16.3 The case of arbitrary d

Concept links · terms present in this machine draft; source roles are unverified: vector space · linear map · Lie algebra

For the general case of arbitrary $d,$ the group $S p (2 d, R)$ will act by automorphisms on $H_{2 d + 1}$ and $h_{2 d + 1}$ , both of which can be identified with $M \oplus R$ . The group acts by linear transformations on the $M$ factor, preserving Ω. The infinitesimal version of this action is computed as in the $d = 1$ case to be

L \cdot ((c _{p} c _{q}), c) = \frac{d}{d t} (ϕ_{e^{t L}} ((c _{p} c _{q}), c))_{∣ t = 0} = (L (c _{p} c _{q}), 0)

where $L \in sp (2 d, R)$ . This action, as in the $d = 1$ case, is given by taking Poisson brackets of a quadratic function with a linear function:

Theorem 16.3

The $sp (2 d, R)$ action on $ℏ_{2 d + 1} = M \oplus R$ by derivations is

$L \cdot (c_{q} \cdot q + c_{p} \cdot p + c) = {μ_{L}, c_{q} \cdot q + c_{p} \cdot p + c} = c_{q}^{'} \cdot q + c_{p}^{'} \cdot p (16.22)$
where
$(c _{p}^{'} c _{q}^{'}) = L (c _{p} c _{q})$

or, equivalently (see section $4.1)$ , on basis vectors of M one has

{μ_{L}, (p q)} = L^{T} (p q)

Proof. One can first prove 16.22 for the cases when only one of $A, B, C$ is nonzero, then the general case follows by linearity. For instance, taking the special case

L = (00 B 0), μ_{L} = \frac{1}{2} q \cdot B q

the action on coordinate functions (the basis vectors of $M)$ is

{\frac{1}{2} q \cdot B q, (p q)} = L^{T} (p q) = (B q 0)

since

{\frac{1}{2} j, k \sum q_{j} B_{j k} q_{k}, p_{l}} = \frac{1}{2} j, k \sum (q_{j} {B_{j k} q_{k}, p_{l}} + {q_{j} B_{j k}, p_{l}} q_{k}) = \frac{1}{2} (j \sum q_{j} B_{j l} + k \sum B_{l k} q_{k}) = j \sum B_{l j} q_{j} (since B = B^{T})

Repeating for A and C gives in general

{μ_{L}, (p q)} = L^{T} (p q)

We can now prove theorem 16.2 as follows:

Proof.

L \to μ_{L}

is clearly a vector space isomorphism of matrices and of quadratic polynomials. To show that it is a Lie algebra isomorphism, the Jacobi identity for the Poisson bracket can be used to show

{μ_{L}, {μ_{L^{'}}, c_{q} \cdot q + c_{p} \cdot p}} - {μ_{L^{'}}, {μ_{L}, c_{q} \cdot q + c_{p} \cdot p}} = {{μ_{L}, μ_{L^{'}}}, c_{q} \cdot q + c_{p} \cdot p}

The left-hand side of this equation is $c_{q}^{''} \cdot q + c_{p}^{''} \cdot p .$ where

(c _{p}^{''} c _{q}^{''}) = (L L^{'} - L^{'} L) (c _{p} c _{q})

As a result, the right-hand side is the linear map given by

{μ_{L}, μ_{L^{'}}} = μ_{[L, L^{'}]}

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF · 印刷页 185、186、187、188、189、190、191、192、193、194、195、196

来源版本：2025-10-20

来源 PDF SHA-256：5a1941b2443b54d5db3d055f1e5ba390429b7a728475258017aaac87ee85a837

OCR 来源 SHA-256：58acd80919915604a65a16cae6fad6397aac629a3d5ce87f6135279c21c576a1

OCR 产物 SHA-256：58acd80919915604a65a16cae6fad6397aac629a3d5ce87f6135279c21c576a1

数学物理阅读

书架

16.3 The case of arbitrary d

16.3 The case of arbitrary d

关系图谱