TITLE: Fock space used in Quantum mechanic : how can we have direct sum of spaces of different dimensions?
QUESTION [11 upvotes]: 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 !

REPLY [11 votes]: 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.