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...