$ell^2$ as colimit in $mathbf{TopVect}_{mathbb{R}}$












1












$begingroup$


Let $mathbf{TopVect}_{mathbb{R}}$ be the category of topological vector spaces with continuous linear maps as morphisms. Is it ineed true that $ell^2 cong varinjlim_{n}oplus_{i=1}^nmathbb{R}$?






vector-spaces category-theory lp-spaces topological-vector-spaces limits-colimits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why $ell^2$ and not $ell^1$ then?
    $endgroup$
    – Mindlack
    Dec 17 '18 at 13:02










  • $begingroup$
    I've removed the tag "abelian-categories" because the category of topological vector spaces is not abelian.
    $endgroup$
    – Arnaud D.
    Dec 17 '18 at 13:11










  • $begingroup$
    True, I made that mistake... also I put $ell^2$ because both are Hilbert...so it seemed like the natural choice for an ansatz
    $endgroup$
    – N00ber
    Dec 17 '18 at 13:53








  • 1




    $begingroup$
    I suppose you mean $$limlimits_{substack{longrightarrow\ {color{red}n}}}oplus_{i=1}^nBbb R.$$
    $endgroup$
    – user593746
    Dec 17 '18 at 14:52






  • 1




    $begingroup$
    You can prove that the claim is false on purely categorical grounds: write down the colimit of the functor that sends $n$ to $oplus_{i=1}^nmathbb{R}$; it is a quotient of an obvious disjoint union, and will be a vector space of uncountable dimension.
    $endgroup$
    – Matematleta
    Dec 17 '18 at 16:47


















1












$begingroup$


Let $mathbf{TopVect}_{mathbb{R}}$ be the category of topological vector spaces with continuous linear maps as morphisms. Is it ineed true that $ell^2 cong varinjlim_{n}oplus_{i=1}^nmathbb{R}$?






vector-spaces category-theory lp-spaces topological-vector-spaces limits-colimits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why $ell^2$ and not $ell^1$ then?
    $endgroup$
    – Mindlack
    Dec 17 '18 at 13:02










  • $begingroup$
    I've removed the tag "abelian-categories" because the category of topological vector spaces is not abelian.
    $endgroup$
    – Arnaud D.
    Dec 17 '18 at 13:11










  • $begingroup$
    True, I made that mistake... also I put $ell^2$ because both are Hilbert...so it seemed like the natural choice for an ansatz
    $endgroup$
    – N00ber
    Dec 17 '18 at 13:53








  • 1




    $begingroup$
    I suppose you mean $$limlimits_{substack{longrightarrow\ {color{red}n}}}oplus_{i=1}^nBbb R.$$
    $endgroup$
    – user593746
    Dec 17 '18 at 14:52






  • 1




    $begingroup$
    You can prove that the claim is false on purely categorical grounds: write down the colimit of the functor that sends $n$ to $oplus_{i=1}^nmathbb{R}$; it is a quotient of an obvious disjoint union, and will be a vector space of uncountable dimension.
    $endgroup$
    – Matematleta
    Dec 17 '18 at 16:47
















1












1








1





$begingroup$


Let $mathbf{TopVect}_{mathbb{R}}$ be the category of topological vector spaces with continuous linear maps as morphisms. Is it ineed true that $ell^2 cong varinjlim_{n}oplus_{i=1}^nmathbb{R}$?






vector-spaces category-theory lp-spaces topological-vector-spaces limits-colimits







share|cite|improve this question











$endgroup$




Let $mathbf{TopVect}_{mathbb{R}}$ be the category of topological vector spaces with continuous linear maps as morphisms. Is it ineed true that $ell^2 cong varinjlim_{n}oplus_{i=1}^nmathbb{R}$?






vector-spaces category-theory lp-spaces topological-vector-spaces limits-colimits




vector-spaces category-theory lp-spaces topological-vector-spaces limits-colimits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 14:56







user593746

















asked Dec 17 '18 at 13:01









N00berN00ber

1629




1629












  • $begingroup$
    Why $ell^2$ and not $ell^1$ then?
    $endgroup$
    – Mindlack
    Dec 17 '18 at 13:02










  • $begingroup$
    I've removed the tag "abelian-categories" because the category of topological vector spaces is not abelian.
    $endgroup$
    – Arnaud D.
    Dec 17 '18 at 13:11










  • $begingroup$
    True, I made that mistake... also I put $ell^2$ because both are Hilbert...so it seemed like the natural choice for an ansatz
    $endgroup$
    – N00ber
    Dec 17 '18 at 13:53








  • 1




    $begingroup$
    I suppose you mean $$limlimits_{substack{longrightarrow\ {color{red}n}}}oplus_{i=1}^nBbb R.$$
    $endgroup$
    – user593746
    Dec 17 '18 at 14:52






  • 1




    $begingroup$
    You can prove that the claim is false on purely categorical grounds: write down the colimit of the functor that sends $n$ to $oplus_{i=1}^nmathbb{R}$; it is a quotient of an obvious disjoint union, and will be a vector space of uncountable dimension.
    $endgroup$
    – Matematleta
    Dec 17 '18 at 16:47




















  • $begingroup$
    Why $ell^2$ and not $ell^1$ then?
    $endgroup$
    – Mindlack
    Dec 17 '18 at 13:02










  • $begingroup$
    I've removed the tag "abelian-categories" because the category of topological vector spaces is not abelian.
    $endgroup$
    – Arnaud D.
    Dec 17 '18 at 13:11










  • $begingroup$
    True, I made that mistake... also I put $ell^2$ because both are Hilbert...so it seemed like the natural choice for an ansatz
    $endgroup$
    – N00ber
    Dec 17 '18 at 13:53








  • 1




    $begingroup$
    I suppose you mean $$limlimits_{substack{longrightarrow\ {color{red}n}}}oplus_{i=1}^nBbb R.$$
    $endgroup$
    – user593746
    Dec 17 '18 at 14:52






  • 1




    $begingroup$
    You can prove that the claim is false on purely categorical grounds: write down the colimit of the functor that sends $n$ to $oplus_{i=1}^nmathbb{R}$; it is a quotient of an obvious disjoint union, and will be a vector space of uncountable dimension.
    $endgroup$
    – Matematleta
    Dec 17 '18 at 16:47


















$begingroup$
Why $ell^2$ and not $ell^1$ then?
$endgroup$
– Mindlack
Dec 17 '18 at 13:02




$begingroup$
Why $ell^2$ and not $ell^1$ then?
$endgroup$
– Mindlack
Dec 17 '18 at 13:02












$begingroup$
I've removed the tag "abelian-categories" because the category of topological vector spaces is not abelian.
$endgroup$
– Arnaud D.
Dec 17 '18 at 13:11




$begingroup$
I've removed the tag "abelian-categories" because the category of topological vector spaces is not abelian.
$endgroup$
– Arnaud D.
Dec 17 '18 at 13:11












$begingroup$
True, I made that mistake... also I put $ell^2$ because both are Hilbert...so it seemed like the natural choice for an ansatz
$endgroup$
– N00ber
Dec 17 '18 at 13:53






$begingroup$
True, I made that mistake... also I put $ell^2$ because both are Hilbert...so it seemed like the natural choice for an ansatz
$endgroup$
– N00ber
Dec 17 '18 at 13:53






1




1




$begingroup$
I suppose you mean $$limlimits_{substack{longrightarrow\ {color{red}n}}}oplus_{i=1}^nBbb R.$$
$endgroup$
– user593746
Dec 17 '18 at 14:52




$begingroup$
I suppose you mean $$limlimits_{substack{longrightarrow\ {color{red}n}}}oplus_{i=1}^nBbb R.$$
$endgroup$
– user593746
Dec 17 '18 at 14:52




1




1




$begingroup$
You can prove that the claim is false on purely categorical grounds: write down the colimit of the functor that sends $n$ to $oplus_{i=1}^nmathbb{R}$; it is a quotient of an obvious disjoint union, and will be a vector space of uncountable dimension.
$endgroup$
– Matematleta
Dec 17 '18 at 16:47






$begingroup$
You can prove that the claim is false on purely categorical grounds: write down the colimit of the functor that sends $n$ to $oplus_{i=1}^nmathbb{R}$; it is a quotient of an obvious disjoint union, and will be a vector space of uncountable dimension.
$endgroup$
– Matematleta
Dec 17 '18 at 16:47












2 Answers
2






active

oldest

votes


















3












$begingroup$

While Zvi's given a good answer for the exact question the OP asked. I suspect the OP might be interested in whether or not it is possible to view $ell^2$ as a colimit.



The problem with trying this with $mathbf{TopVect}_Bbb{R}$ is that it's too big. In order to end up with an uncountably infinite dimensional vector space, we need a restriction on the spaces allowed in the category that forces the space to be e.g. complete.



Hence, as one example of a category making $ell^2$ the colimit, we can take the category of real Hilbert spaces with unitary maps as morphisms.



Then if we have unitary maps $T_n : Bbb{R}^n to H$ for some Hilbert space $H$, we can define $T:ell^2 to H$ by $T(e_i) = T_i(e_i)$ (and this is the only possible valid definition of $T$). It's fairly obvious that this is well defined, since all of the maps $T_i$ are unitary, so we'll end up with ${T_i(e_i)}$ being a countably infinite orthonormal set in $H$.



It might be possible to show that $ell^2$ is the colimit in a slightly larger appropriate category, like Banach spaces with norm preserving maps, but I think this is enough already.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for showing how to fix the problem :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13



















5












$begingroup$

Any direct limit of $Bbb R^n$ is a countable dimensional vector space over $Bbb R$. But any $ell^p_{Bbb R}$ space for any $pge 1$ is uncountable dimensional, so it cannot be a direct limit of $Bbb R^n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the reference :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043905%2fell2-as-colimit-in-mathbftopvect-mathbbr%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

While Zvi's given a good answer for the exact question the OP asked. I suspect the OP might be interested in whether or not it is possible to view $ell^2$ as a colimit.



The problem with trying this with $mathbf{TopVect}_Bbb{R}$ is that it's too big. In order to end up with an uncountably infinite dimensional vector space, we need a restriction on the spaces allowed in the category that forces the space to be e.g. complete.



Hence, as one example of a category making $ell^2$ the colimit, we can take the category of real Hilbert spaces with unitary maps as morphisms.



Then if we have unitary maps $T_n : Bbb{R}^n to H$ for some Hilbert space $H$, we can define $T:ell^2 to H$ by $T(e_i) = T_i(e_i)$ (and this is the only possible valid definition of $T$). It's fairly obvious that this is well defined, since all of the maps $T_i$ are unitary, so we'll end up with ${T_i(e_i)}$ being a countably infinite orthonormal set in $H$.



It might be possible to show that $ell^2$ is the colimit in a slightly larger appropriate category, like Banach spaces with norm preserving maps, but I think this is enough already.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for showing how to fix the problem :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13
















3












$begingroup$

While Zvi's given a good answer for the exact question the OP asked. I suspect the OP might be interested in whether or not it is possible to view $ell^2$ as a colimit.



The problem with trying this with $mathbf{TopVect}_Bbb{R}$ is that it's too big. In order to end up with an uncountably infinite dimensional vector space, we need a restriction on the spaces allowed in the category that forces the space to be e.g. complete.



Hence, as one example of a category making $ell^2$ the colimit, we can take the category of real Hilbert spaces with unitary maps as morphisms.



Then if we have unitary maps $T_n : Bbb{R}^n to H$ for some Hilbert space $H$, we can define $T:ell^2 to H$ by $T(e_i) = T_i(e_i)$ (and this is the only possible valid definition of $T$). It's fairly obvious that this is well defined, since all of the maps $T_i$ are unitary, so we'll end up with ${T_i(e_i)}$ being a countably infinite orthonormal set in $H$.



It might be possible to show that $ell^2$ is the colimit in a slightly larger appropriate category, like Banach spaces with norm preserving maps, but I think this is enough already.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for showing how to fix the problem :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13














3












3








3





$begingroup$

While Zvi's given a good answer for the exact question the OP asked. I suspect the OP might be interested in whether or not it is possible to view $ell^2$ as a colimit.



The problem with trying this with $mathbf{TopVect}_Bbb{R}$ is that it's too big. In order to end up with an uncountably infinite dimensional vector space, we need a restriction on the spaces allowed in the category that forces the space to be e.g. complete.



Hence, as one example of a category making $ell^2$ the colimit, we can take the category of real Hilbert spaces with unitary maps as morphisms.



Then if we have unitary maps $T_n : Bbb{R}^n to H$ for some Hilbert space $H$, we can define $T:ell^2 to H$ by $T(e_i) = T_i(e_i)$ (and this is the only possible valid definition of $T$). It's fairly obvious that this is well defined, since all of the maps $T_i$ are unitary, so we'll end up with ${T_i(e_i)}$ being a countably infinite orthonormal set in $H$.



It might be possible to show that $ell^2$ is the colimit in a slightly larger appropriate category, like Banach spaces with norm preserving maps, but I think this is enough already.






share|cite|improve this answer









$endgroup$



While Zvi's given a good answer for the exact question the OP asked. I suspect the OP might be interested in whether or not it is possible to view $ell^2$ as a colimit.



The problem with trying this with $mathbf{TopVect}_Bbb{R}$ is that it's too big. In order to end up with an uncountably infinite dimensional vector space, we need a restriction on the spaces allowed in the category that forces the space to be e.g. complete.



Hence, as one example of a category making $ell^2$ the colimit, we can take the category of real Hilbert spaces with unitary maps as morphisms.



Then if we have unitary maps $T_n : Bbb{R}^n to H$ for some Hilbert space $H$, we can define $T:ell^2 to H$ by $T(e_i) = T_i(e_i)$ (and this is the only possible valid definition of $T$). It's fairly obvious that this is well defined, since all of the maps $T_i$ are unitary, so we'll end up with ${T_i(e_i)}$ being a countably infinite orthonormal set in $H$.



It might be possible to show that $ell^2$ is the colimit in a slightly larger appropriate category, like Banach spaces with norm preserving maps, but I think this is enough already.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 19:50









jgonjgon

13.6k22041




13.6k22041












  • $begingroup$
    Thanks for showing how to fix the problem :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13


















  • $begingroup$
    Thanks for showing how to fix the problem :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13
















$begingroup$
Thanks for showing how to fix the problem :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13




$begingroup$
Thanks for showing how to fix the problem :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13











5












$begingroup$

Any direct limit of $Bbb R^n$ is a countable dimensional vector space over $Bbb R$. But any $ell^p_{Bbb R}$ space for any $pge 1$ is uncountable dimensional, so it cannot be a direct limit of $Bbb R^n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the reference :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13
















5












$begingroup$

Any direct limit of $Bbb R^n$ is a countable dimensional vector space over $Bbb R$. But any $ell^p_{Bbb R}$ space for any $pge 1$ is uncountable dimensional, so it cannot be a direct limit of $Bbb R^n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the reference :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13














5












5








5





$begingroup$

Any direct limit of $Bbb R^n$ is a countable dimensional vector space over $Bbb R$. But any $ell^p_{Bbb R}$ space for any $pge 1$ is uncountable dimensional, so it cannot be a direct limit of $Bbb R^n$.






share|cite|improve this answer











$endgroup$



Any direct limit of $Bbb R^n$ is a countable dimensional vector space over $Bbb R$. But any $ell^p_{Bbb R}$ space for any $pge 1$ is uncountable dimensional, so it cannot be a direct limit of $Bbb R^n$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 19:32

























answered Dec 17 '18 at 14:55







user593746



















  • $begingroup$
    Thank you for the reference :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13


















  • $begingroup$
    Thank you for the reference :)
    $endgroup$
    – N00ber
    Dec 20 '18 at 11:13
















$begingroup$
Thank you for the reference :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13




$begingroup$
Thank you for the reference :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043905%2fell2-as-colimit-in-mathbftopvect-mathbbr%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more