In the branch of mathematics called functional analysis, a complemented subspace of a topological vector space is a vector subspace for which there exists some other vector subspace of called its (topological) complement in , such that is the direct sum in the category of topological vector spaces. Formally, topological direct sums strengthen the algebraic direct sum by requiring certain maps be continuous; the result retains many nice properties from the operation of direct sum in finite-dimensional vector spaces.

Every finite-dimensional subspace of a Banach space is complemented, but other subspaces may not. In general, classifying all complemented subspaces is a difficult problem, which has been solved only for some well-known Banach spaces.

The concept of a complemented subspace is analogous to, but distinct from, that of a set complement. The set-theoretic complement of a vector subspace is never a complementary subspace.

If is a vector space and and are vector subspaces of then there is a well-defined addition map The map is a morphism in the category of vector spaces — that is to say, linear.

The vector space is said to be the algebraic direct sum (or direct sum in the category of vector spaces) when any of the following equivalent conditions are satisfied:

1. The addition map is a vector space isomorphism.^[1][2]
2. The addition map is bijective.
3. and ; in this case is called an algebraic complement or supplement to in and the two subspaces are said to be complementary or supplementary.^[2][3]

When these conditions hold, the inverse is well-defined and can be written in terms of coordinates as The first coordinate is called the canonical projection of onto ; likewise the second coordinate is the canonical projection onto ^[4]

Equivalently, and are the unique vectors in and respectively, that satisfy As maps, where denotes the identity map on .^[2]

Suppose that the vector space is the algebraic direct sum of . In the category of vector spaces, finite products and coproducts coincide: algebraically, and are indistinguishable. Given a problem involving elements of , one can break the elements down into their components in and , because the projection maps defined above act as inverses to the natural inclusion of and into . Then one can solve the problem in the vector subspaces and recombine to form an element of .

In the category of topological vector spaces, that algebraic decomposition becomes less useful. The definition of a topological vector space requires the addition map to be continuous; its inverse may not be.^[1] The categorical definition of direct sum, however, requires and to be morphisms — that is, continuous linear maps.

The space is the topological direct sum of and if (and only if) any of the following equivalent conditions hold:

1. The addition map is a TVS-isomorphism (that is, a surjective linear homeomorphism).^[1]
2. is the algebraic direct sum of and and also any of the following equivalent conditions:
   1. The inverse of the addition map is continuous.
   2. Both canonical projections and are continuous.
   3. At least one of the canonical projections and is continuous.
   4. The canonical quotient map is an isomorphism of topological vector spaces (i.e. a linear homeomorphism).^[2]
3. is the direct sum of and in the category of topological vector spaces.
4. The map is bijective and open.
5. When considered as additive topological groups, is the topological direct sum of the subgroups and

The topological direct sum is also written ; whether the sum is in the topological or algebraic sense is usually clarified through context.

Every topological direct sum is an algebraic direct sum ; the converse is not guaranteed. Even if both and are closed in , may still fail to be continuous. is a (topological) complement or supplement to if it avoids that pathology — that is, if, topologically, . (Then  is likewise complementary to .)^[1] Condition 2(d) above implies that any topological complement of is isomorphic, as a topological vector space, to the quotient vector space .

is called complemented if it has a topological complement (and uncomplemented if not). The choice of can matter quite strongly: every complemented vector subspace has algebraic complements that do not complement topologically.

Because a linear map between two normed (or Banach) spaces is bounded if and only if it is continuous, the definition in the categories of normed (resp. Banach) spaces is the same as in topological vector spaces.

The vector subspace is complemented in if and only if any of the following holds:^[1]

• There exists a continuous linear map with image such that . That is, is a continuous linear projection onto . (In that case, algebraically , and it is the continuity of that implies that this is a complement.)
• For every TVS the restriction map is surjective.^[5]

If in addition is Banach, then an equivalent condition is

• is closed in , there exists another closed subspace , and is an isomorphism from the abstract direct sum to .

• If is a measure space and has positive measure, then is complemented in .
• , the space of sequences converging to , is complemented in , the space of convergent sequences.
• By Lebesgue decomposition, is complemented in .

For any two topological vector spaces and , the subspaces and are topological complements in .

Every algebraic complement of , the closure of , is also a topological complement. This is because has the indiscrete topology, and so the algebraic projection is continuous.^[6]

If and is surjective, then .^[2]

Suppose is Hausdorff and locally convex and a free topological vector subspace: for some set , we have (as a t.v.s.). Then is a closed and complemented vector subspace of .^{[proof 1]} In particular, any finite-dimensional subspace of is complemented.^[7]

In arbitrary topological vector spaces, a finite-dimensional vector subspace is topologically complemented if and only if for every non-zero , there exists a continuous linear functional on that separates from .^[1] For an example in which this fails, see § Fréchet spaces.

Not all finite-codimensional vector subspaces of a TVS are closed, but those that are, do have complements.^[7][8]

In a Hilbert space, the orthogonal complement of any closed vector subspace is always a topological complement of . This property characterizes Hilbert spaces within the class of Banach spaces: every infinite dimensional, non-Hilbert Banach space contains a closed uncomplemented subspace, a deep theorem of Joram Lindenstrauss and Lior Tzafriri.^[9][3]

Let be a Fréchet space over the field . Then the following are equivalent:^[10]

1. is not normable (that is, any continuous norm does not generate the topology)
2. contains a vector subspace TVS-isomorphic to
3. contains a complemented vector subspace TVS-isomorphic to .

A complemented (vector) subspace of a Hausdorff space is necessarily a closed subset of , as is its complement.^{[1][proof 2]}

From the existence of Hamel bases, every infinite-dimensional Banach space contains unclosed linear subspaces.^{[proof 3]} Since any complemented subspace is closed, none of those subspaces is complemented.

Likewise, if is a complete TVS and is not complete, then has no topological complement in ^[11]

If is a continuous linear surjection, then the following conditions are equivalent:

1. The kernel of has a topological complement.
2. There exists a "right inverse": a continuous linear map such that , where is the identity map.^[5]

(Note: This claim is an erroneous exercise given by Trèves. Let and both be where is endowed with the usual topology, but is endowed with the trivial topology. The identity map is then a continuous, linear bijection but its inverse is not continuous, since has a finer topology than . The kernel has as a topological complement, but we have just shown that no continuous right inverse can exist. If is also open (and thus a TVS homomorphism) then the claimed result holds.)

Topological vector spaces admit the following Cantor-Schröder-Bernstein–type theorem:

Let and be TVSs such that and Suppose that contains a complemented copy of and contains a complemented copy of Then is TVS-isomorphic to

The "self-splitting" assumptions that and cannot be removed: Tim Gowers showed in 1996 that there exist non-isomorphic Banach spaces and , each complemented in the other.^[12]

Understanding the complemented subspaces of an arbitrary Banach space up to isomorphism is a classical problem that has motivated much work in basis theory, particularly the development of absolutely summing operators. The problem remains open for a variety of important Banach spaces, most notably the space .^[13]

For some Banach spaces the question is closed. Most famously, if then the only complemented infinite-dimensional subspaces of are isomorphic to and the same goes for Such spaces are called prime (when their only infinite-dimensional complemented subspaces are isomorphic to the original). These are not the only prime spaces, however.^[13]

The spaces are not prime whenever in fact, they admit uncountably many non-isomorphic complemented subspaces.^[13]

The spaces and are isomorphic to and respectively, so they are indeed prime.^[13]

The space is not prime, because it contains a complemented copy of . No other complemented subspaces of are currently known.^[13]

An infinite-dimensional Banach space is called indecomposable whenever its only complemented subspaces are either finite-dimensional or -codimensional. Because a finite-codimensional subspace of a Banach space is always isomorphic to indecomposable Banach spaces are prime.

The most well-known example of indecomposable spaces are in fact hereditarily indecomposable, which means every infinite-dimensional subspace is also indecomposable.^[14]

• Direct sum – Operation in abstract algebra composing objects into "more complicated" objects
• Direct sum of modules – Operation in abstract algebra
• Direct sum of topological groups

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