Problem about frame bundle in Kobayashi's book
I'm reading Kobayashi's book "Transformation Groups in Differential Geometry" and I have a problem in the proof of this lemma:
At the converse part he says this: 
My question is about $f,$ namely, how is $f$ defined?
fiber-bundles principal-bundles
asked Nov 18 '18 at 19:50
Hurjui Ionut
473211
add a comment |
I'm reading Kobayashi's book "Transformation Groups in Differential Geometry" and I have a problem in the proof of this lemma:
At the converse part he says this:
My question is about $f,$ namely, how is $f$ defined?
fiber-bundles principal-bundles
asked Nov 18 '18 at 19:50
Hurjui Ionut
473211
add a comment |
I'm reading Kobayashi's book "Transformation Groups in Differential Geometry" and I have a problem in the proof of this lemma:
At the converse part he says this:
My question is about $f,$ namely, how is $f$ defined?
fiber-bundles principal-bundles
asked Nov 18 '18 at 19:50
Hurjui Ionut
473211
I'm reading Kobayashi's book "Transformation Groups in Differential Geometry" and I have a problem in the proof of this lemma:
At the converse part he says this:
My question is about $f,$ namely, how is $f$ defined?
fiber-bundles principal-bundles
fiber-bundles principal-bundles
asked Nov 18 '18 at 19:50
Hurjui Ionut
473211
asked Nov 18 '18 at 19:50
Hurjui Ionut
473211
asked Nov 18 '18 at 19:50
Hurjui Ionut
473211
asked Nov 18 '18 at 19:50
Hurjui Ionut
473211
asked Nov 18 '18 at 19:50
Hurjui Ionut
473211
473211
add a comment |
add a comment |
1 Answer
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Let $pi:L(M)to M$ by the projection onto the base. Define $f:Mto M$ by $f(x) := pi(F(p))$, where $pin pi^{-1}(x)subset L(M)$ is arbitrary. Since $F:L(M)to L(M)$ is fibre-preserving (and so maps the fibre $pi^{-1}(x)$ into some other fibre $pi^{-1}(y)$), this definition is independent of the choice of $pin pi^{-1}(x)$, and so gives a function satisfying $fcirc pi = picirc F$.
answered Nov 19 '18 at 4:24
user17945
31318
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
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Let $pi:L(M)to M$ by the projection onto the base. Define $f:Mto M$ by $f(x) := pi(F(p))$, where $pin pi^{-1}(x)subset L(M)$ is arbitrary. Since $F:L(M)to L(M)$ is fibre-preserving (and so maps the fibre $pi^{-1}(x)$ into some other fibre $pi^{-1}(y)$), this definition is independent of the choice of $pin pi^{-1}(x)$, and so gives a function satisfying $fcirc pi = picirc F$.
answered Nov 19 '18 at 4:24
user17945
31318
add a comment |
Let $pi:L(M)to M$ by the projection onto the base. Define $f:Mto M$ by $f(x) := pi(F(p))$, where $pin pi^{-1}(x)subset L(M)$ is arbitrary. Since $F:L(M)to L(M)$ is fibre-preserving (and so maps the fibre $pi^{-1}(x)$ into some other fibre $pi^{-1}(y)$), this definition is independent of the choice of $pin pi^{-1}(x)$, and so gives a function satisfying $fcirc pi = picirc F$.
answered Nov 19 '18 at 4:24
user17945
31318
add a comment |
Let $pi:L(M)to M$ by the projection onto the base. Define $f:Mto M$ by $f(x) := pi(F(p))$, where $pin pi^{-1}(x)subset L(M)$ is arbitrary. Since $F:L(M)to L(M)$ is fibre-preserving (and so maps the fibre $pi^{-1}(x)$ into some other fibre $pi^{-1}(y)$), this definition is independent of the choice of $pin pi^{-1}(x)$, and so gives a function satisfying $fcirc pi = picirc F$.
answered Nov 19 '18 at 4:24
user17945
31318
Let $pi:L(M)to M$ by the projection onto the base. Define $f:Mto M$ by $f(x) := pi(F(p))$, where $pin pi^{-1}(x)subset L(M)$ is arbitrary. Since $F:L(M)to L(M)$ is fibre-preserving (and so maps the fibre $pi^{-1}(x)$ into some other fibre $pi^{-1}(y)$), this definition is independent of the choice of $pin pi^{-1}(x)$, and so gives a function satisfying $fcirc pi = picirc F$.
answered Nov 19 '18 at 4:24
user17945
31318
answered Nov 19 '18 at 4:24
user17945
31318
answered Nov 19 '18 at 4:24
user17945
31318
answered Nov 19 '18 at 4:24
user17945
31318
31318
add a comment |
add a comment |
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