This is a really basic question sorry, I just need to make sure I have my understanding correct.

Given an infinite dimensional Hilbert space $\mathcal{H}$ and two subspaces, also Hilbert spaces, $\mathcal{H}_1$ and $\mathcal{H}_2$. If one can write:

$\mathcal{H}=\mathcal{H}_1 \bigoplus \mathcal{H}_2$

This means that $\mathcal{H}$ is isomorphic to the direct sum, not 'equal', right?

I'm just a little confused as the definiton of the direct sum of two Hilbert spaces (see http://en.wikipedia.org/wiki/Hilbert_space#Direct_sums) does not look to have the same 'form' as $\mathcal{H}$, so it seems a little strange to say they're equal.

I'm fairly sure, if they are isomorphic, the definition used for the direct sum isn't all that important, but I feel like I may be missing something.


2 Answers 2


The $\oplus$ sign can mean several things in several contexts, and I think one should always say what is meant. I usually indicate this by writing things like

$H=H_1\oplus H_2$ (internal algebraic direct sum)

$H=H_1\oplus H_2$ (external topological direct sum)

$H=H_1\oplus H_2$ (orthogonal direct sum)


Anyway, I think your confusion is an algebraic one, and is about the difference between the 'internal direct sum' and 'external direct sum'. So let's forget about Hilbert spaces and just work with vector spaces. Correct me if I am wrong about this.

For two vector spaces $A,B$ one can define the vector space $A\oplus B$ as $A\times B$ (set-product) with componentwise operations; this is the 'external direct sum'.

If $A,B$ happen to be subspaces of a bigger vector space $V$, then to say that $V$ is the internal direct sum of $A,B$ means precisely that the map $A\oplus B\to V$ given by $(a,b)\mapsto a+b$ (where the domain is the external direct sum just constructed) is a bijection. Equivalently, this means $A,B$ intersect trivially (the injective part) and $A+B=V$ (the surjective part).

In conclusion: if by $A\oplus B$ you mean the external direct sum (e.g. constructed as $A\times B$),then the $=$ sign in $V=A\oplus B$ (with $A,B$ subspaces of $V$) should really be an $\cong$ sign: set-theoretically there is no equality, since $V$ does not consists of pairs $(a,b)$.

  • $\begingroup$ Both answers here were good, but the last paragraph in this answer was exactly what I was looking for. Thanks for the clarification. $\endgroup$ Jun 29, 2012 at 0:24

In the setting of Hilbert spaces, the notation $\mathcal{H}=\mathcal{H}_1\oplus\mathcal{H}_2$ usually means that $\mathcal{H}_1$ and $\mathcal{H}_2$ are closed subspaces of $\mathcal{H}$ such that (1) $x\perp y$ whenever $x\in\mathcal{H}_1$ and $y\in\mathcal{H}_2$ and (2) for every $z\in\mathcal{H}$ there exist $x\in\mathcal{H}_1$ and $y\in\mathcal{H}_2$ such that $z=x+y$. Note that (1) implies that $x$ and $y$ are uniquely determined by $z$. One says that $\mathcal{H}$ is the (internal, orthogonal) direct sum of $\mathcal{H}_1$ and $\mathcal{H}_2$.

A different notion is that of an external direct sum: Given two Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$, one constructs a (new) Hilbert space $\mathcal{H}=\mathcal{H}_1\oplus\mathcal{H}_2$ by equipping the set $\mathcal{H}_1\times\mathcal{H}_2 = \{(x,y)\,|\,x\in\mathcal{H}_1,\,y\in\mathcal{H}_2\}$ with coordinate-wise addition and scalar multiplication and the inner product $$\langle (x_1,y_1),(x_2,y_2)\rangle = \langle x_1,x_2\rangle+\langle y_1,y_2\rangle.$$ The two notions of direct sums are tied together by the fact that this newly constructed Hilbert space is the (internal) direct sum of the closed subspaces $\{(x,0)\,|\,x\in\mathcal{H}_1\}$ and $\{(0,y)\,|\,y\in\mathcal{H}_2\}$, which we identify naturally with $\mathcal{H}_1$ and $\mathcal{H}_2$. Conversely, an internal direct sum is clearly isomorphic to the corresponding external direct sum.


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