Mixing
Show that any affine connection $nabla$ on $mathbb{R}^n$ is of the form $nabla=D+Gamma$.
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
asked Nov 21 '18 at 21:49
bbwbbw
47038
add a comment |
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
asked Nov 21 '18 at 21:49
bbwbbw
47038
add a comment |
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
asked Nov 21 '18 at 21:49
bbwbbw
47038
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 '18 at 21:49
bbwbbw
47038
asked Nov 21 '18 at 21:49
bbwbbw
47038
asked Nov 21 '18 at 21:49
bbwbbw
47038
asked Nov 21 '18 at 21:49
bbwbbw
47038
asked Nov 21 '18 at 21:49
bbwbbw
47038
47038
add a comment |
add a comment |
1 Answer
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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.
answered Nov 21 '18 at 22:49
Alex ProvostAlex Provost
15.2k22250
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 '18 at 23:19
add a comment |
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1 Answer
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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.
answered Nov 21 '18 at 22:49
Alex ProvostAlex Provost
15.2k22250
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 '18 at 23:19
add a comment |
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.
answered Nov 21 '18 at 22:49
Alex ProvostAlex Provost
15.2k22250
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 '18 at 23:19
add a comment |
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.
answered Nov 21 '18 at 22:49
Alex ProvostAlex Provost
15.2k22250
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.
answered Nov 21 '18 at 22:49
Alex ProvostAlex Provost
15.2k22250
edited Nov 22 '18 at 12:32
answered Nov 21 '18 at 22:49
Alex ProvostAlex Provost
15.2k22250
answered Nov 21 '18 at 22:49
Alex ProvostAlex Provost
15.2k22250
answered Nov 21 '18 at 22:49
Alex ProvostAlex Provost
15.2k22250
15.2k22250
Oh how could I forgot the equivalence $nabla=D+Gamma Leftrightarrow nabla - D=Gamma$. Thank you for you answer.
– bbw
Nov 21 '18 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 '18 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 '18 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 '18 at 23:19
add a comment |
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