# Quasimorphisms and Bounded Cohomology: Quantitative Version?

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 $$x,y \in \Gamma$$, $$|f(xy)-f(x)-f(y)| In other words, $$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 $$|g(x)| for every $$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.

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 $$|g(x) + g(y) - g(xy)|= |f(x) + f(y) - f(xy)|.$$
$$2D - |g(x^2)| \le\left|2|g(x)| - |g(x^2)|\right| \le|2g(x) - g(x^2)| = |2f(x) - f(x^2)| \le C,$$
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 $$C + 2\varepsilon \ge D$$ for every $$\varepsilon > 0$$. Therefore $$C$$ is a bound of $$g$$.