# Detail on seeing the double cover of $SO^{+}(1, 3)$ as $SL(2, \mathbb{C})$

We can identify Minkowski space-time $$M^4$$, of metric signature $$(1, -1, -1, -1)$$, with the (real) space of $$2 \times 2$$ (complex) Hermitian matrices under the map $$(v_0, v_1, v_2, v_3) \mapsto v_0 I + v_1 \sigma_x + v_2 \sigma_y + v_3 \sigma_z$$ where $$\sigma$$'s are Pauli matrices.

Given a Hermitian matrix $$X$$ and some $$A \in SL(2, \mathbb{C})$$, the matrix $$A X A^{\dagger}$$ is again Hermitian, and $$\det(AXA^{\dagger}) = \det(X)$$, so (Hermitian) conjugation by $$A$$ is a linear transformation of Minkowski space-time which preserves the metric and is thus a Lorentz transformation. It is orthochronous because $$AIA^{\dagger} = AA^{\dagger}$$ is a positive operator, and thus $$\text{tr}(AA^{\dagger}I) = \text{tr}(AA^{\dagger})$$ (which, up to a factor of $$1/2$$, is the $$I$$-coefficient of $$AA^{\dagger}$$ in the $$I, \sigma$$ basis) is positive.

How can I check it has determinant $$1$$, though? It seems painful to compute the other coefficients of the matrix corresponding to this transformation. Is there some slick way?

• Maybe I am missing something, but isn't this obvious from the equation you wrote: $det(AXA^{\dagger})=det(X)$. Using the determinant identity $det(ABC)=det(A)det(B)det(C)$, we arrive at $det(A)det(A^{\dagger})=1$, from which it follows that determinant of $A$ is $1$? – Anonjohn Oct 16 '20 at 5:16
• It is obvious that its squared determinant is 1 since the conjugate transpose has the same determinant as the original matrix – daydreamer Oct 16 '20 at 6:49
• @Anonjohn, the question is not the determinant of $A$, which is 1 by assumption, but the determinant of the linear transformation $X \mapsto AXA^{\dagger}$. – Pedro Oct 16 '20 at 16:07

OP has already argued that it is a Lorentz transformation. It is well-known that Lorentz matrices always have determinant $$\pm 1$$. Let us reformulate OP's question as follows.
How can I check it has determinant $$+1$$ (as opposed to $$-1$$)?
Proof: Since $$\rho: 𝑆𝐿(2,\mathbb{C})\to O(1,3;\mathbb{R})$$ 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. Hence the image $$\rho(𝑆𝐿(2,\mathbb{C}))\subseteq SO^+(1,3;\mathbb{R})$$ must be a subset of the connected component of $$O(1,3;\mathbb{R})$$ that contains the identity, i.e. the restricted Lorentz group $$SO^+(1,3;\mathbb{R})$$. $$\Box$$
• Well, that's slick! I somehow hoped an answer would involve some way of figuring out what the matrix for $\rho(A)$ is in terms of the $I, \sigma$ basis, but perhaps it's unwieldy. Do you know how can one see $\rho$ is surjective (onto $SO^{+}(1, 3; \mathbb{R})$)? – Pedro Oct 16 '20 at 16:16