Complemented subspace


In the branch of mathematics called functional analysis, when a topological vector space admits a direct sum decomposition , the spaces and are called complements of each other.
The concept of a complemented subspace in functional analysis should never be confused with that of a set complement in set theory, which is completely different.

Definition

Algebraic direct sum

If is a vector space and and are vector subspaces of then is the algebraic direct sum of and or the direct sum of and in the category of vector spaces if any of the following equivalent conditions is satisfied:

  1. the canonical map defined by is a vector space isomorphism;
  2. the canonical map is bijective;
  3. and ;
in this case is called an algebraic complementary or algebraic supplementary to in.
If is the algebraic direct sum of and and if is the canonical map defined by, then the inverse of the canonical map S can be written as where and are called the canonical projections of onto and, respectively.
Note that,, and for all.

Topological direct sum

If is the algebraic direct sum of and and if is also a topological vector space then the canonical map defined by is necessarily continuous but its inverse may fail to be continuous;
when it is continuous then is the direct sum of and in the category of TVSs.
If is a discontinuous linear functional on a locally convex Hausdorff TVS, n is any element of such that,, and, then is the algebraic direct sum of and but not the topological direct sum of and.
If is a topological vector space and if and are vector subspaces of, then is the topological direct sum of and or the direct sum of and in the category of TVSs if any of the following equivalent conditions is satisfied:

  1. the canonical map defined by is a TVS-isomorphism;
  2. the canonical map is a bijective open map;
  3. X is the algebraic direct sum of and and the inverse of the canonical map,, is continuous;
  4. X is the algebraic direct sum of and and the canonical projections and are continuous ;
  5. X is the algebraic direct sum of and and at least one of the two canonical projections and is continuous;
  6. X is the algebraic direct sum of and and the restriction to of the canonical quotient map is a TVS-isomorphism of onto ;
  7. The canonical vector space isomorphism is bicontinuous ;
    • Note that this implies that any topological complement of in is TVS-isomorphic to.
in this case is called a complement, complementary, or supplementary to in and we say that is a complemented subspace of.

Complemented subspaces

Let be a topological vector space and let be a vector subspace of.
We say that is a complemented subspace of if any of the following equivalent conditions hold:

  1. there exists a vector subspace of such that is the direct sum of and in the category of TVSs ;
  2. there exists a continuous linear map with range such that ;
  3. there exists a continuous linear projection with range such that is the algebraic sum of and the null space of P.
  4. for every TVS, the restriction map is surjective, where is defined by sending a continuous linear operator to the restriction.
It can be shown that if is a continuous linear bijection between two TVSs, then the following conditions are equivalent:

  1. the kernel of has a topological complement;
  2. there exists a continuous linear map such that, where is the identity map.

For Banach spaces

For the special case of a Banach space, the following definition is equivalent to the one given above for TVSs.
Let be a Banach space and a closed subspace of.
Then is called a complemented subspace of whenever there exists another closed subspace of such that is isomorphic to the direct sum ;
in this case is also a complemented subspace, and and are called complements of each other.

Examples and sufficient conditions

This led to the Schroeder-Bernstein problem: If and are Banach spaces and each is TVS-isomorphic to a complemented subspace of the other, then is TVS-isomorphic to ?
In 1996, Gowers provided a negative answer to this question.

Classifying complemented subspaces of a Banach space

Consider a Banach space.
What are the complemented subspaces of, up to isomorphism?
Answering this question is one of the complemented subspace problems that remains open for a variety of important Banach spaces, most notably the space .
For some Banach spaces the question is closed. Most famously, if then the only complemented subspaces of are isomorphic to, and the same goes for.
Such spaces are called prime. These aren't the only prime spaces, however, as we see in the next section.
The spaces are not prime whenever ; in fact, they admit uncountably many non-isomorphic complemented subspaces.
The spaces and are isomorphic to and, respectively, so they are indeed prime.
The space isn't prime due to the fact that it contains a complemented copy of, however no other complemented subspaces of are currently known.
It's undoubtedly one of the most interesting open problems in functional analysis whether admits other complemented subspaces.

Indecomposable Banach spaces

An infinite-dimensional Banach space is called indecomposable whenever its only complemented subspaces are either finite-dimensional or finite-codimensional.
Due to the fact that a finite-codimensional subspace of a Banach space is always isomorphic to, that makes indecomposable Banach spaces prime.
The most well-known example of indecomposable spaces are in fact hereditarily indecomposable, which means every infinite-dimensional subspace is also indecomposable.
Such spaces are fairly nasty, and have been constructed specifically to defy the sort of behavior we typically desire in a Banach space.