$ell^2$ as colimit in $mathbf{TopVect}_{mathbb{R}}$
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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
asked Dec 17 '18 at 13:01
N00berN00ber
326111
$endgroup$
add a comment |
$begingroup$
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
asked Dec 17 '18 at 13:01
N00berN00ber
326111
$endgroup$
$begingroup$
Why $ell^2$ and not $ell^1$ then?
$endgroup$
– Mindlack
Dec 17 '18 at 13:02
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I've removed the tag "abelian-categories" because the category of topological vector spaces is not abelian.
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– Arnaud D.
Dec 17 '18 at 13:11
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True, I made that mistake... also I put $ell^2$ because both are Hilbert...so it seemed like the natural choice for an ansatz
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– 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
add a comment |
$begingroup$
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
asked Dec 17 '18 at 13:01
N00berN00ber
326111
$endgroup$
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
asked Dec 17 '18 at 13:01
N00berN00ber
326111
asked Dec 17 '18 at 13:01
N00berN00ber
326111
edited Dec 17 '18 at 14:56
user593746
asked Dec 17 '18 at 13:01
N00berN00ber
326111
asked Dec 17 '18 at 13:01
N00berN00ber
326111
asked Dec 17 '18 at 13:01
N00berN00ber
326111
326111
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 17 '18 at 19:50
jgonjgon
15.6k32143
$endgroup$
$begingroup$
Thanks for showing how to fix the problem :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13
add a comment |
$begingroup$
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 $Bbb R^n$.
$endgroup$
$begingroup$
Thank you for the reference :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
answered Dec 17 '18 at 19:50
jgonjgon
15.6k32143
$endgroup$
$begingroup$
Thanks for showing how to fix the problem :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13
add a comment |
$begingroup$
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.
answered Dec 17 '18 at 19:50
jgonjgon
15.6k32143
$endgroup$
$begingroup$
Thanks for showing how to fix the problem :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13
add a comment |
$begingroup$
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.
answered Dec 17 '18 at 19:50
jgonjgon
15.6k32143
$endgroup$
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.
answered Dec 17 '18 at 19:50
jgonjgon
15.6k32143
answered Dec 17 '18 at 19:50
jgonjgon
15.6k32143
answered Dec 17 '18 at 19:50
jgonjgon
15.6k32143
answered Dec 17 '18 at 19:50
jgonjgon
15.6k32143
15.6k32143
$begingroup$
Thanks for showing how to fix the problem :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $Bbb R^n$.
$endgroup$
$begingroup$
Thank you for the reference :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13
add a comment |
$begingroup$
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 $Bbb R^n$.
$endgroup$
$begingroup$
Thank you for the reference :)
$endgroup$
– N00ber
Dec 20 '18 at 11:13
add a comment |
$begingroup$
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 $Bbb R^n$.
$endgroup$
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 $Bbb R^n$.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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