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Let $V_1 , V_2 , \dots , V_n $ be vector subspaces of $ \mathbb{C}^m$ and let $$\alpha = \sum_{r=1}^n (-1)^{r+1} \sum_{ \ i_1 < i_2 < \cdots < i_r } \dim(V_{i_1} \cap \cdots \cap V_{i_r})$$

For $n = 2$ we have the equality $ \alpha = \dim(\sum_{i = 1}^{n} V_i) $; it's false for $n>2$, see this answer.
For $n=3$, we have only the inequality $ \alpha \ge \dim(\sum_{i = 1}^{n} V_i) $; it's false for $n>3$, see this post.
For $n>5$, the inequality $\alpha \ge 0$ is false in general, see the comment of Darij Grinberg below.

Question: Is it true that $\alpha \ge 0$, in the case $n \le 5$?

Remark: I think this question interesting for itself; it admits also applications in the interaction between representations theory and subgroups lattice.

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Sebastien Palcoux
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2 Answers2

5

Here is a purely combinatorial proof for $n=5$. We first generalize the problem as follows:
Let $B_5$ be the boolean lattice of rank $5$, i.e. the subsets lattice of $\{1,2,3,4,5\}$.

Lemma: Let $\phi: B_5 \to \mathbb{R}_{\ge 0}$ be a map satisfying that $\forall a, b \in B_5$:

$(1)$ $ \ $ $a \le b \Rightarrow \phi(a) \le \phi(b)$ [poset morphism]
$(2)$ $ \ $ $\phi(a \vee b) + \phi(a \wedge b) \ge \phi(a ) + \phi(b)$

and let $a_i= \{i \}^\complement$ the complement of $\{i \}$ in $\{1,2,3,4,5\}$, then $$\sum_{r=1}^5 (-1)^{r+1}\sum_{i_1 < i_2 < \cdots < i_r} \phi( a_{i_1} \wedge \cdots \wedge a_{i_r}) \ge 0$$ proof: we reorganize the alternative sum into the sum of the following components:

  • $\phi(\{1,2,3,4\}) - \phi(\{1,2,3\}) - \phi(\{1,2,4\}) + \phi(\{1,2\})$
  • $\phi(\{1,3,4,5 \}) - \phi(\{1,3,4\}) - \phi(\{1,3,5\}) + \phi(\{1,3\})$
  • $\phi(\{2,3,4,5\}) - \phi(\{2,3,4\}) - \phi(\{3,4,5\}) + \phi(\{3,4\})$
  • $\phi(\{1,2,4,5 \}) - \phi(\{2,4,5\}) - \phi(\{1,4,5\}) + \phi(\{4,5\})$
  • $\phi(\{1,2,3,5\}) - \phi(\{1,2,5\}) - \phi(\{2,3,5\}) + \phi(\{2,5\})$
  • $ \phi(\{1,5\}) - \phi(\{1\})$
  • $ \phi(\{2,4\}) - \phi(\{2\})$
  • $ \phi(\{2,3\}) - \phi(\{3\})$
  • $ \phi(\{1,4\}) - \phi(\{4\})$
  • $ \phi(\{3,5\}) - \phi(\{5\})$
  • $ \phi(\emptyset) $

but the first five components are positive by $(2)$, the next five components are positive by $(1)$, and the last is positive by definition $\square$.

Now the answer of the question is yes by observing that the map $\phi$ defined by $$\phi( a_{i_1} \wedge \cdots \wedge a_{i_r}) = \dim (V_{i_1} \cap \cdots \cap V_{i_r})$$ checks $(1)$ and $(2)$. For $(1)$ it is immediate. For $(2)$ we use the following equality and inclusion: $\dim(U+V) = \dim(U) + \dim(V) - \dim(U \cap V)$ and $(A\cap B ) + (A\cap C) \subseteq A$.

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Sebastien Palcoux
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  • Nice work! To restate your argument without the use of $B_5$ (and with a cyclically symmetric choice of inequalities): You sum the inequality $\dim\left(V_1\right) - \dim\left(V_1 \cap V_2\right) - \dim\left(V_1 \cap V_4\right) + \dim\left(V_1 \cap V_2 \cap V_4\right) \geq 0$ and the four inequalities obtained from it by cycling through the indices. Then, you add the inequality $\dim\left(V_1 \cap V_2 \cap V_3\right) - \dim\left(V_1 \cap V_2 \cap V_3 \cap V_4\right) \geq 0$ and the four inequalities obtained from it by cycling through the indices. The result is the claim you wanted to prove. – darij grinberg Mar 27 '16 at 22:26
  • Actually, this tactic gives an even simpler proof of the $n=4$ case: You sum the inequality $\dim\left(V_1\right) - \dim\left(V_1 \cap V_2\right) - \dim\left(V_1 \cap V_3\right) + \dim\left(V_1 \cap V_2 \cap V_3\right) \geq 0$ and the three inequalities obtained from it by cycling through the indices. Then, you add the inequality $\dim\left(V_1 \cap V_3\right) + \dim\left(V_2 \cap V_4\right) \geq \dim\left(V_1 \cap V_2 \cap V_3 \cap V_4\right)$, which is completely obvious (indeed, each of the two addends on its left hand side on its own is greater or equal to the right hand side). Done. – darij grinberg Mar 27 '16 at 22:29
  • That said, the $n=4$ case of course also follows from the $n=5$ case (since we can just take $V_5 = 0$). – darij grinberg Mar 27 '16 at 22:31
  • Actually, I've just realized that in the $n=5$ case, both you and I have forgotten the $\dim\left(V_1\cap V_2\cap V_3\cap V_4\cap V_5\right)$ term. But that's fine -- since $n=5$ is odd, this term appears with a plus sign on the left hand side, so the inequality we obtain without it is better than the one with it. For $n=4$, the addend $\dim\left(V_1\cap V_2\cap V_3\cap V_4\right)$ is necessary, however. – darij grinberg Mar 27 '16 at 22:35
  • @darijgrinberg: I've edited the proof. Moreover, we can always assume $\dim(\bigcap_{i=1}^n V_i) = 0$, see the beginning of my answer for $n=4$. – Sebastien Palcoux Mar 28 '16 at 07:48
1

Here is proof by hand for $n=4$:

Let $X = \bigcap_{i=1}^n V_i$ then by taking $V_i' = V_i \ominus X$ we get that $$\sum_{r=1}^n (-1)^{r+1} \sum_{ \ i_1 < i_2 < \cdots < i_r } \dim(V_{i_1} \cap \cdots \cap V_{i_r}) = \dim(X) + \sum_{r=1}^n (-1)^{r+1} \sum_{ \ i_1 < i_2 < \cdots < i_r } \dim(V'_{i_1} \cap \cdots \cap V'_{i_r})$$ So we can assume $\dim(X) = 0$.

First we apply four times the inequality: $ \dim(U +V + W) \le $ $$ \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W) $$ and we get $$\alpha \ge \sum_{ \ i_1 < i_2 < i_3 } \dim(V_{i_1} + V_{i_2} + V_{i_3})-2\sum_i\dim(V_i) + \sum_{ \ i_1 < i_2 } \dim(V_{i_1} \cap V_{i_2} )$$ Next we apply six times the equality: $$\dim(U+V) = \dim U + \dim V - \dim (U \cap V)$$ and we get $$\alpha \ge \sum_{ \ i_1 < i_2 < i_3 } \dim(V_{i_1} + V_{i_2} + V_{i_3}) - \sum_{ \ i_1 < i_2 } \dim(V_{i_1} + V_{i_2} ) + \sum_i\dim(V_i)$$ Finally we observe that:
$\dim(V_1+V_2+V_3) \ge \dim(V_1+V_3)$
$\dim(V_1+V_2+V_4) \ge \dim(V_2+V_4)$
$\dim(V_1+V_3+V_4) \ge \dim(V_1+V_4)$
$\dim(V_2+V_3+V_4) \ge \dim(V_2+V_3)$
$\sum_i\dim(V_i) \ge \dim(V_1+V_2) + \dim(V_3+V_4)$

It follows that $\alpha \ge 0$ $\square$

Sebastien Palcoux
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