An exact sequence of vector bundles.
$begingroup$
Page 376, Prop 15.6.7: Let $p:E rightarrow M$ be a vector bundle. There exists a canonical exact sequence
$$ 0 rightarrow p^*E xrightarrow{alpha} TE xrightarrow{beta} p^*TMrightarrow 0 $$
I am confused by how the map $alpha$ is defined and why it is "canonical". It writes:
We consider $p^*E rightarrow E$ as the projection of $E oplus E$ onto the first summand.
Let $(v,w) in E_x oplus E_x$. We define $alpha(v,w)$ as the derivative of the curve $t mapsto v+tw$ at $t=0$.
Also, I would like to understand how one computes $alpha$ and show $betaalpha=0$.
My attempt is shown below.
Compute $alpha(v,w)$. My method of computation would to be to consider a local trivialization of
$$ E|_U rightarrow U times Bbb R^n rightarrow Bbb R^m times Bbb R^n$$
So that we have a local frame, $(partial_k, partial_i)_{k=1}^m, _{i=1}^n$ for $TE|_{E|_U}$
Let $gamma(t)$ be the curve, then it has coordinate representation,
$$ gamma(t) = (x, v^i+tw^i)$$
Thus, $gamma'(0)$ is the vector, whose representation under coordinate choice is
$$ 0^k partial_k + w^i partial_i$$
in the tangent space of $gamma(0)=v$.
^This shows injectivity.
Compute $beta$: Now the induce map $(d pi )_v: T_pE rightarrow T_{pi(p)}M$ is given by projection on to the first $m$ coordinates, under the choice of local triviliazation, $E|_U$ such that $U$ is a local chart of $M$.
^This gives us $beta alpha =0$.
differential-geometry algebraic-topology differential-topology smooth-manifolds vector-bundles
asked Jan 2 at 3:01
CL.CL.
2,4163925
$endgroup$
add a comment |
$begingroup$
Page 376, Prop 15.6.7: Let $p:E rightarrow M$ be a vector bundle. There exists a canonical exact sequence
$$ 0 rightarrow p^*E xrightarrow{alpha} TE xrightarrow{beta} p^*TMrightarrow 0 $$
I am confused by how the map $alpha$ is defined and why it is "canonical". It writes:
We consider $p^*E rightarrow E$ as the projection of $E oplus E$ onto the first summand.
Let $(v,w) in E_x oplus E_x$. We define $alpha(v,w)$ as the derivative of the curve $t mapsto v+tw$ at $t=0$.
Also, I would like to understand how one computes $alpha$ and show $betaalpha=0$.
My attempt is shown below.
Compute $alpha(v,w)$. My method of computation would to be to consider a local trivialization of
$$ E|_U rightarrow U times Bbb R^n rightarrow Bbb R^m times Bbb R^n$$
So that we have a local frame, $(partial_k, partial_i)_{k=1}^m, _{i=1}^n$ for $TE|_{E|_U}$
Let $gamma(t)$ be the curve, then it has coordinate representation,
$$ gamma(t) = (x, v^i+tw^i)$$
Thus, $gamma'(0)$ is the vector, whose representation under coordinate choice is
$$ 0^k partial_k + w^i partial_i$$
in the tangent space of $gamma(0)=v$.
^This shows injectivity.
Compute $beta$: Now the induce map $(d pi )_v: T_pE rightarrow T_{pi(p)}M$ is given by projection on to the first $m$ coordinates, under the choice of local triviliazation, $E|_U$ such that $U$ is a local chart of $M$.
^This gives us $beta alpha =0$.
differential-geometry algebraic-topology differential-topology smooth-manifolds vector-bundles
asked Jan 2 at 3:01
CL.CL.
2,4163925
$endgroup$
1
$begingroup$
I think you are going the right direction. You have to use local coordinates, then use the standard coordinatization of the tangent space, and interpret the answer.
$endgroup$
– Charlie Frohman
Jan 2 at 14:56
add a comment |
$begingroup$
Page 376, Prop 15.6.7: Let $p:E rightarrow M$ be a vector bundle. There exists a canonical exact sequence
$$ 0 rightarrow p^*E xrightarrow{alpha} TE xrightarrow{beta} p^*TMrightarrow 0 $$
I am confused by how the map $alpha$ is defined and why it is "canonical". It writes:
We consider $p^*E rightarrow E$ as the projection of $E oplus E$ onto the first summand.
Let $(v,w) in E_x oplus E_x$. We define $alpha(v,w)$ as the derivative of the curve $t mapsto v+tw$ at $t=0$.
Also, I would like to understand how one computes $alpha$ and show $betaalpha=0$.
My attempt is shown below.
Compute $alpha(v,w)$. My method of computation would to be to consider a local trivialization of
$$ E|_U rightarrow U times Bbb R^n rightarrow Bbb R^m times Bbb R^n$$
So that we have a local frame, $(partial_k, partial_i)_{k=1}^m, _{i=1}^n$ for $TE|_{E|_U}$
Let $gamma(t)$ be the curve, then it has coordinate representation,
$$ gamma(t) = (x, v^i+tw^i)$$
Thus, $gamma'(0)$ is the vector, whose representation under coordinate choice is
$$ 0^k partial_k + w^i partial_i$$
in the tangent space of $gamma(0)=v$.
^This shows injectivity.
Compute $beta$: Now the induce map $(d pi )_v: T_pE rightarrow T_{pi(p)}M$ is given by projection on to the first $m$ coordinates, under the choice of local triviliazation, $E|_U$ such that $U$ is a local chart of $M$.
^This gives us $beta alpha =0$.
differential-geometry algebraic-topology differential-topology smooth-manifolds vector-bundles
asked Jan 2 at 3:01
CL.CL.
2,4163925
$endgroup$
Page 376, Prop 15.6.7: Let $p:E rightarrow M$ be a vector bundle. There exists a canonical exact sequence
$$ 0 rightarrow p^*E xrightarrow{alpha} TE xrightarrow{beta} p^*TMrightarrow 0 $$
I am confused by how the map $alpha$ is defined and why it is "canonical". It writes:
We consider $p^*E rightarrow E$ as the projection of $E oplus E$ onto the first summand.
Let $(v,w) in E_x oplus E_x$. We define $alpha(v,w)$ as the derivative of the curve $t mapsto v+tw$ at $t=0$.
Also, I would like to understand how one computes $alpha$ and show $betaalpha=0$.
My attempt is shown below.
Compute $alpha(v,w)$. My method of computation would to be to consider a local trivialization of
$$ E|_U rightarrow U times Bbb R^n rightarrow Bbb R^m times Bbb R^n$$
So that we have a local frame, $(partial_k, partial_i)_{k=1}^m, _{i=1}^n$ for $TE|_{E|_U}$
Let $gamma(t)$ be the curve, then it has coordinate representation,
$$ gamma(t) = (x, v^i+tw^i)$$
Thus, $gamma'(0)$ is the vector, whose representation under coordinate choice is
$$ 0^k partial_k + w^i partial_i$$
in the tangent space of $gamma(0)=v$.
^This shows injectivity.
Compute $beta$: Now the induce map $(d pi )_v: T_pE rightarrow T_{pi(p)}M$ is given by projection on to the first $m$ coordinates, under the choice of local triviliazation, $E|_U$ such that $U$ is a local chart of $M$.
^This gives us $beta alpha =0$.
differential-geometry algebraic-topology differential-topology smooth-manifolds vector-bundles
differential-geometry algebraic-topology differential-topology smooth-manifolds vector-bundles
asked Jan 2 at 3:01
CL.CL.
2,4163925
asked Jan 2 at 3:01
CL.CL.
2,4163925
asked Jan 2 at 3:01
CL.CL.
2,4163925
asked Jan 2 at 3:01
CL.CL.
2,4163925
asked Jan 2 at 3:01
CL.CL.
2,4163925
2,4163925
1
$begingroup$
I think you are going the right direction. You have to use local coordinates, then use the standard coordinatization of the tangent space, and interpret the answer.
$endgroup$
– Charlie Frohman
Jan 2 at 14:56
add a comment |
1
$begingroup$
I think you are going the right direction. You have to use local coordinates, then use the standard coordinatization of the tangent space, and interpret the answer.
$endgroup$
– Charlie Frohman
Jan 2 at 14:56
1
1
$begingroup$
I think you are going the right direction. You have to use local coordinates, then use the standard coordinatization of the tangent space, and interpret the answer.
$endgroup$
– Charlie Frohman
Jan 2 at 14:56
$begingroup$
I think you are going the right direction. You have to use local coordinates, then use the standard coordinatization of the tangent space, and interpret the answer.
$endgroup$
– Charlie Frohman
Jan 2 at 14:56
add a comment |
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$begingroup$
I think you are going the right direction. You have to use local coordinates, then use the standard coordinatization of the tangent space, and interpret the answer.
$endgroup$
– Charlie Frohman
Jan 2 at 14:56