# Tag Info

1

I think your confusion (like mine) is simply over technical English usage. As you rightly state "vectors look like passive elements on which the group matrices act, and do not contain the structure of the group". To my mind, a representation of a group is a triple $(\mathfrak{G},\,V,\,\rho:\mathfrak{G}\to GL(V))$: the group $\mathfrak{G}$ being ...

1

Lie algebras are not a group w.r.t. to the commutator (the Lie bracket). The first reason is that the commutator is not associative. Another is that they almost always lack an identity element, since the identity matrix is, for example, not in $\mathfrak{su}(2)$, and Schur's lemma would, in the fundamental representation, guarantee that only multiples of ...

3

Short answer: complexifications facilitate representation theory. In physics, we typically want to find representations of a Lie algebra $\mathfrak g$, and often times determining the representations of its complexification $\mathfrak g_\mathbb C$ is easier. Moreover, we have the following theorem (see ref 1. Proposition 4.6) which tells us that ...

2

From a mathematical perspective, to develop Lie algebra representation theory most efficiently, we need the field $\mathbb{F}$ of the Lie algebra to be algebraically closed. See e.g. Ref. 1, where this assumption is used already in the beginning of Chapter II. The situation for Lie algebras is similar to when we in linear algebra try to diagonalize, say, a ...

3

1) One way to show that a map constitutes a 1-1 correspondence is by showing that it has an inverse. This is what is done here: elements of $T_eG$ are tangent vectors to $G$ at the unit element, and one-parameter subgroups of $G$ are smooth homomorphisms $\mathbb R\to G$. Note that this is a notion more general than a closed 1-dimensional subgroup, for which ...

1

To add to Qmechanic's Answer and TwoBs's Answer and answer "....what the heck is h then? Any arbitrary function?": $h$ is pretty much arbitrary. It is wontedly taken to be at least differentiability class $C^1$ (all first derivatives continuous) so that the Lie bracket of vector fields is defined as in Qmechanic's answer. You need to assume it is of class ...

3

Think of an infinitesimal Diff as of a translation where the the shift is space dependent, $x^\mu\rightarrow x^\mu+\epsilon^\mu(x)$. Now, you get that the generators are $L_\epsilon=\epsilon^\nu(x)\partial_\nu$ since $L_\epsilon x^\mu=\epsilon^\mu(x)$. They form an infinite space since $\epsilon^\mu(x)$ is a function that can be expanded in infinitely many ...

1

Formally speaking, given a (differentiable, finite dimensional) manifold $M$, then the (infinite dimensional) Lie group of (globally defined) diffeomorphisms (with composition $\circ$ as group structure) has the set $\Gamma(TM)$ of (globally defined, differentiable) vector fields as corresponding Lie algebra. This (infinite dimensional) Lie algebra ...

1

In this answer we will consider a Lie algebra $L$ (rather than a Lie group). Then: If $M$ is a manifold, let there be a Lie algebra homomorphism $\rho:L\to \Gamma(TM)$ into the Lie algebra of vector fields on $M$. The map $\rho$ is called an anchor. If the manifold $(M,\{\cdot,\cdot\}_{PB})$ is a Poisson manifold, it is natural to require that the vector ...

4

The basic idea is the following. For the shake of simplicity, I henceforth assume that every function does not depend explicitly on time (with a little effort, everything could be generalized dealing with a suitable fiber bundle over the axis of time whose fibers are spaces of phases at time $t$). On a symplectic 2n dimensional manifold (a space of ...

0

Great question! One way to look at what is going on is to use the Hamiltonian version of Noether's theorem. The Noether procedure generates a conserved charge $Q$ associated with the symmetry with parameter $\theta$. It turns out that $Q$ is the generator of that symmetry, in the sense that for some function $A$ of phase space variables ...

3

Your idea 1) is the right idea: It's just the law of transformation of matrices generalized from the transformation of matrices: If we apply a general linear transformation $U : V \to V$ on a vector space, the matrices/operators on it transform as $M \mapsto U^\dagger MU$ For unitary operators $U^\dagger = U^{-1}$, so the transformation law becomes $M ... 3 Ghostly Lie algebra cohomology Let$\mathfrak{g}$be our Lie algebra and$V_\rho$a representation space with representation map$\rho : \mathfrak{g} \to \mathrm{End}(V_\rho)$.$V_\rho$is, by the action through the representation, naturally a$\mathfrak{g}$-module (people missing the ring structure in$\mathfrak{g}\$ - just embed it into the universal ...

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