$ell^2$ 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

19511




19511












  • $begingroup$
    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











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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

14.7k22042




14.7k22042












  • $begingroup$
    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


















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