Show that $t-(v,k,\lambda)$ design with $v\leq k+t$ is trivial

Question

Let $D=(V,B)$ be a $t-(v,k,\lambda)$ design where $V$ is a finite set and $B$ is a collection of subsets of $V$. A trivial $t$-design is defined as a design with one block that contains all the points or a design that has all the $k$-subsets of the set. When $t=0$, $D$ is trivial. I set $t>0$, and then I tried to show that the number of blocks is $n \choose k$. It seems that I need to approach in other way.

Answer 1

Unfortunately, I do not have a solution to the problem; just a few observations (which are too long for a comment, so I am posting them here).

First, when the problem talks about $B$ being a collection of subsets of $V$, this could either mean a set of subsets or a "multiset" of subsets of $V$. The latter is not uncommon in combinatorial design theory; however, the statement of the problem would be clearly false, so for now we chose to interpret collection as set.

Second, the seemingly two different cases in the problem ("one block that contains all the points" vs. "a design that has all the $k$-subsets of the set") can really be merged into one because the first case is simply a special case of the second one (with $v=k$).

So, now let's give the scenario a name and call a $t-(v,k,\lambda)$ design rigid if $v\le k+t$. Then the claim is that every rigid design is trivial.

The third observation is that it suffices to show the claim for designs with $v=k+t$ (we might call these strongly rigid) because then it follows for all rigid designs by induction over $t$.

Now in order to do this, let $D=(V,B)$ be a strongly rigid $t-(v,k,\lambda)$ design. Then $D$ is trivial if and only if $\lambda={k\choose t}$.

Now this is the point where things start to get interesting: I have tried to find a counterexample (by finding parameters for which we can choose $\lambda$ to be smaller than $k\choose t$), but haven't managed to find one. For example, consider the case $t=6, k=9, v=15$. Then in order for $$b=\lambda\cdot\frac{15\cdot14\cdot13\cdot12\cdot11\cdot10}{9\cdot8\cdot7\cdot6\cdot5\cdot4}=\lambda\cdot\frac{715}{12}$$ to be an integer, we could simply choose $\lambda=12$, but we also need to consider the number $b''$ of blocks in the second derivative $D''$ of $D$, and here $$b''=\lambda\cdot\frac{13\cdot12\cdot11\cdot10}{7\cdot6\cdot5\cdot4}=12\cdot\frac{143}{7}=\frac{1716}{7}$$ which is not an integer. So, in this example, $12$ is too small and we really need to choose $\lambda=7\cdot 12=84={k\choose t}$.

I have examined several sets of parameters like this, and they all end up with $\lambda={k\choose t}$ in one way or the other. On the other hand, I can't seem to find a systematic approach to a general proof. In the special cases of small values of $t$ (i.e. for $t\le3$), it can be shown with a few number theoretic arguments, but for larger values of $t$ (particularly for compound ones), the number and complexity of the case distinctions grows uncomfortably, so I really don't know how to go on at this point.

If anyone can shed some light on the situation, I'd be very grateful.

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Update, Feb-12-2018:

Sometimes it is rewarding to consult the literature: I found a solution to your problem in Albrecht Beutelspacher's "Einführung in die endliche Geometrie I". The following is based on proposition 4.1.1:

Let $D=(V,B)$ be a $t-(v,k,\lambda)$ design with $t\ge2$, and let $\mathcal{I}$ and $\mathcal{J}$ be disjoint subsets of $V$ or orders $i$ and $j$, respectively, such that $i+j\le t$.
Then the number of blocks of $B$ that contain $\mathcal{I}$ and are disjoint from $\mathcal{J}$ is $$b_{i,j}:=\lambda\cdot\frac{{v-i-j}\choose{k-i}}{{v-t}\choose{k-t}}.$$

In particular, the number $b_{i,j}$ does not depend on the choices of $\mathcal{I}$ and $\mathcal{J}$, (or, in other words, designs do not only behave regularly with respect to vertex sets that are contained in their blocks, but also with respect to vertex sets that their blocks miss – at least as long as these vertex sets are small enough).

The proposition can relatively easily be shown by using induction on $j$.

The original claim then follows as a corollary: Let $D=(V,B)$ be a strongly rigid $t-(v,k,\lambda)$ design, i.e. $v=k+t$. Then the proposition yields that $b_{0,t}$ is positive, i.e. for any set $T$ of $t$ points of $V$, there exists a block $S$ of $B$ that completely misses $T$. But $|V\setminus T|=k$, so $S=V\setminus T$. But since $T$ was arbitrary, this means that every $k$-subset of $V$ is a block, i.e. $D$ is trivial.