Topological vector space completion with respect to a symplectic form?
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Suppose we have an infinite-dimensional vector space $V$ with a symplectic form $omega:Vtimes Vtomathbb R$. It can be given a weak topology that makes $omega$ continuous.
Does it make sense to complete $V$ with respect to this topology? (Should something stronger be used?)
Will $omega$ still be non-degenerate on the completion? (Meaning: if $forall a,omega(a,b)=0$, then $b=0$.) ...Obviously, yes. In fact, even if $omega$ was originally degenerate on $V$, the process of completion would "quotient out" the degenerate subspace, as any sequence in it converges to $0$ according to $omega$.
Is $omega$ necessarily "strongly symplectic", meaning $amapstoomega(a,cdot)$ is a bijection between $V$ and its topological dual $V^*$? If not, can we make it so by changing the topology on $V$ (thus changing $V^*$)?
In general, I'm wondering what the conditions and implications are for a symplectic space to be complete and self-dual, similar to a Hilbert space (but without a norm).
general-topology functional-analysis topological-vector-spaces complete-spaces symplectic-linear-algebra
asked Nov 18 at 11:37
mr_e_man
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down vote
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Suppose we have an infinite-dimensional vector space $V$ with a symplectic form $omega:Vtimes Vtomathbb R$. It can be given a weak topology that makes $omega$ continuous.
Does it make sense to complete $V$ with respect to this topology? (Should something stronger be used?)
Will $omega$ still be non-degenerate on the completion? (Meaning: if $forall a,omega(a,b)=0$, then $b=0$.) ...Obviously, yes. In fact, even if $omega$ was originally degenerate on $V$, the process of completion would "quotient out" the degenerate subspace, as any sequence in it converges to $0$ according to $omega$.
Is $omega$ necessarily "strongly symplectic", meaning $amapstoomega(a,cdot)$ is a bijection between $V$ and its topological dual $V^*$? If not, can we make it so by changing the topology on $V$ (thus changing $V^*$)?
In general, I'm wondering what the conditions and implications are for a symplectic space to be complete and self-dual, similar to a Hilbert space (but without a norm).
general-topology functional-analysis topological-vector-spaces complete-spaces symplectic-linear-algebra
asked Nov 18 at 11:37
mr_e_man
1,2351423
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have an infinite-dimensional vector space $V$ with a symplectic form $omega:Vtimes Vtomathbb R$. It can be given a weak topology that makes $omega$ continuous.
Does it make sense to complete $V$ with respect to this topology? (Should something stronger be used?)
Will $omega$ still be non-degenerate on the completion? (Meaning: if $forall a,omega(a,b)=0$, then $b=0$.) ...Obviously, yes. In fact, even if $omega$ was originally degenerate on $V$, the process of completion would "quotient out" the degenerate subspace, as any sequence in it converges to $0$ according to $omega$.
Is $omega$ necessarily "strongly symplectic", meaning $amapstoomega(a,cdot)$ is a bijection between $V$ and its topological dual $V^*$? If not, can we make it so by changing the topology on $V$ (thus changing $V^*$)?
In general, I'm wondering what the conditions and implications are for a symplectic space to be complete and self-dual, similar to a Hilbert space (but without a norm).
general-topology functional-analysis topological-vector-spaces complete-spaces symplectic-linear-algebra
asked Nov 18 at 11:37
mr_e_man
1,2351423
Suppose we have an infinite-dimensional vector space $V$ with a symplectic form $omega:Vtimes Vtomathbb R$. It can be given a weak topology that makes $omega$ continuous.
Does it make sense to complete $V$ with respect to this topology? (Should something stronger be used?)
Will $omega$ still be non-degenerate on the completion? (Meaning: if $forall a,omega(a,b)=0$, then $b=0$.) ...Obviously, yes. In fact, even if $omega$ was originally degenerate on $V$, the process of completion would "quotient out" the degenerate subspace, as any sequence in it converges to $0$ according to $omega$.
Is $omega$ necessarily "strongly symplectic", meaning $amapstoomega(a,cdot)$ is a bijection between $V$ and its topological dual $V^*$? If not, can we make it so by changing the topology on $V$ (thus changing $V^*$)?
In general, I'm wondering what the conditions and implications are for a symplectic space to be complete and self-dual, similar to a Hilbert space (but without a norm).
general-topology functional-analysis topological-vector-spaces complete-spaces symplectic-linear-algebra
general-topology functional-analysis topological-vector-spaces complete-spaces symplectic-linear-algebra
asked Nov 18 at 11:37
mr_e_man
1,2351423
asked Nov 18 at 11:37
mr_e_man
1,2351423
asked Nov 18 at 11:37
mr_e_man
1,2351423
asked Nov 18 at 11:37
mr_e_man
1,2351423
asked Nov 18 at 11:37
mr_e_man
1,2351423
1,2351423
add a comment |
add a comment |
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