Math 2.0

Order, lattices, ordered algebraic structures Lattices: basic notions Example

Lattice of subspaces

Builds on Vector spaceLinear subspace

Example

Let VV be a vector space over a field FF, and let L(V)L(V) be the set of all linear subspaces of VV, ordered by inclusion. The meet of two subspaces UU and WW is U∩WU \cap W, and their join is the sum U+W={u+w:u∈U,w∈W}U + W = \set{u + w : u \in U, w \in W}. For any family (Ui)i∈I(U_i)_{i \in I} of subspaces, the intersection ⋂iUi\bigcap_i U_i and the span of ⋃iUi\bigcup_i U_i are subspaces. They are the greatest lower bound and the least upper bound of the family. So L(V)L(V) is a complete lattice, with least element {0}\set{0} and greatest element VV. It is a modular lattice: if U⊆WU \subseteq W, then U+(X∩W)=(U+X)∩WU + (X \cap W) = (U + X) \cap W for every subspace XX. The inclusion U+(X∩W)⊆(U+X)∩WU + (X \cap W) \subseteq (U + X) \cap W holds because U⊆WU \subseteq W. For the other inclusion, an element w=u+xw = u + x of the right side with u∈Uu \in U, x∈Xx \in X has x=w−u∈Wx = w - u \in W, so x∈X∩Wx \in X \cap W.

Remarks

The lattice L(V)L(V) is not distributive once dim⁡V≥2\dim V \ge 2. In V=ℝ2V = \R^2, take three distinct lines aa, bb, cc through 00. Then a∩(b+c)=a∩ℝ2=aa \cap (b + c) = a \cap \R^2 = a, while (a∩b)+(a∩c)={0}(a \cap b) + (a \cap c) = \set{0}. When VV is finite-dimensional, the subspaces of dimension 11 and 22 are the points and lines of the projective space of VV, and L(V)L(V) encodes its incidence geometry.

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To understand this page from the ground up, read these 3 pages in order. The foundations come first.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Projective_geometry
About this page
TagH5G6TGRQH1
URIm2://H5G6TGRQH1
Kindexample
WritingW0 stub
MSC06C05, 06B23, 51A05
Instance ofComplete lattice, Modular lattice
UsesVector space, Linear subspace

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