Ngô Quốc Anh

October 11, 2009

An example of sectional curvature of sphere

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 15:34

In \mathbb S^n we consider the metric

{g_{ij}} = {\left( {\displaystyle\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}{\delta _{ij}}.

We now find the sectional curvature of g.


{g_{ij}} = \displaystyle{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}{\delta _{ij}}


{g^{ij}} = \displaystyle{\left( {\frac{{1 + {{\left| y \right|}^2}}}{2}} \right)^2}{\delta _{ij}}.

Now we need to calculate

\Gamma _{ij}^k = \displaystyle\frac{1}{2}{g^{kl}}\left( {\frac{{\partial {g_{il}}}}{{\partial {y_j}}} + \frac{{\partial {g_{lj}}}}{{\partial {y_i}}} - \frac{{\partial {g_{ij}}}}{{\partial {y_l}}}} \right).


\displaystyle\frac{{\partial {g_{il}}}}{{\partial {y_j}}} = \frac{\partial }{{\partial {y_j}}}\left( {{{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)}^2}{\delta _{il}}} \right) = 2\frac{2}{{1 + {{\left| y \right|}^2}}}\frac{{ - 2}}{{{{\left( {1 + {{\left| y \right|}^2}} \right)}^2}}}\left( {2{y_j}} \right){\delta _{il}} = - 2{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^3}{y_j}{\delta _{il}}.

Similarly, one has the following

\displaystyle\frac{{\partial {g_{lj}}}}{{\partial {y_i}}} = - 2{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^3}{y_i}{\delta _{lj}}, \quad \frac{{\partial {g_{ij}}}}{{\partial {y_l}}} = - 2{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^3}{y_l}{\delta _{ij}}.


\displaystyle \Gamma _{ij}^k= \frac{{ - 2}}{{1 + {{\left| y \right|}^2}}}\left( {{y_j}{\delta _{ik}} + {y_i}{\delta _{kj}} - {y_k}{\delta _{ij}}} \right).

Next we need to calculate coefficients R^m_{lij} of the curvature tensor. To this purpose, we use

\displaystyle R_{lij}^m = \frac{{\partial \Gamma _{lj}^m}}{{\partial {y_i}}} - \frac{{\partial \Gamma _{li}^m}}{{\partial {y_j}}} + \Gamma _{in}^m\Gamma _{jl}^n - \Gamma _{jn}^m\Gamma _{il}^n.

We have

\displaystyle\frac{{\partial \Gamma _{lj}^m}}{{\partial {y_i}}} = \frac{\partial }{{\partial {y_i}}}\left( {\frac{{ - 2}}{{1 + {{\left| y \right|}^2}}}\left( {{y_j}{\delta _{ml}} + {y_l}{\delta _{mj}} - {y_m}{\delta _{lj}}} \right)} \right)

which yields

\displaystyle\frac{{\partial \Gamma _{lj}^m}}{{\partial {y_i}}} ={\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}\left( {{y_i}{y_j}{\delta _{ml}} + {y_i}{y_l}{\delta _{mj}} - {y_i}{y_m}{\delta _{lj}}} \right) + \frac{{ - 2}}{{1 + {{\left| y \right|}^2}}}\left( {{\delta _{ij}}{\delta _{ml}} + {\delta _{il}}{\delta _{mj}} - {\delta _{im}}{\delta _{lj}}} \right).

Similarly, one has

\displaystyle\frac{{\partial \Gamma _{li}^m}}{{\partial {y_j}}} = {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}\left( {{y_i}{y_j}{\delta _{ml}} + {y_j}{y_l}{\delta _{mi}} - {y_j}{y_m}{\delta _{li}}} \right) + \frac{{ - 2}}{{1 + {{\left| y \right|}^2}}}\left( {{\delta _{ji}}{\delta _{ml}} + {\delta _{jl}}{\delta _{mi}} - {\delta _{jm}}{\delta _{li}}} \right).


\displaystyle\frac{{\partial \Gamma _{lj}^m}}{{\partial {y_i}}} - \frac{{\partial \Gamma _{li}^m}}{{\partial {y_j}}} = - \frac{4}{{1 + {{\left| y \right|}^2}}}\left( {{\delta _{il}}{\delta _{mj}} - {\delta _{im}}{\delta _{lj}}} \right).

On the other hands,

\displaystyle\Gamma _{in}^m\Gamma _{jl}^n - \Gamma _{jn}^m\Gamma _{il}^n = {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}\left\{ \begin{gathered}  {y_i}{y_j}{\delta _{ml}} + {y_i}{y_l}{\delta _{mj}} - {y_i}{y_m}{\delta _{jl}} + \hfill \\  {y_l}{y_j}{\delta _{im}} + {y_j}{y_l}{\delta _{im}} - y_n^2{\delta _{jl}}{\delta _{im}} - \hfill \\ {y_m}{y_j}{\delta _{il}} - {y_m}{y_l}{\delta _{ij}} + {y_m}{y_i}{\delta _{jl}} \hfill \\ \hfill \\ - {y_i}{y_j}{\delta _{ml}} - {y_j}{y_l}{\delta _{mi}} + {y_j}{y_m}{\delta _{il}} - \hfill \\ {y_l}{y_i}{\delta _{jm}} - {y_i}{y_l}{\delta _{jm}} + y_n^2{\delta _{il}}{\delta _{jm}} + \hfill \\ {y_m}{y_i}{\delta _{jl}} + {y_m}{y_l}{\delta _{ij}} - {y_m}{y_j}{\delta _{il}} \end{gathered} \right\}

which yields

\displaystyle\Gamma _{in}^m\Gamma _{jl}^n - \Gamma _{jn}^m\Gamma _{il}^n = {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}{\left| y \right|^2}\left( {{\delta _{il}}{\delta _{jm}} - {\delta _{jl}}{\delta _{im}}} \right).


\displaystyle R_{lij}^m = - \frac{4}{{1 + {{\left| y \right|}^2}}}\left( {{\delta _{il}}{\delta _{mj}} - {\delta _{im}}{\delta _{lj}}} \right) + {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}{\left| y \right|^2}\left( {{\delta _{il}}{\delta _{jm}} - {\delta _{jl}}{\delta _{im}}} \right) = {\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)^2}\left( {{\delta _{im}}{\delta _{lj}} - {\delta _{il}}{\delta _{mj}}} \right).

Finally, one obtains

\displaystyle K\left( {{e_i},{e_j}} \right) = \frac{{\left\langle {R\left( {{e_i},{e_j}} \right){e_j},{e_i}} \right\rangle }}{{\left\langle {{e_i},{e_i}} \right\rangle \left\langle {{e_j},{e_j}} \right\rangle - {{\left\langle {{e_i},{e_j}} \right\rangle }^2}}} = \frac{{R_{jij}^m\left\langle {{e_m},{e_i}} \right\rangle }}{{{{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)}^4}}} = \frac{{R_{jij}^m{{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)}^2}{\delta _{mi}}}}{{{{\left( {\frac{2}{{1 + {{\left| y \right|}^2}}}} \right)}^4}}} = 1.



  1. How did you calculate the riemannian metric gij?

    Comment by Aaron — May 4, 2015 @ 4:47

    • Hi, thanks for your interest in my post.

      In this note, the metric g is somehow already given; then we proceed to find its associated sectional curvature.

      If you are interested in the form of the metric g, just think about g being as the pull back of the standard Euclidean metric in \mathbb R^n under the stereographic projection \mathbb S^n \to \mathbb R^n. (See my posts:

      Comment by Ngô Quốc Anh — May 4, 2015 @ 6:23

  2. Your calculations are not correct! There are more terms in the difference of the derivatives of the Christoffels, but luckely they are killing with extra terms in the difference of the sum of the Christoffels, where are again more terms are not zero in general. So all in all your curvature tensor is correct again.

    Comment by noname — June 29, 2017 @ 16:55

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at

%d bloggers like this: