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I thought that this would be a simpe question, and placed it here at the Mathematics Stackexchange. Now have to elevate it to Mathoverflow.

LANGUAGE

TVS = topological vector space. Any subspace of a TVS is a TVS in the induced-topology sense.

DEFINITION

For TVS spaces ${\mathbb{V}}_1\subset\mathbb V$, a TVS subspace ${\mathbb{V}}_2\subset\mathbb V$ is a topological complement of ${\mathbb{V}}_1$ in $\mathbb V$, if $\mathbb V$ is their direct sum both algebraically and topologically. This implies that $$ {\mathbb{V}}_1\oplus {\mathbb{V}}_2\,=\,{\mathbb{V}}~\,, $$ and the following addition map is a homeomorphism: $$ {\mathbb{V}}_1\times{\mathbb{V}}_2\,\longrightarrow\,{\mathbb{V}}~~,\qquad \left(\, v\in{{\mathbb V}}_1\,,~\,v^{\,\prime}\in{{\mathbb V}}_2\,\right)\,\longmapsto~ (v+v^{\,\prime})\in\mathbb V~~.\\ $$

QUESTION A

For a COUNTABLY infinite splitting $$ {\mathbb V}=\bigoplus_{i\in\cal I}{\mathbb V}_i $$ to be not only algebraic but also topological, would it be sufficient to impose the condition that a natural map $$ \bigoplus \mathbb{V}_i\longrightarrow\mathbb{V} $$ exists and is a homeomorphism?

Here $\bigoplus \mathbb{V}_i$ is a set of all families $\left(\mathbb{v}_i\right)_{i\in\cal I}$ with $v_i\in {\mathbb V}_i$ and only finitely many non-zero $v_i$. $~~~~~\\\\$

QUESTION B

Would this work also for an uncountable sum (direct integral) of subspaces?

Stated alternatively, can we always be sure that there always exists a necessary measure on the set of these subspaces?

If there is no general answer to question B, can this question be answered for Hilbert spaces?$~~~~~\\\\$

KIND REQUEST

This question is of interest mainly to physicists. Could you please make your answer sufficiently detailed and, if possible, understandable to a layman? Thank you!

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    $\begingroup$ Could you explain what is "the map $\{\mathbb{V}_i\}_{i\in I}\rightarrow \mathbb{V} $"? $\endgroup$
    – abx
    Commented Jul 25, 2022 at 13:46
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    $\begingroup$ I guess that you mean by $\bigoplus V_i$ the set of all families $(v_i)_{i\in I}$ with $v_i\in V_i$ and only finitely many non-zero $v_i$. Then there is a natural map $\bigoplus V_i \to V$, $(v_i)_{i\in I}\mapsto \sum v_i$. If you want this map to be a homeomrphism, you need a topology on the direct sum. If you insist to work in TVS, the natural one would be the finest vector space topology on $\bigoplus V_i$ such that all inclusions $V_j\to \bigoplus V_i$, $v_j\mapsto (v_i)_{i\in I}$ with all $v_i=0$ for $i\neq j$ are continuous. (This is indeed the coproduct in the category TVS.) ... $\endgroup$
    – Jochen Wengenroth
    Commented Jul 25, 2022 at 14:07
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    $\begingroup$ With this interpretation, infinite dimensional Banach spaces never split into a (countable) sum of $V_i\neq \{0\}$. $\endgroup$
    – Jochen Wengenroth
    Commented Jul 25, 2022 at 14:07
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    $\begingroup$ Maybe of interest to you: planetmath.org/… $\endgroup$
    – user7427029
    Commented Jul 25, 2022 at 23:37
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    $\begingroup$ (a) For example in Hilbert space, you want countable direct sum decomposition $\oplus V_i$ to consist of sequences $v_i \in V_i$ with $\sum \|v_i\|^2 <+\infty$. (b) I do not know a definition of "direct integral" other than in Hilbert space. (c) Even in Hilbert space, you may want direct integral with respect to a measure, so that countable direct sum is automatically included, no need for separate cases. For help on this setting, see the mathematical formulation of the spectral theorem. $\endgroup$
    – Gerald Edgar
    Commented Jul 26, 2022 at 0:34

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