Recent content by kakarukeys

try to derive [AB,C] = ? then use [x,p] = i\hbar to find [p^2,x]

Is it right to say? Being able to find a gauge-invariant measure in loop quantization is a big achievement because if we performed a traditional canonical quantization we would be using the ill-defined measure dA^i_a, inner product: \int\Phi^*[A^i_a]\Psi[A^i_a]dA^i_a They are not...

if there is no gauge-invariant measure available, can a gauge-invariant inner product be constructed?

Could someone explain to me why we use a gauge-invariant and diffeomorphism-invariant measure on the quantum configuration space? Is it because we want the inner product to be invariant under gauge transformations. What is a gauge-invariant measure anyway? see http://arxiv.org/abs/hep-th/9305045

yes I read that one too...

I believe you are giving me an internal gauge group of GR, made manifest because of a reformulation. E.g. for spinor formulation, we have SL(2,C), LQG, we have either SO(3), or SU(2). I just want to know the status of Diff(M). I saw the thread before...can't find my answer Here I talk about...

I tried to convince myself it is Diff(M), but I failed. Most books say Bianchi Identities reduce the independent equations in Einstein's equations by 4, therefore there are some redundancies in the metric variables. As a result, there could be many solutions that correspond to one physical...

I couldn't find any theorem which guarantees that. closests two are: (1) if G is compact and F is Hausdorff, \rho_f is open (2) if G is locally compact, F is locally compact and Hausdorff, F is a topological group under the induced group operations, \rho_f is open

Yes, I'm asking if we drop the assumption of positive definiteness of inner product, will it work?

http://en.wikipedia.org/wiki/Gram-Schmidt_process Can Gram–Schmidt process be used to orthonormalize a finite set of linearly independent vectors in a space with any nondegenerate sesquilinear form / symmetric bilinear form not necessarily positive definite? For example in R^2 define \langle...

yes, a restriction is always continuous under the subspace topology. And the subspace topology is actually the topology of B (or A). So If f is continuous then f is "continuous in the first variable and second variable". But is it true if f is "continuous in the first variable and second...

Is it true that f is a continuous function from A \times B to C (A, B, C are topological spaces) if and only if f_{a}: \{a\}\times B \longrightarrow C and f_{b}: A\times \{b\} \longrightarrow C are continuous for all a\in A, b\in B ? f_a(b) = f_b(a) = f(a,b)

At one stage of my calculations, I had ()^2 = 0 implies () = 0 so a 4th order eq became quadratic eq

How did you calculate the determinant? I calculated it with some formulae, and I got only two eigenvalues. Maybe you should try using a software like mathematica to calculate it by brute force. Let just hope that your Dirac matrices are same as mine. I used...

No, there is no typo, I have typed a little too fast. Let me use Latex and state my question clearer. There is one point I don't understand about G-torsor. A Lie group G acts freely and transitively on a manifold F. \rho: F \times G \longrightarrow F \rho(f, g) = fg f(g_1g_2) = (fg_1)g_2 fix...