This is essentially nothing more than the orthogonal diagonalisation of a symmetric matrix that you probably did loads of times when you first learned about matrices. The coordinate transformation can be calculated from the matrix of eigenvectors. This gives a diagonal matrix, which should have...
Firstly, notice that the RHS is antisymmetric in exchange of any pair of the i's and any pair of the j's. So all you have to check is that you get the right answer for i_1=1, i_2=2, \ldots and the same for the j's, which is straightforward.
To fix notation: a Mobius map is a map of the form
z\mapsto \frac{az+b}{cz+d}
for some complex constants a,b,c,d satisfying ad-bc\neq0. It is this number ad-bc that is meant by the 'determinant' in the problem (though that name is not justified quite yet, see later).
The constants are not...
Yes that sounds pretty much right. It might help a bit if I express it in a slightly different way:
We start with some algebraic relations, in this case the anticommutation relations \alpha_i\alpha_j+\alpha_j\alpha_i=2\delta_{ij}. We would like to represent these relations as linear operators...
Sorry I'm being a bit terse, I'll try to explain what I'm doing more fully.
We start with a set of two by two matrices satisfying the algebra (I'm not going to require them to be hermitian, but you could). What we want to prove is that there is a similarity transformation S (again, not...
Nicely done, that's the trick!
When I say 'residual symmetry' I mean that when we diagonalised in part (4) the choice of S wasn't actually unique, and we still have some freedom left over. Specifically, we can choose another S as long as it commutes with the third Pauli matrix. If you work...
Hi,
It's not quite true but it's quite close to being true. (Notice that your second condition is redundant since it's contained in the first).
If the matrices are 2x2, then they must be the Pauli matrices up to a similarity transformation. That is to say, there exists an invertible matrix S...
dz won't be a KVF because a flat disc is not symmetric under z-translation: it would have to be an 'infinite cylinder' sort of object to satisfy this. Depending on what you're doing, dt might be a KVF though (if the setup is stationary).
Hi, and welcome to PF!
The clearest way to see GR as a gauge theory is to formulate it in terms of an orthonormal basis of vectors e_\alpha, satisfying g(e_\alpha,e_\beta)=\eta_{\alpha\beta}. [The components e_\alpha^\mu are sometimes called vierbeins. I use Greek letters from the start of the...
Bear in mind that for arbitrary exponents, the operation of taking powers is multivalued. This is an extension of the familiar case of the square root: for example, 1^{1/2} could be plus or minus 1.
Arbitrary powers are defined by a^b=\exp(b\log(a)). But log is multivalued on the complex...
The problem with this argument is that you assume that the inverse is analytic, which must be proved.
The normal complex inverse function theorem assumes nonzero derivative, and proves existence of a continuous inverse, and then analyticity of the inverse and the formula for the inverse. You...
For a commutative ring, a matrix is invertible if and only if its determinant is a unit.
If that ring is a field, every nonzero element is a unit so we recover the well-known result that a matrix is invertible if and only if it has nonvanishing determinant.
Example: a matrix over the integers...
OK, I now see what's going on here, but it takes quite a bit to develop it. In outline, it goes like this:
You start with a Lie group, which is a group described by n continuous parameters.
You can investigate the Lie group by looking at its local structure, near the identity element. This...
You actually need a little more information. The following simple question should bring it into relief: what's the answer for the case n=0? You need a probability before you have started looking at the balls. (This is the basis of 'Bayesian inference' in which this probability is called a...
Are you worried about the step of finding h given g? You don't quite need to use the fact that a disc is simply connected, basically because we only need to be able to work in a tiny disc around z0.
In more detail:
Pick some disc around g(z0) not containing zero. It should be clear that we...