# Fock space used in Quantum mechanic : how can we have direct sum of spaces of different dimensions?

In physics we work with the Fock space when we have to deal with an undefinite number of particles.

But there is something I misunderstand : how can we have a direct sum of spaces that are not of the same dimension ?

Indeed we have :

$$F=\bigoplus_{n=0}^{\infty}S_\nu H^{\otimes n}$$

Each space in the direct sum hasn't the same dimension as another one.

To make the question more simple lets just take :

$$A=\mathbb{R} \oplus \mathbb{R}^2$$

I don't understand how such a space is defined as we would need a law of addition between $\mathbb{R}$ and $\mathbb{R}^2$.

Let $z$ be in $A$, we would have $z=x+(y,z)$ but what does that + symbol means ?

I just know basics linear algebra so simple answers would be nice !

• I see what you mean but for example I could have $z=(x,0)+(y,z)$ or $z=(0,x)+(y,z)$. Thus how to choose between them ? – StarBucK Apr 14 '17 at 18:01
• You are using $z$ in two different ways: unwise. The typical element of $A$ has the form $(x,(y,z))$ for real $x$, $y$ and $z$. One adds as follows: $(x,(y,z))+(x',(y',z'))=(x+x',(y+y',z+z'))$. It's clear that making a distinction between $(x,(y,z))$ and $(x,y,z)$ is idle. – Lord Shark the Unknown Apr 14 '17 at 18:26
• Ok, so when I work in $U \oplus V$, in fact, I work in $U \times V$, BUT with the specificity here to have a unique decomposition $z \in U \times V$ such as $z=x_u+y_v=(x,0)+(0,y)$. Am I right ? – StarBucK Apr 14 '17 at 18:42

I will answer your simple question about $\mathbb{R} \oplus \mathbb{R}^2$.

That space is the set of ordered pairs $(x,v)$ where $x \in \mathbb{R}$ and $v$ is itself an ordered pair $(y,z) \in \mathbb{R}^2$ . You never add $x$ to $v$, although you can think of $$(x,(y,z)) = (x, (0,0)) + (0, (y,z)).$$ The addition takes place in each component separately, just the way it does for $n$-tuples of numbers.

And, of course $\mathbb{R} \oplus \mathbb{R}^2$ is naturally identified with $\mathbb{R}^3$.

The $\oplus$ notation is historical, and in this context, unfortunately but understandably confusing. That may in fact be what puzzles the OP. The laws for exponents suggest that we should use a product here, not a sum. The addition makes sense only when we think of $\mathbb{R}$ and $\mathbb{R}^2$ as the subspaces $\{(x,0,0)\}$ and $\{(0,y,z)\}$ of $\mathbb{R}^3$. That is precisely the distinction between the internal and external direct sums @HenrySwanson discusses in his answer.

When someone says "direct sum", they mean one of two things, and it's usually clear from context. In particular, the internal direct sum requires a parent space, and the external one does not.

Internal: Let $V$ be a vector space, and $U$ and $W$ be subspaces of $V$. We say that $V$ is an (internal) direct sum of $U$ and $W$ if $U \cap W$ is trivial and $U + W = V$. For example, $V = \Bbb R^3$, $U$ is spanned by $(1, 0, 0)$ and $W$ is spanned by $(0, 1, 0)$ and $(1, 1, 1)$.

External: Let $U$ and $W$ be arbitrary vector spaces. Then $U \oplus W$, called the (external) direct sum of $U$ and $W$, is the space of pairs $\{ (u, w) \mid u \in U, w \in W \}$, where addition is defined componentwise.

These two notions are related as follows: let $V$ be a vector space, and $U$ and $W$ be subspaces. Then $V$ is the (internal) direct sum of $U$ and $W$ iff $V$ is isomorphic to the (external) direct sum of $U$ and $W$ (in a way respecting the inclusion of $U, W$ into $V$).

• And in neither way of defining direct sums must both spaces have the same dimension. – Lord Shark the Unknown Apr 14 '17 at 18:11