Why is the vector representation of Lorentz group $O(3,1)$ a direct sum of the spin-0 and spin-1 representations of the rotation group $SO(3)$? How to understand the statement that the vector representation of Lorentz group $O(3,1)$ is the direct sum of the spin-0 and spin-1 representations of the rotation group $SO(3)$?
 A: The 4-dimensional spacetime $\mathbb{R}^4$ decomposes into a singlet (=time) and a triplet (=space).
A: Vector doesn't go with the word representation. I think the vector you refer to is a 4-vector in the "carrier space" that $4 \times 4$ Lorentz matrices operate on.  The $4 \times 4$ matrices are what is called a "representation" of the Lorentz group.  The $4 \times 4$ representation is the one you are most familiar with. Rotation matrix elements occuppy $3 \times 3$ and $1 \times 1$ blocks on the diagonal. For example, a 4x4 representation of a Lorentz group rotation of a 4-vector about the y-axis is:
$$
\text{Rot}_y(\theta)\begin{bmatrix} x \\ y \\ z \\ t \\ \end{bmatrix}
=\begin{bmatrix}   \cos(\theta) & 0 & \sin(\theta) & 0 \\
                             0 & 1 &           0 & 0 \\
                  -\sin(\theta) & 0 & \cos(\theta) & 0 \\
                             0 & 0 &           0 & 1 \\
\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ t \\ \end{bmatrix}  
$$
Notice that all rotations transform $(x,y,z)$ as a three vector (=spin-$1$) and transform $t$ into itself (ie: as a one vector = spin-$0$).  So, the $4 \times 4$ representation of the Lorentz group is the direct sum of the spin-1 and spin-0 representations of the rotation group $SO(3)$.
