Problems with differential geometry

Question

I have problems with this statement:

"Let $\pi: \mathbb{R} \to S^1$ given by $\pi (x) = (\cos(x), \sin(x))$. Let $f\colon[0, l] \to S^1$ a differentiable function where $f(t)=(f_1(t), f_2(t))$ with $f_1$ and $f_2$ real-valued functions on $[0, l]$. Let $x_0 \in \pi^{-1} (f(0)) \subset \mathbb{R}$ and it defines $$\tilde{f} (t) = x_0 +\int_0^t (f_1 f_2' - f_2 f_1')du$$ Proof that $ \pi \circ \tilde{f} =f $ (this means that it is lifting of f)"

My questions:

1. What is the definition of lifting?
2. Which procedure you would suggest to me to do this proof?

Sorry for my English but I'm from latinoamerica

Answer 1

A lifting of a map $f : X \to S^1$ is a map $\tilde f : X \to \mathbb R$ such that $\pi \circ \tilde f = f$. This was explained in your question.

As a digression let me mention that $\pi$ is a covering projection which is an important concept in algebraic topology (but you need not know it here). Covering projections have a nice theory of liftings.

Let us first observe that $\tilde f$ is the unique differentiable function such that $$\tilde f(0) = x_0 \tag{1}$$ $$\tilde f'(x) = f_1(x)f'_2(x) - f_2(x)f'_1(x) \tag{2}$$ To prove $\pi \circ \tilde f = f$ we need some complex analysis.

We can write $\pi(x) = e^{ix}$. Let $P_+$ und $P_-$ be the sliced planes obtained from $\mathbb C$ by removing the non-negative half $L_+ = \{ x \in \mathbb R \mid x \ge 0 \}$ and the non-positive half $L_- = \{ x \in \mathbb R \mid x \le 0 \}$ of the real axis, respectively. These are open subsets of $\mathbb C$ whose union is $\mathbb C \setminus \{0\}$.

Choose a partition $t_0 = 0 < t_1 < \ldots < t_n = l$ of $[0,l]$ such that each interval $J_i = [t_{i-1},t_i]$ is mapped by $f$ into one of the open sets $P_+ ,P_-$.

We shall show that $\pi(\tilde f(x)) = e^{i\tilde f(x)} = f(x)$ for $x \in J_1$. Let us consider the case that $f(J_1) \subset P_+$ (the other case is treated similarly).

We have $f(0) \in P_+$ and $e^{ix_0} = f(0)$. Take a branch $\ln$ of the complex logarithm on $P_+$. Then $e^{\ln f(0)} = f(0) = e^{ix_0}$, thus $\ln f(0) = i x_0 + 2 k\pi i$ for some $k \in \mathbb Z$. W.l.o.g. we may assume that $k = 0$ (otherwise replace $\ln$ by $\ln + 2 k \pi i$ which is also a branch of the complex logarithm on $P_+$). Since $e^{\ln z} = z$ on $P_+$, it suffices to prove that $\ln f(x) = i\tilde f(x)$ for $x \in J_1$. Since $\ln f(0) = ix_0 = i \tilde f(0)$, this is equivalent to proving $$(\ln \circ f)'(x) = i \tilde f'(x) \tag{3}$$ Noting $f(x) \in S^1$, i.e. $\lvert f(x)\rvert = 1$, and using $(2)$ we get $$(\ln \circ f)'(x) = \dfrac{f'(x)}{f(x)} = \dfrac{f'(x) \overline{f(x)}}{f(x)\overline{f(x)}} = \dfrac{f'(x) \overline{f(x)}}{\lvert f(x)\rvert ^2} = f'(x) \overline{f(x)} \\ = f'_1(x)f_1(x) + f'_2(x)f_2(x) + i(f_1(x)f'_2(x) - f_2(x)f'_1(x)) \\ = \langle f'(x), f(x) \rangle + i \tilde f'(x) . $$ Here $\langle -, - \rangle$ denotes the usual (real!) scalar product on $\mathbb R^2 = \mathbb C$. But $f'(x)$ is a tangential vector to $S^1$ at $f(x)$, thus $f'(x)$ is orthogonal to $f(x)$ and we get $\langle f'(x), f(x) \rangle = 0$.

We can now repeat the above argument for $J_2$: To do so, let $x_1 = \tilde f(t_1)$ and note that $e^{ix_1} = e^{i\tilde f(t_1)} = f(t_1)$.

Do this successively for all $J_i$. This shows that $\pi(\tilde f(x)) = e^{i\tilde f(x)} = f(x)$ for $x \in [0,l]$.

Update:

The values of the function $\ln \circ f : J_1 \to \mathbb C$ are purely imaginary because $e^{\ln(f(x))} = f(x) \in S^1$ for all $x \in J_1$. Hence the values of $(\ln \circ f)' : J_1 \to \mathbb C$ are also purely imaginary. This implies that $(\ln \circ f)'(x) = i \tilde f'(x)$ without invoking the fact $\langle f'(x), f(x) \rangle = 0$. Or, if we change perspective, it shows that $\langle f'(x), f(x) \rangle = 0$.