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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 !

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    $\begingroup$ 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 ? $\endgroup$ – StarBucK Apr 14 '17 at 18:01
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    $\begingroup$ 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. $\endgroup$ – Lord Shark the Unknown Apr 14 '17 at 18:26
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    $\begingroup$ 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 ? $\endgroup$ – StarBucK Apr 14 '17 at 18:42
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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.

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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 $V$).

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

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