Talk:Microbundles (Ex)

In case of daubts You should get familiar with the definition of microbundle.

Exercise 0.1 [Milnor1964, Lemma 2.1]. Let $M$$<wikitex>; Let us begin with the definition of [[Microbundle|microbundle]]. {{beginthm|Definition|}} An $n$-dimensional microbundle is a quadruple $(E,B,i,j)$ such that there is a sequence $$B\xrightarrow{i} E\xrightarrow{j} B$$ and the following conditions hold. #$j\circ i=\id_B$ #for all $x\in B$ there exist open neigbourhood $U\subset B$ and an open neighbourhood $V\subset E$ of $i(b)$ and a homeomorphism $$h\colon V\to U\times \mathbb{R}^n.$$ Moreover, the homeomorphism above must make the following diagram commute: $$ \xymatrix{ U \ar[d]^{i}\ar[r]& U\times\mathbb{R}^n \ar[d]^{p_1}\ V\ar[r]^{j} \ar[ur]^{h} & U,} $$ where $p_1$ is projection on the first factor and $U$ is included as a <script type="math/tex">M$ be a topological manifold. Show that $\xi_M : = (M \times M, M, \Delta_M, p_1)$$\xi_M : = (M \times M, M, \Delta_M, p_1)$ is a microbundle.

Proof.

Let $M$$M$ be a topological manifold. Then the composition $p_1\circ\Delta_M$$p_1\circ\Delta_M$ sends $x\mapsto (x,x)\mapsto x$$x\mapsto (x,x)\mapsto x$, so the first condition in the definition is satisfied.

To prove that the second condition is satisfied we need to use local chart around $x$$x$. Choose $U$$U$ to be one of the open sets coming from atlas of $M$$M$ and let $\phi\colon U\to \mathbb{R}^n$$\phi\colon U\to \mathbb{R}^n$ be associated chart. The obvious choice for neighbourhood $V\subset M\times M$$V\subset M\times M$ is to take $U\times U$$U\times U$. The first naive candidate for $h\colon V=U\times U\to U\times\mathbb{R}^n$$h\colon V=U\times U\to U\times\mathbb{R}^n$ would be map $\id\times \phi$$\id\times \phi$. However such $h$$h$ fails to make the following diagram commute

$\displaystyle \xymatrix{ &U\times U\ar[rd]^{p_1}\ar[dd]^h&\\ U\ar[ru]^{\Delta_M}\ar[rd]_{\id\times\{0\}} & & U\\ &U\times \Rr^n\ar[ru]_{p_1}&}$

since $(u,u)$$(u,u)$ is mapped to $(u,\phi(u))$$(u,\phi(u))$ and $\phi(u)$$\phi(u)$ doesn't necessarily be $0$$0$ (well, it depends on our definition of an atlas of a manifold?). We need just a small adjustment: $h(u,v)=(u,h(u)-h(v))$$h(u,v)=(u,h(u)-h(v))$.

$\square$$\square$

Exercise 0.2 [Milnor1964, Theorem 2.2]. Let $M$$M$ be a (paracompact!) smooth manifold. Show that $TM$$TM$ and $\xi_M$$\xi_M$ are isomorphic microbundles.

Proof. We have two concurring definitions of micro-tangent bundle. The first is given by the exercise above, the second by forgeting about the vector bundle structure on $TM$$TM$ and treating it just as a microbundle $(TM, M, \pi,s_0)$$(TM, M, \pi,s_0)$ where $M\xrightarrow{s_0} TM$$M\xrightarrow{s_0} TM$ is the zero section.

To fix the notation please consult the definition of microbundle isomorphism on page on microbundles .

In our case we have

$\displaystyle \xymatrix{ & V\ar[dd]^H \ar[rd]^{p_1}&\\ M\ar[dr]_{\Delta_M}\ar[ur]^{s_0} & & M\\ & M\times M \ar[ru]&}$

where $V\subset TM$$V\subset TM$ is an open neighbourhood of the zero section.

$\square$$\square$

We need to find a map $H\colon V\to

$-section in $U\times \mathbb{R}^n$. {{endthm}} {{beginthm|Exercise|{{citeD|Milnor1964|Lemma 2.1}}}} Let $M$ be a topological manifold. Show that $\xi_M : = (M \times M, M, \Delta_M, p_1)$ is a microbundle. {{endthm}} {{beginproof}} Let $M$ be a topological manifold. Then the composition $p_1\circ\Delta_M$ sends $x\mapsto (x,x)\mapsto x$, so the first condition in the definition is satisfied. To prove that the second condition is satisfied we need to use local chart around $x$. Choose $U$ to be one of the open sets coming from atlas of $M$ and let $\phi\colon U\to \mathbb{R}^n$ be associated chart. The obvious choice for neighbourhood $V\subset M\times M$ is to take $U\times U$. The first naive candidate for $h\colon V=U\times U\to U\times\mathbb{R}^n$ would be map $\id\times \phi$. However such $h$ fails to make the following diagram commute $$ \xymatrix{ U \ar[d]^{\Delta_M}\ar[r]^{\id\times \{0\}}& U\times\mathbb{R}^n \ar[d]^{p_1}\ V\ar[r]^{p_1} \ar[ur]^{h} & U,} $$ since $(u,u)$ is mapped to $(u,\phi(u))$ and $\phi(u)$ doesn't necessarily be $M$ be a topological manifold. Show that $\xi_M : = (M \times M, M, \Delta_M, p_1)$$\xi_M : = (M \times M, M, \Delta_M, p_1)$ is a microbundle.