Let $n$ be a natural number, and consider the discrete cube $2^{[n]} := \{ A: A \subset \{1,\ldots,n\}\}$ consisting of all subsets of the $n$-element set $[n] := \{1,\ldots,n\}$. Define a *downset* in $2^{[n]}$ to be a collection ${\mathcal D}$ of elements $A$ in $2^{[n]}$ with the property that if $A \in {\mathcal D}$ and $B \subset A$, then $B \in {\mathcal D}$.

My question is: what are the largest and smallest possible values for the alternating sum $\sum_{A \in {\mathcal D}} (-1)^{|A|}$, as ${\mathcal D}$ ranges over downsets in $2^{[n]}$, as a function of $n$? (Here $|A|$ denotes the cardinality of $A$.)

The trivial bounds here are $\pm 2^{n-1}$, by taking only the positive or negative values of ${\mathcal D}$, but of course these values are attained on a "checkerboard" set which is very far from being a downset, and this suggests that significant improvement is possible.

By taking ${\mathcal D}$ to be the set of all subsets $A$ of $2^{[n]}$ of cardinality at most $r$ for some $1 \leq r \leq n-1$, this gives a value of $(-1)^r \binom{n-1}{r}$; setting $r$ close to $(n-1)/2$ then seems to give reasonably good extremals (of size about $2^n/\sqrt{n}$ asymptotically). In the spirit of Sperner's lemma, one might tentatively conjecture that these are the extremal examples, but I was unable to prove or disprove this. (I feel like I'm missing some obvious application of downset isoperimetric inequalities or something.)

One motivation for this question is from analytic number theory: partial divisor sums $\sum_{d|a: d \leq x} \mu(d)$ of the Mobius function (which show up from time to time in this subject) can be viewed as an alternating sum over a downset, where $n$ is the number of prime factors $p_1,\ldots,p_n$ of $a$, and ${\mathcal D}$ is the collection of subsets $A$ of $[n]$ for which $\prod_{i \in A} p_i \leq x$. So any bounds on the general alternating-sum-of-downset problem would imply bounds on partial divisor sums of the Mobius function that depend only on the number of prime factors.

One small observation (using the shifting technology of Frankl) which may or may not be of use: given two natural numbers $1 \leq i < j \leq n$ and a downset ${\mathcal D}$, define the *$ij$-shift* of ${\mathcal D}$ to be the set formed by replacing any element of ${\mathcal D}$ of the form $A \cup \{j\}$ with $A \cup \{i\}$, if $A$ is disjoint from $\{i,j\}$ and $A \cup \{i\}$ is not already in A. Note that this is again a downset. Call a downset ${\mathcal D}$ *shift-minimal* if it is equal to all of its $ij$-shifts. Then one can reduce without loss of generality to the shift-minimal case, because shifting does not affect the sum $\sum_{A \in {\mathcal D}} (-1)^{|A|}$. In other contexts, the reduction to the shift-minimal case can be very powerful, but for some strange reason I was unable to exploit it here.