Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$ and suppose that ${\cal U}$ is a free ultrafilter on $\omega$. We write $f \leq\_{\cal U} g$ if $$\{n\in\omega: f(n) \leq g(n)\}\in{\cal U}.$$ 


Similar to the [usual bounding and dominating numbers](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum), we define $${\frak b}\_{\cal U} = \min\{|B|: B\subseteq \omega^\omega \land \forall f\in\omega^\omega \exists g\in B(g\not \leq\_{\cal U} f)\},$$
and 
$${\frak d}\_{\cal U} = \min\{|D|: D\subseteq \omega^\omega \land \forall f\in\omega^\omega \exists g\in D(f \leq\_{\cal U} g)\}.$$


Do we have ${\frak b} = {\frak b}\_{\cal U}$ and ${\frak d} = {\frak d}\_{\cal U}$?



Bounding and dominating numbers ${\frak b}, {\frak d}$ via ultrafilters
For any two functions $f,g$ the sets $\{n\in\omega:f(n)\leq g(n)\}$ and $\{n\in\omega:g(n)\leq f(n)\}$ cover $\omega$, so one of them must be in $\mathcal U$. Hence we have a dichotomy $f\leq\_{\mathcal U}g$ or $f\leq\_{\mathcal U}g$.


It follows that every unbounded family is dominating: if $B$ is an unbounded with respect to $\leq\_{\mathcal U}$family, then for any $f\in\omega^\omega$ there is $g\in B$ such that $f\not\geq\_{\mathcal U}g$ and hence $f\leq\_{\mathcal U}g$, hence $B$ is dominating. Since every dominating set is clearly unbounded, it follows that $\frak b\_{\mathcal U}=\frak d\_{\mathcal U}$. However, it is consistent that $\frak b<\frak d$, so it is consistent that $\frak b\_{\mathcal U}\neq\frak b$ or $\frak d\_{\mathcal U}\neq\frak d$.


We can say something more: every set unbounded with respect to $\leq\_{\mathcal U}$ is unbounded with respect to $\leq^\*$ (because $f\not\leq\_{\mathcal U}g\Rightarrow f\not\leq^\*g$) and every set dominating with respect to $\leq^\*$ is dominating with respect to $\leq\_{\mathcal U}$ (because $f\leq^\*g\Rightarrow f\leq\_{\mathcal U}g$), so $\frak b\leq\frak b\_{\mathcal U}=\frak d\_{\mathcal U}\leq\frak d$, hence it's consistent that one of these inequalities is strict.


Of course assuming $CH$ we would have $\frak b=\frak b\_{\mathcal U}=\frak d\_{\mathcal U}=\frak d=\frak c$, hence the only open question left is whether we can (consistently) have $\frak b<\frak b\_{\mathcal U}=\frak d\_{\mathcal U}<\frak d$, which is a question I can't answer.


