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5

The group $U(N)\times U(N)$ acts on your original space, but some group elements act in the same way: the action is not faithful, in other words, this groups maps onto the symmetry group (this is implicitly assumed by what you wrote), but not injectively. Specifically, the diagonal subgroup of scalar matrices (i.e. matrices that are scalar multiples of the ...


3

In quantum mechanics, the relevant representations of symmetry groups on the space of states are not our usual linear representation, but projective representations on the Hilbert space. The projective representations of a semi-simple Lie group - such as the rotation group $\mathrm{SO}(n)$ - are in bijection to linear representations of its universal cover. ...


3

The Poincaré group is the semi-direct product of the six-dimensional Lorentz group and the four-dimensional translations and hence ten-dimensional (or "has ten parameters" is less precise diction). Since in a global inertial coordinate system you have to have the Minkowski metric by definition, only those transformations (diffeomorphisms) which preserve the ...


3

In the $SU(4)$ language, the 10-dimensional representation is the symmetric spintensor $T_{(ab)}$ with $4\times 5 / (2\times 1) = 10$ components. In the $SO(6)$ representation, it is the self-dual 3-form with $$ \frac 12 \cdot \frac{ 6\times 5 \times 4}{3\times 2 \times 1} = 10$$ components. It's the tensor $T_{[kmn]}$ that also obeys $$ T_{kmn} = \frac{\...


3

Representation of $SU(2)$ is pseudo-real. Which means, if $\mathbf{[2]}$ and $\mathbf{[\bar{2}]}$ are the fundamental and anti-fundamental representation of $SU(2)$, then there exists an anti-symmetric matrix $\cal{C}$, which connect both of them, as $\cal{C}\mathbf{[2]}\cal{C}^{-1}=\mathbf{[\bar{2}]}$. Another way of saying this, both $\mathbf{[2]}$ and $\...


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A 3D cube with pacman topology is translationally invariant and not rotationally invariant. A space like this is a possible (but unlikely) flat space part of a cosmological spacetime


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The issue is that the "spin representation of $SO(3)$" is not a representation of $SO(3)$ at all, but a representation of its double cover $SU(2)$ (see https://en.wikipedia.org/wiki/Spin_group). Since we sometimes write down representations in terms of infinitesimal generators (in other words, as a representation of the Lie algebra of the Lie group in ...


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It seems OP's main question is how to understand the representation of the matter fields of YM theory. The matter fields can in principle transform in any representation $\rho:G\to {\rm End}(V)$ of the local gauge group $G=SU(N)$, e.g. the fundamental, or adjoint representation. Here ${\rm End}(V)$ denotes the algebra of endomorphisms on the vector space $...


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The spin group is related to spin-half objects, called spinors. If you rotate a spinor by 360 degrees, you get back the negative of the spinor you started with. Now it would be nice if you could represent the action of this rotation by saying that an element of $SO(n)$ is acting on the spinor. However, this cannot be done because a rotation by 360 degrees is ...


1

One reason there are more possible eigenvalues of the Casimir operator of the rotations than appear in the spherical harmonics is that the spherical harmonics are proper representations of $\mathrm{SO}(n)$ while the possible values for the Casimir operator classify the possible irreducible representations of $\mathfrak{so}(n)$. By general Lie theoretic ...


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The sentence above C.11 explicitly says that they talk about 3-forms under $SO(6)$, i.e. antisymmetric tensors $T_{[abc]}$ where $a,b,c=1,2,3,4,5,6$. Those have $$ \frac{6\times 5\times 4}{3\times 2 \times 1} = 20 $$ components. By the Dirac matrix calculus, all differential forms may be obtained from the tensor product of two spinors and the Dirac spinor is ...



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