Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$.

I) First recall the fact that

$SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$.

This follows partly because:

There is a bijective isometry from the Minkowski space $(\mathbb{R}^{1,3},||\cdot||^2)$ to the space of $2\times2 $ Hermitian matrices $(u(2),\det(\cdot))$,
$$\mathbb{R}^{1,3} ~\cong ~ u(2)
~:=~\{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \sigma^{\dagger}=\sigma \}
~=~ {\rm span}_{\mathbb{R}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$
$$\mathbb{R}^{1,3}~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ u(2), $$
$$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma), \qquad \sigma_{0}~:=~{\bf 1}_{2 \times 2}.\tag{1}$$

There is a group action $\rho: SL(2,\mathbb{C})\times u(2) \to u(2)$ given by
$$g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^{\dagger},
\qquad g\in SL(2,\mathbb{C}),\qquad\sigma\in u(2), \tag{2}$$
which is length preserving, i.e. $g$ is a pseudo-orthogonal (or Lorentz) transformation.
In other words, there is a Lie group homomorphism

$$\rho: SL(2,\mathbb{C}) \quad\to\quad O(u(2),\mathbb{R})~\cong~ O(1,3;\mathbb{R}) .\tag{3}$$

Since $\rho$ is a continuous map and $SL(2,\mathbb{C})$ is a connected set, the image $\rho(SL(2,\mathbb{C}))$ must again be a connected set. In fact, one may show so there is a surjective Lie group homomorphism$^1$

$$\rho: SL(2,\mathbb{C}) \quad\to\quad SO^+(u(2),\mathbb{R})~\cong~ SO^+(1,3;\mathbb{R}) , $$
$$\rho(\pm {\bf 1}_{2 \times 2})~=~{\bf 1}_{u(2)}.\tag{4}$$

The Lie group $SL(2,\mathbb{C})=\pm e^{sl(2,\mathbb{C})}$ has Lie algebra
$$ sl(2,\mathbb{C})
~=~ \{\tau\in{\rm Mat}_{2\times 2}(\mathbb{C}) \mid {\rm tr}(\tau)~=~0 \}
~=~{\rm span}_{\mathbb{C}} \{\sigma_{i} \mid i=1,2,3\}.\tag{5}$$

The Lie group homomorphism $\rho: SL(2,\mathbb{C}) \to O(u(2),\mathbb{R})$ induces a Lie algebra homomorphism
$$\rho: sl(2,\mathbb{C})\to o(u(2),\mathbb{R})\tag{6}$$
given by
$$ \rho(\tau)\sigma ~=~ \tau \sigma +\sigma \tau^{\dagger},
\qquad \tau\in sl(2,\mathbb{C}),\qquad\sigma\in u(2), $$
$$ \rho(\tau) ~=~ L_{\tau} +R_{\tau^{\dagger}},\tag{7}$$
where we have defined left and right multiplication of $2\times 2$ matrices
$$L_{\sigma}(\tau)~:=~\sigma \tau~=:~ R_{\tau}(\sigma),
\qquad \sigma,\tau ~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}).\tag{8}$$

II) Note that the Lorentz Lie algebra $so(1,3;\mathbb{R}) \cong sl(2,\mathbb{C})$ does *not$^2$* contain two perpendicular copies of, say, the real Lie algebra $su(2)$ or $sl(2,\mathbb{R})$. For comparison and completeness, let us mention that for other signatures in $4$ dimensions, one has

$$SO(4;\mathbb{R})~\cong~[SU(2)\times SU(2)]/\mathbb{Z}_2,
\qquad\text{(compact form)}\tag{9}$$

$$SO^+(2,2;\mathbb{R})~\cong~[SL(2,\mathbb{R})\times SL(2,\mathbb{R})]/\mathbb{Z}_2.\qquad\text{(split form)}\tag{10}$$

The compact form (9) has a nice proof using quaternions

$$(\mathbb{R}^4,||\cdot||^2) ~\cong~ (\mathbb{H},|\cdot|^2)\quad\text{and}\quad SU(2)~\cong~ U(1,\mathbb{H}),\tag{11}$$

see also this Math.SE post and this Phys.SE post. The split form (10) uses a bijective isometry

$$(\mathbb{R}^{2,2},||\cdot||^2) ~\cong~({\rm Mat}_{2\times 2}(\mathbb{R}),\det(\cdot)).\tag{12}$$

To decompose Minkowski space into left- and right-handed Weyl spinor representations, one must go to the complexification, i.e. one must use the fact that

$SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ is (the double cover of) the complexified proper Lorentz group $SO(1,3;\mathbb{C})$.

Note that Refs. 1-2 do not discuss complexification$^2$. One can more or less repeat the construction from section I with the real numbers $\mathbb{R}$ replaced by complex numbers $\mathbb{C}$, however with some important caveats.

There is a bijective isometry from the complexified Minkowski space $(\mathbb{C}^{1,3},||\cdot||^2)$ to the space of $2\times2 $ matrices $({\rm Mat}_{2\times 2}(\mathbb{C}),\det(\cdot))$,
$$\mathbb{C}^{1,3} ~\cong ~ {\rm Mat}_{2\times 2}(\mathbb{C})
~=~ {\rm span}_{\mathbb{C}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$
$$ M(1,3;\mathbb{C})~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}) , $$
$$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma).\tag{13}$$
Note that forms are taken to be bilinear rather than sesquilinear.

There is a surjective Lie group homomorphism$^3$

$$\rho: SL(2,\mathbb{C}) \times SL(2,\mathbb{C}) \quad\to\quad
SO({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})~\cong~ SO(1,3;\mathbb{C})\tag{14}$$
given by
$$(g_L, g_R)\quad \mapsto\quad\rho(g_L, g_R)\sigma~:= ~g_L\sigma g^{\dagger}_R, $$
$$ g_L, g_R\in SL(2,\mathbb{C}),\qquad\sigma~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}).\tag{15} $$

The Lie group
$SL(2,\mathbb{C})\times SL(2,\mathbb{C})$
has Lie algebra $sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$.

The Lie group homomorphism

$$\rho: SL(2,\mathbb{C})\times SL(2,\mathbb{C})
\quad\to\quad SO({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})\tag{16}$$
induces a Lie algebra homomorphism
$$\rho: sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})\quad\to\quad
so({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})\tag{17}$$
given by
$$ \rho(\tau_L\oplus\tau_R)\sigma ~=~ \tau_L \sigma +\sigma \tau^{\dagger}_R,
\qquad \tau_L,\tau_R\in sl(2,\mathbb{C}),\qquad
\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}), $$
$$ \rho(\tau_L\oplus\tau_R) ~=~ L_{\tau_L} +R_{\tau^{\dagger}_R}.\tag{18}$$

The left action (acting from left on a two-dimensional complex column vector) yields by definition the (left-handed Weyl) spinor representation $(\frac{1}{2},0)$, while the right action (acting from right on a two-dimensional complex row vector) yields by definition the right-handed Weyl/complex conjugate spinor representation $(0,\frac{1}{2})$. The above shows that

The complexified Minkowski space $\mathbb{C}^{1,3}$ is a $(\frac{1}{2},\frac{1}{2})$ representation of the Lie group $SL(2,\mathbb{C}) \times SL(2,\mathbb{C})$, whose action respects the Minkowski metric.

References:

Anthony Zee, *Quantum Field Theory in a Nutshell,* 1st edition, 2003.

Anthony Zee, *Quantum Field Theory in a Nutshell,* 2nd edition, 2010.

$^1$ It is easy to check that it is *not* possible to describe discrete Lorentz transformations, such as, e.g. parity $P$, time-reversal $T$, or $PT$ with a group element $g\in GL(2,\mathbb{C})$ and formula (2).

$^2$ For a laugh, check out the (in several ways) wrong second sentence on p.113 in Ref. 1: *"The mathematically sophisticated say that the algebra $SO(3,1)$ is isomorphic to $SU(2)\otimes SU(2)$."* The corrected statement would e.g. be *"The mathematically sophisticated say that the group $SO(3,1;\mathbb{C})$ is locally isomorphic to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$."* Nevertheless, let me rush to add that Zee's book is overall a very nice book. In Ref. 2, the above sentence is removed, and a subsection called *"More on $SO(4)$, $SO(3,1)$, and $SO(2,2)$"* is added on page 531-532.

$^3$ It is not possible to mimic an improper Lorentz transformations $\Lambda\in O(1,3;\mathbb{C})$ [i.e. with negative determinant $\det (\Lambda)=-1$] with the help of two matrices $g_L, g_R\in GL(2,\mathbb{C})$ in formula (15); such as, e.g., the spatial parity transformation
$$P:~~(x^0,x^1,x^2,x^3) ~\mapsto~ (x^0,-x^1,-x^2,-x^3).\tag{19}$$
Similarly, the Weyl spinor representations are representations of (the double cover of) $SO(1,3;\mathbb{C})$ but *not* of (the double cover of) $O(1,3;\mathbb{C})$. E.g. the spatial parity transformation (19) intertwine between left-handed and right-handed Weyl spinor representations.

OP wrote (v1):

*What does "the ${\bf N}$ of a group" mean?*

1) Physicists are referring to an irreducible representation (irrep) for whatever group $G$ we are talking about. The number ${\bf N}$ refers to the dimension of the irrep. The point is that irreps are so rare that irreps are often uniquely specified by their dimension (modulo isomorphisms). (This is not quite true in general, and physicists then start to decorate the bold-faced dimension symbol with other ornaments, e.g. ${\bf 3}$ and $\bar{\bf 3}$, or e.g. ${\bf 8}_v$ and ${\bf 8}_s$ and ${\bf 8}_c$, etc, to distinguish.)

2) By the way, concerning a group representation $\rho: G \to GL(V,\mathbb{F})$, where $G$ is a group, where $\mathbb{F}$ is a field (typically $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$), where $V$ is a $\mathbb{F}$-vector space, and where $\rho$ is a group homomorphism; be aware that physicists refer to both the map $\rho$ and the vector space $V$ as "a representation".

## Best Answer

It's not that hard to see how a rotation can end up being represented by a matrix of dimension $(2j+1)\times(2j+1)$. The key concept is that this matrix acts on a subspace $V$ of the Hilbert space $\mathcal H$; that is, $V$ contains state vectors (kets). Generally, $V$ is required to be an invariant subspace in the sense that if $v\in V$, then under a rotation $v$ will in general go to some different vector $v'$ but it will nevertheless stay in $V$.

The easiest way to see this is by way of example, so let me show how this works for $j=2$. There are in general many possible realizations of $V$, but the cleanest realization is as the vector space of functions $f:\mathbb R^3\to \mathbb C$ which are homogeneous polynomials of degree 2, and which are 'traceless' in the sense that $$ ⟨f⟩=\int_{S^2} f(\hat{\mathbf{r}})\,\mathrm d \Omega=0.\tag1 $$ This vector space is best analysed in a convenient basis, and the cleanest one is $$ B=\{x^2+y^2-2z^2, xz, yz, xy, x^2-y^2\}. $$ It is fairly easy to see that $V$ is closed under rotations, because each vector component will go into a linear combination of $x,y$ and $z$, and multiplying any two such combinations will again give a homogeneous polynomial. Rotations will also not affect the tracelessness condition (1).

To calculate the effect of a rotation $R\in\mathrm{SO}(3)$, you simply take a given $f\in V$ to the function $G(R)f\in V$ which is given by $$(G(R)f)(\mathbf r)=f(R^{-1}\mathbf r).$$ (The reason for the inverse is so that the operators $G(R)$ have the nice property that $G(R_1\circ R_2)=G(R_1)\circ G(R_2)$, so that $G$ itself is a homomorphism between $\mathrm{SO}(3)$ and the group of unitary transformations on $V$, $\mathrm{U}(V)$.)

For any given $R$, $G(R)$ is a geometrical transformation but it is also, at a simpler level, a linear transformation in a finite-dimensional vector space $V$ with basis $B$, so you can simply represent it by its matrix with respect to this basis. Thus, for example, a rotation by 90° about the $+x$ axis would be represented by the matrix $$ \begin{pmatrix} -\tfrac12&0&0&0&\tfrac12\\ 0&0&0&-1&0\\ 0&0&-1&0&0\\ 0&1&0&0&0\\ \tfrac32&0&0&0&\tfrac12\\ \end{pmatrix}. $$ (Work it out!)

The others have given more detail on how this works mathematically - the function $G$ being a

representationof the group $\mathrm{SO}(3)$ - but I think that examples of this sort help a lot in visualizing what's going on.