# plane curve and compact Riemann surface

Riemann surfaces are one dimension complex manifold. Yet, apart from the elementary examples, like $\mathbb{P}^1,\mathbb{C},\mathbb{H}$ and their quotients by some groups, it is not easy to construct directly some compact Riemann surfaces(of course, according to the Uniformization theorem, all compact Riemann surfaces arise in as quotients of these three Riemann surfaces). Here we get inspirations from the theory of manifolds. In fact, we know that some important examples of manifolds arise as particular subsets of the Euclidean spaces. In particular, we can construct some manifolds from the loci of some polynomials. This is the way we are going to use in the following.

For a simple of dimension, we consider a polynomial of two variables non constant, $P\in \mathbb{C}[X,Y]$. We write it as $P(X,Y)=a_0(X)Y^d+a_1(X)Y^{d-1}+...+a_d(X)$ where $a_0(X)$ is not the zero polynomial. We set $C_P=\{(x,y)\in\mathbb{C}^2|P(x,y)=0\}$, the zero set of $P$. For the present, we pretend that $C_P$ is a Riemann surface(we can see that, it is not a Riemann surface because of some points). Then we should try to find these ‘bad’ points. One way is to find a holomorphic map from $C_P$ to some Riemann surface that we know, and use this map to find which points are not ‘good’.

Consider the projection $\pi_1:C_P\rightarrow \mathbb{C},(x,y)\mapsto x$. Note that, for those $x\in\mathbb{C}$ such that $a_0(x)\neq 0$, we have that $P_x(Y)=P(x,Y)$ has $d$ solutions for the variable $Y$. So, if we note $S=\{x\in\mathbb{C}|a_0(x)=0\}$, so $\pi_1: C_P-\pi_1^{-1}(S)\rightarrow \mathbb{C}-S$ is a ramified covering of degree $d$. so, in this way $C_P'=C_P-\pi_1^{-1}(S)$ is a Riemann surface.

If $a_0(X)$ is a non-zero constant, then $S=\emptyset$, we can say something more. In fact, we can compactify $\mathbb{C}$ to $\mathbb{P}^1$, and add also one point to $C_P$ to make it compact. So, in this way, we get a compact Riemann surface. In this case, we can even calculate the genus of $C_P$ using Riemann-Hurwitz formula.

Here we consider a simple example. Suppose that $P(X,Y)=Y^d-\prod_{i=1}^l(X-x_i)$ where $x_i$ are distincts points and $gcd(d,l)=1$. Then, the ramified points of $\pi_1$ are $x_1,...,x_l,\infty$ of ramification degree $d$(the point difficult is the infinity). So, according to the Riemann-Hurwitz formula, we have that

$2-2g(C_P\bigcup \{\infty\})=d(2-2g(\mathbb{P}^1))-(d-1)(l+1)$

Thus we get that

$g(C_P\bigcup \{\infty\})=(l-1)(d-1)/2$.

This formula shows that, for any genus $g$, there is a polynomial such that the Riemann surface defined by this polynomial(with compactification) is of genus $g$.

This method is, in some sense, more explicit than the quotient method.

# youth subcultures

Most part of this post is taken from wikipedia articles.

(1)the Pop. First we start by saying something about the conformism. The fifties-sixties(of the twentieth century, of course) are usually called the conformism. It is a mixture of two contradictory things: the first is originated from the time before the world war, when people tend to be rigid, reactionary, and out-dated(always around work, family and hometown); the second is after the world war, when the Europe starts to mimic or cope the American model, during which people tend to be artificiel and inhumain, like a machine automatic, technocracy. The pop defines the tendance of anti-conformism. It takes place during the seventies(from about 1965 to 1975). Here the pop itself have two variations, opposite one from the other: the minet-pop and the hippies.

(2)The hippie. The first signs of hippie emerged in German, around 1896 to 1908 when a youth movement arose as a countercultural reaction to the organized social and cultural clubs that centered around German folk music. This movement is known as ‘der wandervogel’. During the early twentieth century, the Germans settled in united states, and brought with them the values of this movement. Gradually this led to the youth movement in United States. By 1965, the hippie became an established social group in USA, and eventually spread to other countries, like Europe, Australia, Japan, etc.

The common value of the hippie: rejet the established institutions, criticize middle class values, oppose nuclear weapons, oppose Vietnam wars, champion sexual liberation, vegetarians, eco-friendly, use psychedelic drugs, oppose political and social orthodoxy, choose a gentle and non-doctrinaire ideology to express peace, love and personal freedom(For example, in the song ‘all you need is love‘ of Beatles).

The early hippies(1960-1966):Merry Prankster and others, promote drugs. There is a song ‘that is for the other one‘, written by the Grateful Dead, about the trips of Merry Prankster. During this period, New York, Berkeley, California anchored the American folk music,

Then we will say something related to politics.(Ronald Creagh) it is an utopian socialism, desire for the transformation of society not through political revolution, or through reformist action pushed forward by the state, but through the creation of a counter-society of a socialistcharacter in the midst of the current system, which will be made up of ideal communities of a more or less libertarian social form. The hippies are often pacifists, participate in non-violent political demonstrations, such as civil rights marches, anti-Vietnam war demonstrations(organize political action group to oppose the war, refuse to serve in the military and conduct ‘teach-ins’ on college campuses covering the Vietnamese history and the large political context of the war). Of course, the degree of political involvement varies from person to person among the hippies. Amon the the most active are the Yippies, a subgroup of the hippies.

Some music works help in this movement. For example, the song ‘San Francisco(be sure to wear flowers in your hair)’ by Scott McKenwie in 1967, is a homecoming song for all Vietnam veterans arriving in SF from 1967. What is more, McKenwie dedicates every American performance of ‘SF’ to Vietnam veterans, and he sings in 2002 at the 20th anniversary of the dedication of the Vietnam veterans memorial.

Hippie political expression often takes the form of ‘dropping out’ of society to implement the changes they sought.

There are other movements in which the hippies involve, like organic farming.

These movements are carried through anti-authoritarian and non-violent means. ‘The way of the hippie is antithetical to all repressive hierarchical power structures since they are adverse to the hippie goals of peace, love and freedom. Hippie do not impose their beliefs on others. Instead, hippies seek to change the world through reason and by living what they believe’.

The political ideals of the hippie are influential, like anarcho-punk, New-Age movement and others.

# elliptic curves over the complex number field

The most fruitful field over which the theory of elliptic curves is developed is perhaps $\mathbb{C}$.

First we shall explain the origin of the name, ‘elliptic curve’. This comes from elliptic integrals. Suppose that $(x/a)^2+(y/b)^2=1$ is an ellipse, then the length function with respect to $x$ is just $F(x)=\int_0^x\sqrt{1+(dy/dx)^2}dx$. After some easy calculations, we see that $F(x)=a\int_0^{x/a}\sqrt{(1-k^2x^2)/(1-x^2)}dx$ with $k^2=1-b^2/a^2$(we suppose that $b). This leads to the study of the following kind of integrals

$G(x)=\int_0^x\frac{1}{\sqrt{(1-k^2x^2)(1-x^2)}}dx$

which is an elliptic integral(of the first kind, to be more precise). It is not this integral which interests us, it is the inverse function of the elliptic integral that leads to the study of elliptic curves.

In fact, when we allow $x$ to take complex values, we can show that the inverse $G^{-1}$ is double-periodic. This is rather interesting. This, in some sense, means that, the above integral is not well defined if we allow $x$ to take complex values. Just like the logarithm function, $log(z)=log|z|+i\theta+2ki\pi(k\in\mathbb{Z})$, the elliptic integral $G$ has similar property: the value of $G(x)$ is unique up to an additive factor in $\mathbb{Z}\omega_1+\mathbb{Z}\omega_2$. This is the essence of the elliptic integrals.

Apart from constructing elliptic functions from elliptic integrals, do we have other methods? Indeed, we do have. One famous example is the Weierstrass $\wp$ function. For a lattice $\Gamma=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2$ in $\mathbb{C}$, we define

$\wp(z,\Gamma)=1/z^2+\sum_{w\in \Gamma-0}(1/(z-w)^2-1/w^2)(z\in\mathbb{C})$

It is easily shown that $\wp(z,\Gamma)$(meromorphic function) is indeed $\omega_1,\omega_2$-periodic. What is more, it has a double pole on $0\in \mathbb{C}/\Gamma$, it is an even function. Its derivative, $\wp(z,\Gamma)$, has a pole on $0$ of order $3$, it is an odd function. We can show that, the field of meromorphic functions $\mathfrak{M}(\mathbb{C}/\Gamma)$(that is, the elliptic functions with periods in $\Gamma$) can be expressed using only these two functions, that is

$\mathfrak{M}(\mathbb{C}/\Gamma)=\mathbb{C}(\wp(z,\Gamma),\wp'(z,\Gamma))$

In fact, we can show that, if $f\in\mathfrak{M}(\mathbb{C}/\Gamma)$, then it is a rational function of $\wp, \wp'$. Suppose that $f$ has a pole at $P$ of order $n$. If $P=0$ on the torus, then we can find $m,p\in\mathbb{Z}$ such that $\wp^m\wp'^p$ has also a pole on $P$ of order $n$. Then choosing some constant $c$, we have that $f-c\wp^m\wp'^p$ has a pole on $P$ of order less than $n$. In this way, we eliminate the pole of $f$ on $P=0$ of order $n$(we can continue this procedure to eliminate all the poles on $P=0$). If $f$ does not have a pole on $0$, we suppose that $f$ has $n$ poles altogether(on the torus, of course, not on the complex plane). Then consider $f(z)-f(0)$, this function has also $n$ poles altogether ans still lies in $\mathfrak{M}(\mathbb{C}/\Gamma)$. So, the function $1/(f(z)-f(0))$ has a pole on $0$ of order at least $1$. So, we can use the above procedure to eliminate its pole on $0$, and so get an elliptic function having less than $n$ poles altogether. So, with this method, we can reduce the number of poles of these rational functions of $f$. At last, we will surely get an elliptic function having no poles, which means that it must be a constant, thus proving the statement.

Another surprising result is that, $\wp,\wp'$ are not independent. In fact, we have that

$\wp'^2=4\wp^3-g_2\wp-g_3$

where $g_2=g_2(\Gamma)=60G_4(\Gamma),g_3=g_3(\Gamma)=140G_6(\Gamma)$ with $G_k(\Gamma)=\sum_{w\in \Gamma-0}1/w^k(k>2)$.

This formula ressembles much to the polynomial defining an elliptic curve. In fact, we can show that

For any elliptic curve $E(\mathbb{C})$, we can find some lattice $\Gamma$ such that up tp isomorphism $E$ is defined by a polynomial $y^2=4x^3-g_2(\Gamma)x-g_3(\Gamma)$.

In this way, we establish a one-to-one correspondance between the set of elliptic curves and the set of lattice up to scalar factor. Here the ‘scalar factor’ needs to be explained. In fact, this comes from an easy fact, $\Gamma$ should be the same as $a\Gamma$($\mathbb{C}/\Gamma$ should be the same as $\mathbb{C}/a\Gamma$) where $a$ is a non-zero complex number. In this way, we can always assume that $\Gamma=\mathbb{Z}+\mathbb{Z}\tau$ with $Im(\tau)>0$. The most important result about this correspondance is perhaps the following:

Suppose an elliptic curve $E: y^2=4x^3-g_2(\Gamma)x-g_3(\Gamma)$, then the map $\phi:\mathbb{C}/\Gamma\rightarrow E, z\mapsto (\wp(z),\wp'(z))(0\mapsto e_E)$ is a bi-holomorphism and isomorphism of groups.

In this way, we use the natural group structure on $\mathbb{C}/\Gamma$ to induce a group structure on the elliptic curve, and surprisingly, this induced group structure coincides with the more well-known group structure on the elliptic curve defined using intersection of straight lines with the elliptic curve.

In summary, we have three objets: an elliptic curve $E$, a lattice $\Gamma$(and the associated torus $\mathbb{C}/\Gamma$), and the function field $\mathbb{C}(x,y)/(y^2-(4x^3-g_2x-g_3))$.

# representation of Lie groups-complexification

Suppose that $G$ is a (real) Lie group with Lie algebra $\mathfrak{g}$. The complexification of Lie groups and Lie algebras is part of the Weyl’s trick which I will talk about in the following.

Note that $\mathfrak{g}$ is only a real vector space, not a complex one. One naive way of giving it a complex structure is to get it tensor with $\mathbb{C}$. That is, $\mathfrak{g}_c=\mathfrak{g}\bigotimes_{\mathbb{R}}\mathbb{C}$. Then the problem is to give this new vector space a $\mathbb{C}$-linear Lie bracket. One natural way is this, suppose that $v\bigotimes a,u\bigotimes b\in \mathbb{C}$, we set $[v\bigotimes a,u\bigotimes b]=[v,u]\bigotimes ab$ and then expand this Lie bracket linearly to the whole space.

Complexifying a vector space is always easy, yet the problem is to complexify a Lie group. Here we give one definition of the complexification of a Lie group(note here I use the word ‘the’ to indicate the uniqueness of complexification, which can be seen from the definition):

Definition: a complex Lie group $G'$ is called complexification of $G$, if there is a morphism of (real) Lie groups, $i:G\rightarrow G'$, such that, for any complex Lie group $H$ and any morphism of (real) Lie groups $h:G\rightarrow H$, there is a unique morphism of complex Lie groups $h':G'\rightarrow H$ such that $h'\circ i=h$.

(We will often note the complexification of a Lie group $G$ as $G_c$)A priori, there is no reason that the complexification of a Lie group exists. Indeed, this can happen. For example, the universal cover of $SL_2(\mathbb{R})$(often written as $\tilde{SL_2(\mathbb{R})}$) does not have a complexification(see here).

Yet as for compact Lie groups, there always exist complexifications. This is really an important result in the representation theory, which will lead to the Weyl’s trick:

Theorem: $G$ is a compact Lie group, then there exists a complexification $(G_c,i)$ such that (1)$i:G\rightarrow G_c$ is injective;(2)$i(G)$ is a maximal compact subgroup of $G_c$;(3)$\mathfrak{g}(G_c)=\mathfrak{g}(G)\bigotimes \mathbb{C}$.

Note in the second phrase, we use the word ‘subgroup’, which is a very strong result in Lie group theory, as is known that a subgroup of a Lie group is by definition closed.

The essential ingredient in the proof of this result is the fact that, the complexification of the unitary group $U_n$ is GL_n(\mathbb{C}). If this is established, then by Peter-Weyl theorem we know that every compact Lie group $G$ can be injected into some $GL_n(\mathbb{C})$(write it as $f:G\rightarrow GL_n(\mathbb{C})$). What is more, $U_n$ is a maximal compact subgroup of $GL_n(\mathbb{C})$, so $f(G)$ is included in some $gU_ng^{-1}$ for some $g\in GL_n(\mathbb{C})$. Then we construct $G_c$ as $G_c=\{g e^{iX}|g\in G,X\in \mathfrak{g}\}$. Then we should verify that $G_c$ is indeed a complex Lie group and is a complexification of $G$.

Remark: the construction of $G_c$ ressembles very much to the polar decomposition of a matrix.

As for why the complexification of $U_n$ is $GL_n(\mathbb{C})$, this can be better seen using their Lie algebras. It is not hard to construct an isomorphism between $\mathfrak{u}_{n c}$ and $\mathfrak{gl}_n(\mathbb{C})$. For example, for $v\bigotimes a\in \mathfrak{u}_{n c}$, we send it to $av$ in the latter space. For any $M\in \mathfrak{gl}_n(\mathbb{C})$, we send it to $(M-M^{h})/2\bigotimes 1+(-i)(M+M^h)/2\bigotimes i$ where $M^h$ means the complex transpose of $M$. This in some sense means that their corresponding complex Lie groups are isomorphic.

Now we can say something about the Weyl’s trick. For example, we want to study the representations of some Lie group, say, $SL_n(\mathbb{R})$. We do not know much about this. Then we can study the representations of its Lie algebra, $\mathfrak{sl}_n(\mathbb{R})$. Perhaps it is still a bit difficult. Then we lift to the complex representations of its complexification $\mathfrak{sl}_n(\mathbb{C})$(this can be easily verified). Here we can easily construct an isomorphism between $\mathfrak{sl}_n(\mathbb{C})$ and $\mathfrak{su}_n$. So the complex representations of $\mathfrak{sl}_n(\mathbb{C})$ are equivalent to the representations of $\mathfrak{su}_n$, which are again equivalent to the representations of $SU_n$. Since $SL_n(\mathbb{R})$ is connected, this means that the representations of $SL_n(\mathbb{R})$ are equivalent to the complex representations of $SL_n(\mathbb{C})$. The latter are equivalent to the complex representations of its Lie algebra, $\mathfrak{sl}_n(\mathbb{C})$. So, in this way we see that the representations of $SL_n(\mathbb{R})$ are equivalent to the representations of $SU_n$. In other words,

$Rep(SL_n(\mathbb{R}))\approx Rep(\mathfrak{sl}_n(\mathbb{R}))\approx Rep_c(\mathfrak{sl}_n(\mathbb{C})) \approx Rep_c(\mathfrak{su}_n(\mathbb{C}))\approx Rep(\mathfrak{su}_n)\approx Rep(SU_n)$

where the subscript $c$ means complex representations. This equivalence also means that, any irreducible representation in one category corresponds to an irreducible representation in another category. So, since $SU_n$ is compact, which means that all its finite dimensional representations are semi-simple. This implies that all the finite dimensional representations of $SL_n(\mathbb{R})$ are also semi-simple. This result is not so easily obtained if we don’t use Weyl’s trick.

# Cauchy problems in general relativity

The Einstein field equation

$G_{\alpha\beta}=T_{\alpha\beta}$

, as we have said, is highly non-linear. One other problem is, how to construct a time-space with some matter distribution on it such that the above equation is satisfied. This is not an easy problem. Note that there is a term ‘time’ in the equation, this indicates us that perhaps we can start from some low-dimension manifold(hyper-surface) and consider the evolution problem. This is the main idea for the Cauchy problem in general relativity.

Here for simplicity, we consider the case where there is no matter in the time-space, that is to say, $T=0$.

In general situation for the Cauchy problem, we seek some equation, say, $f:\mathbb{R}^{1+3}\rightarrow \mathbb{R}$ such that $Pf(t,x)=0,f(0,x)=f_0(x),\partial_tf(0,x)=f_1(x)$ where $Pf$ just represents an operator. What about the situation for a time-space? Here the $f$ should be the Riemann metric $g$, or something equivalent, operator $P$ correspond to the Einstein tensor, that is $G(g)$. What about the initial conditions? Since the starting point is a hyper-surface, this hyper-surface itself has a Riemann metric, $g'$, so $g(0,x)=g'(x)$ corresponds to $f(0,x)=f_0(x)$. As for the second condition, note that $\partial_t$ in some sense here means the normal vector for this hyper-surface. This reminds us to consider the second fundamental form $k$ which involves the normal vector field. So, the second fundamental form corresponds to the condition that $\partial_tf(0,x)=f_1(x)$.

In mathematics, we use the Ricci tensor instead of the Einstein tensor to express the Einstein field equation. So, in the vacuum case, the field equation becomes

$Ric_{\alpha\beta}=0$

Yet we can show that, if we pose the condition that $Ric_{ij}=0$ for $i,j=1,2,3$ on the whole time-space $M$and the initial condition $Ric_{0\alpha}=0$ on the hyper-surface $\Sigma$, then $Ric_{0\alpha}=0$ on all the $M$.

This result, in some sense, resembles exactly to the Cauchy problem in general.

Next I will try to count the degree of liberties. Note that, the differential equation $Ric_{ij}(g)=0$ is of second order. So, to have exactly finitely many possible solutions, we should set the initial values of $g$ and its first derivative with respect to time $\partial_tg$(thus on $\Sigma$). In the above, in fact we just set six set of values, that is, $g'_{ij}$ and $k_{ij}$. But $g$ has ten independent components, which means that to fully resolve the differential equation $Ric_{\alpha\beta}(g)=0$(note here I write deliberately the greek letters as indice, which range $0,1,2,3$, to indicate that the full Einstein equation in the vacuum is the same as $Ric_{\alpha\beta}$). That is why we pose directly the other four condition as $Ric_{0\alpha}=0$ on the whole $M$. Thus, in this way, we can say that this differential equation with the initial conditions are exactly-determined, or admits finitely many solutions.

Can we say something more about the nature of Einstein field equation? In fact, we can show that, yet under some conditions on the choice of coordinates.

Here is one example. We say that a local coordinate $(t=x^0,x^1,x^2,x^3)$ is a wave coordinate if each axis satisfies $\Box x^{\alpha}=0$. The operator $\Box$ is the d’Alembertian(in Riemannian manifold, we replace this by the Laplace operator, $\Delta x^{\alpha}=0$). We can show, that under this kind of coordinate, the Ricci tensor has a particular expression

$Ric_{\alpha\beta}=-1/2g^{\gamma\nu}\frac{\partial^2g_{\alpha\beta}}{\partial x^{\gamma}\partial x^{\nu}}+F(g)(\partial g)^2$

where $F(g)$ is a function on $g$ and $(\partial g)^2$ is a quadratic form on the first derivatives of $g$. So, we see easily, that essentially, the Einstein field equation in the vacuum is a wave equation, or a hyperbolic differential equation.

Yet this kind of choice(which is generally called, the gauge) is a bit special, if we choose another type of coordinate, then the nature of the differential equation will change, too.

Meanwhile, a choice of coordinate is necessary, in that the Einstein field equation is invariant under diffeomorphisms. So in order to choose a representative from the isomorphism class, we need to pose conditions on the coordinates.

One last remark. A priori, we can pose conditions on the coordinates only on the initial manifold, $\Sigma$. So, there is a problem of coherence. That is to say, if we pose some condition about the coordinates on $\Sigma$, then we should verify that this condition is also satisfied on the whole $M$.

There is a very nice review on the Cauchy problem in general relativity, see here. The chapter 7 of the book ‘the large scale structure of space-time’ of Hawking and Ellis discusses this problem, too.

# Hopf fiber

Basically, the Hopf fiber is a fiber structure on the $2$-sphere $S^2$. Or, it describes a decomposition of $S^3$ into $S^2$ and $S^1$.

Start from the definition of $S^3\subset \mathbb{C}^2$, it consists of points $(z,z')\in\mathbb{C}^2$ such that $|z|^2+|z'|^2=1$. On defines a projection $p:S^3\rightarrow S^2,(z,z')\mapsto (2\bar{z}z',|z|^2-|z'|^2)$ where $2\bar{z}z'$ stands for the first coordinates of points in $S^2$. Note that, the image of $p$ indeed lies in $S^2$. We will show that $p$ is surjective and for each $P\in S^2$, the fiber $p^{-1}(P)$ is a circle. And this describes a fiber structure on $S^2$.

Indeed, for any $(x+iy,w)\in S^2$, we want to find $(z,z')$ such that

$|z|^2+|z'|^2=1,|z|^2-|z'|^2=w,2\bar{z}z'=x+iy$

Yet this is almost trivial, since we have that

$|z|^2=(1+w)/2,|z'|^2=(1-w)/2,2\bar{z}z'=x+iy$

This system is always solvable. Moreover, if $p(z,z')=(x+iy,w)$, then $p^{-1}(x+iy,w)=(e^{i\theta}z,e^{i\theta}z')$. This can also be seen from the above system of equations. Note that sometimes we are likely to take this projection, $p':S^3\rightarrow S^2,(z,z')\mapsto (z,|z'|)$. One problem with this projection is that, its image is not the entire $S^2$($|z'|$ is nover negative), another point is that $p'^{-1}(p'(z,z'))=p'^{-1}(z,|z'|)=(z,e^{i\alpha}z')$, yet for the point $(z,0)\in S^2$, its fiber is only itself, only one point, not a circle.

Another interesting way to describe this fiber is to use some Lie groups. Note that, for $S^3$, the Lie group $SU(2)$ acts transitively and freely on $S^3$(because each matrix in $SU(2)$ can be written as $\begin{pmatrix}a &b \\ -\bar{b} & \bar{a} \end{pmatrix}$, in general, a matrix in $U(2)$ can be written as $e^{i\theta}\begin{pmatrix} a &b\\-\bar{b}&\bar{a}\end{pmatrix}$).

But, $SU(2)$ acts also mysteriously on $\mathbb{R}^3$! This is a rather important fact. Note that, in some sense, this is also reasonable: that is $SU(2)$ acts also on the tangent space of some point $P\in S^3$ on $S^3$, and $S^3$ has a group structure(that is exactly $SU(2)$). After giving this tangent space $T_P(S^3)$ a metric(induced from that of $S^3$, of course), $SU(2)$ acts thus on $T_P(S^3)=\mathbb{R}^3$ preserving the metric, thus acts also on $S^2\subset \mathbb{R}^3$. Yet, we know that, $SO(3)$ acts on $S^2$, we can show that the action of $SU(2)$ on $S^2$ induces a homomorphism $SU(2)\rightarrow SO(3)$, with kernel $\pm Id$. Moreover, $SO(3)$ acts transitively on $S^2$, and for each point $P\in S^2$, the stabilizer of $SO(3)$ on $P$ is a circle. Note also that, for the opposite point $-P\in S^2$, the stabilizer of $SO(3)$ on $-P$ is the same as the previous, $Stab(SO(3),P)=Stab(SO(3),-P)$. This shows that, not rigorously, that $SU(2)$ acts on $S^2$ with for each point, a stabilizer group the same as one copy of circle.

More concretely, since $SU(2)$ consists of matrices of the form $\begin{pmatrix}a&b\\-\bar{b}&\bar{a}\end{pmatrix}$ with $|a|^2+|b|^2=1$, we consider the conjugate action of $SU(2)$ on itself. This is indeed a group action, and we note this as

$Ad_g:SU(2)\rightarrow SU(2),h\mapsto ghg^{-1}$

Now take derivative with respect to the variable $h$, and transform this action to the Lie algebra

$ad_g:\mathfrak{g}(SU(2))\rightarrow \mathfrak{g}(SU(2)),H\mapsto gHg^{-1}$

Note that, the Lie algebra of $SU(2)$ consists of matrices $H\in M_{2\times 2}(\mathbb{C})$ such that $Tr(H)=0,\bar{H}^t+H=0$. More specifically, we can find a $\mathbb{R}$-basis for $\mathfrak{su}(2)=\mathfrak{g}(SU(2))$(note that, $SU(2)$ is not a complex Lie group, only a real Lie group, thus its Lie algebra is only a real vector space, not a complex one):

$e_1=\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, e_2=\begin{pmatrix}0&i\\i&0\end{pmatrix},e_3=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$

We can show that the metric induced from $SU(2)=S^3$ on $\mathfrak{su}(2)$ is preserved by $SU(2)$, and $e_1,e_2,e_3$ forms an orthonormal basis. Now, we want to show that the stabilizer of each point $P\in S^2=S^2(\mathfrak{su}(2))$ is a circle, since $SU(2)$ acts transitively on $S^2$, we can verify this for just one point, say $e_3$. Note that, if $g\in SU(2)$ preserves $e_3$, then $g=r_{\theta}=\begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} &\cos{\theta}\end{pmatrix}(\theta\in\mathbb{R})$. So, after some routine yet simple calculations, we find that, under the basis $e_1,e_2$, $ad_{r_{\theta}}=\begin{pmatrix} \cos{2\theta} & \sin{2\theta} \\ -\sin{2\theta} &\cos{2\theta}\end{pmatrix}$(why, you may wonder, there is an increase in the ‘speed’? This can be explained by taking another derivative with respect to $g$ in $ad_g$, thus we get that $ad_G(H)=[G,H]=GH-HG$. For $g=r_{\theta}$, we have that $G=\frac{dr_{\theta}}{d\theta}|_{\theta=0}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=e_3$. It is readily verified that $ad_{e_3}(e_1)=-2e_2,ad_{e_3}(e_2)=2e_1$, this little $2$ is the origin for this speed increase! And this $2$ accounts also for the fact that $\pi_1(S^3)=\pi_1(SU(2))=0,\pi_1(SO(3))=\pi_1(\mathbb{P}^3_{\mathbb{R}})=\mathbb{Z}/2$).

So, in this way, we describe another way of the Hopf fiber on $S^2$.

The second method enables us to generalize to higher dimension case.

A nice demonstration of the Hopf fiber is here.

# elliptic curves over $latex \mathbb{Q}_p$ and modulo $latex p$

Suppose that $E:y^2=x^3+ax+b$ is an elliptic curve over $\mathbb{Q}_p$. Note, a change of variable $x=p^{2n}x',y=p^{3n}y'$ gives $E:y'^2=x'^3+ap^{-4n}+bp^{-6n}$. So, we can always assume that $a,b\in\mathbb{Z}_p$(sometimes we require also that the discriminant $\Delta=4a^3+27b^2$ has maximal absolute value $|\Delta|_p$, but here we don’t need it).

For each point $P\in E(\mathbb{Q}_p)$, we can choose a representative $P=[x:y:z]$ such that $a,y,z\in\mathbb{Z}_p$ yet not all of them lie in $p\mathbb{Z}_p$(just to make sure that after modulo $p$, this point lies in $\mathbb{P}^2_{\mathbb{F}_p}$). And, we associate $[x:y:z]$ to the point $[x':y':z']\in\mathbb{P}^2_{\mathbb{F}_p}$. We see easily that this $P'=[x':y':z']$ also satisfies the equation $y^2=x^3+a'x+b'$. Thus we have defined an application $f:E(\mathbb{Q}_p)\rightarrow E(\mathbb{F}_p)=E',P\mapsto P'$. We can verify that $f$ is a group homomorphism. Yet we are not sure that, $E'$ is still an elliptic curve because we do not know, a priori, whether $\Delta'=0$ or not in $latex\mathbb{F}_p$. In any case, we define

$E^0(\mathbb{Q}_p)=\{P\in E(\mathbb{Q}_p)|P'\text{ is not singular in }E'\}$.

We can verify that if $P,Q\in E^0(\mathbb{Q}_p)$, then $P+Q\in E^0(\mathbb{Q}_p)$. What is more, since $e_E=[0:1:0]$ is not singular in $E'$, which shows that $E^0(\mathbb{Q}_p)$ is a subgroup of $E(\mathbb{Q}_p)$.

Moreover, we consider the preimage of $[0:1:0]$ in $E(\mathbb{Q}_p)$. In other words, we set

$E^1(\mathbb{Q}_p)=f^{-1}([0:1:0])$.

This shows that $E^1(\mathbb{Q}_p)$ is a subgroup of $E^0(\mathbb{Q}_p)$. In a similar way, we can define,

$E^n(\mathbb{Q}_p)=\{P=[x:y:z]\in E^1(\mathbb{Q}_p)|x/y\in p^n\mathbb{Z}_p\}$.

So, we get a filtration:

$E(\mathbb{Q}_p)\supset E^0(\mathbb{Q}_p)\supset E^1(\mathbb{Q}_p)\supset...$

Then there is an important theorem:

(1)$E(\mathbb{Q}_p)/E^0(\mathbb{Q}_p)$ is a finite set;

(2)We have an isomorphism $E^0(\mathbb{Q}_p)/E^1(\mathbb{Q}_p)\rightarrow E'^{ns}(\mathbb{F}_p)$;

(3)For each $n>0$, $E^n(\mathbb{Q}_p)$ is a subgroup of $E(\mathbb{Q}_p)$, and the map $f_n:E^n(\mathbb{Q}_p)\rightarrow\mathbb{F}_p,P=[x:y:z]\mapsto p^{-n}x/y(mod p)$ is an isomorphism;

(4) $\bigcap_nE^n(\mathbb{Q}_p)=\{0\}$.

An important corollary is

If $gcd(m,p)=1$, then the multiplication map $E^1(\mathbb{Q}_p)\rightarrow E^1(\mathbb{Q}_p),P\mapsto mP$ is a bijection(hence an isomorphism).

The corollary is a direct consequence of the third proposition in the theorem above and the Hensel’s lemma. If $P\in E^n(\mathbb{Q}_p)-E^{n+1}(\mathbb{Q}_P)$, then the isomorphism between $E^n(\mathbb{Q}_p)/E^{n+1}(\mathbb{Q}_p)=\mathbb{F}_p$ induces a multiplication map $\mathbb{F}_p\rightarrow\mathbb{F}_p,x\mapsto mx$. So, the image of $P$ in $\mathbb{F}_p$ is not trivial, and $gcd(m,p)=1$, which shows that the image of $mP$ in $\mathbb{F}_p$ is not trivial, either, thus this multiplication map is injective. To show the surjectivity, suppose again that $Q\in E^n(\mathbb{Q}_p)-E^{n+1}(\mathbb{Q}_p)$, then there is an element $x\in\mathbb{F}_p$ such that $mx=f_n(Q)$. So, there is a $P_n\in E^n(\mathbb{Q}_p)$ such that $x=f_n(P_n)$, and thus $Q-mP_n\in E^{n+1}(\mathbb{Q}_p)$. We can continue this process, and find a series of $P_m\in E^{m}(\mathbb{Q}_p)$ such that $Q-m(P_n+P_{n+1}+...+P_k)\in E^{k+1}(\mathbb{Q}_p)$. Push this process to infinity, and using the fact that $\mathbb{Q}_p$ is a complet space, we find $Q=m\sum_nP_n$. so, we conclude the proof.

So, we see that the isomorphisms $f_n:E^n(\mathbb{Q}_p)/E^{n+1}(\mathbb{Q}_p)\rightarrow\mathbb{F}_p$ are rather important maps. We first prove (3). Suppose that $E^n(\mathbb{Q}_p)$ is a subgroup of $E(\mathbb{Q}_p)$. Then for any $P=[x:y:z]\in E^n(\mathbb{Q}_p)$, we can write $x=p^Ax',y=p^By',z=p^Cz'$ with $x',y',z'$ units in $\mathbb{Z}_p$, yet since $y^2z=x^3+axz^2+bz^3$, we have that $p^{2B+C}y'^2z'=p^{3A}x'^3+ap^{A+2C}+bz'^3p^{3C}$. We know that $A-B\geq n$. It is easy to see that we must have $B\leq C$. If $C\leq A$, by taking $ord_p$ on both sides, we have that $2B+C=3C$, thus $B=C$. But  this contradicts the definition of $E^1(\mathbb{Q}_p)$(here we use the essential hypothesis that we are working in $E^1(\mathbb{Q}_p)$, not $E^0$,nor $E$, and that is why this proposition starts from $E^1$ instead of these latter two). So, we must $B\leq C\leq A$. Yet we also have $2B+C=A$, so we can set $A-B=r,C-A=2r$. So, if $P\in E^n(\mathbb{Q}_p)-E^{n+1}(\mathbb{Q}_p)$, we can assume that $P=[p^nx':y':p^{3n}z']$. Thus the equation becomes $p^{3n}y'^2z'=p^{3n}x'^3+ap^{7n}x'z'^2+bp^{9n}z'^3$. We can divide both sides by $p^{3n}$ and modulo $p$, thus we see that $P'=[x',y',z']$ satisfies the equation $Y^2Z=X^3$ in $\mathbb{P}^2_{\mathbb{F}_p}$.

A very important fact about this singular curve $Y^2Z=X^3$ is that, it has an affine group structure! We can assume $Y=1$ and the equation becomes $Z=X^3$. There is an important parametrization of this curve. That is $[x:1:x^3](x\in K)$. Using this parametrization, we can show that, the map $E(K)-\{[0:0:1]\}\rightarrow K,[x:1:x^3]\mapsto x$ is a group morphisme.

So, now we can conclude the proof for (2) by noting that the curve $Y^2Z=X^3$ over $\mathbb{F}_p$ is just isomorphic to $\mathbb{F}_p$.

The proof of the first point (1) is a bit tricky. Since $\mathbb{P}^2_{\mathbb{F}_p}$ is compact, so is $E(\mathbb{Q}_p)$. What is more, $E^0(\mathbb{Q}_p)$ is an open subgroup of $E(\mathbb{Q}_p)$, then $E(\mathbb{Q}_p)/E^0(\mathbb{Q}_p)$ is compact, discret, and thus finite.

The other two propositions (2) and (4) are not hard to prove.