Weak symplectic forms and differential calculus in Banach spaces

1 Introduction

There are scarcely any decent expositions of infinite dimensional symplectic vector spaces. One good basic exposition is by Marsden and Ratiu. The Darboux theorem for a real reflexive Banach space is proved in Lang and probably in fewer other places than one might guess. (Other references.)

2 Bilinear forms

Let $E$ be a real Banach space. For a bilinear form $B:E\times E\to ℝ$, define

$$\parallel B\parallel =\underset{\parallel e\parallel \le 1,\parallel f\parallel \le 1}{sup}|B(e,f)|.$$

One proves that $B$ is continuous if and only if . Namely, a bilinear form is continuous if and only if it is bounded.

If $B:E\times E\to ℝ$ is a continuous bilinear form, we define $B^{\mathrm{♭}}:E\to E^{*}$ by

$$B^{\mathrm{♭}}(e)(f)=B(e,f),e\in E,f\in E;$$

indeed, for $e\in E$, $\parallel B^{\mathrm{♭}}(e)f\parallel =\parallel B(e,f)\parallel \le \parallel B\parallel \parallel e\parallel \parallel f\parallel$, showing that $\parallel B^{\mathrm{♭}}(e)\parallel \le \parallel B\parallel \parallel e\parallel$, so that $B^{\mathrm{♭}}(e)$ is continuous $E\to ℝ$. Moreover, it is apparent that $B^{\mathrm{♭}}$ is linear, and

$\parallel B^{\mathrm{♭}}\parallel$ $=\underset{\parallel e\parallel \le 1}{sup}\parallel B^{\mathrm{♭}}(e)\parallel$

$=\underset{\parallel e\parallel \le 1}{sup}\underset{\parallel f\parallel \le 1}{sup}|B^{\mathrm{♭}}(e)(f)|$

$=M,$

so $B^{\mathrm{♭}}:E\to E^{*}$ is continuous.

We call a continuous bilinear form $B:E\times E\to F$ weakly nondegenerate if $B^{\mathrm{♭}}:E\to E^{*}$ is one-to-one. Since $B^{\mathrm{♭}}$ is linear, this is equivalent to the statement that $B^{\mathrm{♭}}(e)=0$ implies that $e=0$, which is equivalent to the statement that if $B(e,f)=0$ for all $f$ then $e=0$.

An isomorphism of Banach spaces is a linear isomorphism $T:E\to F$ that is continuous such that $T^{-1}F\to E$ is continuous. Equivalently, to say that $T:E\to F$ is an isomorphism of Banach spaces means that $T:E\to F$ is a bijective bounded linear map such that $T^{-1}:F\to E$ is a bounded linear map. It follows from the open mapping theorem that if $T:E\to F$ is an onto bounded linear isomorphism, hence is an isomorphism of Banach spaces.

We say that a continuous bilinear form $B:E\times E\to ℝ$ is strongly nondegenerate if $B^{\mathrm{♭}}:E\to E^{*}$ is an isomorphism of Banach spaces.

For a real vector space $V$ and a bilinear form $B:V\times V\to ℝ$, we say that $B$ is alternating if $B(v,v)=0$ for all $v\in V$. We say that $B$ is skew-symmetric if $B(u,v)=-B(v,u)$ for all $u,v\in V$. It is straightforward to check that $B$ is alternating if and only if $B$ is skew-symmetric.

For Banach spaces $E_1,\mathrm{\dots },E_p$ and $F$, let $ℒ(E_1,\mathrm{\dots },E_p;F)$ denote the set of continuous multilinear maps $E_1\times \mathrm{\cdots }{E}_p\to F$. For a multilinear map $T:E_1\times \mathrm{\cdots }\times E_p\to F$ to be continuous it is equivalent that

namely that it is bounded with the operator norm. With this norm, $ℒ(E_1,\mathrm{\dots },E_p;F)$ is a Banach space. We write

$${ℒ}_p(E;F)=ℒ(E_1,\mathrm{\dots },E_p;F).$$

For Banach spaces $E$ and $F$, we denote by $\mathrm{GL}(E;F)$ the set of isomorphisms $E\to F$. One proves that $\mathrm{GL}(E;F)$ is an open set in the Banach space $ℒ(E;F)$ and that with the subspace topology, $u\mapsto u^{-1}$ is continuous $\mathrm{GL}(E;F)\to \mathrm{GL}(F;E)$.

For Banach spaces $E,F,G$, define

$$\varphi :ℒ(E,F;G)\to ℒ(E;ℒ(F,G))$$

by $\varphi (f)(x)(y)=f(x,y)$ for $f\in ℒ(E,F;G)$, $x\in E$, and $y\in F$. One proves that $\varphi$ is an isometric isomorphism.

3 Differentiable functions

Let $E$ and $F$ be Banach spaces and let $U$ be a nonempty open subset of $E$. For $a\in U$, a function $f:U\to F$ is said to be differentiable at $a$ if (i) $f$ is continuous at $a$ and (ii) there is a linear mapping $g:E\to F$ such that

$${\parallel f(x)-f(a)-(g(x)-g(a))\parallel }_F=o({\parallel x-a\parallel }_E),$$

as $x\to a$ in $E$. We prove that there is at most one such linear mapping $g$ and write $f^{\prime }(a)=g$, and call $f^{\prime }(a)$ the derivative of $f$ at $a$. We also prove that if $f$ is differentiable at $a$ then $f^{\prime }(a):E\to F$ is continuous at $a$ and therefore, being linear, is continuous on $E$, namely $f^{\prime }(a)\in ℒ(E;F)$.

If $f:U\to F$ is differentiable at each $a\in U$, we say that $f$ is differentiable on $U$. We call $f^{\prime }:U\to ℒ(E;F)$ the derivative of $f$. We also write $Df=f^{\prime }$.

We say that $f:U\to F$ is $C^1$, also called continuously differentiable, if (i) $f$ is differentiable on $U$ and (ii) $f^{\prime }:U\to ℒ(E;F)$ is continuous.

Let $E,F,G$ be Banach spaces, let $U$ be an open subset of $E$, let $V$ be an open subset of $F$, and let $f:U\to F$ and $g:V\to G$ be continuous. Suppose that $a\in U$ and that $f(a)\in V$. We define $g\circ f:f^{-1}(V)\to G$ on $f^{-1}(V)$. One proves that if $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, then $h=g\circ f:f^{-1}(V)\to F$ is differentiable at $a$ and satisfies

$$h^{\prime }(a)=g^{\prime }(f(a))\circ f^{\prime }(a).$$

For Banach spaces $E$ and $F$, let $\varphi :\mathrm{GL}(E;F)\to ℒ(F;E)$ be defined by $\varphi (u)=u^{-1}$. $\mathrm{GL}(E;F)$ is an open subset of the Banach space $ℒ(E;F)$ and $\varphi$ is continuous. It is proved that $\varphi$ continuously differentiable, and that for $u\in \mathrm{GL}(E;F)$, the derivative of $\varphi$ at $u$,

$${\varphi }^{\prime }(u)\in ℒ(ℒ(E;F);ℒ(F;E)),$$

satisfies

$${\varphi }^{\prime }(u)(h)=-u^{-1}\circ h\circ u^{-1},h\in ℒ(E;F).$$

4 Symplectic forms

A weak symplectic form on a Banach space $E$ is a continuous bilinear form $\mathrm{\Omega }:E\times E\to ℝ$ that is weakly nondegenerate and and alternating.

A strong symplectic form on a Banach space $E$ is a continuous bilinear form $\mathrm{\Omega }:E\times E\to ℝ$ that is strongly nondegenerate and alternating. If $\mathrm{\Omega }$ is a strong symplectic form on a Banach space $E$, we define ${\mathrm{\Omega }}^{\mathrm{♯}}:E^{*}\to E$ by ${\mathrm{\Omega }}^{\mathrm{♯}}={({\mathrm{\Omega }}^{\mathrm{♭}})}^{-1}$, which is an isomorphism of Banach spaces.

5 Hamiltonian functions

Let $E$ be a real Banach space $E$, let $𝒟(A)$ be a linear subspace of $E$, and let $A:𝒟(A)\to E$ be a linear map, called an operator in $E$. Write $ℛ(A)=A𝒟(A)$. For a weak symplectic form $\omega$ on $E$, we say that $A$ is $\omega$-skew if

$$\omega (Ae,f)=-\omega (e,Af),e,f\in 𝒟(A).$$

If $ℛ(A)\subset 𝒟(A)$ and $A^2=-I$, then for $e,f\in 𝒟(A)$ we have $\omega (Ae,Af)=-\omega (e,A^{2}f)=-\omega (e,-f)=\omega (e,f)$.

For an $\omega$-skew operator $A$ in $E$, we define $H:𝒟(A)\to ℝ$, called the Hamiltonian function of $A$, by

$$H(u)=\frac{1}{2}\omega (Au,u),u\in 𝒟(A).$$

For a linear operator $A$ in $E$, we define

$$𝒢(A)=\{(u,Au):u\in 𝒟(A)\}.$$

$𝒢(A)$ is a linear subspace of $E\times E$. We say that $A$ is closed if $𝒢(A)$ is a closed subset of $E\times E$. One proves that a linear operator $A$ in $E$ is closed if and only if the linear space $𝒟(A)$ with the norm

$${\parallel e\parallel }_A=\parallel e\parallel +\parallel Ae\parallel ,e\in 𝒟(A)$$

is a Banach space.

For $T\in ℒ(E)$, we define $T^{*}\omega :E\times E\to ℝ$ by

$$(T^{*}\omega )(e,f)=\omega (Te,Tf),(e,f)\in E\times E;$$

$T^{*}\omega$ is called the pullback of $\omega$ by $T$. It is apparent that $T^{*}\omega$ is bilinear. We have

$\parallel T^{*}\omega \parallel$ $=\underset{\parallel e\parallel \le 1,\parallel f\parallel \le 1}{sup}|\omega (Te,Tf)|$

$\le \underset{\parallel e\parallel \le 1,\parallel f\parallel \le 1}{sup}\parallel \omega \parallel \parallel Te\parallel \parallel Tf\parallel$

$\le \underset{\parallel e\parallel \le 1,\parallel f\parallel \le 1}{sup}\parallel \omega \parallel \parallel T\parallel \parallel e\parallel \parallel T\parallel \parallel f\parallel$

$=\parallel \omega \parallel {\parallel T\parallel }^2,$

showing that $T^{*}\omega$ is continuous. For $e\in E$, because $\omega$ is alternating we have

$$(T^{*}\omega )(e,e)=\omega (Te,Te)=0,$$

i.e. $T^{*}\omega$ is alternating. For $e\in E$, suppose that $(T^{*}\omega )(e,f)=0$ for all $f\in E$. That is, $\omega (Te,Tf)=0$ for all $f\in E$, and thus, to establish that $T^{*}\omega$ is weakly nondegenerate it suffices that $T$ be onto. In the case that $T^{*}\omega =\omega$, we say that $T\in ℒ(E)$ is a canonical transformation.

Suppose that $A$ is a closed $\omega$-skew operator in $E$, with Hamiltonian function $H:𝒟(A)\to ℝ$. $𝒟(A)$ is a Banach space with the norm ${\parallel e\parallel }_A=\parallel e\parallel +\parallel Ae\parallel$. For $u\in 𝒟(A)$ and $v\in 𝒟(A)$, using the fact that $A$ is $\omega$-skew we check that

$$H(v)-H(u)-\omega (Au,v-u)=\frac{1}{2}\omega (A(v-u),v-u),$$

hence

$$|H(v)-H(u)-\omega (Au,v-u)|\le \frac{1}{2}\parallel \omega \parallel \parallel A(v-u)\parallel \parallel v-u\parallel \le \frac{1}{2}\parallel \omega \parallel {\parallel v-u\parallel }_A^2.$$

This shows that $H$ is differentiable on the Banach space $𝒟(A)$, with derivative $H^{\prime }:𝒟(A)\to 𝒟{(A)}^{*}$ defined by

$$H^{\prime }(u)(e)=\omega (Au,e),u\in 𝒟(A),e\in 𝒟(A).$$

Moreover, for $u,v\in 𝒟(A)$ we have

$\parallel H^{\prime }(v)-H^{\prime }(u)\parallel$ $=\underset{{\parallel e\parallel }_A\le 1}{sup}|H^{\prime }(v)(e)-H^{\prime }(u)(e)|$

$=\underset{{\parallel e\parallel }_A\le 1}{sup}|\omega (Av,e)-\omega (Au,e)|$

$=\underset{{\parallel e\parallel }_A\le 1}{sup}|\omega (A(v-u),e)|$

$\le \underset{{\parallel e\parallel }_A\le 1}{sup}\parallel \omega \parallel \parallel A(v-u)\parallel \parallel e\parallel$

$\le \parallel \omega \parallel \parallel A(v-u)\parallel$

$\le \parallel \omega \parallel {\parallel v-u\parallel }_A,$

showing that $H^{\prime }:𝒟(A)\to 𝒟{(A)}^{*}$ is continuous, namely that $H$ is $C^1$. (We also write $DH=H^{\prime }$.)

Suppose that $A$ is a closed operator in $E$ and that $H:𝒟(A)\to ℝ$ is some function such that $H^{\prime }(u)e=\omega (Au,e)$ for all $u\in 𝒟(A)$ and $e\in 𝒟(A)$. On the one hand, because $H^{\prime }$ is continuous and linear, the second derivative $D^{2}H:𝒟(A)\to ℒ(𝒟(A),𝒟{(A)}^{*})$ is

$$(D^{2}H)(u)(e)(f)=H^{\prime }(e)(f)=\omega (Ae,f),u,e,f\in 𝒟(A).$$

On the other hand, because $D^{2}H$ is continuous, for each $u\in 𝒟(A)$, the bilinear form $(D^{2}H)(u):𝒟(A)\times 𝒟(A)\to ℝ$ is symmetric. That is, $(D^{2}H)(u)(e)(f)=(D^{2}H)(u)(f)(e)$, which by the above means

$$\omega (Ae,f)=\omega (Af,e),e,f\in 𝒟(A),$$

showing that $A$ is $\omega$-skew. Let $G:𝒟(A)\to ℝ$ be the Hamiltonian function of $A$, i..e

$$G(u)=\frac{1}{2}\omega (Au,u),u\in 𝒟(A).$$

What we established earlier tells us that

$$G^{\prime }(u)(e)=\omega (Au,e),u\in 𝒟(A),e\in 𝒟(A).$$

Then we have that for $G^{\prime }=H^{\prime }$. Let $K=G-H$, which is $C^1$ with $K^{\prime }=0$. The mean value theorem tells us that for any $x,y\in 𝒟(A)$,

$$K(x+y)-K(x)={\int }_0^1{K}^{\prime }(x+ty)(y)𝑑t=0,$$

and thus $K(u)=K(0)=C$ for all $u\in 𝒟(A)$. Therefore, $G=H+C$.

6 Semigroups

Let $E$ be a real Banach space, let $\omega$ be a weak symplectic form on $E$, and let $A$ be a closed densely defined $\omega$-skew linear operator in $E$. Suppose that $A$ is the infinitesimal generator of a strongly continuous one-parameter semigroup $\{U_t:t\ge 0\}$, where $U_t\in ℒ(E)$ for each $t$, and let $H$ be the Hamiltonian function of $A$.

Theorem 1.

For each $t\ge 0$, $U_t$ is a canonical transformation.

For each $t\ge 0$ and for each $x\in 𝒟(A)$,

$$H(U_{t}x)=H(x).$$

Proof.

For $u,v\in 𝒟(A)$ and $t\ge 0$, using the chain rule and the fact that $\omega$ is a bilinear form,

$$\frac{d}{dt}\omega (U_{t}u,U_{t}v)=\omega (\frac{d}{dt}{U}_{t}u,U_{t}v)+\omega (U_{t}u,\frac{d}{dt}{U}_{t}v).$$

Because $A$ is the infinitesimal generator of $\{U_t:t\ge 0\}$, it follows that $\frac{d}{dt}(U_{t}w)=A{U}_{t}w$ for each $w\in 𝒟(A)$. Using this and the fact that $A$ is $\omega$-skew,

${\displaystyle \frac{d}{dt}}\omega (U_{t}u,U_{t}v)$ $=\omega (A{U}_{t}u,U_{t}v)+\omega (U_{t}u,A{U}_{t}v)$

$=-\omega (U_{t}u,A{U}_{t}v)+\omega (U_{t}u,A{U}_{t}v)$

$=0.$

This implies that $\omega (U_{t}u,U_{t}v)=\omega (U_{0}u,U_{0}v)=\omega (u,v)$ for all $t\ge 0$, which means that $U_t$ is a canonical transformation for each $t\ge 0$.

For any $t\ge 0$ and $x\in 𝒟(A)$, $A{U}_{t}x=U_{t}Ax$. (The infinitesimal generator of a one-parameter semigroup commutes with each element of the semigroup.) Then, using the fact that $U_t$ is a canonical transformation,

$H(U_{t}x)$ $={\displaystyle \frac{1}{2}}\omega (A(U_{t}x),U_{t}x)$

$={\displaystyle \frac{1}{2}}\omega (U_{t}Ax,U_{t}x)$

$={\displaystyle \frac{1}{2}}\omega (Ax,x)$

$=H(x).$

∎

Suppose that there is some $c>0$ such that $H(u)\ge c{\parallel u\parallel }_A^2$ for all $u\in 𝒟(A)$, namely that $H$ is coercive on the Banach space $𝒟(A)$. Let $t\ge 0$ and let $u\in 𝒟(A)$. Then $U_{t}u\in 𝒟(A)$, so using the hypothesis and Theorem 1,

$${\parallel U_{t}u\parallel }_A^2\le \frac{1}{c}H(U_{t}u)=\frac{1}{c}H(u)=\frac{1}{2c}\omega (Au,u)\le \frac{1}{2c}\parallel \omega \parallel \parallel Au\parallel \parallel u\parallel \le \frac{\parallel \omega \parallel }{2c}{\parallel u\parallel }_A^2.$$

Therefore, for each $t\ge 0$ and $u\in 𝒟(A)$,

$${\parallel U_{t}u\parallel }_A\le \sqrt{\frac{\parallel \omega \parallel }{2c}}{\parallel u\parallel }_A.$$

7 Hilbert spaces

For a real vector space $V$, a complex struture on $V$ is a linear map $J:V\to V$ such that $J^2=-I$. For $v\in V$, define $iv=Jv\in V$, for which on the one hand,

$(\alpha +i\beta )(\gamma +i\delta )v$ $=(\alpha +i\beta )(\gamma v+\delta Jv)$

$=\alpha \gamma v+\alpha \delta Jv+J(\beta \gamma v)+J(\beta \delta Jv)$

$=\alpha \gamma v+(\alpha \delta +\beta \gamma )Jv+\beta \delta J^{2}v$

$=(\alpha \gamma -\beta \delta )v+(\alpha \delta +\beta \gamma )Jv,$

and on the other hand,

$$(\alpha +i\beta )(\gamma +i\delta )v=(\alpha \gamma -\beta \delta +(\alpha \delta +\beta \gamma )i)v.$$

It follows that $V$ with $iv=Jv$ is a complex vector space. We emphasize that the complex vector space $V$ contains the same elements as the real vector space $V$. The following theorem connects symplectic forms, real inner products, and complex inner products. By a complex inner product on a complex vector space $W$, we mean a function $h:W\times W\to ℂ$ that is conjugate symmetric, complex linear in the first argument, $h(w,w)\ge 0$ for all $w\in W$, and $h(w,w)=0$ implies $w=0$.

Theorem 2.

Let $H$ be a real Hilbert space with inner product $⟨\cdot ,\cdot ⟩:H\times H\to ℝ$ and let $\omega$ be a weak symplectic form on $H$. Then there is a complex structure $J:H\to H$ and a real inner product $s$ on $H$ such that

$$s(x,y)=-\omega (Jx,y),x,y\in H$$

is a real inner product on the real vector space $H$, and

$$h(x,y)=s(x,y)-i\omega (x,y),x,y\in H$$

is a complex inner product on $H$ with the complex structure $J$.

Furthermore, the following are equivalent:

1. The norm induced by $h$ is equivalent with the norm induced by $⟨\cdot ,\cdot ⟩$.

2. The norm induced by $s$ is equivalent with the norm induced by $⟨\cdot ,\cdot ⟩$.

3. $\omega$ is a strong symplectic form on the real Hilbert space $H$.

Proof.

By the Riesz representation theorem, because $\omega$ is a bounded bilinear form there is a unique $A\in ℒ(H)$ such that

$$\omega (x,y)=⟨Ax,y⟩,x,y\in H.$$ (1)

Because $\omega$ is skew-symmetric,

$$⟨Ax,y⟩=\omega (x,y)=-\omega (y,x)=-⟨Ay,x⟩=⟨(-A)y,x⟩.$$

On the other hand, because $⟨\cdot ,\cdot ⟩$ is a real inner product, $⟨Ax,y⟩=⟨x,A^{*}y⟩=⟨A^{*}y,x⟩$. Therefore $A^{*}=-A$.

$A^{*}A=(-A)A=-A^2$ and $A{A}^{*}=A(-A)=-A^2$, so $A$ is normal. Therefore $A$ has a polar decomposition: there is a unitary $U\in ℒ(H)$ and some $P\in ℒ(H)$ with $P\ge 0$, such that

$$A=UP,$$

and such that $A,U,P$ commute; a fortiori, $P$ is self-adjoint. If $Ax=0$, then $\omega (x,y)=⟨Ax,y⟩=⟨0,y⟩=0$ for all $y\in H$, and because $\omega$ is weakly nondegenerate this implies that $x=0$, hence $A$ is one-to-one, which implies that $P$ is one-to-one (this implication does not use that $U$ is unitary). We have

$$A^{*}={(UP)}^{*}=P^{*}{U}^{*}=P{U}^{*},A^{*}=-A=-UP=-PU,$$

hence

$$P{U}^{*}=P(-U).$$

Because $P$ is one-to-one, this yields $U^{*}=-U$. But $U$ is unitary, i.e. $U^{*}U=I$ and $U{U}^{*}=I$. Therefore $(-U)U=I$, i.e. $-U^2=I$. This means that $U$ is a complex structure on the real Hilbert space $H$. We write $J=U$.

The complex structure $J$ satisfies, for $x,y\in H$,

$$\omega (Jx,Jy)=⟨AJx,Jy⟩=⟨JAx,Jy⟩=⟨Ax,J^{*}Jy⟩=⟨Ax,y⟩=\omega (x,y),$$

showing that $J$ is a canonical transformation.

$s:H\times H\to ℝ$ is defined, for $x,y\in H$, by

$$s(x,y)=-\omega (Jx,y)=-⟨AJx,y⟩=⟨(-J)Ax,y⟩=⟨J^{-1}Ax,y⟩=⟨Px,y⟩.$$

It is apparent that $s$ is bilinear. Because $P$ is self-adjoint and $⟨\cdot ,\cdot ⟩$ is symmetric,

$$s(x,y)=⟨Px,y⟩=⟨x,Py⟩=⟨Py,x⟩=s(y,x),$$

showing that $s$ is symmetric. Because $P\ge 0$, for any $x\in H$ we have $s(x,x)=⟨Px,x⟩\ge 0$, namely $s$ is positive. Also because $P\ge 0$, there is a unique $S\in ℒ(H)$, $S\ge 0$, satisfying $S^2=P$. If $s(x,x)=0$, we get

$$0=⟨Px,x⟩=⟨S^{2}x,x⟩=⟨Sx,Sx⟩={\parallel Sx\parallel }^2,$$

hence $Sx=0$ and so $Px=0$, and because $P$ is one-to-one, $x=0$. Therefore $s$ is positive definite, and thus is a real inner product on $H$.

$h:H\times H\to ℂ$ is defined, for $x,y\in H$, by

$$h(x,y)=s(x,y)-i\omega (x,y)=⟨Px,y⟩-i\omega (x,y)=⟨Px,y⟩-i⟨Ax,y⟩.$$

For $x_1,x_2,y\in H$,

$$h(x_1+x_2,y)=h(x_1,y)+h(x_2,y).$$

For $\alpha +i\beta \in ℂ$,

$h((\alpha +i\beta )x,y)$ $=h(\alpha x+\beta Jx,y)$

$=h(\alpha x,y)+\beta h(Jx,y)$

$=\alpha h(x,y)+\beta ⟨PJx,y⟩-i\beta ⟨AJx,y⟩$

$=\alpha h(x,y)+\beta ⟨Ax,y⟩-i\beta ⟨A(-J^{-1})x,y⟩$

$=\alpha h(x,y)+\beta \omega (x,y)+i\beta ⟨Px,y⟩$

$=\alpha h(x,y)+\beta \omega (x,y)+i\beta s(x,y)$

$=\alpha h(x,y)+i\beta (s(x,y)-i\omega (x,y))$

$=\alpha h(x,y)+i\beta h(x,y)$

$=(\alpha +i\beta )h(x,y).$

Therefore $h$ is complex linear in its first argument. Because $s$ is symmetric and $\omega$ is skew-symmetric, $h(x,y)=s(x,y)-i\omega (x,y)$ satisfies

$$h(y,x)=s(y,x)-i\omega (y,x)=s(x,y)+i\omega (x,y)=\overline{h(x,y)},$$

showing that $h$ is conjugate symmetric. For $x\in H$,

$$h(x,x)=s(x,x)-i\omega (x,x)=s(x,x)\ge 0.$$

If $h(x,x)=0$, then $s(x,x)=0$, which implies that $x=0$. Therefore $h$ is a complex inner product on $H$ with the complex structure $J$.

Suppose that $\omega$ is a strong symplectic form on the real Hilbert space $H$. That is, ${\omega }^{\mathrm{♭}}:H\to H^{*}$ is an isomorphism of Banach spaces. We shall show that $A$, from (1), is onto. For $y\in H$, define $\lambda :H\to ℝ$ by $\lambda (x)=⟨x,y⟩$. Then $\lambda \in H^{*}$, so there is some $v\in H$ for which ${\omega }^{\mathrm{♭}}(v)=\lambda$. That is, $\omega (v,x)=\lambda (x)=⟨x,y⟩=⟨y,x⟩$ for all $x\in H$. But $\omega (v,x)=⟨Av,x⟩$, so $⟨Av,x⟩=⟨y,x⟩$ for all $x\in H$, which implies that $Av=y$, and thus shows that $A$ is onto, and hence invertible in $ℒ(H)$. Because $A=UP$ and $A,U$ are invertible in $ℒ(H)$, $P$ is invertible in $ℒ(H)$. Therefore $S$, $P=S^2$, $S\ge 0$, is invertible in $ℒ(H)$, whence

${\parallel x\parallel }^2$ $={\parallel S^{-1}Sx\parallel }^2$

$\le {\parallel S^{-1}\parallel }^2{\parallel Sx\parallel }^2$

$={\parallel S^{-1}\parallel }^2⟨Sx,Sx⟩$

$={\parallel S^{-1}\parallel }^2⟨Px,x⟩$

$={\parallel S^{-1}\parallel }^{2}s(x,x)$

$=\parallel S^{-1}\parallel {\parallel x\parallel }_s^2,$

and on the other hand

$${\parallel x\parallel }_s^2=s(x,x)=⟨Px,x⟩\le \parallel Px\parallel \parallel x\parallel \le \parallel P\parallel {\parallel x\parallel }^2={\parallel S\parallel }^2{\parallel x\parallel }^2.$$

so

$$\parallel x\parallel \le \parallel S^{-1}\parallel {\parallel x\parallel }_s,{\parallel x\parallel }_s\le \parallel S\parallel \parallel x\parallel .$$

Namely this establishes that the norms ${\parallel x\parallel }^2=⟨x,x⟩$ and ${\parallel x\parallel }_s^2=s(x,x)$ are equivalent. ∎

8 Hamiltonian vector fields

Let $E$ be a real Banach space and let $k\ge 1$; if we do not specify $k$ we merely suppose that it is $\ge 1$. A $C^k$ vector field on $U$, where $U$ an open subset of $E$, is a $C^k$ function $v:U\to E$.

Let $v$ be a $C^k$, $k\ge 1$, vector field on $E$. For $x\in E$, an integral curve of $v$ through $x$ is a differentiable function $\varphi :J\to E$, where $J$ is some open interval in $ℝ$ containing $0$, that satisfies

$${\varphi }^{\prime }(t)=(v\circ \varphi )(t),t\in J,\varphi (0)=x.$$

If $\psi :I\to E$ and $\varphi :J\to E$ are integral curves of $v$ through $x$, it is proved that for $t\in I\cap J$, $\psi (t)=\varphi (t)$. An integral curve of $v$ through $x$, $\varphi :J\to E$, is said to be maximal if there is no integral curve of $v$ through $x$ whose domain strictly includes $J$. If $X:E\to E$ is a $C^1$ vector field, for each $x\in E$ it is proved that there is a unique maximal integral curve of $v$ through $x$, denoted ${\varphi }_x:J_x\to E$. A vector field $v:E\to E$ is called complete when $J_x=ℝ$ for each $x\in E$. For a vector field $v:E\to E$, a $C^1$ function $f:E\to ℝ$ is called a first integral of $v$ if for any integral curve $\varphi :J\to E$ of $v$, $f\circ \varphi :J\to E$ is constant. It is proved that if a vector field has a first integral $f:E\to ℝ$ such that $f^{-1}(c)$ is a compact subset of $E$ for each $c\in ℝ$, then $v$ is a complete vector field.

The flow of $v$ is the function $\sigma :{\mathrm{\Sigma }}_v\to E$, where

$${\mathrm{\Sigma }}_v=\bigcup _{x\in E}{J}_x\times \{x\},$$

such that for each $x\in E$, $\sigma (t,x)={\varphi }_x(t)$, $t\in J_x$. It is proved that ${\mathrm{\Sigma }}_v$ is an open subset of $ℝ\times E$, and that $\sigma :{\mathrm{\Sigma }}_v\to E$ is continuous. It is also proved that for any $k\ge 1$, if $v$ is $C^k$ then $\sigma :{\mathrm{\Sigma }}_v\to E$ is $C^k$. If $(s,x),(t,\sigma (s,x)),(t+s,x)\in {\mathrm{\Sigma }}_v$, then

$$\sigma (t+s,x)=\sigma (t,\sigma (s,x)).$$

When $v$ is a complete vector field, its flow is called a global flow. In this case, for $t\in ℝ$ we define ${\sigma }_t:E\to E$ by ${\sigma }_t(x)=\sigma (t,x)$. Then ${\sigma }_t^{-1}={\sigma }_{-t}$, and thus each ${\sigma }_t$ is a $C^k$ diffeomorphism $E\to E$.

9 Differential forms

For vector spaces $V$ and $W$ and for $p\ge 1$, a function $f:V^p\to W$ is called alternating if $(v_1,\mathrm{\dots },v_p)\in V^p$ and $v_i=v_{i+1}$ for some $1\le i\le p-1$ imply that $f(v_1,\mathrm{\dots },v_p)=0$.

For Banach spaces $E$ and $F$ and for $p\ge 1$, we denote by ${𝒜}_p(E;F)$ the set of alternating elements of ${ℒ}_p(E;F)$. In particular, ${𝒜}_1(E;F)={ℒ}_1(E;F)=ℒ(E;F)$. ${𝒜}_p(E;F)$ is a closed linear subspace of the Banach space ${ℒ}_p(E;F)$. We define

$${𝒜}_0(E;F)={ℒ}_0(E;F)=F.$$

Let ${\mathrm{\Sigma }}_n$ be the set of permutation $\{1,\mathrm{\dots },n\}$, which has $n!$ elements. Let ${\mathrm{Sh}}_{p,q}$ be the set of permutations $\sigma$ of $\{1,\mathrm{\dots },p,p+1,\mathrm{\dots },p+q\}$ for which

The set ${\mathrm{Sh}}_{p,q}$ has $\left(\genfrac{}{}{0pt}{}{p+q}{p}\right)=\left(\genfrac{}{}{0pt}{}{p+q}{q}\right)$ elements.

For $f\in {𝒜}_p(E;ℝ)$ and $g\in {𝒜}_q(E;ℝ)$, we define $f\wedge g:E^p\times E^q\to ℝ$ by

It is proved that $f\wedge g\in {𝒜}_{p+q}(E;ℝ)$.

For $f\in {𝒜}_p(E;ℝ)$ and $g\in {𝒜}_q(E;ℝ)$,

$\parallel f\wedge g\parallel$ $=\underset{\parallel x_1\parallel \le 1,\mathrm{\dots },\parallel x_{p+q}\parallel \le 1}{sup}|(f\wedge g)(x_1,\mathrm{\dots },x_p,x_{p+1},\mathrm{\dots },x_{p+q})|$

$\le \underset{\parallel x_1\parallel \le 1,\mathrm{\dots },\parallel x_{p+q}\parallel \le 1}{sup}{\displaystyle \sum _{\sigma \in {\mathrm{Sh}}_{p,q}}}|f(x_{\sigma (1)},\mathrm{\dots },x_{\sigma (p)})g(x_{\sigma (p+1)},\mathrm{\dots },x_{\sigma (p+q)})|$

$\le \underset{\parallel x_1\parallel \le 1,\mathrm{\dots },\parallel x_{p+q}\parallel \le 1}{sup}{\displaystyle \sum _{\sigma \in {\mathrm{Sh}}_{p,q}}}\parallel f\parallel \parallel g\parallel$

$=\left({\displaystyle \genfrac{}{}{0pt}{}{p+q}{p}}\right)\parallel f\parallel \parallel g\parallel ,$

showing that the operator norm of the bilinear map $(f,g)\mapsto f\wedge g$, ${𝒜}_p(E;ℝ)\times {𝒜}_q(E;ℝ)$ is $\le \left(\genfrac{}{}{0pt}{}{p+q}{p}\right)$, and thus is continuous.

One proves that for $f\in {𝒜}_p(E;ℝ)$ and $g\in {𝒜}_q(E;ℝ)$, then

$$g\wedge f={(-1)}^{pq}f\wedge g.$$

It is also proved that for $f\in {𝒜}_p(E;ℝ)$, $g\in {𝒜}_q(E;ℝ)$, and $h\in {𝒜}_r(E;ℝ)$, then

$$(f\wedge g)\wedge h=f\wedge (g\wedge h).$$

It thus makes sense to speak about $f_1\wedge \mathrm{\cdots }\wedge f_n$. We remind ourselves that ${𝒜}_1(E;ℝ)=ℒ(E;ℝ)=E^{*}$. It is proved that if $f_1,\mathrm{\dots },f_n\in E^{*}$, then $f_1\wedge \mathrm{\cdots }\wedge f_n\in {𝒜}_n(E;ℝ)$ satisfies

$$f_1\wedge \mathrm{\cdots }\wedge f_n(x_1,\mathrm{\dots },x_n)=\sum _{\sigma \in {\mathrm{\Sigma }}_n}\mathrm{sgn}(\sigma )f_1(x_{\sigma (1)})\mathrm{\cdots }{f}_n(x_{\sigma (n)}),(x_1,\mathrm{\dots },x_n)\in E^n,$$

and that $f_1,\mathrm{\dots },f_n\in E^{*}$ are linearly independent if and only if $f_1\wedge \mathrm{\cdots }\wedge f_n=0$.

Let $U$ be an open subset of the Banach space $E$. For $k\ge 0$ and $p\ge 0$, a $C^k$ differential form of degree $p$ on $U$ is a $C^k$ function

$$\alpha :U\to {𝒜}_p(E;ℝ).$$

We abbreviate “differential form of degree $p$” as “differential $p$-form”. In particular, a $C^k$ differential $0$-form is a $C^k$ function $U\to {𝒜}_0(E;ℝ)=ℝ$. We denote by ${\mathrm{\Omega }}_p^{(k)}(U,ℝ)$ the set of $C^k$ differential $p$-forms on $U$. It is apparent that this is a real vector space.

For a $C^k$ function $f:U\to ℝ$, with $k\ge 1$, the derivative $f^{\prime }$ is $C^{k-1}$ function $U\to ℒ(E;ℝ)={𝒜}_1(E;ℝ)$, hence $f^{\prime }\in {\mathrm{\Omega }}_p^{(k-1)}(U)$.

For $\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ)$ and $\beta \in {\mathrm{\Omega }}_q^{(k)}(U,ℝ)$, we define $\alpha \wedge :U\to {𝒜}_{p+q}(E;ℝ)$ by

$$(\alpha \wedge \beta )(x)=(\alpha (x))\wedge (\beta (x)),x\in U.$$

It is proved that $\alpha \wedge \beta \in {\mathrm{\Omega }}_{p+q}^{(k)}(U,ℝ)$.

Suppose that $k\ge 1$ and $\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ)$, i.e. $\alpha :U\to {𝒜}_p(U;ℝ)$ is a $C^k$ function. Then the derivative is the $C^{k-1}$ function

$${\alpha }^{\prime }:U\to ℒ(E;{𝒜}_p(E;ℝ).$$

We define $d\alpha :U\to {𝒜}_{p+1}(E;ℝ)$ by

$$(d\alpha )(x)({\xi }_0,{\xi }_1,\mathrm{\dots },{\xi }_p)=\sum _{i=0}^p{(-1)}^i{\alpha }^{\prime }(x)({\xi }_i)({\xi }_0,\mathrm{\dots },{\widehat{\xi }}_i,\mathrm{\dots },{\xi }_p)$$

It is proved that $d\alpha \in {\mathrm{\Omega }}_{p+1}^{(k-1)}(U,ℝ)$.

In particular, if $f:U\to ℝ$ is a $C^k$ function, then $df\in {\mathrm{\Omega }}_1^{(k-1)}(U,ℝ)$ is the function $df:U\to {𝒜}_1(E;ℝ)=ℒ(E;ℝ)$ defined by

$$(df)(x)(\xi )=f^{\prime }(x)(\xi ),x\in U,\xi \in E.$$

Thus, $df=f^{\prime }$.

For $\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ)$ and $\beta \in {\mathrm{\Omega }}_q^{(k)}(U,ℝ)$ with $k\ge 1$, it is a fact that

$$d(\alpha \wedge \beta )=(d\alpha )\wedge \beta +{(-1)}^p\alpha \wedge (d\beta ).$$

In particular, an element $f$ of ${\mathrm{\Omega }}_0^{(k)}(U,ℝ)$ is a $C^k$ function $U\to ℝ$, for which, because $f\wedge \beta =f\beta$,

$$d(f\beta )=(df)\wedge \beta +f(d\beta ).$$

For $\alpha \in {\mathrm{\Omega }}_p^{(}k)(U,ℝ)$, with $k\ge 2$,

$$d(d\alpha )=0.$$

Let $\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ)$, let $V$ be an open subset of a Banach space $F$, and let $\varphi :V\to U$ be a $C^{k+1}$ function. The the pullback of $\alpha$ by $f$, denoted ${\varphi }^{*}\alpha :V\to {𝒜}_p(F;ℝ)$, is an element of ${\mathrm{\Omega }}_p^{(k)}(V,ℝ)$ satisfying

$$({\varphi }^{*}\alpha )(y)({\eta }_1,\mathrm{\dots },{\eta }_p)=\alpha (\varphi (y))({\varphi }^{\prime }(y)({\eta }_1),\mathrm{\dots },{\varphi }^{\prime }(y)({\eta }_p)),({\eta }_1,\mathrm{\dots },{\eta }_p)\in F^p.$$

The pullback satisfies, for $\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ)$ and $\beta \in {\mathrm{\Omega }}_q^{(k)}(U,ℝ)$,

$${\varphi }^{*}(\alpha \wedge \beta )=({\varphi }^{*}\alpha )\wedge ({\varphi }^{*}\beta ),$$

which is an element of ${\mathrm{\Omega }}_{p+q}^{(k)}(V,ℝ)$. It also satisfies, if $\varphi :V\to U$ and $f:U\to ℝ$ are $C^1$,

$${\varphi }^{*}(df)=d({\varphi }^{*}f),$$

where $({\varphi }^{*}f)(y)=f(\varphi (y))$.

10 Contractions and Lie derivatives

Let $U$ be an open subset of a Banach space $E$, let $k\ge 1$, $p\ge 1$, let $v$ be a $C^k$ vector field on $U$, and let $\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ)$. We define ${\iota }_v\alpha :U\to {𝒜}_{p-1}(E;ℝ)$ by

$$({\iota }_v\alpha )(x)(v_1,\mathrm{\dots },v_{p-1})=\alpha (v(x),v_1,\mathrm{\dots },v_{p-1}),(v_1,\mathrm{\dots },v_{p-1})\in E^{p-1}.$$

(It is straightforward to check that indeed $({\iota }_v\alpha )(x)\in {𝒜}_{p-1}(E;ℝ)$.) It is proved that ${\iota }_v\alpha :U\to {𝒜}_{p-1}(E;ℝ)$ is $C^k$, and thus ${\iota }_v\alpha \in {\mathrm{\Omega }}_{p-1}^{(k)}(U,ℝ)$. For $p=0$, with $f\in {\mathrm{\Omega }}_0^{(k)}(U,ℝ)$, i.e. $f$ is a $C^k$ function $U\to ℝ$, we define ${\iota }_{v}f=0$. We call ${\iota }_v\alpha$ the contraction of $\alpha$ by $v$.

It can be proved that if $\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ)$ and $\beta \in {\mathrm{\Omega }}_q^{(k)}(U,ℝ)$,

$${\iota }_v(\alpha \wedge \beta )=({\iota }_v\alpha )\wedge \beta +{(-1)}^p\alpha \wedge {\iota }_v\beta .$$

Also, for a $C^k$ vector field $w$ on $U$,

$${\iota }_v({\iota }_w\alpha )=-{\iota }_w({\iota }_v\alpha ),$$

and hence ${\iota }_v^2\alpha =0$. And $(v,\alpha )\mapsto {\iota }_v\alpha$ is bilinear.

For a $C^k$ vector field $v$ on $U$ and $\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ)$, the Lie derivative of $\alpha$ with respect to $v$ is

$${ℒ}_v\alpha =d({\iota }_v\alpha )+{\iota }_{v}d\alpha \in {\mathrm{\Omega }}_p^{(k)}(U,ℝ).$$

The Lie derivative satisfies

$${ℒ}_v(\alpha \wedge \beta )=({ℒ}_v\alpha )\wedge \beta +\alpha \wedge {ℒ}_v\beta .$$

If $\omega$ is a weak symplectic form on a Banach space $E$ and $v$ is a $C^1$ vector field on $E$, we say that $v$ is a symplectic vector field if

$${ℒ}_v\omega =0.$$

If there is some $C^1$ function $H:E\to E$ such that

$${\iota }_v\omega =-dH,$$

we say that $v$ is a Hamiltonian vector field with Hamiltonian function $H$. If $v$ is a Hamiltonian vector field with Hamiltonian function $H$, then

$${ℒ}_v\omega =d({\iota }_v\omega )+{\iota }_{v}d\omega =d({\iota }_v\omega )=d(-dH)=-d^{2}H=0,$$

showing that if a vector field is Hamiltonian then it is symplectic. (This is analogous to the statement that if a differential form is exact then it is closed.)