Is this a weaker version of Simonyi's Conjecture

Question

So, I'm aware of Simonyi's conjecture which says that if $\mathcal{A}, \mathcal{B} \subset \mathcal{P}(n)$ satisfy the conditions: $$\forall A,A'\in\mathcal{A} \mbox{ and } \forall B, B' \in\mathcal{B}, (A \backslash B)=(A'\backslash B') \implies A=A' $$ $$\forall A,A'\in\mathcal{A} \mbox{ and } \forall B, B' \in\mathcal{B}, (B \backslash A)=(B'\backslash A') \implies B=B' $$ Then $|\mathcal{A}||\mathcal{B}|\leq2^{n}$. But, my teacher mentioned the other day something like if $\mathcal{A}, \mathcal{B} \subset \mathcal{P}(n)$ are such that $|A \cap B|$ is always even, then we also had that $|\mathcal{A}||\mathcal{B}|\leq2^{n}$. Firstly, is this true? And, secondly, is this a weaker version of the conditions in Simonyi's conjecture?

Answer 1

I was just looking into Simonyi's conjecture and noticed that your post has gone unanswered for quite some time. As for your question, yes, if $|A\cap B|$ is always even, then $|\mathcal{A}||\mathcal{B}|\leq 2^n$. To prove this, let $S_{\cal{A}}=\text{span}\{\mathbf{1}_A:A\in\cal{A}\}$ where $\mathbf{1}_A$ is the indicator vector of the set $A$ and the span is taken over $\mathbb{F}_2$. Similarly define $S_{\cal{B}}$. For any $A\in\cal{A},B\in\cal{B}$, we have $\langle\mathbf{1}_A,\mathbf{1}_B\rangle=|A\cap B|\pmod 2=0$. Therefore $S_\cal{A}$ and $S_\cal{B}$ are orthogonal subspaces of $\mathbb{F}_2^n$, so, if $k=\dim(S_\cal{A})$ and $h=\dim(S_\cal{B})$, then $k+h\leq n$. As we are over $\mathbb{F}_2$, we find that $|\cal{A}|\leq 2^k$ and $|\cal{B}|\leq 2^h$, so $|\cal{A}||\cal{B}|\leq 2^n$.

You can reach the same conclusion if $|A\cap B|$ is always odd by considering $S_\cal{A}$ and $S_\cal{B}'$ where $S_{\cal{B}}'=\text{span}\{\mathbf{1}_B-\mathbf{1}_{B^*}:B\in\cal{B}\}$ where $B^*$ is any fixed element of $\cal{B}$.

However, I don't see how this is related to Simonyi's conjecture.