# Under which conditions: dim(W1 + W2 + W3) = dim(W1) + dim(W2) + dim(W3) − dim(W1 ∩ W2) − dim(W2 ∩ W3) − dim(W3 ∩ W1) + dim(W1 ∩ W2 ∩ W3) [closed]

Let $$V$$ be a finite dimensional vector space over a field $$K$$, and let $$W_1$$, $$W_2$$ and $$W_3$$ be subspaces of $$V$$. By analogy with the inclusion-exclusion principle for sets, and taking into account the dimension formula for a sum of 2 subspaces, we can ask whether the following equality holds:

$$\dim(W_1 + W_2 + W_3) = \dim(W_1) + \dim(W_2) + \dim(W_3) − \dim(W_1 \cap W_2) − \dim(W_2 \cap W_3) − \dim(W_3 \cap W_1) + \dim(W_1 \cap W_2 \cap W_3)$$ (†) it is false for the sum of 3 subspaces.

This formula does not always hold: for example take three distinct lines in $$\mathbb R^2$$ as $$U$$, $$V$$, $$W$$. All intersections have 0 dimensions. The LHS is $$2$$, the RHS is $$3$$.

I would like to state general assumptions on the subspaces of $$W_1$$, $$W_2$$, $$W_3$$ which guarantee that the formula does hold and prove it under these assumptions.

But for which triples of (finite-dimensional) subspaces does this hold? Do you have any reference for reading? I studied a few books but could not seem to find anything. To sum up, my question I am trying to prove that the formula is True even though it is False.

Many thanks

• You can find also some posts on Mathematics where a different formula was suggested, such as: Dimension of the sum of subspaces and The dimension of the sum of subspaces $(U_1,\ldots,U_n)$ And there is also this related post on MathOverflow: Is there a version of inclusion/exclusion for vector spaces? Nov 20 '21 at 7:50
• @MartinSleziak Thank you for your sources. However, I would like to explore an assumption to prove this. (even if the formula does not always hold).
– user468543
Nov 20 '21 at 8:16
• One possible answer would be: If there is a basis such that each $W_i$ is generated by a subset of this basis then the formula holds. Nov 20 '21 at 8:54
• I think you can answer this question in the spirit of indecomposable representations of quivers/posets. Any datum $(V; W_1, W_2, W_3)$ of three subspaces $W_i$ of a f.d. vector space $V$ is isomorphic to the sum of five indecomposable things: $(K;0,0,0)$, $(K; K, 0, 0)$, $(K; 0, K, 0)$, $(K; 0, 0, K)$ and $(K^2; K(1,0), K(0,1), K(1,1))$. Your formula holds if and only your datum doesn't contain the last indecomposable example. Nov 20 '21 at 9:33
• I have rolled back the yesterday’s anonymous suggested edit that claimed to “improved formatting and spelling”, but in reality just deleted two paragraphs that are setting the context for the question. Nov 22 '21 at 15:32

According to the theory of indecomposable modules of the Dynkin quiver $$D_4$$ with all arrows pointing to the central node there are $$12$$ cases corresponding to the $$12$$ positive roots of the Lie algebra of type $$D_4$$. These cases for $$(V;W_1,W_2,W_3)$$ are
1. $$(0;K,0,0), (0;0,K,0), (0;0,0,K)$$,
2. $$(K;0,0,0), (K;K,0,0), (K;0,K,0), (K;0,0,K), (K;K,K,0), (K;K,0,K),(K;0,K,K),(K;K,K,K)$$
3. $$(K^2,K,K,K)$$
All maps $$W_i\to V$$ in the first group are not injective. So, they are irrelevant. The second group consists precisely of those cases, where there is a basis of $$V$$ such that all $$W_i$$ are spanned by a subset of the basis. The last is case is causing the problems. In this case the difference $$\mathrm{LHS}-\mathrm{RHS}=-1$$. So we arrive at the following statement:
In the formula, one always has $$\mathrm{LHS}\le \mathrm{RHS}$$ with equality if and only if $$V$$ has a basis such that all $$W_i$$ are spanned by a subset of that basis.