Lorentz Tensors Restricted to the Rotation Subgroup

1. The starting point: restricting Lorentz to rotations

We begin with representations of the Lorentz group \(SO(1,3)\), and then restrict them to the rotation subgroup \(SO(3)\).

The key first step is:

\[ \mathbf{4} \longrightarrow \mathbf{3} \oplus \mathbf{1} \]

Interpretation:

- \(A_0\) is a scalar under rotations, so it transforms as \(\mathbf{1}\).

- \(A_i\), with \(i=1,2,3\), are the spatial components, so they transform as an ordinary vector \(\mathbf{3}\).

2. Build the tensor product

Now consider a rank-2 Lorentz tensor:

\[ T_{\mu\nu} = A_\mu B_\nu \]

Under restriction, this becomes:

\[ 4 \otimes 4 \longrightarrow (3+1)\otimes(3+1) \]

Expanding gives:

\[ (3+1)\otimes(3+1)=3\otimes3\oplus3\otimes1\oplus1\otimes3\oplus1\otimes1 \]

3. Decompose each piece under \(SO(3)\)

(a) The spatial tensor \(3\otimes 3\)

This is the place where symmetry and antisymmetry matter:

\[ 3\otimes 3 = 5\oplus3\oplus1 \]

This corresponds to decomposing \(T_{ij}\) into:

- a symmetric traceless part, which gives \(5\),

- an antisymmetric part, which gives \(3\),

- and the trace, which gives \(1\).

(b) The mixed components

\[ 3\otimes1 = 3 \qquad (T_{i0}) \]

\[ 1\otimes3 = 3 \qquad (T_{0i}) \]

These are just vectors.

(c) The scalar part

\[ 1\otimes1 = 1 \qquad (T_{00}) \]

4. Full decomposition

Putting everything together:

\[ 4\otimes4 \longrightarrow 5\oplus3\oplus1\oplus3\oplus3\oplus1 \]

5. Understand each piece concretely

Component	Object	Representation under \(SO(3)\)
\(T_{00}\)	scalar	\(1\)
\(T_{0i}\)	vector	\(3\)
\(T_{i0}\)	vector	\(3\)
\(T_{ij}\) symmetric traceless	spin-2 part	\(5\)
\(T_{ij}\) antisymmetric	vector-like part	\(3\)
trace \(T_{ii}\)	scalar	\(1\)

6. Connect this to the Lorentz decomposition

We also know that as a Lorentz representation,

\[ 4\otimes4 = 6\oplus10 \]

where:

- \(6\) is the antisymmetric tensor \(T_{[\mu\nu]}\),

- \(10\) is the symmetric tensor \(T_{(\mu\nu)}\).

7. Restrict each Lorentz piece to \(SO(3)\)

(A) The antisymmetric tensor \(6\)

Split it into:

- \(T_{0i}\): 3 components, transforming as a vector,

- \(T_{ij}\): antisymmetric in 3 dimensions, also 3 components.

A key fact in three dimensions is:

\[ T_{ij} \longleftrightarrow \epsilon_{ijk} B_k \]

So an antisymmetric spatial 2-tensor is equivalent to a vector. Therefore:

\[ \mathbf{6} \longrightarrow \mathbf{3}\oplus\mathbf{3} \]

Important clarification. Both \(3\)'s come from the same antisymmetric Lorentz tensor. The split is because when we restrict to rotations, the antisymmetric tensor breaks into time-space components and space-space components, and each of those transforms as a \(3\).

(B) The symmetric tensor \(10\)

Now split the symmetric tensor into:

- \(T_{00}\), giving \(1\),

- \(T_{0i}=T_{i0}\), giving \(3\),

- symmetric spatial \(T_{ij}\), giving \(5\oplus1\).

Hence:

\[ \mathbf{10} \longrightarrow \mathbf{5}\oplus\mathbf{3}\oplus\mathbf{1}\oplus\mathbf{1} \]

8. Final structure

\[ (6\oplus10) \longrightarrow (3\oplus3)\oplus(5\oplus3\oplus1\oplus1) \]

9. Key conceptual insights

(1) Why symmetry matters only sometimes

Symmetry and antisymmetry only apply when the two indices are of the same kind. So:

- \(T_{ij}\) can be split into symmetric and antisymmetric parts,

- but \(T_{0i}\) and \(T_{i0}\) cannot.

(2) Why antisymmetric \(T_{ij}\) becomes a vector

In three dimensions, an antisymmetric 2-tensor has three independent components, and using \(\epsilon_{ijk}\) it can be identified with a vector.

(3) Why \(6=3+3\)

The reason is that after separating time from space, the antisymmetric Lorentz tensor splits into:

- time-space components, transforming as a vector, and

- space-space components, dual to a vector.

(4) Physical interpretation

For the electromagnetic tensor \(F_{\mu\nu}\), we have:

\[ F_{0i} \sim E_i \]

\[ F_{ij} \sim \epsilon_{ijk} B_k \]

So:

\[ 6 \rightarrow 3\oplus3 \]

corresponding to the electric and magnetic fields.

(5) The big picture

Restricting from the Lorentz group to the rotation subgroup separates time from space and reorganizes everything into familiar spin representations:

\[ \text{spin }0 \rightarrow 1, \qquad \text{spin }1 \rightarrow 3, \qquad \text{spin }2 \rightarrow 5 \]

10. Final intuition

A good way to think about the whole process is:

Take a relativistic object and ask what it looks like to someone who is only allowed to rotate space. Then the Lorentz representations naturally reorganize into \(SO(3)\) spin multiplets.

That viewpoint leads directly to:

\[ 4\rightarrow3+1, \qquad 4\otimes4 \rightarrow 5+3+1+1+3+3 \]

This is additional material to complement page ? of Group Theory in a Nutshell by Anthony Zee.