Recent content by synoe

Im not familiar with the categories so I couldn't understand precisely what you mean. But your statements seem to be different from things written in the textbooks I read. My statement is as follows. If a map f:M\to M^\prime is a diffeomorphism, then the two diffeomorphic manifolds (M,g) and...

I can't believe reparameterizations in GR are isometries since spacetime does not have isometries in general. At least Lagrangian level, arbitrary manifolds can be allowed. To be an isometry, the map f must be to itself f:M\to M because we can't compare tensors T_p at p with pullbacked tensors...

I may resolved the question. Could you give me your opinions? Coordinate transformations on a manifold can be seen by two different ways, passive and active. The passive one is just a reparameterization, which maps a local coordinate to another local coordinate at a same point on the manifold...

Is it true although dx^\mu transforms as a vector under the general coordinate transformation ? dx^\mu\to dx^{\prime\mu}(x)=\frac{\partial x^{\prime\mu}}{\partial x^\nu}dx^\nu

Let f:p\mapsto f(p) be a diffeomorphism on a m dimensional manifold (M,g). In general this map doesn't preserve the length of a vector unless f is the isometry. g_p(V,V)\ne g_{f(p)}(f_\ast V,f_\ast V). Here, f_\ast:T_pM\to T_{f(p)}M is the induced map. In spite of this fact why...

How about the Lorentz invariant bilinear form \bar{\Psi}\Psi? If \Psi takes the form \Psi= \begin{pmatrix} \psi_+\\ 0 \end{pmatrix} in a frame, its Dirac adjoint takes the form \bar{\Psi}= \begin{pmatrix} 0&\bar{\psi}_+ \end{pmatrix}. Then, the bilinear \bar{\Psi}\Psi equals to zero. This...

I can't understand why a commutable matrix can decompose a representation into a block diagonal form. Could you tell me the concrete way of decomposition? And I would like to realize the proposition in matrix form. I think the five dimensional spinors which is written as \Psi= \begin{pmatrix}...

In four dimensions, left and right chiral fermion can be written as \psi_L= \begin{pmatrix} \psi_+\\ 0 \end{pmatrix},\qquad \psi_R= \begin{pmatrix} 0\\ \psi_- \end{pmatrix}, respectively, where \psi_+ and \psi_- are some two components spinors(Weyl spinors?). In this representation, the...

I have a question about chirality. When a spinor \psi have plus chirality, namely \gamma_5\psi=+\psi, how can I write this condition for the Dirac adjoint \bar{\psi}=\psi^\dagger i\gamma^0? Let me choose the signature as \eta_{\mu\nu}=\mathrm{diag}(-,+,+,+) and define \gamma_5\equiv...

In the way of defining the adjoint representation, \mathrm{ad}_XY=[X,Y], where X,Y are elements of a Lie algebra, how to determine the components of its representation, which equals to the structure constant?

In string theory, the Neveu-Schwarz B-field appears in the action: S_{NS}=\frac{1}{4\pi\alpha^\prime}\int d^2\xi\;\epsilon^{\mu\nu}B_{ij}\partial_\mu X^i\partial_\nu X^j. In Polchinski's text, the antisymmetric tensor appears in the form of \frac{1}{4\pi\alpha^\prime}\int...

\frac{\partial}{\partial X^a} commutes with \frac{\partial}{\partial\sigma^\mu}? Is it trivial? I couldn't follow this equation. Could you explain in more detail?

Thank you ! It was very helpful to understand. It seems that for the \sigma-action, the variation accidentally coincides with the Lie derivative of the metric. I have a feeling that this coincidence is necessary in view of the geometry of the target space. Can I understand in more geometrical way?

Thank you samalkhaiat. Your calculation seems to be just a definition of the Lie derivative, the variation at the same "coordinate point". My question is why the Lie derivative appears in the \sigma-action in the form of \delta G_{ij}\partial_\mu X^i\partial_\nu X^j when acting X\to...

For the non-linear sigma action, S_G=\frac{1}{4\pi\alpha^\prime}\int d^2\sigma\sqrt{-\gamma(\sigma)}\gamma^{\mu\nu}(\sigma)G_{ij}(X)\partial_\mu(\sigma) X^i\partial_\nu X^j(\sigma), Let us consider an infinitesimal target space transformation X^\mu\to X^{\prime\mu}(X)=X+\epsilon\xi^\mu(X). The...