# Computation of sectional curvature on torus

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.

• 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 – Yu Ding Apr 26 at 15:48
• @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. – khrenb Apr 27 at 3:19

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 $$K=0$$.