Quasimorphisms and Bounded Cohomology: Quantitative Version?












9












$begingroup$


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.










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$endgroup$

















    9












    $begingroup$


    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.










    share|cite|improve this question











    $endgroup$















      9












      9








      9


      2



      $begingroup$


      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.










      share|cite|improve this question











      $endgroup$




      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






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      edited 11 hours ago







      BharatRam

















      asked 12 hours ago









      BharatRamBharatRam

      369213




      369213






















          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 $g$.






          share|cite|improve this 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 $g$.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              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 $g$.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                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 $g$.






                share|cite|improve this answer











                $endgroup$



                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 $g$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 8 hours ago









                Max

                5341617




                5341617










                answered 10 hours ago









                HYLHYL

                974516




                974516






























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