2 added 969 characters in body; edited body

I should add that if we think of $u _{m,n}$ as saying "$\dot{g}(m) = n$" and replace the Boolean operations with the corresponding quantifiers, then the left side says "$\dot{g}$ is a real," and the right side says "$\dot{g}$ doesn't dominate every real in the ground model." This suggests how we can characterize forcings that don't add any unbounded reals, for example, namely the following identity holds:

$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} \sum _{k < f(m)} u _{m,k}$

Forcings that don't add any reals are precisely those that satisfy the following identity:

$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} u _{m,f(m)}$

You can easily generalize this to talking about functions $\kappa \to \lambda$; the above two results so generalized are precisely Theorem 15.38 and Lemma 15.39 in Jech, "Set Theory".

1

Stefan's answer pointed me in the right direction, and then talking it over with prof. Leo Harrington we've got an answer:

A complete Boolean algebra $\mathbb{B}$ never adds a dominating real iff for any collection $\{ u _{m,k} : m, k \in \omega \} \subset \mathbb{B}^+$ the following weaker form of weak $(\omega ,\omega )$-distributivity holds:

$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{n \in \omega} \sum _{n < m < \omega} \sum _{k < f(m)} u _{m,k}$

For the reverse implication, suppose "weak weak $(\omega ,\omega )$-distributivity" holds, and for contradiction let $u \neq 0$ be the Boolean value of the sentence "there exists a dominating real," and let $\dot{g}$ be a name witnessing this, i.e. the sentence "$\dot{g}$ is a dominating real" has Boolean value $u$. Define $u _{m,k} = || \dot{g} (m) = k ||$. Now if $G$ is any $\mathbb{B}$-generic filter containing $u$, noting that the left side of the distributivity identity is (at least) $u$, we know that the right side belongs to $G$. It's then not hard to see that:

$(\exists f \in (\omega ^{\omega})^V )(\forall n \in \omega )(\exists m > n)(\exists k < f(m))(u _{m,k} \in G)$

which is to say that there's a real $f$ in the ground model such that:

$(\forall n)(\exists m > n)(\dot{g}^G (m) < f(m))$

so $f$ is not dominated by $\dot{g}$, contradiction.

For the forward implication, it should suffice to show it in the case where for each $m$, the set $\{ u _{m,k} : k \in \omega \}$ is an antichain with least upper bound $u$ independent of $m$ (I haven't checked this detail personally). So let $\{ u _{m,k}\}$ be such a collection for which the identity fails. Consider the name:

$\dot{g} = \{ (u _{m,k}, (m,k)) : m, k \in \omega \}$

It's not hard to see that the right side of the identity is at most $u$, so assuming the identity fails it's strictly less than $u$, so since $\mathbb{B}$ is separative there's a generic $G$ containing $u$ avoiding the right side of the identity. It's not hard to see from here that $\dot{g}^G$ will dominate all the ground model's reals.