MATH 422 Lecture Note #16 (2018 Spring)
Orientation and degree

Jae Choon Cha
POSTECH $\def\id{\operatorname{id}} \def\rel{\text{ rel }} \def\Z{\mathbb{Z}} \def\Q{\mathbb{Q}} \def\R{\mathbb{R}} \def\C{\mathbb{C}} \def\sm{\smallsetminus} \def\rank{\operatorname{rank}} \def\vp{{\vphantom{1}}} \def\Ker{\operatorname{Ker}} \def\Im{\operatorname{Im}} \def\Real{\operatorname{Re}} \def\inte{\operatorname{int}} \def\sign{\operatorname{sign}} \def\deg{\operatorname{deg}}$


We begin with the notion of an orientation of a vector space.

Orientation of a vector space

Definition. Suppose $V$ is a vector space over $\R$ with dimension $n$. Suppose $(b_1,\ldots,b_n)$ and $(b_1',\ldots,b_n')$ are bases for $V$. As a convention, a basis is always ordered. Express each $b_i'$ as a linear combination of the $b_i$, that is, $b_i'=\sum_j a_{ij} b_j$ with $a_{ij}\in \R$. We say that the bases $(b_i)$ and $(b_i')$ are equivalent if the $n\times n$ matrix $A=[a_{ij}]$ has positive determinant.

It is easily seen that this is an equivalence relation on the set of all bases for $V$. Also, since the determinant of the above basis change matrix $A$ is either positive or negative, there are exactly two equivalence classes.

Definition. An orientation on $V$ is an equivalence class of a basis. An oriented vector space is a pair $(V,o)$ of a vector space $V$ and an orientation $o$ of $V$. When the choice of the orientation is understood from the context, we often denote $(V,o)$ by $V$.

As a special case, when $\dim V=0$, an orientation for $V$ is defined to be either $+$ or $-$.

The orientation which is the equivalence class of a basis $(b_i)$ is often denoted by $(b_i)$.

Example. The standard basis $(e_i)$ of the vector space $\R^n$ is defined by $e_i = (0,\ldots,1,\ldots,0)$, where all the coordinates are zero but the $i$th is $1$. The equivalence class of $(e_i)$ is called the standard orientation for $\R^n$. Unless stated otherwise, when we regard $\R^n$ as an oriented vector space, it is equipped with the standard orientation.

If $\phi\colon V\to W$ is a linear isomorphism of vector spaces, then an orientation $(b_i)$ for $V$ gives rise to an orientation $(\phi(b_i))$ for $W$.

Definition. A linear isomorphism $\phi\colon V \to W$ between oriented vector spaces $V$ and $W$ is said to be orientation preserving if the orientation of $V$ gives rise to the orientation of $W$ under $\phi$.

Orientation of a manifold

Now, we are ready to define the orientation of a smooth manifold.

Definition. Suppose $M$ is a smooth manifold of dimension $n$ (with or without boundary). An orientation on $M$ consists of a choice of an orientation on $TM_x$ for each $x\in M$ with the following property: for each $x\in M$, there is a parametrization $g\colon U\cap \R^n_+\to M$ such that $g(U)$ is an open neighborhood of $x$ and $dg_p\colon \R^n\to TM_{g(p)}$ is orientation preserving for all $p\in U\cap\R^n_+$.

If there exists an orientation on $M$, then we say that $M$ is orientable. A pair $(M,o)$ of a smooth manifold $M$ and an orientation $o$ on $M$ is called an oriented manifold. As usual, we denote an oriented manifold $(M,o)$ by $M$, when the choice of the orientation $o$ is understood from the context.

Lemma. Suppose $M$ is a smooth manifold of dimension $n$ with nonempty boundary. If $M$ is orientable, then $\partial M$ is orientable.

Indeed, in the proof of the lemma, we will describe a rule to construct an orientation of $\partial M$ from a given orientation of $M$. For this purpose, we first consider the local model $\R^n_+$ of a manifold with boundary. Suppose $x\in \partial\R^n_+$. We have $T(\R^n_+)_x = \R^n$ and $T(\partial\R^n_+)_x = \R^{n-1} = \R^{n-1}\times 0 \subset \R^n$. For a tangent vector $v=(v_1,\ldots,v_n)\in T(\R^n_+)_x = \R^n$ which does not lie in $T(\partial\R^n_+)_x = \R^{n-1}$, there are two cases: we say that $v$ is inward if $v_n>0$, and $v$ is outward if $v_n<0$.

Now, for the general case of a smooth manifold $M$ with boundary, let $x\in \partial M$ be a boundary point. By definition, there is a parametrization $g\colon U\cap \R^n_+\to M$ such that $0\in U\subset \R^n$, $g(0)=x$. A tangent vector $v\in TM_x$ is inward (respectively outward) if $v=dg_0(w)$ for some inward (respectively outward) vector $w\in T(\R^n_+)_0 = \R^n$. Note that if $v\in TM_x$ is either inward or outward, then $v$ is not contained in the subspace $T(\partial M)_x = dg_0(\R^{n-1})$ of $TM_x$.

Proof of the lemma. Choose an orientation for $M$. Let $x\in \partial M$. Choose an outward vector $v\in TM_x$. Orient $T(\partial M)_x$ by choosing a basis $(b_1,\ldots,b_{n-1})$ for $T(\partial M)_x$ such that $(v,b_1,\ldots,b_{n-1})$ is the orientation on $TM_x$. This gives an orientation on $\partial M$. $\quad\square$

The orientation on $\partial M$ constructed in the proof of the lemma is called the orientation induced by the orientation of $M$.

Example. The following cases illustrate that our definition gives formal versions of familiar conventions.

  1. Consider the 1-dimensional manifold $M=[0,1]$. The standard orientation of $\R$ restricts to an orientation on $M=[0,1]$. What is the induced orientation on $\partial M=\{0,1\}$? At $x=1$, $v=1\in TM_x=\R$ is an outward tangent vector, and $v$ agrees with the preferred orientation of $TM_x$. So, the induced orientation at $x=1$ is $+$. At $x=1$, $v=-1$ is an outward tangent vector but this is the opposite of the preferred orientation of $TM_x$. So, the induced orientation at $x=0$ is $-$. Our convention agrees with the standard convention used for integrals: over an interval $[a,b]$, $\int_a^b f(x)\,dx = F(b)-F(a)$ if $F'(x)=f(x)$.

  2. Consider a 2-dimensional manifold $M$ in $\R^3$. At a point $x\in M$, an orientation of the tangent space $TM_x$ determines a "positive" normal direction for $M$ at $x$: if $(b_1,b_2)$ orients $TM_x$, then the cross product $n=b_1\times b_2$ is called a positive normal vector. Recall the standard right hand rule convention which determines the direction of $b_1\times b_2$: it says, indeed, $n$ is positive normal if $(b_1,b_2,n)$ agrees with the standard orientation of $\R^3$. Conversely, a nonzero vector $n\in \R^3$ perpendicular to $TM_x$ determines an orientation $(b_1,b_2)$ of $TM_x$, by the same rule. Note that if we reverse the orientation of $TM_x$, namely if we use $(b_2,b_1)$ instead, then the associated positive normal $b_2\times b_1$ is the opposite direction.

    When the surface $M\subset \R^3$ has nonemtpy boundary, the standard rule to orient the 1-dimensional manifold $\partial M$ used in multivariable calculus is as follows: at a point in $\inte M$ which is close to $\partial M$, align the thumb of your right hand along the positive normal vector, and encircle the normal vector by the remaining four fingers. Then the direction of the four fingers is the orientation of $\partial M$. This agrees with our definition of the induced orientation on $\partial M$.

Degree of a smooth map

Suppose $M$ and $N$ are oriented smooth manifolds of the same dimension $n$, without boundary. Suppose $M$ is compact and $N$ is connected.

Definition. Suppose $f\colon M \to N$ is a smooth map. For a regular point $x\in M$, define the local degree of $f$ at $x$ by

\[\deg_x f = \begin{cases} 1 & \text{if }df_x \colon TM_x\to TN_{f(x)} \text{ is orientation preserving}, \\ -1 & \text{otherwise}. \end{cases}\]

Choose a regular value $y\in N$ for $f$, using the Sard-Brown theorem. Define the degree of $f$ by

\[\deg f = \sum_{x\in f^{-1}(y)} \deg_x f.\]

Note that the above sum is finite, since the compactness of $M$ implies $\#f^{-1}(y) <\infty$. Indeed, $\deg f \equiv \#f^{-1}(y) \bmod 2$.

Our main result is the following statement.


  1. $\deg f$ is well-defined, independent of the choice of the regular value $y$.

  2. If two smooth maps $f$ and $g\colon M\to N$ are smoothly homotopic, then $\deg f = \deg g$.

Here, we say that $f$ and $g$ are smoothly homotopic if there is a smooth map $F\colon M\times I \to N$ such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$ for all $x\in M$.

Temporarily, until we prove the first part of the theorem, we denote $\sum_{x\in f^{-1}(y)} \deg_x f$ by $\deg(f,y)$.

Lemma. For a fixed smooth map $f\colon M\to N$, the subset of regular values of $f$ is open in $N$, and $\deg(f,y)$ is a locally constant function on the subset of regular values.

Proof. By our earlier argument used for $\#f^{-1}(y)$, a regular value $y\in N$ has an open neighborhood $V$ such that $f^{-1}(V)$ is a disjoint union of open subsets $U_1,\ldots,U_k \subset M$ on each of which the restriction $f|_{U_i} \colon U_i \to V$ is a diffeomorphism. (We use that $M$ is compact to show this.) From this, it follows that every $y'\in V$ is a regular value. So the set of regular values is open.

We will show that $\deg_x f$ is locally constant. The conclusion that $\deg(f,y)$ is locally constant is immediately obtained from this. By the definition of an orientation, we may assume that there is a diffeomorphism $g_i\colon U_i' \to U_i$ of a connected open subset $U_i'\subset \R^m$ such that $d(g_i)_p\colon \R^m \to TM_{g_i(p)}$ is orientation preserving for all $p\in U_i'$, and there is a diffeomorphism $h\colon V' \to V$ of an open subset $V'\subset \R^m$ such that $dh_q\colon \R^m \to TM_{h(q)}$ is orientation preserving for all $q\in V'$. Then $df_x\colon TM_x \to TM_y$ is orientation preserving at $g_i(p)=x'\in U_i$ if and only if $d(h^{-1}fg_i)_p\colon \R^n \to \R^n$ is orientation preserving. Namely, $\deg_{x'} f = \deg_p h^{-1}fg_i$. Observe that $\det d(h^{-1}fg_i)_p$ is continuous on $p$. Since $d(h^{-1}fg_i)$ is nonsingular on $U_i$ which is connected, it follows that the sign of $\det d(h^{-1}fg_i)p$ does not change on $U_i$. This shows that $\deg_p h^{-1}fg_i$ is constant on $U_i$. $\quad\square$

Lemma A. Suppose $f\colon M\to N$ is smooth, and $M$ is the boundary of another manifold $X$. If $f$ extends to a smooth map $F\colon X\to N$, then $\deg(f,y)=0$ for all regular values $y\in N$ for $f$.

Proof. First, we claim that we may assume that $y$ is a regular value of both $f$ and $F$. To show this, recall that there is an open neighborhood $V$ of $y$ in $N$, which consists of regular values for $f$, on which $\deg(f,-)$ is constant. By the Sard-Brown theorem, there is a regular value for $F$, say $y'$, in $V$. Now we have that $y'$ is a regular value for both $f$ and $F$, and that $\deg(f,y')=\deg(f,y)$.

When $y$ is a regular value of both $f$ and $F$, $P:=F^{-1}(y)$ is a compact manifold with $\partial P = F^{-1}(y)\cap M = f^{-1}(y)$. The dimension of $P$ is 1, and thus $P$ is a disjoint union of finitely many arcs and circles.

Suppose $A$ is an arc component of $P$. The boundary $\partial A$ consists of two points, say $\partial A = \{p,q\}$.

Claim. $\deg_p f = – \deg_q f$.

Proof of the claim. Recall that $TA_x = \Ker\{dF_x\colon TX_x \to TN_y\} \subset TX_x$ for $x\in A$, and thus $dF_x$ induces an isomorphism $TX_x/TA_x \cong TN_y$, since $dF_x$ is onto. Choose vectors $v_1,\ldots,v_n\in TX_x$ such that $(dF_x(v_1),\ldots,dF_x(v_n))$ is a basis which determines the given orientation for $TN_y$. Orient the one-dimensional vector space $TA_x$ by choosing a nonzero vector $v=v(x)\in TA_x$ such that $(v,v_1,\ldots,v_n)$ is the orientation of $TX_x$. It is a straightforward exercise that this orientation on $TA_x$ is well-defined, independent of the choice of the $v_i$. Moreover, this gives an orientation for $A$. This can be seen using the fact that $F$ composed with appropriate parametrizations is of the form $(x_1,\ldots,x_{n+1}) \mapsto (x_1,\ldots,x_n)$.

Note that an orientation of an arc is outward at an endpoint, and inward at the other endpoint. So, we may assume $v(p)\in TA_p$ is inward and $v(q)\in TA_q$ is outward for $A$. At $q$, a basis of $TM_q$ which determines the orientation of $M=\partial X$ can be taken as the above $(v_1,\ldots,v_n)$ used to choose $v$, since $v$ is outward and thus $(v,v_1,\ldots,v_n)$ is the orientation of $X$ by the definition of the induced orientation on $\partial X = M$. It follows that $df_q \colon TM_q \to TN_y$ is orientation preserving. On the other hand, at $p$, the same argument now shows that $df_p \colon TM_p \to TN_y$ is orientation reserving, since $v$ is inward in this case. This shows the claim.

By the claim, the two summands $\deg_p f$ and $\deg_q f$ are eliminated in the sum $\deg(f,y)=\sum_{x\in f^{-1}(y)} \deg_x f$. Since all points in $f^{-1}(y)$ are paired up in this way, it follows that $\deg(f,y)=0$. $\quad\square$

Lemma B. If $N$ is connected and $y,y'\in N$, then there is an orientation preserving diffeomorphism $h\colon N \to N$ which is homotopic to the identity and satisfies $h(y)=y'$.

Proof. Choose a path $\alpha\colon I\to N$ such that $\alpha(0)=y$ and $\alpha(1)=y'$. Since $I$ is compact and $\alpha(s)$ has an open neighborhood diffeomorphic to $\R^n$ for each $s\in [0,1]$, there is a subdivision $0=s_0 < s_1 < \cdots < s_k=1$ such that $y_{i-1}=\alpha(s_{i-1})$ and $y_i=\alpha(s_i)$ are contained in an open set $U_i$ diffeomorphic to $\R^n$ for each $i$. So, there is a diffeomorphism $g_i\colon U_i\to U_i$ such that $g_i(y_{i-1})=y_i$, $g_i|_{U_i-\overline V_i}$ is equal to the identity for some open subset $V_i\subset U_i$, and $g_i\simeq \id_{U_i}$ rel $U_i-\overline V_i$. Extend $g_i\colon U_i\to U_i$ to a diffeomorphism $h_i\colon N\to N$ by defining $g_i$ to be the identity on $N-U_i$. Then $h_i\simeq \id_N$. Let $h=h_k \circ \cdots \circ h_2\circ h_1$ be the composition. Then $h\simeq \id_N$, and $h(y)=h(y_0)=y_k = y'$. $\quad\square$

Now, we are ready to prove the theorem.

Proof of the theorem. First we claim that if $f$ and $g\colon M\to N$ are smoothly homotopic and $y\in N$ is a regular value for $f$ and $g$, then $\deg(f,y)=\deg(g,y)$. To show this, let $F\colon I\times M \to N$ be a smooth map such that $F(0,x)=f(x)$, $F(1,x)=g(x)$. Since $T(I\times M)_{(t,x)}=TI_t \times TM_x$, the standard orientation of $I$ and the given orientation of $M$ gives rise to an orientaiton of $I\times M$. Also, this orientation on $I\times M$ induces an orientation on the boundary $(\partial I\times M)= \{0\}\times M \cup \{1\}\times M$. Since the orientation of $I$ is outward at $1$ and inward at $0$, the orientation of $\{1\}\times M$ agrees with that of $M$, while the orientation of $\{0\} \times M$ is the opposite. So,

\[\deg(F|_{\partial M},y)=\deg(g,y)-\deg(f,y)\]

if $y$ is a regular value for $F|_ {\partial M}$, or equivalently if $y$ is a regular value for both $f$ and $g$. By Lemma A, $\deg(F|_ {\partial M},y)=0$. The claim follows from this.

Suppose $y$ and $y'\in N$ are two regular values of a smooth map $f\colon M\to N$. By Lemma B, $h(y)=y'$ for some diffeomorphism $h\colon N \to N$ homotopic to the identity. Since $h$ is a diffeomorphism, $y'=h(y)$ is a regular value for $h\circ f$. Moreover, we have

\[\begin{aligned} \deg(f,y) & =\deg(h\circ f, h(y)) &&\text{by the definition of }\deg, \\ & =\deg(h\circ f, y') &&\text{since }y'=h(y), \\ &= \deg(\id\circ f,y') = \deg(f,y') &&\text{by the claim}. \end{aligned}\]

This shows that $\deg f=\deg(f,y)$ is well-defined, independent of the choice of $y$.

Suppose $f$ and $g$ are smoothly homotopic. Since the set of regular values for $f$ is open and since the set of regular values for $g$ is dense, there is a common regular value $y$ for $f$ and $g$. Now, by the claim, $\deg f = \deg(f,y)=\deg(g,y)=\deg g$. This shows the smooth homotopy invariance of the degree. $\quad\square$


Our first application is an alternative proof of Brouwer's fixed point theorem. Recall from Lecture 15 that Brouwer's theorem is a consequence of the following statement:

Theorem. There is no smooth retraction $r\colon D^n \to \partial D^n$.

Indeed we will given an alternative proof of this, which is very short, using the properties of degree.

Proof. Suppose $r\colon D^n\to \partial D^n$ is a smooth retraction. By Lemma A, $r|_ {\partial D^n}=\id_{\partial D^n}$ has degree zero. But, by definition, $\deg \id_{\partial D^n}=1$. This is a contradiction. $\quad\square$

Our second application concerns tangent vector fields on a manifold.

Definition. A smooth tangent vector field on a smooth manifold $M\subset \R^k$ is a smooth map $v\colon M\to \R^k$ such that $v(x)\in TM_x$ for each $x\in M$.


  1. If $M$ is an open subset of $\R^n$, a smooth tangent vector field on $M$ is just a vector field $v\colon M\to \R^n$ which is smooth.

  2. For $x\in S^n\subset \R^{n+1}$, the tangent space $TM_x$ is given by $TM_x = \{v\in \R^{n+1}\mid x\cdot v = 0\}$. That is, $TM_x$ is the orthogonal complement $\langle x \rangle ^\perp$ of the subspace $\langle x\rangle \subset \R^{n+1}$ spanned by $x$. So, a smooth map $v\colon S^n\to \R^{n+1}$ is a tangent vector field on $S^n$ if and only if $v(x)\cdot x=0$ for all $x\in S^n$.

We say that a tangent vector field $v$ on $M$ is nowhere vanishing if $v(x)\ne 0$ for all $x\in M$.

Theorem. There exists a nowhere vanishing smooth tangent vector field on $S^n$ if and only if $n$ is odd.

Proof. Suppose $n$ is odd. Define $v\colon S^n\to \R^{n+1}$ by

\[v(x_1,\ldots,x_{n+1}) = (x_2,-x_1,\ldots,x_{n+1},-x_n)\]

for $x=(x_1,\ldots,x_{n+1}) \in S^n\subset \R^{n+1}$. Then, since $v(x)\cdot x = 0$, $v$ is a tangent vector field on $S^n$. Obviously $v$ is nowhere vanishing on $S^n$.

Conversely, suppose $v\colon S^n\to \R^{n+1}$ is a nowhere vanishing smooth tangent vector field on $S^n$. Define $f\colon S^n \to S^n$ by $f(x)=\frac{v(x)}{|v(x)|}$. Define $F(x,t)=\cos(\pi t)x + \sin (\pi t) f(x)$. Then, $F(x,t)\in S^n$ for $x\in S^n$, since $x$ and $f(x)$ are orthogonal unit vectors. That is, $F\colon S^n\times I \to S^n$ is smooth. Moreover, $F(x,0)=x$ and $F(x,1)=-x$. We will use the following lemma.


  1. Let $r\colon S^n \to S^n$ be the reflection map $r(x_1,\ldots,x_{n+1}) = (-x_1,x_2,\ldots,x_{n+1})$. Then $\deg r = -1$.

  2. Let $a\colon S^n \to S^n$ be the antipodal map $a(x)=-x$. Then $\deg a = (-1)^{n+1}$.

Proof. Consider $y=(0,\ldots,0,1)\in S^n$, the north pole. Then $r^{-1}(y)=\{y\}$, $T(S^n)_y = \R^n$, and $dr_y\colon \R^n\to \R^n$ is the linear isomorphism $(x_1,\ldots,x_n)\to (-x_1,\ldots,x_n)$. So, $\deg r = \deg(r,y)=-1$. By the well-definedness of degree, it follows that for any $x\in S^n$, $dr_x \colon T(S^n)_ x \to T(S^n)_ {r(x)}$ is orientation reversing. The same argument shows that other reflections $(x_1,\ldots,x_i,\ldots,x_{n+1}) \mapsto (x_1,\ldots,-x_i,\ldots,x_{n+1})$ have the same property. Now, since the antipodal map $a\colon S^n \to S^n$ is the composition of $n+1$ reflections, we have that $da_x$ is the composition of the $(n+1)$ derivatives of the reflections which are all orientation reversing. It follows that $\deg a = (-1)^{n+1}$. $\quad\square$

Now, returning to the proof of the theorem, recall that we showed that the identity and the antipodal map on $S^n$ are homotopic via $F$. So, they have the same degree. By the above lemma, it follows that $1=(-1)^{n+1}$. So $n$ is odd. $\quad\square$