Direct sum in physics

The x-y plane is of dimension 2. While each line is of dimension 1. 1+1=2.

Let me tell you a bit more about direct sums. Given a vector space $U$ and two subspaces $V$ and $W$, we say $U$ is the direct sum of $V$ and $W$ (and write $U=V\oplus W$) if the only vector which $V$ and $W$ share is $0$, i.e. $V\cap W=\{0\}$, and every vector in $U$ can be written as a sum of a vector in $V$ and a vector in $W$, i.e. $U=V+W$. This notion of direct sum is called inner direct sum. It is very visual and easy to understand. For example, consider the $x$-axis $V:=\text{span}\{(1,0)\}=\{(x,0)\in\mathbb{R}^2\}$ and the $y$-axis $W:=\text{span}\{(0,1)\}=\{(0,y)\in\mathbb{R}^2\}$ in the plane. We then quickly check $U:=\mathbb{R}^2=V\oplus W$. Notice however that, unless we add more structure, the direct sum does not encode the orthogonality of the $x$ and $y$ axes. Indeed, if one considers the diagonal $Z:=\text{span}\{(1,1)\}=\{(x,x)\in\mathbb{R}^2\}$ we also have $U=V\oplus Z$ and $U=Z\oplus W$.

Now, suppose you have two (unrelated as far as we care) vector spaces $V$ and $W$. We may ask if there is some bigger vector space $U$ which contains $V$ and $W$ and $U=V\oplus W$. This is always possible and the construction is called the outer direct sum. We define the space $V\oplus W$ as the set of all tuples $(v,w)$ where $v\in V$ and $w\in W$. This forms a vector space with the addition $(v_1,w_1)+(v_2,w_2):=(v_1+v_2,w_1+w_2)$ and scalar multiplication $k(v,w):=(kv,kw)$. We may then consider the spaces $\tilde{V}:=\{(v,0)\in U\oplus W\}$ and $\tilde{W}:=\{(0,w)\in U\oplus W\}$. $V$ and $W$ are isomorphic to $\tilde{V}$ and $\tilde{W}$ as vector spaces respectively. Both are subspaces of $V\oplus W$ and in fact $V\oplus W=\tilde{V}\oplus\tilde{W}$. In the left of this equation we have the outer direct sum just defined while in the second we have the inner direct sum defined in the previous paragraph. It is thus costumary to identify $V$ and $W$ with $\tilde{V}$ and $\tilde{W}$ repectively. Moreover, we use the notation $v+w:=(v,w)$ for all $v\in V$ and $w\in W$. Then both notions of inner product are virtually undistinguishable and most authors don't care to distinguish between them. Notice that with the outer notion of direct sum we truly have $\mathbb{R}^2=\mathbb{R}\oplus\mathbb{R}$.

In physics the direct sum pops up whenever vector spaces do. In classical mechanics, for example, the position of a particle is described by the vector space (well, not really but let us avoid the problem of an origin for the moment) $V=\mathbb{R}^3$. Now consider we have a second particle whose position is described by the vector space $W=\mathbb{R}^3$. Then the configuration space of the whole system is $V\oplus W$ (in the outer notion). Its elements are the pairs of vectors $(\vec{r}_1,\vec{r}_2)$, where $\vec{r}_1$ is the position of the first particle while $\vec{r}_2$ the position of the second. This is of course isomorphic to $\mathbb{R}^6$, in which the first three elements of $(x_1,y_1,z_1,x_2,y_2,z_2)$ are interpreted to be the position of the first particle while the second three the position of the second.

There is another way that the direct sum appears in physics. Suppose that there is a particle that is moving in a space that has two connected components $M=U\sqcup V$. Then the Hilbert space is a direct sum $\mathcal{H}=L^2(M)=L^2(U)\oplus L^2(V)$. Actually the two spaces don't have to be disjoint, their intersection only need to have measure 0.