# The relation between Lie algebra and conservative quantities

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1. Sep 21, 2015

### EnigmaticField

In quantum mechanics, a physical quantity is expressed as an operator G, then the unitary transformation coresponding to the physical quantity is expressed as exp(-iG/ħt), being also an operator, where t is the tranformation parameter. G is actually the conservative quantity corresponding to the transformation generated by G.

In classical mechanics, an one-parameter Lie group action is expressed as exp(tX), X being an element of the Lie algebra of the Lie group in question. According to the Noether theorem, the invariance of a physical system under a Lie group action gives a conservative quantity. However, I found it's like X is not the correspponding conservative quantity, like the situation in quantum mechanics. For example, you can see an element of the Lie algebra of SO(3) is not (a quantity proportional to) the angular momentum. So what is the relation between the Lie algebra elements and the corresponding conservative quantity? Is there a generic formula to express the relation?

Last edited: Sep 21, 2015
2. Sep 21, 2015

### samalkhaiat

In both CM and QM, Noether theorem provides a representation of the symmetry algebra in terms of dynamical quantities. In CM, the Lie algebra is realised by Poisson algebra, while in QM it is commutator algebra. So, if the algebra of the symmetry group is given by
$$[X_{i} , X_{j}] = C_{ij}{}^{k}X_{k} ,$$
then Noether theorem allows you to construct the following constants of motion $$G_{i} = P_{a} \ (X_{i})^{a}{}_{b} \ Q^{b} .$$ Now, you can use the fundamental Poisson brackets (or commutation relations) $$\{Q^{a},P_{b}\} = \delta^{a}_{b},$$
$$\{Q^{a},Q^{b}\} = \{P_{a},P_{b}\}= 0 ,$$ to show that $G_{i}$ generate the correct infinitesimal transformations on the dynamical variables $$\{G_{i},Q^{c}\} \equiv \delta_{i}Q^{c} = - (X_{i})^{c}{}_{b} \ Q^{b} ,$$ $$\{G_{i},P_{c}\} \equiv \delta_{i}P_{c} = P_{a} \ (X_{i})^{a}{}_{c} ,$$ and satisfy the Lie algebra of the symmetry group $$\{G_{i},G_{j}\} = P_{a} \ (X_{i})^{a}{}_{c} (X_{j})^{c}{}_{b} \ Q^{b} - P_{a} \ (X_{j})^{a}{}_{c} (X_{i})^{c}{}_{b} \ Q^{b} ,$$ which gives $$\{G_{i},G_{j}\} = P_{a} \ [X_{i},X_{j}]^{a}{}_{b} \ Q^{b} = C_{ij}{}^{k} \ G_{k} .$$ You can also show that $$\frac{dG_{i}}{dt} = \delta_{i} H = \{G_{i} , H \} .$$ So, if the transformations are canonical and leave the Hamiltonian invariant, then the $G_{i}$’s form a set of conserved quantities.

3. Sep 23, 2015

### EnigmaticField

Wow! It turns out there is really such a mysterious formula relating the Lie algebra and the dynamical quantities. Are P_a and Q^b the general coordinates and their conjugate momenta? So to get the dynamical quantities from their corresponding Lie algebra, we need to find a set of the coordinate-conjugate momentum pairs first? Does this formula apply to field mechanics, where the general coordinates are replaced by the field variables?

Can you tell me where this kind of technical material can be found? Books or papers or websites of which kind of subjects?

Last edited: Sep 23, 2015
4. Sep 23, 2015

### samalkhaiat

Yes.
Yes. That is to say, you need a Lagrangian.
Yes. In field theory, the field functions are coordinates in some function space, the field space. They are functions on the space-time manifold. So, instead of the $Q^{a}(t)$ of point particles, the field coordinates are $Q^{a}(t , \vec{x})$ and similarly, the conjugate momentum is given by $$P_{a}(t,\vec{x}) = \frac{\partial \mathcal{L}}{\partial (\partial_{t}Q^{a})} ,$$ where $\mathcal{L}(Q^{a}, \partial_{t}Q^{a},\partial_{x}Q^{a})$ is the Lagrangian density of the fields. The constants of motion (also called Noether charges) are obtained by integrating over the spatial volume of the fields $$G_{i} = \int d^{3}x \ P_{a}(t,\vec{x}) \ (X_{i})^{a}{}_{b} \ Q^{b}(t,\vec{x}) .$$
That is the subject of Noether Theorem in Classical Field Theory, just Google it. Also, almost all text books on quantum field theory start with explaining classical fields and Noether theorem.

Last edited: Sep 24, 2015
5. Sep 24, 2015

### sams_rhythm

6. Sep 27, 2015

### EnigmaticField

I find a simple test of your formula to convince me that your formula makes sense: I take $X_i$ as $E_{12}$, the basis vector along the z axis of the Lie algebra of SO(3) and get $G_{12}=Q^1P_2-Q^2P_1$, which, as expected, is just the usual angular momentum along the z axis.
On the other hand, now I find how your mysterious formula comes from. I actually have read Noether theorem in several different places but none of them expresses the conservative quantities in terms of $Q^a$ and $P_b$. After the identification of $Q^a$ and $P_b$, I understand this formula is not as mysterious as I considered.
An expression of the conservative quantities which bears the closest similarity with yours is the one I read from a book of geometrical physics, where the conservative current is $J^k=\frac{\partial\mathcal{L}}{\partial\nabla\phi^a}({E^k})^a{}_b\phi^b$, which satisfies $divJ^k=[\frac{\partial\mathcal{L}}{\partial\phi^a{}_{/j}}({E^k})^a{}_b\phi^b]_{/j}=0$, where / denotes the covariant derivative. In this case the conjugate momentum is more general than yours, being $P_a=\frac{\partial\mathcal{L}}{\partial\nabla\phi^a}$. Then $\int_{M}divJ^k vol^4=\int_{\partial M}J^k\bullet Ni_Nvol^4$. If we choose $N$ to be timelike we get a 3-dim spatial integral.

One example puzzling me is the Pauli–Lubanski pseudovector, which has one expression $S_a=1/(2M)\epsilon_{abcd}J^{bc}P^d$ in one place (Adam D. Helfer's paper), where $J^{bc}$ is the relativistic angular momentum and $P^d$ is the relativistic 4-momentum, and another expression $L_a=1/2\epsilon_{abcd}\theta^{bc}\otimes\theta^{d}$ in another place (G .A. J. Sparling's paper), where $\theta^{bc}$ is the Levi-Civita connection 1-form, which inhabits in the Lie algebra of the Lorentz group, and $\theta^{d}$ is the solder form, which inhibits in the Lie algebra of the translation group. In the latter case, $L_a$ is also called Fefferman tensor. I think these two expressions seem to talk about the same thing in two different versions. So it's like in this case, the two dynamical quantities can be represented as the two geometrical quantities in general relativity: the angular momentum is represented by the Levi-Civita connection 1-form, and the 4-momentum is represented by the solder form. What is the explanation of this representation?

Last edited: Sep 27, 2015
7. Sep 30, 2015

### haushofer

As a nice exercise you consider the action of a non-rel. point particle, and calculate the noether charges of translations and boosts. You have to be careful of the fact that the Lagrangian goes to a total derivative under boosts. Plugging these charges into the Poisson brackets reveals a central extension which is just the mass. So the underlying Lie algebra allows for this central extension,

[P,B]~M

In quantum mechanics this is an important fact; the algebra is known as the bargmann algebra.

8. Sep 30, 2015

### samalkhaiat

I don't know why should the index on L be the same as the space-time index on the Pauli-Lubanski psedovector! Any way, if you want to study Noether theorem in field theory in some detail, have a look at the PDF below

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9. Oct 3, 2015

### EnigmaticField

What does this mean?
In my impression I have seen the term "central extension" several times in some papers but I keep not quite knowing what it means. So can you elaborate a little bit what is "central extension"?

If the index on L isn't the spacetime index, what is it? Fiber index? But Sparling considered the principle bundle of the tangent bundle of the spacetime manifold, so the fiber index is just the spacetime index.
I just wonder if there is a certain relation between the connection 1-form and angular momentum as well as that between the solder 1-form and the 4-momentum because Sparling in his 1982 paper called $L_a$ the Pauli–Lubanski spin tensor though in his 2001 paper he called it the Fefferman tensor. And if there is indeed a certain relationship, the angular momentum and 4-momentum of what? Because Sparling's paper was talking about the gravitational field formulated by the spacetime geometry, there are no particles (Recall Pauli-Lubanski pseudotensor is the spin angular momentum of some particles) there, I wonder what angular momentum and 4-momentum are referred to? The angular momentum and 4-momentum of something like gravitons? In Sparling's paper there are vectors $D_{gh}$ and $D_{f}$ conjugate to $\theta^{bc}$ and $\theta^d$, respectively, satisfying
$<\theta^d,D_f>=\delta^d{}_f, \\ <\theta^d, D_{gh}>=0, \\ <\theta^{bc}, D_{f}>=0, \\ <\theta^{bc}, D_{gh}>=\frac{1}{2}(\delta^b{}_g\delta^c{}_h-\delta^b{}_h\delta^c{}_g),$
so I think if the Levi-Civita connection and solder form really represent certain mysterious angular momentum and 4 momentum, respectively, the Pauli-Lubanski pseudotensor can be expressed as $S^a=\frac{1}{2}\epsilon^{abcd}D_{bc}D_d$. Or actually $L_a$ is called the Pauli–Lubanski spin tensor just because its expression bears a similarity with that of $S_a$, not because $\theta^{bc}$ has some relation with some mysterious angular momentum and $\theta^d$ has some relation with some mysterious 4-momentum?

Thank you very much. I hope it will broaden my understanding about Noether theorem.

Last edited: Oct 3, 2015
10. Oct 5, 2015

### samalkhaiat

I have not seen the paper, so I don't know what sort of manifold his spacetime is. Knowing that G. Sparling is obsessed with the Twister formalism of GR, I can only guess that the indices are the "dotted" and "undotted" indices of the principal $SL(2,C)$-bundle over some 4-demensional (Lorenzian) curved spacetime, i.e., the triple $(M^{4} , S , \bar{S})$ with $S$ being the bundle of Weyl spinors, and $S \otimes \bar{S}$ is isomorphic to the complixification of the cotangent bundle of $M^{4}$.
Let me ask you this: If I claim that $V_{a} \sim \epsilon_{abcd} P^{b}J^{cd}$ is the Pauli-Lubanski psedovector. What do I need to do in order to prove my claim?

11. Oct 21, 2015

### EnigmaticField

I just read Sparling's paper and found he indeed had used the bundle $S\otimes\bar{S}$ isomorphic to the complexification of the cotangent bundle of $M^4$, but when he talked about the Pauli-Lubanski pseudovector, he used the principal bundle of the orthonormal frames of $M^4$. You can see Eq. (9) and the texts prior to Eq. (9) in my attachment, in which he used the name Fefferman tensor instead of the Pauli-Lubanski pseudovector, which was used in his paper in 1982 to refer to the same expression.
On the other hand, Sparling in his section involved with $S\otimes\bar{S}$ did define an expression, $L=\frac{1}{2}i(\theta^{AA'}\wedge\theta_{A'}\bar{\psi}_A-\theta^{AA'}\wedge\psi_{A'}\bar{\theta}_A)$, where $\theta^{AA'}=\theta(\delta^{A'}\otimes\bar{\delta}^A)$ with $\delta^{A'}$ being the normalized spin frame, $\theta_{A'}=d\psi_{A'}-\Gamma_{A'}{}^{B'}\psi_{B'}, \bar{\theta}_A=d\bar{\psi}_A-\bar{\Gamma}_A{}^B\bar{\psi}_B$, in which the indices $A$ and $A'$ are indeed the dotted and undotted spinor indices you guessed. This $L$ on the spinor bundle satisfies an equality (which is equivalent to the Einstein vacuum field equations) the Pauli-Lubanski pseudovector on the principal orthonormal-frame bundle of $M^4$ satisfies but Sparling didn't refer to it as the Pauli-Lubanski pseudovector. I wonder why you guessed the indices in the $L_a=\frac{1}{2}\epsilon_{abcd}P^bJ^{cd}$ are the spinor indices. Can the Pauli-Lubanski pseudovector be also defined on the spinor bundle?

I think you need to make sure $P^b$ and $J^{cd}$ you give satisfy the following equations:
$V_aP^a=0 \quad(1),\\$
$[V_a, P_b]=0\quad(2),\\$
$[V_a,J_{cd}]=V_cg_{ad}-V_dg_{ca}\quad(3),\\$
$[V_a, V_b]=\epsilon_{abcd}V^cP^d\quad(4)$.
It's like $P^b$ and $J^{cd}$ given in Sparling's paper, namely $D^b$ and $D^{cd}$, respectively, don't satisfy (2) and (4) above, because Sparling considered a curved spacetime, whose curvature doesn't vanish, leading to $[P_a, P_b]=R_{ab}{}^{cd}J_{cd}$ instead of 0 for a flat spacetime.
Well, the above 4 equations are for the Pauli-Lubanski pseudovector in a flat spacetime. Maybe the Pauli-Lubanski pseudovector can be generalized for a curved spacetime, in which the right-hand sides of (2) and (4) need to be generalized to include the terms involved in the curvature. How do you think?

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12. Oct 26, 2015

### samalkhaiat

If the Poincare’ group is the symmetry group of your manifold, then you can use its generators $(P^{a}, M^{ab})$ to construct the Pauli-Lubanski vector whose square is a Casimir operator. It determines the spin of irreducible representations. If your manifold is Reimanian with Lorentz signature, the Pauli-Lubanski vector can only live in the flat tangent space at a point. Depending on the spin groups ($SL(2,C)$ or $SO(1,3)$) the vector can be given either by $$\mathbb{V}_{\alpha \dot{\alpha}} = \mathbb{P}^{\beta}{}_{\dot{\alpha}} \ \mathbb{J}_{\beta \alpha} - \mathbb{P}_{\alpha}{}^{\dot{\beta}} \bar{\mathbb{J}}_{\dot{\beta} \dot{\alpha}} ,$$ or $$V_{a} = \frac{1}{2} \epsilon_{abcd}P^{b}M^{cd} .$$

13. Nov 2, 2015

### EnigmaticField

Why is so? What obstruct the definition of the Pauli-Lubanski vector on a curved spacetime?

Do your generators $P^a$ and $M^{ab}$ represent the dynamical quantities, the angular momentum and the 4-momentum, or the Lie algebra basis of the Poincare group? The dynamical quantities and the Lie algebra basis are different in the case of the SO(1,3) representation while it's like in the spinor representation they just differ by a constant.
What does $\mathbb{P}^\beta{}_{\dot{\alpha}}$ and $\mathbb{J_{\beta\alpha}}$ stand for? The Lie algebra basis of the $S\otimes\bar{S}$ representation of the translation group and the Lie algebra basis of the $S\otimes\bar{S}$ representation of Lorentz group? So the translation group also has the spinor representation?

14. Nov 2, 2015

### samalkhaiat

I’ve already indirectly answered this. The object $V \sim (\epsilon) \ (P) \ (M)$ is called Pauli-Lubanski vector if and only if $(P) \ \mbox{and} \ (M)$ generate a closed algebra isomorphic to the Poincare’ algebra. In curved space-time, Poincare’ group can only be defined in the flat tangent space at a point.
The absence of Poincare’ symmetry.

If you have a Poincare’ symmetric field theory, then the abstract Poincare generators can be realized in terms of the dynamical variables (I said this in my first post). So, whatever $(P,M)$ are, they need to satisfy the Poincare algebra. Then, and only then, you can call $(\epsilon \ P \ M)$ the Pauli-Lubanski vector.

Have you studied the representation theory of the proper spinorial group? These are the Poincare’ generators written using the spinor indices of the group $SL(2,C)$:
$$\mathbb{P}_{\alpha \dot{\alpha}} = (\sigma^{a})_{\alpha \dot{\alpha}} \ P_{a}$$ $$\mathbb{J}_{\alpha \beta} = (1/2) (\sigma^{ab})_{\alpha \beta} \ M_{ab} , \ \ \ \mathbb{J}_{\dot{\alpha} \dot{\beta}} = -(1/2) (\bar{\sigma}^{ab})_{\dot{\alpha} \dot{\beta}} \ M_{ab} .$$
No, translation does not have spinor representation, but the proper spinorial group (a semi-direct product of the $SL(2,C)$ group and the translation group) does have spinor representations.

15. Dec 8, 2015

### EnigmaticField

Does your closed algebra mean $C_{ij}{}^k$ in $[X_i,X_j]=C_{ij}{}^kX_k$ are constants, where $X_i, i=1, ..., 10$ are the Lie algebra basis of the Poincare group? Consider that for a curved spacetime in $[P_a,P_b]=R_{ab}{}^{cd}J_{cd}$, $R_{ab}{}^{cd}$ are not constants.

Why must the Pauli-Lubanski vector be defined on the Poincare algebra? Is that because the original purpose of building the Pauli-Lubanski vector is to construct the unitary irreducible representations of the Poincare group? (In such a construction, we need a (maybe two) Casimir operator(s) of the Poincare group so that its eigenvalues can be used to label the unitary irreducible representations, and the Pauli-Lubanski vector serves such a purpose: the square of the Pauli-Lubanski vector is a Casimir operator of the Poincare group.) If we consider a curved spacetime whose curvature doesn't vanish, the square of $V \sim (\epsilon) \ (P) \ (M)$ is not a Casimir operator of the group (which shouldn't be called the Poincare group because it's Lie algebra doesn't satisfy the Poincare algebra) consisting of the translation group and rotation-boost group of the curved spacetime. Is this the very reason you say the Pauli-Lubanski vector can only be defined on a flat spacetime?
So does $V_a=\frac{1}{2}\epsilon_{abcd}P^bM^{cd}$ on a flat spacetime represent any physical quantity besides serving the function of labeling the unitary irreducible representations of the Poincare group? I ask this because if it represents a certain physical quantity, the definition of it may still serve the function of representing that physical quantity when the spacetime becomes curved though it can't serve the function of labeling the unitary irreducible representations of a certain group.
Maybe that's the case for the definition of $L_i=\frac{1}{2}\epsilon_{ijkm}\theta^j\otimes\theta^{km}$ in Sparling's paper, in whose 1982 version the object was called the Pauli-Lubanski spin tensor (You can see this in
Eq. (8) and the text immediately prior to it of my attachment), which should be a misnomer (because the spacetime considered in Sparling's paper is curved). And that's why Sparling in his 2001 paper, which is a republication of his 1982 paper, changed to call it the Fefferman tensor (See Eq. (8) and the text immediately prior to it in the attachment of the 11th post of this thread). Thus $V \sim (\epsilon) \ (P) \ (M)$ can actually be defined on a curved spacetime and indeed has functions (Sparling's paper said its symmetric part gives the conformal structure of the Cauchy-Riemann structure of the hypersurface in question and its skew symmetric part controls the Einstein equations and explains the origin of the Witten positive energy proof) though it can't serve the function of labeling the unitary irreducible representations of a certain group and thus can't be called the Pauli-Lubanski vector.

And, is $\mathbb{V}_{\alpha\dot{\alpha}}=\mathbb{P}^\beta{}_{\dot{\alpha}}\mathbb{J}_{\beta\alpha}-\mathbb{P}_\alpha{}^{\dot{\beta}}\bar{\mathbb{J}}_{\dot{\beta}\dot{\alpha}}$ built to label the unitary irreducible representations of the proper spinorial group? If so, do $\mathbb{P}, \mathbb{J}, \bar{\mathbb{J}}$ also satisfy some spinor version of the Poincare algebra, $[\mathbb{X}_i,\mathbb{X}_j]=\mathbb{C}_{ij}{}^k\mathbb{X}_k$ (the commutator algebra version) (If so, now there are only 3 such equations!)? On the other hand, does it also represent any physical quantity?

In classical mechanics, though both the dynamical variables and the Lie algebra basis vectors satisfy the Poincare algebra, what the dynamical variables satisfy is the Poisson algebra version of the Pincare algebra while what the Lie algebra basis vectors satisfy is the commutator version of the Poincare algebra. Doesn't this difference matter?

If which version of the Poincare algebra is satisfied really doesn't matter, why don't the numerical values of $(P,M)$ matter in the definition of the Pauli-Lubanski vector? Is that due to that the only purpose of building the Pauli-Lubanski vector is to label the irreducible unitary representations of the Poincare group so what matters is only whether $(P,M)$ satisfies the Poincare algebra? I just wonder if it also represents a physical quantity, how could its numerical values unmatter?
I had never heard about the proper spinorial group until you mentioned it to me. I googled "the proper spinorial group" and read some related material on web to have a little bit of understanding of it. But I still have some questions about your expressions for $\mathbb{P}_{\alpha \dot{\beta}}, \mathbb{J}_{\alpha \beta}, \bar{\mathbb{J}}_{\dot{\alpha} \dot{\beta}}$.

I wonder about the choice of the spinor indices of $\mathbb{P}, \mathbb{J}, \bar{\mathbb{J}}$. Does the choice of the undotted spinor indices or the dotted indices of them as well as putting these indices as covariant or contravariant ones imply the desired spinor transformation rules under the Lorentz transformation? I have long learnt the transformation rule for $\mathbb{P}=\sigma_aP^a$ is $A\mathbb{P}A^\dagger=\Lambda^a{}_b(A)P^b\sigma_a \Rightarrow A\sigma_bA^\dagger=\Lambda^a{}_b(A)\sigma_a$, where $A\in SL(2,C)$, $\Lambda \in SO(1,3)$ and $\Lambda^a{}_b(A)$ denotes $\Lambda^a{}_b$ is a function of $A$, so I think the spinor indices for $\mathbb{P}$ should be put as $\mathbb{P}^\alpha{}_\dot{\beta}$, instead of $\mathbb{P}_{\alpha\dot{\beta}}$, which shall transform as $A^{-1T}\mathbb{P}A^\dagger$. I think the rationalization for the transformation rule $A\mathbb{P}A^\dagger$ is that a 4-vector transforms as the $(\frac{1}{2}, \frac{1}{2})$ spinor representation of the Lorentz group. On the other hand, I also learnt an anti-symmetric 2-rank tensor transforms as the $(0,1)\oplus(1,0)$ spinor representation of the Lorentz group. Does this entail the transformation rules for $\mathbb{J}$ and $\bar{\mathbb{J}}$ to be, respectively, $A\mathbb{J}A^{T}=\Lambda^a{}_c(A)M^{cd}\Lambda^b{}_d(A)\sigma_{ab} \Rightarrow A\sigma_{cd}A^{T}=\frac{1}{2}\Lambda^a{}_c(A)\sigma_{ab}\Lambda^b{}_d(A)$ and $A^{\dagger-1}\bar{\mathbb{J}}\bar{A}^{-1}=-\frac{1}{2}\Lambda^a{}_c(A)M^{cd}\Lambda^b{}_d(A)\bar{\sigma}_{ab} \Rightarrow A^{\dagger-1}\bar{\sigma}_{cd}\bar{A}^{-1}=-\frac{1}{2}\Lambda^a{}_c(A)\bar{\sigma}_{ab}\Lambda^b{}_d(A)$, and thus justify the spinor indices for $\mathbb{J}$ and $\bar{\mathbb{J}}$ to be put as $\mathbb{J}^{\alpha\beta}$ and $\bar{\mathbb{J}}^{\dot{\alpha}\dot{\beta}}$, respectively?

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