# How to compute $B(X, Y) = \bar{\nabla}_\bar{X}\bar{Y} - \nabla_XY$

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.

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

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 $$e_1, e_2$$.