PhyslibAlpha.ClassicalMechanics.MomentMap

Given a Lie algebra $\mathfrak{g}$, an element $Z \in \mathfrak{g}$, and a linear functional $\nu \in \mathfrak{g}^*$ (represented as a function $\mathfrak{g} \to \mathbb{R}$), this function computes the value of the coadjoint vector field $Z_{\mathfrak{g}^*}(\nu)$ at an element $Z' \in \mathfrak{g}$. It is defined by the pairing $\langle \nu, [Z, Z'] \rangle$, or simply $\nu([Z, Z'])$.

Given an affine symplectic action of a Lie algebra $\mathfrak{g}$ on a symplectic vector space $E$, the affine vector field $Z_E$ associated with an element $Z \in \mathfrak{g}$ is defined at a point $x \in E$ as $$Z_E(x) = A(Z)x + b(Z)$$ where $A(Z)$ is the linear part of the infinitesimal action and $b(Z)$ is the translation part.

For a symplectic vector space $E$ with a symplectic form $\sigma$ and an affine action of a Lie algebra $\mathfrak{g}$—where each element $Z \in \mathfrak{g}$ acts via the affine vector field $Z_E(x) = A_Z x + b_Z$—the moment $\mu_Z(x)$ is the real-valued function of a point $x \in E$ and an element $Z \in \mathfrak{g}$ defined by: $$\mu_Z(x) = -\frac{1}{2} \sigma(A_Z x, x) - \sigma(b_Z, x)$$ where $A_Z$ is the infinitesimally symplectic linear map and $b_Z$ is the translation vector associated with $Z$.

For a real Lie algebra $\mathfrak{g}$ acting affinely on a symplectic vector space $(E, \sigma)$, the Souriau cocycle $f$ is a 2-form on $\mathfrak{g}$ defined by $f(Z)(Z') = \sigma(b(Z), b(Z'))$ for $Z, Z' \in \mathfrak{g}$, where $b(Z) \in E$ is the translation component of the infinitesimal action of $Z$ on $E$.

In the context of an affine symplectic action of a Lie algebra $\mathfrak{g}$ on a symplectic vector space $(E, \sigma)$, let $A(Z)$ denote the linear part of the infinitesimal action corresponding to $Z \in \mathfrak{g}$. For any $Z \in \mathfrak{g}$ and any vectors $v, w \in E$, the identity $$\sigma(A(Z)v, w) = -\sigma(v, A(Z)w)$$ holds, expressing the fact that $A(Z)$ is an infinitesimally symplectic linear operator.

Let $E$ be a symplectic vector space with symplectic form $\sigma$. For an affine symplectic action $\rho$ of a Lie algebra $\mathfrak{g}$ on $E$, let $A(Z)$ denote the linear part of the infinitesimal action of an element $Z \in \mathfrak{g}$. Then for any vectors $v, w \in E$, the identity $$\sigma(v, A(Z)w) = -\sigma(A(Z)v, w)$$ holds, expressing that $A(Z)$ is an infinitesimally symplectic operator.

Let $(E, \sigma)$ be a symplectic vector space and $\mathfrak{g}$ be a Lie algebra with an affine symplectic action $\rho$ on $E$. For any $Z \in \mathfrak{g}$, let $A_Z$ denote the linear part of the infinitesimal action of $Z$ on $E$. Then for all $v, w \in E$, it holds that $$\sigma(A_Z v, w) = \sigma(A_Z w, v).$$ This symmetry follows from the fact that $A_Z$ is infinitesimally symplectic and $\sigma$ is an alternating form.

Let $(E, \sigma)$ be a symplectic vector space and $\mathfrak{g}$ be a Lie algebra with an affine symplectic action $\rho$ on $E$. For any point $x \in E$ and Lie algebra elements $Z, Z' \in \mathfrak{g}$, the moment $\mu_Z(x)$ is additive in its Lie algebra argument: $$\mu_{Z+Z'}(x) = \mu_Z(x) + \mu_{Z'}(x)$$ where $\mu_Z(x)$ is defined as $-\frac{1}{2} \sigma(A_Z x, x) - \sigma(b_Z, x)$, with $A_Z$ and $b_Z$ being the linear and translation components of the infinitesimal action of $Z$.

Let $(E, \sigma)$ be a symplectic vector space and $\mathfrak{g}$ a Lie algebra with an affine symplectic action $\rho$. For any point $x \in E$, scalar $c \in \mathbb{R}$, and Lie algebra element $Z \in \mathfrak{g}$, the moment map $\mu_Z(x)$ satisfies the homogeneity property: $$\mu_{c \cdot Z}(x) = c \mu_Z(x).$$

Let $(E, \sigma)$ be a real symplectic vector space and $\mathfrak{g}$ be a Lie algebra acting affinely on $E$. For any $Z, Z' \in \mathfrak{g}$ and any point $x \in E$, the symplectic form $\sigma$ evaluated on the associated affine vector fields $Z_E(x)$ and $Z'_E(x)$ satisfies the identity: $$\sigma(Z_E(x), Z'_E(x)) = \mu_{\llbracket Z, Z' \rrbracket}(x) + f(Z, Z')$$ where $\mu$ is the moment map of the action and $f$ is the Souriau cocycle.

For a symplectic vector space $(E, \sigma)$ and a Lie algebra $\mathfrak{g}$ acting affinely on $E$, the Souriau cocycle $f: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ (defined as $f(Z, Z') = \sigma(b(Z), b(Z'))$) is antisymmetric. That is, for any $Z, Z' \in \mathfrak{g}$, it holds that $$f(Z', Z) = -f(Z, Z').$$

Let $\sigma$ be a symplectic form on a vector space $E$. For an affine symplectic action of a Lie algebra $\mathfrak{g}$ on $E$ with Souriau cocycle $f$, and for any elements $Z, Z', Z'' \in \mathfrak{g}$, the following cyclic identity holds: $$f(Z, [Z', Z'']) + f(Z', [Z'', Z]) + f(Z'', [Z, Z']) = 0$$ where $[\cdot, \cdot]$ denotes the Lie bracket on $\mathfrak{g}$.

Let $(E, \sigma)$ be a symplectic vector space and $\mathfrak{g}$ be a real Lie algebra acting affinely on $E$. Let $f: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ be the Souriau cocycle associated with this action. For any $Z, Z', Z'' \in \mathfrak{g}$, the cocycle satisfies the identity: $$f(\llbracket Z, Z' \rrbracket, Z'') = f(Z, \llbracket Z', Z'' \rrbracket) - f(Z', \llbracket Z, Z'' \rrbracket),$$ where $\llbracket \cdot, \cdot \rrbracket$ denotes the Lie bracket on $\mathfrak{g}$. This identity characterizes $f$ as a Lie algebra cocycle for the coadjoint representation.

Let $(E, \sigma)$ be a real symplectic vector space and $\mathfrak{g}$ be a real Lie algebra acting affinely on $E$. For any element $Z \in \mathfrak{g}$, let $Z_E(x) = A_Z x + b_Z$ be the corresponding affine vector field at $x \in E$, where $A_Z$ is the infinitesimally symplectic linear part and $b_Z$ is the translation part. Let $\mu_Z(x) = -\frac{1}{2} \sigma(A_Z x, x) - \sigma(b_Z, x)$ be the moment map and $f(Z)(Z') = \sigma(b_Z, b_{Z'})$ be the Souriau cocycle. For any constant linear functional $\mu_0 \in \mathfrak{g}^*$ (a constant torsor), the following identity holds for all $Z, Z' \in \mathfrak{g}$ and $x \in E$: $$\sigma(Z_E(x), Z'_E(x)) = (\mu_{[Z, Z']}(x) - \mu_0([Z, Z'])) + (f(Z)(Z') + \mu_0([Z, Z']))$$ where $\mu_0([Z, Z'])$ represents the coadjoint action $Z_{\mathfrak{g}^*}(\mu_0)$ evaluated at $Z'$. This identity demonstrates that shifting the moment map by a constant $\mu_0$ results in a corresponding shift of the Souriau cocycle while preserving the underlying relationship between the symplectic form on vector fields and the moment.

Let $(E, \sigma)$ be a symplectic vector space and $\mathfrak{g}$ be a real Lie algebra acting on $E$ by affine symplectic vector fields $Z_E(x) = A_Z x + b_Z$. Let $\mu_Z(x)$ be the moment map and $f(Z, Z')$ be the Souriau cocycle associated with this action. If the cocycle $f$ is a coboundary, meaning there exists a linear functional $\mu_0 \in \mathfrak{g}^*$ such that $f(Z, Z') = \mu_0([Z, Z'])$ for all $Z, Z' \in \mathfrak{g}$, then for any $x \in E$ and $Z, Z' \in \mathfrak{g}$, the following identity holds: $$\sigma(Z_E(x), Z'_E(x)) = \mu_{[Z, Z']}(x) + \mu_0([Z, Z'])$$ This identity signifies that the adjusted moment map $\mu + \mu_0$ is exactly equivariant.

For an affine symplectic action of a Lie algebra $\mathfrak{g}$ on a symplectic vector space $E$ with translation component $b : \mathfrak{g} \to E$, if the action is linear (i.e., $b = 0$), then the Souriau cocycle $f(Z, Z') = \sigma(b Z, b Z')$ vanishes for all $Z, Z' \in \mathfrak{g}$.

Given a real normed vector space $E$, a bilinear form $\sigma$ on $E$, and a function $u : E \to \mathbb{R}$, a vector field $X : E \to E$ is defined to be the **symplectic gradient** of $u$ with respect to $\sigma$ if for all $x, v \in E$, the following identity holds: $$\sigma(X(x), v) = -du_x(v)$$ where $du_x$ denotes the Fréchet derivative of $u$ at the point $x$ evaluated on the vector $v$.

For an element $Z$ of the Lie algebra $\mathfrak{g}$ acting on a symplectic vector space $E$, the function $\rho.\text{sigmaA}(Z)$ is the continuous linear map $E \to \mathcal{L}(E, \mathbb{R})$ that maps a vector $y \in E$ to the linear form $v \mapsto \sigma(A_Z y, v)$. Here $A_Z$ represents the linear part of the infinitesimal affine action of $Z$ on $E$, and $\sigma$ is the symplectic form. For any $y, v \in E$, the identity is: $$(\rho.\text{sigmaA}(Z))(y)(v) = \sigma(A_Z y, v)$$

For an element $Z$ of the Lie algebra $\mathfrak{g}$ acting affinely on a symplectic vector space $E$ with symplectic form $\sigma$, let $A_Z$ be the linear part of the infinitesimal action. For any vectors $y, v \in E$, the value of the map $\rho.\text{sigmaA}(Z)$ satisfies the identity $$(\rho.\text{sigmaA}(Z))(y)(v) = \sigma(A_Z y, v).$$

Given a Lie algebra $\mathfrak{g}$ acting affinely on a symplectic vector space $(E, \sigma)$, let $b(Z) \in E$ denote the translation component of the infinitesimal action of an element $Z \in \mathfrak{g}$. This definition represents the continuous linear form on $E$ defined by the mapping $v \mapsto \sigma(b(Z), v)$.

Let $(E, \sigma)$ be a symplectic vector space and $\rho$ be an affine symplectic action of a Lie algebra $\mathfrak{g}$ on $E$. For any $Z \in \mathfrak{g}$ and $v \in E$, the value of the linear form $\rho.\sigma_B(Z)$ at $v$ is given by $\sigma(b(Z), v)$, where $b(Z)$ is the translation component of the infinitesimal action of $Z$ on $E$.

Let $E$ be a real symplectic vector space with symplectic form $\sigma$. Let $\mathfrak{g}$ be a Lie algebra acting on $E$ by affine vector fields $Z_E(x) = A_Z x + b_Z$, where $A_Z$ is the linear part of the action of $Z \in \mathfrak{g}$. For any $x \in E$, the Fréchet derivative of the mapping $y \mapsto \sigma(A_Z y, y)$ at the point $x$ is the continuous linear map $v \mapsto \sigma(A_Z x, v) + \sigma(A_Z v, x)$.

Let $E$ be a vector space equipped with a bilinear form $\sigma$. Consider an affine action of a Lie algebra $\mathfrak{g}$ on $E$, where for each $Z \in \mathfrak{g}$, the action is given by the affine vector field $Z_E(x) = A_Z x + b_Z$. If the linear map $A_Z$ satisfies the symmetry condition $\sigma(A_Z v, w) = \sigma(A_Z w, v)$ for all $v, w \in E$, then the Fréchet derivative of the moment map $\mu_Z(x) = -\frac{1}{2} \sigma(A_Z x, x) - \sigma(b_Z, x)$ at a point $x \in E$ is the linear map $v \mapsto -\sigma(Z_E(x), v)$.

Let $(E, \sigma)$ be a symplectic vector space and $\mathfrak{g}$ be a Lie algebra acting on $E$ via an affine symplectic action. For any $Z \in \mathfrak{g}$, let $\mu_Z: E \to \mathbb{R}$ be the moment map and $Z_E: E \to E$ be the associated affine vector field. For any $x \in E$, the function $y \mapsto \mu_Z(y)$ is Fréchet differentiable at $x$ with derivative $$v \mapsto -\sigma(Z_E(x), v).$$

Let $(E, \sigma)$ be a symplectic vector space. For any element $Z$ in a Lie algebra $\mathfrak{g}$ acting on $E$ via an affine symplectic action, let $Z_E$ be the associated affine vector field and $\mu_Z: E \to \mathbb{R}$ be the moment map. Then $Z_E$ is the symplectic gradient of $\mu_Z$ with respect to $\sigma$. That is, for all $x, v \in E$, the identity $$\sigma(Z_E(x), v) = -d(\mu_Z)_x(v)$$ holds, where $d(\mu_Z)_x$ denotes the Fréchet derivative of the function $y \mapsto \mu_Z(y)$ at the point $x$.

Let $E$ be a real symplectic vector space with symplectic form $\sigma$. For an element $Z$ of a Lie algebra $\mathfrak{g}$ acting on $E$ via an affine symplectic action $\rho$, let $Z_E(y) = A_Z y + b_Z$ be the associated affine vector field, where $A_Z$ is the linear part and $b_Z$ is the translation part. Let $\rho.\text{sigmaA}(Z)$ be the continuous linear map $y \mapsto \sigma(A_Z y, \cdot)$. The Fréchet derivative of the map $y \mapsto -\sigma(Z_E(y), \cdot)$ at any point $x \in E$ is the linear map $-\rho.\text{sigmaA}(Z)$.

Let $E$ be a real finite-dimensional symplectic vector space with symplectic form $\sigma$. Suppose an affine symplectic action of a Lie algebra $\mathfrak{g}$ on $E$ is given, where for each $Z \in \mathfrak{g}$, $Z_E(x)$ is the associated affine vector field and $\mu_Z(x)$ is the corresponding moment map. If $\psi : E \to \mathfrak{g} \to \mathbb{R}$ is a function such that for every $Z \in \mathfrak{g}$ and $x \in E$, the Fréchet derivative of $y \mapsto \psi(y, Z)$ at $x$ is the linear map $v \mapsto -\sigma(Z_E(x), v)$, then for each $Z \in \mathfrak{g}$, there exists a constant $c \in \mathbb{R}$ such that $\psi(x, Z) = \mu_Z(x) + c$ for all $x \in E$.

Let $(E, \sigma)$ be a symplectic vector space and $\mathfrak{g}$ be a Lie algebra acting affinely on $E$. For any $Z, Z' \in \mathfrak{g}$ and $x \in E$, the Fréchet derivative of the moment function $\mu_{Z'}$ at $x$ in the direction of the affine vector field $Z_E(x)$ is given by: $$D\mu_{Z'}(x) \cdot Z_E(x) = \mu_{[Z, Z']}(x) + f(Z, Z')$$ where $[Z, Z']$ is the Lie bracket in $\mathfrak{g}$, $\mu_{[Z, Z']}(x)$ is the moment of the bracket evaluated at $x$, and $f(Z, Z')$ is the Souriau cocycle associated with the affine action.

Let $(E, \sigma)$ be a symplectic vector space and let $\mathfrak{g}$ be a Lie algebra acting on $E$ via affine symplectic vector fields. For any elements $Z, Z' \in \mathfrak{g}$, let $Z_E$ and $Z'_E$ be the corresponding vector fields on $E$. Then, for any points $x, v \in E$, the following identity holds: $$\sigma([Z, Z']_E(x), v) = -d_x (\sigma(Z_E, Z'_E))(v)$$ where $[Z, Z']_E$ is the vector field associated with the Lie bracket $\llbracket Z, Z' \rrbracket \in \mathfrak{g}$, and $d_x (\sigma(Z_E, Z'_E))(v)$ denotes the Fréchet derivative of the function $y \mapsto \sigma(Z_E(y), Z'_E(y))$ at the point $x$ in the direction $v$.

Let $E$ be a real symplectic vector space with a symplectic form $\sigma$. Let $\mathfrak{g}$ be a Lie algebra with an affine symplectic action $\rho$ on $E$. For any $Z \in \mathfrak{g}$, let $Z_E$ denote the associated affine vector field and $\mu_Z: E \to \mathbb{R}$ be the corresponding moment map. Let $H: E \to \mathbb{R}$ be a Hamiltonian function and $X: E \to E$ be its symplectic gradient, which satisfies $\sigma(X(x), v) = -dH_x(v)$ for all $x, v \in E$. Suppose $H$ is invariant under the action of $Z$, such that the Fréchet derivative $dH_x(Z_E(x)) = 0$ for all $x \in E$. If $\gamma: \mathbb{R} \to E$ is a trajectory satisfying the Hamiltonian equations of motion $\dot{\gamma}(t) = X(\gamma(t))$, then the derivative of the moment map component $\mu_Z$ along the trajectory is zero: $$\frac{d}{dt} \mu_Z(\gamma(t)) = 0$$

Let $(E, \sigma)$ be a real symplectic vector space. Let $H: E \to \mathbb{R}$ be a Hamiltonian function and $X: E \to E$ be its symplectic gradient, which satisfies the identity $\sigma(X(x), v) = -dH_x(v)$ for all $x, v \in E$. Suppose $Z \in \mathfrak{g}$ is an element of a Lie algebra acting on $E$ via an affine symplectic action, where $Z_E$ is the associated affine vector field. If $H$ is invariant under this action, meaning that the Fréchet derivative of $H$ along the vector field $Z_E$ vanishes ($dH_x(Z_E(x)) = 0$ for all $x \in E$), and if $\gamma: \mathbb{R} \to E$ is an integral curve of the vector field $X$ (i.e., $\dot{\gamma}(t) = X(\gamma(t))$), then the moment map $\mu_Z$ is constant along the motion $\gamma$: $$\mu_Z(\gamma(t)) = \mu_Z(\gamma(s))$$ for any times $t, s \in \mathbb{R}$.

The plane $\mathbb{R} \times \mathbb{R}$ is equipped with the structure of a Lie ring, where the Lie bracket $[x, y]$ is defined as the commutator $xy - yx$ of the product ring. Since $\mathbb{R} \times \mathbb{R}$ is a commutative ring under component-wise multiplication, the Lie bracket is identically zero, making it an abelian Lie ring.

The plane $\mathbb{R}^2$ (represented as $\mathbb{R} \times \mathbb{R}$) is equipped with the structure of a Lie algebra over the real numbers $\mathbb{R}$. Since the plane is treated as a commutative associative algebra under component-wise multiplication, the Lie bracket is defined by the commutator $[x, y] = xy - yx = 0$ for all $x, y \in \mathbb{R}^2$, making it an abelian Lie algebra.

The standard symplectic form $\sigma$ on the plane $\mathbb{R}^2$ is a bilinear form defined by $\sigma(v, w) = v_1 w_2 - v_2 w_1$ for vectors $v = (v_1, v_2)$ and $w = (w_1, w_2)$ in $\mathbb{R}^2$.

This definition constructs the affine symplectic action of the abelian Lie algebra $\mathfrak{g} = \mathbb{R}^2$ on the plane $E = \mathbb{R}^2$ equipped with the standard symplectic form $\sigma(v, w) = v_1 w_2 - v_2 w_1$. For any element $Z \in \mathfrak{g}$, the action is given by the constant vector field $Z_E(x) = Z$. In Souriau's framework of affine vector fields $Z_E(x) = A(Z)x + b(Z)$, this corresponds to the linear maps $A = 0$ and the translation part $b = \text{id}_E$.

The standard bilinear form $\sigma$ on the plane $\mathbb{R}^2$, defined by $\sigma(v, w) = v_1 w_2 - v_2 w_1$ for vectors $v = (v_1, v_2)$ and $w = (w_1, w_2)$, is a symplectic form (i.e., it is alternating and non-degenerate).

Consider the plane $\mathbb{R}^2$ equipped with the standard symplectic form $\sigma(v, w) = v_1 w_2 - v_2 w_1$ for $v, w \in \mathbb{R}^2$. For the affine symplectic action of the abelian Lie algebra $\mathfrak{g} = \mathbb{R}^2$ on the plane by translations, the Souriau cocycle $f: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ (defined by $f(Z, Z') = \sigma(b(Z), b(Z'))$ where $b$ is the translation component) satisfies $f((1, 0), (0, 1)) = 1$.

Consider the abelian Lie algebra $\mathfrak{g} = \mathbb{R}^2$ acting by translations on the symplectic plane $(\mathbb{R}^2, \sigma)$ equipped with the standard symplectic form $\sigma(v, w) = v_1 w_2 - v_2 w_1$. The Souriau cocycle $f$ associated with this action, defined by $f(Z, Z') = \sigma(Z, Z')$, is not a coboundary. That is, there does not exist a linear functional $\mu_0 \in \mathfrak{g}^*$ such that for all $Z, Z' \in \mathfrak{g}$, the relation $f(Z, Z') = \mu_0([Z, Z'])$ holds.