I am learning Riemannian geometry. I received the following problem. $f:\mathbb{R}^2\to\mathbb{R}^4$ is defined by $$f(\theta,\phi)=\frac{1}{\sqrt{2}}(\cos\,\theta,\sin\,\theta,\cos\,\phi,\sin\,\phi).$$ The map $f$ is an immersion and its image is a torus $\mathbb{T}^2$. The problem asks to show the sectional curvature of the image of $f$ is zero.

Since $\mathbb{T}^2$ is $2$-dimensional, we choose linearly independent $x,y\in T_p\mathbb{T}^2$, where $p$ is a point in $\mathbb{T}^2$. Then the sectional curvature is $$K(x,y)=\frac{(x,y,x,y)}{|x|^2|y|^2-\langle x,y\rangle^2}.$$ Let $X,Y$ be vector fields on $\mathbb{T}^2$ that extend $x$ and $y$ respectively, then $$K(x,y)=\frac{\langle R(X,Y)X,Y\rangle}{|x|^2|y|^2-\langle x,y\rangle^2},$$ where $$\langle R(X,Y)X,Y\rangle=\langle\nabla_Y\nabla_X X-\nabla_X\nabla_Y X+\nabla_{[X,Y]}X,Y\rangle.$$ I know the definition of induced metric and Levi-Civita connection, but I don't know how to compute them.

  • 1
    $\begingroup$ Since on this surface there are two global parallel vector fields, the curvature tensor vanishes everywhere. See my answer to a similar question math.stackexchange.com/q/3201187 $\endgroup$
    – Yuval
    Apr 26, 2019 at 15:48
  • $\begingroup$ @YuDing Thank you for your comment. Now I understand it. It turns out that the problem needs knowledge in second fundamental form but I thought it does not. I also found that my question duplicates this question. $\endgroup$
    – 1615244
    Apr 27, 2019 at 3:19

1 Answer 1


Another way to do it: compute $$\partial_\theta = \frac{1}{\sqrt{2}}(-\sin\theta\,\partial_x+\cos\theta\,\partial_y) \quad\mbox{and}\quad \partial_\phi = \frac{1}{\sqrt{2}}(-\sin\phi\,\partial_z + \cos\phi\,\partial_w).$$We have that $\partial_\theta \cdot \partial_\theta = \partial_\phi\cdot\partial_\phi = 1$ and $\partial_\theta \cdot \partial_\phi = 0$. This means that all the Christoffel symbols vanish. So $R = 0$, hence $K=0$.


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