Understanding Matrix Inversion in SL(2,C)

[LAHLH]

Hi,

I'm just reading about the group SL(2,C). In the book that I'm using(Jones, groups reps and physics), he defines a 2x2 matrix from a generic 4 vector v_{\mu} and a vector \sigma_{\mu}:=(1,\vec{\sigma}), as V:=v_{\mu}\sigma^{\mu}

He nows wants to invert this equation to solve for v_{\mu}, and he suggests tracing with another vector of matrices defined as \tilde{\sigma_{\mu}}:=(1,-\vec{\sigma}), and he obtains v_{\mu}=\tfrac{1}{2}Tr(\tilde{\sigma_{\mu}}V)

I can't seem to get this, starting with V:=v_{\mu}\sigma^{\mu} and then multiplying by \tilde{\sigma_{\nu}}, leads to \tilde{\sigma_{\nu}}V:=v_{\mu}\sigma^{\mu}\tilde{\sigma_{\nu}}

Now I'm not sure what indices I'm supposed to trace with?

 

[chrispb]

Your V is a 2x2 matrix. If you want to recover v_{\nu} (i.e. the nu'th component of the vector v), you should take \frac{1}{2} Tr(\tilde{\sigma}_{\nu}V), where nu is not summed over. In other words, if you want the first component, take the 2x2 matrix \tilde{\sigma}_1, left-multiply (though inside a trace order doesn't matter) it with the 2x2 matrix V, and take half the trace of the resultant 2x2 matrix. Try a simple example if this isn't clear; pick (1,0,0,0) and try to recover v_0.

 

[Fredrik]

I'm going to write all indices as subscripts. \{I,\sigma_1,\sigma_2,\sigma_3\}, is a basis for the real vector space of complex 2×2 self-adjoint matrices. I'll call that space V. If we define an inner product by

\langle A,B\rangle=\frac{1}{2}\operatorname{Tr}(A^\dagger B)

for all A,B in V, it's an orthonormal basis. So if we define \sigma_0=I, any x \in V can be expressed as x=x_\mu \sigma_\mu, with x_\mu=\langle\sigma_\mu,x\rangle. Note that all the x_\mu are real. (This is implied by the facts that V is a real vector space and that \{\sigma_0,\sigma_1,\sigma_2,\sigma_3\} is a basis of V).

x_\mu=\langle\sigma_\mu,x\rangle=\frac{1}{2}\operatorname{Tr}(\sigma_\mu^\dagger x)=\frac{1}{2}(\sigma_\mu x)_{\nu\nu}

The map \mathbb R^4\ni (x_0,x_1,x_2,x_3)\mapsto x_\mu\sigma_\mu\in V is an isomorphism. So V is isomorphic to \mathbb R^4.

 

[LAHLH]

Thanks for the help.

I convinced myself of this in the end.

<br /> \tilde{\sigma_{\mu}}V:=\tilde{\sigma_{\mu}}(v_{\beta}\sigma^{\beta})<br />
Tracing (on the 2x2 matrix indices, not 4 vec indices, which is I think what was confusing me):
Tr( \tilde{\sigma_{\mu}}V)=( \tilde{\sigma_{\mu}}V)_{ii}=(\tilde{\sigma_{\mu}})_{ij}(v_{\beta}\sigma^{\beta})_{ji}=v_{\beta} (\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji}

Now we have that, \tilde{\sigma_{\mu}}=(1,-\vec{\sigma}) and \sigma^{\mu}=(1,-\vec{\sigma}). Therefore if \mu=\beta : (\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji}=1_{ii}=2 (since e.g. 1x1=1, \sigma_{x}\sigma_{x}=1, \sigma_{y}\sigma_{y}=1, \sigma_{z}\sigma_{z}=1 etc, no summation)

On the other hand if \mu\neq\beta, we end up with (\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji} equal to the trace on another Pauli matrix by virtue of the cyclic identity \sigma_{i}\sigma_{j}=i\epsilon_{ijk}\sigma_{k}, and the since the Pauli matrices are traceless, this trace of the product is zero.

Combining these facts:

Tr(\tilde{\sigma}_{\mu}V)=v_{\beta}\delta^{\beta}_{\mu}2 which implies v_{\beta}=\tfrac{1}{2}Tr(\tilde{\sigma}_{\mu}V)