Let $g : \mathbb{R}^2 \rightarrow \mathbb{R}^4$ be the immersion defined by $$ g(x, y) = (\cos(x), \sin(x), \cos(y), \sin(y)). $$ Let $e_1 = \frac{\partial}{\partial x}$ and $e_2 = \frac{\partial}{\partial y}$ and $\bar{M} = \mathbb{R}^4$.

Compute $B(e_i, e_j)$ for $1 \leq, i, j \leq 2$ by direct computation.

I know that $B(X, Y) = \bar{\nabla}_\bar{X}\bar{Y} - \nabla_XY$. where $\nabla_XY = (\bar{\nabla}_\bar{X}\bar{Y})^T$. I know that $\bar{\nabla}$ is the Riemannian connection on $\bar{M}$, and $\bar{X}$ and $\bar{Y}$ are local extensions to $\bar{M}$. However, I can't seem to find in my book how to actually carry out the computations of $\bar{\nabla}_\bar{X}\bar{Y}$ and $\nabla_XY$ since most of my book is theory based.

  • $\begingroup$ $e_1$ and $e_2$ are the basis of the tangente bundle of $M=g(\mathbb{R}^2)$. Is it right? $\endgroup$
    – DiegoMath
    Commented Apr 24, 2019 at 19:28
  • $\begingroup$ Do you mean $e_1=\frac{\partial}{\partial x}$ (and similarly $e_2$) instead of $\frac{\delta}{\delta u}$? $\endgroup$ Commented Apr 24, 2019 at 19:29
  • $\begingroup$ @user10354138: Yes. Fixed. $\endgroup$
    – user525033
    Commented Apr 24, 2019 at 19:43

1 Answer 1


We see $e_1=[-\sin x, \cos x, 0,0]$, $e_2=[0, 0, -\sin y, \cos y]$, then $\bar\nabla_{e_1}e_1=[-\cos x, -\sin x, 0, 0]$: this is the directional derivative of $e_1$ in the $e_1$ direction, therefore we just need to take derivative with respect to $x$. Similarly, $\bar\nabla_{e_1}e_2=\bar\nabla_{e_2}e_1=0$, $\bar \nabla_{e_2}e_2=[0, 0, -\cos y, -\sin y]$. Now both $\bar\nabla_{e_1}e_1$ and $\bar\nabla_{e_2}{e_2}$ are both $\perp$ the surface, so $\nabla_{e_i}e_j=0$ for all $i, j$.

This surface is a flat torus in ${\mathbb R}^4$ with global paralled vector fields $e_1, e_2$.


You must log in to answer this question.