Talk:Microbundles (Ex)

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<wikitex>;
<wikitex>;
Let us begin with the definition of [[Microbundle|microbundle]].
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First, You should get familiar with the definition of [[Microbundle|microbundle]].
{{beginthm|Definition|}}
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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.
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#$j\circ i=\id_B$
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#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 $$h\colon V\to U\times \mathbb{R}^n.$$
+
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Moreover, the homeomorphism above must make the following diagram commute:
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$$
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\xymatrix{
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U \ar[d]^{i}\ar[r]& U\times\mathbb{R}^n \ar[d]^{p_1}\\
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V\ar[r]^{j} \ar[ur]^{h} & U,}
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$$
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where $p_1$ is projection on the first factor and $U$ is included as a $0$-section in $U\times \mathbb{R}^n$.
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{{endthm}}
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{{beginthm|Exercise|{{citeD|Milnor1964|Lemma 2.1}}}}
{{beginthm|Exercise|{{citeD|Milnor1964|Lemma 2.1}}}}
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To prove that the second condition is satisfied we need to use local chart around $x$.
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
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{
\xymatrix{
U \ar[d]^{\Delta_M}\ar[r]^{\id\times \{0\}}& U\times\mathbb{R}^n \ar[d]^{p_1}\\
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&U\times U\ar[rd]^{p_1}\ar[dd]^h&\\
V\ar[r]^{p_1} \ar[ur]^{h} & U,}
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U\ar[ru]^{\Delta_M}\ar[rd]_{\id\times\{0\}} & & U\\
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&U\times \Rr^n\ar[ru]_{p_1}&}
$$
$$
since $(u,u)$ is mapped to $(u,\phi(u))$ and $\phi(u)$ doesn't necessarily be $0$ (well, it depends on our definition of an atlas of a manifold?). We need just a small adjustment: $h(u,v)=(u,h(u)-h(v))$.
since $(u,u)$ is mapped to $(u,\phi(u))$ and $\phi(u)$ doesn't necessarily be $0$ (well, it depends on our definition of an atlas of a manifold?). We need just a small adjustment: $h(u,v)=(u,h(u)-h(v))$.
Line 39: Line 28:
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.
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.
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To fix the notation please consult the definition of microbundle isomorphism on page on [[Microbundle|microbundles]] .
{{beginthm|Definition}}
+
In our case we have
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{
\xymatrix{
U \ar[d]^{i_1}\ar[r]^{i_2}& V_2 \ar[d]^{p_2}\\
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& V\ar[dd]^H \ar[rd]^{\pi}&\\
V_1\ar[r]^{p_1} \ar[ur]^{H} & U,}
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M\ar[dr]_{\Delta_M}\ar[ur]^{s_0} & & M\\
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& M\times M \ar[ru]_{p_1}&}
$$
$$
{{endthm|Definition}}
In our case we have
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where $V\subset TM$ is an open neighbourhood of the zero section.
$$
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\xymatrix{
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We need to find a neighbourhood $V$ and a map $H\colon V\to U\times U$ such that points in the zero section ($\{(x,0)\}$ in local coordinates) are mapped to the diagonal $\{(x,x)\}$.
U \ar[d]^{\Delta_M}\ar[r]^{s_0}& V \ar[d]^{p_2}\\
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U\times U\ar[r]^{p_1} \ar[ur]^{H} & U,}
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At each point this is easy: Fix $b\in M$ and let $V'\subset TM$ be a neighbourhood of $i(b)$ coming from the vector bundle structure. Choose a trivialization $V'\to M\times \Rr^n$ and then set $H\colon M\times \Rr^n\to M\times M$, $$H(x,v)=(x,\exp(b,v)).$$By definition of $\exp$ we have $H(b,0)=(b,\exp(b,0))=(b,b)$.
$$
+
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.
+
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}}
{{endproof}}
</wikitex>
</wikitex>

Latest revision as of 09:16, 30 May 2012

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

Exercise 0.1 [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.

Proof.

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

\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) is mapped to (u,\phi(u)) and \phi(u) doesn't necessarily be 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)).

\square


Exercise 0.2 [Milnor1964, Theorem 2.2]. Let M be a (paracompact!) smooth manifold. Show that TM and \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 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 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 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,
\displaystyle 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
\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 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.
\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 be a topological manifold. Show that \xi_M : = (M \times M, M, \Delta_M, p_1) is a microbundle.

Proof.

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

\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) is mapped to (u,\phi(u)) and \phi(u) doesn't necessarily be 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)).

\square


Exercise 0.2 [Milnor1964, Theorem 2.2]. Let M be a (paracompact!) smooth manifold. Show that TM and \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 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 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 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,
\displaystyle 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
\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 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.
\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) is a microbundle.

Proof.

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

\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) is mapped to (u,\phi(u)) and \phi(u) doesn't necessarily be 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)).

\square


Exercise 0.2 [Milnor1964, Theorem 2.2]. Let M be a (paracompact!) smooth manifold. Show that TM and \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 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 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 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,
\displaystyle 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
\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 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.
\square


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