Talk:Microbundles (Ex)

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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.
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.
#$j\circ i=\id_B$
#$j\circ i=\id_B$
#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 $$h \colon V \to U\times \mathbb{R}^n.$$
+
#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:
Moreover, the homeomorphism above must make the following diagram commute:
$$ \xymatrix{& V \ar[dr]^{j|_V} \ar[dd] \\ U \ar[dr]_{\times 0} \ar[ur]^{i|_U} & & U \\ & U \times \Rr^n \ar[ur]_{p_1}} $$
+
$$
+
\xymatrix{
+
U \ar[d]^{i}\ar[r]& U\times\mathbb{R}^n \ar[d]^{p_1}\\
+
V\ar[r]^{j} \ar[ur]^{h} & U,}
+
$$
+
where $p_1$ is projection on the first factor and $U$ is included as a $0$-section in $U\times \mathbb{R}^n$.
{{endthm}}
{{endthm}}
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Let $M$ be a topological manifold. Show that $\xi_M : = (M \times M, M, \Delta_M, p_1)$ is a microbundle.
Let $M$ be a topological manifold. Show that $\xi_M : = (M \times M, M, \Delta_M, p_1)$ is a microbundle.
{{endthm}}
{{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.
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 $x$.
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 candidate for $V\subset M\times M$ is to take $U\times U$. Now the first naive candidate for $h\colon V=U\times U\to U\times\mathbb{R}^n$ would be map $\id\times \phi$. However
+
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 candidate for $V\subset M\times M$ 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 $0$ (well, it depends on our definition of an atlas of a manifold?). We need just a small adjustment: $h(u,v)=(u,h(u)-h(v))$.
+
{{endproof}}
+
{{beginthm|Exercise|{{citeD|Milnor1964|Theorem 2.2}}}}
{{beginthm|Exercise|{{citeD|Milnor1964|Theorem 2.2}}}}
Let $M$ be a smooth manifold. Show that $TM$ and $\xi_M$ are isomorphic microbundles.
+
Let $M$ be a (paracompact!) smooth manifold. Show that $TM$ and $\xi_M$ are isomorphic microbundles.
{{endthm}}
{{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
+
+
+
{{endproof}}
+
</wikitex>
</wikitex>

Revision as of 20:25, 29 May 2012

Let us begin with the definition of microbundle.

Definition 0.1.

An n-dimensional microbundle is a quadruple (E,B,i,j) such that there is a sequence
\displaystyle B\xrightarrow{i} E\xrightarrow{j} B
and the following conditions hold.
  1. j\circ i=\id_B
  2. 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
    \displaystyle h\colon V\to U\times \mathbb{R}^n.

Moreover, the homeomorphism above must make the following diagram commute:

\displaystyle  \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 0-section in U\times \mathbb{R}^n.

Exercise 0.2 [Milnor1964, Lemma 2.1].

Let
Tex syntax error
be a topological manifold. Show that \xi_M : = (M \times M, M, \Delta_M, p_1) is a microbundle.

Proof.

Let
Tex syntax error
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
Tex syntax error
and let \phi\colon U\to \mathbb{R}^n be associated chart. The obvious candidate for 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 \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 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 0.3 [Milnor1964, Theorem 2.2].

Let
Tex syntax error
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.

However to show that these two definition agree we need a notion of microbundle isomorphism.

Definition 0.4. 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.

\displaystyle  \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,}

In our case


\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 candidate for $V\subset M\times M$ 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 -dimensional microbundle is a quadruple (E,B,i,j) such that there is a sequence
\displaystyle B\xrightarrow{i} E\xrightarrow{j} B
and the following conditions hold.
  1. j\circ i=\id_B
  2. 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
    \displaystyle h\colon V\to U\times \mathbb{R}^n.

Moreover, the homeomorphism above must make the following diagram commute:

\displaystyle  \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 0-section in U\times \mathbb{R}^n.

Exercise 0.2 [Milnor1964, Lemma 2.1].

Let
Tex syntax error
be a topological manifold. Show that \xi_M : = (M \times M, M, \Delta_M, p_1) is a microbundle.

Proof.

Let
Tex syntax error
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
Tex syntax error
and let \phi\colon U\to \mathbb{R}^n be associated chart. The obvious candidate for 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 \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 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 0.3 [Milnor1964, Theorem 2.2].

Let
Tex syntax error
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.

However to show that these two definition agree we need a notion of microbundle isomorphism.

Definition 0.4. 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.

\displaystyle  \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,}

In our case


\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 {{endproof}} n-dimensional microbundle is a quadruple (E,B,i,j) such that there is a sequence
\displaystyle B\xrightarrow{i} E\xrightarrow{j} B
and the following conditions hold.
  1. j\circ i=\id_B
  2. 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
    \displaystyle h\colon V\to U\times \mathbb{R}^n.

Moreover, the homeomorphism above must make the following diagram commute:

\displaystyle  \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 0-section in U\times \mathbb{R}^n.

Exercise 0.2 [Milnor1964, Lemma 2.1].

Let
Tex syntax error
be a topological manifold. Show that \xi_M : = (M \times M, M, \Delta_M, p_1) is a microbundle.

Proof.

Let
Tex syntax error
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
Tex syntax error
and let \phi\colon U\to \mathbb{R}^n be associated chart. The obvious candidate for 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 \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 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 0.3 [Milnor1964, Theorem 2.2].

Let
Tex syntax error
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.

However to show that these two definition agree we need a notion of microbundle isomorphism.

Definition 0.4. 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.

\displaystyle  \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,}

In our case


\square


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