Let $\Delta(\kappa, \mu)$ be the statement: "let $F$ be a family of cardinality $\kappa$ of sets of cardinality less than $\mu$. Then there is a family $G \subset F$ of cardinality $\kappa$ and a set $r$ such that $a \cap b=r$ for every $a,b \in G$".


We know that if $\kappa$ is a regular cardinal and $\lambda^{<\mu} < \kappa$, for every $\lambda < \kappa$ then $\Delta(\kappa, \mu)$ holds. My question is:



> 
> Does $\Delta(\kappa, \mu)$ for some regular uncountable $\kappa$ and some uncountable $\mu<\kappa$ imply some weak form of the GCH below $\kappa$?
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> 
> 



Does the generalized $\Delta$-system lemma imply some weak version of the GCH?
It is a very nice question! The answer is yes, natural instances of the $\Delta$ system property, which hold under GCH, are in fact equivalent to the GCH. 


**Theorem.** $\Delta(\omega\_2,\omega\_1)$ is equivalent to CH.


Proof: You've pointed out that CH implies the principle, since the
hypothesis you mention for this case amounts to
$\omega\_1^{\lt\omega\_1}<\omega\_2$, which amounts to CH. So let us
consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the
tree of all finite binary sequences, and label the nodes of $T$
with distinct natural numbers. Let $F$ be the subsets of $\omega$
arising as the sets of labels occuring on any of $\omega\_2$ many
branches through $T$. Thus, $F$ has size $\omega\_2$, and any two
elements of $F$ have finite intersection. I claim that this family
of sets can have no $\Delta$-system of size $\omega\_2$, and
indeed, it can have no $\Delta$-system even with three elements.
If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap
b=a\cap c$, and so $a$ and $b$ branch out at the same node that
$a$ and $c$ do, in which case $b$ and $c$ must agree one step
longer, so $b\cap c\neq r$. QED


The same idea works for higher cardinals as follows: 


**Theorem.** For any infinite cardinal $\delta$, we have 
$\Delta(\delta^{++},\delta^+)$ is equivalent to
$2^\delta=\delta^+$.


Proof. If $2^\delta=\delta^+$, then your criterion, which amounts to 
$(\delta^+)^{\lt\delta^+}<\delta^{++}$, is fulfilled, and so the
$\Delta$ property holds. Conversely, consider the tree
$T=2^{\lt\delta}$, the binary sequences of length less than $\delta$. Let $F$ be a family of
$\delta^{++}$ many branches through $T$, regarding each branch $b$ as a subset of $T$, the set of its initial segments. Each such branch has size $\delta$, since the tree has height $\delta$. But for the same reason as before, there can be no $\Delta$ system with even three elements, since the tree is merely binary branching, and so three distinct branches cannot have a common root. This contradicts $\Delta(\delta^{++},\delta^+)$, as desired. QED


**Corollary.** The full GCH is equivalent to the assertion that $\Delta(\delta^{++},\delta^+)$ for every infinite cardinal $\delta$. 




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**Update.** The same idea shows that the hypothesis you mention
is optimal: one can reverse the lemma from the conclusion to the hypothesis. 


**Theorem.** The following are equivalent, for regular $\kappa$
and $\mu\lt\kappa$:


1. $\Delta(\kappa,\mu)$
2. $\lambda^{\lt\mu}\lt\kappa$ for every $\lambda\lt\kappa$.


Proof. You mentioned that 2 implies 1, and this is how one usually
sees the $\Delta$ system lemma stated. For the converse, suppose
that $\lambda^{\lt\mu}\geq\kappa$ for some $\lambda\lt\kappa$.
Since $\kappa$ is regular and $\mu\lt\kappa$, this implies $\lambda^\eta\geq\kappa$
for some $\eta\lt\mu$. Let $T$ be the $\lambda$-branching tree
$\lambda^{\lt\eta}$, which has height $\eta$. Let $F$ be a family
of $\kappa$ many branches through this tree, where we think of a
branch as the set of nodes in the tree that lie on it, a maximal linearly
ordered subset of the tree $T$. Each such branch is a set of size $\eta$. I claim that this family has no
subfamily that is $\Delta$ system of size $\lambda^+$. The reason is
that because the tree is $\lambda$-branching, if we have $\lambda^+$
many branches with a common root, then at least two of them must
extend that root to the next level in the same way, a contradiction to it being a
root. Thus, the failure of 2 implies the failure of 1, as desired.
QED


