N-dimensional sphere and degree problem

Question

Let $f,g:\mathbb{S}^{n}\rightarrow\mathbb{S}^{n}$ be two differential mappings such that $\langle f(x),g(x)\rangle\neq 0$ for all $x\in\mathbb{S}^{n}$. Prove that $deg(f) = \pm deg(g)$. Here it is worth mentioning that $\mathbb{S}^{n} = \{\textbf{x}\in\mathbb{R}^{n+1}; |\textbf{x}| = 1\}$. Moreover, the notation $deg(f)$ denotes the degree of the application $f$. I've tried to solve it but did not succeed so far. Thank you in advance.

Answer 1

I take $S^n$ to be the unit sphere in $\Bbb R^{n + 1}$. Thus

$\vert y \vert = 1 \tag 0$

for all $y \in S^n$.

Since for all $x \in S^n$,

$\langle f(x), g(x) \rangle \ne 0, \tag 1$

we can assume that $\langle f(x), g(x) \rangle$ does not change sign on $S^n$; so suppose we assume to begin that

$\langle f(x), g(x) \rangle > 0. \tag 2$

For $t \in [0. 1]$, we define

$L(x, t) = (1 - t) f(x) + t g(x); \tag 3$

then

$\vert L(x, t) \vert^2 = \langle L(x, t), L(x, t) \rangle = \langle (1 - t) f(x) + t g(x), (1 - t) f(x) + t g(x) \rangle$ $= (1 - t)^2 \langle f(x), f(x) \rangle + 2t(1 - t) \langle f(x), g(x) \rangle + t^2 \langle g(x), g(x) \rangle$ $= (1 - t)^2 \vert f(x) \vert^2 + 2t(1 - t)\langle f(x), g(x) \rangle + t^2 \vert g(x) \vert^2 ; \tag 4$

by virtue of (0) and (2), we thus have

$\vert L(x, t) \vert^2 = (1 - t)^2 + 2t(1 - t) \langle f(x), g(x) \rangle + t^2 \ge (1 - t)^2 + t^2 = 2t^2 - 2t + 1; \tag 5$

since

$\dfrac{d(2t^2 - 2t + 1)}{dt} = 4t - 2, \tag 6$

which is zero for

$t = \dfrac{1}{2}, \tag 7$

$2t^2 - 2t + 1$ takes its minimum value of $1/2$ at $t = 1/2$; thus we have that

$\vert L(x, t) \vert^2 \ge \dfrac{1}{2}, \tag 8$

so

$\vert L(x, t) \vert \ge \dfrac{1}{\sqrt 2}; \tag 9$

in the light of (9) we see that

$H(x, t) = \dfrac{L(x, t)}{\vert L(x, t) \vert} \tag{10}$

is well-defined for all $x \in S^n$ and $t \in [0, 1]$; also

$\vert H(x, t) \vert = \dfrac{\vert L(x, t) \vert}{\vert L(x, t) \vert} = 1, \tag{11}$

$H(x, 0) = f(x), \tag{12}$

$H(x, 1) = g(x); \tag{13}$

we see via (11)-(13) that $H(x, t)$ is in fact a continuous homotopy 'twixt $f(x)$ and $g(x)$ in $S^n$; thus

$\deg f = \deg g. \tag{14}$

In the event that

$\langle f(x), g(x) \rangle < 0, \tag{15}$

we set

$L(x, t) = (1 - t)(-f(x)) + tg(x); \tag{16}$

then in a manner similar to the above we find once again that

$\vert L(x, t) \vert^2 = (1 - t)^2 -2t(1 - t)\langle f(x), g(x) \rangle + t^2 \ge 2t^2 - 2t + 1; \tag{17}$

again we have

$\vert L(x, t) \vert \ge \dfrac{1}{\sqrt 2}; \tag{18}$

so again we may set

$H(x, t) = \dfrac{L(x, t)}{\vert L(x, t) \vert} \tag{19}$

and now we obtain a homotopy 'twixt $-f(x)$ and $g(x)$; thus

$\deg g = - \deg f. \tag{20}$

As long as (1) binds, we have

$\deg g = \pm \deg f, \tag{21}$

which was to be demonstrated.

Answer 2

I think Lewis's answer has a small mistake, taking $f(x)=x,\ g(x)=-x$ i.e. Identity map and an antipodal map on $S^n$. Now we have $(f(x),g(x))=(x,-x)\equiv -1\ on\ S^n$. However we know that: $\deg (f)=1,\ \deg(g)=(-1)^{n+1}$. It seems to be an error to say $\deg f=-\deg g$, due to it should be dependent on dimision.

Now check out that from the homotopic Lewis constructed above, we obtain $$ \deg (-f)=\deg g,\ \forall (f,g)<0\ on\ S^n$$

Denote the antipodal map as $\alpha:S^n\rightarrow S^n,\ x\mapsto -x$, then we have:$$ \deg(-f)=\deg(\alpha\circ f)=\deg\alpha\cdot\deg f=(-1)^{n+1}\deg f $$ Hence the result maintain the same but a little bit different when $(f,g)<0\ on\ S^n$.