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Let $V,W$ be topological vector spaces and fix continuous antisymmetric bilinear forms $\omega_1:V\times V\to \mathbb{R}$, $\omega_2:W\times W\to\mathbb{R}$. Since $\omega_1$ is a 2-cocycle (in fact an alternating bihomomorphism) for $(V,+)$ we can form the group extension $\tilde{V}=(V\times\mathbb{R},\cdot)$, where $$(u,\alpha)\cdot(v,\beta)=(u+v,\alpha+\beta-\omega_1(u,v))$$ and $\tilde{V}$ just has the product topology. Similarly we can form an extension $\tilde{W}$ of $W$.

These arise as invariants in my work and I have a (rather tedious) proof of the following crucial result: $$ \tilde{V}\cong \tilde{W}~\Rightarrow~V\cong W \quad (\text{as topological groups}).$$


  • Are there some good references for this or similar results?

  • Is this result just a consequence of some general cohomological machinery?

Related to the first question:

  • Given abelian groups $G,H$, and extensions $\tilde{G},\tilde{H}$ by $K$ when does $\tilde{G}\cong\tilde{H}$ imply $G\cong H$?

My feeling is that this is a small chunk of a more general result, but I don't know enough group theory/cohomolgy to see it. For instance if $G,H$ are torsion-free abelian groups and $\tilde{G},\tilde{H}$ extensions of $G,H$ by alternating bihomomorphisms with values in $\mathbb{Z}$ then an almost identical proof to the TVS case shows that $\tilde{G}\cong\tilde{H}\Rightarrow G\cong H$

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Do you assume the $\omega_i$ to be non degenerate ? (aka symplectic) If yes, then the center of $\tilde{V}$ is simply $0\times\mathbb{R}$, and you recover $V$ by taking the quotient. Otherwise, the center is $K\times\mathbb{R}$ with $K$ the kernel of $\omega_1$ and the situation not so clear. – BS. Jun 6 '12 at 13:54
In the application the forms are non-degenerate, so your comment simplifies things considerably! However my proof (and hence the question) is for arbitrary $\omega_i$. – Ollie Margetts Jun 6 '12 at 14:19
Then, using commutator subgroup instead of center should do the trick when the $\omega_i$ are nonzero (i.e. the "tilde" groups are non abelian) : 0\times\mathbb{R} is then recovered as the commutator subgroup. If forms are $0$, you can only conclude that $V$ and $W$ are (closed) hyperplanes in the same TVS, and if they are not equal, they are isomorphic, considering the (continuous) automorphism that is identity on their intersection and exchanges the two vectors in a basis of a 2-dim complementary subspace. – BS. Jun 6 '12 at 14:31
Yes, you're right. It's much simpler than I had thought. Unless you want to post your comments as an answer I'll close it as no longer relevant. – Ollie Margetts Jun 6 '12 at 16:38

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