differential one form example on $S^1$

Question

I am studying differential geometry and I came across this example in the lecture notes. Which is said to be a good exercise.

Let $\omega=xdx+ydy \in \Omega^1(\mathbb{R}^2)$, and $S^1 \subset \mathbb{R}^2$. then $\forall v \in T_pS^1$, $\omega(v)=0.$

My attempt: On $S^1$, we have: $x^2+y^2=1$, thus $d(x^2+y^2)=d(1) \Rightarrow d(x^2)+d(y^2)=0 \Rightarrow2xdx+2ydy=0$, hence $xdx+ydy=0$.

I don't think this argument is correct since I have not used the $v \in T_pS^1$.

Answer 1

Your argument is actually correct, but the reason why it is correct is hidden in the simple words "On $S^1$, we have $x^2+y^2=1$...".

If you want to write this argument more formally, here's how we'd do it. $x,y$ are the coordinate functions on $\Bbb{R}^2$, so $x:\Bbb{R}^2\to\Bbb{R}$ is the function $(a,b)\mapsto a$, and $y:\Bbb{R}^2\to\Bbb{R}$ is the function $(a,b)\mapsto b$. Hence, $\omega=x\,dx+y\,dy$ is a $1$-form on $\Bbb{R}^2$. You're asked to show that for every $p\in S^1$ and every $v\in T_pS^1$, $\omega_p(v)=0$. In other words, if $\iota:S^1\to \Bbb{R}^2$ denotes the canonical inclusion, then you're supposed to show that the pullback differential form $\iota^*\omega=0$, where this is now an equality of differential $1$-forms on $S^1$.

The nice thing about pullback is that it plays very nicely with all the operations (addition, multiplication, exterior derivatives, wedge products etc). So, \begin{align} \iota^*\omega&=\iota^*(x\,dx+y\,dy)\\ &=(\iota^*x)\cdot (\iota^*dx)+(\iota^*y)\cdot (\iota^*dy)\\ &=(\iota^*x)\cdot (d(\iota^*x)) + (\iota^*y)\cdot (d(\iota^*y))\\ &=\frac{1}{2}d\left[[\iota^*x]^2+[\iota^*y]^2\right]\\ &=\frac{1}{2}d\left[(x\circ \iota)^2+(y\circ \iota)^2\right]\\ &=\frac{1}{2}d\left[1\right]\\ &=0 \end{align} Here, $\iota^*x$ means $x\circ \iota:S^1\to\Bbb{R}$, i.e it is simply the restriction $x|_{S^1}:S^1\to\Bbb{R}$. So, this is a function on $S^1$, and when you said "On $S^1$, we have $x^2+y^2=1$", what is meant is that $(x\circ \iota)^2+(y\circ \iota)^2=1$. In each and every single equal sign above, we're dealing with differential $1$-forms on the manifold $S^1$.

(because otherwise, if you take the symbols literally, you'll see that on the one hand, $\omega=x\,dx+y\,dy$, which is clearly not the zero differential form on $\Bbb{R}^2$, yet later on we're suddenly claiming $x\,dx+y\,dy=0$). So, long story short, it is this pullback via the inclusion which is enforcing the fact that we're only looking at points $p\in S^1$ and vectors $v\in T_pS^1$. With some practice, the role played by the inclusion map becomes "obvious", which is why almost no one explicitly writes it out.

Answer 2

If $v$ belongs to $T_pS^1$ then $v$ can be represented by a smooth path $\alpha:(-\varepsilon,\varepsilon)\to S^1$ where $\alpha(0)=p$ and we denote this $v=v_\alpha$. Now $v_\alpha$ acts on smooth functions as $v_\alpha(f)=(f\circ\alpha)'(0)$.

The path $\alpha$ must in the standard coordinates on $\mathbb{R}^2$ satisfy $\alpha_1(t)^2+\alpha_2(t)^2=1$ where $\alpha(t)=(\alpha_1(t),\alpha_2(t))$. Moreover for all $t$ we must have $$0=(\alpha_1(t)^2+\alpha_2(t)^2)'=2\alpha_1(t)\alpha_1'(t)+2\alpha_2(t)\alpha_2'(t)$$ and $$\alpha_1(t)\alpha_1'(t)+\alpha_2(t)\alpha_2'(t)=0$$ for all $t$.

Consider now how $w_p$ acts on $v=v_\alpha$. We get (in the local coordinates) $$w_p(v_\alpha)=p_1dx(v_\alpha)+p_2dy(v_\alpha)=p_1d\pi_1(v_\alpha)+p_2d\pi_2(v_\alpha)=$$$$p_1v_\alpha(\pi_1)+p_2v_\alpha(\pi_2)=p_1(\pi_1\circ\alpha)'(0)+p_2(\pi_2\circ\alpha)'(0)=$$$$p_1\alpha_1'(0)+p_2\alpha_2'(0)=\alpha_1(0)\alpha_1'(0)+\alpha_2(0)\alpha_2'(0)=0.$$ Here $p=(p_1,p_2)$ and $\pi_1$ and $\pi_2$ are the projections on the first and second coordinates in $\mathbb{R}^2$ respectively.

Answer 3

I don't know what facts, definitions, etc you want to use but here are several possible approaches that I can think of:

$1).\ $ One fast way is to switch to polar coordinates. Then, $\omega_p=(\cos t(-\sin t)+\sin t(\cos t))dt=0.$ There are some details to fill in, though.

$2).\ $ Another, from scratch approach might be to fix $p\in S^1$ and a chart $(p,U).$ Note that $U$ is an open in $\mathbb R^2.$ With the inclusion $i:S^1\to \mathbb R^2$ $T_pS^1,$ we may regard $T_pS^1$ as a subspace of $T_p\mathbb R^2$ via $i_*: T_pS^1 \to T_p \mathbb R^2.$ Then, a tangent vector $v\in T_pS^1$ has the form

$\tag1 v=a\frac{\partial}{\partial x}\bigg |_p+b\frac{\partial}{\partial y}\bigg |_p$

for some real numbers $a,b.$ You want to show that

$\tag2\omega_p(v)=x(p)dx_p(v)+y(p)dy_p(v)=0$

We have

$\tag3 dx\left(\frac{\partial }{\partial x}\right )=1;\ dx\left(\frac{\partial }{\partial y}\right )=0;\ dy\left(\frac{\partial }{\partial y}\right )=1;\ dy\left(\frac{\partial }{\partial x}\right )=0$

Using $(1)-(3),$ we get

$\tag4 \omega_p(v)=x(p)a+y(p)b$

But we also have the standard isomorphism $T_p\mathbb R^2\cong \mathbb R^2:$

$\tag5 \frac{\partial}{\partial x}\bigg |_p \to \vec i;\ \frac{\partial}{\partial y}\bigg |_p \to \vec j$

and so $(x(p),y(p))$ and $(a,b)$ are orthogonal in $\mathbb R^2.$ And since $(4)$ is the dot product of these vectors, it is zero.

$3).\ $ Still another way to show this would be to note that $v_pf=0$ for $f(x,y)=x^2+y^2.$ It follows from $(1)$ in item $2).,$ that $2xa+2yb=0.$ The claim now follows immediately on substituting this into $(4)$ of the same item.