f(x) + f(y) - f(xy)|.$$
Since
$$2D - |g(x^2)| leleft|2|g(x)| - |g(x^2)|right| le|2g(x) - g(x^2)| = |2f(x) - f(x^2)| le C,$$
we have $|g(x^2)| ge 2D - C$. On the other hand $D ge |g(x^2)|$ by assumption, so $C ge D$. Therefore $g$ is bounded by $C$.
In the general case, that is if we don't assume that $|g(x)| = D$ for some $x in Gamma$, then for every $varepsilon > 0$ we can still find $x in Gamma$ such that $|g(x)| ge D - varepsilon$. A similar argument shows that $C + 2varepsilon ge D$ for every $varepsilon > 0$. Therefore $C$ is a bound of $g$.
$endgroup$
add a comment |
$begingroup$
Suppose that $f : Gamma to mathbf{R}$ is a trivial quasi-isomorphism and write $f = phi + g$ where $phi in Hom(Gamma,mathbf{R})$ and $g : Gamma to mathbf{R}$ is a bounded function as you did. Here we prove that if
$$|f(x) + f(y) - f(xy)| le C,$$
for all $x,y in Gamma$, then $g$ is bounded by $C$.
Suppose that $g$ is bounded by $D$ and that the bound is optimal, namely $D = sup_{x in Gamma} |g(x)|$. For simplicity, let us assume for the moment that there exists $x in Gamma$ such that $|g(x)| = D$. First of all, as $phi$ is a group homomorphism, we have
$$|g(x) + g(y) - g(xy)|= |f(x) + f(y) - f(xy)|.$$
Since
$$2D - |g(x^2)| leleft|2|g(x)| - |g(x^2)|right| le|2g(x) - g(x^2)| = |2f(x) - f(x^2)| le C,$$
we have $|g(x^2)| ge 2D - C$. On the other hand $D ge |g(x^2)|$ by assumption, so $C ge D$. Therefore $g$ is bounded by $C$.
In the general case, that is if we don't assume that $|g(x)| = D$ for some $x in Gamma$, then for every $varepsilon > 0$ we can still find $x in Gamma$ such that $|g(x)| ge D - varepsilon$. A similar argument shows that $C + 2varepsilon ge D$ for every $varepsilon > 0$. Therefore $C$ is a bound of $g$.
$endgroup$
add a comment |
$begingroup$
Suppose that $f : Gamma to mathbf{R}$ is a trivial quasi-isomorphism and write $f = phi + g$ where $phi in Hom(Gamma,mathbf{R})$ and $g : Gamma to mathbf{R}$ is a bounded function as you did. Here we prove that if
$$|f(x) + f(y) - f(xy)| le C,$$
for all $x,y in Gamma$, then $g$ is bounded by $C$.
Suppose that $g$ is bounded by $D$ and that the bound is optimal, namely $D = sup_{x in Gamma} |g(x)|$. For simplicity, let us assume for the moment that there exists $x in Gamma$ such that $|g(x)| = D$. First of all, as $phi$ is a group homomorphism, we have
$$|g(x) + g(y) - g(xy)|= |f(x) + f(y) - f(xy)|.$$
Since
$$2D - |g(x^2)| leleft|2|g(x)| - |g(x^2)|right| le|2g(x) - g(x^2)| = |2f(x) - f(x^2)| le C,$$
we have $|g(x^2)| ge 2D - C$. On the other hand $D ge |g(x^2)|$ by assumption, so $C ge D$. Therefore $g$ is bounded by $C$.
In the general case, that is if we don't assume that $|g(x)| = D$ for some $x in Gamma$, then for every $varepsilon > 0$ we can still find $x in Gamma$ such that $|g(x)| ge D - varepsilon$. A similar argument shows that $C + 2varepsilon ge D$ for every $varepsilon > 0$. Therefore $C$ is a bound of $g$.
$endgroup$
Suppose that $f : Gamma to mathbf{R}$ is a trivial quasi-isomorphism and write $f = phi + g$ where $phi in Hom(Gamma,mathbf{R})$ and $g : Gamma to mathbf{R}$ is a bounded function as you did. Here we prove that if
$$|f(x) + f(y) - f(xy)| le C,$$
for all $x,y in Gamma$, then $g$ is bounded by $C$.
Suppose that $g$ is bounded by $D$ and that the bound is optimal, namely $D = sup_{x in Gamma} |g(x)|$. For simplicity, let us assume for the moment that there exists $x in Gamma$ such that $|g(x)| = D$. First of all, as $phi$ is a group homomorphism, we have
$$|g(x) + g(y) - g(xy)|= |f(x) + f(y) - f(xy)|.$$
Since
$$2D - |g(x^2)| leleft|2|g(x)| - |g(x^2)|right| le|2g(x) - g(x^2)| = |2f(x) - f(x^2)| le C,$$
we have $|g(x^2)| ge 2D - C$. On the other hand $D ge |g(x^2)|$ by assumption, so $C ge D$. Therefore $g$ is bounded by $C$.
In the general case, that is if we don't assume that $|g(x)| = D$ for some $x in Gamma$, then for every $varepsilon > 0$ we can still find $x in Gamma$ such that $|g(x)| ge D - varepsilon$. A similar argument shows that $C + 2varepsilon ge D$ for every $varepsilon > 0$. Therefore $C$ is a bound of $g$.
edited 3 hours ago
Max
5341617
5341617
answered 5 hours ago
HYLHYL
974516
974516
add a comment |
add a comment |
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