If X {\displaystyle X} is a vector space and M {\displaystyle M} and N {\displaystyle N} are vector subspaces of X {\displaystyle X} then there is a well-defined addition map S : M × N → X ( m , n ) ↦ m + n {\displaystyle {\begin{alignedat}{4}S:\;&&M\times N&&\;\to \;&X\\&&(m,n)&&\;\mapsto \;&m+n\\\end{alignedat}}} The map S {\displaystyle S} is a morphism in the category of vector spaces — that is to say, linear.
Main articles: Direct sum and Direct sum of modules
The vector space X {\displaystyle X} is said to be the algebraic direct sum (or direct sum in the category of vector spaces) M ⊕ N {\displaystyle M\oplus N} when any of the following equivalent conditions are satisfied:
When these conditions hold, the inverse S − 1 : X → M × N {\displaystyle S^{-1}:X\to M\times N} is well-defined and can be written in terms of coordinates as S − 1 = ( P M , P N ) . {\displaystyle S^{-1}=\left(P_{M},P_{N}\right){\text{.}}} The first coordinate P M : X → M {\displaystyle P_{M}:X\to M} is called the canonical projection of X {\displaystyle X} onto M {\displaystyle M} ; likewise the second coordinate is the canonical projection onto N . {\displaystyle N.} 5
Equivalently, P M ( x ) {\displaystyle P_{M}(x)} and P N ( x ) {\displaystyle P_{N}(x)} are the unique vectors in M {\displaystyle M} and N , {\displaystyle N,} respectively, that satisfy x = P M ( x ) + P N ( x ) . {\displaystyle x=P_{M}(x)+P_{N}(x){\text{.}}} As maps, P M + P N = Id X , ker P M = N , and ker P N = M {\displaystyle P_{M}+P_{N}=\operatorname {Id} _{X},\qquad \ker P_{M}=N,\qquad {\text{ and }}\qquad \ker P_{N}=M} where Id X {\displaystyle \operatorname {Id} _{X}} denotes the identity map on X {\displaystyle X} .6
See also: Coproduct, Direct sum § Direct sum in categories, and Direct sum of topological groups
Suppose that the vector space X {\displaystyle X} is the algebraic direct sum of M ⊕ N {\displaystyle M\oplus N} . In the category of vector spaces, finite products and coproducts coincide: algebraically, M ⊕ N {\displaystyle M\oplus N} and M × N {\displaystyle M\times N} are indistinguishable. Given a problem involving elements of X {\displaystyle X} , one can break the elements down into their components in M {\displaystyle M} and N {\displaystyle N} , because the projection maps defined above act as inverses to the natural inclusion of M {\displaystyle M} and N {\displaystyle N} into X {\displaystyle X} . Then one can solve the problem in the vector subspaces and recombine to form an element of X {\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 S {\displaystyle S} to be continuous; its inverse S − 1 : X → M × N {\displaystyle S^{-1}:X\to M\times N} may not be.7 The categorical definition of direct sum, however, requires P M {\displaystyle P_{M}} and P N {\displaystyle P_{N}} to be morphisms — that is, continuous linear maps.
The space X {\displaystyle X} is the topological direct sum of M {\displaystyle M} and N {\displaystyle N} if (and only if) any of the following equivalent conditions hold:
The topological direct sum is also written X = M ⊕ N {\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 X = M ⊕ N {\displaystyle X=M\oplus N} ; the converse is not guaranteed. Even if both M {\displaystyle M} and N {\displaystyle N} are closed in X {\displaystyle X} , S − 1 {\displaystyle S^{-1}} may still fail to be continuous. N {\displaystyle N} is a (topological) complement or supplement to M {\displaystyle M} if it avoids that pathology — that is, if, topologically, X = M ⊕ N {\displaystyle X=M\oplus N} . (Then M {\displaystyle M} is likewise complementary to N {\displaystyle N} .)10 Condition 2(d) above implies that any topological complement of M {\displaystyle M} is isomorphic, as a topological vector space, to the quotient vector space X / M {\displaystyle X/M} .
M {\displaystyle M} is called complemented if it has a topological complement N {\displaystyle N} (and uncomplemented if not). The choice of N {\displaystyle N} can matter quite strongly: every complemented vector subspace M {\displaystyle M} has algebraic complements that do not complement M {\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.
The vector subspace M {\displaystyle M} is complemented in X {\displaystyle X} if and only if any of the following holds:11
If in addition X {\displaystyle X} is Banach, then an equivalent condition is
For any two topological vector spaces X {\displaystyle X} and Y {\displaystyle Y} , the subspaces X × { 0 } {\displaystyle X\times \{0\}} and { 0 } × Y {\displaystyle \{0\}\times Y} are topological complements in X × Y {\displaystyle X\times Y} .
Every algebraic complement of { 0 } ¯ {\displaystyle {\overline {\{0\}}}} , the closure of 0 {\displaystyle 0} , is also a topological complement. This is because { 0 } ¯ {\displaystyle {\overline {\{0\}}}} has the indiscrete topology, and so the algebraic projection is continuous.13
If X = M ⊕ N {\displaystyle X=M\oplus N} and A : X → Y {\displaystyle A:X\to Y} is surjective, then Y = A M ⊕ A N {\displaystyle Y=AM\oplus AN} .14
Suppose X {\displaystyle X} is Hausdorff and locally convex and Y {\displaystyle Y} a free topological vector subspace: for some set I {\displaystyle I} , we have Y ≅ K I {\displaystyle Y\cong \mathbb {K} ^{I}} (as a t.v.s.). Then Y {\displaystyle Y} is a closed and complemented vector subspace of X {\displaystyle X} .15 In particular, any finite-dimensional subspace of X {\displaystyle X} is complemented.16
In arbitrary topological vector spaces, a finite-dimensional vector subspace Y {\displaystyle Y} is topologically complemented if and only if for every non-zero y ∈ Y {\displaystyle y\in Y} , there exists a continuous linear functional on X {\displaystyle X} that separates y {\displaystyle y} from 0 {\displaystyle 0} .17 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.1819
In a Hilbert space, the orthogonal complement M ⊥ {\displaystyle M^{\bot }} of any closed vector subspace M {\displaystyle M} is always a topological complement of M {\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.2021
Let X {\displaystyle X} be a Fréchet space over the field K {\displaystyle \mathbb {K} } . Then the following are equivalent:22
A complemented (vector) subspace of a Hausdorff space X {\displaystyle X} is necessarily a closed subset of X {\displaystyle X} , as is its complement.2324
From the existence of Hamel bases, every infinite-dimensional Banach space contains unclosed linear subspaces.25 Since any complemented subspace is closed, none of those subspaces is complemented.
Likewise, if X {\displaystyle X} is a complete TVS and X / M {\displaystyle X/M} is not complete, then M {\displaystyle M} has no topological complement in X . {\displaystyle X.} 26
If A : X → Y {\displaystyle A:X\to Y} is a continuous linear surjection, then the following conditions are equivalent:
(Note: This claim is an erroneous exercise given by Trèves. Let X {\displaystyle X} and Y {\displaystyle Y} both be R {\displaystyle \mathbb {R} } where X {\displaystyle X} is endowed with the usual topology, but Y {\displaystyle Y} is endowed with the trivial topology. The identity map X → Y {\displaystyle X\to Y} is then a continuous, linear bijection but its inverse is not continuous, since X {\displaystyle X} has a finer topology than Y {\displaystyle Y} . The kernel { 0 } {\displaystyle \{0\}} has X {\displaystyle X} as a topological complement, but we have just shown that no continuous right inverse can exist. If A : X → Y {\displaystyle A:X\to Y} is also open (and thus a TVS homomorphism) then the claimed result holds.)
Topological vector spaces admit the following Cantor-Schröder-Bernstein–type theorem:
The "self-splitting" assumptions that X = X ⊕ X {\displaystyle X=X\oplus X} and Y = Y ⊕ Y {\displaystyle Y=Y\oplus Y} cannot be removed: Tim Gowers showed in 1996 that there exist non-isomorphic Banach spaces X {\displaystyle X} and Y {\displaystyle Y} , each complemented in the other.28
Understanding the complemented subspaces of an arbitrary Banach space X {\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 L 1 [ 0 , 1 ] {\displaystyle L_{1}[0,1]} .29
For some Banach spaces the question is closed. Most famously, if 1 ≤ p ≤ ∞ {\displaystyle 1\leq p\leq \infty } then the only complemented infinite-dimensional subspaces of ℓ p {\displaystyle \ell _{p}} are isomorphic to ℓ p , {\displaystyle \ell _{p},} and the same goes for c 0 . {\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.30
The spaces L p [ 0 , 1 ] {\displaystyle L_{p}[0,1]} are not prime whenever p ∈ ( 1 , 2 ) ∪ ( 2 , ∞ ) ; {\displaystyle p\in (1,2)\cup (2,\infty );} in fact, they admit uncountably many non-isomorphic complemented subspaces.31
The spaces L 2 [ 0 , 1 ] {\displaystyle L_{2}[0,1]} and L ∞ [ 0 , 1 ] {\displaystyle L_{\infty }[0,1]} are isomorphic to ℓ 2 {\displaystyle \ell _{2}} and ℓ ∞ , {\displaystyle \ell _{\infty },} respectively, so they are indeed prime.32
The space L 1 [ 0 , 1 ] {\displaystyle L_{1}[0,1]} is not prime, because it contains a complemented copy of ℓ 1 {\displaystyle \ell _{1}} . No other complemented subspaces of L 1 [ 0 , 1 ] {\displaystyle L_{1}[0,1]} are currently known.33
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 X {\displaystyle X} is always isomorphic to X , {\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.34
Grothendieck 1973, pp. 34–36. - Grothendieck, Alexander (1973). Topological Vector Spaces. Translated by Chaljub, Orlando. New York: Gordon and Breach Science Publishers. ISBN 978-0-677-30020-7. OCLC 886098. https://archive.org/details/topologicalvecto0000grot ↩
Fabian, Marián J.; Habala, Petr; Hájek, Petr; Montesinos Santalucía, Vicente; Zizler, Václav (2011). Banach Space Theory: The Basis for Linear and Nonlinear Analysis (PDF). New York: Springer. pp. 179–181. doi:10.1007/978-1-4419-7515-7. ISBN 978-1-4419-7515-7. 978-1-4419-7515-7 ↩
Brezis, Haim (2011). Functional Analysis, Sobolev Spaces, and Partial Differential Equations. Universitext. New York: Springer. pp. 38–39. ISBN 978-0-387-70913-0. 978-0-387-70913-0 ↩
Schaefer & Wolff 1999, pp. 19–24. - Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135. https://search.worldcat.org/oclc/840278135 ↩
Trèves 2006, p. 36. - Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322. https://search.worldcat.org/oclc/853623322 ↩
Wilansky 2013, p. 63. - Wilansky, Albert (2013). Modern Methods in Topological Vector Spaces. Mineola, New York: Dover Publications, Inc. ISBN 978-0-486-49353-4. OCLC 849801114. https://search.worldcat.org/oclc/849801114 ↩
Y {\displaystyle Y} is closed because K I {\displaystyle \mathbb {K} ^{I}} is complete and X {\displaystyle X} is Hausdorff. Let f = ( f i ) i ∈ I : Y → K I {\displaystyle f=\left(f_{i}\right)_{i\in I}:Y\to \mathbb {K} ^{I}} be a TVS-isomorphism; each f i : Y → K {\displaystyle f_{i}:Y\to \mathbb {K} } is a continuous linear functional. By the Hahn–Banach theorem, we may extend each f i {\displaystyle f_{i}} to a continuous linear functional F i : X → K {\displaystyle F_{i}:X\to \mathbb {K} } on X . {\displaystyle X.} The joint map F : X → K I {\displaystyle F:X\to \mathbb {K} ^{I}} is a continuous linear surjection whose restriction to Y {\displaystyle Y} is f {\displaystyle f} . The composition P = f − 1 ∘ F : X → Y {\displaystyle P=f^{-1}\circ F:X\to Y} is then a continuous continuous projection onto Y {\displaystyle Y} .Q.E.D. /wiki/Complete_(topology) ↩
Rudin 1991, p. 106. - Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277. https://archive.org/details/functionalanalys00rudi ↩
Serre, Jean-Pierre (1955). "Un théoreme de dualité". Commentarii Mathematici Helvetici. 29 (1): 9–26. doi:10.1007/BF02564268. S2CID 123643759. /wiki/Commentarii_Mathematici_Helvetici ↩
Lindenstrauss, J., & Tzafriri, L. (1971). On the complemented subspaces problem. Israel Journal of Mathematics, 9, 263-269. ↩
Jarchow 1981, pp. 129–130. - Jarchow, Hans (1981). Locally convex spaces. Stuttgart: B.G. Teubner. ISBN 978-3-519-02224-4. OCLC 8210342. https://search.worldcat.org/oclc/8210342 ↩
In a Hausdorff space, { 0 } {\displaystyle \{0\}} is closed. A complemented space is the kernel of the (continuous) projection onto its complement. Thus it is the preimage of { 0 } {\displaystyle \{0\}} under a continuous map, and so closed.Q.E.D. /wiki/Q.E.D. ↩
Any sequence { e j } j = 0 ∞ ∈ X ω {\displaystyle \{e_{j}\}_{j=0}^{\infty }\in X^{\omega }} defines a summation map T : l 1 → X ; T ( { x j } j ) = ∑ j x j e j {\displaystyle T:l^{1}\to X;T(\{x_{j}\}_{j})=\sum _{j}{x_{j}e_{j}}} . But if { e j } j {\displaystyle \{e_{j}\}_{j}} are (algebraically) linearly independent and { x j } j {\displaystyle \{x_{j}\}_{j}} has full support, then T ( x ) ∈ span { e j } j ¯ ∖ span { e j } j {\displaystyle T(x)\in {\overline {\operatorname {span} {\{e_{j}\}_{j}}}}\setminus \operatorname {span} {\{e_{j}\}_{j}}} . Q.E.D. /wiki/Q.E.D. ↩
Schaefer & Wolff 1999, pp. 190–202. - Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135. https://search.worldcat.org/oclc/840278135 ↩
Narici & Beckenstein 2011, pp. 100–101. - Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834. https://search.worldcat.org/oclc/144216834 ↩
Albiac, Fernando; Kalton, Nigel J. (2006). Topics in Banach Space Theory. GTM 233 (2nd ed.). Switzerland: Springer (published 2016). pp. 29–232. doi:10.1007/978-3-319-31557-7. ISBN 978-3-319-31557-7. 978-3-319-31557-7 ↩
Argyros, Spiros; Tolias, Andreas (2004). Methods in the Theory of Hereditarily Indecomposable Banach Spaces. American Mathematical Soc. ISBN 978-0-8218-3521-0. 978-0-8218-3521-0 ↩