44 votes
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Lie algebra in simple terms

It's an enormous subject, but briefly: Lie groups are smooth groups. Technically, Lie groups are sets that are both a smooth manifold, like a sphere for instance, and also have a group structure (...
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  • 2,950
35 votes
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Why are relativistic quantum field theories so much more restrictive than non-relativistic ones?

One of the reasons relativistic theories are so restrictive is because of the rigidity of the the symmetry group. Indeed, the (homogeneous part) of the same is simple, as opposed to that of non-...
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32 votes

What does it mean for particles to "be" the irreducible unitary representations of the Poincare group?

Irreducible representations of the Poincare group are the smallest subspaces that are closed under the action of the Poincare group, which includes boosts, rotations, and translations. The point is ...
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  • 95.5k
31 votes

Why do all fields in a QFT transform like *irreducible* representations of some group?

Gell-Mann's totalitarian principle provides one possible answer. If a physical system is invariant under a symmetry group $G$ then everything not forbidden by $G$-symmetry is compulsory! This means ...
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  • 172k
27 votes
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Why is there this relationship between quaternions and Pauli matrices?

At the level of formulas, the three quaternionic units $i_a$, $a\in~\{1,2,3\}$, in $\mathbb{H}\cong \mathbb{R}^4$ satisfy $$i_a i_b ~=~ -\delta_{ab} + \sum_{c=1}^3\varepsilon_{abc} i_c, \qquad\qquad ...
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  • 172k
25 votes
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Why are band maxima / minima often (always?) at high-symmetry points?

$\renewcommand{ket}[1]{|#1\rangle}$ The basic logical connection here is $$\text{symmetry} \rightarrow \text{degeneracy} \rightarrow \text{avoided crossing} \rightarrow \text{band gap} \, .$$ $\...
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24 votes

Comprehensive book on group theory for physicists?

Here is my extensively review of various books I had read. For meta discussion, see I have several book reviews. How should I answer in the book request?. Books being reviewed: Wu-Ki Tung, Group ...
24 votes

Why do we need complex representations in Grand Unified Theories?

Charge conjugation is extremely slippery because there are two different versions of it; there have been many questions on this site mixing them up (1, 2, 3, 4, 5, 6, 7, 8, 9), several asked by myself ...
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23 votes
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Integrating the generator of the infinitesimal special conformal transformation

In order to determine the finite SCT from its infinitesimal version, we need to solve for the integral curves of the special conformal killing field $X$ defined by \begin{align} X(x) = 2(b\cdot x) x ...
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  • 54.8k
21 votes
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Why Lie algebras if what we care about in physics are groups?

The objects which matter in physics are Lie groups and not Lie algebras. Lie algebras approximate only infinitesimal group transformations and in quantum mechanics the finite and global properties of ...
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21 votes

Can we do better than "a spinor is something that transforms like a spinor"?

The proper analogous formalization of spinors is not to view them as some sort of different functions from tensors on the same underlying vector space $V$, but instead to expand our idea of the ...
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  • 108k
20 votes
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Difference between Cartesian product $\times$ and tensor product $\otimes$ on groups

I) The main point is that we usually only consider tensor products $V \otimes W$ of vector spaces $V$, $W$ (as opposed to general sets $V$, $W$). But groups (say $G$, $H$) are often not vector spaces. ...
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  • 172k
20 votes
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Invariance under boosts but not rotations?

Every quantum field theory has a symmetry group under which its Lagrangian is invariant. Like every group, it must be closed. The boosts are not closed under composition, so they cannot form a ...
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20 votes
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Why are the generators of rotation in the 4-dimensional Euclidean space correspond to rotations in a plane?

Does it mean that a given rotation in 4-dimensional Euclidean space cannot be associated with a unique axis ($\hat{\textbf{n}}$) of rotation? If yes, why is that the case? Yes, this is absolutely ...
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19 votes
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Is $SU(2)\times U(1) = U(2)$?

The relevant Lie group isomorphism reads $$\begin{align} U(2)~\cong~&[U(1)\times SU(2)]/\mathbb{Z}_2, \cr Z(SU(2))~\cong~&\mathbb{Z}_2.\end{align}\tag{1a} $$ In detail, the Lie group ...
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  • 172k
19 votes

What defines a large gauge transformation, really?

Bundles and compactified spacetime A gauge theory cannot be looked at purely locally, it has inherently global features one cannot see locally. The proper mathematical formalization of a Yang-Mills ...
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  • 108k
19 votes
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Why do all fields in a QFT transform like *irreducible* representations of some group?

This is only semantics. A reducible representation $\mathbf R$ of the symmetry group can be decomposed into a direct sum $\mathbf R_1 \oplus \cdots \oplus \mathbf R_N$ of irreducible representations. ...
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  • 6,326
17 votes
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Diffeomorphism group vs. $\operatorname{GL}(4,\mathbb{R})$ in General Relativity

Let there be given a 4-dimensional real manifold$^1$ $M$. As OP says, the set ${\rm Diff}(M)$ is the group of globally defined $C^{\infty}$-diffeomorphisms $f:M\to M$. The set ${\rm Diff}(M)$ is an ...
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  • 172k
17 votes

Why is there this relationship between quaternions and Pauli matrices?

1. Pauli matrices-Rotations-Special Unitary matrices $\:\mathrm{SU}(2)\:$ Any vector in $\mathbb{R}^3$ can be represented by a $2\times2$ hermitian traceless matrix and vice versa. So, there exists ...
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  • 13.5k
17 votes

Relation between the Dirac Algebra and the Lorentz group

It's pretty annoying that P&S just give you $$S^{\mu \nu} = \frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]$$ from thin air, here is a way to derive it similar to Bjorken-Drell's derivation (who start ...
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  • 3,651
17 votes
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Conformal field theory does not have... conformal symmetry?

The Virasoro algebra is a true symmetry of the theory, in the sense that the action of a conformal field theory is conformally invariant if it exists, and in the sense that the algebra elements map ...
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  • 108k
16 votes

$su(1,1) \cong su(2)$?

The ladder operators do belong to the real Lie algebras$^1$ $$\begin{align} su(1,1)~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid m^{\dagger}\sigma_3=-\sigma_3m,~ {\rm tr}(m)=0\}\cr ~=~&{\...
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  • 172k
16 votes

The anticommutator of $SU(N)$ generators

Indeed, the anticommutator $$ S^{AB}\equiv \{T^A,T^B\} $$ is not in the Lie algebra, but, rather, in the universal enveloping algebra (formed by sums of products of generators); and, as you ...
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16 votes

Are all representations of a finite group unitary?

Every representation $(D,V)$ of a finite group $G$ is equivalent to a unitary representation. It is often termed as Weyl's unitary trick. This works by simply redefining your inner product by ...
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  • 3,192
15 votes
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$su(1,1) \cong su(2)$?

You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for ...
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14 votes

Comprehensive book on group theory for physicists?

There is a new book called Physics From Symmetry which is written specifically for physicists and includes a long, very illustrative introduction to group theory. I especially liked that here concepts ...
14 votes

Comprehensive book on group theory for physicists?

A rather recent book is An Introduction to Tensors and Group Theory for Physicists. It also speaks of vectors and tensors at a good level. In my opinion it clears up the confusion physicists tend to ...
14 votes
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Why $SU(3)$ and not $U(3)$?

Suppose that $\text{U}(3)$ was the gauge group. We can decompose this as $$\text{U}(3)=\text{U}(1)\times\text{SU}(3),$$ which implies that in addition to the $\text{SU}(3)$ that has eight generators ...
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