# Talk:Microbundles (Ex)

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First, You should get familiar with the definition of microbundle.

Exercise 0.1 [Milnor1964, Lemma 2.1]. Let $M$$; First, You should get familiar with the definition of [[Microbundle|microbundle]]. {{beginthm|Exercise|{{citeD|Milnor1964|Lemma 2.1}}}} Let M be a topological manifold. Show that \xi_M : = (M \times M, M, \Delta_M, p_1) is a microbundle. {{endthm}} {{beginproof}} Let M be a topological manifold. Then the composition p_1\circ\Delta_M sends x\mapsto (x,x)\mapsto x, so the first condition in the definition is satisfied. To prove that the second condition is satisfied we need to use local chart around x. Choose U to be one of the open sets coming from atlas of M and let \phi\colon U\to \mathbb{R}^n be associated chart. The obvious choice for neighbourhood V\subset M\times M is to take U\times U. The first naive candidate for h\colon V=U\times U\to U\times\mathbb{R}^n would be map \id\times \phi. However such h fails to make the following diagram commute \xymatrix{ &U\times U\ar[rd]^{p_1}\ar[dd]^h&\ U\ar[ru]^{\Delta_M}\ar[rd]_{\id\times\{0\}} & & U\ &U\times \Rr^n\ar[ru]_{p_1}&} since (u,u) is mapped to (u,\phi(u)) and \phi(u) doesn't necessarily be be a topological manifold. Show that $\xi_M : = (M \times M, M, \Delta_M, p_1)$$\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/var/www/vhost/map.mpim-bonn.mpg.de/tmp/AppWikiTex/tex_vxPwc4$$x$. Choose $U$$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

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

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

We need to find a neighbourhood $V$$V$ and a map $H\colon V\to U\times U$$H\colon V\to U\times U$ such that points in the zero section ($\{(x,0)\}$$\{(x,0)\}$ in local coordinates) are mapped to the diagonal $\{(x,x)\}$$\{(x,x)\}$.

At each point this is easy: Fix $b\in M$$b\in M$ and let $V'\subset TM$$V'\subset TM$ be a neighbourhood of $i(b)$$i(b)$ coming from the vector bundle structure. Choose a trivialization $V'\to M\times \Rr^n$$V'\to M\times \Rr^n$ and then set $H\colon M\times \Rr^n\to M\times M$$H\colon M\times \Rr^n\to M\times M$,
$\displaystyle H(x,v)=(x,\exp(b,v)).$
By definition of $\exp$$\exp$ we have $H(b,0)=(b,\exp(b,0))=(b,b)$$H(b,0)=(b,\exp(b,0))=(b,b)$. However, we may now let $b$$b$ vary as $x$$x$ does and define
$\displaystyle H(x,v)=(x,\exp(x,v)).$
As checked before this map maps the zero section to the diagonal. By definition of the expotential map the derivative of $H/var/www/vhost/map.mpim-bonn.mpg.de/tmp/AppWikiTex/tex_I73PBF$$H$ is non-vanishing along the zero section, so by the inverse function theorem there exist a neighbourhood $V\subset TM$$V\subset TM$ of $M$$M$ on which $H$$H$ is a diffeomorphism.
$\square$$\square$

$(well, it depends on our definition of an atlas of a manifold?). We need just a small adjustment:$h(u,v)=(u,h(u)-h(v))$. {{endproof}} {{beginthm|Exercise|{{citeD|Milnor1964|Theorem 2.2}}}} Let$M$be a (paracompact!) smooth manifold. Show that$TM$and$\xi_M$are isomorphic microbundles. {{endthm}} {{beginproof}} We have two concurring definitions of micro-tangent bundle. The first is given by the exercise above, the second by forgeting about the vector bundle structure on$TM$and treating it just as a microbundle$(TM, M, \pi,s_0)$where$M\xrightarrow{s_0} TM$is the zero section. To fix the notation please consult the definition of microbundle isomorphism on page on [[Microbundle|microbundles]] . In our case we have $$\xymatrix{ & V\ar[dd]^H \ar[rd]^{\pi}&\ M\ar[dr]_{\Delta_M}\ar[ur]^{s_0} & & M\ & M\times M \ar[ru]_{p_1}&}$$ where$V\subset TM$is an open neighbourhood of the zero section. We need to find a neighbourhood$V$and a map$H\colon V\to U\times U$such that points in the zero section ($\{(x,0)\}$in local coordinates) are mapped to the diagonal$\{(x,x)\}$. At each point this is easy: Fix$b\in M$and let$V'\subset TM$be a neighbourhood of$i(b)$coming from the vector bundle structure. Choose a trivialization$V'\to M\times \Rr^n$and then set$H\colon M\times \Rr^n\to M\times M$, $$H(x,v)=(x,\exp(b,v)).$$By definition of$\exp$we have$H(b,0)=(b,\exp(b,0))=(b,b)$. However, we may now let$b$vary as$x$does and define $$H(x,v)=(x,\exp(x,v)).$$As checked before this map maps the zero section to the diagonal. By definition of the expotential map the derivative of$H$is non-vanishing along the zero section, so by the inverse function theorem there exist a neighbourhood$V\subset TM$of$M$on which$H\$ is a diffeomorphism. {{endproof}} 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/var/www/vhost/map.mpim-bonn.mpg.de/tmp/AppWikiTex/tex_vxPwc4$$x$. Choose $U$$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

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

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

We need to find a neighbourhood $V$$V$ and a map $H\colon V\to U\times U$$H\colon V\to U\times U$ such that points in the zero section ($\{(x,0)\}$$\{(x,0)\}$ in local coordinates) are mapped to the diagonal $\{(x,x)\}$$\{(x,x)\}$.

At each point this is easy: Fix $b\in M$$b\in M$ and let $V'\subset TM$$V'\subset TM$ be a neighbourhood of $i(b)$$i(b)$ coming from the vector bundle structure. Choose a trivialization $V'\to M\times \Rr^n$$V'\to M\times \Rr^n$ and then set $H\colon M\times \Rr^n\to M\times M$$H\colon M\times \Rr^n\to M\times M$,
$\displaystyle H(x,v)=(x,\exp(b,v)).$
By definition of $\exp$$\exp$ we have $H(b,0)=(b,\exp(b,0))=(b,b)$$H(b,0)=(b,\exp(b,0))=(b,b)$. However, we may now let $b$$b$ vary as $x$$x$ does and define
$\displaystyle H(x,v)=(x,\exp(x,v)).$
As checked before this map maps the zero section to the diagonal. By definition of the expotential map the derivative of $H/var/www/vhost/map.mpim-bonn.mpg.de/tmp/AppWikiTex/tex_I73PBF$$H$ is non-vanishing along the zero section, so by the inverse function theorem there exist a neighbourhood $V\subset TM$$V\subset TM$ of $M$$M$ on which $H$$H$ is a diffeomorphism.
$\square$$\square$