Timeline for Splitting of a topological vector space (TVS) into an (a) countable sum and (b) direct integral of subspaces
Current License: CC BY-SA 4.0
18 events
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| Jul 26, 2022 at 15:47 | comment | added | Michael_1812 | @GeraldEdgar Thank you! This is really helpful. Could you please give me a reference on this? | |
| Jul 26, 2022 at 14:02 | comment | added | Gerald Edgar | The $l^1$ direct sum takes sequences $v_i \in V_i$ with $\sum \|v_i\|<+\infty$, as described by @JochenWengenroth . Some Banach spaces can be split that way, but not Hilbert space. For Hilbert space, take the $l^2$ direct sum where you use $\sum \|v_i\|^2 < + \infty$. | |
| Jul 26, 2022 at 13:23 | comment | added | Michael_1812 | @GeraldEdgar Now I am lost. Above, you told me that, yes, a Hilbert space permits for a countable direct sum $\oplus V_i$ consisting of sequences $v_i\in V_i$ with $\sum||v_i||^2 < \infty$. And now you are saying that a Hilbert space is never an infinite direct sum in the $l^1$ sense. What is $l^1$ sense? Also, is my understanding correct that a Banach space can be split into an infinite sum consisting of vectors obeying $\sum||v_i||^2 < \infty$ ? (For the needs of a physicist that would be sufficient.) | |
| Jul 26, 2022 at 13:14 | comment | added | Michael_1812 | @JochenWengenroth Could you please advise me a reference on this? Preferably, a simple textbook, not a serious monograph. | |
| Jul 26, 2022 at 10:18 | comment | added | Gerald Edgar | ... so Hilbert space is never the infinite direct sum $\oplus V_i$ in the $l^1$ sense mentioned by @JochenWengenroth . A nontrivial infinite $l^1$ direct sum is never reflexive. Von Neumann's "direct integral" was not so much for Hilbert spaces alone but more for operators on Hilbert space. | |
| Jul 26, 2022 at 7:39 | comment | added | Jochen Wengenroth | The Hilbert space direct sum described by Gerald Edgar is the categrorical direct sum for the category of Hilbert spaces and linear contractions as morphisms (sometimescalled short maps), this works similarly in the category of Banach spaces and linear contractions where $\bigoplus V_i$ is the spaces of $(v_i)_{i\in I}$ such that $\sum_i \|v_i\| <\infty$. | |
| Jul 26, 2022 at 2:02 | comment | added | Michael_1812 | @GeraldEdgar Thanks for your comment. And yes, I also saw direct integral only in the Hilbert space context. So my Question A remains in force, for a general-type Banach space. | |
| Jul 26, 2022 at 1:58 | history | edited | Michael_1812 | CC BY-SA 4.0 |
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| Jul 26, 2022 at 0:34 | comment | added | Gerald Edgar | (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. | |
| Jul 25, 2022 at 23:47 | comment | added | Michael_1812 | @user7427029 Thank you for the link. I see that an infinite-dim Banach does not possess a countable Hamel basis. Does it always posses an uncountable one? If yes, would this warrant that a Banach can always split into a direct integral of subspaces? | |
| Jul 25, 2022 at 23:37 | comment | added | user7427029 | Maybe of interest to you: planetmath.org/… | |
| Jul 25, 2022 at 22:44 | comment | added | Michael_1812 | @JochenWengenroth Many thanks! Both of your comments are extremely useful. You are saying that an infinite-dimensional Banach space will never split into a countable sum sum of spaces ${\mathbb V}_i$. Are you talking about a Banach space of uncountable dimensions? Can it be split into an uncountable sum? Also, how about a countable-dim Banach space? | |
| Jul 25, 2022 at 22:38 | history | edited | Michael_1812 | CC BY-SA 4.0 |
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| Jul 25, 2022 at 22:30 | comment | added | Michael_1812 | @abx My statement about the map was indeed obscure, but fortunately, Jochen Wengenroth in his comment explained this point in more detail. I will now make a correction in my question, to clarify this. | |
| Jul 25, 2022 at 14:07 | comment | added | Jochen Wengenroth | With this interpretation, infinite dimensional Banach spaces never split into a (countable) sum of $V_i\neq \{0\}$. | |
| Jul 25, 2022 at 14:07 | comment | added | Jochen Wengenroth | 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.) ... | |
| Jul 25, 2022 at 13:46 | comment | added | abx | Could you explain what is "the map $\{\mathbb{V}_i\}_{i\in I}\rightarrow \mathbb{V} $"? | |
| Jul 25, 2022 at 13:22 | history | asked | Michael_1812 | CC BY-SA 4.0 |