# Talk:Microbundles (Ex)

(Difference between revisions)

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

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

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

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$$x$ does and define
$\displaystyle H(x,v)=(x,\exp(x,v)).$
As checked before this map maps the zero section to the diagonal. By definition of the expotential map the derivative of $H$$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$

$-section in$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)$$\xi_M : = (M \times M, M, \Delta_M, p_1)$ is a microbundle. Proof. Let $M$$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$$M\xrightarrow{s_0} TM$ is the zero section. To fix the notation please consult the definition of microbundle isomorphism on page on microbundles . In our case we have $\displaystyle \xymatrix{ & V\ar[dd]^H \ar[rd]^{\pi}&\\ M\ar[dr]_{\Delta_M}\ar[ur]^{s_0} & & M\\ & M\times M \ar[ru]_{p_1}&}$ where $V\subset TM$$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$$x$ does and define $\displaystyle H(x,v)=(x,\exp(x,v)).$ As checked before this map maps the zero section to the diagonal. By definition of the expotential map the derivative of $H$$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$$ (well, it depends on our definition of an atlas of a manifold?). We need just a small adjustment: $h(u,v)=(u,h(u)-h(v))$. {{endproof}} {{beginthm|Exercise|{{citeD|Milnor1964|Theorem 2.2}}}} Let $M$ be a (paracompact!) smooth manifold. Show that $TM$ and $\xi_M$ are isomorphic microbundles. {{endthm}} {{beginproof}} We have two concurring definitions of micro-tangent bundle. The first is given by the exercise above, the second by forgeting about the vector bundle structure on $TM$ and treating it just as a microbundle $(TM, M, \pi,s_0)$ where $M\xrightarrow{s_0} TM$ is the zero section. However to show that these two definition agree we need a notion of microbundle isomorphism. {{beginthm|Definition}} Two microbundles $(E_n,X,i_n,j_n)$, $n=1,2$ over the same space $X$ are isomorphic if there exist neighbourhoods $V_1\subset E_1$ of $i_1(B)$ and $V_2\subset E_2$ of $i_2(B)$ and a homeomorphism $H\colon V_1\to V_2$ making the following diagram commute. $$\xymatrix{ U \ar[d]^{i_1}\ar[r]^{i_2}& V_2 \ar[d]^{p_2}\ V_1\ar[r]^{p_1} \ar[ur]^{H} & U,}$$ {{endthm|Definition}} In our case we have $$\xymatrix{ U \ar[d]^{\Delta_M}\ar[r]^{s_0}& V \ar[d]^{p_2}\ U\times U\ar[r]^{p_1} \ar[ur]^{H} & U,}$$ where $V\subset TM$ is an open neighbourhood of the zero section. Because of the vector bundle structure we may identify $V\cong U\times \mathbb{R}^n$ via local trivialisation. {{endproof}} M be a topological manifold. Show that $\xi_M : = (M \times M, M, \Delta_M, p_1)$$\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$$x$. Choose $U$$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$$x$ does and define
$\displaystyle H(x,v)=(x,\exp(x,v)).$
As checked before this map maps the zero section to the diagonal. By definition of the expotential map the derivative of $H$$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$