In classical mechanics on a flat 3D vector space (which we will refer to as configuration space), we can define position and momentum vectors $x$ and $p$. We can then define angular momentum as $L = r \times p$. Choose an orthonormal basis $\{e_i\}$ such that vectors are represented as $v^ie_i$ with respect to this basis ($i,j,k \in \{1, 2, 3\}$). Then in component form we can denote the components of angular momentum as $L_i = \epsilon_{ij}^{\;\;k}x^jp_k$.

The Hamiltonian formalism formulates classical mechanics on a symplectic manifold (which I believe is equivalently a vector space in this context), the cotangent bundle of the configuration space, and we take the canonical symplectic form. Then in symplectic space, angular momentum is defined as a function.

It's not clear to me how this is defined in symplectic space, especially given the presence of the cross product in it's definition. The only thing I can think of is to use the component form of the angular momentum, and then we would define the function as e.g. $L_x = yp_z - zp_y$, so we would really be defining 3 functions.

I think it would be useful for me to illustrate the context in which this arose to perhaps make my confusions clearer. I was looking at motion of a point particle of mass $m$ in a potential of the form $V(r) = -\frac{k}{r}$. Angular momentum is conserved, but there is also conservation of the Runge-Lenz vector, $A = p\times L - mke_r$, where $e_r$ is the radial unit vector. Note that both these quantities have components in configuration space. Define the Hamiltonian in the usual way, $H = \frac{p^2}{2m} +V(r)$.

Using the canonical symplectic form, we can write the inverse symplectic form in matrix form as \begin{align} \omega = \begin{bmatrix} 0&I_3 \\ -I_3 & 0 \end{bmatrix}, \end{align} where $I_3$ is the $3\times 3$ identity matrix, and we can write the classical equations of motion as ($a, b \in \{1, 2, 3, 4, 5, 6\}$) \begin{align} \dot{y}^a = \omega^{ab}\frac{\partial H}{\partial y^b}, \end{align} where we have identified the position components $x^i$ with $y^i$ and momentum components $p^i$ with $y^{i+3}$, the components of our symplectic space (we call this the symplectic equation of motion). Because of this identification, it would seem that we can then identify a basis of symplectic space to be $\{e_1, e_2, e_3, b_1, b_2, b_3\}$, where the $b_i$ are the 'basis of momentum' (I have struggled to find resources that formalise this, so this may be one point where I am going wrong).

We can then define a 'normalised' RL vector, $B = \frac{A}{\sqrt{2m|H|}}$. For $H<0$ (the other cases just give rise to a different algebra but the concept is the same), one can show that these normalised components of the RL vector and the components of angular momentum form the following algebra under the Poisson bracket: \begin{align} &\{L_i, L_j\} = \epsilon_{ij}^{\;\;k}L_k\\ &\{L_i, B_j\} = \epsilon_{ij}^{\;\;k}B_k\\ &\{B_i, B_j\} = \epsilon_{ij}^{\;\;k}L_k, \end{align} which is exactly the Lie algebra $\mathfrak{so(4)}$.

What is confusing me then is that it seems to me the functions (the conserved quantities) form a basis of symplectic space (since the Lie algebra is a vector space). I don't know how to interpret this, as it then seems that we should be able to express these quantities in terms of the basis used in the symplectic equation of motion. However, this comes back to my initial point - I'm unsure how angular momentum is defined as a vector quantity in symplectic space.

It may be clear from my outline of the question that I am quite new to differential geometry, so please correct me where I have misidentified anything.


3 Answers 3


How is angular momentum defined on symplectic space?

The broadest definition of angular momentum is that it is the infinitesimal generator of spatial rotations. Concretely, let $\mathbf x$ be a point in phase space - in your case, $\mathbf x = (x,y,z,p_x,p_y,p_z)$. Let $\mathbf x(\lambda)$ be the image of that point after rotating by an angle $\theta$. Then one can write

$$\frac{d}{d\lambda} \mathbf x(\lambda)\bigg|_{\lambda=0} = \left\{\mathbf x, L\right\}$$ for some function $L$.

Example: Rotations on $\mathbb R^2$

Let $$\mathbf x(\lambda) = \pmatrix{x(\lambda)\\y(\lambda)\\p_x(\lambda)\\p_y(\lambda)} = \pmatrix{x\cos(\lambda) - y \sin(\lambda) \\ x\sin(\lambda) + y\cos(\lambda) \\ p_x\cos(\lambda) - p_y \sin(\lambda) \\ p_x\sin(\lambda) + p_y\cos(\lambda)}$$ $$\implies \mathbf x'(0) = \pmatrix{-y\\x\\-p_y\\p_x}= \pmatrix{\partial L/\partial p_x\\\partial L/\partial p_y \\ -\partial L/\partial x \\ -\partial L/\partial y}$$ $$\implies L = x p_y - yp_x $$ where we set the arbitrary integration constant to zero.

From this perspective, we define angular momentum by first specifying a group action of the rotation group $\mathrm{SO}(n)$ on the phase space. In $n$ spatial dimensions, the rotation group is $n(n-1)/2$ dimensional which means that a generic rotation is specified by $n(n-1)/2$ independent parameters. Each parameter defines a family of rotations, and which in turn gives rise to $n(n-1)/2$ functions $L_i$; if the Hamiltonian is invariant under the action of a family of rotations, then the corresponding $L_i$ is a constant of the motion.

Example: Rotations on $\mathrm S^1$

Let $$\mathbf x(\lambda) = \pmatrix{\theta(\lambda) \\ \ell(\lambda)}= \pmatrix{\theta_0 + \lambda\\\ell_0}$$ define a rotation for a particle confined to a circle, where $\theta$ and $\ell$ are the angular coordinate and its conjugate momentum, respectively. As per the previous example, $$\mathbf x'(0) = \pmatrix{1\\0} = \pmatrix{\partial L/\partial \ell \\ -\partial L/\partial \theta}$$ $$\implies L = \ell$$ which yields the fairly obvious result that $\ell$ is the appropriate angular momentum for the system.

Because of this identification, it would seem that we can then identify a basis of symplectic space to be $\{e_1,e_2,e_3,b_1,b_2,b_3\}$, where the $b_i$ are the 'basis of momentum'.

Contrary to your assertion, symplectic manifolds are not generally equipped with a vector space structure. The tangent space to a point in a symplectic manifold is vector space, which in your example is spanned by e.g. $\left\{\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z},\frac{\partial}{\partial p_x},\frac{\partial}{\partial p_y},\frac{\partial}{\partial p_z}\right\}$, which is what I think you mean when you describe a basis for the symplectic space.

What is confusing me then is that it seems to me the functions (the conserved quantities) form a basis of symplectic space (since the Lie algebra is a vector space).

Every smooth function $F$ corresponds to a Hamiltonian vector field $\mathbf X_F = \{\cdot, F\}$. In my first example of rotations on $\mathbb R^2$, for example, we would have $$\mathbf X_L = - y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y} -p_y \frac{\partial}{\partial p_x} + p_x \frac{\partial}{\partial p_y}$$

For rotations in $\mathbb R^3$ there are $3(3-1)/2=3$ components of angular momentum corresponding to $3$ such Hamiltonian vector fields. If you wish to use these as part of a basis for the tangent space, you are welcome to do so; however, you cannot form a basis using $L_i$ and $B_i$ because the six corresponding vector fields are not linearly independent, as noted by Qmechanic.

To see this explicitly, note that $L_iB_i \mapsto \mathbf X_{L_iB_i}=L_i\mathbf X_{B_i} + B_i\mathbf X_{L_i}$. As a result, $$0 = \sum_i L_i B_i \mapsto \sum_i L_i\mathbf X_{B_i} + B_i\mathbf X_{L_i} = \vec 0$$ which means that the set $\{\mathbf X_{L_i}, \mathbf X_{B_i}\}$ are not linearly independent.

Is it then the case that the Lie algebra formed by the conserved quantities under the Poisson bracket is an entirely different space from the tangent space at a point? So in the Lie algebra, the functions form a basis, but the corresponding Hamiltonian vector fields do not form a basis of the tangent space?

Yes. The space of all conserved quantities - that is, smooth functions $f:M\rightarrow \mathbb R$ which Poisson-commute with the Hamiltonian $H$ - constitute a vector space, which becomes a Lie algebra when equipped with the product operation provided by the Poisson bracket. To each such quantity we can associate a Hamiltonian vector field $\mathbf X_f$ which commutes with $\mathbf X_H$, and the set of all such vector fields (equipped with the commutator bracket) is a Lie algebra which is isomorphic to the previous one by construction. Once again, note that the elements of this Lie algebra are vector fields. In contrast, the tangent space $T_pM$ is the space of all tangent vectors (not vector fields, but individual vectors) attached to the specific point $p\in M$.

The space of all vector fields $\Gamma(T_pM)$, equipped with pointwise addition and scalar multiplication, can also be thought of as a vector space. Unfortunately, however, it would constitute an (uncountably) infinite dimensional vector space. To see why, consider $\hat x$ and $\hat y$ as vector fields on $\mathbb R^2$. A linear combination of these two vector fields can reproduce any vector at a point, but not a general vector field. To do the latter, we must allow multiplication by smooth functions rather than merely by real numbers, but this turns $\Gamma(T_pM)$ into a module rather than a vector space.

With this bit of background, your question becomes "does the set of Hamiltonian vector fields $\mathbf X_f$ which commute with $\mathbf X_H$ form a basis for the module $\Gamma(T_pM)$?" The answer to this is generally no; firstly because $\Gamma(T_pM)$ generically has no basis in the first place (modules that do are called free), and secondly because for a $2n$-dimensional symplectic manifold, there are at most $2n-1$ linearly-independent Hamiltonian vector fields which commute with $\mathbf X_H$, including $\mathbf X_H$ itself (though in most cases there are far fewer).

  • $\begingroup$ This has clarified a lot. Is it then the case that the Lie algebra formed by the conserved quantities under the Poisson bracket is an entirely different space from the tangent space at a point? So in the Lie algebra, the functions form a basis, but the corresponding Hamiltonian vector fields do not form a basis of the tangent space? $\endgroup$
    – Bedge
    Jul 12, 2021 at 17:18
  • $\begingroup$ @Bedge I've updated my answer to address your follow-up question. $\endgroup$
    – J. Murray
    Jul 12, 2021 at 19:07
  • $\begingroup$ @Bedge I've written a second answer which, after some reflection and reading your other comments, I think more directly addresses your fundamental question. If this is the case, you can un-accept this answer and I can subsequently delete it. If my new answer misses the mark, feel free to let me know. $\endgroup$
    – J. Murray
    Jul 13, 2021 at 20:42

Elementary Mechanics

Consider the elementary mechanics of a point particle moving through the space of possible positions $Q:=\mathbb R^n$. This space has a natural affine structure (see also my answer here) with the vector space given by $V:= (\mathbb R^n,+,\cdot_\mathbb R)$. It's important to distinguish between $Q$ (the set of points) and $V$ (the vector space of displacements), even when the underlying sets for $Q$ and $V$ are the same; the failure to clearly do so is often a cause of confusion.

If $q(t)=\big(q^1(t),\ldots,q^n(t)\big)\in Q$ is the position of a particle at time $t$, its velocity $$\mathbf v(t) := \lim_{\epsilon\rightarrow 0}\frac{q(t+\epsilon)-q(t)}{\epsilon}$$ is an element of $V$, as are the momentum $\mathbf p(t):= m\mathbf v(t)$ and displacement $\mathbf r(t) := q(t)-\mathcal O$ where $\mathcal O$ is the (arbitrarily chosen) origin. From the latter two objects, we can construct the bivector $\mathbf L(t) := \mathbf r(t) \wedge \mathbf p(t)$ which is not an element of $V$ but rather a higher-degree element of the exterior algebra over $V$. In the special case of $n=3$ dimensions, we can use the Hodge duality between vectors and bivectors to construct the pseudovector $\tilde{\mathbf L}(t) = \mathbf r(t) \times \mathbf p(t) \equiv \star \mathbf L(t)$; this is what we typically call the angular momentum of a point particle about $\mathcal O$.

Because of the relationship between $Q$ and $V$, a change of coordinates $q\mapsto q'$ induces a corresponding linear transformation on the vectors. If we implement a rotation $q^i \mapsto q'^i = R^i_{\ \ j} q^j$ with $\mathbf R$ an element of $\mathrm{SO}(n)$, then the components of the vectors change in precisely the same way, i.e. $v^i \mapsto v'^i = R^i_{\ \ j} v^j$. This is often taken to be the definition of a vector in an elementary context.

Lagrangian Mechanics

If the configuration space of possible points is no longer $\mathbb R^n$, then this natural affine structure goes out the window. This is what happens when we incorporate constraints into our choice of $Q$ - for the simple pendulum, it makes sense to take $Q=S^1$ to define the angular position of the pendulum bob, for example. In such cases, we can't define vectors as displacements between points in $Q$.

Instead, we turn to differential geometry and the notion of tangent vectors to $Q$, a notion I assume you are familiar with and will not discuss here. The salient point is that unlike before, there is no notion of a displacement vector $\mathbf r$, which means that the previous definition of angular momentum needs to be updated.

In Lagrangian mechanics, we do this as follows. The action functional for the system is given by $$S[q] := \int_{t_1}^{t_2} \mathcal L\big(q(t),\dot q(t)\big) \mathrm dt$$ where $\mathcal L: Q\times TQ \rightarrow \mathbb R$ is the system's Lagrangian function and $q:\mathbb R\rightarrow Q$ is a trajectory through the configuration space. If $q$ is a solution to the Lagrange equations of motion, then (using cartesian coordinates, for example) an infinitesimal rotation $\mathbf R \simeq \mathbb I + \epsilon \mathbf A$ - with $\mathbf A \in \mathfrak{so}(n)$ an antisymmetric matrix - induces a change in the action given by $$\delta S= \epsilon L\big(q(t),\dot q(t)\big)\bigg|^{t_f}_{t_i} \qquad L\big(q(t),\dot q(t) \big) := \frac{\partial \mathcal L}{\partial \dot q^i} A^i_{\ \ j} q^j\equiv p_i A^i_{\ \ j} q^j$$

If $\mathbf R$ is a symmetry of the system, then $\delta S=0$ and so $L$ is conserved on-shell. In this way, we define the angular momentum $L$ as the Noether charge which is conserved in the presence of rotational symmetry. Note that because the Lie algebra $\mathfrak{so}(n)$ is $\frac{n(n-1)}{2}$-dimensional, there are $\frac{n(n-1)}{2}$ linearly-independent $\mathbf A$'s which generate rotations, and so there are $\frac{n(n-1)}{2}$ Noether charges $L_i$. Once again for $3$-dimensional systems we can arrange them into a (pseudo)vector $\vec L\equiv(L_1,L_2,L_3)$, but we must note that $\vec L$ is a vector only insofar as its components transform "like a vector does" under coordinate transformations. In particular, it is emphatically not a tangent vector to $Q$.

Hamiltonian Mechanics

In Hamiltonian mechanics, the picture changes once again. Here the space of states is the so-called phase space $M$, which is most generally taken to be a $2n$-dimensional symplectic manifold. For any non-singular Lagrangian system (i.e. one such that the kinetic matrix $\frac{\partial^2 \mathcal L}{\partial \dot q^i \dot q^j}$ is invertible), the corresponding phase space is given by the cotangent bundle $M:=T^*Q$ equipped with the symplectic form $\omega:=\sum_{i=1}^n \mathrm dp_i\wedge \mathrm dq^i$.

This $\omega$ is a $2$-form (an antisymmetric $(0,2)$-tensor) which eats two vectors and spits out a real number. However, it is critical to note that these vectors are not tangent vectors to $Q$, but rather tangent vectors to $M=T^*Q$. That is, when I talk about tangent vectors in this section, I am referring to tangent vectors to the cotangent bundle of $Q$. It is very easy to tie oneself in knots here, as is probably becoming obvious.

Let $x:\mathbb R\rightarrow M$ be the phase trajectory of the system. The Hamilton equations can be expressed as $x'(t) = \mathbf X_H\big|_{x(t)}$ where $\mathbf X_H:= \{\cdot, H\}$ is the Hamiltonian vector field associated to the Hamiltonian function $H$ via the Poisson bracket. This is a vector equation, where again vector means tangent vector to $M=T^*Q$. The canonical basis for this space (at a point), which is $2n$-dimensional, is $\{\frac{\partial}{\partial x^i}\}$ where $x \equiv (q^1,\ldots, q^n,p_1,\ldots, p_n)$.

If we implement a family of rotations $R_\lambda$ (i.e. "rotation by an angle $\lambda$") as a diffeomorphism (in fact, a symplectomorphism) on $M$, we can ask whether there exists a smooth function $L:M\rightarrow \mathbb R$ such that $$\frac{d}{d\lambda} R_\lambda\bigg|_{\lambda=0} = \mathbf X_L \equiv \{\cdot, L\}$$ The function $L$ would then be called the infinitesimal generator of that family of rotations. The answer is yes, and indeed these functions are the angular momenta in Hamiltonian mechanics.

Once again, $(L_1,L_2,L_3)$ is not a tangent vector to $Q$, but rather a collection of functions (in this case, functions $M\rightarrow \mathbb R$) which transforms under coordinate rotations like the components of tangent vectors to $Q$ would (though they get an extra sign flip under inversions, hence the term pseudovector). It is certainly not a tangent vector to $M=T^*Q$.

  • $\begingroup$ Fantastic, this is a great explanation of a lot of formalism I haven't seen before that makes it clearer to me where my confusions are. I think, as you have picked up on, my misunderstandings are a step beyond what I thought they were when I wrote my question. I think I will leave your other answer as accepted, as I think it addresses my original post quite directly. Please leave this answer here too, as it addresses a lot of my further questions that arose from the answers provided, even though they have strayed to a broader topic. $\endgroup$
    – Bedge
    Jul 14, 2021 at 9:57

In the Kepler-problem, note that only 5 of the 6 integrals of motion $(\vec{L},\vec{B})$ are algebraically independent, since $\vec{L}\cdot \vec{B}\equiv 0$. They parametrize a 5-dimensional submanifold $M$ of the 6-dimensional canonical phase space $\mathbb{R}^6$ with symplectic 2-form $\omega=\sum_{i=1}^3\mathrm{d}p_i \wedge \mathrm{d}q^i$.

$C^{\infty}(M)$ is the induced Poisson algebra of conserved quantities. The canonical Poisson algebra $$ C^{\infty}(\mathbb{R}^6) ~\supseteq~V~:=~{\rm span}_{\mathbb{R}} \{\vec{L},\vec{B}\}$$ contains a 6-dimensional Lie algebra representation $V$ [of the 6-dimensional Lie algebra $so(4)$], which is spanned by 6 linearly independent functions $(\vec{L},\vec{B}):\mathbb{R}^6\to \mathbb{R}$.

In particular,

  • Not every function $f(\vec{r},\vec{p})$ of phase space can be re-written as a function $g(\vec{L},\vec{B})$ of integrals of motion.

  • The canonical Poisson structure on the phase space $\mathbb{R}^6$ is not entirely determined by the induced Poisson structure on $M$.

  • $\begingroup$ Following from my discussion with J. Murray, are you able to be more specific about what the full Poisson structure is? The conserved quantities form a closed Lie algebra under the Poisson structure, so is this some sort of invariant subspace of the full Poisson structure? $\endgroup$
    – Bedge
    Jul 13, 2021 at 10:43
  • $\begingroup$ If we were to then restrict to the induced Poisson algebra of smooth functions that commute with the Hamiltonian (i.e. conserved quantities), would that then be a faithful representation of $so(4)$? I ask this because I was under the impression that the Lie algebra of the conserved quantities was isomorphic to $so(4)$, but perhaps that is not the case. $\endgroup$
    – Bedge
    Jul 13, 2021 at 13:24
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jul 13, 2021 at 13:55
  • $\begingroup$ Thank you for clarifying. When you say "6 linearly independent functions $(L, B)$ of the 6-dimensional Lie algebra $so(4)$", does this mean that the Poisson algebra of $(L, B)$ being the same as that of $so(4)$ is enough to call them isomorphic as Lie algebras? $\endgroup$
    – Bedge
    Jul 13, 2021 at 14:50

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