This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Since momentum is treated as a covector in physics, it is natural to consider the cotangent bundle \(T^\ast M\) of a manifold \(M\) when describing phase space mathematically.
Even aside from physical reasons, \(T^\ast M\) is far more natural than the tangent bundle \(TM\) when describing a symplectic manifold, because \(T^\ast M\) carries a natural symplectic form. Before examining what a symplectic form is, we first study symplectic geometry in a linear algebra setting.
Symplectic form
Definition 1 A vector space \((V,\omega)\) is called a symplectic vector space if \(\omega:V\times V\rightarrow \mathbb{R}\) satisfies the following two conditions.
- (Skew-symmetry) For any \(v,w\in V\), we have \(\omega(v,w)=-\omega(w,v)\).
- (Nondegeneracy) If \(\omega(v,w)=0\) for all \(w\in V\), then \(v=0\).
In this case, \(\omega\) is called a symplectic form.
It is not difficult to see that every symplectic vector space must have even dimension.
Lemma 2 Let a skew-symmetric bilinear map \(\omega:V\times V\rightarrow \mathbb{R}\) be given. Then there exists a basis \(u_1,\ldots, u_k, e_1, \ldots,e_n,f_1,\ldots, f_n\) of \(V\) such that the following conditions are all satisfied.
- For all \(v\in V\), we have \(\omega(u_i,v)=0\).
- For all \(i,j\), we have \(\omega(e_i,e_j)=\omega(f_i,f_j)=0\).
- For all \(i,j\), we have \(\omega(e_i,f_j)=\delta_{ij}\).
Proof
First, one can easily check that the set
\[\{u\in V\mid \omega(u,v)=0\text{ for all $v\in V$}\}\]is a subspace of \(V\). Thus, choosing a basis of this subspace, we obtain \(u_1,\ldots, u_k\). Now, write \(V=U\oplus W\). Then we can find a basis \(e_1,\ldots, e_n,f_1,\ldots, f_n\) of \(W\) as follows.
Choose an arbitrary vector \(e_1\in W\). Since \(\omega\) is non-degenerate on \(W\), there exists \(f_1\in W\) such that \(\omega(e_1,f_1)\neq 0\), and by an appropriate scalar multiplication we may assume \(\omega(e_1,f_1)=1\). Since \(\omega\) is skew-symmetric, it is obvious that \(\omega(e_1,e_1)=\omega(f_1,f_1)=0\).
Repeating this process, suppose vectors \(e_1,\ldots, e_k, f_1,\ldots, f_k\in W\) satisfying the two conditions
- For all \(i,j\), we have \(\omega(e_i,e_j)=\omega(f_i,f_j)=0\).
- For all \(i,j\), we have \(\omega(e_i,f_j)=\delta_{ij}\).
are given, and choose an arbitrary vector \(e_{k+1}\) not belonging to \(\span\{e_1,\ldots, e_k,f_1,\ldots, f_k\}\leq W\). If for all \(i=1,\ldots, k\),
\[\omega(e_{k+1}, e_i)=\lambda_i,\qquad\omega(e_{k+1},f_i)=\eta_i\]then by considering the vector
\[e_{k+1}-\sum_{i=1}^k(\lambda_i f_i+\eta_i e_i)\]instead of \(e_{k+1}\), we may assume that \(e_{k+1}\) satisfies
\[\omega(e_{k+1},e_i)=\omega(e_{k+1},f_i)=0\qquad\text{for all $i=1,\ldots, k$}\]Meanwhile, since \(\omega\) is non-degenerate on \(W\), there exists a vector \(f_{k+1}\in W\) such that \(\omega(e_{k+1},f_{k+1})\neq 0\). Likewise, if \(f_{k+1}\) satisfies
\[\omega(f_{k+1}, e_i)=\lambda_i',\qquad\omega(f_{k+1},f_i)=\eta_i'\]then by considering the vector
\[f_{k+1}-\sum_{i=1}^k(\lambda_i' f_i+\eta_i' e_i)\]instead of \(f_{k+1}\), we may assume that \(f_{k+1}\) satisfies
\[\omega(f_{k+1},e_i)=\omega(f_{k+1},f_i)=0\qquad\text{for all $i=1,\ldots, k$}\]and then by an appropriate scalar multiplication we may assume \(\omega(e_{k+1},f_{k+1})=1\).
In the above lemma, the subspace \(U=\span\{u_1,\ldots, u_k\}\) is the space where \(\omega\) vanishes identically, and thus on the complement \(W\) of this subspace, \(\omega\) becomes a symplectic form. Conversely, if a symplectic form is given on an arbitrary vector space, then since \(\omega\) is non-degenerate we must have \(U=0\), and therefore any symplectic vector space must have even dimension. In this case, the basis
\[e_1,\ldots, e_n, f_1,\ldots, f_n\]obtained from the above lemma is called a symplectic basis. If a linear map preserving the symplectic form is called a (linear) symplectomorphism, then by choosing a symplectic basis one can verify that any symplectic vector space is symplectomorphic to the space \((\mathbb{R}^{2n},\omega_0)\) that we studied in the previous post.
Definition 3 Let \((V,\omega)\) be a symplectic vector space and let \(W\leq V\) be an arbitrary subspace. Then the symplectic complement of \(W\) is the space defined by
\[W^\omega=\{v\in V\mid\omega(v,w)=0\text{ for all $w\in W$}\}\]- If \(W\subseteq W^\omega\), then \(W\) is called an isotropic subspace.
- If \(W^\omega\subseteq W\), then \(W\) is called a coisotropic subspace.
- If \(W\cap W^\omega=\{0\}\), then \(W\) is called a symplectic subspace.
- If \(W=W^\omega\), then \(W\) is called a Lagrangian subspace.
Lemma 4 For a symplectic vector space \((V,\omega)\) and its subspace \(W\), the following hold.
- We have \(\dim W+\dim W^\omega=\dim V\), and \((W^\omega)^\omega=W\).
- \(W\) is a symplectic subspace if and only if \(W^\omega\) is a symplectic subspace.
- \(W\) is isotropic if and only if \(W^\omega\) is coisotropic. Also, \(W\) is coisotropic if and only if \(W^\omega\) is isotropic.
- \(W\) is Lagrangian if and only if \(W\) is isotropic and \(\dim W=\frac{1}{2}\dim V\).
Proof
-
Since \(\omega\) is a non-degenerate pairing, the map \(v\mapsto \omega(v,-)\) defines an isomorphism from \(V\) to \(V^\ast\). ([Linear Algebra], §Dual Spaces, ⁋Proposition 4)
Let \(W^\perp\subseteq V^\ast\) be the annihilator of \(W\). ([Linear Algebra], §Dual Spaces, ⁋Definition 7) For any \(u\in W^\omega\),
\[\omega(u,w)=0\qquad\text{for all $w\in W$}\]holds, and therefore \(\omega(u,-)\) always lies in \(W^\perp\). Conversely, whenever an arbitrary \(\varphi\in V^\ast\) is given, there exists a unique \(u\in V\) such that \(\varphi=\omega(u,-)\), and if \(\varphi\in W^\perp\) then
\[0=\varphi(w)=\omega(u,w)\qquad\text{for all $w\in W$}\]holds, so \(u\in W^\omega\). That is, via the above isomorphism we know that the two spaces \(W^\perp\) and \(W^\omega\) are isomorphic. Now the first equality of 1 is obvious from
\[\dim V=\dim W+\dim W^\perp=\dim W+\dim W^\omega\]and the equality \((W^\omega)^\omega=W\) follows from \(W\subseteq(W^\omega)^\omega\) and the fact that we must have \(\dim (W^\omega)^\omega=\dim W\) by the first equality.
- Since \((W^\omega)^\omega=W\), we have \(W\cap W^\omega=(W^\omega)^\omega\cap W^\omega\).
- If \(W\subseteq W^\omega\), then \((W^\omega)^\omega\subseteq W^\omega\), so \(W^\omega\) is coisotropic.
- If \(W\) is Lagrangian, then \(W=W^\omega\), so from \(\dim W+\dim W^\omega\) we obtain \(\dim W=\frac{1}{2}\dim V\) and \(W\) is an isotropic subspace. Conversely, if \(\dim W=\frac{1}{2}\dim V\), then \(\dim W^\omega\) is also \(\frac{1}{2}\dim V\), and therefore an isotropic subspace satisfying such a dimension condition is Lagrangian.
Symplectic quotient
Meanwhile, for an \(\mathbb{R}\)-vector space \(V\) and any subspace \(W\), the quotient space \(V/W\) is always well defined. However, even if \(V\) is a symplectic vector space, \(V/W\) need not carry the structure of a symplectic vector space in general. For example, if \(W\) is an odd-dimensional subspace, then \(V/W\) also becomes odd-dimensional, so it certainly cannot have the structure of a symplectic vector space.
Even if \(W\) is an even-dimensional subspace, the same is true: we may attempt to define a symplectic form on the quotient space \(V/W\) by
\[\overline{\omega}([v_1],[v_2])=\omega(v_1,v_2)\]but for this to be well defined, the value of
\[\omega(v_1+w_1,v_2+w_2)=\omega(v_1,v_2)+\omega(v_1,w_2)+\omega(w_1,v_2)+\omega(w_1,w_2)\]must coincide with \(\omega(v_1,v_2)\). If no condition is imposed on \(W\), there is no reason for the last three terms on the right-hand side to be \(0\), so a symplectic form on \(V/W\) is not generally well defined. Moreover, even if \(W\) is a symplectic subspace, only the last term on the right-hand side, namely \(\omega(w_1,w_2)\), vanishes, so the condition is still insufficient.
To make this work, one can define the quotient space in the following manner.
Lemma 5 Let \((V,\omega)\) be a symplectic vector space and let \(W\) be a coisotropic subspace. Then there exists a unique symplectic structure \(\overline{\omega}\) on \(W/W^\omega\) such that the pullback of \(\overline{\omega}\) by the projection \(W\rightarrow W/W^\omega\) coincides with the restriction of \(\omega\) to \(W\).
Proof
For arbitrary \([w_1],[w_2]\in W\), define \(\overline{\omega}([w_1],[w_2])=\omega(w_1,w_2)\). It is obvious that if this formula yields a well-defined symplectic form, then the desired property holds. First, for any \(u_1,u_2\in W^\omega\),
\[\omega(w_1+u_1,w_2+u_2)=\omega(w_1,w_2)+\omega(w_1,u_2)+\omega(u_1,w_2)+\omega(u_1,u_2)\]holds, and since \(w_1,w_2,u_1,u_2\) are all elements of \(W\) and \(u_1,u_2\) are elements of \(W^\omega\), the last three terms on the right-hand side all become 0. Therefore \(\overline{\omega}\) is well defined. That \(\overline{\omega}\) is skew-symmetric is obvious by definition, so it suffices to show that \(\overline{\omega}\) is non-degenerate. If \([w]\in W\) satisfies \(\overline{\omega}([w],[w'])=0\) for all \([w']\in W\), then
\[0=\overline{\omega}([w],[w'])=\omega(w,w')\qquad\text{for all $w'\in W$}\]so by definition \(w\in W^\omega\) and therefore \([w]=0\). That is, \(\overline{\omega}\) is non-degenerate.
Since every 1-dimensional subspace of a symplectic vector space is an isotropic subspace, by Lemma 4 every codimension-1 subspace \(W\) is a coisotropic subspace. Applying Lemma 5 to this space, we obtain a new symplectic vector space with dimension reduced by 2 from the original vector space.
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