What happens if we vary with respect to an arbitrary connection with torsion? I assume from your writing that it won't yield metric compatibility, and this confuses me.
On a heuristic level (I haven't done the calculation yet, just curious): S. Jensen (pdf given above) writes
<<Requiring that...
Imagine we are talking about the group SO(4). The second rank antisymmetric representation is reducible into self-dual and antiself-dual representations. I think a good way to visualise this is by noticing that the projection of \Lambda^{2}V into self and antiself dual subspaces commutes with...
Nevermind. While the tool needed was clear from the beginning, i.e. an identity about products of traces, I was completely oblivious to the existence of one. It turns out that
\sum_{\mu}Tr(G\sigma_{\mu})Tr(\overline{\sigma}^{\mu}H) = 2 Tr(GH)
While I still don't know the proof of this, this...
Ok, I think I confused myself again. I thought I was thinking about this the right way, but I can't apply my reasoning to a slightly harder case: the proper orthochronous Lorentz group, SO(3,1), or equivalently, SL(2,C).
Here the commutators read
\begin{split}
[J_{i}, J_{j}] &=...
We have that
\Lambda^{\mu}_{\nu} = \frac{1}{2}Tr(\overline{\sigma}^{\mu}A\sigma^{\nu}A^{\dagger})
I would like to make sense of the statement that this is a homomorphism because the correspondence above is preserved under multiplication.
Can someone clarify how I could see this?
Hi,
I am confused on a very basic fact. I can write \xi = (\xi_{1}, \xi_{2}) and a spin rotation matrix as
U =
\left( \begin{array}{ccc}
e^{-\frac{i}{2}\phi} & 0 \\
0 & e^{\frac{i}{2}\phi}
\end{array} \right)
A spinor rotates under a 2\pi rotation as
\xi ' =
\left( \begin{array}{ccc}...
Sure, the rank of SU(3) is 2, so is the dimension of the Cartan subalgebra. There one could see more clearly from the (modified) commutation relations that there are indeed two commuting generators.
Ok, what I said above was correct; I see now that from the commutation relations there are no commuting generators, but I can always pick one at random and see that it commutes trivially with itself. This is one way to see that the rank is 1. And from this I can infer there will be only one...
Thank you very much, very clear. So, I suppose there is a way to find out how many Casimir operators I can construct, given an algebra with its commutation relations. I know SO(3) has J^2, but it is not clear to me how I know that it is the only Casimir there is.
Also, the rank of the Lie group...
I am reading in my group theory book the well known commutation relations of the Lie algebra of SO(3), i.e. [J,J]=i\epsilon J.
What I don't understand is the statement that "from the relations we can infer that the algebra has rank 1".
Any ideas?
Hi,
Can anyone clarify what precisely is meant by "nonlocal transformations" in the BV formalism?
Specifically, they claim that for any gauge theory with an open algebra, it is possible to go to a different basis for the algebra whereby one achieves closure (and even one where the gauge...
Thanks, very clear. Two things:
- They definitely are perfect absorbers, by definition. Wiki, though, states that at any temperature and frequency, they also are the most efficient emitters possible emissivity=1). Is this inaccurate?
Because of this, and because I really meant "black to our...