Complemented subspace

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

$${\displaystyle {\begin{alignedat}{4}S:\;&&M\times N&&\;\to \;&X\\&&(m,n)&&\;\mapsto \;&m+n\\\end{alignedat}}}$$

is a linear map or said differently, it is a morphism in the category of vector spaces. The vector space ${\displaystyle X}$ is said to be the algebraic direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ or the direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ in the category of vector spaces if 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 complements or an algebraic supplements to ${\displaystyle M}$ in ${\displaystyle X}$ and these spaces are said to be complementary or supplementary.[1]

The condition ${\displaystyle M\cap N=\{0\}}$ can be expressed by saying that ${\displaystyle M}$ and ${\displaystyle N}$ are linearly independent vector subspaces.[3] More generally, a collection ${\displaystyle M_{1},\ldots ,M_{d}}$ of subspaces of ${\displaystyle X}$ are said to be linearly independent if ${\textstyle M_{i}\cap \sum _{k\neq i}M_{k}=\{0\}}$ for every index ${\displaystyle i,}$ where ${\textstyle \sum _{k\neq i}M_{k}={\Big \{}\sum _{k\neq i}m_{k}:m_{k}\in M_{k}{\text{ for all }}k{\Big \}}=\operatorname {span} \cup _{\stackrel {1\leq k\leq d,}{k\neq i}}M_{k}.}$[3]

Suppose that the vector space ${\displaystyle X}$ is the algebraic direct sum of ${\displaystyle M}$ in ${\displaystyle N.}$ Then algebraically, ${\displaystyle X}$ may be treated as if it were ${\displaystyle M\oplus N}$ or ${\displaystyle M\times N.}$ Assuming now that ${\displaystyle X}$ is also a topological vector space, such as any normed space, then the addition map

$${\displaystyle {\begin{alignedat}{4}S:\;&&M\times N&&\;\to \;&X\\&&(m,n)&&\;\mapsto \;&m+n\\\end{alignedat}}}$$

will necessarily be a continuous[note 3] bijection but its inverse ${\displaystyle S^{-1}:X\to M\times N}$ may fail to be continuous. However, if (and only if) this map ${\displaystyle S^{-1}}$ is continuous will ${\displaystyle X}$ be the topological direct sum of ${\displaystyle M}$ and ${\displaystyle N,}$ which by definition is just another way of say that ${\displaystyle X}$ is the (categorical) direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ in the category of topological vector spaces (rather than merely the direct sum in the category of vector spaces). This is equivalent to the addition map ${\displaystyle S:M\times N\to X}$ being a linear homeomorphism, which when true, allows for ${\displaystyle X}$ to be treated as if it were ${\displaystyle M\oplus N}$ or ${\displaystyle M\times N}$ both algebraically and topologically. That this is not always true is the reason why definitions such as "complemented subspace" are needed, where a subspace ${\displaystyle M}$ is called (topologically) complement in ${\displaystyle X}$ if there exists some other subspace ${\displaystyle N}$ that allows ${\displaystyle X}$ to be treated as if it were ${\displaystyle M\oplus N}$ or ${\displaystyle M\times N}$ both algebraically and topologically.

For an example where ${\displaystyle X}$ is an algebraic, but not topological, direct sum of two vector subspaces ${\displaystyle M}$ and ${\displaystyle N,}$ suppose ${\displaystyle f}$ is a discontinuous linear functional on a normed space ${\displaystyle X}$ (where a linear functional ${\displaystyle f}$ is discontinuous if and only if ${\displaystyle \ker f}$ is not closed in ${\displaystyle X}$) and let ${\displaystyle n\in X}$ be such that ${\displaystyle f(n)\neq 0.}$ Then ${\displaystyle X}$ is the algebraic direct sum of ${\displaystyle M:=\ker f}$ and ${\displaystyle N:=\operatorname {span} \{n\}}$ but ${\displaystyle X}$ is not the topological direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ because ${\displaystyle N}$ is not closed in ${\displaystyle X.}$ This subspace being closed is necessary when ${\displaystyle X}$ is Hausdorff[1] because then ${\displaystyle M\times \{0\}}$ is closed in the product topology on ${\displaystyle M\times N}$ so that ${\displaystyle S:M\times N\to X}$ being a homeomorphism would force ${\displaystyle S\left(M\times \{0\}\right)=M}$ to be closed in ${\displaystyle X}$ (and similarly for ${\displaystyle N}$). This proof generalized to show that a topologically complemented subspace of a Hausdorff topological vector space (TVS) ${\displaystyle X}$ must be closed in ${\displaystyle X,}$ which is why many authors require this property (that is, being closed) as part of the definition of a complemented subspace. This counter example also generalizes to any Hausdorff locally convex TVS that admits a discontinuous linear functional and moreover, the kernel of such a discontinuous linear functional cannot be topologically complemented in ${\displaystyle X}$ (because it is not a closed subspace of ${\displaystyle X}$).

Because a linear map between two normed (or Banach) spaces is bounded if and only if it is continuous, if ${\displaystyle X}$ is a normed space (resp. a Banach space) then ${\displaystyle X}$ is the direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ in the category of normed spaces (resp. the category of Banach spaces) if and only if ${\displaystyle X}$ is the topological direct sum of ${\displaystyle M}$ and ${\displaystyle N.}$

If ${\displaystyle X}$ is the algebraic direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ then by definition, the addition map

$${\displaystyle {\begin{alignedat}{4}S:\;&&M\times N&&\;\to \;&X\\&&(m,n)&&\;\mapsto \;&m+n\\\end{alignedat}}}$$

is a linear bijection so its inverse ${\displaystyle S^{-1}:X\to M\times N}$ is well-defined and it can be written in terms of coordinates as

$${\displaystyle S^{-1}:=\left(P_{M},P_{N}\right)}$$

where the first coordinate ${\displaystyle P_{M}:X\to M}$ is called the canonical projection of ${\displaystyle X}$ onto ${\displaystyle M}$ while the second coordinate ${\displaystyle P_{N}:X\to N}$ is called the canonical projection of ${\displaystyle X}$ onto ${\displaystyle N.}$[2] To clarify, for any ${\displaystyle x\in X,}$ the defining property of ${\displaystyle P_{M}(x)}$ and ${\displaystyle P_{N}(x)}$ is that they are the unique vectors in ${\displaystyle M}$ and ${\displaystyle N,}$ respectively, that satisfy

$${\displaystyle S^{-1}(x):=\left(P_{M}(x),P_{N}(x)\right),\qquad {\text{ or equivalently }}\qquad x=P_{M}(x)+P_{N}(x).}$$

These maps satisfy

$${\displaystyle P_{M}+P_{N}=\operatorname {Id} _{X},\qquad \ker P_{M}=N,\qquad {\text{ and }}\qquad \ker P_{N}=M}$$

where ${\displaystyle \operatorname {Id} _{X}}$ denotes the identity map on ${\displaystyle X.}$

If ${\displaystyle X}$ is a topological vector space (TVS) and if ${\displaystyle M}$ and ${\displaystyle N}$ are vector subspaces of ${\displaystyle X,}$ then ${\displaystyle X}$ is the topological direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ if any of the following equivalent conditions are satisfied:

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 (categorical) direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ in the category of topological vector spaces (whose morphisms are continuous linear operators).
3. The addition map ${\displaystyle S:M\times N\to X}$ is a bijective open map.
4. ${\displaystyle X}$ is the algebraic direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ (that is, the addition map ${\displaystyle S:M\times N\to X}$ a bijection) and also any of the following equivalent conditions hold:
   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 two canonical projections ${\displaystyle P_{M}:X\to M}$ and ${\displaystyle P_{N}:X\to N}$ is continuous.
   4. The map ${\displaystyle p:N\to X/M}$ defined by ${\displaystyle n\mapsto n+M}$ is a TVS-isomorphism.[4]
      • This map is the restriction to ${\displaystyle N}$ of the canonical quotient map ${\displaystyle Q:X\to X/M}$ defined by ${\displaystyle Q(x):=x+M.}$ The space ${\displaystyle X/M}$ has quotient topology induced by ${\displaystyle Q.}$
   5. The canonical linear map ${\displaystyle p:N\to X/M}$ is a bicontinuous vector space isomorphism (meaning ${\displaystyle p}$ and its inverse are continuous);
      • This implies, in particular, that any topological complement of ${\displaystyle M}$ in ${\displaystyle X}$ is TVS-isomorphic to the quotient space ${\displaystyle X/M.}$
5. When considered as additive topological groups (so scalar multiplication is ignored), ${\displaystyle X}$ is the topological direct sum of the topological subgroups ${\displaystyle M}$ and ${\displaystyle N.}$

If any of the above conditions hold then ${\displaystyle M}$ and ${\displaystyle N}$ are said to be topological supplements[1] or topological complements in ${\displaystyle X,}$ they are said to be (topologically) complementary or supplementary to each other, and each subspace is said to be a (topologically) complemented subspace of ${\displaystyle X.}$

Let ${\displaystyle X}$ be a topological vector space and let ${\displaystyle M}$ be a vector subspace of ${\displaystyle X.}$ Then ${\displaystyle M}$ is called a complemented subspace of ${\displaystyle X}$ and is said to admit a topological supplement[1] if any of the following equivalent conditions hold:

1. There exists a vector subspace ${\displaystyle N}$ of ${\displaystyle X}$ such that ${\displaystyle X}$ is the topological direct sum of ${\displaystyle M}$ and ${\displaystyle N}$ (or said differently, ${\displaystyle M}$ is topologically complemented in ${\displaystyle X}$);
2. There exists a continuous linear map ${\displaystyle P:X\to X}$ with image ${\displaystyle P(X)=M}$ such that ${\displaystyle P\circ P=P}$;
   • Such a map is called a continuous linear projection from ${\displaystyle X}$ onto ${\displaystyle M.}$
3. There exists a continuous linear projection ${\displaystyle P:X\to X}$ with image ${\displaystyle P(X)=M}$ such that ${\displaystyle X}$ is the algebraic sum of ${\displaystyle M}$ and the null space of ${\displaystyle P.}$
4. For every TVS ${\displaystyle Y,}$ the restriction map ${\displaystyle R:L(X;Y)\to L(M;Y)}$ is surjective, where ${\displaystyle R}$ is defined by sending a continuous linear operator ${\displaystyle u:X\to Y}$ to its restriction ${\displaystyle u{\big \vert }_{M}:M\to Y}$ (that is, ${\displaystyle R(u):=u{\big \vert }_{M},}$).[4]

where if in addition ${\displaystyle X}$ is a Banach space then this list can be extended to include:

5. ${\displaystyle M}$ is a closed in ${\displaystyle X}$ and there exists another closed subspace ${\displaystyle N}$ of ${\displaystyle X}$ such that ${\displaystyle X}$ is isomorphic to the direct sum ${\displaystyle M\oplus N}$ via the addition map.

In this case ${\displaystyle N}$ is also a complemented subspace, and ${\displaystyle M}$ and ${\displaystyle N}$ are called complements of each other in ${\displaystyle X.}$

Topological complements of kernels

It can be shown that if ${\displaystyle f:X\to Y}$ is a continuous linear surjection between two TVSs, then the following conditions are equivalent:

1. The kernel of ${\displaystyle f}$ has a topological complement.
2. There exists a continuous linear map ${\displaystyle g:Y\to X}$ such that ${\displaystyle f\circ g=\operatorname {Id} _{Y},}$ where ${\displaystyle \operatorname {Id} _{Y}:Y\to Y}$ is the identity map.[4]

Let ${\displaystyle X}$ be a non-normable Fréchet space on which there exists a continuous norm. Then ${\displaystyle X}$ contains a closed vector subspace that has no topological complement.[10]

If ${\displaystyle X}$ is a complete TVS and ${\displaystyle M}$ is a closed vector subspace of ${\displaystyle X}$ such that ${\displaystyle X/M}$ is not complete, then ${\displaystyle H}$ does not have a topological complement in ${\displaystyle X.}$[10]