Degree of a map when restricting it to a submanifold

Question

Let $X, Y$ be closed oriented smooth manifolds of dimension $(n+1)$, where $Y$ is also connected, and let $F: X → Y$ be a smooth map. As usual, the degree $\deg{F}$ is defined as

$$ \deg{F} = ∑_{x ∈ F^{-1}(y)} ±1 \; , $$

where $y ∈ Y$ is any regular value of $F$ and the sign is chosen depending on whether $F$ preserves or reverses orientation at the given point $x$.

Let $N^n ⊂ Y$ be an oriented, embedded and connected submanifold such that $N$ is closed as a subset of $Y$. Suppose $M ≔ F^{-1}(N) ⊂ X$ is also an oriented embedded submanifold of dimension $n$. By construction, $M$ and $N$ are both compact and $N$ is connected, so the degree of the restriction $\tilde{F} ≔ F|_M: M → N$ is well-defined. Is there anything that can be said about the degree of $\tilde{F}$? Do we have $\deg{\tilde{F}} = ± \deg{F}$? Notice that if $y \in Y$ is a regular value of $F$ (and thus of $\tilde{F}$) and happens to lie in $N$, then $F^{-1}(y) = \tilde{F}^{-1}(y)$ and the claim that $\deg{\tilde{F}} = \pm \deg{F}$ follows if it can be shown that:

Let $x \in F^{-1}(y) = \tilde{F}^{-1}(y)$. Then $F$ is orientation-preserving at $x$ if and only if $\tilde{F}$ is orientation-preserving (or orientation-reversing – in the case of the minus sign).

If it helps, I'm particularly interested in the special case where $X, Y$ are Riemannian manifolds and $M$ and $N$ are constructed as follows: Let $p: Y → ℝ$ be a smooth map and suppose $t ∈ ℝ$ is a regular value of both $p$ and $p ∘ F$. Then define $N ≔ p^{-1}(t)$ and $M ≔ F^{-1}(t) = (p ∘ F)^{-1}(t)$ and assume that $N$ is connected. Note that $N$ and $M$ are automatically two-sided (and thus oriented) because the gradients of $p$ and $p ∘ F$ give rise to a trivialization of their normal bundles.

My issue in proving the claim even in this special case is that I'm having trouble relating the normal vectors on $M$ and $N$ to each other as I don't know what $F$ and the two Riemannian metrics are doing.

Answer 1

Is there anything that can be said about the degree of $\tilde{F}$? Do we have $\deg \tilde{F} = \pm \deg{F}$?

I will show below that without further assumptions, the answer to this is no.

For each $n\in \mathbb{Z}$, there is a smooth map $F:S^2\rightarrow S^2$ of degree $0$ for which the restriction to the equator $\tilde{F}:S^1\rightarrow S^1$ has degree $n$.

If $p:S^2\rightarrow \mathbb{R}$ is the height function, then $N = S^1 = p^{-1}(0)$, and $0$ is a regular value for $p$. But $0$ will not be a regular value of $p\circ F$, so this doesn't answer your specific case of interest.

The map $F$ is constructed as a composition $S^ 2\xrightarrow{f} D^2\xrightarrow{\rho_n} D^2\xrightarrow{g} S^2$.

The map $f$ is projection from $S^2\subseteq \mathbb{R}^3$ to the unit disk in $\mathbb{R}^2$. In terms of coordinates, it is $f(x,y,z) = (x,y)$. When restricted to the equatorial $S^1 = \{(x,y,z)\in S^2: z = 0\}$, $f|_{S^1}:S^1\rightarrow S^1$ is the identity map (so has degree $1$).

The map $\rho_n$ is a $n$-fold rotation. In polar coordinates, it is $\rho_n(r,\theta) = (r, n\theta)$. When restricted to $S^1$, it is the canonical degree $n$ self-map of $S^1$.

The map $g$ maps $D^2$ to the northern hemisphere of $S^2$. In coordinates, it is $g(x,y) = (x,y,\sqrt{1-x^2 - y^2})$. When restricted to $S^1$, it is the identity map, so degree $1$.

Viewed as a map from $S^1$ to itself, the degree is $\deg(f)\deg(\rho_n)\deg(g) = n$. Viewed as a map from $S^2$ to itself, it is not surjective, so has degree $0$.

Answer 2

While Jason DeVito posted a nice counterexample to my question in the general case, I've managed to prove the special case I was interested in:

Let $X^{n+1}, Y^{n+1}$ be smooth, compact and oriented manifolds, with $Y$ also being connected, and $F: X → Y$ be a smooth map. Clearly, the degree $\deg{F}$ of $F$ is well-defined.

Let $p: Y → ℝ$ be another smooth map and suppose $t ∈ ℝ$ is a regular value of both $p$ and $p ∘ F$. (By Sard's theorem this is fulfilled for almost every $t$ in $ℝ$.) Define $N ≔ p^{-1}(t)$ and $M ≔ (p ∘ F)^{-1}(t) = F^{-1}(N)$. By the regular-value theorem, $N$ and $M$ are embedded submanifolds of codimension 1. Furthermore, they are compact and – as will be shown – they can be oriented canonically. Assume now in addition that $N$ is connected. Then the degree of $\tilde{F} ≔ F|_M: M → N$ is well-defined and $\deg{\tilde{F}} = \deg{F}$.

Idea: As mentioned before, the proof is based on the following insight: Suppose for a second that $y ∈ N$ is not only a regular value of $\tilde{F}$ but also of $F$. Then, since $\tilde{F}^{-1}(y) = F^{-1}(y)$,

$$ \begin{align} \deg{\tilde{F}} &= ∑_{x ∈ \tilde{F}^{-1}(y)} σ_x(\tilde{F}) = ∑_{x ∈ F^{-1}(y)} σ_x(\tilde{F}) \\ \deg{F} &= ∑_{x ∈ F^{-1}(y)} σ_x(F) \; , \end{align} $$

where $\sigma_x(F) = \pm 1$ depending on whether $F$ preserves or flips the orientation at $x$. So the claim follows if it can be shown that $σ_x(\tilde{F}) = σ_x(F)$ for all $x ∈ F^{-1}(y)$, i.e. that $\tilde{F}$ is orientation-preserving at $x$ if and only if $F$ is.

Preliminaries: To carry out the proof, equip both $X$ and $Y$ with some Riemannian metric (which exists by a partition-of-unity argument). Let $ν$ be the normal vector field on $M$ defined as

$$ \DeclareMathOperator{\grad}{grad} ν ≔ \grad(p ∘ F) \; . $$

Notice that, since $ν ≠ 0$ everywhere on $M$, this provides a trivialization of the normal bundle of $M$ (put differently, $M$ is two-sided), and so $X$ induces an orientation on $M$ in the usual way: For any $x ∈ M$, a basis $e_1, …, e_n$ of $T_x M$ is called positively oriented if and only if $e_1, …, e_n, ν$ is a positively oriented basis of the restricted bundle $T_x X$. Similarly, $N$ inherits an orientation from $Y$ by considering its normal vector field $\grad{p}$.

Since $Y$ is oriented, there exists a differential form $ω ∈ Ω^{n+1}(Y)$ of top degree such that $ω ≠ 0$ everywhere and $ω(v_1, …, v_{n+1}) > 0$ for every basis $(v_1, …, v_{n+1})$ of $TY$ which is positively oriented. Again, by definition of the induced orientation on $N$, the $n$-form $\tilde{ω} ∈ Ω^n(N)$, defined as

$$ \tilde{ω}(·, …, ·) ≔ ω(·, …, ·, \grad{p}) \; , $$

is the nowhere vanishing top-form associated with the orientation on $N$.

Step 1: Compute at any $x ∈ M$:

$$ \begin{align} ⟨F_* ν |_x, \grad{p}|_{F(x)}⟩ &= dp(F_* ν) = dp(F_* \grad(p ∘ F)) \\ &= [F_* \grad(p ∘ F)](p) \\ &= [\grad(p ∘ F)](p ∘ F) \\ &= ⟨\grad(p ∘ F), \grad(p ∘ F)⟩ \\ &= ‖\grad(p ∘ F)‖² > 0 \end{align} $$

since $M = (p ∘ F)^{-1}(t)$ and $t$ was a regular value of $p ∘ F$. In particular, this implies that $F_* ν ≠ 0$ everywhere and so if $D_x \tilde{F} = D_x F\big|_{T_xM}$ has full rank, $D_x F$ will have full rank, too. Hence, if $y ∈ N$ is a regular value of $\tilde{F}$, it will be a regular value of $F$, too. Moreover, this shows that the orientations on $M$ and $N$ given by the gradient fields (which depend on the previously chosen Riemannian metrics) are in fact independent of the choice of metric.

Step 2: So let $y ∈ N$ now be a regular value of $\tilde{F}$ and therefore, by the previous step, also of $F$, and let $x ∈ F^{-1}(y) ∈ M$. Suppose $(e_1, …, e_n)$ is any oriented basis of $T_x M$ (so that $(e_1, …, e_n, ν)$ is an oriented basis of $T_x X$). Due to the regular-value property, $(F_* e_1, …, F_* e_n)$ is a basis of $T_y N$ and so $(F_* e_1, …, F_* e_n, \grad{p})$ is a basis of $T_y Y$. Write

$$ F_* ν = ∑_i V^i F_* e_i + W \grad{p} $$

for some $V^1, …, V^n, W ∈ ℝ$. Since $\grad{p} ∈ (T_y N)^⟂$ and $F_* e_i = \tilde{F}_* e_i ∈ T_y N$, it follows that

$$ W = \frac{1}{‖\grad{p}‖²} ⟨F_* ν, \grad{p}⟩ \; , $$

where $⟨F_* ν, \grad{p}⟩ > 0$ as shown previously. Compute:

$$ \begin{align} ω(F_* e_1, …, F_* e_n, F_* ν) &= W · ω(F_* e_1, …, F_* e_n, \grad{p}) \\ &= W · \tilde{ω}(\tilde{F}_* e_1, …, \tilde{F}_* e_n) \; . \end{align} $$

Therefore, $ω(F_* e_1, …, F_* e_n, F_* ν)$ and $\tilde{ω}(\tilde{F}_* e_1, …, \tilde{F}_* e_n)$ have the same sign. Put differently, $(F_* e_1, …, F_* e_n, F_* ν)$ is an oriented basis of $T_y Y$ if and only if $(\tilde{F}_* e_1, …, \tilde{F}_* e_n)$ is one for $T_y N$. This proves the claim.

Side note: $p$ need not be defined on all of $Y$. It suffices for it to be defined on some open set $U \subset Y$ and to demand that $N \subset U$ still be compact. To see that the proof still goes through, simply replace all occurrences of $p ∘ F$ above with $p ∘ F_U$ where $F_U ≔ F|_{F^{-1}(U)}$.

Answer 3

A good night's sleep later, I think I can give the most general answer to my question. As it turns out, the particular case I was interested in is actually quite general.

Let $X^{n+1}, Y^{n+1}$ be smooth, closed and oriented manifolds, with $Y$ also being connected, and $F: X → Y$ be a smooth map. Clearly, the degree $\deg{F}$ of $F$ is well-defined.

Let $N ⊂ Y$ be a smooth, closed, connected and oriented submanifold and suppose $M ≔ F^{-1}(N)$ is a smooth oriented submanifold, too. Assume $\DeclareMathOperator{\img}{img} \img{DF} ⊄ TN$ everywhere on $M$. Then $M$ is canonically oriented (so the degree of $\tilde{F} ≔ F|_M: M → N$ is well-defined) and $\deg{\tilde{F}} = \deg{F}$.

(Note that the assumption $\img{DF} ⊄ TN$ rules out the counterexample by Jason DeVito.)

The proof rests on a reduction to the case of my previous answer. To this end, choose any Riemannian metric on $N$ (which exists by a partition-of-unity argument). By orientability of $N$, there is a unit normal field $ν$ on $N$ w.r.t. this Riemannian metric. Then for small enough $ε > 0$,

$$ U ≔ \{ \exp_y(ν t) \,|\, y ∈ N, t ∈ (-ε, ε) \} $$

is an open set diffeomorphic to $N × (-ε, ε)$ (i.e. a tubular neighborhood of $N$) and so

$$ p: U → ℝ, p(z) = p((y(z), t(z))) ≔ t(z) $$

is well-defined and smooth. Observe that $N = p^{-1}(0)$ and that for all $y ∈ N$, $t ∈ (-ε, ε)$: $∂_t p(\exp_y(ν t)) = 1$ and hence $dp ≠ 0$. Hence, $0$ is a regular value of $p$. Moreover, for every $x ∈ M = (p ∘ F_U)^{-1}(0))$ the assumption $\img{D_x F} ⊄ T_{F(x)}N$ implies that there exists some $v ∈ T_x X$ such that $⟨D_x F(v), ν⟩ ≠ 0$. Hence $d(p ∘ F_U)(v) = ⟨ν, DF(v)⟩ ≠ 0$, so $0$ is also a regular value of $p ∘ F_U$.

Therefore, the requirements of my previous answer are fulfilled and orientability of $M$ and the formula for degree follow.