manifold – Mathtuition88

SO(3) diffeomorphic to RP^3

$SO(3)\cong\mathbb{R}P^3$ }

Proof:

We consider $SO(3)$ as the group of all rotations about the origin of $\mathbb{R}^3$ under the operation of composition. Every non-trivial rotation is determined by its axis of rotation (a line through the origin) and its angle of rotation.

We consider $\mathbb{R}P^3$ as the unit 3-sphere $S^3$ with antipodal points identified.

Consider the map from the unit ball $D$ in $\mathbb{R}^3$ to $SO(3)$ , which maps $(x,y,z)$ to the rotation about $(x,y,z)$ through the angle $\pi\sqrt{x^2+y^2+z^2}$ (and maps $(0,0,0)$ to the identity rotation). This mapping is clearly smooth and surjective. Its restriction to the interior of $D$ is injective since on the interior $\pi\sqrt{x^2+y^2+z^2}<\pi$ . On the boundary of $D$ , two rotations through $\pi$ and through $-\pi$ are the same. Hence the mapping induces a smooth bijective map from $D/\sim$ , with antipodal points on the boundary identified, to $SO(3)$ . The inverse of this map, $\displaystyle ((x,y,z),\pi\sqrt{x^2+y^2+z^2})\mapsto(x,y,z)$ is also smooth. (To see that the inverse is smooth, write $\theta=\pi\sqrt{x^2+y^2+z^2}$ . Then $x=\sqrt{\frac{\theta^2}{\pi^2}-y^2-z^2}$ , and so $\frac{\partial^k x}{\partial\theta^k}$ exists and is continuous for all orders $k$ . Similar results hold for the variables $y$ and $z$ , and also mixed partials. By multivariable chain rule, one can see that all component functions are indeed smooth, so the inverse is smooth as claimed.)

Hence $SO(3)\cong D/\sim$ , the unit ball $D$ in $\mathbb{R}^3$ with antipodal points on the boundary identified.

Next, the mapping $\displaystyle (x,y,z)\mapsto (x,y,z,\sqrt{1-x^2-y^2-z^2})$ is a diffeomorphism between $D/\sim$ and the upper unit hemisphere of $S^3$ with antipodal points on the equator identified. The latter space is clearly diffeomorphic to $\mathbb{R}P^3$ . Hence, we have shown $\displaystyle SO(3)\cong D/\sim\cong\mathbb{R}P^3.$

SU(2) diffeomorphic to S^3 (3-sphere)

$SU(2)\cong S^3$ (diffeomorphic)

Proof:
We have that $\displaystyle SU(2)=\left\{\begin{pmatrix}\alpha &-\overline{\beta}\\ \beta &\overline{\alpha}\end{pmatrix}: \alpha, \beta\in\mathbb{C}, |\alpha|^2+|\beta|^2=1\right\}.$

Since $\mathbb{R}^4\cong\mathbb{C}^2$ , we may view $S^3$ as $\displaystyle S^3=\{(\alpha,\beta)\in\mathbb{C}^2: |\alpha|^2+|\beta|^2=1\}.$

Consider the map
$\begin{aligned}f: S^3&\to SU(2)\\ f(\alpha,\beta)&=\begin{pmatrix}\alpha &-\overline{\beta}\\ \beta &\overline{\alpha}\end{pmatrix}. \end{aligned}$

It is clear that $f$ is well-defined since if $(\alpha,\beta)\in S^3$ , then $f(\alpha,\beta)\in SU(2)$ .

If $f(\alpha_1,\beta_1)=f(\alpha_2,\beta_2)$ , it is clear that $(\alpha_1, \beta_1)=(\alpha_2,\beta_2)$ . So $f$ is injective. It is also clear that $f$ is surjective.

Note that $SU(2)\subseteq M(2,\mathbb{C})\cong\mathbb{R}^8$ , where $M(2,\mathbb{C})$ denotes the set of 2 by 2 complex matrices.

When $f$ is viewed as a function $\widetilde{f}: \mathbb{R}^4\to\mathbb{R}^8$ , it is clear that $\widetilde{f}$ and $\widetilde{f}^{-1}$ are smooth maps since their component functions are of class $C^\infty$ . Since $SU(2)$ and $S^3$ are submanifolds, the restrictions to these submanifolds (i.e.\ $f$ and $f^{-1}$ ) are also smooth.

Hence $f$ is a diffeomorphism.

De Rham Cohomology

De Rham Cohomology is a very cool sounding term in advanced math. This blog post is a short introduction on how it is defined.

Also, do check out our presentation on the relation between De Rham Cohomology and physics: De Rham Cohomology.

Definition:
A differential form $\omega$ on a manifold $M$ is said to be closed if $d\omega=0$ , and exact if $\omega=d\tau$ for some $\tau$ of degree one less.

Corollary:
Since $d^2=0$ , every exact form is closed.

Definition:
Let $Z^k(M)$ be the vector space of all closed $k$ -forms on $M$ .

Let $B^k(M)$ be the vector space of all exact $k$ -forms on $M$ .

Since every exact form is closed, hence $B^k(M)\subseteq Z^k(M)$ .

The de Rham cohomology of $M$ in degree $k$ is defined as the quotient vector space $\displaystyle H^k(M):=Z^k(M)/B^k(M).$

The quotient vector space construction induces an equivalence relation on $Z^k(M)$ :

$w'\sim w$ in $Z^k(M)$ iff $w'-w\in B^k(M)$ iff $w'=w+d\tau$ for some exact form $d\tau$ .

The equivalence class of a closed form $\omega$ is called its cohomology class and denoted by $[\omega]$ .

CP1 and S^2 are smooth manifolds and diffeomorphic (proof)

Proposition: $\mathbb{C}P^1$ is a smooth manifold.

Proof:
Define $U_1=\{[z^1, z^2]\mid z^1\neq 0\}$ and $U_2=\{[z^1, z^2]\mid z^2\neq 0\}$ . Also define $g_i: U_i\to\mathbb{C}$ by $g_1([z^1, z^2])=\frac{z^2}{z^1}$ and $g_2([z^1, z^2])=\frac{\overline{z^1}}{\overline{z^2}}$ .

Let $f:\mathbb{C}\to\mathbb{R}^2$ be the homeomorphism from $\mathbb{C}$ to $\mathbb{R}^2$ defined by $f(x+iy)=(x,y)$ and define $\phi_i: U_i\to\mathbb{R}^2$ by $\phi_i=f\circ g_i$ .

Note that $\{U_1, U_2\}$ is an open cover of $\mathbb{C}P^1$ , and $\phi_i$ are well-defined homeomorphisms (from $U_i$ onto an open set in $\mathbb{R}^2$ ). Then $\{(U_1,\phi_1), (U_2,\phi_2)\}$ is an atlas of $\mathbb{C}P^1$ .

The transition function $\displaystyle \phi_2\circ\phi_1^{-1}: \phi_1(U_1\cap U_2)\to\mathbb{R}^2,$
$\begin{aligned} \phi_2\phi_1^{-1}(x,y)&=\phi_2 g_1^{-1}(x+iy)\\ &=\phi_2([1,x+iy])\\ &=f(\frac{1}{x-iy})\\ &=f(\frac{x+iy}{x^2+y^2})\\ &=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) \end{aligned}$
is differentiable of class $C^\infty$ . Similarly, $\phi_1\circ\phi_2^{-1}:\phi_2(U_1\cap U_2)\to\mathbb{R}^2$ is of class $C^\infty$ . Hence $\mathbb{C}P^1$ is a smooth manifold.

Proposition:
$S^2$ is a smooth manifold.

Proof:
Define $V_1=S^2\setminus\{(0,0,1)\}$ and $V_2=S^2\setminus\{(0,0,-1)\}$ . Then $\{V_1, V_2\}$ is an open cover of $S^2$ .

Define $\psi_1: V_1\to\mathbb{R}^2$ by $\psi_1(x,y,z)=(\frac{x}{1-z},\frac{y}{1-z})$ and $\psi_2: V_2\to\mathbb{R}^2$ by $\psi_2(x,y,z)=(\frac{x}{1+z},\frac{y}{1+z})$ .

We can check that $\psi_1^{-1}(x,y)=(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{-1+x^2+y^2}{1+x^2+y^2})$ . Hence $\psi_1$ is a homeomorphism from $V_1$ onto an open set in $\mathbb{R}^2$ . Similarly, $\psi_2$ is a homeomorphism from $V_2$ onto an open set in $\mathbb{R}^2$ . Thus $\{V_1,V_2\}$ is an atlas for $S^2$ .

The composite $\psi_2\circ\psi_1^{-1}: \psi_1(V_1\cap V_2)\to\mathbb{R}^2$ is differentiable of class $C^\infty$ since both $\psi_2$ , $\psi_1^{-1}$ are of class $C^\infty$ . Similarly, $\psi_1\circ\psi_2^{-1}$ is of class $C^\infty$ . Thus $S^2$ is a smooth manifold.

We can also compute the transition function explicitly:
$\displaystyle \psi_2\psi_1^{-1}(x,y)=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}).$
Note that $\psi_2\psi_1^{-1}=\phi_2\phi_1^{-1}$ .

Define $h:\mathbb{C}P^1\to S^2$ by $h(\phi_1^{-1}(x,y))=\psi_1^{-1}(x,y)$ and $h(\phi_2^{-1}(x,y))=\psi_2^{-1}(x,y)$ .

We see that $h$ is well-defined since if $\phi_1^{-1}(x,y)=\phi_2^{-1}(u,v)$ then $\displaystyle (u,v)=\phi_2\phi_1^{-1}(x,y)=\psi_2\psi_1^{-1}(x,y)$ so that $\psi_2^{-1}(u,v)=\psi_1^{-1}(x,y)$ .

Similarly, we have a well-defined inverse $h^{-1}: S^2\to\mathbb{C}P^1$ defined by $h^{-1}(\psi_1^{-1}(x,y))=\phi_1^{-1}(x,y)$ and $h^{-1}(\psi_2^{-1}(x,y))=\phi_2^{-1}(x,y)$ .

We check that (from our previous workings)
$\begin{aligned} \psi_1 h\phi_1^{-1}(x,y)&=(x,y)\\ \psi_2 h\phi_1^{-1}(x,y)&=\psi_2\psi_1^{-1}(x,y)\\ \psi_1 h\phi_2^{-1}(x,y)&=\psi_1\psi_2^{-1}(x,y)\\ \psi_2 h\phi_2^{-1}(x,y)&=(x,y) \end{aligned}$
are of class $C^\infty$ . So $h$ is a smooth map. Similarly, $h^{-1}$ is smooth. Hence $h$ is a diffeomorphism.

Tangent space (Derivation definition)

Let $M$ be a smooth manifold, and let $p\in M$ . A linear map $v: C^\infty(M)\to\mathbb{R}$ is called a derivation at $p$ if it satisfies $\displaystyle v(fg)=f(p)vg+g(p)vf\qquad\text{for all}\ f,g\in C^\infty(M).$
The tangent space to $M$ at $p$ , denoted by $T_pM$ , is defined as the set of all derivations of $C^\infty(M)$ at $p$ .

Smooth/Differentiable Manifold

Smooth Manifold
A smooth manifold is a pair $(M,\mathcal{A})$ , where $M$ is a topological manifold and $\mathcal{A}$ is a smooth structure on $M$ .

Topological Manifold
A topological $n$ -manifold $M$ is a topological space such that:
1) $M$ is Hausdorff: For every distinct pair of points $p,q\in M$ , there are disjoint open subsets $U,V\subset M$ such that $p\in U$ and $q\in V$ .
2) $M$ is second countable: There exists a countable basis for the topology of $M$ .
3) $M$ is locally Euclidean of dimension $n$ : Every point of $M$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n$ . For each $p\in M$ , there exists:
– an open set $U\subset M$ containing $p$ ;
– an open set $\widetilde{U}\subset\mathbb{R}^n$ ; and
– a homeomorphism $\varphi: U\to\widetilde{U}$ .

Smooth structure
A smooth structure $\mathcal{A}$ on a topological $n$ -manifold $M$ is a maximal smooth atlas.

Smooth Atlas
$\mathcal{A}=\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in J}$ is called a smooth atlas if $M=\bigcup_{\alpha\in J}U_\alpha$ and for any two charts $(U,\varphi)$ , $(V,\psi)$ in $\mathcal{A}$ (such that $U\cap V\neq\emptyset$ ), the transition map $\displaystyle \psi\circ\varphi^{-1}:\varphi(U\cap V)\to\psi(U\cap V)$ is a diffeomorphism.

Source:
Introduction to Smooth Manifolds (Graduate Texts in Mathematics, Vol. 218) by John Lee

Differentiable Manifolds (Modern Birkhäuser Classics) by Lawrence Conlon

These two books are highly recommended books for Differentiable Manifolds. John Lee’s book has almost become the standard book. Its style is similar to Hatcher’s Algebraic Topology, it can be wordy but it has detailed description and explanation of the ideas, so it is good for those learning the material for the first time.

Lawrence Conlon’s book is more concise, and has specialized chapters that link to Algebraic Topology.

RP^n Projective n-space

Define an equivalence relation on $S^n\subset\mathbb{R}^{n+1}$ by writing $v\sim w$ if and only if $v=\pm w$ . The quotient space $P^n=S^n/\sim$ is called projective $n$ -space. (This is one of the ways that we defined the projective plane $P^2$ .) The canonical projection $\pi: S^n\to P^n$ is just $\pi(v)=\{\pm v\}$ . Define $U_i\subset P^n$ , $1\leq i\leq n+1$ , by setting $\displaystyle U_i=\{\pi(x^1, \dots, x^{n+1})\mid x^i\neq 0\}.$

Prove
1) $U_i$ is open in $P^n$ .
2) $\{U_1, \dots, U_{n+1}\}$ covers $P^n$ .
3) There is a homeomorphism $\varphi_i: U_i\to\mathbb{R}^n$ .
4) $P^n$ is compact, connected, and Hausdorff, hence is an $n$ -manifold.

Proof:
1) $\pi^{-1}U_i=\{(x^1, \dots, x^{n+1})\mid x^i\neq 0\}$ is open in $S^n$ , so $U_i$ is open in $P^n$ .
2) Let $y=\pi(x^1,\dots, x^{n+1})\in P^n$ . Then since $(x^1,\dots, x^{n+1})\neq(0,\dots,0)$ , so $y\in\bigcup_{i=1}^{n+1}U_i$ . Hence $P^n\subset\bigcup_{i=1}^{n+1}U_i$ .
3) Consider $A=\{(x^1,\dots, x^{n+1})\mid x^i+1\}\cong\mathbb{R}^n$ . Define $\displaystyle \varphi_i(\pi(x^1,\dots, x^{n+1}))=(\frac{x^1}{\|x^i\|},\dots,\frac{x^{i-1}}{\|x^i\|},1,\dots,\frac{x^{n+1}}{\|x^i\|})$ for $x^i>0$ . If $x^i<0$ , then $\varphi_i(\pi(x^1,\dots, x^{n+1}))=\varphi_i(\pi(-x^1,\dots, -x^{n+1}))$ . Then $\varphi_i$ is well-defined.

$\displaystyle \varphi_i^{-1}(x^1,\dots,1,\dots,x^{n+1})=\pi(\frac{x^1}{\|v\|},\dots,\frac{1}{\|v\|},\dots,\frac{x^{n+1}}{\|v\|}),$ where $v=(x^1,\dots, 1,\dots, x^{n+1})$ . Both $\varphi_i$ and $\varphi_i^{-1}$ are continuous, so $\varphi_i: U_i\to A$ is a homeomorphism.
4) Since $S^n$ is compact and connected, so is $P^n=S^n/\sim$ . $P^n$ is a CW-complex with one cell in each dimension, i.e.\ $P^n=\bigcup_{i=0}^n e^n$ . Since CW-complexes are Hausdorff, so is $P^n$ .

Equivalence of C^infinity atlases

Equivalence of $C^\infty$ atlases is an equivalence relation. Each $C^\infty$ atlas on $M$ is equivalent to a unique maximal $C^\infty$ atlas on $M$ .

Proof:

Reflexive: If $A$ is a $C^\infty$ atlas, then $A\cup A=A$ is also a $C^\infty$ atlas.

Symmetry: Let $A$ and $B$ be two $C^\infty$ atlases such that $A\cup B$ is also a $C^\infty$ atlas. Then certainly $B\cup A$ is also a $C^\infty$ atlas.

Transitivity: Let $A, B, C$ be $C^\infty$ atlases, such that $A\cup B$ and $B\cup C$ are both $C^\infty$ atlases.

Notation:
$\begin{aligned} A&=\{(U_\alpha,\varphi_\alpha)\}\\ B&=\{(V_\beta, \psi_\beta)\}\\ C&=\{(W_\gamma, f_\gamma)\}. \end{aligned}$

Then $\displaystyle \varphi_\alpha\circ f_\gamma^{-1}=\varphi_\alpha\circ\psi_\beta^{-1}\circ\psi_\beta\circ f_\gamma^{-1}: f_\gamma(U_\alpha\cap W_\gamma)\to\varphi_\alpha(U_\alpha\cap W_\gamma)$ is a diffeomorphism since both $\varphi_\alpha\circ\psi_\beta^{-1}$ and $\psi_\beta\circ f_\gamma^{-1}$ are diffeomorphisms due to $A\cup B$ and $B\cup C$ being $C^\infty$ atlases. Also, $M=\bigcup U_\alpha$ , $M=\bigcup W_\gamma$ implies $M=(\bigcup U_\alpha)\cup(\bigcup W_\gamma)$ so $A\cup C$ is also a $C^\infty$ atlas.

Let $A$ be a $C^\infty$ atlas on $M$ . Define $B$ to be the union of all $C^\infty$ atlases equivalent to $A$ . Then $B\sim A$ . If $B'\sim A$ , then $B'\subseteq B$ , so that $B$ is the unique maximal $C^\infty$ atlas equivalent to $A$ .

Tangent Space is Vector Space

Prove that the operation of linear combination, as in Definition 2.2.7, makes $T_p(U)$ into an $n$ -dimensional vector space over $\mathbb{R}$ . The zero vector is the infinitesimal curve represented by the constant $p$ . If $\langle s\rangle_p\in T_p(U)$ , then $-\langle s\rangle_p=\langle s^-\rangle_p$ where $s^-(t)=s(-t)$ , defined for all sufficiently small values of $t$ .

Proof:

We verify the axioms of a vector space.

Multiplicative axioms:
* $1\langle s_1\rangle_p=\langle 1s_1+0-(1+0-1)p\rangle_p=\langle s_1\rangle_p$
* $(ab)\langle s_1\rangle_p=\langle abs_1-(ab-1)p\rangle_p$
$\begin{aligned} a(b\langle s_1\rangle_p)&=a\langle bs_1-(b-1)p\rangle_p\\ &=\langle abs_1-(ab-a)p-(a-1)p\rangle_p\\ &=\langle abs_1-(ab-1)p\rangle_p\\ &=(ab)\langle s_1\rangle_p \end{aligned}$

Additive Axioms:
* $\langle s_1\rangle_p+\langle s_2\rangle_p=\langle s_2\rangle_p+\langle s_1\rangle_p=\langle s_1+s_2-p\rangle_p$
* $\begin{aligned} (\langle s_1\rangle_p+\langle s_2\rangle_p)+\langle s_3\rangle_p&=\langle s_1+s_2-p\rangle_p+\langle s_3\rangle_p\\ &=\langle s_1+s_2-p+s_3-p\rangle_p\\ &=\langle s_1+s_2+s_3-2p\rangle_p \end{aligned}$
$\begin{aligned} \langle s_1\rangle_p+(\langle s_2\rangle_p+\langle s_3\rangle_p)&=\langle s_1\rangle_p+\langle s_2+s_3-p\rangle_p\\ &=\langle s_1+s_2+s_3-2p\rangle_p \end{aligned}$
* $\langle s\rangle_p+\langle s^-\rangle_p=\langle s+s^- -p\rangle_p$

$\frac{d}{dt}f(s(t)+s(-t)-p)|_{t=0}=0=\frac{d}{dt}f(p)|_{t=0}$

Hence $\langle s\rangle_p+\langle s^-\rangle_p=\langle p\rangle_p$ .
* $\langle s_1\rangle_p+\langle p\rangle_p=\langle s_1+p-p\rangle_p=\langle s_1\rangle_p$

Distributive Axioms:
* $\begin{aligned} a(\langle s_1\rangle_p+\langle s_2\rangle_p)&=a\langle s_1+s_2-p\rangle_p\\ &=\langle a(s_1+s_2-p)-(a-1)p\rangle_p \end{aligned}$
$\begin{aligned} a\langle s_1\rangle_p+a\langle s_2\rangle_p&=\langle as_1-(a-1)p\rangle_p+\langle as_2-(a-1)p\rangle_p\\ &=\langle a(s_1+s_2)-2(a-1)p-p\rangle_p\\ &=\langle a(s_1+s_2-p)-(a-1)p\rangle_p\\ &=a(\langle s_1\rangle_p+\langle s_2\rangle_p) \end{aligned}$
* $(a+b)\langle s_1\rangle_p=\langle (a+b)s_1-(a+b-1)p\rangle_p$

$a\langle s_1\rangle_p+b\langle s_1\rangle_p=\langle as_1+bs_1-(a+b-1)p\rangle_p=(a+b)\langle s_1\rangle_p$

Hence $T_p(U)$ is a vector space over $\mathbb{R}$ . Since $U\subseteq\mathbb{R}^n$ , $T_p(U)$ is $n$ -dimensional.

Multivariable Derivative and Partial Derivatives

If $L_p(x)=c+\sum_{i=1}^n b_ix^i$ is a derivative of $f$ at $p$ , then $\displaystyle b_i=\frac{\partial f}{\partial x^i}(p),$ $1\leq i\leq n$ . In particular, if $f$ is differentiable at $p$ , these partial derivatives exist and the derivative $L_p$ is unique.

Proof:
Let $h=x-p$ , then $\lim_{x\to p}\frac{f(x)-L_p(x)}{\|x-p\|}=0$ becomes $\displaystyle \lim_{h\to 0}\frac{f(p+h)-f(p)}{\|h\|}-\lim_{h\to 0}\frac{L_p(h)-c}{\|h\|}=0$ since $L_p(p+h)=L_p(p)+L_p(h)-c=f(p)+L_p(h)-c$ .

By choosing $h=(0,\dots,h^i,\dots, 0)$ (all zeroes except in $i$ th position), then as $h\to 0$ , $\displaystyle \lim_{h\to 0}\frac{f(p+h)-f(p)}{\|h\|}=\frac{\partial f}{\partial x^i}(p)$ and $\displaystyle \lim_{h\to 0}\frac{L_p(h)-c}{\|h\|}=\lim_{h^i\to 0}\frac{b_ih^i}{h^i}=b_i.$ So $b_i=\frac{\partial f}{\partial x^i}(p)$ .