Show that any affine connection $nabla$ on $mathbb{R}^n$ is of the form $nabla=D+Gamma$.
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Show that any affine connection $nabla$ on $mathbb{R}^n$ is of the form $nabla=D+Gamma$, where $D$ is the Euclidean connection and $Gamma:mathcal{X}(mathbb{R}^n) times mathcal{X}(mathbb{R}^n) rightarrow mathcal{X}(mathbb{R}^n)$ is any $C^{infty}(mathbb{R}^n)$-bilinear map.
Showing that $D+Gamma$ is an affine connection is easy, but I don't know how to show the converse. I mean, what's special about $mathbb{R}^n$?
manifolds riemannian-geometry vector-fields connections
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up vote
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Show that any affine connection $nabla$ on $mathbb{R}^n$ is of the form $nabla=D+Gamma$, where $D$ is the Euclidean connection and $Gamma:mathcal{X}(mathbb{R}^n) times mathcal{X}(mathbb{R}^n) rightarrow mathcal{X}(mathbb{R}^n)$ is any $C^{infty}(mathbb{R}^n)$-bilinear map.
Showing that $D+Gamma$ is an affine connection is easy, but I don't know how to show the converse. I mean, what's special about $mathbb{R}^n$?
manifolds riemannian-geometry vector-fields connections
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Show that any affine connection $nabla$ on $mathbb{R}^n$ is of the form $nabla=D+Gamma$, where $D$ is the Euclidean connection and $Gamma:mathcal{X}(mathbb{R}^n) times mathcal{X}(mathbb{R}^n) rightarrow mathcal{X}(mathbb{R}^n)$ is any $C^{infty}(mathbb{R}^n)$-bilinear map.
Showing that $D+Gamma$ is an affine connection is easy, but I don't know how to show the converse. I mean, what's special about $mathbb{R}^n$?
manifolds riemannian-geometry vector-fields connections
Show that any affine connection $nabla$ on $mathbb{R}^n$ is of the form $nabla=D+Gamma$, where $D$ is the Euclidean connection and $Gamma:mathcal{X}(mathbb{R}^n) times mathcal{X}(mathbb{R}^n) rightarrow mathcal{X}(mathbb{R}^n)$ is any $C^{infty}(mathbb{R}^n)$-bilinear map.
Showing that $D+Gamma$ is an affine connection is easy, but I don't know how to show the converse. I mean, what's special about $mathbb{R}^n$?
manifolds riemannian-geometry vector-fields connections
manifolds riemannian-geometry vector-fields connections
asked Nov 21 at 21:49
bbw
38617
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This follows from the general result that the difference of two affine connections provide such a bilinear map (tensor) $Gamma$. The nontrivial part is to explain why it is $C^infty$-linear in the second argument (which is not the case for the connections themselves). But by the Leibniz rule, for any $C^infty$ function $f$ and vector fields $X,Y$ we have
$$begin{align}(nabla - D)(X,fY) &= nabla(X,fY) - D(X,fY) \&= (Xf)Y + fnabla(X,Y) - (Xf)Y - fD(X,Y) \ &= f(nabla - D)(X,Y),end{align}$$
so the $C^infty$-bilinearity is a consequence of the fact that the "constant terms" (those independent of the connection) cancel each other out in the difference.
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 at 23:19
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This follows from the general result that the difference of two affine connections provide such a bilinear map (tensor) $Gamma$. The nontrivial part is to explain why it is $C^infty$-linear in the second argument (which is not the case for the connections themselves). But by the Leibniz rule, for any $C^infty$ function $f$ and vector fields $X,Y$ we have
$$begin{align}(nabla - D)(X,fY) &= nabla(X,fY) - D(X,fY) \&= (Xf)Y + fnabla(X,Y) - (Xf)Y - fD(X,Y) \ &= f(nabla - D)(X,Y),end{align}$$
so the $C^infty$-bilinearity is a consequence of the fact that the "constant terms" (those independent of the connection) cancel each other out in the difference.
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 at 23:19
add a comment |
up vote
2
down vote
accepted
This follows from the general result that the difference of two affine connections provide such a bilinear map (tensor) $Gamma$. The nontrivial part is to explain why it is $C^infty$-linear in the second argument (which is not the case for the connections themselves). But by the Leibniz rule, for any $C^infty$ function $f$ and vector fields $X,Y$ we have
$$begin{align}(nabla - D)(X,fY) &= nabla(X,fY) - D(X,fY) \&= (Xf)Y + fnabla(X,Y) - (Xf)Y - fD(X,Y) \ &= f(nabla - D)(X,Y),end{align}$$
so the $C^infty$-bilinearity is a consequence of the fact that the "constant terms" (those independent of the connection) cancel each other out in the difference.
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 at 23:19
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This follows from the general result that the difference of two affine connections provide such a bilinear map (tensor) $Gamma$. The nontrivial part is to explain why it is $C^infty$-linear in the second argument (which is not the case for the connections themselves). But by the Leibniz rule, for any $C^infty$ function $f$ and vector fields $X,Y$ we have
$$begin{align}(nabla - D)(X,fY) &= nabla(X,fY) - D(X,fY) \&= (Xf)Y + fnabla(X,Y) - (Xf)Y - fD(X,Y) \ &= f(nabla - D)(X,Y),end{align}$$
so the $C^infty$-bilinearity is a consequence of the fact that the "constant terms" (those independent of the connection) cancel each other out in the difference.
This follows from the general result that the difference of two affine connections provide such a bilinear map (tensor) $Gamma$. The nontrivial part is to explain why it is $C^infty$-linear in the second argument (which is not the case for the connections themselves). But by the Leibniz rule, for any $C^infty$ function $f$ and vector fields $X,Y$ we have
$$begin{align}(nabla - D)(X,fY) &= nabla(X,fY) - D(X,fY) \&= (Xf)Y + fnabla(X,Y) - (Xf)Y - fD(X,Y) \ &= f(nabla - D)(X,Y),end{align}$$
so the $C^infty$-bilinearity is a consequence of the fact that the "constant terms" (those independent of the connection) cancel each other out in the difference.
edited Nov 22 at 12:32
answered Nov 21 at 22:49
Alex Provost
15.1k22250
15.1k22250
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 at 23:19
add a comment |
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 at 23:19
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 at 23:19
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 at 23:19
add a comment |
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