Vector field on an odd sphere. $X = \sum_{i=1}^n -y^i \frac{\partial}{\partial x^i} + x^i \frac{\partial}{\partial y^i}$ is smooth.

Question

This is problem 14.2 from Loring Tu's Introduction to Manifolds.

Vector field on an odd sphere.

Let $x^1,y^1,\dots,x^n,y^n$ be the standard coordinates on $R^{2n}$. The unit sphere $S^{2n-1}$ in $R^{2n}$ is defined by the equation $\sum (x^i)^2 + (y^i)^2=1$. Show that $$X = \sum_{i=1}^n -y^i \frac{\partial}{\partial x^i} + x^i \frac{\partial}{\partial y^i}$$ is a nowhere-vanishing smooth vector field on $S^{2n-1}$.

I have shown that it is nowhere-vanishing, but I do not know how to show that this is smooth. From Proposition 14.2 of the text, it is equivalent to showing that there is an atlas on the unit sphere such that on any chart $(U,\phi) = (U,z^1,\dots z^n)$ of the atlas, the coefficients $a^i$ of $X=\sum a^i \partial/\partial z^i$ relative to the frame $\partial/\partial z^i$ are all smooth. I know of two charts for the unit sphere : the projection charts onto each hemisphere and the stereographic projection. However, for each atlas has only $2n-1$ coordinates, so there are $2n-1$ basis vectors $\partial / \partial z^i$, whereas the standard coordinates give us $2n$ bases. How can we make a transition between two coordinates of different number of basis here?

This is a solution I found for this problem.

However, I can't figure out how we can formally justify $$\frac{\partial}{\partial t^i} = \sum \frac{\partial z_k}{\partial t^i}\frac{\partial}{\partial z_k}.$$

Here we are expressing the standard tangent vectors on $R^{2n}$, i.e. $\partial / \partial t^i$ in terms of the tangent vectors $\partial / \partial z_k$ on the unit sphere $S^{2n-1}$ given by the stereographic projection. However, these two coordinate maps belong to spaces of different dimensions. So we cannot use, say the following proposition from the text. As we can see from the proof, the transition relationship depends on the fact that the two frames are with respect to the same tangent space, hence one is a linear combination of the others. However, here they belong to different spaces, so how can we come up with such a linear combination?

Answer 1

You don't need to change coordinates, because the map $X:S^{n-1}\to T_pS^{n-1}$ defined by $p=(x_1,\cdots,x_n)\mapsto ( x_2, −x_1, x_4, −x_3,\cdots , −x_n)$ is evidently smooth, non-vanishing and well-defined since $\langle (x_1,\cdots,x_n),( x_2, −x_1, x_4, −x_3,\cdots , −x_n)\rangle=0.$ So $X$ is the required vector field.

Answer 2

In order to obtain the desired relation above between vectors on $S^{2n-1}$ and $\mathbb{R}^{2n-1}$ you want to use the pushforward of the charts corresponding to stereographic projections.

Given a smooth map $f:M\rightarrow N$ between two manifolds, this induces another map $f_\ast:T_pM\rightarrow T_{f(p)}N$, mapping a vector from the tangent space at $p\in M$ to one at $f(p)\in N$. Note that $M$ and $N$ do not necessarily have the same dimension. This map is defined in the following way:

Let $g\in C^{\infty}(N,\mathbb{R})$, so $g\circ f\in C^{\infty}(M,\mathbb{R})$. A vector $X\in T_pM$ acts on $g\circ f$ and takes it to a number. Now, we define $f_\ast(X)\in T_{f(p)}N$ such that $$f_\ast(X)[g]=X[g\circ f].$$ In terms of charts $(U,\phi),(V,\psi)$ on $M,N$ respectively, this becomes $$f_\ast(X)[g\circ\psi^{-1}(y)]=X[g\circ f\circ\phi^{-1}(x)]$$ with $x=\phi(p)$ and $y=\psi(f(p))$. If we rewrite $X=X^{i}\frac{\partial}{\partial x^i}$ and $f_\ast(X)=Y^j\frac{\partial}{\partial y^j}$ then the above yields $$Y^j=X^i\frac{\partial y^j(x)}{\partial x^i}.$$ But this is precisely what is used in your solution. In your case, $M=S^{2n-1}$ and $N=\mathbb{R}^{2n-1}$ with e.g. $$f:S^{2n-1}\backslash\{(0,\dots,1)\}\rightarrow\mathbb{R}^{2n-1}, (t_1,\dots,t_{2n})\mapsto\frac{1}{1-t_{2n}}(t_1,\dots,t_{2n-1})$$ the chart excluding the north pole. Applying the procedure above to this, you can rewrite a vector given on the sphere in terms of the local coordinates given via stereographic projections.

Answer 3

While it might be a good exercise, there's no need here to work in coordinate charts. If $F:\mathbb{R}^{2n}\to\mathbb{R}$ is $F(p) = \|p\|^2$, the sphere $S^{2n-1}$ is defined as $F^{-1}(1)$.

Define a vector field $V$ on $\mathbb{R}^{2n}$ by $$V_{(x,y)} = \sum_i -y^i\partial x_i + x^i\partial y_i$$ As the restriction of a smooth function to a smooth submanifold is again smooth, it will suffice to show that $X$ is the restriction to $S^{2n-1}$ of $V$. We will see this by showing that $X$ is perpendicular to the gradient of $F$ at a point $(x,y)\in S^{2n-1}$.

Note that $\nabla F_p = \sum 2p_i\partial_i$. Then: $$\begin{align*} \langle V_{(x,y)}, \nabla F_{(x,y)}\rangle &= \bigg\langle \sum_i -y_i\partial x_i + x_i \partial y_i , \sum_i 2x_i \partial x_i + 2y_i \partial y_i\bigg\rangle \\ &= \sum_i -2y_ix_i + 2x_iy_i \\ &= 0 \end{align*}$$

As $V$ is nonvanishing away from zero, $X$ is manifestly a nonvanishing smooth vector field on $S^{2n-1}$.