John

# 161 Actions

2016
Apr
4
awarded  Yearling
2015
Apr
4
awarded  Yearling
Jan
29
comment Generalized spin connection and dreibein in higher spin gravity
Ok, then it is not exactly about HS theory and related ads/cft duality, which requires matter fields to be present in a theory and they cannot couple to sl(n) Chern-Simons. You probably need to google higher teichmuller theory, which is a way to talk about sl(n) structures
Jan
28
answered Generalized spin connection and dreibein in higher spin gravity
2014
Aug
14
comment General relativity from helicity 2 massless field theory by using Deser's arguments
You can freely download Thomas Ortin's Gravity and Strings book, where you find a nice section on deser's argument and how to reconstruct gravity. Just the description by Deser himself is not the best one:-)
Jul
25
comment I want to decompose a tensor product using Littlewood-Richardson rule, How do I find the component of this in each irreducible space?
If the multiplicity is greater than one than there is no canonical way of indentifying tensors of the same symmetry type, you have a linear subspace corresponding to a given Young diagram and the dimension of this space is multiplicity. All bases in this linear space are equivally good. If the case is so simple that the mupltiplicity is always one, then the problem is just to order indices (like in the example you gave when two tensors are identically the same after using Young properties).
Jul
25
comment I want to decompose a tensor product using Littlewood-Richardson rule, How do I find the component of this in each irreducible space?
Ok, but what exactly you need to prove? from very general properties of tensor products you know that given reducible tensor can be decomposed into irreducibles. Irreducible tensors will contribute according to the multiplicity they have (in your example it is always one, but in general it can be greater). So you cannot have the number of irreducible tensors of some type be greater (or less) than the multiplicty. The rest is how to write all this with indices, which can be ambigous unless you order indices in some way. You can refer to stackexchange, but it is ok for me not to do that :)
Jul
24
comment I want to decompose a tensor product using Littlewood-Richardson rule, How do I find the component of this in each irreducible space?
To avoid doublecounting you need to order indices. For example, you can try alphabetical order, i.e. to have as many $a$'s in front of $b$'s as you can, etc. Somehow what you see is a little bit more complicated example than the question how to distinguish $\epsilon_{abcd}$ and $\epsilon_{bacd}$, where $\epsilon$ is rank-four anti-symmetric. You know the answer - to order indices.
Jul
24
comment I want to decompose a tensor product using Littlewood-Richardson rule, How do I find the component of this in each irreducible space?
Ok, I got it. The thing is that the two tensors are identically the same. Remember that $a[p+k],ab[p-k-1]\equiv0$, which is a Young symmetry property that tells us that the tensor has $[p+k,k]$-diagram. Now we can use this defining property to prove that $a[p-1]b[k+1],ab[p-k-1]$ is proportional to $a[p+k],ab[p-k-1]$. Indeed, $a[p-1]b[k+1],ab[p-k-1]$ is almost a Young condition if we add $a[p][b[k],b]b[p-k-1]$, where I antisymmetrize $[b[k],b]$, we get zero, so the two tensors are identically the same. Ok?
Jul
23
comment I want to decompose a tensor product using Littlewood-Richardson rule, How do I find the component of this in each irreducible space?
Ok, let's try one more time to understand. I do not understand your/my notation a little bit. If the tensor is $a[p]|b[p]$ then the projection is $a[p-k]b[k]|b[p]$ while what you wrote does not have $p$ indices on both sides.
Jul
22
comment I want to decompose a tensor product using Littlewood-Richardson rule, How do I find the component of this in each irreducible space?
So, the relatioship is true in general, but of course (SA)T will never be equal (AS)T for some tensor T, which is not a scalar, so please compare, expand or do anything else always in the same basis, symmetric or anti-symmetric, or some else basis.
Jul
22
comment I want to decompose a tensor product using Littlewood-Richardson rule, How do I find the component of this in each irreducible space?
First comment, the phenomenon you observed exists even for a trivial tensor product when you have tensor product of an irreducible tensor $T^A$ given in anti-symmetric basis, as in your example, and a trivial scalar tensor. Then if you apply Young symmetrizer in the form (SA) you will change the basis from antisymmetric to symmetric. Applying in another order (AS) does nothing.
Jun
25
comment $\langle TT\rangle$ correlator of the boundary CFT from metric fluctuations in the bulk gravity
There is nothing special about this computation, just stress-tensor/graviton have more indices than scalar. For scalar, you can look at appendix A of arxiv.org/pdf/hep-th/9804058v2.pdf, where it is explained why it is the coefficient of $k^d$ that needs to be looked at (it is the only one surviving actually and gives the right dependence when Fourier transformed back). I cannot help with Mathematica file
Jun
24
comment $\langle TT\rangle$ correlator of the boundary CFT from metric fluctuations in the bulk gravity
you take arbitary Lagrangian of gravity+something, the empty AdS is an exact solution. So you can reexpand your action around this solution. Then, one defines propagators, verties as usual. The only point is that the linearized equations are $\square_{AdS}\phi=0$ necessarily, so AdS plane waves are solutions and propagators are of the same form. The details of your Lagrangian are interaction vertices. The last question - it is the AdS/CFT conjecture, and the procedure for getting $<TT>$ this way is a part of it, you can consult any AdS-CFT review
Jun
22
comment $\langle TT\rangle$ correlator of the boundary CFT from metric fluctuations in the bulk gravity
the boundary term, so boundary terms bacome important. In principle boundary terms are important when there are boundaries in order to have a well-defined variation problem, see Gibbons-Hawking boundary terms. For AdS CFT the following paper arxiv.org/pdf/hep-th/9806216.pdf addresses both questions, boundary terms and stress-tensor two-point function, I think it is the best.
Jun
22
comment $\langle TT\rangle$ correlator of the boundary CFT from metric fluctuations in the bulk gravity
(1) I meant the common wisdom that the CFT partition function is an AdS path integral over configurations with presribed boundary conditions (to be identified with the sources to CFT operators). The first approximation is just the classical action evaluated on the solutions and the first approximation to that is quadratic action evaluated on the solutions to the linearized equations. (2) Bessel function is a solution to the boundary problem in AdS, so you just solve $\square_{AdS} \phi=0$. (3) There is a problem with 2-pt functions that the action is not bulk integral but reduces to ...
Jun
20
comment $\langle TT\rangle$ correlator of the boundary CFT from metric fluctuations in the bulk gravity
Appendix B.5 in qpt.physics.harvard.edu/p234.pdf addresses this derivation. In principle what you are saying is OK, but it is the most non-invariant recipe I have ever seen. One takes the quadratic part of the action, evaluates it on the solutions to linearized equations (propagators). By definition the result is a total derivative, so it resides on the boundary. Careful regularization is needed since the boundary is not a true boundary, so one moves it to the finite distance $\epsilon$ and takes the limit at the end. arxiv.org/pdf/hep-th/0209067.pdf should be useful.
Jun
16
comment How does one expand gravity Lagrangians about an $AdS$ background?
It does not derive strictly speaking, but contains all relevant formulas. How to derive I explained above - it is a unique second-order action (equations of motion) that is invariant under the gauge symmetry above. For example, you can check first with vector field $\delta A_m=\nabla_m\xi$ by writing all second order terms (there are only two, $\square A_m$ and $\nabla_m \nabla^n A_n$) and fixing the coefficient from gauge invariance. The spin-two follows the same logic but there are more indices and terms to consider. In checking gauge invariance you will have to use $R=gg$
Jun
15
comment How does one expand gravity Lagrangians about an $AdS$ background?
I found arxiv.org/pdf/1105.3735.pdf which contains a lot about F-P action on flat and AdS, so you can find out how it is fixed by gauge invariance
Jun
15
comment How does one expand gravity Lagrangians about an $AdS$ background?
In general one can say that the expansion over background is always covariant with respect to the symmetries of this background. As for the F-P action, it is fully fixed by gauge invariance. To simplify problem you start with ansatz for second order equations of motion with gauge symmetry $\delta h_{\mu\nu} =\nabla_\nu\xi_\mu+\nabla_\mu\xi_\nu$ and find that the solution is unique. The mass-like term comes exactly from $R=gg$. (One can consider first a simpler problem over flat space and then turn on cosmological constant)