$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for some $K in C^{infty}(M)$
$begingroup$
Let $(M, langle , rangle )$ be a $2$-dimensional Riemannian manifold, $nabla$ it's Levi-Civita connection. Show that there is a function $K in C^{infty}(M)$ such that:
$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for all $X,Y,Z in Gamma(TM)$
I don't really have a strategy, thus far I've only used the definition and tried to use that $nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?
differential-geometry riemannian-geometry
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
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add a comment |
$begingroup$
Let $(M, langle , rangle )$ be a $2$-dimensional Riemannian manifold, $nabla$ it's Levi-Civita connection. Show that there is a function $K in C^{infty}(M)$ such that:
$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for all $X,Y,Z in Gamma(TM)$
I don't really have a strategy, thus far I've only used the definition and tried to use that $nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?
differential-geometry riemannian-geometry
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
$endgroup$
add a comment |
$begingroup$
Let $(M, langle , rangle )$ be a $2$-dimensional Riemannian manifold, $nabla$ it's Levi-Civita connection. Show that there is a function $K in C^{infty}(M)$ such that:
$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for all $X,Y,Z in Gamma(TM)$
I don't really have a strategy, thus far I've only used the definition and tried to use that $nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?
differential-geometry riemannian-geometry
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
$endgroup$
Let $(M, langle , rangle )$ be a $2$-dimensional Riemannian manifold, $nabla$ it's Levi-Civita connection. Show that there is a function $K in C^{infty}(M)$ such that:
$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for all $X,Y,Z in Gamma(TM)$
I don't really have a strategy, thus far I've only used the definition and tried to use that $nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
edited Dec 17 '18 at 15:03
user593746
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
1,24311530
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2 Answers
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Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.7k767147
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Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
$endgroup$
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.7k767147
$endgroup$
add a comment |
$begingroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.7k767147
$endgroup$
add a comment |
$begingroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.7k767147
$endgroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.7k767147
answered Dec 17 '18 at 17:22
TravisTravis
60.7k767147
answered Dec 17 '18 at 17:22
TravisTravis
60.7k767147
answered Dec 17 '18 at 17:22
TravisTravis
60.7k767147
60.7k767147
add a comment |
add a comment |
$begingroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
$endgroup$
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
$begingroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
$endgroup$
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
$begingroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
$endgroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
714
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
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