Third Homotopy Group of the Sphere and Hopf Fibration

The idea of this note comes from the chapter $\pi_3(S^2)$ , H. Hopf, W.K. Clifford, F. Klein written by H. Samelson in I.M. James’ book, History of Topology, and more precisely from the quoted letter from Hopf to Freudhental:

In case you are still interested in the question of the [homotopy] classes of maps of the 3-sphere $S^3$ onto the 2-sphere $S^2$ I want to tell you that I now can answer this question: there exist infinitely many classes. Namely there is a class invariant of the given map the counter image of $x$ consists of finitely many simple closed oriented polygons $P_1$ , $P_2$ , …, $P_a$ and likewise the counter image of $y$ consists of polygons $Q_1$ , $Q_2$ , …, $Q_b$ . If $v_{ij}$ denotes the linking number of $P_i$ and $Q_j$ , then $\sum_{i,j} v_{ij}= \gamma$ is independent of $x, ~ y$ and of the approximation and does not change under continuous change of the map. For every $\gamma$ there exists maps. Whether to every $\gamma$ there is only one map, I do not know. If the whole $S^2$ is not covered by the image, then $\gamma$ is $=0$ . A consequence is that one cannot sweep the line elements on $S^2$ continuously into a point.

Our aim is to developp this nice geometric argument following problems 13, 14 and 15 of Milnor’s book, Topology from differential viewpoint. Thus, our main result is:

Theorem: The group $\pi_3(\mathbb{S}^2)$ is infinite.

In fact, it may be proved that $\pi_3(\mathbb{S}^2) \simeq \mathbb{Z}$ . As a consequence that $\pi_n(\mathbb{S}^m)$ may be non-trivial when $n>m$ is that the homotopy group is not determined by the CW complex structure, unlike homology and cohomology groups.

The first step is to define the linking number of two submanifolds of $\mathbb{R}^{k+1}$ with total dimension $k$ . For that purpose, we use degree theory; for more information, see Milnor’s book.

Definition: Let $f : M \to N$ be a smooth map between two $n$ -dimensional smooth compact oriented manifolds. If $p \in M$ is a regular point, let $\mathrm{sign} (df(p))$ be $+1$ if $df(p)$ is orientation-preserving and $-1$ otherwise. The degree of $f$ with respect to a regular value $q \in N$ is defined by

$\displaystyle \mathrm{deg}(f,q)= \sum\limits_{p \in f^{-1}(q)} \mathrm{sign}(df(p))$ .

Property 1: The degree $\mathrm{deg}(f,q)$ does not depend on the regular value $q$ .

In particular, property 1 allows us to define the degree $\mathrm{deg}(f)$ of a smooth map $f$ without reference to any regular value.

Property 2: If two smooth maps $f$ and $g$ are smoothly homotopic, then $\mathrm{deg}(f)= \mathrm{deg}(g)$ .

Property 3: Let $M$ , $N$ be two smooth manifolds and $f : M \to N$ be a smooth map. If there exists a smooth manifold $X$ whose boundary is $M$ and such that $f$ extends to a smooth map $X \to N$ , then $\mathrm{deg}(f)=0$ .

Properties 1 and 2 are fundamental in degree theory; property 3 is a rather technical lemma used to prove the two previous properties, but it will be useful later.

Linking number:

Definition: Let $M,N \subset \mathbb{R}^{k+1}$ be two submanifolds of total dimension $m+n=k$ . The linking number $l(M,N)$ is defined as the degree of the linking map

$\lambda : \left\{ \begin{array}{ccc} M \times N & \to & \mathbb{S}^k \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$ .

Property 4: $l(M,N)= (-1)^{(m+1)(n+1)} l(N,M)$ .

Proof. Let $\lambda_1, \lambda_2$ be the linking maps

$\lambda_1 : \left\{ \begin{array}{ccc} M \times N & \to & \mathbb{S}^k \\ (a,b) & \mapsto & \frac{a-b}{ \| a - b \| } \end{array} \right.$ and $\lambda_2 : \left\{ \begin{array}{ccc} N \times M & \to & \mathbb{S}^k \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$ ,

and let for convenience

$I : \left\{ \begin{array}{ccc} M \times N & \to & N \times M \\ (a,b) & \mapsto & (b,a) \end{array} \right.$ .

Then $\lambda_1 = - \lambda_2 \circ I$ . It is an easy exercice to show that $\deg(f \circ g)= \deg(f) \cdot \deg(g)$ when $g$ is a diffeomorphism, and that $\deg(-f)= (-1)^{k+1} \deg(f)$ . So

$l(M,N)= \deg(\lambda_1)= (-1)^{k+1} \cdot \deg(I) \cdot \deg(\lambda_2)= (-1)^{m+n+1} (-1)^{mn} l(N,M)$ . $\square$

Property 5: If $M$ bounds an oriented manifold disjoint from $N$ , then $l(M,N)=0$ .

Proof. If $X \subset \mathbb{R}^{k+1}$ is an oriented submanifold such that $\partial X = M$ , then the map

$\Lambda : \left\{ \begin{array}{ccc} X \times N & \to & \mathbb{S}^k \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$

extends the linking map $\lambda : M \times N \to \mathbb{S}^k$ , and $\partial (X \times N)= \partial X \times N= M \times N$ , hence $l(M,N)= \deg(\lambda)=0$ according to property 3. $\square$

If $M$ and $N$ are two knots in $\mathbb{R}^3$ , it is possible to compute $l(M,N)$ from a regular diagram.

More precisely, if $v \in \mathbb{S}^2$ and $P$ is the plane normal to $v$ , then the projection of $M \cup N$ on $P$ is a regular diagram $D$ if $v$ is a regular value of $\lambda$ ; moreover, $\lambda^{-1}(v)$ is exactly the crossing points of $D$ when $M$ is over $N$ . Therefore, $l(M,N)=r-s$ where $r$ (resp. $s$ ) is the number of crossing points where $\lambda$ is orientation-preserving (resp. orientation-reversing); on $D$ , such crossing points correspond respectively to

For example, the linking number computed on the following diagramm is two:

Before introducing Hopf invariant, let us mention some results about cobordism, needed in the sequel:

Some cobordism theory:

Definition: Two oriented compact manifolds $M, N$ are cobordant if there exists an oriented compact manifold $X$ , called a cobordism, such that $\partial X= M \coprod (-N)$ where $-N$ denotes $N$ with the reversed orientation.

Lemma 1: Let $f : M \to \mathbb{S}^p$ be a smooth map and $y \in \mathbb{S}^p$ be a regular value. If $z \in \mathbb{S}^p$ is a regular value sufficently close to $y$ , then there exists a cobordism $X \subset M \times [0,1]$ between $f^{-1}(y)$ and $f^{-1}(z)$ .

Proof. If $C$ is the set of singular points of $f$ , $f(C)$ is compact so there exists an open neighborhood $V$ of $y$ containing only regular values. Let $z \in V$ .

Let $\gamma : [0,1] \to \mathbb{S}^p$ be a great circle from $y$ to $z$ , and let $r_t$ be a rotation whose restriction to $\mathrm{span}(y,z)$ is a rotation sending $y$ to $\gamma(t)$ . In particular, if $r_0= \mathrm{Id}$ , $(r_t)$ defines an isotopy between $\mathrm{Id}$ and $r_1$ . Let

$F : \left\{ \begin{array}{ccc} M \times [0,1] & \to & \mathbb{S}^p \\ (x,t) & \mapsto & r_t \circ f(x) \end{array} \right.$ .

Because $r_t^{-1}(z) \in V$ , $z$ is a regular value of $F(t, \cdot)$ and finally to $F$ since

$\displaystyle dF(m,s)= r_s \circ df(m) + \frac{\partial r_t}{\partial t}_{|t=s} \circ f(m) dt$ .

Thus, $F^{-1}(z) \subset M \times [0,1]$ defines a cobordism between $f^{-1}(z)$ and $f^{-1}(r_1^{-1}(z))=f^{-1}(y)$ . $\square$

Lemma 2: If $f,g : M \to \mathbb{S}^p$ are smoothly homotopic and $y \in \mathbb{S}^p$ is a regular value for both, then there exists a cobordism $X \subset M \times [0,1]$ between $f^{-1}(y)$ and $g^{-1}(y)$ .

Proof. Let $H : M \times [0,1] \to \mathbb{S}^p$ be a homotopy between $f$ and $g$ . Let $z$ be a regular value of $H$ close to $y$ so that $f^{-1}(y)$ and $f^{-1}(z)$ , and $g^{-1}(y)$ and $g^{-1}(z)$ , are cobordant (using lemma 1). Then $H^{-1}(z)$ defines a cobordism between $f^{-1}(z)$ and $g^{-1}(z)$ . Lemma 2 follows since cobordism is an equivalent relation. $\square$

Hopf invariant:

From now on, let $f : \mathbb{S}^{2p-1} \to \mathbb{S}^p$ be a smooth map and $y \neq z$ be two regular values. Applying a stereographic projection, we may view $f^{-1}(y)$ and $f^{-1}(z)$ as submanifolds of $\mathbb{R}^{2p-1}$ . Then the linking number $l(f^{-1}(y),f^{-1}(z))$ is well-defined.

Claim 1: The linking number $l(f^{-1}(y),f^{-1}(z))$ does not depend on the stereographic projection.

Proof. A stereographic projection is an orientation-preserving diffeomorphism, so if $p$ and $q$ are two stereographic projections, then $q \circ p^{-1} \times q \circ p^{-1}$ induces an orientation-preserving diffeomorphism between $pf^{-1}(y) \times pf^{-1}(z)$ and $qf^{-1}(y) \times qf^{-1}(z)$ , hence

$l(pf^{-1}(y),pf^{-1}(z))= l(qf^{-1}(y),qf^{-1}(z))$ . $\square$

Claim 2: The linking number $l(f^{-1}(y),f^{-1}(z))$ is locally constant as a function of $y$ .

Proof. Let the mapping maps

$\lambda_1 : \left\{ \begin{array}{ccc} f^{-1}(y) \times f^{-1}(z) & \to & \mathbb{S}^{2p-2} \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$ and $\lambda_2 : \left\{ \begin{array}{ccc} f^{-1}(x) \times f^{-1}(z) & \to & \mathbb{S}^{2p-2} \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$ .

According to lemma 1, there exists a cobordism $X \subset \mathbb{R}^{2p-1} \times [0,1]$ between $f^{-1}(y)$ and $f^{-1}(x)$ when $x$ is close to $y$ . Let

$\Phi : \left\{ \begin{array}{ccc} X \times f^{-1}(z) & \to & \mathbb{S}^{2p-2} \times [0,1] \\ (a,t,b) & \to & \left( \frac{a-b}{ \| a-b \| },t \right) \end{array} \right.$ .

For convenience, let $\partial \Phi$ be the restriction of $\Phi$ on $\partial X \times f^{-1}(z)$ , and $\partial_1\Phi$ (resp. $\partial_2 \Phi$ ) be the restriction of $\partial \Phi$ on $f^{-1}(x) \times \{ 0 \} \times f^{-1}(z)$ (resp. on $f^{-1}(y) \times \{ 1 \} \times f^{-1}(z)$ ). According to property 3,

$0 = \deg(\partial \Phi)= \deg( \partial_1 \Phi)+ \deg(\partial_2 \Phi)$ .

If $\varphi_1, \varphi_2$ are the obvious diffeomorphisms

$\varphi_1 : f^{-1}(x) \times \{ 0 \} \times f^{-1}(z) \to f^{-1}(x) \times f^{-1}(z)$

and

$\varphi_2 : f^{-1}(y) \times \{1\} \times f^{-1}(z) \to f^{-1}(y) \times f^{-1}(z)$ ,

then $\partial_i \Phi= \lambda_i \circ \varphi_i$ ( $i=1,2$ ), hence

$0= \deg( \lambda_1 ) \cdot \deg(\varphi_1)+ \deg( \lambda_2) \cdot \deg(\varphi_2)$ .

But $\varphi_1$ is orientation-preserving whereas $\varphi_2$ is not. Therefore,

$l( f^{-1}(y), f^{-1}(z))= \deg( \lambda_1)= \deg(\lambda_2)= l(f^{-1}(x), f^{-1}(z))$ . $\square$

Claim 3: If $y \neq z$ are regular values of $g : \mathbb{S}^{2p-1} \to \mathbb{S}^p$ and $\| f-g \|_{\infty} <1$ , then

$l(f^{-1}(y),f^{-1}(z)) = l(g^{-1}(y), f^{-1}(z))$ .

Proof. Let $H : \mathbb{S}^{2p-1} \times [0,1] \to \mathbb{S}^p$ be the homotopy

$\displaystyle H(x,t)= \frac{(1-t)f(x)+tg(x)}{ \| (1-t)f(x)+g(x) \| }$ .

$H$ is well-defined since $(1-t)f(x)+tg(x)=0$ implies

$1= \| f(x) \| =t \| f(x)-g(x) \| \leq \| f-g \|_{\infty} <1$ ,

a contradiction. Therefore, $f$ and $g$ are smoothly homotopic. Using lemma 2, we prove in the same way as for claim 2 that the degrees of the associated linking maps are equal so that

$l(f^{-1}(y) , f^{-1}(z))= l(g^{-1}(y),f^{-1}(z))$ . $\square$

Property 6: The linking number $l(f^{-1}(y),f^{-1}(z))$ depends only on the homotopy class of $f$ .

Proof. According to claim 1, $l(f^{-1}(y),f^{-1}(z))$ does not depend on the stereographic projection. According to claim 2 and property 4, the linking number does not depend on the regular values $y$ and $z$ by connectedness of $\mathbb{S}^{2p-1}$ . Finally, if $H$ is a smooth homotopy between $f$ and $g$ , then we deduce that

$l(f^{-1}(y),f^{-1}(z))= l(g^{-1}(y),g^{-1}(z))$

from claim 3 and property 4, taking a sequence $0=t_0 < t_1 < \cdots < t_k=1$ satisfying $\| H(t_i , \cdot ) - H ( t_{i+1} , \cdot ) \|_{\infty} < 1$ for all $0 \leq i \leq k-1$ . $\square$

Definition: We define the Hopf invariant $H(f)$ of $f$ as the linking number $l(f^{-1}(y),f^{-1}(z))$ for some regular values $y \neq z$ .

We just showed that Hopf invariant is a homotopic invariant.

Property 7: If $f : \mathbb{S}^{2p-1} \to \mathbb{S}^p$ and $g : \mathbb{S}^p \to \mathbb{S}^p$ , then $H(g \circ f )= H(f) \cdot \deg(g)^2$ .

Proof. Let $y \neq z$ be two regular values of $g \circ f$ and $g$ , and let $g^{-1}(y)= \{ y_1, \dots, y_r \}$ and $g^{-1}(z)= \{ z_1, \dots, z_s \}$ . The inclusions

$f^{-1}(y_i) \times f^{-1}(z_i) \hookrightarrow (g \circ f)^{-1}(y) \times (g \circ f)^{-1} (z)$

induce an orientation-preserving diffeomorphism between

$\displaystyle \coprod\limits_{i=1}^r \coprod\limits_{j=1}^s \left( \mathrm{sign}(dg(y_i)) f^{-1}(y_i) \right) \times \left( \mathrm{sign}(dg(z_j)) f^{-1}(z_j) \right)$

and

$\displaystyle (g \circ f)^{-1}(y) \times (g \circ f )^{-1}(z)$ ,

so, if $\lambda_{ij} : \left\{ \begin{array}{ccc} f^{-1}(y_i) \times f^{-1}(z_j) & \to & \mathbb{S}^{2p-2} \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$ , we have:

$\begin{array}{lcl} \deg(\lambda) & = & \displaystyle \sum\limits_{i=1}^r \sum\limits_{j=1}^s \mathrm{sign}(dg(y_i)) \mathrm{sign}(dg(z_j)) \cdot \underset{= H(f)}{\underbrace{\deg(\lambda_{ij})}} \\ \\ & = & \displaystyle H(f) \left( \sum\limits_{i=1}^r \mathrm{sign}(dg(y_i)) \right) \left( \sum\limits_{j=1}^s \mathrm{sign}(dg(z_j)) \right) \\ \\ & = & H(f) \cdot \deg(g)^2 \hspace{1cm} \square \end{array}$

Hopf fibration:

The sphere $\mathbb{S}^3$ can be viewed as the unit sphere $\{ (z_1,z_2) \mid |z_1|^2+ |z_2|^2=1 \}$ in $\mathbb{C}^2 \simeq \mathbb{R}^4$ . Noticing that the intersection between $\mathbb{S}^3$ and any complexe line is a circle, we can say that $\mathbb{S}^3$ is covered by a family of circles $\mathbb{S}^1$ indexed by $\mathbb{C}P^1 \simeq \mathbb{S}^2$ .

More precisely, if $(x,y,z,t) \in \mathbb{S}^3 \subset \mathbb{R}^4$ then $(x+iy,z+it) \in \mathbb{S}^3 \subset \mathbb{C}^2$ belongs to the complex line $(z+it)z_2= (x+iy)z_1$ ; to this line (in $\mathbb{C}P^1$ ) is associated the complex number $\frac{x+iy}{z+it} \in \mathbb{C} \cup \{ \infty\}$ . Finally, if $h : \mathbb{S}^2 \to \mathbb{C} \cup \{\infty\}$ denotes the diffeomorphism induced by the stereographic projection, we get a point $h^{-1} \left( \frac{x+iy}{z+it} \right)$ of the sphere $\mathbb{S}^2$ .

Definition: Hopf map $\pi : \left\{ \begin{array}{ccc} \mathbb{S}^3 & \to & \mathbb{S}^2 \\ (x_1,x_2,x_3,x_4) & \mapsto & h^{-1} \left( \frac{x_1+ix_2}{x_3+ix_4} \right) \end{array} \right.$ induces the Hopf fibration $\mathbb{S}^1 \to \mathbb{S}^3 \overset{\pi}{\longrightarrow} \mathbb{S}^2$ .

It is possible to visualize the decomposition of $\mathbb{S}^3$ by circles in $\mathbb{R}^3$ using stereographic projection. It gives something like that, a decomposition in concentric torii each covered by their Villarceau circles:

Very nice animations about Hopf fibration can be found in Etienne Ghys, Joe Leys and Aurélien Alvarèz’s movie, Dimensions, and in Niles Johnson’s lecture, Visualizations of Hopf fibration. Now,

$\pi^{-1}(1,0,0)= \{ (x,y,x,y) \mid x^2+y^2= \frac{1}{2} \}$ and $\pi^{-1}(-1,0,0) = \{ (x,y,-x,-y) \mid x^2+y^2= \frac{1}{2} \}$

are two circles in $\mathbb{S}^3$ . If $\varphi : \mathbb{S}^3 \subset \mathbb{R}^4 \to \mathbb{R}^3$ is the stereographic projection with respect to $(0,0,0,1)$ , then $\varphi(x,y,z,t)= \left( \frac{x}{1-t}, \frac{y}{1-t}, \frac{z}{1-t} \right)$ , so the previous circles become

$\left\{ \left( \frac{x}{1-y}, \frac{y}{1-y}, \frac{x}{1-y} \right) \mid x^2+y^2= \frac{1}{2} \right\}$ and $\left\{ \left( \frac{x}{1+y}, \frac{y}{1+y}, - \frac{x}{1+y} \right) \mid x^2+y^2= \frac{1}{2} \right\}$ .

In fact, it is just a Hopf link; precisely, if $C_1$ and $C_2$ denotes the projections of the two previous circles on the plane $z=0$ , that is

$C_1=\left\{ \left( \frac{x}{1-y}, \frac{y}{1-y} \right) \mid x^2+y^2= \frac{1}{2} \right\}$ and $C_2= \left\{ \left( \frac{x}{1+y}, \frac{y}{1+y} \right) \mid x^2+y^2= \frac{1}{2} \right\}$ ,

then $C_1 \cap C_2= \left\{ \left( \pm \frac{1}{\sqrt{2}}, 0 \right) \right\}$ where $C_1$ is over $C_2$ at one point, and vice-versa at the other. Therefore, $H(\pi)=1$ (be careful to the orientations!).

On the other hand, $H(f)=0$ for every constant map $f : \mathbb{S}^3 \to \mathbb{S}^2$ . Indeed, let

$p : \left\{ \begin{array}{ccc} \mathbb{S}_3 & \to & D^2 \\ (x,y,z,t) & \mapsto & (z,t) \end{array} \right.$

where $D^2$ is the unit two-dimensional disk viewed as a subspace of $\mathbb{S}^2$ ; in particular, since $D^2$ is contractible, $p$ is homotopic to a constant map $f$ and $H(p)=H(f)$ . But the circles $p^{-1}(0,0)= \{ (x,y,0) \mid x^2+y^2=1 \}$ and $p^{-1}(1/2,0)= \{ (x,y,1/2) \mid x^2+y^2= 3/4 \}$ (viewed in $\mathbb{R}^3$ thanks to the stereographic projection with respect to $(0,0,0,1)$ ) are clearly unlinked, hence $H(p)=0$ according to property 5.

Moreover, using property 7, we deduce that the applications

$p_n : \left\{ \begin{array}{ccc} \mathbb{S}^{3} \subset \mathbb{C}^2 & \to & \mathbb{S}^2= \mathbb{C} \cup \{ \infty \} \\ (z_1,z_2) & \mapsto & \left( z_1/z_2 \right)^n \end{array} \right.$

define a family of pairwise non-homotopic maps, since $H(p_n)= n^2$ . We just proved our main theorem!

Hopf fibration in higher dimensions:

Hopf fibration $\mathbb{S}^1 \to \mathbb{S}^3 \to \mathbb{S}^2$ is obtained using Hopf map $\left\{ \begin{array}{ccc} \mathbb{S}^3 \subset \mathbb{C}^2 & \to & \mathbb{S}^2= \mathbb{C} \cup \{ \infty \} \\ (z_1,z_2) & \mapsto & z_1/z_2 \end{array} \right.$ . But the same thing can be done by replacing $\mathbb{C}$ with any real division algebra, for example the quaternions $\mathbb{H}$ or the octonions $\mathbb{O}$ .

The associated maps $p : \mathbb{S}^7 \subset \mathbb{H}^2 \to \mathbb{S}^4 = \mathbb{H} \cup \{ \infty \}$ and $q : \mathbb{S}^{15} \subset \mathbb{O}^2 \to \mathbb{S}^8 = \mathbb{O} \cup \{ \infty \}$ induce respectively the Hopf fibrations $\mathbb{S}^3 \longrightarrow \mathbb{S}^7 \overset{p}{\longrightarrow} \mathbb{S}^4$ and $\mathbb{S}^7 \longrightarrow \mathbb{S}^{15} \overset{q}{\longrightarrow} \mathbb{S}^8$ .

It can be shown (for example in Hatcher’s book, Algebraic topology, using a homological interpretation of Hopf invariant) that $H(p)=H(q)=1$ , hence:

Theorem: $\pi_3(\mathbb{S}^2)$ , $\pi_7(\mathbb{S}^4)$ and $\pi_{15}(\mathbb{S}^8)$ are infinite.

However, Hopf fibrations do not exist in other dimensions since a finite-dimensional real division algebra has dimension one, two, four or eight.

Algebraic topology, Topology

Cobordism, Homotopy groups, Hopf fibration, Hopf invariant, Topological degree

January 4, 2014

4 Comments

aaronmaroja

Why do $\varphi_1$ and $\varphi_2$ have opposite orientations?

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February 17, 2017 8:22 pm

chiasme

This follows from the definition of a cobordism: $f^{-1}(x) \times \{ 0 \}$ is a copy of $f^{-1}(x)$ and $f^{-1}(y) \times \{ 1 \}$ a copy of $-f^{-1}(y)$ , ie., the orientation of $f^{-1}(y)$ is reversed.

Permalink, Reply

February 18, 2017 3:47 pm
- aaronmaroja
  
  Thanks for your time to answer. See, I’m having trouble understanding why is that is your definition. Milnor just says the boundary $\partial X$ is the union of the two manifolds product $\{0\}$ and $\{1\}$ , without any regards to opposite orientation. Could you, please, explain where doe the minus sign come from?
  
  
  February 18, 2017 4:04 pm
- chiasme
  
  There exist several cobordisms, and the one I am interested in is oriented cobordism. This is because fixing an orientation is necessary to define a degree. Maybe this question can help to understand why this is a good definition.
  
  
  March 4, 2017 8:10 pm

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