Mixing
Lin Independent Sections iff Trivial Bundle
This is from Hatcher: Let $(E,B,p)$ is a vector bundle.
If one has $n$-linearly independent sections, the map $h:B times mathbb{R}^n to E$ given by $h(b,t_1,···,t_n)= sum_i t_i s_i(b)$ is a linear isomorphism in each fiber, and is continuous since its composition with a local trivialization $p^{-1}(U) to U times mathbb{R}^n$ is continuous.
I've had no trouble up til now. This part just seems vague to me. How is the composition continuous?
general-topology geometry algebraic-topology vector-bundles
asked Nov 20 '18 at 20:11
Emilio Minichiello
3297
add a comment |
This is from Hatcher: Let $(E,B,p)$ is a vector bundle.
If one has $n$-linearly independent sections, the map $h:B times mathbb{R}^n to E$ given by $h(b,t_1,···,t_n)= sum_i t_i s_i(b)$ is a linear isomorphism in each fiber, and is continuous since its composition with a local trivialization $p^{-1}(U) to U times mathbb{R}^n$ is continuous.
I've had no trouble up til now. This part just seems vague to me. How is the composition continuous?
general-topology geometry algebraic-topology vector-bundles
asked Nov 20 '18 at 20:11
Emilio Minichiello
3297
1
I would rather say it's continuous because addition and multiplication and also $s_i$ are continuous. Doesn't that suffice?
– Berci
Nov 20 '18 at 23:31
add a comment |
This is from Hatcher: Let $(E,B,p)$ is a vector bundle.
If one has $n$-linearly independent sections, the map $h:B times mathbb{R}^n to E$ given by $h(b,t_1,···,t_n)= sum_i t_i s_i(b)$ is a linear isomorphism in each fiber, and is continuous since its composition with a local trivialization $p^{-1}(U) to U times mathbb{R}^n$ is continuous.
I've had no trouble up til now. This part just seems vague to me. How is the composition continuous?
general-topology geometry algebraic-topology vector-bundles
asked Nov 20 '18 at 20:11
Emilio Minichiello
3297
This is from Hatcher: Let $(E,B,p)$ is a vector bundle.
If one has $n$-linearly independent sections, the map $h:B times mathbb{R}^n to E$ given by $h(b,t_1,···,t_n)= sum_i t_i s_i(b)$ is a linear isomorphism in each fiber, and is continuous since its composition with a local trivialization $p^{-1}(U) to U times mathbb{R}^n$ is continuous.
I've had no trouble up til now. This part just seems vague to me. How is the composition continuous?
general-topology geometry algebraic-topology vector-bundles
general-topology geometry algebraic-topology vector-bundles
asked Nov 20 '18 at 20:11
Emilio Minichiello
3297
asked Nov 20 '18 at 20:11
Emilio Minichiello
3297
edited Nov 20 '18 at 23:02
asked Nov 20 '18 at 20:11
Emilio Minichiello
3297
asked Nov 20 '18 at 20:11
Emilio Minichiello
3297
asked Nov 20 '18 at 20:11
Emilio Minichiello
3297
3297
1
I would rather say it's continuous because addition and multiplication and also $s_i$ are continuous. Doesn't that suffice?
– Berci
Nov 20 '18 at 23:31
add a comment |
1
I would rather say it's continuous because addition and multiplication and also $s_i$ are continuous. Doesn't that suffice?
– Berci
Nov 20 '18 at 23:31
1
1
I would rather say it's continuous because addition and multiplication and also $s_i$ are continuous. Doesn't that suffice?
– Berci
Nov 20 '18 at 23:31
I would rather say it's continuous because addition and multiplication and also $s_i$ are continuous. Doesn't that suffice?
– Berci
Nov 20 '18 at 23:31
add a comment |
1 Answer
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oldest
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With regards to your specific query, the local trivialisations commute with the fibrewise linear structure on $E$. So if $varphi:p^{-1}(U)xrightarrow{cong}Utimes mathbb{R}^n$ is such a local trivialisation then for each $bin U$ we have
$$varphileft(sum_{i=1}^n t_icdot s_i(b)right)=sum_{i=1}^n t_icdot varphi (s_i(b))$$
where the right-hand side is interpreted as the sum in $mathbb{R}^ncong{b}timesmathbb{R}^n$. The point is that the sections $s_i$ are continuous with respect to the variable $bin U$ so if for each $i=1,dots,n$ we let $hat s_i:Urightarrow mathbb{R}^n$ be the composite
$$hat s_i:Uxrightarrow{s_i} E_Uxrightarrow{varphi}Utimes mathbb{R}^nxrightarrow{pr_2} mathbb{R}^n$$
then we get a family of continuous $mathbb{R}^n$-valued maps on $U$, and our first equation tells us that the composite $varphicirc h|_{Utimesmathbb{R}^n}:Utimesmathbb{R}^nrightarrow Utimes mathbb{R}^n$ is equal to the composite
$$Utimes mathbb{R}^nxrightarrow{Deltatimes 1}U^{n+1}timesmathbb{R}^nxrightarrow{shuf}Utimes (mathbb{R}times U)^nxrightarrow{1timesprod(1timeshat s_i)}Utimes (mathbb{R}times mathbb{R}^n)^nxrightarrow{1times m^n}Utimes (mathbb{R}^n)^nxrightarrow{1timesoplus}Utimesmathbb{R}^n$$
where $Delta:Urightarrow U^{n+1}$ is the $(n+1)$-fold diagonal, the second map $shuf$ shuffles the coordinates appropriately, $m:mathbb{R}timesmathbb{R}^nrightarrow mathbb{R}^n$ is scalar multiplication and $oplus:mathbb{R}^ntimesdotstimesmathbb{R^n}rightarrowmathbb{R}^n$ $(x_1,dots,x_n)mapsto x_1+dots+x_n$ is the iterated vector addition in $mathbb{R}^n$.
It should be clear from this presentation that $varphicirc h$ is continuous (and even smooth if you work in the $C^infty$ category).
answered Nov 21 '18 at 10:18
Tyrone
4,37511225
Wow, this is a great explanation, thank you.
– Emilio Minichiello
Nov 21 '18 at 12:31
add a comment |
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1 Answer
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With regards to your specific query, the local trivialisations commute with the fibrewise linear structure on $E$. So if $varphi:p^{-1}(U)xrightarrow{cong}Utimes mathbb{R}^n$ is such a local trivialisation then for each $bin U$ we have
$$varphileft(sum_{i=1}^n t_icdot s_i(b)right)=sum_{i=1}^n t_icdot varphi (s_i(b))$$
where the right-hand side is interpreted as the sum in $mathbb{R}^ncong{b}timesmathbb{R}^n$. The point is that the sections $s_i$ are continuous with respect to the variable $bin U$ so if for each $i=1,dots,n$ we let $hat s_i:Urightarrow mathbb{R}^n$ be the composite
$$hat s_i:Uxrightarrow{s_i} E_Uxrightarrow{varphi}Utimes mathbb{R}^nxrightarrow{pr_2} mathbb{R}^n$$
then we get a family of continuous $mathbb{R}^n$-valued maps on $U$, and our first equation tells us that the composite $varphicirc h|_{Utimesmathbb{R}^n}:Utimesmathbb{R}^nrightarrow Utimes mathbb{R}^n$ is equal to the composite
$$Utimes mathbb{R}^nxrightarrow{Deltatimes 1}U^{n+1}timesmathbb{R}^nxrightarrow{shuf}Utimes (mathbb{R}times U)^nxrightarrow{1timesprod(1timeshat s_i)}Utimes (mathbb{R}times mathbb{R}^n)^nxrightarrow{1times m^n}Utimes (mathbb{R}^n)^nxrightarrow{1timesoplus}Utimesmathbb{R}^n$$
where $Delta:Urightarrow U^{n+1}$ is the $(n+1)$-fold diagonal, the second map $shuf$ shuffles the coordinates appropriately, $m:mathbb{R}timesmathbb{R}^nrightarrow mathbb{R}^n$ is scalar multiplication and $oplus:mathbb{R}^ntimesdotstimesmathbb{R^n}rightarrowmathbb{R}^n$ $(x_1,dots,x_n)mapsto x_1+dots+x_n$ is the iterated vector addition in $mathbb{R}^n$.
It should be clear from this presentation that $varphicirc h$ is continuous (and even smooth if you work in the $C^infty$ category).
answered Nov 21 '18 at 10:18
Tyrone
4,37511225
Wow, this is a great explanation, thank you.
– Emilio Minichiello
Nov 21 '18 at 12:31
add a comment |
With regards to your specific query, the local trivialisations commute with the fibrewise linear structure on $E$. So if $varphi:p^{-1}(U)xrightarrow{cong}Utimes mathbb{R}^n$ is such a local trivialisation then for each $bin U$ we have
$$varphileft(sum_{i=1}^n t_icdot s_i(b)right)=sum_{i=1}^n t_icdot varphi (s_i(b))$$
where the right-hand side is interpreted as the sum in $mathbb{R}^ncong{b}timesmathbb{R}^n$. The point is that the sections $s_i$ are continuous with respect to the variable $bin U$ so if for each $i=1,dots,n$ we let $hat s_i:Urightarrow mathbb{R}^n$ be the composite
$$hat s_i:Uxrightarrow{s_i} E_Uxrightarrow{varphi}Utimes mathbb{R}^nxrightarrow{pr_2} mathbb{R}^n$$
then we get a family of continuous $mathbb{R}^n$-valued maps on $U$, and our first equation tells us that the composite $varphicirc h|_{Utimesmathbb{R}^n}:Utimesmathbb{R}^nrightarrow Utimes mathbb{R}^n$ is equal to the composite
$$Utimes mathbb{R}^nxrightarrow{Deltatimes 1}U^{n+1}timesmathbb{R}^nxrightarrow{shuf}Utimes (mathbb{R}times U)^nxrightarrow{1timesprod(1timeshat s_i)}Utimes (mathbb{R}times mathbb{R}^n)^nxrightarrow{1times m^n}Utimes (mathbb{R}^n)^nxrightarrow{1timesoplus}Utimesmathbb{R}^n$$
where $Delta:Urightarrow U^{n+1}$ is the $(n+1)$-fold diagonal, the second map $shuf$ shuffles the coordinates appropriately, $m:mathbb{R}timesmathbb{R}^nrightarrow mathbb{R}^n$ is scalar multiplication and $oplus:mathbb{R}^ntimesdotstimesmathbb{R^n}rightarrowmathbb{R}^n$ $(x_1,dots,x_n)mapsto x_1+dots+x_n$ is the iterated vector addition in $mathbb{R}^n$.
It should be clear from this presentation that $varphicirc h$ is continuous (and even smooth if you work in the $C^infty$ category).
answered Nov 21 '18 at 10:18
Tyrone
4,37511225
Wow, this is a great explanation, thank you.
– Emilio Minichiello
Nov 21 '18 at 12:31
add a comment |
With regards to your specific query, the local trivialisations commute with the fibrewise linear structure on $E$. So if $varphi:p^{-1}(U)xrightarrow{cong}Utimes mathbb{R}^n$ is such a local trivialisation then for each $bin U$ we have
$$varphileft(sum_{i=1}^n t_icdot s_i(b)right)=sum_{i=1}^n t_icdot varphi (s_i(b))$$
where the right-hand side is interpreted as the sum in $mathbb{R}^ncong{b}timesmathbb{R}^n$. The point is that the sections $s_i$ are continuous with respect to the variable $bin U$ so if for each $i=1,dots,n$ we let $hat s_i:Urightarrow mathbb{R}^n$ be the composite
$$hat s_i:Uxrightarrow{s_i} E_Uxrightarrow{varphi}Utimes mathbb{R}^nxrightarrow{pr_2} mathbb{R}^n$$
then we get a family of continuous $mathbb{R}^n$-valued maps on $U$, and our first equation tells us that the composite $varphicirc h|_{Utimesmathbb{R}^n}:Utimesmathbb{R}^nrightarrow Utimes mathbb{R}^n$ is equal to the composite
$$Utimes mathbb{R}^nxrightarrow{Deltatimes 1}U^{n+1}timesmathbb{R}^nxrightarrow{shuf}Utimes (mathbb{R}times U)^nxrightarrow{1timesprod(1timeshat s_i)}Utimes (mathbb{R}times mathbb{R}^n)^nxrightarrow{1times m^n}Utimes (mathbb{R}^n)^nxrightarrow{1timesoplus}Utimesmathbb{R}^n$$
where $Delta:Urightarrow U^{n+1}$ is the $(n+1)$-fold diagonal, the second map $shuf$ shuffles the coordinates appropriately, $m:mathbb{R}timesmathbb{R}^nrightarrow mathbb{R}^n$ is scalar multiplication and $oplus:mathbb{R}^ntimesdotstimesmathbb{R^n}rightarrowmathbb{R}^n$ $(x_1,dots,x_n)mapsto x_1+dots+x_n$ is the iterated vector addition in $mathbb{R}^n$.
It should be clear from this presentation that $varphicirc h$ is continuous (and even smooth if you work in the $C^infty$ category).
answered Nov 21 '18 at 10:18
Tyrone
4,37511225
With regards to your specific query, the local trivialisations commute with the fibrewise linear structure on $E$. So if $varphi:p^{-1}(U)xrightarrow{cong}Utimes mathbb{R}^n$ is such a local trivialisation then for each $bin U$ we have
$$varphileft(sum_{i=1}^n t_icdot s_i(b)right)=sum_{i=1}^n t_icdot varphi (s_i(b))$$
where the right-hand side is interpreted as the sum in $mathbb{R}^ncong{b}timesmathbb{R}^n$. The point is that the sections $s_i$ are continuous with respect to the variable $bin U$ so if for each $i=1,dots,n$ we let $hat s_i:Urightarrow mathbb{R}^n$ be the composite
$$hat s_i:Uxrightarrow{s_i} E_Uxrightarrow{varphi}Utimes mathbb{R}^nxrightarrow{pr_2} mathbb{R}^n$$
then we get a family of continuous $mathbb{R}^n$-valued maps on $U$, and our first equation tells us that the composite $varphicirc h|_{Utimesmathbb{R}^n}:Utimesmathbb{R}^nrightarrow Utimes mathbb{R}^n$ is equal to the composite
$$Utimes mathbb{R}^nxrightarrow{Deltatimes 1}U^{n+1}timesmathbb{R}^nxrightarrow{shuf}Utimes (mathbb{R}times U)^nxrightarrow{1timesprod(1timeshat s_i)}Utimes (mathbb{R}times mathbb{R}^n)^nxrightarrow{1times m^n}Utimes (mathbb{R}^n)^nxrightarrow{1timesoplus}Utimesmathbb{R}^n$$
where $Delta:Urightarrow U^{n+1}$ is the $(n+1)$-fold diagonal, the second map $shuf$ shuffles the coordinates appropriately, $m:mathbb{R}timesmathbb{R}^nrightarrow mathbb{R}^n$ is scalar multiplication and $oplus:mathbb{R}^ntimesdotstimesmathbb{R^n}rightarrowmathbb{R}^n$ $(x_1,dots,x_n)mapsto x_1+dots+x_n$ is the iterated vector addition in $mathbb{R}^n$.
It should be clear from this presentation that $varphicirc h$ is continuous (and even smooth if you work in the $C^infty$ category).
answered Nov 21 '18 at 10:18
Tyrone
4,37511225
answered Nov 21 '18 at 10:18
Tyrone
4,37511225
answered Nov 21 '18 at 10:18
Tyrone
4,37511225
answered Nov 21 '18 at 10:18
Tyrone
4,37511225
4,37511225
Wow, this is a great explanation, thank you.
– Emilio Minichiello
Nov 21 '18 at 12:31
add a comment |
Wow, this is a great explanation, thank you.
– Emilio Minichiello
Nov 21 '18 at 12:31
Wow, this is a great explanation, thank you.
– Emilio Minichiello
Nov 21 '18 at 12:31
Wow, this is a great explanation, thank you.
– Emilio Minichiello
Nov 21 '18 at 12:31
add a comment |
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I would rather say it's continuous because addition and multiplication and also $s_i$ are continuous. Doesn't that suffice?
– Berci
Nov 20 '18 at 23:31