If ${\displaystyle X}$  is a vector space and ${\displaystyle M}$  and ${\displaystyle N}$  are vector subspaces of ${\displaystyle X}$  then there is a well-defined addition map

Algebraic direct sum edit

The vector space ${\displaystyle X}$  is said to be the algebraic direct sum (or direct sum in the category of vector spaces) ${\displaystyle M\oplus N}$  when any of the following equivalent conditions are satisfied:

1. The addition map ${\displaystyle S:M\times N\to X}$  is a vector space isomorphism.^[1][2]
2. The addition map is bijective.
3. ${\displaystyle M\cap N=\{0\}}$  and ${\displaystyle M+N=X}$ ; in this case ${\displaystyle N}$  is called an algebraic complement or supplement to ${\displaystyle M}$  in ${\displaystyle X}$  and the two subspaces are said to be complementary or supplementary.^[2][3]

When these conditions hold, the inverse ${\displaystyle S^{-1}:X\to M\times N}$  is well-defined and can be written in terms of coordinates as

Equivalently, ${\displaystyle P_{M}(x)}$  and ${\displaystyle P_{N}(x)}$  are the unique vectors in ${\displaystyle M}$  and ${\displaystyle N,}$  respectively, that satisfy

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

In the category of topological vector spaces, that algebraic decomposition becomes less useful. The definition of a topological vector space requires the addition map ${\displaystyle S}$  to be continuous; its inverse ${\displaystyle S^{-1}:X\to M\times N}$  may not be.^[1] The categorical definition of direct sum, however, requires ${\displaystyle P_{M}}$  and ${\displaystyle P_{N}}$  to be morphisms — that is, continuous linear maps.

The space ${\displaystyle X}$  is the topological direct sum of ${\displaystyle M}$  and ${\displaystyle N}$  if (and only if) any of the following equivalent conditions hold:

1. The addition map ${\displaystyle S:M\times N\to X}$  is a TVS-isomorphism (that is, a surjective linear homeomorphism).^[1]
2. ${\displaystyle X}$  is the algebraic direct sum of ${\displaystyle M}$  and ${\displaystyle N}$  and also any of the following equivalent conditions:
   1. The inverse of the addition map ${\displaystyle S^{-1}:X\to M\times N}$  is continuous.
   2. Both canonical projections ${\displaystyle P_{M}:X\to M}$  and ${\displaystyle P_{N}:X\to N}$  are continuous.
   3. At least one of the canonical projections ${\displaystyle P_{M}}$  and ${\displaystyle P_{N}}$  is continuous.
   4. The canonical quotient map ${\displaystyle p:N\to X/M;p(n)=n+M}$  is an isomorphism of topological vector spaces (i.e. a linear homeomorphism).^[2]
3. ${\displaystyle X}$  is the direct sum of ${\displaystyle M}$  and ${\displaystyle N}$  in the category of topological vector spaces.
4. The map ${\displaystyle S}$  is bijective and open.
5. When considered as additive topological groups, ${\displaystyle X}$  is the topological direct sum of the subgroups ${\displaystyle M}$  and ${\displaystyle N.}$ 

The topological direct sum is also written ${\displaystyle X=M\oplus N}$ ; whether the sum is in the topological or algebraic sense is usually clarified through context.

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

${\displaystyle M}$  is called complemented if it has a topological complement ${\displaystyle N}$  (and uncomplemented if not). The choice of ${\displaystyle N}$  can matter quite strongly: every complemented vector subspace ${\displaystyle M}$  has algebraic complements that do not complement ${\displaystyle M}$  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.

Equivalent characterizations edit

The vector subspace ${\displaystyle M}$  is complemented in ${\displaystyle X}$  if and only if any of the following holds:^[1]

• There exists a continuous linear map ${\displaystyle P_{M}:X\to X}$  with image ${\displaystyle P_{M}(X)=M}$  such that ${\displaystyle P\circ P=P}$ . That is, ${\displaystyle P_{M}}$  is a continuous linear projection onto ${\displaystyle M}$ . (In that case, algebraically ${\displaystyle X=M\oplus \ker {P}}$ , and it is the continuity of ${\displaystyle P_{M}}$  that implies that this is a complement.)
• For every TVS ${\displaystyle Y,}$  the restriction map ${\displaystyle R:L(X;Y)\to L(M;Y);R(u)=u|_{M}}$  is surjective.^[5]

If in addition ${\displaystyle X}$  is Banach, then an equivalent condition is

• ${\displaystyle M}$  is closed in ${\displaystyle X}$ , there exists another closed subspace ${\displaystyle N\subseteq X}$ , and ${\displaystyle S}$  is an isomorphism from the abstract direct sum ${\displaystyle M\oplus N}$  to ${\displaystyle X}$ .

• If ${\displaystyle Y}$  is a measure space and ${\displaystyle X\subseteq Y}$  has positive measure, then ${\displaystyle L^{p}(X)}$  is complemented in ${\displaystyle L^{p}(Y)}$ .
• ${\displaystyle c_{0}}$ , the space of sequences converging to ${\displaystyle 0}$ , is complemented in ${\displaystyle c}$ , the space of convergent sequences.
• By Lebesgue decomposition, ${\displaystyle L^{1}([0,1])}$  is complemented in ${\displaystyle \mathrm {rca} ([0,1])\cong C([0,1])^{*}}$ .

For any two topological vector spaces ${\displaystyle X}$  and ${\displaystyle Y}$ , the subspaces ${\displaystyle X\times \{0\}}$  and ${\displaystyle \{0\}\times Y}$  are topological complements in ${\displaystyle X\times Y}$ .

Every algebraic complement of ${\displaystyle {\overline {\{0\}}}}$ , the closure of ${\displaystyle 0}$ , is also a topological complement. This is because ${\displaystyle {\overline {\{0\}}}}$  has the indiscrete topology, and so the algebraic projection is continuous.^[6]

If ${\displaystyle X=M\oplus N}$  and ${\displaystyle A:X\to Y}$  is surjective, then ${\displaystyle Y=AM\oplus AN}$ .^[2]

Finite dimension edit

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

In arbitrary topological vector spaces, a finite-dimensional vector subspace ${\displaystyle Y}$  is topologically complemented if and only if for every non-zero ${\displaystyle y\in Y}$ , there exists a continuous linear functional on ${\displaystyle X}$  that separates ${\displaystyle y}$  from ${\displaystyle 0}$ .^[1] For an example in which this fails, see § Fréchet spaces.

Finite codimension edit

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

Hilbert spaces edit

In a Hilbert space, the orthogonal complement ${\displaystyle M^{\bot }}$  of any closed vector subspace ${\displaystyle M}$  is always a topological complement of ${\displaystyle M}$ . 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]

Fréchet spaces edit

Let ${\displaystyle X}$  be a Fréchet space over the field ${\displaystyle \mathbb {K} }$ . Then the following are equivalent:^[10]

1. ${\displaystyle X}$  is not normable (that is, any continuous norm does not generate the topology)
2. ${\displaystyle X}$  contains a vector subspace TVS-isomorphic to ${\displaystyle \mathbb {K} ^{\mathbb {N} }.}$ 
3. ${\displaystyle X}$  contains a complemented vector subspace TVS-isomorphic to ${\displaystyle \mathbb {K} ^{\mathbb {N} }}$ .

A complemented (vector) subspace of a Hausdorff space ${\displaystyle X}$  is necessarily a closed subset of ${\displaystyle X}$ , 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 ${\displaystyle X}$  is a complete TVS and ${\displaystyle X/M}$  is not complete, then ${\displaystyle M}$  has no topological complement in ${\displaystyle X.}$ ^[11]

If ${\displaystyle A:X\to Y}$  is a continuous linear surjection, then the following conditions are equivalent:

1. The kernel of ${\displaystyle A}$  has a topological complement.
2. There exists a "right inverse": a continuous linear map ${\displaystyle B:Y\to X}$  such that ${\displaystyle AB=\mathrm {Id} _{Y}}$ , where ${\displaystyle \operatorname {Id} _{Y}:Y\to Y}$  is the identity map.^[5]

(Note: This claim is an erroneous exercise given by Trèves. Let ${\displaystyle X}$  and ${\displaystyle Y}$  both be ${\displaystyle \mathbb {R} }$  where ${\displaystyle X}$  is endowed with the usual topology, but ${\displaystyle Y}$  is endowed with the trivial topology. The identity map ${\displaystyle X\to Y}$  is then a continuous, linear bijection but its inverse is not continuous, since ${\displaystyle X}$  has a finer topology than ${\displaystyle Y}$ . The kernel ${\displaystyle \{0\}}$  has ${\displaystyle X}$  as a topological complement, but we have just shown that no continuous right inverse can exist. If ${\displaystyle A:X\to Y}$  is also open (and thus a TVS homomorphism) then the claimed result holds.)

The Method of Decomposition edit

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

Let ${\displaystyle X}$  and ${\displaystyle Y}$  be TVSs such that ${\displaystyle X=X\oplus X}$  and ${\displaystyle Y=Y\oplus Y.}$  Suppose that ${\displaystyle Y}$  contains a complemented copy of ${\displaystyle X}$  and ${\displaystyle X}$  contains a complemented copy of ${\displaystyle Y.}$  Then ${\displaystyle X}$  is TVS-isomorphic to ${\displaystyle Y.}$ 

The "self-splitting" assumptions that ${\displaystyle X=X\oplus X}$  and ${\displaystyle Y=Y\oplus Y}$  cannot be removed: Tim Gowers showed in 1996 that there exist non-isomorphic Banach spaces ${\displaystyle X}$  and ${\displaystyle Y}$ , each complemented in the other.^[12]

Understanding the complemented subspaces of an arbitrary Banach space ${\displaystyle X}$  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 ${\displaystyle L_{1}[0,1]}$ .^[13]

For some Banach spaces the question is closed. Most famously, if ${\displaystyle 1\leq p\leq \infty }$  then the only complemented infinite-dimensional subspaces of ${\displaystyle \ell _{p}}$  are isomorphic to ${\displaystyle \ell _{p},}$  and the same goes for ${\displaystyle c_{0}.}$  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 ${\displaystyle L_{p}[0,1]}$  are not prime whenever ${\displaystyle p\in (1,2)\cup (2,\infty );}$  in fact, they admit uncountably many non-isomorphic complemented subspaces.^[13]

The spaces ${\displaystyle L_{2}[0,1]}$  and ${\displaystyle L_{\infty }[0,1]}$  are isomorphic to ${\displaystyle \ell _{2}}$  and ${\displaystyle \ell _{\infty },}$  respectively, so they are indeed prime.^[13]

The space ${\displaystyle L_{1}[0,1]}$  is not prime, because it contains a complemented copy of ${\displaystyle \ell _{1}}$ . No other complemented subspaces of ${\displaystyle L_{1}[0,1]}$  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 ${\displaystyle X}$  is always isomorphic to ${\displaystyle X,}$  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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